Tewari - Organic Chemistry Vi

ORGANIC CHEMISTRY A Modern Approach Volume-I

ABOUT THE AUTHOR Nimai Tewari is a retired associate Professor, Department of Chemistry, Katwa College (affiliated to The University of Burdwan), West Bengal. A PhD in Organic Chemistry from Calcutta University, he has taught the subject for a period of more than three decades. He has published various research papers in national and international journals. Apart from Organic Chemistry—A Modern Approach, Dr Tewari has authored three more books on Organic Chemistry for undergraduate and postgraduate students. His research interest includes Organic Synthesis and Heterocyclic Chemistry.

ORGANIC CHEMISTRY A Modern Approach Volume-I

Nimai Tewari Associate Professor (Retired) Department of Chemistry Katwa College (affiliated to The University of Burdwan) West Bengal

McGraw Hill Education (India) Private Limited CHENNAI McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai 600 116 Organic Chemistry—A Modern Approach (Volume-I) Copyright © 2017, by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. ISBN (13): 978-93-5260-565-1 ISBN (10): 93-5260-565-9 Managing Director: Kaushik Bellani Director—Science & Engineering Portfolio: Vibha Mahajan Lead—UG Portfolio: Suman Sen Content Development Lead: Shalini Jha Specialist—Product Development: Amit Chatterjee Production Head: Satinder S Baveja Sr. Manager—Production: Piyaray Pandita General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

Typeset at Text-o-Graphics, B-1/56, Aravali Apartment, Sector-34, Noida 201 301, and printed at Cover Printer:

Visit us at: www.mheducation.co.in

Dedicated to my daughter Aindrila and my son-in-law Ritam Mukherjee whose forbearance, constant encouragement and inspiration made this work possible

CONTENTS

Preface 1.

xv

Structure, Bonding and Properties of Organic Molecules 1.1

1.2

1.1–1.313

Hybridization, Bond Lengths, Bond Strengths or Bond Dissociation Enthalpies, Bond Angles and VSEPR Theory 1.2 1.1.1 Hybridization 1.2 1.1.2 Bond Length 1.9 1.1.3 Bond Dissociation Enthalpy or Bond Dissociation Energy 1.9 1.1.4 Bond Angle 1.10 1.1.5 VESPR Theory and Molecular Geometry 1.11 Solved Problems 1.13 Study Problems 1.21 Electronegativity and Bond Polarity 1.22 Solved Problems 1.33 Study Problems 1.42

1.3

Molecular Formula as a Clue to Structure: Double Bond Equivalent (DBE) or Index of Hydrogen Deficiency (IHD) 1.44 Solved Problems 1.44 Study Problems 1.46

1.4

Acids and Bases 1.46 1.4.1 Brönsted-Lowry Theory of Acids and Bases 1.46 1.4.2 Lewis Acid-Base Theory 1.52 Solved Problems 1.53 Study Problems 1.57 Inductive and Electrometric Effects 1.60 1.5.1 Inductive Effect 1.60 1.5.2 Field Effect 1.67 1.5.3 Electromeric Effect 1.67 Solved Problems 1.67 Study Problems 1.73

1.5

Contents

viii

1.6

Resonance and Resonance Effect or Mesomeric Effect 1.75 1.6.1 Resonance Energy 1.77 1.6.2 Rules for Writing Meaningful Resonance Structures 1.78 1.6.3 Relative Contribution of Resonance Structures towards Resonance Hybrid 1.78 1.6.4 Resonance or Mesomeric Effect 1.80 1.6.5 Isovalent and Heterovalent Resonance 1.81 1.6.6 Effect of Resonance on the Properties of Molecules 1.81 Solved Problems 1.98 Study Problems 1.112

1.7

Hyperconjugation 1.117 1.7.1 Sacrificial and Isovalent Hyperconjugation 1.119 1.7.2 Effect of Hyperconjugation on the Physical and Chemical Properties of Molecules and on the Stabilities of Intermediates 1.119 Solved Problems 1.122 Study Problems 1.124 Steric Effect 1.125 1.8.1 Properties of Molecules Influenced by Steric Effect 1.126 1.8.2 Proton Sponges 1.132 1.8.3 Face Strain or F-Strain 1.133 1.8.4 Steric Acceleration and Steric Retardation 1.134 1.8.5 Bredt’s Rule 1.135 Solved Problems 1.138 Study Problems 1.142 Intermolecular Forces 1.147 1.9.1 Dipole–Dipole Interactions 1.147 1.9.2 van der Waals Forces 1.149 1.9.3 Effect of Intermolecular Forces on Different Properties of Compounds 1.150 Solved Problems 1.161 Study Problems 1.167

1.8

1.9

1.10

1.11

Reactive Intermediates 1.170 1.10.1 Nonclassical Carbocation 1.177 1.10.2 Carbonium Ion and Carbenium Ion or Carbocation 1.178 Solved Problems 1.193 Study Problems 1.212 Tautomerism 1.218 1.11.1 Mechanism of Keto-enol Tautomerism 1.221 1.11.2 Difference between Resonance and Tautomerism 1.222 1.11.3 Position of the Tautomeric Equilibrium 1.223 1.11.4 Ring-chain Tautomerism 1.228 1.11.5 Valence Tautomerism 1.228 Solved Problems 1.228 Study Problems 1.236

Contents

1.12

1.13

1.14

2.

ix

Aromaticity 1.240 1.12.1 Criteria for Aromaticity 1.240 1.12.2 Antiaromatic Compounds 1.242 1.12.3 Nonaromatic Compounds 1.242 1.12.4 Classification of Compounds as Aromatic, Antiaromatic and Nonaromatic by Comparing their Stabilities with that of the Corresponding Open Chain Compounds 1.243 1.12.5 Modern Definition of Aromaticity 1.243 1.12.6 Molecular Orbital Energy Diagram of Some Ions and Molecules 1.244 1.12.7 Use of Inscribed Polygon Method to Determine the Relative Energies of p Molecular Orbitals for Cyclic Planar and Completely Conjugated Compounds and to Classify Them as Aromatic and Antiaromatic 1.245 1.12.8 Classification of Some Molecules and Ions as Aromatic, Antiaromatic and Nonaromatic 1.247 1.12.9 Homoaromatic Compounds 1.251 1.12.10 Some Chemical and Physical Consequences of Aromaticity 1.251 Solved Problems 1.262 Study Problems 1.267 Thermodynamics, Energy Diagrams and Kinetics of Organic Reactions 1.276 1.13.1 Thermodynamics 1.276 1.13.2 Energy Diagram 1.278 1.13.3 Kinetics 1.281 1.13.4 Catalysis 1.283 1.13.5 Hammond Postulate 1.284 1.13.6 Kinetic Control versus Thermodynamic Control of a Chemical Reaction 1.286 Solved Problems 1.287 Study Problems 1.295 Methods of Determining Mechanisms of Reactions 1.298 1.14.1 Kinetic Isotope Effects 1.305 Solved Problems 1.307 Study Problems 1.312

Principles of Stereochemistry

2.1–2.230

Introduction 2.2 2.1

Projection Formulas of Stereoisomers 2.3 2.1.1 Flying-Wedge Projection Formula 2.3 2.1.2 Fischer Projection Formula 2.4 2.1.3 Sawhorse Projection Formula 2.10 2.1.4 Newman Projection Formula 2.11 2.1.5 Interconversion of Projection Formulas 2.12 Solved Problems 2.16 Study Problems 2.25 2.2 Symmetry Elements 2.29 2.2.1 Simple Axis of Symmetry or Rotational Axis of Symmetry (Cn) 2.2.2 Plane of Symmetry (s) 2.33

2.29

Contents

x

2.3

2.4

2.2.3 Centre of Symmetry (i) 2.36 2.2.4 Alternating Axis of Symmetry (Sn) 2.37 2.2.5 Symmetric, Asymmetric and Dissymmetric Molecules 2.38 Solved Problems 2.39 Study Problems 2.56 Isomerism: Constitutional Isomers and Stereoisomers 2.58 2.3.1 Constitutional Isomers 2.58 2.3.2 Stereoisomers 2.59 2.3.3 Enantiomers and Diastereoisomers 2.59 Solved Problems 2.64 Study Problems 2.79 Molecular Chirality 2.84 2.4.1 Chiral and Achiral Molecules 2.84 2.4.2 Source of Chirality 2.85 2.4.3 Stereocentre or Stereogenic Centre 2.88 2.4.4 Meso Compounds 2.89 Solved Problems 2.91 Study Problems 2.104

2.5

Configuration and Configurational Nomenclature 2.108 2.5.1 D, L-System of Configurational Designation 2.109 2.5.2 Specification of Configuration: The R, S-System 2.111 2.5.3 Erythro and Threo Nomenclature of Compounds with Two Adjacent Chiral Centres 2.124 2.5.4 The E-Z System of Designating Alkene Diastereoisomers 2.125 2.5.5 R, S and E, Z Assignment in the Same Molecule (Geometric Enantiomerism) 2.127 2.5.6 Syn-anti Nomenclature for Aldols 2.128 2.5.7 Number of Stereoisomers for Compounds with Chiral Centres 2.129 2.5.8 Chirotopic and Achirotopic Atom in a Molecule 2.131 2.5.9 Prostereoisomerism and Topicity 2.133 Solved Problems 2.141 Study Problems 2.175

2.6

Optical Activity of Chiral Compounds 2.182 2.6.1 Plane-polarized Light 2.182 2.6.2 Optical Activity 2.182 2.6.3 Polarimeter 2.183 2.6.4 Specific Rotation 2.184 2.6.5 The Necessary and Sufficient Condition (the ultimate criterion) for Optical Activity 2.186 2.6.6 Racemic Modification 2.187 2.6.7 Enantiomeric Excess (EE) or Optical Purity (EE) 2.187 2.6.8 Racemization 2.188 2.6.9 Resolution of Racemic Modification 2.192 Solved Problems 2.199 Study Problems 2.208

Contents

2.7

3.

xi

Conformation of Acyclic Organic Molecules 2.211 2.7.1 Conformations of Ethane (CH3—CH3) 2.213 2.7.2 Conformations of Propane (CH3CH2CH3) 2.214 2.7.3 Conformations of Butane (CH3CH2CH2CH3) 2.215 2.7.4 Conformations of Chloroethane (CH3CH2Cl) 2.218 2.7.5 Conformations of 1,2-dichloroethane (ClCH2CH2Cl) 2.218 2.7.6 Conformations of Some Typical Acyclic Molecules 2.219 2.7.7 Invertomerism 2.220 Solved Problems 2.221 Study Problems 2.227

Nucleophilic Substitution Reactions at Saturated Carbon Atom

3.1–3.180

Introduction 3.2 3.1 The SN2 reaction 3.3 3.1.1 Example of SN2 Reaction 3.3 3.1.2 Kinetics of SN2 Reaction 3.3 3.1.3 Mechanism of SN2 Reaction 3.3 3.1.4 Stereochemistry of SN2 Reaction 3.5 3.1.5 Evidence in Favour of SN2 Mechanism 3.6 3.1.6 Factors Influencing SN2 Reaction Rate or SN2 Reactivity 3.7 Solved Problems 3.35 Study Problems 3.74 3.2

The SN1 Reaction 3.86 3.2.1 Example of SN1 Reaction 3.86 3.2.2 Kinetics of SN1 Reaction 3.86 3.2.3 Mechanism of SN1 Reaction 3.86 3.2.4 Stereochemistry of SN1 Reaction 3.87 3.2.5 Evidence in Favour of SN1 Mechanism 3.90 3.2.6 Factors Influencing SN1 Reaction Rate or SN1 Reactivity 3.91 3.2.7 Carbocation Rearrangements in SN1 Reactions 3.105 3.2.8 Comparison of the SN2 and SN1 Reactions 3.108 3.2.9 Summary of Reactivity of Alkyl Halide in Necleophilic Substitution Reactions 3.108 3.2.10 Factors Favouring SN1 and SN2 Reactions 3.108 3.2.11 SNi and SNi¢ Mechanisms 3.109 3.2.12 SN1¢ Mechanism 3.112 3.2.13 Isotope Effects and Salt Effects (Methods used to Distinguish between SN1 and SN2 Type Reactions) 3.113 Solved Problems 3.114 Study Problems 3.135 3.3 Neighbouring Group Participation (NGP) 3.146 3.3.1 Definition 3.146 3.3.2 Mechanism of NGP 3.146 3.3.3 Example of NGP 3.146 3.3.4 Anchimeric Assistance 3.147

Contents

xii

3.3.5 Evidence for Participation by a Neighbouring Group 3.148 3.3.6 Various Cases of Neighbouring Group Participation 3.148 Solved Problems 3.163 Study Problems 3.175 4.

Elimination Reactions

4.1–4.128

Introduction 4.2 4.1 The E2 Reaction 4.4 4.1.1 Example of E2 Reaction 4.4 4.1.2 Kinetics of E2 Reaction 4.4 4.1.3 Mechanism of E2 Reaction 4.4 4.1.4 Stereochemistry of E2 Reaction 4.5 4.1.5 Evidence in Favour of the E2 Mechanism 4.32 4.1.6 Factors Influencing E2 Reaction Rate or E2 Reactivity 4.33 4.1.7 Factors that Govern the Proportions of E2 and SN2 Reactions 4.34 4.1.8 Regioselectivity in b-elimination Reactions (Orientation of p Bond in the Product Alkene) 4.37 4.1.9 Hofmann Exhaustive Methylation or Hofmann Degradation 4.43 4.1.10 Fragmentations 4.45 4.1.11 Summary of the E2 Reaction 4.47 Solved Problems 4.47 Study Problems 4.74 4.2 The E1 Reaction 4.86 4.2.1 Example of E1 Reaction 4.86 4.2.2 Kinetics of E1 Reaction 4.87 4.2.3 Mechanism of E1 Reaction 4.87 4.2.4 Stereochemistry of E1 Reaction 4.88 4.2.5 Evidence in Favour of the E1 Mechanism 4.90 4.2.6 Factors Influencing E1 Reaction Rate or E1 Reactivity 4.90 4.2.7 Regioselectivity of E1 Reactions 4.90 4.2.8 Rearrangement of the Carbocation Intermediate Involved in an E1 Reaction 4.93 4.2.9 Acid-Catalyzed Dehydration of Alcohols 4.95 4.2.10 Dehydration using POCl3 and Pyridine 4.99 4.2.11 Factors Influencing the Extent of E1 and E2 Reactions 4.100 4.2.12 Factors that Govern the Proportions of E1 and SN1 Reactions 4.101 Solved Problems 4.102 Study Problems 4.114 4.3 The E1CB Reaction 4.118 4.3.1 Example of E1cB Reaction 4.118 4.3.2 Kinetics of E1cB Reaction 4.118 4.3.3 Mechanism of E1cB Reaction 4.118 4.3.4 The Nature of the Substrate 4.119

Contents

xiii

4.3.5 To Distinguish between E1cB and E2 Mechanisms 4.119 Solved Problems 4.121 Study Problems 4.125 4.4 a- or 1,1-Elimination 4.126 4.4.1 Example of a- or 1,1-Elimination Reaction 4.126 4.4.2 Kinetics of a- or 1,1-Elimination Reaction 4.126 4.4.3 Mechanism of a- or 1,1-Elimination Reaction 4.126 4.4.4 Structure of the Substrate Involved in a-Elimination 4.127 Solved Problems 4.127 Study Problems 4.128 Index

I.1–I.6

PREFACE

In the course of teaching Organic Chemistry to undergraduate students, I have been constantly feeling the need of a concise volume that deals with their important topics on the subject matter. Students have often expressed their difficulty caused by the absence of such a compact book. My present effort is to meet this long-felt need and the book has been designed primarily for the students who have taken a basic course in Organic Chemistry at the undergraduate level. The book covers some important topics on Organic Chemistry in four chapters. It starts with a chapter on Structure, Bonding and Properties of Organic Molecules, highlighting concepts like hybridization, electronegativity and bond polarity, acids and bases, inductive effect, resonance, steric effect, intermolecular forces, etc. Stereochemistry is an essential part of the organic chemistry courses. Chapter 2 deals with the Principles of Stereochemistry. Chapter 3 covers Nucleophilic Substitution Reactions at Saturated Carbon Atom and Chapter 4 covers the Elimination Reactions. By following a modern methodology of learning, the book presents a large number of reactions with discussions supported with mechanistic explanations and diagrams, wherever needed. Organic chemistry is best learned by solving problems. Each article of every chapter concludes with a number of Solved as well as Study Problems to provide an opportunity to the students for self-evaluation. I believe that this book will be of great utility for the students who have taken a basic course of Organic Chemistry in B.Sc. (Chemistry) Hons., besides being equally effective for advance students of Chemistry because of the in-depth discussion in a readerfriendly language. The book will also be useful for the students preparing for competitive examinations like NET, SLET, etc.

Acknowledgements I offer my sincere gratitude to Mr. Kaushik Bellani, MD, McGraw Hill Education (India) Pvt. Ltd. and Mrs. Vibha Mahajan, Director, Science & Engineering Portfolio for successful

xvi

Preface

publication of this book. I also wish to thank Mr. Sumen Sen, Mr. Amit Chatterjee and Mr. P L Pandita for taking keen interest in publishing this book. I am grateful to all of them. I also owe a debt of gratitude to my colleagues for constructive suggestions and to my students who encouraged me constantly. I appreciate the interest and enthusiasm shown by my wife Mrs. Dali Tewari and my daughter Andrila Tewari (Mukherjee) during the long period of preparation of the manuscript. Valuable suggestions from the readers for the improvement of the book will be most welcome.

Nimai Tewari

1

CHAPTER

STRUCTURE, BONDING AND PROPERTIES OF ORGANIC MOLECULES Chapter Outline

1.1

1.2 1.3

1.4

1.5

1.6

Hybridization, Bond Lengths, Bond Strengths or Bond Dissociation Enthalpies, Bond Angles and VSEPR Theory 1.1.1 Hybridization 1.1.2 Bond Length 1.1.3 Bond Dissociation Enthalpy or Bond Dissociation Energy 1.1.4 Bond Angle 1.1.5 VESPR Theory and Molecular Geometry Electronegativity and Bond Polarity Molecular Formula as a clue to structure: Double Bond Equivalent (DBE) or Index of Hydrogen Deficiency (IHD) Acids and Bases 1.4.1 Brönsted-Lowry Theory of Acids and Bases 1.4.2 Lewis Acid-Base Theory Inductive and Electrometric Effects 1.5.1 Inductive Effect 1.5.2 Field Effect 1.5.3 Electromeric Effect Resonance and Resonance Effect or Mesomeric Effect 1.6.1 Resonance Energy 1.6.2 Rules for Writing Meaningful Resonance Structures 1.6.3 Relative Contribution of Resonance Structures towards Resonance Hybrid 1.6.4 Resonance or Mesomeric Effect

1.6.5

1.7

1.8

1.9

1.10

1.11

Isovalent and Heterovalent Resonance 1.6.6 Effect of Resonance on the Properties of Molecules Hyperconjugation 1.7.1 Sacrificial and Isovalent Hyperconjugation 1.7.2 Effect of Hyperconjugation on the Physical and Chemical Properties of Molecules and on the Stabilities of Intermediates Steric Effect 1.8.1 Properties of Molecules Influenced by Steric Effect 1.8.2 Proton Sponges 1.8.3 Face Strain or F-Strain 1.8.4 Steric Acceleration and Steric Retardation 1.8.5 Bredt’s Rule Intermolecular Forces 1.9.1 Dipole–Dipole Interactions 1.9.2 van der Waals Forces 1.9.3 Effect of Intermolecular Forces on Different Properties of Compounds Reactive Intermediates 1.10.1 Nonclassical Carbocation 1.10.2 Carbonium Ion and Carbenium Ion or Carbocation Tautomerism 1.11.1 Mechanism of Keto-enol Tautomerism 1.11.2 Difference between Resonance and Tautomerism

1.2

1.12

Organic Chemistry—A Modern Approach

1.11.3 Position of the Tautomeric Equilibrium 1.11.4 Ring-chain Tautomerism 1.11.5 Valence Tautomerism Aromaticity 1.12.1 Criteria for Aromaticity 1.12.2 Antiaromatic Compounds 1.12.3 Nonaromatic Compounds 1.12.4 Classification of Compounds as Aromatic, Antiaromatic and Nonaromatic by Comparing their Stabilities with that of the Corresponding Open Chain Compounds 1.12.5 Modern Definition of Aromaticity 1.12.6 Molecular Orbital Energy Diagram of Some Ions and Molecules 1.12.7 Use of Inscribed Polygon Method to Determine the Relative Energies of p Molecular Orbitals for Cyclic Planar and Completely Conjugated

1.1 1.1.1

1.13

1.14

Compounds and to Classify Them as Aromatic and Antiaromatic 1.12.8 Classification of Some Molecules and Ions as Aromatic, Antiaromatic and Nonaromatic 1.12.9 Homoaromatic Compounds 1.12.10 Some Chemical and Physical Consequences of Aromaticity Thermodynamics, Energy Diagrams and Kinetics of Organic Reactions 1.13.1 Thermodynamics 1.13.2 Energy Diagram 1.13.3 Kinetics 1.13.4 Catalysis 1.13.5 Hammond Postulate 1.13.6 Kinetic Control versus Thermodynamic Control of a Chemical Reaction Methods of Determining Mechanisms of Reactions 1.14.1 Kinetic Isotope Effects

HYBRIDIZATION, BOND LENGTHS, BOND STRENGTHS OR BOND DISSOCIATION ENTHALPIES, BOND ANGLES AND VSEPR THEORY Hybridization

The process of intermixing of atomic orbitals of the same atom having slightly different energies so as to redistribute their energies and give new orbitals of equal energies and identical shapes and sizes is called hybridization. The number of hybrid orbitals is equal to the number of pure atomic orbitals reshuffled. The hybrid orbitals are more effective in forming stable bonds as compared to the pure atomic orbitals and this is because they can undergo more effective overlapping. The relative overlap of orbitals decreases in the following order: sp > sp2 > sp3 >> p. The hybrid orbitals are oriented in space in some preferred directions to have a stable arrangement in which there is minimum repulsion among themselves. Therefore, the type of hybridization governs the geometrical shapes of the molecules. Hybrid orbitals

Hybridization

Geometry

Bond angle

2 3 4

s + p = sp s + p + p = sp2 s + p + p + p = sp3

Linear Trigonal planar Tetrahedral

180° 120° 109.5°

Structure, Bonding and Proper es of Organic Molecules

1.3

In fact, the number of groups surrounding a particular atom determines its geometry. A group is either an atom or a lone pair of electrons. Any atom surrounding by two, three and four groups are linear, trigonal planar and tetrahedral, respectively, and they have bond angles of 180°, 120° and 109.5°, respectively. The hybridization of a C, O or N atom can be determined by the number of p bonds it forms. If it forms no p bond, one p bond and two p bonds, it is sp3, sp2 and sp hybridized, respectively. All single bonds are s bonds. A double bond consists of one s bond and one p bond. A triple bond consists of one s bond and two p bonds. “Hybrid orbital number” method can be used to determine the hybridized state of an atom within a molecule. Hybrid orbital number = (number of s bonds) + (number of unshared pair of electrons) If the hybrid orbital number is 2, the atom is sp hybridized, if it is 3, the atom is sp2 hybridized and if it is 4, the atom is sp3 hybridized. The electronic configuration of carbon atom in its ground state is 1s2 2s2 2px1 2py1 2pz°, i.e., one odd electron is present in each of 2px and 2py orbitals of carbon atom. The number of odd electrons present in the valence shell of an atom generally gives the measure of covalency of that atom. So, the valency of carbon should be two. However, the valency of carbon in almost all organic compounds is 4, except a few extremely unstable compounds, where its valency is 2, like methylene (:CH2), dichloromethylene (:CCl2), etc. During chemical reaction, the two electrons present in 2s orbital become unpaired by absorbing energy and one of them is promoted to 2pz orbital. This is the excited state of carbon atom and the electronic configuration of carbon atom in this state is 1s2 2s1 2px1 2py1 2pz1. Thus, in the excited state, four odd electrons are present in the outermost shell of carbon atom. The presence of these four unpaired electrons accounts for the tetravalency of carbon.

sp3-Hybridization: When one s and three p orbitals of the valence shell of a carbon atom merge together to form four new equivalent orbitals having the same energy and shape, it results in tetrahedral or sp3 –hybridization. The resulting orbitals are called sp3 hybrid orbitals

1.4

Organic Chemistry—A Modern Approach

The four sp3 hybrid orbitals each containing one electron are directed towards the four corners of a regular tetrahedron, making an angle of 109°28′ with one another and the atom lies at the centre of the tetrahedron. The hybrid orbitals are oriented in such a fashion in space that there occurs minimum repulsion between them. The formation of sp3 hybrid orbitals by the combination of s, px, py and pz atomic orbitals may be shown as follows:

Formation of methane (CH4) molecule: During the formation of methane molecule, one 2s orbital and three 2p orbitals of excited carbon atom undergo hybridization to form four equivalent sp3 hybrid orbitals. The hybrid orbitals are directed towards the four corners of a regular tetrahedron. Each hybrid orbitals containing an unpaired electron overlaps with the 1s orbital of a hydrogen atom resulting in the formation of four C—H s bonds. Thus, methane molecule possesses a highly stable tetrahedral geometry with each H—C—H bond angle equal to 109°28′.

Structure, Bonding and Proper es of Organic Molecules

1.5

It is to be noted that if the four atoms linked covalently to the carbon atom are not the same, the geometry of the molecule would still be tetrahedral but it may not be regular in shape, e.g., methyl bromide (CH3Br), bromoform (CHBr3), etc. In these cases, the bond angles differ slightly from the normal value of 109°28′.

sp2-Hybridization: When one s orbital and two p orbitals of the valence shell of a carbon atom merge together and redistribute their energies to form three equivalent new orbitals of equal energy and identical shape, the type of hybridization occurs is called sp2-hybridization. The new orbitals formed as a result of this hybridization are called sp2 hybrid orbitals.

1.6

Organic Chemistry—A Modern Approach

All three hybrid orbitals each containing one electron lie in one plane and make an angle of 120° with each other, i.e., they are directed towards the three corners of an equilateral triangle with the carbon atom in the centre of the triangle. The unhybridized 2pz orbital (containing one electron) remains perpendicular to the plane of the triangle with its two lobes above and below that plane. Therefore, a molecule in which the central atom is sp2-hybridized possesses triangular planar shape and the hybridization is called planar trigonal hybridization. The formation of sp2 hybrid orbitals by combination of s, px and py atomic orbitals is shown below.

Formation of ethylene (C2H4) molecule: During the formation of ethylene molecule, each of the two carbon atoms undergoes sp2-hybridization, leaving the 2pz orbitals unhybridized. The three sp2 hybrid orbitals of each carbon atom are planar and oriented at an angle of 120° to each other. The unhybridized 2pz orbitals are perpendicular to the plane of sp2 hybrid orbitals. One sp2 orbital of one carbon overlaps axially with one sp2 orbital of the other carbon to form a C—C s bond. The remaining two sp2 orbitals of each carbon overlap with the half-filled 1s orbitals of two hydrogen atoms resulting in the formation of a total of three C—H s bonds. The unhybridized 2pz orbitals of one carbon overlap with that of the other carbon in a sideways fashion to from a p bond between the two carbon atoms. The p bond consists of two equal electron clouds distributed above and

Structure, Bonding and Proper es of Organic Molecules

1.7

below the plane of carbon and hydrogen atoms. Since all the six atoms in the molecule lie in one plane, ethylene is a planar molecule. Each C—C—H or H—C—H bond angle is nearly equal to 120°.

sp-Hybridization: When one s and one p orbital of the valence shell of a carbon atom merge together and redistribute their energies to form two equivalent hybrid orbitals of equal energy and identical shape, it results in sp-hybridization or diagonal hybridization.

1.8

Organic Chemistry—A Modern Approach

Formation of acetylene (C2H2) molecule: In acetylene molecule, the two carbon atoms are sp-hybridized. There are two unhybridized orbitals (2py and 2pz) on each C atom. Two sp hybrid orbitals are linear and directed at an angle of 180°. The unhybridized p orbitals are perpendicular to the sp hybrid orbitals and also perpendicular to each other. One sp hybrid orbital of one carbon overlaps axially with the similar orbital of the other carbon to form a C—C s bond. The remaining hybrid orbital of each C atom overlaps with halffilled 1s orbital of H atom to form a total of two C—H s bond. Thus, acetylene molecule is linear. The unhybridized py orbitals of two carbons and the unhybridized pz orbitals of two carbons overlap sideways separately to form two different p bonds. Electron clouds of one p bond lie above and below the internuclear axis representing the s bond while the electron clouds of the other p bond lie in front and backside of the internuclear axis. These two sets of p electron cloud merge into one another to form a cylindrical cloud of electrons around the internuclear axis surrounding the C—C s bond. Each C—C—H bond angle is equal to 180°. p p

2px 2py

s H 1s

sp

C

2px 2py

s sp

sp

C

s sp

H

H

s

C

p s

C

s

1s 180° H H C ∫∫ C Linear acetylene molecule

H

Structure, Bonding and Proper es of Organic Molecules

1.9

1.1.2 Bond Length Bond length is defined as the equilibrium distance between the centres of the nuclei of two bonded atoms in a covalent molecule. Bond length depends on the factors such as (i) size of atoms, (ii) multiplicity of bonds, (iii) s-character of the orbitals and (iv) resonance, hyperconjugation, etc. Bond length increases with increase in the size of atoms but it decreases with increase in bond multiplicity. Thus, a triple bond is shorter than a double bond which in turn is shorter than a single bond. An s orbital is closer to the nucleus than a p orbital. So, electrons in the s orbital is more tightly held by the nucleus than the electrons in the p orbital. For this reason, with increase in s-character of the hybrid orbital, the attractive force on the electron(s) increases and so, the size of the hybrid orbital decreases. As a consequence, the length of the bond obtained by overlapping of the hybrid orbitals with the s orbitals of hydrogen, for example, decreases. The s-character of sp3, sp2 and sp hybrid orbitals are 25 percent, 33.33 percent and 50 percent respectively. Thus, the lengths of the C—H bonds involving C atoms with different hybridization follow the order: Csp3 —H (1.093 Å) > Csp2 —H (1.078 Å) > Csp — H (1.057 Å).

1.1.3

Bond Dissociation Enthalpy or Bond Dissociation Energy

The bond dissociation enthalpy or bond dissociation energy which is a measure of bond strength may be defined as the amount of energy required to break one mole of a particular type of bond between two atoms in the gaseous state so as to produce neutral gaseous atoms or free radicals. Bond dissociation energies are usually abbreviated by the symbol DH° and are usually expressed in kJ mol–1 or kcal mol–1. The greater the bond dissociation energy, stronger the bond. The factors affecting bond dissociation energy are: (i) size of the bonded atom, (ii) multiplicity of bonds, (iii) s-character of the hybrid orbital involved in bond formation, etc. Larger the size of the bonded atoms, greater the bond length and lesser is the bond dissociation energy. The bond dissociation energy increases with increase in bond multiplicity. A C ∫∫ C bond is stronger than a C == C bond which in turn is stronger than a C – C bond. The bond dissociation energy also increases with increase in the s-character of the hybrid orbital and this is because with increase in s-character of the hybrid orbital, the electron density in the region of overlap increases. Bond type

Bond dissociation energy kJ mol–1 (kcal mol–1)

C(sp3) — C(sp3) C(sp3) — C(sp2)

346.3 (82.76) 357.6 (85.48)

C(sp3) — C(sp) C(sp2) — C(sp2) C(sp2) — C(sp)

382.5 (91.42) 383.2 (91.58) 403.7 (96.48)

C(sp) — C(sp)

433.5 (103.6)

1.10

Organic Chemistry—A Modern Approach

Bond Dissociation Energies (kcal mol–1) of Some Chemical Bonds Bond

DH°

Bond

DH°

Bond

DH°

H—H

104

F—F

38

Ph—H

110

D—D

106

Cl—Cl

58

CH2==CH CH2—H

88

C—C

83.1

Br—Br

46

Ph CH2—H

85

N—N

38.4

I—I

36

CH3—F

107

O—O

33.2

CH3—H

104

CH3—Cl

84

N—F

136

C2H5—H

98

CH—Br

70

H — Cl

103

(CH3)2 CH—H

95

CH3—I

56

H — Br

88

CH2==CH—H

108

CH2==CH—Cl

82

H—I

71

Another measure of bond strength is Bond Energy. This is actually the average bond dissociation energy, denoted by E. For example, cleavage of a C—H bond in methane requires different bond dissociation energies in successive steps as indicated below and energy of the C—H bond is determined by an average. CH Æ .CH + H. DH ∞ = 104 Kcal mol –1 4

.CH 3

3

1

CH2 Æ .CH Æ

CH2 + H. DH 2∞ = 106 Kcal mol -1 .CH + H. DH ∞ = 106 Kcal mol -1 3 .C . + H. DH ∞ = 81 Kcal mol -1

CH4

.C. + 4H. Total: 397 Kcal mol -1

Æ

Æ

4

Therefore, EC—H = 397/4 = 99.25 kcal mol–1. However, bond dissociation energy (DH°) is more convenient than bond energy (E) for our purpose.

1.1.4

Bond Angle

The bond angle is the angle between two bonds around the central atom in a molecule. For example, H—C—H bond angle in methane (CH4) is 109.5° and H—O—H bond angle in water is 104.5°. Bond angle is affected by the factors such as (i) type of hybridization, (ii) number of lone pair of electrons on the central atom, (iii) electronegativity of the central atom and (iv) electronegativily of the atom bonded to the central atom. The greater the amount of s-character of the orbital used by carbon to form the bond, the larger the bond angle. For example, sp-hybridized carbons have bond angles of 180°, sp2-hybridized carbons have bond angles of 120° and sp3-hydridized carbon have bond angles of 109.5°.

Structure, Bonding and Proper es of Organic Molecules

1.1.5

1.11

VESPR Theory and Molecular Geometry

We can predict the arrangement of atom in molecules and ions on the basis of a relatively simple idea called the valence shell electron pair repulsion (VSEPR) theory. We can apply this theory on the basis of the following considerations: 1. In a molecule or ion, the central atom is covalently bonded to two or more atoms or groups. 2. Covalent bonds contain shared pair of electrons which are often called bond pairs or bonding pairs. The unshared electrons of the central atom are called nonbonding pairs or unshared pairs or lone pairs. 3. Since electron pairs repel each other, the electron pairs of the valence shell tend to stay as far apart as possible to avoid electronic repulsion. 4. If the central atom is surrounded by bond pairs as well as lone pairs of electrons, the repulsions among themselves are different. As a result, the molecule possesses an irregular or distorted geometry. The repulsive interactions of various electron pairs decrease in the order: lone pair–lone pair (lp – lp)> lone pair–bond pair (lp – bp) > bond pair–bond pair (bp – bp). 5. The geometry of a molecule is to be settled by considering all of the electron pairs, bonding and nonbonding. However, the shape of the molecule is to be described by referring to the positions of the atoms and not by the position of the electron pairs. Let us consider the following examples. Methane (CH4) molecule: In CH4 molecule, the total number of electrons surrounding the central carbon atom = 4 valence electrons of C atom + 4 electrons of four singly-bonded H atom = 8 electron or 4 electrons pairs = 4 s bond pairs. The four bond pairs experience minimum repulsion if they are tetrahedrally oriented, i.e., if all the H—C—H bond angles are of 109.5°. Hence, the shape of CH4 molecule is tetrahedral. H

H

C

109.5°

H H Tetrahedral methane molecule

Ammonia (NH3) molecule: In NH3 molecule, the total number of electrons in the valence shell of the central N atom = 5 valence electrons of N atom + 3 electrons of three singly-bonded H atom = 8 electron or 4 electron pairs = 3 s bond pairs + 1 lone pair. All four electron pairs experience minimum repulsion if they occupy the four corners of a tetrahedron. As lone pair–bond pair repulsion is greater than the bond pair–bond pair repulsion, the H—N—H bond angle is slightly deviated from the normal tetrahedral angle

1.12

Organic Chemistry—A Modern Approach

(109.5°) and is reduced to 107°, i.e., the tetrahedron is somewhat distorted. Excluding the lone pair, the shape of the molecule is trigonal pyramidal.

Water (H2O) molecule: In water molecule, the total number of electrons surrounding the central O atom = 6 valence electron of oxygen atom + 2 electrons of two singly-bonded H atoms = 8 electrons or 4 electron pairs = 2 s bond pairs + 2 lone pairs. In order to minimise the extent of mutual repulsion, these four electron pairs are oriented towards the four corners of a tetrahedron. However, the tetrahedron is somewhat distorted due to the strong repulsive forces exerted by the lone pairs on each bond pair of electrons. In fact, the H—O—H bond angle is reduced to 104.5° from the normal tetrahedral angle of 109.5°. Excluding the lone pairs, the shape of the molecule is angular or V-shaped.

H 104.5° H

or

O

O H 104.5° H

Angular or V-shaped water molecule

Boron tifluoride (BF3) molecule: In boron trifluoride molecule, the total number of electrons in the valence shell of the central boron atom = 3 valence electrons of B atom + 3 electrons of three singly-bonded F atom = 6 electrons or 3 electrons pairs = 3 s bond pairs. The three bond pairs experience minimum propulsion if they remain at 120° angle with respect to each other. Therefore, the geometrical shape of BF3 molecule is trigonal planar. F 120°

B

120°

F 120° F Trigonal planar boron trifluoride molecule

Acetylene (HCCH) molecule: The number of electrons surrounding each carbon atom of acetylene molecule = 4 valence electron of carbon + 3 electrons of one triply-bonded C atom + 1 electron of one singly-bonded H atom = 8 electrons = 4 electron pairs = 2 s bond

Structure, Bonding and Proper es of Organic Molecules

1.13

pairs + 2p bond pairs. In order to minimise the repulsive forces between the bond pairs, the shape of acetylene molecule is linear. The effect of electrons involved in the formation of a p bond is not generally considered in determining the geometrical shape of a molecule. 180° 180° H——C∫∫C——H Linear acetylene molecule

1.

Give the state of hybridization of the central atom of each of the following species and predict their shapes. @

!

(b) BF3

(c) C H 3

(d) C H 3

(e) :N H 2

(f)

(g) NH 3

(h)

(i) H3O≈

(j) HCN

(l) CO2

(m) BeH2

H 2S:

(k) CCl4 Solution (a)

BF4*

≈ (n) NH 4

(b)

(c) F

B

F

(d)

(e) N

H H

H

(g)

(h) H

H

B

H

S

H H

sp3;

H H

sp3; bent or angular

(i) O

–

F

sp3; trigonal pyramidal

H—C∫∫N

–

F

N

H

H

sp2; trigonal planar

sp3; bent or angular

sp3; trigonal pyramidal

C

(f)

–

C

+

H

F

sp2; trigonal planar

(j)

@

(a) H2O

F F

H

+

H H

sp3; trigonal pyramidal

tetrahedral

(k)

(l) Cl

O==C==O sp; linear

sp; linear

C Cl

Cl Cl

sp3; tetrahedral

1.14

Organic Chemistry—A Modern Approach

(m)

(n) H—Be—H

H

sp; linear

N H

+

H H

sp3; tetrahedral

2.

Draw the structure of a hydrocarbon which contains: (a) Three sp3-hybridized carbon atoms; (b) One sp3 and two sp2-hybridized carbon atoms; (c) One sp and two sp2-hybridized carbon atoms; (d) Two sp3 and two sp-hybridized carbon atoms.

Solution

(a)

(b)

(c)

(d)

3.

Which orbitals are used to form each bond in methylamine, CH 3 NH 2?

Solution

Each C—H bond is formed by the overlap of an sp3 orbital of carbon with the s orbital of hydrogen ( Csp3 —H1s). The C—N bond is formed by the overlap of an sp3 orbital of carbon with sp3 orbital of nitrogen ( Csp3 — N sp3 ) and each N—H bond is formed by the overlap of an sp3 orbital of nitrogen with the s orbital of hydrogen ( N sp3 —H1s).

Structure, Bonding and Proper es of Organic Molecules

1.15 : O:

4.

|| Answer the following question for the acetaldehyde (CH 3 — C — H) molecule: (a) Determine the hybridization of oxygen and the two carbon atoms. (b) Which orbitals are involved in forming the carbon–oxygen double bond? (c) Mention the type of orbital in which the lone pairs reside.

Solution (a)

(b)

(c) 5.

The s bond is formed by the end-on overlap of an sp2 orbital of carbon with an sp2 orbital of oxygen and the p bond is formed by the side-by-side overlap of the 2p orbital of carbon with the 2p orbital of oxygen. The two sp2 hybrid orbitals are occupied by the two lone pairs of oxygen. Mention the state of hybridization of the starred (*) carbon atoms in each of the following compounds: * (b) CH 3 C N

(a) (d)

*

CH2 == C == CH2

(e)

*

*

(c) HC ∫∫ C — CHO (f)

–

*

Solution

(a) Three groups around the starred carbon: sp2-hybridized; (b) two groups around the starred carbon: sp-hybridized; (c) three groups around the starred carbon: sp2-hybridized; (d) two groups around the starred carbon: sp-hybridized; (e) three groups around the starred carbon: sp2-hybridized; (f) four groups around the starred carbon; sp3-hybridized. 6.

How many s and p bonds are present in each of the following molecules? (a) CH3—C ∫∫ C—CH == CH—CH3 (b) –CH2CH3 (c) CH3 CH == C == CH CH2 CH3

1.16

Organic Chemistry—A Modern Approach

Solution

(a) (b) (c) 7.

s bond = 13; p bond = 3; s bond = 18; p bond = 3; s bond 15; p bond 2 Designate the state of hybridization of all the atoms of the following molecule:

Solution The hybridized atoms within the molecule is designated as a, b and c according to their hybridization status: a = sp3; b = sp2 and c = sp.

8.

Draw the orbital picture for each of the following molecules: (a) ethylene (b) acetylene (c) ketene (CH2 == C == O) (d) acraldehyde (e) acrylonitrile (f) but-1, 2, 3-triene.

Solution

(a) Ethylene:

(b) Acetylene:

Structure, Bonding and Proper es of Organic Molecules

(c) Ketene:

(d) Acraldehyde:

(e) Acrylonitrile:

1.17

1.18

Organic Chemistry—A Modern Approach

(f) But-1, 2, 3-triene:

9.

Which atoms in each of the following molecules always remain in the same plane and why? (a) CH3CH==CH CH3

(b) C6H5C ≡≡ CCH3

(c) CH3CH==C==C==CH2

(d) CH2==CH—C ≡≡ CH

2

Solution The two sp -hybridized carbon atoms and the atom directly attached to them always remain in the same plane. This is applicable also when the two sp2-hybridzed carbon atoms are linked through an even number of sp-hybridized carbon atoms. Therefore, in (a), all atoms excluding the six methyl hydrogens remain in the same plane. In (b), all atoms excluding the three methyl hydrogens remain in the same plane. In (c), all atoms excluding the three methyl hydrogens remain in the same plane and in (d), all atoms remain in the same plane. This is also because the two sp-hybridized carbon atoms and the atoms directly attached to them remain in the same line.

10.

Explain why a p bond is weaker than a s bond.

Solution A p bond is weaker than a s bond because the end-on overlap that forms s bonds is better than the side-to-side overlap that forms p bonds. This is also because the electron density in a p bond is farther from the two nuclei as compared to that in a s bond.

11.

In CH3CH3, the C—H bond is shorter and stronger than the C—C bond – explain.

Solution The s orbital of hydrogen is closer to the nucleus than the sp3 orbital of carbon having less s character. So, the carbon and hydrogen nuclei are closer together in sp3-s overlap than the carbon nuclei in sp3-sp3 overlap. Because of this, the C—H bond in ethane (CH3—CH3) is shorter and stronger than the C—C bond. Again, as the percentage of s character of the overlapping orbitals increases, the electron density in the region of overlap increases and as a result, the strength of the bond increases. Since there is greater electron density in the region of sp3-s overlap than in the region of sp3-sp3 overlap, therefore, the C—H bond is stronger than the C—C bond.

Structure, Bonding and Proper es of Organic Molecules

12.

1.19

Arrange the indicated bonds in each of the following compounds in order of increasing bond strength and increasing bond length: bond 1 HC∫∫ C—CH==CH—CH2—CH3 (a) (b) H3C ≠ ≠ ≠ NH—CH2CH3 bond 3 bond 1 bond 2 N CH2—C∫∫ N bond 2 ≠ bond 3

Solution Greater the bond multiplicity, shorter the bond and greater the bond strength. Therefore, in compound (a), bond length increases in the order: bond 1 < bond 2 < bond 3 and bond strength increase in the order: bond 3 < bond 2 < bond 1. In compound (b), the bond length increases in the order: bond 3 < bond 2 < bond 1 and the bond strength increases in the order: bond 1 < bond 2 < bond 3.

13.

Which of the indicated bonds in each pair of compounds is shorter and why? (a) (b) ==

O

(c) CH3—C—H and H—CH2OH ≠ ≠ Solution As the s-character of the overlapping orbitals increases, the nuclei involved in bond formation becomes closer and the bond becomes shorter. Therefore, (a) the Csp—H bond in the second compound is shorter than C 2 —H bond in the first compound; sp

(b) the N sp2 —H bond in the second compound is shorter than the N sp3 —H bond in the first compound, and (c) the Csp2 —H bond in the first compound is shorter than the C 3 —H bond in the second compound. Determine the state of hybridization of the indicated atom in each of the following species: (a)

(b)

(c) –

–

NH2 ≠

O2N

(f) O

–

Ø

O

Ø

(e) O ==

==

(d)

O2N –

O ==

O

==

O

C6H5—C

–

CH2 ≠

==

sp

14.

H—C—NH2 ≠

1.20

Organic Chemistry—A Modern Approach

Solution In any system of the type X = Y – Z:, Z is sp2-hybridized and the unshared electron pair on it occupies a p orbital to delocalize the electron pair and make the system conjugated. Therefore, the indicated atoms in (a), (b), (c), (d) and (f) are sp2-hybridized. However, due to violation of Bredt’s rule (introduction of a double bond is not possible at the bridgehead position in bridged bicyclic compounds with small rings) delocalization of the unshared electron pair is not possible in the compound (e) and therefore, the indicated carbon atom is sp3-hybridized.

15.

“Bromination of methane is less exothermic than that of chlorination” — explain with DH° calculation.

[Bond dissociation energies for C—H = 104 kcal/mol; Br—Br = 46 kcal/mol; H—Br = 87.5 kcal/mol; Cl—Cl = 58 kcal/mol; H—Cl = 103 kcal/mol; C—Cl = 83.5 kcal/mol and C—Br = 70 kcal/mol] Solution

hn CH 4 + Br – Br ææÆ CH 3 - Br + HBr

DH° (bromination) = [(–70) + (–87.5)] – [(–104) + (–46)] = (–157.5) – (–150) = –7.5 kcal/mol hn CH 4 + Cl – Cl ææÆ CH 3 - Cl + HCl

DH° (chlorination) = [(–83.5) + (103)] – [(–104) + (–58)] = (–186.5) – (–162) = – 24.5 kcal/mol Thus, bromination of methane is less exothermic than chlorination of methane. 16.

The H—X bond in hydrogen halides becomes longer and weaker as the atomic mass of the halogen increases—explain.

Solution A p orbital of halogen overlaps with the s orbital of hydrogen to form the hydrogen–halogen (H—X) bond in hydrogen halides. For bond formation, fluorine uses the p orbital that belongs to the second shell of electrons while chlorine uses the p orbital that belongs to the third shell of electrons. Since the average distance from the nucleus is greater for an electron in the third shell than for an electron in the second shell, therefore, the average electron density is less in a 3p orbital than in a 2p orbital. Consequently, the electron density in the region where the s and p orbitals overlap decreases as the size of the halogen increases. Thus, the hydrogen–halogen bond becomes longer and weaker as the atomic mass of the halogen increases.

Structure, Bonding and Proper es of Organic Molecules

1. 2. 3.

1.21

A carbon-carbon bond formed by sp2–sp2 overlap is stronger than the one formed by sp3–sp3 overlap. Explain. Which atom in the ammonium ion (NH≈4 ) has the least electron density and why? What is the state of hybridization of each of the C, O and N atom in the following compound? C∫∫ C—CH—CH2—O—CH3

C

==

H3 C

NH2

O

4.

Predict the geometry around each of the indicated atoms: H3 (a) B ≠ (d) H 3 O ≠

H3 (b) H 3N Æ B ≠

≠

!

!

!

(e) C (CH 3 )3

(f) N H 4

N (g) H C ≠

H (h) (CH 3 )2 N ≠

(i) Et 2 N ≠

(j) CH 3 C H 2CH 3

(k) CH 2 = CH– C H 2

≠

≠

5.

O || (c) CH3 —C—OCH3

≠

@

@

≠

Predict the mentioned bond angles: (a) C—O—C bond angle in CH3COCH3 (b) F—B—F bond angle in BF4@ (c) C—C—N bond angle in C2H5CN

(d) H—N—H bond angle in NH ≈4 !

(e) C—O—H bond angle in CH3CH2OH (f) C—N—C bond angle in (CH 3 )2 N H 2 6.

Which of the indicated bonds is shorter and why? (a) (c)

(b) CH 3 CH = CH – C ∫ C - CH 3 ≠

9

Cl ≠

≠

(d)

≠ ≠

NH2 OH

≠

(e) CH 3 CH = CH – CH = CH - CH 3 ≠

≠

O

(f)

||¨

CH 3 — O — CH 2 — C — CH 3 ≠

(g)

N—CH2CH2N==CHCH2CH3 ≠ ≠

1.22

Organic Chemistry—A Modern Approach

7.

Give the state of hybridization of the indicated atoms: (a)

≈

≈

(CH 3 )2 N H 2 ≠

≠

1.2

(c) B F4@

(e) H3 N Æ BF3

(f) CH2 == CH — CH2

≠

(d) CH 3 N O2 (g)

(b) (CH 3 )2 O H ≠

CH3CH == CH — NH2 ≠

≠

@

≠

== CH2 (h) CH2 == C ≠

ELECTRONEGATIVITY AND BOND POLARITY

Electronegativity is the tendency of an atom to pull the bonding electrons towards itself. It increases across a row of the periodic table (excluding the noble gases) and it decreases down a column of the periodic table. The electronegativity values of some common elements are given in the following table.

A bond with the electrons shared equally between the two atoms, i.e., a bond in which each electron spends as much time in the vicinity of one atom as in the other, is called a nonpolar covalent bond. For example, the H—H covalent bond, the Br—Br covalent bond and the C—C covalent bond in ethane etc. are nonpolar covalent bonds. When two atoms of different electronegativities form a covalent bond, the electrons are not shared equally between them. The atom with greater electronegativity draws the electron pair closer to it. As a result of this unequal distribution of the bonding electrons, the bond acquires a slight positive charge (indicated by the symbol d+) on the end that has the less electronegative atom and a slight negative charge (indicated by the symbol d–) on the other end that has the more electronegative atom, i.e., a polarity developes within a bond. Such a bond is called a polar covalent bond. An example of a polar covalent bond is

Structure, Bonding and Proper es of Organic Molecules

1.23

the one in hydrogen fluoride (H—F). The fluorine atom, with its greater electronegativity, pulls the bonding electrons towards itself. As a consequence, the hydrogen atom becomes somewhat electron deficient and acquires a partial positive charge (d+) while the fluorine atom becomes somewhat electron rich and acquires a partial negative charge (d–). So, the hydrogen fluoride molecule is a dipole (Hd+— Fd–). The direction of bond polarity is indicated by an arrow, the head of which indicates the negative end of the bond. A short perpendicular line is drawn near the tail of the arrow to indicate the positive end of the bond.

The extent of polarity in a covalent bond is expressed quantitatively by the physical property known as ‘dipole moment’ m, which is, in fact, the product of the magnitude of charge e on any polar end and the distance d between the centres of positive and negative charges, i.e., m = e × d. Dipole moment is expressed in Debye unit (D). The charge e and the distance d are of the order 10–10 esu and 10–8 cm, respectively. Therefore, m is of the order of 10–10 × 10–8 = 10–18 esu.cm (in C.G.S. system). This value of dipole moment is equal to one Debye unit, i.e., 1D = 10–18 esu. cm. The SI unit of dipole moment is coulomb-meter (C.m.) and 1 C.m. = 2.9962 × 1029 D. The value of m for a nonpolar molecule is zero. For example, m = 0D for the molecules like H2, N2, O2, etc. For polar molecules, m has a definite value, e.g., for HF molecule m = 1.91 D. Polarities of molecules increase with increase in the value of m. Dipole moment is a vector quantity. The dipole moment of a molecule is the result of the vector sum of the individual bond moments present in the molecule. When a molecule is formed by two atoms of different electronegativities (i.e., by two different atoms), the molecule must possess a dipole moment because there is no question to cancellation of the moment of this only polar bond. However, when a molecule contains two or more polar bonds, i.e., when a molecule contains more than two atoms, the molecule may or may not possess dipole moment and in that case the polarity of the molecule depends on the shape of the molecule. A symmetrical polyatomic molecule possesses no dipole moment because the individual dipole moments of the bonds cancel each other. It is to be noted that the symmetry in calculating dipole moment is not the molecular symmetry. H2O molecule, for example, has a symmetrical structure because it has two planes of symmetry (su planes). But, it possesses dipole moment (1.85 D) because of its angular structure

.

The value of dipole moment of a molecule is equal to the product of the charge e and the distance, d, between the positive and negative charge centres, i.e., m = e × d, it can be well understood by the following examples:

1.24

(i)

(ii)

Organic Chemistry—A Modern Approach

Although an H—F (0.92 Å) molecule is smaller than an H—Cl (1.27 Å molecule, it has a dipole moment larger than HCl (1.75 D compared to 1.03 D). Since fluorine (electronegativity: 4.0) is more electronegative than chlorine (electronegativity: 3.0), therefore, the magnitude of e in H—F is much higher than that in H—Cl and for this reason although d in HF is smaller than in HCl (because of smaller size of F compared to Cl), the product e × d, i.e., m, is larger. Chloromethane (CH3Cl) has a larger dipole moment (1.87 D) than fluoromethane CH3F (1.81 D), even though fluorine is more electronegative than chlorine. Due to higher electronegativity of fluorine than chlorine, the separation of charge in the C—X bond, i.e., the magnitude of e in CH3F, is somewhat higher than that in CH3Cl. However, the C—F bond is shorter than the C—Cl bond (1.42 Å compared to 1.77 Å) and because of this, the value of the product, i.e., m, is larger for chloromethane than for fluoromethane.

That the polarity of a molecule depends upon the shape of the molecule can be well understood by the following example. NH3 possesses considerable dipole moment and although the N—F bonds are more polar than N—H bonds, NF3 (0.26 D) has a much smaller dipole moment than NH3 (1.46 D). Both NH3 and NF3 with sp3 hybridized N atom are pyramidal in shape. The unshared electron pair on nitrogen occupying an sp3 orbital contributes a large dipole moment (because an unshared pair has no other atom attached to it to partially neutralize its negative charge) in the direction opposite to the triangular base of the pyramid. In NH3, the net moment resulting from three N—H bond moments adds to the moment contributed by the unshared pair because they act in the same direction and for this reason, it has considerable dipole moment. In NF3, on the other hand, the vectorial sum of these N—F bond moments acts in the direction opposite to that of the moment caused by the unshared pair. Since these moments are of about the same size, therefore, NF3 has a much smaller dipole moment than NH3.

Resonance often plays an important role in determining the polarities of molecules. For example: (i) The dipole moment of ethylchloride, CH3 CH2Cl (2.05 D), is larger than that of vinyl chloride, CH2 == CH—Cl (1.44 D). Chlorine is more electronegative than carbon and so, it attracts the C—Cl bonding electrons more towards itself. As a result,

Structure, Bonding and Proper es of Organic Molecules

1.25

polarity develops in the C—Cl bond in ethyl chloride and the compound shows considerable dipole moments. In vinyl chloride, on the other hand, the unshared electron pair on chlorine becomes involved in resonance interaction with the p-orbital of the double bond. This resonance interaction, therefore, tends to oppose the usual displacement of electrons towards chlorine. Also, the sp2 hybridized carbon being more electronegative than the sp3 hybridized carbon is less willing to release electrons to chlorine. So, although there is still a net displacement of electrons towards chlorine, it is less than in ethyl chloride. Hence, ethyl chloride is more polar than vinyl chloride, i.e., the dipole moment of ethyl chloride is greater than that of vinyl chloride.

(ii)

The dipole moment of chlorobenzene is larger than that of fluorobenzene. In chlorobenzene, the Cl atom withdraws electrons from the ring by its –I effect and donates electrons to the ring by its +R effect. Because of large size and high electronegativtiy of chlorine as compared to that of carbon, the resonance electron donation is much weaker than inductive electron withdrawal. So, the moment caused by the –I effect is much stronger than the moment caused by the +R effect and because of this, chlorobenzene possesses a net dipole moment which is relatively high. In fluorobenzene, on the other hand, the +R effect fluorine even being more electronegative than chlorine is much more important because of effective p orbital overlap between the orbitals of carbon and fluorine which are of comparable size. So, the moment due to inductive electron withdrawal is predominantly balanced by the moment due to resonance electron donation and the result is that the dipole moment of fluorobenzene is relatively low.

1.26

Organic Chemistry—A Modern Approach

Aromaticity often plays a role in determining the polarity of a molecule. For example: (i) Azulene (a bicyclic hydrocarbon) has a much higher dipole moment. Azulene contains 10 p-electrons. Redistribution of these electrons between the two rings generates an aromatic system which consists of an aromatic cycloheptatrienyl cation and an aromatic cyclopentadienyl anion. Since this aromatic dipolar form is quite stable and more contributing, azulene has a much higher dipole moment.

(ii)

Pyrrole and furan are polar molecules, but the dipole moment of furan is smaller than and opposite in direction from that of pyrrole. In pyrrole, the unshared electron pair on nitrogen is highly delocalized with the p-electrons of the ring for maintaining aromaticily, i.e., to form a delocalized cyclic system of (4n + 2)p electrons, where n = 1. Since the moment caused by the electron delocalization is much greater than that caused by the –I effect of nitrogen atom, pyrrole has a net dipole momemt of 1.81 D and the dipole points towards the ring. In furan, on the other hand, the unshared electron pair on oxygen is not well delocalized into the ring due to greater electronegativity of oxygen. Hence, the moment due to electron delocalization is small and in fact, it is somewhat smaller than that caused by the –I effect of oxygen. For this reason, the net dipole moment of furan is relatively small (0.70 D) and the dipole points towards oxygen.

Formation of charge-transfer complex is often responsible for exhibiting dipole moment by some molecules. 1,3,5-Trinitrobenzene, for example, shows significant dipole moment in benzene but does not display dipole moment in carbon tetrachloride. 1,3,5-Trinitrobenzene being symmetrical possesses no net dipole moment, i.e., it is a nonpolar molecule. When it dissolves in benzene, a charge-transfer bonding involving some kind of donor–acceptor interaction (benzene is the electron donor and the electron deficient 1,3,5-trinitrobenzene is the electron acceptor) is formed. Due to charge separation in the resulting complex, the compound shows significant dipole moment in benzene. Since carbon tetrachloride cannot act as a donor molecule, the compound does not display dipole moment in carbon tetrachloride.

Structure, Bonding and Proper es of Organic Molecules

1.27

Conformation plays an important role in determining the polarities of some molecules. For example: (i) 1,2-Dichloroethane has a very little dipole moments, where as 1,2-ethanediol has considersable dipole moment. 1,2-Dichloroethane exists in anti and gauche staggered form. Since the two C—Cl dipoles in the anti form are antiparallel, its dipole moment is assumed to be zero. However, the value of dipole moment of the gauche form is approximately 3.2 D (calculated with reference to C2H5Cl). Because of dipolar and steric repulsion, the gauche form is less stable than the anti form by 1.2 kcal/mol and in the vapour phase the compound exists in 88% anti and 12% gauche form. Because of such distribution of the conformers, the overall dipole moment of 1,2-dichloroethane is relatively low (1.2 D).

In the gauche conformer of 1,2-ethanediol (ethylene glycol), the two —OH groups become involved in the formation of intramolecular hydrogen bond. However, in the anti conformer, no intramolecular hydrogen bond is formed because the two —OH groups are oppositely placed. The steric and polar repulsion of the hydroxyl groups in the gauche form is more than outweighed by the energy gained by the formation of hydrogen bond (5 kcal/mol). Because of this, the compound, particularly in the gas phase, exists almost exclusively in the gauche form having a finite dipole moment. For such conformational distribution, 1,2-ethanediol is found to possess considerable dipole moment.

1.28

(ii)

Organic Chemistry—A Modern Approach

The optically inactive (meso) form of 1,2-dichloro-1,2-di-p-tolylethane has a dipole moment lower than that of the active (d or l) form. The meso form of the compound exists in the following three conformations:

In the conformer I, bulky tolyl groups are anti to each other and also the two Cl atoms are anti to each other. So there is no steric interaction caused by the tolyl groups and repulsive interaction caused by the C—Cl dipoles. Therefore, this conformer is relatively more stable and it has practically no dipole moment. The conformers II and III are polar because the two Cl atoms are gauche to each other. However, these are very unstable because the bulky p-tolyl groups which are gauch to each other are involved in steric interaction and there occurs repulsive interaction between the two C—Cl dipoles. Therefore, the most favoured conformer

Structure, Bonding and Proper es of Organic Molecules

1.29

of meso-1,2-dichloro-1,2-di-p-tolylethane is I and as a consequence, the overall dipole moment of the meso compound is relatively low. The optically active form of this compound exists in the following three conformations:

(iii)

The conformer IV in which C—Cl dipoles are antiparallel has no appreciable dipole moment. However, this conformer is unstable because the bulky p-tolyl groups which are gauche to each other are involved in steric interaction. The conformer VI has an appreciable dipole moment but it is unstable due to the same steric interaction and also due to repulsive interaction between the two C—Cl dipoles. In the conformer V, the two p-tolyl groups are anti to each other. So, it is relatively stable and the preferred conformer. Since it has an appreciable dipole moment, the overall dipole moment of active 1,2-dichloro-1,2-ditoylethane is relatively large. The dipole moment of cis-1,2-dichlorocyclohexane is larger than that of its transisomer. Each of the conformational isomers of cis-1,2-dichlorcyclohexane has one axial and one equatorial chlorine atom. As both the forms are equally stable, 50% of the molecules exist in one and 50% in the other form. Since Cl atoms are gauche to each other in both the forms, they have appreciable dipole moment and consequently, the overall dipole moment of the compound is relatively large. The trans-isomer, on the other hand, exists in diequatorial and diaxial forms. The diequatorial form in which the two Cl atom are gauche to each other has appreciable dipole moment. Although favoured sterically, it is disfavoured by dipolar repulsion. The diaxial form with a very small dipole moment, although disfavowred sterically, is free from dipolar repulsion and is relatively stable. This form, therefore, exists predominantly and so, the overall dipole moment of this isomer is much lower than that of the cis-isomer.

1.30

(iv)

Organic Chemistry—A Modern Approach

The dipole moment of hexane-3,4-dione is very small, whereas that of cyclohex3,5-dien-1,2-dione is very large. Hexane-3,4-dione may exist in two conformations such as s-cis (cisoid) and s-trans (transoid). In the s-trans conformer, the two C == O dipoles are antiparallel. Since the C == O bond moment cancels each other, therefore, this conformer has no net dipole moment. Again, as the negatively polarized oxygen atoms are as far as possible from each other, this conformer is free from any repulsive interaction between the two oxygen atoms. In the s-cis conformer, on the other hand, the C == O dipoles remain on the same side of the double bond make an angle of 60° with each other. Consequently, a large net moment of two strong C == O bond moments operates. Again, as the two C == O dipoles are close enough to repeal each other, this conformer is relatively less stable. For this reason, the compound exists almost exclusively, in the more stable s-trans conformation and the overall dipole moment of the compound is very small. The planar and rigid cyclohex-3,5-dien-1,2-dione molecule exists exclusively in the highly polar s-cis conformation in which the two C == O bond moments make an angle of 60° with each other. It cannot exists in the s-trans conformation because it is very much unstable due to severe angle strain. For this reason, the dipole moment of the compound is quite large.

Structure, Bonding and Proper es of Organic Molecules

1.31

Induced moment often plays an important role in determining the dipole moments of molecules. The expected order of dipole moment of various fluormethanes is CH2F2 > CH3F ª CHF3 > CF4 . However, the experimental order is: CF3F > CH2F2 > CHF3 > CF4. This anomaly can be explained in terms of opposing induced moment. In the symmetrical tetrahedral molecule of CF4, the resultant of three C — F bond moments cancels the fourth C — F bond moment and because of this, the molecule possesses no net dipole moment.

In CH3F, the moment due to the C—F bond is not cancelled and the molecule possesses a net dipole moment which is the resultant of C—F and C—H bond moments. The dipole moment of CH2F2 in which the two C—F bonds make an angle of nearly 109°5 is expected to be larger than that of CH3F because the resultant of two C—F bond moments must be greater than one C—F bond moment. Again, the dipole moment of CH3F is expected to be equal to that of CHF3 because the moment of a —CF3 group is equal to that of a C—F bond and the moment of a —CH3 group is equal to that of a C—H bond. Thus, the expected order of dipole moment is CH2F2 Ò CH3F ª CH3F Ò CF4. However, this order does not agree with the experimental dipole moment values. This can be explained by considering the moment acting in the opposite direction induced by each C—F dipole in the other. Since there is only one C—F bond in CH3F, the opposing induced moment is absent in it. So it has the largest dipole moment. In CH2F2, there are two C—F bond. Thus, opposing induced moment operates in this case and that partly cancels the resultant moment of two C—F bonds. Because of this, its dipole moment is smaller than CH3F. In CHF3, there are three C—F bonds and so, the magnitude of the opposing induced moment is relatively large in

1.32

Organic Chemistry—A Modern Approach

this case. In fact, this reduces the resultant of three C—F bond moments considerably and because of this its dipole moment is smaller than that of CH2F2 and much smaller than that of CH3F. Hence, the order of decreasing dipole moments of these four fluoromethanes in CH3F > CH2F2 > CHF3 > CF4. Dipole moment of disubstituted benzenes: The dipole moment of a disubstituted banzene C6H4 AB can be given as m AB = m 2A + m B2 + 2m A m B cos a where mA and mB are the two group moments and a is the angle between their direction. When both the groups are electron donating or both are electron withdrawing, the values of a for ortho-, meta- and para-isomer are 60°, 120° and 180°, respectively. However, when one group is electron donating and the other is electron withdrawing, the value of a for ortho-, meta- and para-isomer are 120°, 60° and 0°, respectively. For example: (i) Dichlorobenzenes: In each of the three isomeric dichlorobenzenes, the angle between the vectors of the two group moment may be shown as follows:

In this case, A == B == Cl and m A = m B = mC6H5Cl = 1.55 D, The group moment of

—Cl is to be used with negative sign because it is directed away from the benzene ring. For o-dichlorobenzene, where a = 60°, the net dipole moment is mO-C6H4Cl2 = (-1.55)2 + (-1.55)2 + 2(-1.55)2 cos60∞ = 2.40525 + 2.4025 + 2 ¥ 2.4025 ¥

1 2

= 7.2075 = 2.68 D

(ii)

1ˆ Ê Similarly, for m-dichlorobenzene, where a = 120∞ Á cos120∞ = ˜ , the net dipole Ë 2¯ moment is 1.55 D and for p-dichlorobenzene, where a = 180° (cos180° = –1), the net dipole moment is 0 D. Nitrotoluenes: In this case, A = CH 3 , B = NO2 , m A = mC H CH = 0.4 D and 6

5

3

mB = mC6H5NO2 = 3.95 D. The group moment of the electron–withdrawing —NO2 group is negative because it is directed away from the ring. The group moment of the electron-releasing —CH3 group is also negative because it is assumed to directed away from the ring.

Structure, Bonding and Proper es of Organic Molecules

1.33

For o-nitrotoluene, where a = 120∞, the net dipole moment is mO-CH3 C6H4 NO2 = (-0.4)2 + (-3.95)2 + 2 ¥ (-0.4) ¥ (-3.95) ¥ cos120∞ Ê 1ˆ = 0.16 + 15.6025 + 2 ¥ (-0.4) ¥ (-3.95) ¥ Á - ˜ Ë 2¯ = 0.16 + 15.6025 - 1.58 = 14.1825 = 3.76 D 1 Similarly, for m-nitrotoluene, where a = 60∞ ÊÁ cos60∞ = ˆ˜ the net dipole moment is 4.16 D Ë 2¯ and for p-nitrotoluene, a = 0° (cos 0° = 1), where the net dipole moment is 4.35 D. Formal charge: A formal charge represents the integer charges (positive or negative) that an atom bears in a molecule or ion, assuming that electrons in a chemical bond are equally shared between atoms irrespective of relative electronegativity. The concept of formal charge helps us to determine which atoms bear most of the charge in a charged molecule and also helps us to find out charged atoms in molecules that are neutral overall. The formal charge of any atom in a molecule can be calculated as follows: Formal Charge = number of valence electrons or periodic table group number – (number of nonbonding electron +1/2 number of bonding electrons) = V – (N + B/2).

1.

Predict the direction of dipole moment of the indicated bond in each of | the following molecules by using the symbol d+ and d– or Æ: (a) H—Cl (b) CH3—Cl (c) CH3—NH2 (d) I—Cl (e) CH3—MgBr (f) CH3CH2—OH (g) F—Br

(h) HO—Cl

Solution

(a)

(b)

(c)

(d)

(i)

— Li

1.34

Organic Chemistry—A Modern Approach

(e)

(f)

(g)

(h)

(i) 2.

Which one of the following compounds has (a) the most positively charged hydrogen, (b) the most negatively charged hydrogen and (c) a nonpolar covalent bond LiH, HCl, H2

Solution (a) H—Cl has the most positively charged hydrogen because chlorine is more electronegative than hydrogen. (b) LiH has the most negatively charged hydrogen because lithium is less elelctronegative than hydrogen. (c) The H—H bond is a nonpolar covalent bond because the atoms that share the bonding electrons are identical.

3.

Give the direction of the important bond moments in each of the following molecules (neglect C—H bonds) and also give the direction of the net dipole moment from the molecule. If the molecule has no dipole moment, write m = 0 D. (a)

cis-BrCH == CH Br

(b) trans-BrCH == CHBr

(c)

CH2 == CF2

(d) Br2C == CCl2

(e)

cis-BrCl C == CClBr Cl

(f) trans-BrCl C == CClBr

(g)

(h) Cl

Br

Cl

Br

Cl

(i)

Br

Br

Cl

Cl H

Br

(j)

(k)

Br

Br

(l)

H

(n)

C Cl

Cl

Cl

Cl

F Cl

Cl

Cl

Cl

Cl

(m)

Cl

Cl

F

Structure, Bonding and Proper es of Organic Molecules

1.35

Solution

Br

Br

(a)

C==C H

H

Br net dipole moment

(b)

H C==C

H

Br

trans(m = 0D)

cis-

(c)

(d)

Br

Cl C==C Cl

Br

net dipole moment

(e)

Cl

Cl C==C Br

Br

(g)

net dipole moment

Br

(f) Cl

Br trans(m = 0D)

Br

Cl

(h) Br

Cl

cis-

Cl

net dipole moment

net dipole moment F

Cl Br

(i)

Br

Cl

Cl

net dipole moment

Cl C==C

net dipole moment

(j) Cl

Cl H

Br

(k)

Cl

Cl

Br

Cl

(l) Cl

Br

H (m = 0D)

Cl (m = 0D)

F

(m)

net dipole moment

C

(n) Cl

Cl

F

1.36

Organic Chemistry—A Modern Approach

4.

How does bond polarity differ from bond polarizability? Explain with suitable examples.

Solution Polarization of a covalent bond (bond polarity) is a permanent phenomenon and is caused by the inductive mechanism of electron displacement in a bond in the normal state of a molecule. This permanent polarization arises due to difference in electronegativities of the two dissimilar bonded atoms. Polarization influences the physical as well as chemical properties of a molecule. The order of decreasing polarization of the C—X (X = halogen) bond in alkyl halides is as follows:

R3C—F > R3C—Cl > R3C—Br > R3C—I Polarizability, on the other hand, is a temporary phenomenon and is caused by the inductomeric mechanism of electron displacement in a covalent bond during a reaction when charged reagents are brought near the bond concerned. When the reagent is removed from the vicinity of the bond, the electrons resume their original position. Polarizability influences mainly the chemical properties of a molecule. Polarizability depends on the strength of the bond. In general, the more electronegative an atom or group, the less will be the effect of polarizability. In alkyl halides, for example, the decreasing order of polarizability of C—X bond is as follows: R3C—I > R3C—Br > R3C—Cl > R3C—F > (R == H or alkyl group) Therefore, the C—I bond is more polarizable than the C—F bond and that is why in nuclephilic displacement reactions (SN2), CH3—I is found to be more reactive than CH3—F, even though F is more electronegative. 5. The dipole moment of a bond gives an idea about the ionic character of that bond. Calculate the percentage of ionic character of HF. Given that the dipole moment of HF is 1.91 D and its bond length is 0.92 Å. Solution If HF were 100% ionic, each atom would carry a charge equal to one unit, i.e., 4.8 × 10–10 esu. As the bond length of HF is 0.92 Å, the dipole moment of completely ionic bond is

mionic = e × d = 4.8 × 10–10 esu × 0.92 × 10–8 cm = 4.416 × 10–18 esu.cm = 4.416D

(

10–18esu.cm = 1 D)

Therefore the percentage of ionic character of the H–F bond = 6.

mobserved 1.91 ¥ 100 ¥ 100 = = 43.25 m ionic 4.416

Explain the following observations: (a) CO2 has no dipole moment but SO2 has. (b) CF4 has no dipole moment but CH3F has. (c) BF3 has no dipole moment but NF3 has. (d) trans-BrCH == CHBr has no dipole moment but cis-BrCH == CHBr has.

Structure, Bonding and Proper es of Organic Molecules

1.37

Solution (a) CO2 has no dipole moment, i.e., it is a nonpolar molecule. This triatomic molecule with the carbon atom in the middle has linear structure (C is sp hybridized). Since the individual C == O bond moments are equal in magnitude but opposite in direction, therefore, they cancel each other and as a result, the molecule becomes nonpolar, i.e., it has no net dipole moment. Viewing alternatively, the centre of positive charge coincides with the centre of negative charge in this molecule, i.e., d = 0 and so m = e × d = e × 0 = 0 D.

(b)

On the other hand, the angular (bond angle 119.5°) sulphur dioxide molecule has appreciable dipole moment. This is because there operates a resultant moment of the two polar S == O bonds and the moment due to the unshared electron pair acting in the direction opposite to that of the resultant moment is insufficient to cancel the resultant moment completely. Although the C—F bond of carbon tetrafluoride (CF4) having very symmetrical tetrahedral structure is polar, CF4 has no net dipole moment and this is due to the fact that the net moment contributed (vectorially) by the three C—F bonds exactly cancels the remaining C—F bond moment. In fact, the centre of positive charge and the centre of negative charge coincide in this tetrahedral molecule, i.e., d = 0 and so m = e × d = e × 0 = 0 D.

Methyl fluoride (CH3F), on the other hand, has an appreciable dipole moment because the C—F bond moment is not cancelled by any opposing moment. The remaining three C—H bonds contribute a small net moment (the vectorial sum) in the same direction and so, the overall dipole moment of the molecule is somewhat increased.

1.38

(c)

(d)

Organic Chemistry—A Modern Approach

Since the boron atom in BF3 molecule is sp2 hybridized, the shape of the molecule is planar trigonal with the boron atom at the centre of an equilateral triangle and the three fluorine atoms at the three corners. In this symmetrical molecule, the vectorial sum of the two B—F bond moments is cancelled by the remaining B—F bond moment. Because of this, the molecule has no net dipole moment. In fact, the centre of positive charge coincides with the centre of negative charge in this triangular molecule, i.e., d = 0 and so m = e × d = e × 0 = 0 D.

Since the nitrogen atom in NF3 is sp3 hybridized, the shape of the molecule is pyramidal with nitrogen at the apex and fluorines at the corners of a triangular base. The net moment contributed (vectorially) by the N—F bonds opposes the moment due to the unshared electron pair. Since these opposing moments are approximately of the same size, therefore, the molecule has a very small dipole moment (0.26 D) of indeterminate direction. The carbon atoms in these two isomeric dibromoethenes are trigonal (sp2 hybridized). So, all the atoms of each molecule lie in the same plane. Since the bromine atoms in the cis-isomer lie on the same side of the double bond, therefore, the molecule has a net dipole moment which is the vectorial sum of two C—Br bond moments. In the trans-isomer, on the other hand, the bromine atoms lie on the opposite sides of the double bond and the two C—Br bond moments are antiparallel. The individual C—Br bond moments, therefore, cancel each other and so, the molecule has no net dipole moment.

Structure, Bonding and Proper es of Organic Molecules

7.

1.39

The molecule of any compound composed of two dissimilar elements is always polar–justify the statement.

Solution The statement is not always true. A diatomic molecule (A—B) consisting of two dissimilar elements with different electronegativities is always polar because there is no possibility of cancellation of moment of this only polar bond. HBr, HF, etc. are examples of such molecules. A polyatomic molecule consisting atoms of two dissimilar elements may or may not be polar. Polarity of such molecules depends on their geometrical shapes. If the molecule is symmetrical, i.e., if the individual bond moments cancel each other, it will be nonpolar and if not it will be polar. The linear triatomic molecule CO2, for example, is nonpolar (m = 0 D) because the two oppositely oriented C==O bond moments cancel each other. However, the angular H2O molecule is polar because the two O—H bond moments and the moments caused by two unshared pairs do not cancel out each other and a resultant moment operates in the molecule.

8.

Identify the three isomeric chlorotoluenes having dipole moments: 1.35 D, 1.9 D and 1.78 D.

Solution Methyl group (—CH3) can exert a +I effect while the Cl atom can exert a —I effect. In p-chlorotoluene, the —CH3 and —Cl group moments act linearly. So, its dipole moment will be equal to the sum of the two group moments and hence the dipole moment of the para-isomer will be the maximum. If the —CH3 group moment is considered to act in a direction away from the ring, then in m-chlorotoluene, these two group moments act at an angle of 60°. Hence, the dipole moment of m-chlorotoluene is expected to be somewhat less than that of the p-isomer. In o-chlorotoluene, these two group moments act an angle of 120° and hence the dipole moment of this isomer is expected to be somewhat less than that of the m-isomer. Therefore, the dipole moment values of 1.35 D, 1.78 D and 1.90 D correspond to o-, m- and p-chlorotoluene, respectively

1.40

Organic Chemistry—A Modern Approach

9.

How can you account for the following fact? (a) p-diacetylbenzene is polar but p-dinitrobenzene is nonpolar. (b) The group moments for —NH2 and —NO2 are 1.53 D and 3.95 D, respectively, but the measured dipole moment of p-nitroaniline is 6.20 D.

Solution (a) In p-diacetylbenzene, the two freely rotating C==O group moments cancel each other only in one conformation in which they are oriented anti to each other. This particular conformation, therefore, is nonpolar. In all other conformations, the two C==O group moments make a particular angle with each other. Thus, a resulting moment operates in all these conformations, i.e., in all these conformations, the compound is polar. Because of this, p-diacetylbenzene possesses an appreciable dipole moment, i.e., it is a polar molecule.

(b)

In p-dinitrobenzene, the group moments of the two freely rotating —NO2 groups act in the same line and thus, cancel each other. For this reason, p-dinitrobenzene does not have net dipole moment, i.e., the molecule is a nonpolar one. The dipole moment of p-nitroaniline is expected to be the sum of the moments caused by the —NH2 and —NO2 groups because these group moments act linearly in the same direction. But the dipole moments of p-nitroaniline are somewhat greater than the sum of the two group moments. In this compound, the —NO2 group with draws electrons by its —R effect while the —NH2 group donates electrons by its +R effect. As a consequence, there occurs a very effective resonance interaction which results in a much higher value of e. For this reason, the dipole moment of p-nitroaniline (6.20 D) is larger than the sum of the values of nitrobenzene (3.95 D) and aniline (1.53 D).

Structure, Bonding and Proper es of Organic Molecules

10.

1.41

Determine the formal charge on each atom in the following species: (a)

È ˘ Í ˙ ÍH — O — H ˙ | Í ˙ H Î ˚

!

(d)

H H | | H—N— B —H | | H H

(g)

:C ∫∫ O:

–

(b) O == O — O: H È ˘ Í ˙ | (e) ÍH3C — N — H ˙ Í ˙ | Í ˙ H Î ˚

(c)

:O: N = O= O:

!

(f) CH 3 - N ∫ C :

Solution !

(a)

(b)

ÈH — O — H ˘ Í ˙ The formal charge on the O atom = V – (N + B/2) = 6 – (2 + 6/2) | Í ˙ H ÍÎ ˙˚ = + 1 and on each H atom = 1 – (0 +2/2) = 0 =O :O = O: The formal charge on the left-hand O atom = 6 – (4 + 4/2) = 0, on the right-hand O atom = 6 – (6 + 2/2) = –1 and on the middle O atom = 6 – (2 + 6/2) = + 1. –

(c)

:O: N = O= O:

The formal charge on the double bonded O atom = 6 – (4 + 4/2) = 0, on each of the single bonded O atom = 6 – (6 + 4/2) = –1 and on the N atom = 5 – (0 + 8/2) = + 1.

(d)

H H | | H—N— B —H | | H H The formal charge on the B atom = 3 – (0 + 8/2) = –1, on the nitrogen atom = 5 – (0 + 8/2)= +1 and on each H atom = 1 – (0 + 2/2) = 0.

1.42

Organic Chemistry—A Modern Approach

H H È ˘ Í ˙ | | ÍH — C — N — H ˙ Í ˙ | | Í ˙ H H Î ˚

(e)

!

The formal charge on the C atom = 4 – (0 + 8/2) = 0, on the N atom = 5 – (0 + 8/2 = +1 and on each H atom = 1 – (0 + 2/2) = 0 H | H — C — N ∫∫ C : | H

(f)

The formal charge on the left-hand C atom = 4 – (0 + 8/2) = 0, on the N atom = 5 – (0 + 8/2) = + 1 and on the right-hand C atom = 4 – (2 + 6/2) = – 1. :C ∫∫ O: The formal charge on the C atom = 4 – (2 + 6/2) = –1 and on the O atom

(g)

= 6 – (2 + 6/2) = + 1.

1.

Explain why the dipole moment p-chlorophenol is higher than that of p-fluorophenol.

2.

Ethylene glycol has a higher value of dipole moment than 1,2-dimethoxy ethane —why ? Boron trifluoride (BF3) has no dipole moment (m = 0 D). Explain how this observation confirms the geometry of boron trifluoride predicted by VSEPR theory. Sulphur dioxide (SO2) has a dipole moment (m = 1.63 D), but carbon dioxide (CO2) has no dipole moment (m = 0D). What do these facts indicate about the geometry of sulphur dioxide? Although the H—F bond length is smaller (0.92 Å) than H—Cl bond length (1.27 Å), HF has a dipole moment larger than HCl (1.75 D compared to 1.03 D). Explain. Although fluorine is more electronegative than chlorine chloromethane (CH3Cl) has larger dipole moment (1.87 D) than fluoromethane, CH3F (1.81 D). Explain.

3. 4.

5. 6. 7. 8. 9. 10.

NF3 (0.26 D) has much smaller dipole moment than NH3 (1.46 D), even though N—F bonds are more polar than N—H bonds. Explain. Explain why the dipole moment of ethyl fluoride (CH3CH2F) is larger than that of vinyl fluoride (CH2 == CH — F). The dipole moment of chlorobenzene is larger than that of fluorobenze, even though fluorine is more electronegative than chlorine. Explain. Azulene is polar but its isomer naphthalene is nonpolar—why?

Structure, Bonding and Proper es of Organic Molecules

11.

1.43

Compare the dipole moments of the following compounds with reasons: (a)

(b)

13.

[Hint: (a) The first compound is more polar because the electron-donating (+R)– OMe group is attached to the positively polarized seven-membered aromatic ring and the electron-withdrawing (–R) —CN group is attached to negatively polarized five-memebered aromatic ring. The first ring is actually a cycloheptatrienyl cation system and the second ring is actually a cyclopentadienyl system.] 1,4-Dioxane has no dipole moment. Predict the form (boat or chair) in which it exists. Oxalic acid has dipole moment of zero in the gas phase. Explain.

14.

4-Aminopyridine

12.

4-cyanopyridine,

has a larger dipole moment (4.4 D) than (1.6 D). Explain.

15.

Which one have lower dipole moment: C2H5CN or C2H5NC? Explain. [Hint: In C2H5—C ∫∫ N, the s- and p-moments operate in the same direction, whereas in C2H5—N==C:, the s- and p-moments operate in the opposite directions.]

16.

The dipole moment of CH3Cl is greater (1.87 D) than the C–Cl bond moment (1.5 D) — why ? Why 1-butyne has a larger dipole moment than 1-butene?

17.

≈

@

18.

Determine the formal charge on C in (a ) a carbocation, CH3 ; (b) a carbanion, CH 3 ; (c) a free radical, CH 3 and a carbene, :CH2.

19.

Find out the correct Lewis structure for N2O (:N ∫∫ N— O: or :N == O == N:) and Cl2 O (:O — Cl — Cl: or :Cl — O — Cl:) by determining formal charges of each atom.

20. 21.

Carbon dioxide molecule has no net dipole moment. How can this observation be used to prove that the shape of the molecule is not angular? The following molecule shows very high dipole moment. Explain the reason.

1.44

Organic Chemistry—A Modern Approach

1.3

MOLECULAR FORMULA AS A CLUE TO STRUCTURE: DOUBLE BOND EQUIVALENT (DBE) OR INDEX OF HYDROGEN DEFICIENCY (IHD)

The molecular formula of a compound provides more information about the structure of the compound than might be apparent at first glance. For determining the structure of an organic compound, it is necessary to ascertain whether the compound is an unsaturated one or not and if it is an unsaturated compound, the degree of unsaturation is also needed to be known. In an organic compound, the degree of unsaturation is expressed in terms of Double Bond Equivalent (DBE). It is also called the Index Hydrogen Deficiency (IHD). If a hydrocarbon contains two hydrogen atoms less than the alkane containing the same number of C atoms, its double bond equivalent is 1, i.e., the compound may contain one double bond or a ring. The compound C4H8, for example, contains two H atoms less than the alkane (butane, C4H10) containing the same number of C atoms. So its double bond equivalent is 1. Thus, the compound may have one double bond or it may be a cyclic one. That is, the compound may be but-1-ene (CH3CH2CH == CH2) or but-2-ene (CH3CH == CHCH3) or cyclobutane ( ). Similarly, the double bond equivalent 2 indicates the presence of two double bond or one triple bond or one double bond and one ring or two rings in a compound. So, the term SODAR (Sum of Double bonds And Rings) is also frequently used. From the molecular formula of a compound, its Double Bond Equivalent (or Index of Hydrogen Deficiency or Sum of Double bonds And Rings) can easily be calculated. Â n(v - 2) + 1, where n is the number The double bond equivalent (DBE) of a compound = 2 of different types of atom present in the compound and v is the valency of each type of atom. It is to be noted that if the value of DBE of a compound is < 4, the compound is not a benzenoid aromatic compound.

1.

Calculate the double bond equivalent of a compound with the molecular formula C6H8. Indicate whether the compound is an aromatic one or not.

6(4 - 2) + 8(1 - 2) 12 - 8 +1 = +1= 3 2 2 The compound is not an aromatic one because its double bond equivalent is less than 4. Solution Double bond equivalent of the compound =

2.

Determine the double bond equivalent of each of the following compounds. Each consumes 2 moles of hydrogen on catalytic hydrogenation. How many rings are present in each of these compounds? (a) C8H8Cl2

(b) C8H10O2

(c) C5H6Br2

(d) C8H9Cl0

Structure, Bonding and Proper es of Organic Molecules

1.45

Solution (a) The DBE of the compound 8(4 - 2) + 8(1 - 2) + 2(1 - 2) 16 - 8 - 2 6 = +1= +1= +1= 4 2 2 2

(b)

The compound consumes 2 moles of hydrogen per mole on catalytic hydrogenation. So, there are two double bonds or one triple bond and (4 – 2) or two rings in the compound. The DBE of the compound =

(c)

(d)

8(4 - 2) + 10(1 - 2) + 2(2 - 2) 16 - 10 6 +1= +1= +1= 4 2 2 2

The compound consumes 2 moles of hydrogen per mole on catalytic hydrogenation. So, the compound contains two double bonds or one triple bond and (4 – 2) or two rings. The DBE of the compound 5(4 - 2) + 6(1 - 2) + 2(1 - 2) 10 - 6 - 2 2 = +1= +1= +1= 2 2 2 2 Since the compound consumes 2 moles of hydrogen per mole on catalytic hydrogenation, therefore, the compound contains two double bonds or one triple bond and no ring. The DBE of the compound 8(4 - 2) + 9(1 - 2) + 1(1 - 2) + 1(2 - 2) 16 - 9 - 1 6 = +1= +1= +1= 4 2 2 2 Since the compound consumes 2 moles of hydrogen per mole on catalytic hydrogenation, therefore, it contains two double bonds or one triple bond and (4 –2) or two rings.

3.

Write structures and the IUPAC name of all the isomeric compounds (molecular formula: C4H6) by determining the double bond equivalent. 4(4 - 2) + 6(1 - 2) 8-6 2 Solution The DBE of the compound = +1= +1= +1= 2 2 2 2 The compound, therefore, contains two double bonds or one triple bond or one double bond and one ring or two rings. The following nine isomers of the compound are possible: 1. CH2==CH—CH==CH2 (Buta-1,3-diene) 2. CH2==C==CH—CH3 (Buta-1,2-diene) 3. CH3CH2C∫∫CH (But-1-yne) 4. CH3C∫∫CCH3 (But-2-yne) 5. 6. 7.

(cyclobutene) (1-Methyleyelopropene) CH3 (3-Methylcyclopropene)

1.46

Organic Chemistry—A Modern Approach

8. 9.

== CH2 (Methylenecyclopropane)

(Bicyle[1.1.0]butane)

1.

Calculate the double bond equivalent (DBE) of each of the following compounds: (b) C12H16N2O4 (a) C13H9BrS

2.

Calculate the double bond equivalent (DBE) of a compound with the molecular formula C6H10. The compound consumes 2 moles of hydrogen on catalytic hydrogenation. Write structures and the IUPAC names of all the possible isomers of the compound. How does one distinguish between rings and double bonds?

3. 4.

5.

1.4

Calculate the double bond equivalent (DBE) for each molecular formula: (a) C6H11Br (b) C5H8O (c) C8H9N and propose one possible structure for each of these compounds. Which atoms can be ignored during the calculation of double bond equivalent?

ACIDS AND BASES

Several organic compounds behave as acids and bases. Two important theories have been discussed below:

1.4.1

Brönsted-Lowry Theory of Acids and Bases

According to the Brönsted-Lowry theory, a substance that can donate a proton is called a Brönsted acid (represented by the symbol H–A) while a substance that can accept a proton is called a Brönsted base (represented by the symbol B:). In the following reaction, hydrochloric acid (HCl) is a Brönsted acid because it donates a proton to water while water is a Brönsted base because it accepts a proton from HCl by using one of its unshared electron pairs. Similarly, in the reverse reaction H3O≈ is a Brönsted acid because it donates proton to Cl① and Cl① is Brönsted base because it accepts a proton from H3O≈.

It is to be noted that all Brönsted acids contain a proton and the net charge may be zero, ≈ or (e.g., HCl, H3O≈, HSO4 , etc.) while all Brönsted bases contain a lone pair of electrons or a p bond and the net charge may be zero or (e.g., NH 3 , CH2 == CH2, OH①, etc.).

Structure, Bonding and Proper es of Organic Molecules

1.47

The molecule or ion that forms when an acid loses its proton is called the conjugate base of that acid. The chloride ion (Cl①) is, therefore, the conjugate base of HCl. The molecule or ion that forms when a base accepts a proton is called the conjugate acid of that base. The hydronium ion (H3O≈) is, therefore, the conjugate acid of water. Therefore, in any Brönsted acid-base reaction, there are two conjugate acid-base pairs. Acid strength and pKa: Acidity is a measure of how easily a compound gives up a proton while basicity is a measure of how well a compound shares its electrons with a proton. A strong acid is one that gives up its proton easily and hence its conjugate base must be weak because it has little affinity for the proton. A weak acid, on the other hand, is one that gives up its proton with difficulty and so, its conjugate base is strong because it has high affinity for the proton. Thus, the important relationship which exists between an acid and its conjugate base is: the stronger the acid, the weaker its conjugate base. For example, HI is a stronger acid than HCl. This means that I① is a weaker base than Cl①. Similarly, the stronger the base, the weaker its conjugate acid. For example, OH① is a stronger base than NH3. This means that H2O is a weaker acid than NH ! 4 . When a strong acid, like HCl, HI, HNO3, H2SO4, etc., is dissociated in water, it dissociates almost completely, which means that products are favoured at equilibrium. When a weaker acid, like CH3COOH, C6H5COOH, etc., is dissociated in water, it dissociates only to a small extent, so reactants are favoured at equilibrium. Acidity is measured by an equilibrium constant, when a Brönsted acid H—A is dissolved in water, an acid-base reaction occurs. The transfer of proton from the acid to water can be represented by the following equilibrium: !

ææ Æ A:@ + H3 O: H - A + H2O: ¨æ æ K eq =

[A:@ ] [H3O:! ] [HA] [H2O]

For dilute aqueous solutions, the concentration of water remains effectively constant and so, the expression for the equilibrium constant can be rewritten in terms of a new constant, Ka, called the acid dissociation constant or acidity constant. K a = K eq .[H 2O] =

[A:@ ] [H 3O] [HA]

A large value of Ka means that the acid is a strong acid and a small value of Ka means that the acid is a weak acid. The strong acid hydrochloric acid, for example, has an acidity constant of 107, whereas the weak acid acetic acid has an acidity constant of only 1.74 × 10–5. For convenience, the strength of an acid is generally indicated by its pKa value rather than its Ka value. pKa = – log Ka

1.48

Organic Chemistry—A Modern Approach

Thus, the stronger the acid, the larger its Ka and consequently, the smaller is its pKa. Very strong acids have pKa values < 1, moderately strong acids have pKa values 1 – 5, weak acids have pKa values 5 – 15 and extremely weak acids have pKa values > 15.

Acid

Ka

pKa

Conjugate base

CH3CH3

10–50

50

CH3CH2@

CH2 == CH2

10–44

44

CH2==CH@

H—H

10–35

35

H@

NH3

10–33

33

NH2@

HC ∫∫ CH

10–25

25

HC ∫∫ C@

CH3CH2OH

10–16

16

CH3CH2O@

H2O

10–15.7

15.7

OH@

4.8

CH3COO@

CH3COOH

–4.8

10

CF3COOH

1

0

CF3COO@

HNO3

20

–1.3

NO3@

H3O

50

–1.7

H2O

HCl

7

10

–7

Cl@

H2SO4

109

–9

HSO4@

HI

1010

–10

I@

HClO4

1010

–10

ClO4@

!

Increasing Strength (basicity) of the conjugate base

Increasing Strength (acidity) of the acid

Scale of acidi es and basici es

Since a strong acid readily donates a proton and a strong base readily accepts a proton, therefore, these two species react to form a weaker conjugate acid and a weaker conjugate base that do not donate or accept a proton readily. Thus, the equilibrium always favours the formation of the weaker acid and weaker base. The order of acidity of some of the important weak acids that we generally encounter is as follow: RH < RCH == CH2 < H2 < NH3 < RC ≡≡ CH < ROH < H2O < RCOOH Acidity increases

Structure, Bonding and Proper es of Organic Molecules

1.49

Like acids, the stronger the base, the larger its Kb and consequently smaller is it pKb. It is, however, more usual to describe the strength of base also in terms of Ka and pKa, thereby establishing a single continuous scale for both acids and bases. To make this possible, the following equilibrium is considered. BH≈ Conjugate acid

ææ Æ B: + H3O + H2O ¨æ æ Base

of the Base

Therefore,

Ka =

[B :][H 3O≈ ] [BH≈ ]

Also, pKa = – log Ka Thus, weaker the base B: or stronger its conjugate acid BH≈, the larger is its Ka and consequently the smaller is its pKa.

1.4.1.1

Factors Affecting Acidity

Acid strength is determined by the stability of the conjugate base that is formed when the acid gives up its proton. Any factor that stabilizes conjugate base A① makes the starting acid HA more acidic. Four factors generally affect the acidity of acid H–A and these are: (1) element effect, (2) inductive effect, (3) resonance effect and (4) hybridization effect. In this article we will discuss the first and last effects. We will discuss the other two in the subsequent articles.

Element effect The most important factor that determines the acidity of a Brönsted acid is the identity of the atom (i.e., the location of the atom in the periodic table) to which the acidic hydrogen is attached. The effect of the attached atom A on the acidity of a Brönsted acid H—A is termed as element effect. Across a row (period) of the periodic table, Brönsted acidity increases as the electronegativity of the atom to which the acidic hydrogen is attached increases. For example: pKa

H — CH3 50

H — NH2 33

H — OH 15.7

H—F 3.2

Increasing electronegativity Increasing acidity

Down the column (group) of the periodic table, Brönsted acidity increases as the size of the atom to which the acidic hydrogen is attached increases. For example:

1.50

pKa

Organic Chemistry—A Modern Approach

H—F 3.2

H — Cl –7

H — Br –9

H—I –10

Increasing size Increasing acidity

Explanation of these Trends: Although the ionization of H–A occurs in one step and involves a base to accept the proton, we may think of the actual process as the sum of three processes for understanding the observed trends in acidity. These three processes are as follows: Bond breaking:

(1)

Electron transfer to A.:

e – + A.Æ A:

(2)

Loss of an electron from H.:

H.Æ H≈ + e –

(3)

@

If the identical terms from either side are cancelled, the sum of these steps is the overall dissociation reaction. Anything that makes any of these steps more fovourable tends to increase the acidity of the Brönsted acid. Let us consider the energetics of each of these steps. In the first step, hemolytic cleavage of the H—A bond occurs. Hence, the energy required for this step is called the bond dissociation enthalpy. Trends in bond dissociation enthalpy are shown by the data given below: Within group 17 (VIII A) of the periodic table Bond H—F H — Cl H — Br –1 569 431 368 Bond dissociation enthalpy (kJ mol )

H—I 297

Within the second period of the periodic table Bond H — CH3 H — NH2 H — OH –1 439 448 498 Bond dissociation enthalpy (kJ mol )

H—F 569

Since the bond dissociation enthalpy is the amount of energy required for the dissociation of a bond, therefore, smaller amounts represent more favorable reactions. In the second step, an atom or group A. accepts an electron from the corresponding anion. The energy released in this step is, in fact, the electron gain enthalpy of .A. Trends in electron gain enthalpy are shown by the data given below:

Structure, Bonding and Proper es of Organic Molecules

1.51

Within group 17 (VIII A) of the periodic table Atom

:F ◊

:Cl ◊

:Br ◊

:I ◊

Electron gain enthalpy (kJ mol–1)

328

349

324

295

Within the second period of the periodic table Group Electron gain enthalpy (kJ mol–1)

◊CH3

◊NH3

◊OH

◊F:

7.7

74

177

328

Since the electron gain enthalpy is the energy released when a group combines with an electron, therefore, larger amounts represent more favourable reactions. In the third step, the hydrogen atom loses an electron. The energy required for this step is the ionization enthalpy of the hydrogen atom. Since this step is the same for all the Brönsted acids, it does not enter into comparison of different acids. From the bond dissociation enthalpy and electron gain enthalpy data, it becomes clear that within a column (group) of the periodic table, electron gain enthalpies do not change as much as bond dissociation enthalpies. For example, the electron gain enthalpies of Cl and I differ by only 54 kJ mol–1 while the bond dissociation enthalpies of H — Cl and H — I differ 134 kJ mol–1. Hence, the greater strength of the Brönsted acids H — A towards higher atomic number or larger size within a column (group) of the periodic table is due primarily to weaker H — A bonds. Across a row (period) of the periodic table, electron gain enthalpies change much more than the bond dissociation enthalpies. The increase in electron affinities from •CH3 to •F is 320 kJ mol–1, whereas the increase in bond dissociation enthalpies from H — CH3 to H — F is only 130 kJ mol–1. The direction of the change in bond dissociation enthalpies is also to be noted. Hence, the greater strength of the Brönsted acid H — A from left to right along a row (period) of the periodic table is due primarily to the ability of the atoms or groups A to attract electrons. This trend, in fact is similar to the trend in electronegativities which is actually the ability to attract electrons within a chemical bond.

Hybridization effect: The hybridization of the atom bonded to hydrogen is an important factor affecting the acidity of H — A. Let us consider ethane (CH3CH3), ethylene (CH2 == CH2) and acetylene (HC ≡≡ CH), the three different compounds containing C—H bonds to illustrate the phenomenon. From the pKa values, it becomes clear that acetylene is more acidic than ethylene which in turn is more acidic than ethane.

1.52

Organic Chemistry—A Modern Approach

This order of acidities can be explained on the basis of the hybridization state of carbon in each compound. Electrons of 2s orbital have lower energy than those of 2p orbitals and this is because 2s electrons tend, on the average, to be much closer to the nucleus than 2p electrons. Therefore, the electrons in an sp hybrid orbital having 50% s character (because they arise from one s orbital and one p orbital) are closer, on the average, to the carbon nucleus than those in an sp2 hybrid orbital having 33.3% s character. In turn, the electrons in an sp2 hybrid orbital are closer to the nucleus than those in an sp3 hybrid orbital having 25% s character. The closer the bonding electrons to the nucleus, the more electronegative the atom. This means, in effect, the sp-hybridized carbon in acetylene is more electronegative than the sp2-hybridized carbon in ethylene, which in turn is more electronegative than the sp3-hybridized carbon in ethane. Because the electronegativity of carbon atoms follows the order Csp > C 2 > C 3 , proton release will be progressively sp

sp

favoured in ethane, ethylene and acetylene, i.e., the order of acidity is HC ≡≡ CH > H2C == CH2 > CH3—CH3. This order may also be explained on the basis of the stability of the carbanion, i.e., the conjugate base of the carbon acid. The more the carbanion is stable, the more the hydrocarbon is acidic. As the amount of s character of the hybrid orbital occupied by the unshared electron pair increases, the stability of the carbanion increases and this is due to the fact that the electron pair in an orbital having good s character is held more tightly by the nucleus and hence of lower energy.

1.4.2

Lewis Acid-Base Theory

According to G.N. Lewis, an acid is a species that can accept an electron pair and a base is a species that can donate an electron pair. All Brönsted acids are Lewis acids because they can lose a proton and the proton can accept an electron pair. The Lewis definition of an acid is much broader than the Brönsted-Lowry definition and this is because according to Lewis definition not only the compounds that can donate protons but also the compounds like aluminium chloride (AlCl3), boron trifluoride (BF3) etc., are acids because they have sextet of electrons (vacant orbitals in their valence shell) and thus can accept a pair of electrons. The latter compounds are not Brönsted acids because they are not proton donors. They react with a compound containing an unshared electron pair just like a proton reacts with ammonia.

Structure, Bonding and Proper es of Organic Molecules

1.53

A curved arrow is drawn from the electron pair of the base to the electron-deficient atom of the acid. The term ‘Lewis acid’ is generally used to refer to non-proton-donating acids like AlCl3, BF3, FeCl3, etc. All Lewis bases are Brönsted bases and vice versa. They have an unshared pair of electrons that they can share. They can share electrons either with an atom like Al, Fe, B or they can share the electrons with H≈ . For example, water, diethyl ether and ammonia are both Lewis bases and Brönsted bases.

1.

Predict the relative acidity of (a) ethyl alcohol (CH3CH2OH) and ethylamine (CH3CH2NH2); (b) ethyl alcohol (CH3CH2OH) and ethanethiol (CH3CH2SH); ≈

(c) CH 3 OH 2

≈

and CH 3NH 3

Solution The nature of the atom holding the proton to be lost, i.e., the atom that will hold the unshared pair of electrons in the conjugate base being formed is to be considered. The better this atom accommodates these electrons, the greater the extent to which the conjugate base is formed, and hence, the stronger is the acid.

1.54

Organic Chemistry—A Modern Approach

H: A Acid (holds proton)

ææ Æ + :B ¨æ æ

:A @ +

H:B≈

Conjugae base (holds the unshared pair of electrons)

Two factors that determine an ability to accommodate electrons are (i) its electronegativity, since a more electronegative atom has a greater avidity for electrons: and (ii) its size, since a larger atom permits greater dispersal of charge of the electrons and this tends to stabilize a charged particle. The electronegativity of atoms increases from left to right across a period of the Periodic Table and size of atoms increases on moving down the group of the Periodic Table. (a) relative acidity: CH3CH2OH > CH3CH2NH2 Nitrogen and oxygen are in the same period or row of the Periodic Table and oxygen is more electronegative than nitrogen. (b) relative acidity: CH3CH2SH > CH3CH2OH Sulphur and oxygen are in the same group of the Periodic Table, and sulphur is larger in size. ≈

≈

(c) relative acidity: CH3 OH2 > CH3 NH2 Oxygen and nitrogen are in the same period or row of the Periodic Table and being more electronegative oxygen is less able to accommodate a positive charge. 2.

≈

Which is the stronger acid of each pair: (a) CH 3 O H 2 or CH 3 OH; ≈

(b) CH 3 N H 2 or CH 3 NH 2 ; (c) H2S or HS①; (d) OH① or H2O Solution The accommodation of the electron pair left behind upon loss of a proton is easiest for the neutral conjugate base formed from a positively charged acid [equation (i)], harder for the negatively charged conjugate base formed from a neutral acid [equation (ii)] and still harder for the doubly charged conjugate base formed from a negatively charged acid [equation (iii)]. ≈

… (i)

ææ Æ H2O + H:B H3O≈ + : B ¨æ æ Positive

ææ Æ H2O + :B ¨æ æ

Neutral

ææ Æ OH@ + :B ¨æ æ

Negative

≈

Neutral

OH

@

≈

… (ii)

+ H:B

Negative

O2 @

Doubly negative

≈

… (iii)

+ H:B ≈

Therefore, (a) CH3 OH2 is more acidic than CH3OH, (b) CH 3 NH 2 is more acidic than CH3NH2, (c) H2S is more acidic than SH① and (d) H2O is more acidic than OH①.

Structure, Bonding and Proper es of Organic Molecules

3.

1.55

Arrange the members of each group in order of decreasing basicity: (a) OH①, SH①, SeH① (b) F①, Cl①, Br①, I① (c) F①, OH①, NH2①, CH3① (d) Cl①, SH①.

Solution In (a) and (b), the atoms of each set are in the same group of the Periodic Table, and thus the negative charge will be better accommodated, i.e., the basicity of the anion decreases with increase in atomic size down the group. Therefore, the decreasing order of basicity is as follows: (a) OH① > SH① > SeH①; (b) F① > Cl① > Br① > I①.

In (c) and (d), the atoms of each set are in the same period of the Periodic Table, and therefore, the accommodation of the negative charge depends upon electronegativity: the more electronegative the element on moving from left to right in the Periodic Table, the weaker the base. Therefore, the decreasing order of basicity is as follows: (a) CH3① > NH2① > OH① > F① ; (b) SH① > Cl①. 4. Arrange the member of each group in order of increasing basicity: (a) H2S, HS①, S2①; (b) H3O⊕, H2O, OH① (c) NH3, NH2①. Solution For a given atom, the availability of electrons is greatest in an electronrich negatively charged species and least in an electron-poor positively charged species. Therefore, the increasing order of basicity is as follows: (a) H2S < HS① < S2①, (b) H3O⊕, < H2O < OH① (c) NH3 < NH2①.

5.

Arrange the following ions in order of decreasing acidity and explain the ≈

≈

order: CH 3 CH 2 N H 3 , CH 3 CH == NH 2 , CH 3 C ∫∫ N H Solution These three ions differ in the hybridization of the nitrogen atom to which the acidic hydrogen is attached. It is known that hybridization of an atom affects its electronegativly (sp is more electronegative than sp2, and sp2 is more electronegative than sp3). Again, a more electronegative atom is less willing to accommodate the positive charge and more willing to releasing a proton. Therefore, the decreasing order of acidity of these ≈

≈

≈

ions is CH 3C ∫ NH > CH 3CH = NH 2 > CH 3CH 2 NH3 6.

Write down the conjugate acids of (a) CH3O①, (b) CH3CH2NH2, (c) [: O — O:]2@ , (d) CH2 = CH2, (e) :CH2 and the conjugate bases of (f) CH3CH2NH2, (g) CH2 = CH2, (h) HNO3.

Solution

(a) CH3OH, ≈

(d) CH3 — CH2 , @

(g) CH2 == CH , 7.

≈

(b) CH 3CH 2 NH 3 , ≈

(c) HO — O:@ , @

(e) CH3 ,

(f) CH3CH2 NH ,

(h) NO3①.

Which of the following species are expected to be amphoteric and why? (a)

HCO3①,

(b) HF,

(c)

≈

N H4 ,

(d)

NH3,

(e)

Cl①

1.56

Organic Chemistry—A Modern Approach

Solution (a) It is amphoteric because it gives H2CO3 (CO2 + H2O) and CO32①; (b) it is amphoteric because it gives H2F⊕ and F① ions ; (c) it cannot accept an H⊕ ion because it has no unshared pair of electrons, and hence, it is not amphoteric; (d) it is amphoteric because ≈ it gives NH4 and NH2① ions; (f) it is not amphoteric because it cannot donate H⊕.

8.

Write down the net ionic reaction taking place when sodium hydride (NaH) is dissolved in methanol.

Solution Sodium hydride contains a sodium ion (Na⊕) and a hydride ion (H①). Although the sodium ion is a spectator ion, the hydride ion (the conjugate base of a very weak acid H2) is a very strong base. Because of this, the hydride ion reacts readily with a molecule of methanol, MeOH (a stronger acid), to produce methoxide ion, MeO① (a weaker base) and H2 (a weaker acid). CH3OH + :H@ Æ CH3O@ + H2≠ Stronger acid (pK a = 15.5)

9.

Stronger base

Weaker base

Weaker acid (pK a = 35)

Explain why propylamine (CH3CH2CH2NH2) is more basic than pyridine (C6H5N), which in turn more basic than ethyl cyanide (CH3CH2CN).

Solution The basicity of these nitrogenous compounds can be explained in terms of the hybridization state of nitrogen. The nitrogen atoms in propylamine, pyridine and ethyl cyanide are sp3, sp2 and sp hybridized respectively, and so the nonbonding pair of electrons on nitrogen occupies sp3 (25 percent s character), sp2 (33.3 percent s character) and sp, (50 percent s character) hybrid orbitals respectively. Since s electrons are held more tightly to the nucleus and hence of lower energy compared to p electrons, the unshared pair of electrons in orbitals having a large amount of s character is attracted towards the nitrogen nucleus with a greater force and is less available for forming a bond with proton. Thus, the availability of the unshared pair of electrons on nitrogen, i.e., the basicity of these nitrogenous compounds decreases in the order given below.

10.

But-2-ynoic acid (CH3C ∫∫ CCOOH) is a stronger acid than but-2-enoic acid (CH3CH == CHCOOH), which in turn is a stronger acid than butanoic acid (CH3CH2CH2COOH). Explain.

Structure, Bonding and Proper es of Organic Molecules

1.57

Solution The electronegativity of various hybridized carbon atoms follows the order:

C sp > C sp2 > C sp3 . The a-carbon of butanoic acid, but-2-enoic acid and but-2-ynoic acid are sp3, sp2 and sp hybridized respectively. Therefore, the conjugate base of but-2-ynoic acid in more stable than the conjugate base of but-2-enoic acid, which in turn is more stable than the conjugate base of butanoic acid. CH3CH2CH2 COO①, CH3CH == CHCOO①, CH3C ∫∫ CCOO①

Stability increases

Hence, the acidity order is CH3C ∫ CCOOH > CH3CH = CH COOH > CH3CH2CH2COOH 11.

How can CH3C ∫∫ CH be converted into CH3C ∫∫ CT?

Solution CH3C ∫∫ CH is first treated with sodium amide in liquid ammonia and then T2O is added to the resulting solution containing CH3C ∫∫ C① Na⊕ to form CH3C ∫∫ CT. The following two acid-base reaction takes place.

1.

Arrange the following compounds in order increasing acidity and explain your answer: CH3CH2OH, CH3CH2NH2, CH3CH2CH3

2.

≈

@

NaH reacts with ethanol (CH3CH2OH) not to produce Na:CH 2CH 2OH plus H2, @

≈

but to produce CH 3CH 2 O Na plus H2 — why? 3.

Arrange the following compounds in order of decreasing basicity and explain your answer: CH3ONa, CH3CH2Li, NaNH2

1.58

Organic Chemistry—A Modern Approach

4.

Indicate which is the stronger acid in each of the following pairs and why? ≈

≈

(a) CH 3CH 2 NH 3 or CH 3CH 2 OH 2 5.

6.

(b) HCl or HBr

(c) CH3OH or CH3SH

Designate the following species as Lewis acids or bases: (a) :H①

(b) H2C == CH2

(e) CH3OH

(f) Me3 C

!

(c) CH3CH2CH2CH3

(d) BF3

(g) :Cl:@

Which of the indicated protons in each of the following molecules is more acidic and why? C∫∫ C—H ¨

(b)

(a)

S—H ¨

(c)

NH—H ¨ CH2O—H ¨

7.

Predict the conjugate acid and the conjugate base of the following compound and give your reasons: NH2 HO

8.

Arrange the following species in order of increasing acidity and explain the order: (a) H2O, H3O⊕ HO①

(b) HC ≡≡ CCH2CH3, CH3CH2 CH == CH2, CH3CH2 CH2CH3

(e) HCl, H2S, NH3 9.

Arrange the following ions in order of increasing basicity and explain the order: (a)

–

CH2CH2 ,

–

–

C∫∫ C,

CH== CH

(b) CH3①, Cl①, HO①, 10.

Which compounds are expected to be formed when OH① is allowed to react with the ≈

carbocation Me3 C as a Brönsted-Lowry base and as a Lewis base? [Hint: 2-methylpropene (CH2 = CMe2) will be formed when OH① will react as a Brönsted-Lowry base and tert-butyl alcohol will be formed when OH① will react as a Lewis base.] 11.

Classify each compound as a Brönsted-Lowry base, a Lewis base, both or neither: (a) Me2C == O CH3 (c)

(b) (d)

C2H5—Br CH3

Structure, Bonding and Proper es of Organic Molecules

12.

1.59

Classify each compound as a Brönsted-Lowry acid, a Lewis acid, both or neither: (a) BF3

(b) BF4①

(c) H3O⊕

≈

(d) (CH 3 )3 C

13.

Explain why acetylene (HC ≡≡ CH) is more acidic than ethane (CH3CH3), even though the C—H bond in acetylene has a higher bond dissociation energy than that the C—H bond in ethane.

14.

Explain why cyclopropane is more acidic than cyclohexane. [Hint: In cyclopropane, imperfectly hybridized sp3 orbitals of carbon are involved in bond formation. The inner orbitals forming C—C bonds have about 17 percent s character (approximately sp5 orbitals), while the outer orbitals forming C—H bonds have about 33% s character (approximately sp2 orbitals). The orbitals forming C—H bonds in cyclohexane have only 25 percent s character (sp3 orbitals). Since the outer orbitals have greater s character, therefore, the conjugate base of cyclopropane is relatively more stable than that of cyclohexane and for this reason, cyclopropane is more acidic than cyclohexane.

15.

Glycine is an amino acid which in solution exists in equilibrium between two forms: ≈ H 3 N CH2COO① H2NCH2COOH Which form is expected to be favoured at equilibrium? [Hint:

The ionic form contains the groups that are the weaker acid and weaker base. The equilibrium, therefore, will favour this form.]

1.60

Organic Chemistry—A Modern Approach

Important fact and concepts at a glance 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

1.5 1.5.1

Brönsted-Lowry acids: Proton donors Brönsted-Lowry bases: Proton acceptors Lewis acids: Electron pair acceptors Lewis bases: Electron pair donors The lower the pKa, the stronger the acid (pKa = –log Ka) In acid–base reactions, the equilibrium favours the weaker acid and weaker base because they are relatively more stable. The conjugate base of any acid having a higher pKa can deprotonate an acid. The acidity of H–A increases both left-to-right across a period and down a group of the periodic table. The acidity of H–A increases as the percent s character of the atom in A① attached to hydrogen increases. The more stable the conjugate base, the more acidic the acid.

INDUCTIVE AND ELECTROMETRIC EFFECTS Inductive Effect

When a covalent bond is formed between two atoms having different electronegativities, the bonding electron pair is not shared absolutely equally by the two atoms. The electron pair is attracted more by the more electronegative atom and so, it becomes shifted more towards that atom. As a consequence, the more electronegative atom acquires a partial negative charge (i.e., d+). When an electronegative (electron-withdrawing) F atom, for example, is attached to the end of a carbon chain (whose carbon atoms are designated as 1, 2, 3, … etc.), the s electrons of the C-1—F bond are attracted by or displaced more towards the F atom. As a result, the F atom acquires a partial negative charge (d-) and the carbon, C-1, acquires a partial positive charge (d+). As C-1 is now somewhat positively polarized, it, in turn, attracts the s electrons of the C-1 — C-2 bond towards itself. This causes C-2 to acquire a partial positive charge (dd+) smaller in magnitude than that on C-1. Similarly, C-3 acquired a partial positive charge (ddd+) even smaller than that on C-2. ddd +

dd +

d+

3

2

1

d-

C Æ— C Æ— C Æ— F [where, d + > dd + > ddd + ]

Similarly, if an element less electronegative than carbon, such as lithium (Li) (or any other electron-releasing group or atom) is attached to the terminal carbon of a carbon chain, then a partial positive charge (d+) is developed on the Li atom and a partial negative charge (d-) is developed or the C-1 atom. The small negative charge developed on C-1, in turn, repels the s electrons of the C-1—C-2 bond towards C-2. As a result, C-2 acquired a partial negative charge (dd-) smaller in magnitude than that on C-1. Similarly, C-3 acquired a partial negative charge (ddd-) which is even smaller than that on C-2.

Structure, Bonding and Proper es of Organic Molecules ddd -

dd -

d-

3

2

1

1.61 d+

C Æ— C Æ— C Æ— Li [where, d - > dd - > ddd - ]

Definition: When some atom or group, either more or less electronegative than carbon remains attached to a carbon chain, a permanent bond polarization caused by displacement of electrons occurs along the carbon chain. This is what is called inductive effect. This type of charge dispersal, which goes on diminishing rapidly as the distance from the source increases and almost dies down after the third carbon atom. Although the inductive effect causes some degree of polarity in the covalent bond, yet the bond is never cleaved due to this effect. The inductive effect is represented by the symbol ( Æ— ). The arrow always points towards the more electronegative atom or group. Inductive effect cannot be expressed by any absolute value. Relative inductive effect of an atom or group is measured by taking H atom of R3C—H molecule as standard. When an atom or group Z of the C—Z bond of R3C—Z molecule attracts the bonding electrons more strongly than hydrogen of the C–H bond in R3C – H molecule, then according to the definition introduced by Ingold, Z is said to have negative inductive effect or electronwithdrawing inductive effect or –I effect. On the other hand, if the atom or group Z pushes the bonding electrons of the C—Z bond more strongly than the hydrogen atom of the C—H bond, then it is said to have positive inductive effect or electron-releasing inductive effect or +I effect. Some +I and –I groups are arranged below in the order of decreasing strength of inductive effect: +I groups: –I group:

@

① ① - NH > –O > –COO > (CH3)3C – > (CH3)2CH – > CH3CH2 – > CH3 – > D ≈

≈

≈

- NR 3 > - SR 2 > - NH 3 > –NO2 > – SO2R > – CN > – COOH > –F > –Cl > –Br > – I > – OR > – OH

The phenomenon of inductive effect may be used to explain some important properties of organic compounds such as acidic property, basic property, bond polarity and chemical reactivity. Stabilities of carbocation and carbanions can also be explained by inductive effect.

1.5.1.1

Acidity of organic compounds

(1) Strength of monocarboxylic acids Relative strengths of monocarboxylic acids can be explained in terms of inductive effect of the substituent present in the carbon chain. If an electron-releasing group or atom is attached directly to the —COOH group or to the carbon chain close to the –COOH group, then the positive inductive effect (+I) of such group increases the electron density on the oxygen atom of the O—H group and as a consequence, the shared pair of electron of the O—H bond is less strongly attracted towards the oxygen atom. As a result, the dissociation of the O—H bond to give H⊕ ion is less favoured. Thus, a +I group, when present in a monocarboxylic acid molecule, decreases the strength of that acid. On the

1.62

Organic Chemistry—A Modern Approach

other hand, if an electron-withdrawing group or atom is attached to the carbon chain close to the —COOH group, the negative inductive effect (–I) of such group decreases the electron density on the oxygen atom of the O—H group and consequently, the O—H bonding electrons are more strongly attracted towards the oxygen atom. As a result, the dissociation of the O—H bond to give H⊕ ion is facilitated. Thus, a –I group, when present in a moncarboxylic acid molecule, increases the strength of the acid. Fluoroacetic acid, for example, is stronger than acetic acid. O || CH3Æ— CÆ— O Æ—H Acetic acid (pK a = 4.76) (Weaker acid due to + I effect of the - CH3 group)

O || F —¨ CH2 —¨ C —¨ O —¨ H Fluoroacetic acid (pK a = 2.66) (Stronger acid due to - I effect of the F atom)

Acid strength can also be explained in terms of relative stabilities of the acid and its conjugate base. The more conjugate base is stable, the more the acid will be acidic. Electron-withdrawing groups (EWGs) disperse the negative charge on the conjugate base, thus stabilise it and hence increase acidity. Electron-donating groups (EDG’s), on the other hand, intensity the negative charge on the conjugate base, thus destabilise it and hence decrease acidity.

The electron-withdrawing F atom disperses the negative charge on the conjugate base of fluoroacetic acid and thereby stabilizes it. In the conjugate base of acetic acid, on the other hand, the negative charge on the carboxylate anion is intensified by the electronreleasing —CH3 group and as a consequence, the anion is destabilized. It thus follows that fluoroacetic which yields a more stable conjugate base is relatively more acidic than acetic acid.

Structure, Bonding and Proper es of Organic Molecules

1.63

The strength of carboxylic acid increases as the extent of –I effect of the substituent increases and this is because the O—H bond becomes progressively more weak or the conjugate base becomes progressively more stable. The –I effect of halogens follow the order: F > Cl > Br > I. So, among the halogen substituted acetic acids, fluoroacetic acid (FCH2COOH) is the strongest while iodoacetic acid (ICH2COOH) is the weakest. F —¨ CH2 —¨ COOH Fluoroacetic acid (p K a = 2.66)

Cl —¨ CH2 —¨ COOH Chloroacetic acid (p K a = 2.86)

Br —¨ CH2 —¨ COOH Bromoacetic acid (p K a = 2.90)

I —¨ CH2 —¨ COOH Iodoacetic acid (p K a = 3.16)

With increase in the number of electron-attracting substituents, the strength of the acid increases and this is because the O—H bond becomes progressively more weak or the conjugate base becomes progressively more stable. For example, trichloroacetic acid is more stronger than dichloroacetic acid which in turn is more stronger than monochloroacetic acid.

Acid strength increases

As the distance of the electron-attracting substituent from the carboxyl group increases, the strength of the acid decreases and this is because the O—H bond becomes progressively more stronger or the conjugate base becomes progressively less stable. For example, 2-chlorobutanoic acid is a stronger acid than 3-chlorobutanoic acid which in turn is a stronger acid than 4-chlorobutanoic acid. Cl | ≠ CH3 - CH2 – CH—¨ COOH 2-Chlorobutanoic acid (pK a = 2.84)

Cl | ≠ CH3 - CH —¨CH2 —¨ COOH 3-Chlorobutanoic acid (pK a = 4.06)

Cl | ≠ CH —¨ CH2 —¨ CH2 —¨ COOH 2

4-Chlorobutanoic acid (pK a = 4.52)

Acid strength increases

(2) Acidic character of alcohols

d-

d+

The acidic nature of alcohols is due to the presence of polar O — H bond. Oxygen being more electronegative than hydrogen withdraws electrons of the O—H bond towards itself. As a result, the O — H bond becomes weak and alcohols exhibit acidic character by donating its proton to strong bases. In fact, alcohols are very weak acids (pKa = 16–18). The alkyl (–R) group by its electron-donating inductive (+I) effect increases the electron

1.64

Organic Chemistry—A Modern Approach

density on oxygen (R Æ— O—H) and as a result, the polarity of O–H bond decreases. As the number of electron-releasing alkyl (–R) group increases, the polarity of the O—H bond progressively decreases and because of this, the acidity of alcohols decreases gradually from primary (containing on alkyl group) to tertiary (containing three alkyl groups).

(3) Gas-phase acidity of alcohols The relative order of acidity of different types of alcohols is reversed in the gas phase compared with the relative order of acidity in solution. Relative gas-phase acidity: (CH3)3COH > (CH3)2CHOH > CH3CH2OH > CH3OH. In the gas phase, the a-alkyl substituents stabilize the charged alkoxide ions by a polarization mechanism. That is, the electron clouds of each alkyl group distort so that the electron density moves away from the negative charge on the aloxide oxygen, leaving a partial positive charge nearby. As a consequence, the anion is stabilized by its favouarble electrostatic interaction (forces of attraction) with the partial positive charges. The stabilization of the alkoxide ion by this polarization effect increases with increase in the number of a-alkyl substituents. Therefore, a tertiary alkoxide ion having three a-alkyl substituents is the most stable one and the methoxide ion with no a-alkyl substituent in the least stable one, consequently, tert-butyl alcohol, (CH3)3COH, is the most acidic in the gas phase and methyl alcohol, CH3OH, is the least acidic in the gas phase.

Structure, Bonding and Proper es of Organic Molecules

1.5.1.2

1.65

Basicity of organic compounds

(1) Basic strength of amines The availability of the unshared pair of electrons on nitrogen for protonation determines the basic strength of nitrogenous compounds. We might expect to see an increase in basic strength in the sequence NH3 Æ CH3NH2 Æ (CH3)2NH Æ (CH3)3N. Due to the increasing inductive effect (+I) of successive —CH3 group making the N atom progressively more electron rich, i.e., making the unshared pair of electrons more readily available. However, this sequence of basic strength of amines agrees with result if measurements of basicity are made in the gas phase or in a solvent in which H-bonding does not take place.

The observed order of basic strength of methylamines in water is (CH3)2NH(2°) > CH3NH2 (1°) > (CH3)3N (3°). In the H-bonding solvent water, the basic strength of an amine is determined not only by the electron availability on the N atom, but also by the extent to which the conjugate acid, i.e., the ammonium ion is stabilized by salvation through hydrogen bonding. As the number of electron-releasing (+I) methyl group increases, the availability of the unshared pair of electrons on nitrogen increases and so, the amines become progressively more basic. Again, as the number of hydrogen attached to nitrogen of the conjugate acid decreases, the extent of salvation by water molecules through H-bonding decreases. As a result, the stability of the conjugate acid decreases, and the amines become progressively less basic. The net effect of these two oppositely acting factors is that the secondary amine (CH3)2NH (with two electron-releasing —CH3 groups and two hydrogen atoms in the conjugate acid for H-bonding) becomes the most basic and the tertiary amine (CH3)3N (with three electron-donating —CH3 groups and only one hydrogen in the conjugate acid for H-bonding) becomes the least basic.

1.66

1.5.1.3

Organic Chemistry—A Modern Approach

Stabilities of carbocations

The increasing order of stability of methyl, primary (1°), secondary (2°) and tertiary (3°) ≈

≈

≈

≈

carbocations is CH 3 < CH 3 CH 2 < (CH 3 )2 CH < (CH 3 )3 C. This order of stability can be explained on the basis of inductive effect. An electron-donating (+I) methyl group tends to stabilize an adjacent carbocation by dispersing the positive charge. Therefore, the greater the number of methyl groups attached to the cationic carbon, the more is the dispersal of positive charge and hence stabler will be the carbocation. Thus, tertiary butyl cation having three methyl groups is more stable than isopropyl cation having two methyl groups which in turn is more stable than ethyl cation having only one methyl group and obviously, methyl cation containing only three hydrogens is the least stable one. Hence, the increasing order of stability is

1.5.1.4

Stabilities of carbanions

An alkyl group intensifies the negative charge on the anionic carbon of a carbanion by its electron-releasing +I effect and thereby, destabilize it. Therefore, methyl anion containing no alkyl group is the most stable one and tert-butyl anion containing three methyl groups is the least stable one. Hence, the order of increasing stability is

Structure, Bonding and Proper es of Organic Molecules

1.5.2

1.67

Field Effect

It is known that inductive effect operates through bonds. However, there is another type of inductive effect operating not through the bond but through space or solvent molecules. This effect is called field effect. It is not very easy to distinguish the inductive effect from the field effect. However, an attempt has been made in this respect by taking advantage of the fact that the inductive effect depends on the nature of the bonds, but the field effect depends on the geometry of the molecule. For example, in the diastereoisomers A and B, the –I effect of the Cl atoms on the position of electron in the —COOH group and therefore, on acidity should be the same since an equal number of bond intervene. Hence, acid strength of the two isomeric acids should be the same. However, the acid strength of the two acids has been found to be different. The acid B is actually more acidic than the acid A. The conjugate base of A is destabilized by repulsive interaction because the negative ends of d+

d-

the two C — Cl dipoles are closer to the —COO① group. Hence, the isomer B is a stronger acid than A. This is also supported by the pKa values of the two acids.

1.5.3

Electromeric Effect

Definition: The temporary effect involving complete transfer of a pair of p electrons of a multiple bond (double bonds such as C == C and C == O and triple bonds such as C ∫∫ C and C ∫∫ N) to one of the multiple bonded atoms (usually the more electronegative one) in the presence of an attacking reagent is called elctromeric effect or E-effect. As soon as the attacking reagent is removed, the transferred p electron pair again from the bond and the molecule reverts back to its original position. Since this effect occurs by the presence of the attacking reagent, it takes places in the direction which facilitates the reaction. The electromeric effect may be represented as follows: in the presence of reagent A==B

+

–

A—B in the absence of reagent

1.68

Organic Chemistry—A Modern Approach

Inductive effect also plays an important role in determining the direction of the electromeric effect. If a +I or –I group is attached to a doubly bonded carbon, the direction of the electromeric effect is the same as that of the inductive effect. For example:

+E effect: If the electron pair of the p bond is transferred to that doubly bonded atom to which the attacking species gets finally attached, then the effect is called +E effect. For example:

–E effect: If the electron pair of the p bond is transferred to that doubly bonded atom to which the attacking species does not get finally attached, the effect is called –E effect. For example:

Distinction between inductive effect and electromeric effect Inductive effect

Electromeric effect

(1) It involves a permanent displacement of the s (1) It involves a temporary but complete transfer of p electron pair to the more electronegative electron pair towards the more electronegative atom, i.e., it is a temporary effect. atom, i.e., it is a permanent effect. (2) In this effect, the displaced electron pair does (2) The transferred electron pair occupies a new not leave its molecular orbital and only a slight orbital. distortion in the shape of orbitals occurs. (3) It does not need any outside reagent for its (3) It operates only in the presence of an outside operation. reagent. (4) There occurs a partial separation of charges.

(4) There occurs a complete separation of charges.

(5) It affects both physical properties and chemical (5) It does not affect the physical properties but it reactivity of the molecule concerned. enhances the chemical reactivity of the molecule concerned. (6) Electronegativity of the bonded atoms controls (6) Its direction is always such that it favours the the direction of inductive effect. reaction. The electronegativity of the bonded atoms or I effect of the attached group may determine the direction of electomeric effect.

Structure, Bonding and Proper es of Organic Molecules

1.

1.69

Explain why CH3CHFCH2OH is more acidic than CH3CHBrCH2OH.

Solution These two compounds differ only in the halogen atom that is attached to C-2 of these alcohols. Since fluorine is more electronegative than bromine, therefore, there occurs greater electron withdrawal from the O—H bond in 2-fluoro-1-propanol than from the O—H bond in 2-bromo-1-propanol and for this reason, the first alcohol is relatively more acidic than the second one. F Br | | ≠ ≠ CH3 —CH—¨ CH2 —¨ O—¨ H CH3 — CH—¨ CH2 —¨ O—¨ H 2- Fluoro -1- propanol 2- Bromo -1- propanol (more acidic) (less acidic)

2.

Which one in each pair is more acidic and why? (a) (b) (c) (d)

CH3CBr2CH2OH or BrCH2CHBrCH2OH CH3CH2CH2CH2OH or CH3O CH2CH2OH (CH3)3 CCH2COOH or (CH3)3SiCH2COOH CH3CHFCH2OH or CH3CHClCH2OH

Solution (a) The two compounds differ in the location of one of the two bromine atoms. Since the bromine atom in the first alcohol (2,2-dibromoprop-1-ol) is closer to the O—H bond than the bromine atom in the second alcohol (2,3-dibromprop-1-ol), therefore, it withdraws O—H bonding electrons more effectively and causes the first alcohol to be more acidic than the second one.

Br | ≠ CH3 — C —¨ CH2 —¨ O—¨ H | Ø Br 2,2- Dibromoprop-1- ol (more acidic) (b)

Br | ≠ Br—¨ CH2 —CH—¨ CH2 —¨ O—¨ H 2,3- Dibromoprop-1- ol (less acidic)

The electronegative ethereal oxygen present in 2-methoxy-1-ethanol (CH3OCH2CH2OH) withdraws O–H bonding electrons and for this reason, this compound is more acidic than 1-butanol (CH3CH2CH2CH2OH) in which less electronegative carbon is present in the place of oxygen. CH3CH2CH2CH2OH 1- Butanol (less acidic)

CH3O—¨ CH2 —¨ CH2 —¨ O—¨ H 2- methoxy -1- ethanol (more acidic)

1.70

Organic Chemistry—A Modern Approach

(c)

Silicon, being electropositive in nature, pushes electron to the O—H bond and makes the bond more stronger. For this reason, the silicon containing carboxylic acid is less acidic than the other containing carbon in the place of silicon. (CH3 )3 CCH2COOH (more acidic)

(d)

These two alcohols differ only in the halogen atom that is attached to the middle carbon of the molecule. Because fluorine is more electronegative than chlorine, there in greater electron withdrawal from the O—H bond in the fluorinated compound, causing it to be the stronger acid. CH3 — CH—¨ CH2 —¨ O—¨ H | Ø F 2- Fluoro -1- propanol (more acidic)

3.

(CH3 )3 SiCH2COOH (less acidic)

CH3 — CH—¨ CH2 —¨ O—¨ H | Ø Cl 2-Chloro-1- propanol (less acidic)

Although HCl is a weaker acid than HBr, chloroacetic acid (ClCH2COOH) is a stronger acid than bromoacetic acid (BrCH2COOH). Explain.

Solution HBr is a stronger acid than HCl because the H—Br bond is weaker than the H—Cl bond and also Br① being larger in size than Cl① is relatively more stable conjugate base. In both carboxylic acids, the same kind of bond is broken (an O—H bond). Therefore, the only factor to be considered is the electronegativities (–I effect) of the halogen atoms that are pulling the O–H bonding electron. Because Cl is more electronegative than Br, Cl is better at pulling electrons away from the O—H bond. Thus, the conjugate base of chloroacetic acid is relatively more stable. Hence, ClCH2COOH is a stronger acid than BrCH2COOH. 4. How do you account for the following facts? (a) Ethanamine is more basic than 2,2,2-trifluroro-1-ethanamine (F3CCH NH2) (b) The amine (C2F5)3N is virtually non-basic. Solution (a) It is the availability of the unshared pair of electrons on nitrogen for protonation which determines the basic strength of nitrogenous compounds. The —C2H5 group in ethanamine (C2H5 N H 2 ) pushes electrons towards the nitrogen atom and makes the unshared pair of electrons on nitrogen easily available. In 2,2,2-trifluoro-1ethanamine (F3C CH 2 N H 2) , on the other hand, the electron-attracting fluorine atom or trifluoromethyl group (F3C –) pulls electrons from nitrogen through carbon and thereby decreases the availability of the unshared pair considerably. Ethanamine is, therefore, more basic than 2,2,2-trifluoro-1-ethanamine.

Structure, Bonding and Proper es of Organic Molecules

1.71

This observation may also be explained in terms of the stability of the ammonium ion, ≈

the conjugate acid of the amine (R - NH 3 ) . The more the ammonium ion in stable, the more the amine is willing to take up a proton, i.e., the more the amine is basic. The electron-releasing —C2H5 group in ethanamine stabilizes the corresponding ≈

ammonium ion (C2 H5Æ— NH3 ) by delocalizing the positive charge, whereas the electron-attracting –CF3 group in 2,2,2-trifluoro-1-ethanamine destabilizes the ≈

corresponding ammonium ion (CF3 —¨ CH2 —¨ NH3 ) by intensifying the positive (b)

5.

charge. Ethanamine is, therefore, more basic than 2,2,-trifluoro-1-ethanamine. Due to the presence of three strongly electron-attracting pentafluoroethyl (–C2F5) groups in (C2F5 )3 N, the availability of the unshared pair of electron on nitrogen is markedly decreased and because of this, this amine is virtually non-basic.

Malonic acid, HO2CCH2CO2H (a dicarboxylic acid) is stronger in its first ionization (pKa1 = 2.83) than acetic acid, CH3COOH (analogous monocarboxylic acid; pKa = 4.76) but weaker in its second ionization (pKa2 = 5.69) than acetic acid. Explain these observations.

==

Solution Malonic acid is stronger in its first ionization than acetic acid because a second carboxyl group (—COOH) in the acid exerts its –I effect on the first and thereby facilitates the release of the first proton by making the O—H bond weaker compared to acetic acid. O –I group HO C CH C O H 2

2

The expulsion of second proton from the conjugate base (HO2CCH2COO①) is difficult because the electron-donating —COO① group makes the O—H bond stronger compared

1.72

Organic Chemistry—A Modern Approach

to acetic acid. Furthermore, the resulting dianion is destabilized by having two negative charges in close proximity.

Therefore, malonic acid is stronger in its first ionization but weaker in its second ionization than acetic acid. 6. In the absence of a solvent (i.e., in the gas phase), most carboxylic acids are far weaker than they are in solution. For example, pKa of acetic acid in aqueous medium is 4.75 but in the gas phase, it is about 130. Explain. Solution When an acetic acid molecule donates a proton to a water molecule in the gas phase, the separation of the oppositely charged ions is very difficult.

In solution, solvent (water) molecules surround the ions, insulating them from one another, stabilizing them and making it far easier to separate them than in the gas phase. It is for this reason, in the gas phase most carboxylic acids are far weaker than they are in solution. 7. The DG∞ values for the dissociation of acetic acid and chloroacetic acid in water at 25°C are +27 and +16 kJ mol–1, respectively, i.e., chloroacetic acid is a stronger acid than acetic acid. Explain these thermodynamic values by considering the effect of solvent. Solution Water molecule solvates both the undissociated acetic acid (CH3COOH) and its anion (CH3COO①) by forming hydrogen bonds to them. Because the water molecules are more attracted by the negative charge, hydrogen bonding to CH3CO2① is much stronger. This differential solvation has important consequences for the entropy change that accompanies the ionization. Because of solvation of any species, the entropy of the solvent decreases and that is due to the fact that solvent molecules become much more ordered as they surround molecules of the solute. Because solvation of acetate ion (CH3CO2①) is stronger, the solvent molecules become more orderly around it. Therefore, the entropy change (DS°) for the ionization of acetic acid is negative and so, the TDS∞ term in the equation DG° = DH° –TDS° makes an acid-weakening positive contribution to DG∞. In fact, the TDS° term contributes more to the value of DG° than does DH°, and accounts for the fact that the free-energy change for the dissociation of acetic aid is positive, i.e., unfavourable (DG° = + 27 kJ mol–1). The Cl atom stabilizes the chloroacetate ion by exerting its –I effect. Thus, it makes the ion less prone to cause an ordering of the solvent because it requires less

Structure, Bonding and Proper es of Organic Molecules

1.73

stabilization through solvation. The entropy change (DS°) for the ionization of chloroacetic acid is, therefore, less negative. Both DH° and TDS° are more favourable for the ionization of chloroacetic acid. The larger contribution is clearly in the entropy term. For this reason, the DG° value for the dissociation of chloroacetic acid is less positive than that of acetic acid. Hence, chloroacetic acid is a stronger, acid than acetic acid. 8.

Arrange 1°, 2° and 3° alcohols in order of increasing rate of reaction with sodium and explain the order. Explain why potassium instead of sodium is used to prepare the alkoxide salt of tert-butyl alcohol.

Solution 1°, 2° and 3° alcohols may be arranged in order of increasing rate of reaction with sodium as follows:

R3COH

R2CHOH

RCH2OH

Tertiary or 3°

Secondary or 2°

Primary or 1° d-

d+

As the number of electron-releasing alkyl (–R) group increases, the polarity of the O — H bond gradually decreases and O—H bond becomes progressively more stronger. Therefore, the decreasing order of acidity of primary, secondary and tertiary alcohols containing one, two and three alkyl groups, respectively, is

Due to this order of acidity, the rate of reaction of alcohols with sodium increases gradually from 3° to 1° alcohol. tert-Butyl alcohol, being a very weak acid, reacts with less electropositive sodium very slowly and therefore, relatively more electropositive and reactive metallic potassium is normally used to prepare its alkoxide.

1.

Which is the weakest base and why? (a)

FCH2COONa

(b)

F2CHCOONa

(c)

F3CCOONa

(d)

CH3COONa

2.

Explain why trifluoromethanol is 1000 times more acidic than ethanol.

3.

Account for the difference in acidities: (a) CH3CO CH2COOH > CH3CH2CH2COOH (b) C6H5CH2COOH > CH3CH2COOH

1.74

Organic Chemistry—A Modern Approach

4.

Account for the fact that

Me3CCH 2CH 2 N H 2

is a stronger base than

≈

Me3 N CH 2CH 2 N H 2 . 5.

6.

Arrange the following compounds in order of increasing acidity and explain the order: CH3COOH, H COOH, CH3CH2COOH, (CH3)3C COOH The chloro acid I is a stronger acid than the acid without the chlorine whereas the chloro acid II is a weaker acid than the corresponding acid with no chlorine. Explain. COOH Cl COOH

Cl I

7.

II

Explain why 4-bromobicyclo[2.2.2]octane-1-carboxylic acid (A) is a considerably stronger acid than 5-bromopentanoic acid (B). Br

COOH

Br COOH

(A)

(B)

[Hint: Consider the possible conformation and modes of transmission of the electrical effect of the C—Br dipole.] 8.

9.

Decide which member in each of the following pairs of compounds is the stronger base. Give your reasoning. (a)

CF3CH2CH2NH2 or CH3CH2CH2 N H 2

(b)

(CH 3 )3 N CH 2CH 2 N H 2 or O2CCH 2 CH 2NH 2

(c)

N, N-dimethylmethanamine or trifluoro-N, N-bis (trifluoromethyl) methanmine

≈

@

Arrange the following dicarboxylic acids in order of increasing pKa1 values and explain the order. COOH COOH COOH HOOC

COOH COOH

10.

Arrange the following carboxylate ions in order of increasing acidity and explain the order HOOC CH2COO①, HOOC CH2CH2COO①, HOOC COO①

11.

Arrange the following carboxylic acids in order of increasing acidity and explain the order.

Structure, Bonding and Proper es of Organic Molecules

1.75

≈

O2NCH2COOH, Me3 NCH2COOH , HOCH2COOH 12.

The free-energy change, DG°, for the ionization of the acid HA is 21 kJ mol–1 and for the acid HB it is –21 kJ mol–1. Which is the stronger acid and why ?

13.

In which C—C bond of CH3CH2CH2Br, the inductive effect is expected to be the least? [Hint:

The magnitude of inductive effect decreases with distance and hence the 2

3

1

effect is least in C-2—C-3 bond (CH3Æ— CH2Æ— CH2Æ— Br) ] 14.

Which one of the following two compounds is more acidic and why? COOH COOH

Cl

C

C

C

C

Cl I

II

[Hint: II is more acidic than I and the reason is field effect.]

1.6

RESONANCE AND RESONANCE EFFECT OR MESOMERIC EFFECT

In a number of cases, it is not possible to describe the electronic structure of a species (molecules, ions or radicals) adequately by a single Lewis structure but by a combination of two or more structures and this is because none of these structures conforms to all the observed properties of these substances.

Examples (1)

2The Lewis structure of the carbonate ion (CO3 ), in which the position of electrons are fixed, shows that there are two C—O single bonds and one C==O double bond.

It has been determined experimentally that all the three carbon-oxygen bond lengths in carbonate ion are equal (130 pm). This bond length is slightly greater than that of the double bond but less than that of the single bond. Thus, the ion is symmetrical. In fact, three structures (I, II and III) can be written for the carbonate ion as shown below:

1.76

(2)

Organic Chemistry—A Modern Approach

These three equivalent structures have the same placement of atom but a different arrangement of nonbonding p electrons. The electron density is shared by all the three oxygen atoms, i.e., the electrons are delocalized over the three oxygen atoms, the actual structure of the ions lies in between these structures. The ion is actually a resonance hybrid (weighted average) of all these structures represented frequently by the non-Lewis structure IV and because of this, all the carbon–oxygen bonds are found to be equal in length. These structures are called resonance structures or canonical structures or simply canonicals and they have no real existence. In the hybrid structure, the negative charge is distributed equally among the oxygen atom. The delocalization of charge leads to greater stability of the ion. Benzene molecule can be represented as a resonance hybrid (III) of two Kekule (Lewis) structures, I and II.

Neither of the two structures can fully explain all the properties of benzene. For example, both the structures I and II contain two types of carbon–carbon bond such as C—C (1.54 Å) and C == C (1.34 Å). But actually, it has been found that all the six carbon–carbon bonds in benzene are equal in length (1.34 Å). This suggests that the actual structure of benzene can neither be represented by I nor by II but by a resonance hybrid of these two structures in which all the six carbon–carbon bond are equal in length (1.39 Å) and lie in between carbon–carbon double bond length of 1.34 Å and carbon–carbon single bond length of 1.54 Å. So, benzene is quite often represented by the non-Lewis structure III. The circle inside the ring indicates completely delocalized 6p electrons. Since I and II are exactly equivalent, they are of exactly the same stability and make equal contribution to the hybrid.

(1)

Resonance theory

This theory states that whenever a molecule or ion can be represented by two or more Lewis structures that differ only in the positions of electrons, two things will always be true: (i) none of these structures, which are called resonance structures or canonical structures

Structure, Bonding and Proper es of Organic Molecules

1.77

or simply canonicals, is a real representation for the molecule or ion and none of them can satisfactorily explain all its physical and chemical properties of the substance; (ii) the actual molecule or ion is better represented by a hybrid (weighted average) of these structures and a resonance hybrid is more stable, i.e., it has lower energy than any of its contributing structures. The resonance structures are represented by means of a double headed arrow (´). Resonance structures are written by shifting the positions of electrons by using curved arrows. In reality, no such movement of electrons takes place and the different structures are only arbitrary.

1.6.1

Resonance Energy

The difference between the enthalpy of formation of the actual molecule (observed value) and that of the resonance structure having the lowest internal energy (obtained by calculation) is called the resonance energy of the molecule. Resonance energy is expressed in kcal mol–1 or kJ mol–1. The resonance energy is greater when (i) the contributing structures are all equivalent and (ii) the number of contributing structures of roughly comparable energy is greater. Greater the resonance energy, more will be the stability of the compound.

1.6.1.1

Calculation of resonance energy

Resonance energy is not a measurable quantity. It can only be estimated from thermochemical data. Resonance energy of benzene, for example, can be calculated from heat of hydrogenation values. Heat of hydrogenation is the quantity of heat evolved when 1 mole of an unsaturated compound is hydrogenated. Cyclohexene containing one double bond has a heat of hydrogenation 28.6 kcal mol–1. We might reasonably expect 1,3,5-cyclohexatriene to have a heat of hydrogenation of about three times as large as cyclohexene, i.e., 3 × 28.6 = 85.8 kcal mol–1 . Actually, the value for benzene is 49.8 kcal mol–1. It is 36 kcal mol–1 less than the expected value. So, benzene evolves 36 kcal less energy per mole than predicted. This can only mean that benzene is more stable than hypothetical cyclohexatriene by 36 kcal mol–1 energy. This 36 kcal mol–1 energy is the resonance energy of benzene.

The resonance energy may also be calculated using heat of combustion value.

1.78

Organic Chemistry—A Modern Approach

1.6.2

Rules for Writing Meaningful Resonance Structures

The following rules are to be followed while writing realistic resonance structures: 1. All atoms in a resonance structure must exhibit their proper valency, i.e., each resonance structure must be a bonafide Lewis structure. So, in any resonance structure, a carbon atom may not have five bonds or a hydrogen atom may not have two and so on. 2. The various resonance structures should differ only in the positions of only p and n electrons and not in the positions of atoms, i.e., the basic structure involving s-bonds should remain unchanged. 3. The number of paired electrons in the resonance structures must be the same. Thus, CH2 — CH2 does not contribute towards the resonance hybrid of ethylene. 4. 5.

1.6.3

All the atoms involved in the process of resonance must be coplanar (or nearly coplanar). All the resonance structure should have nearly the same energy.

Relative Contribution of Resonance Structures towards Resonance Hybrid

All the resonance structures are not equally significant, i.e., all of them do not contribute equally towards the resonance hybrid. Relative contributions of resonance structures towards the resonance hybrid depend on their relative stabilities. The more the resonance structure is stable, the more will be its contribution towards the resonance hybrid. Various factors which govern the stability of a resonance structure and its relative contribution towards the hybrid are as follows: 1. Nonpolar resonance structures are more stable than the dipolar resonance structures and contribute more towards the resonance hybrid. For example: !

@

CH3 — CH == CH — CH == CH2 ´ CH3 CH — CH == CH — CH2 (More stable: major contributor) (Less stable: minor contributor) 2.

Resonance structures with a greater number of covalent bond are more stable and contribute more towards resonance hybrid.

3.

The more stable and more contributing structure of an anion is one in which the negative charge resides on the more electronegative atom while the more stable and more contributing structure of a cation is one in which the positive charge resides on the least electronegative atom. For example:

Structure, Bonding and Proper es of Organic Molecules

1.79

4.

Resonance structures in which octets of all the atoms are fulfilled are relatively more stable and hence more contributing towards the resonance hybrid. For example:

5.

Aromatic resonance structures are more stable and hence more contributing than the non aromatic resonance structure having the same number of covalent bonds. For example:

6.

A resonance structure having two units of charge on the same atom is less stable and hence less contributing. Again, structures having like charges on adjacent atoms are less stable and less contributing. However, a resonance structure having two dissimilar charges close to each other is relatively more stable and more contributing than the structure in which the charges are further apart. For example:

7.

Resonance energy of a system involving monopolar resonance structures is greater than that involving bipolar resonance structures. So, the former type of systems (i.e., molecules or ions) are relatively more stable than the latter type. For example:

1.80

Organic Chemistry—A Modern Approach

8.

1.6.4

Equivalent resonance structures are more important and more contributing than non-equivalent resonance structures. Fro example:

Resonance or Mesomeric Effect

Displacement of non bonding or p-electrons from one part of a conjugate system (having alternate single and double bond) to the other part causing permanent polarity in the system by creating centres of high and low electron density is called resonance effect (R-effect) or mesomeric effect (M-effect). There are two types of resonance or mesomeric effect: (i) +R or +M effect: An atom or a group is said to have +R or +M effect if it causes transfer of electrons away from the atom or group attached to a double bond or a conjugated system. +R or +M groups: –OH, - O R, - SH, - NH 2 - Cl:, – Br:, - I:, etc. + R effect of –Cl in vinyl chloride (Cl – CH = CH2), for example, may be shown as follows:

(ii)

–R or –M Effect: An atom or a group is said to have –R or –M effect if it causes transfer of electrons towards the atom or group attached to a double bond or a conjugate system. –R or –M groups: >C = O, –CHO, –COOR, –CN, –NO2, etc. –R effect of the –CHO group in aeraldehyde (CH2==CH — CHO), for example, may be shown as follows: +

CH2==CH—CH ==O

´ CH 2—CH ==CH—O

–

Structure, Bonding and Proper es of Organic Molecules

1.6.5

1.81

Isovalent and Heterovalent Resonance

When in a resonance hybrid, each of the resonance structures contains the same number of covalent bonds then the resonance is known as isovalent resonance. For example: CH3—C

O == –

´ CH3—C

O ==

–

(Acetate ion)

O

O

When in a resonance hybrid, each of the resonance structures does not contain the same number of covalent bonds it is known as heterovalent resonance. For example: ≈

CH2 == CH — CH == O: ´ CH2 — CH == CH — O:@ (Acraldehyde)

1.6.6

Effect of Resonance on the Properties of Molecules

The following properties of different molecules or ions can be explained by resonance.

(1) Bond length (a) Because of resonance, the single bond present in a molecule or ion may acquire a partial double bond character with consequent decrease in bond length. Similarly, the double bond may acquire some single bond character with consequent increase in bond length. For example, due to resonance, the C—Cl bond (1.69 Å) of vinyl chloride (CH2 == CH — Cl) becomes shorter than the C—Cl bond (1.76 Å) of ethyl chloride and its C == C bond (1.38 Å) becomes longer than C == C bond (1.34 Å) of ethene.

(b)

Because of resonance, some ions may have bonds of equal length. For example, the four sulphur–oxygen bonds in sulphate ion (SO42), the three nitrogen–oxygen bonds in nitrate ion (NO3①), the two nitrogen–oxygen bonds in nitrite ion (NO2①), the two ≈

carbon–carbon bonds in allyl cation (CH2 ==CH — CH2 ) and the two carbon–carbon @

bonds in allyl anion (CH2 ==CH — CH2 ) have the same length. Each of these ions can be represented as a resonance hybrid of two or more equivalent resonance structures.

1.82

Organic Chemistry—A Modern Approach

From the above resonance structures it becomes clear that the two negative charges are distributed evenly over four oxygen atoms in SO24@ and three oxygen atoms in NO23@ , one negative charge is distributed evenly over two oxygen atoms in NO3@ @

and two carbon atoms in CH 2 = CH - CH 2 and one positive charge is distributed ≈ evenly over two carbon atoms in allyl cation, CH2 == CH — CH2 . Therefore, the order of bonds involved in resonance in each species taking individually is same and hence the bonds are equal in length.

(2) Bond dissociation energy The bond dissociation energy of various compounds can be explained in terms of resonance. The bond dissociation energy for a benzylic hydrogen is less than that of a methane hydrogen.

From the enthalpy values it becomes clear that 19 kcal/mol less energy is needed to form the benzyl radical from toluene than to form methyl radical from methane. This difference

Structure, Bonding and Proper es of Organic Molecules

1.83

in energy requirement is actually the difference in energy content between the radical and its precursor. Both toluene (C6H5CH3) and benzyl radical (C6H5CH2) can be represented as resonance hybrids:

Toluene (less effective resonance)

Benzyl radical (more effective resonance)

Since toluene is a resonance hybrid of only two resonance structures (I and II) and the benzyl radical is a resonance hybrid of five resonance structures (III–IV), therefore, the benzyl radical is much more resonance-stabilized than toluene. As a result of this, the difference in energy content between benzyl radical and toluene becomes relatively small. On the other hand, both methane (CH4) and methyl radical CH 3 can be represented satisfactorily by single (localized) structures. CH4 Æ CH3 +H Methane Methyl radical None of these species is resonance-stabilized and thus the difference in energy content between the methyl radical and methane is relatively large. Because of this, it requires higher energy to form methyl radical from methane than to form benzyl radical from toluene, i.e., the C6H5CH2—H bond dissociation energy is less than CH3—H bond dissociation energy.

(3) Dipole moment Because of resonance, both the magnitude of the charge separated (e) and the distance between two charge centres (d) in any molecule may increase and so, the value of dipole moment (m = e × d) may increase. For example, the dipole moment of nitroethene is greater than nitroethane. Due to resonance, the amount of charge separated and the distance between the positive and negative charge centres in nitroethene (CH2 == CHNO2) are greater as compared to nitroethane (CH3CH2NO2) in which the carbon chain is not involved in resonance. Consequently, the dipole moment (m) of nitroethene is greater than that of nitroethane.

1.84

1.6.6.1

Organic Chemistry—A Modern Approach

Acidity of organic compounds

(1) Acidity of carboxylic acids The acidity of carboxylic acid, for example, acetic acid, is partly due to the significant electron-withdrawing inductive effect (–I) of the carbonyl group which makes the hydroxyl oxygen more electron deficient and thereby makes the proton release easier.

Resonance plays an important role in making the carboxylic acids acidic. In an aqueous medium, acetic acid dissociates as follows: :O

: O: || || ææ Æ CH3 — C —O—H + H2O ¨æ CH — C — O:@ + H3 O! æ 3 Acetic acid Acetate ion

Both acetic acid and its conjugate base acetate ion can be represented as resonance hybrids:

Acetic acid is a resonance hybrid of two non equivalent resonance structures, I and II, of which II involves separation of charges. In structure II, the two atoms of similar electronegativtiy carry opposite charges, therefore, II should contain more energy and hence be less stable then I. Consequently, resonance in acetic acid is not very stabilizing. The acetate ion, on the other hand, is a resonance hybrid of two energetically equivalent resonance structures, III and IV. Now it is known that resonance stabilization is greatest when the contributing structures are equivalent. Therefore, the acetate ion is much more stabilized by resonance as compared to acetic acid. This difference in stabilization causes the equilibrium to shift to the right. Acetic acid thus behaves as an acid. Unlike

Structure, Bonding and Proper es of Organic Molecules

1.85

the carboxyl group, the O—H bond in ethanol, C2H5OH (an alcohol), is not weakened by inductive electron withdrawal. Again, both ethanol and its conjugate base, i.e., the ethoxide ion (C2H5O①), can be represented satisfactorily by single (localized) structures, V and VI, i.e., either of these species is not stabilized by resonance. The nonbonded electron pair remains localized on the oxygen atom. Therefore, the driving force, i.e., the relative stabilization of the ethoxide ion with respect to the undissociated ethanol molecule, that promotes dissociation is absent in this case. For this reason, ethanol is much less acidic than acetic acid. ææ Æ C2 H5 — O:@ + H≈ C2 H5 —OH ¨æ æ V

VI

(2) Substituted benzoic acids An electron-releasing group decreases the acidity of benzoic acid, while an electron-attracting group increases it. The acidity also depends on the position of the substituent present. For example, p-nitrobenzoic acid (pKa = 3.43) is somewhat stronger than the m-isomer (pKa = 3.28) but much weaker than the o-isomer (pKa = 2.17). This can be explained as follows. In the conjugate base of a nitrobenzoic acid, the negative charge on oxygen in the carboxylate (—COO①) group is not in proper conjugation with the —NO2 group and so it cannot be stabilized by resonance. The –I effect of the —NO2 group is expected to fall off with distance. Therefore, p-nitrobenzoic acid is expected to be weaker acid than the m-isomer. However, the acidity order is reverse of that predicted on the basis of the –I effect. Although a p-NO2 group cannot enter into resonance with the —COO① group, it exerts its –R effect to reduce the electron density of the carbon bearing the —COO① group. A m-NO2 group also exerts its –R effect, but the carbon bearing the —COO① group does not become positive. The conjugate base of the p-acid is, therefore, somewhat more stabilized by resonance than the conjugate base of the m-acid and for this reason, p-nitrobenzoic acid is somewhat stronger than its m-isomer.

1.86

Organic Chemistry—A Modern Approach

Because of close proximity, the o-NO2 group exerts powerful –I effect and through –R effect it withdraws electron from the —COO① group by making the carbon bearing this group positive. Also, there occurs a more direct electrostatic interaction between the —COO① and —NO2 groups. Because of all these factors, the o-nitrobenzoate ion is much more stabilized with respect to the o-acid than the p-nitrobenzoate ion with respect to p-acid and therefore, o-nitrobenzoic acid is a much stronger acid than its p-isomer.

Unlike halogen substituted saturated monocarboxylic acids, the effect of halogens on the acidity of benzoic acid cannot be explained only by –I effect. Because of strong –I effect p-fluorobenzoic acid is expected to be more acidic than p-bromobenzoic acid. However, it has been found that p-bromobenzoic acid is actually more acidic. Halogens exert their +R effect and tend to decrease the strength of the acid by increasing electron density of the carboxyl group. The +R effect of the two halogens does not follow the expected order based on electronegativity. It is known that the p-orbital overlap is much more effective when the overlapping orbitals are of comparable size and so the +R effect of fluorine (that uses 2p orbital) is stronger than that of bromine (that uses 4p orbital). In fact, the +R and –I effects of fluorine are nearly counter balanced. However, the +R effect of bromine is much weaker as compared to its –I effect. Thus, although both the halogen atoms cause lowering of electron density of the ring, the lowering is much effective in the case of bromo-acid than in the case of fluoro-acid. For this reason, p-fluorobenzoic (pKa = 4.14) is less acid than p-bromobenzoic acid (pKa = 4.00).

(3) Acidic Character of Phenols (Ar-OH) Greater the ease with which a compound releases ! proton (H ) in its aqueous solution, the stronger it will be as an acid. That phenol is acidic in nature and is a stronger acid than alcohol can be well explained in terms of resonance.

Structure, Bonding and Proper es of Organic Molecules

1.87

Phenol in its aqueous solution ionizes to produce phenoxide ion (the conjugate base) as follows: O—H

O + H2O

Phenol

–

+

+ H3O Phenoxide ion

Phenol can be represented as a resonance hybrid of the following five resonance structures (I–V):

Due to contribution of the resonance structure II, III and IV, the O atom becomes positively polarized. Because of this, the polarity of the O–H bond increases and hence the tendency of O—H bond fission (to release a proton) also increases. On the other hand, no such resonance is possible in a molecule of alcohol (R—OH). Instead, a +I effect of R-group operates which increases the electron density of oxygen. So, the alcoholic O—H bond is relatively less prolar and the tendency of bond cleavage resulting in proton release is actually very small. Hence, phenol is more acidic than an alcohol. This relative acidity may also be explained by considering the phenol-phenoxide ion and alcohol-alkoxide ion equilibria. Like phenol, phenoxide ion may also be represented as a resonance hybrid of the following five (VI–X) resonance structures:

Three (II, III and IV) out of five resonance structures of phenol involve charge separation and so, their contribution to the hybrid is relatively small. The resonance-structures of phenoxide ion, on the other brand, involve no charge separation. The negative charge is only delocalized. Because of this, phenoxide ion is more resonance-stabilized than phenol. The effect of this difference in stabilization is that ionization energy of phenol

1.88

Organic Chemistry—A Modern Approach

becomes lower than the hypothetical ionization energy when resonance is absent. As a consequence, the equilibrium of phenol-phenoxide ion tends to shift towards the right, i.e., phenol exhibits acidic property by releasing proton (H!) easily. On the other hand, both alcohol and the alkoxide ion can be satisfactorily represented by single (localized) structures. ææ Æ R — OH ¨æ R — O:@ + H≈ æ Alkoxide ion Alcohol None of these two species is stabilized by resonance. The driving force for ionization, i.e., the relative stabilization of the anion with respect to the undissociated molecule, is absent in this case. Because of this, alcohols are much less acidic than phenols. The acidity of phenols is partly due to the –I effect of the aryl group as compared to the +I effect of the alkyl group in alcohols. The energy diagram for the dissociation of an alcohol and phenol may be represented as follows:

Ionization of both alcohol and phenol requires energy. Acidity of a compound depends on the ionization energy, i.e., the difference in energy content of the acid and its conjugate base. The less the energy required for ionization, the more the compound acidic. Both the phenol and conjugate base are resonance-stabilized, but the stabilization is far greater for the conjugate base, i.e., the phenoxide ion, than for the phenol (a > b). The effect of this difference in stabilization is that ionization energy of the phenol becomes lower than the hypothetical ionization energy (DE2 < DE1) and as a result, the ionization is favoured, i.e., a phenol exhibits acidic character.

Structure, Bonding and Proper es of Organic Molecules

1.89

Since either the alcohol or its conjugate base is not stabilized by resonance, therefore, the energy required for its ionization is considerably high and in fact, it is much higher than that required for the ionization of the phenol (DE1 > DE2). Because of this, the ionization of a phenol is very much favoured as compared to an alcohol, i.e., phenols are much stronger acids than alcohols.

1.6.6.2

Effect of ring substituents on the acid strength of phenols

The acidity of phenols depends on whether a ring substituent is electron-attracting or electron-releasing and also its position with respect to the —OH group. An electron-attracting substituents like —X, —NO2, —CHO, —CN, etc., withdraws electrons from the ring and thereby stabilizes the phenoxide ion more than the phenol. Thus, the energy difference between the phenol and its conjugate base decreases and as a result, the energy required for ionization decreases. Consequently, the ionization equilibrium becomes favoured. Therefore, an electron-attracting substituent increases the acidity of phenol. An electron-releasing substituent, like —CH3, —OCH3, —NH2, etc., donates electron to the ring and thereby destabilizes the phenoxide ion relative to the phenol by intensifying the negative charge on it. For this reason, the difference in stability between the phenol and its conjugate base, i.e., the phenoxide ion increases and consequently, the strength of the phenol as an acid decreases. Therefore, an electronreleasing substituent decreases the acidity of phenol.

A substituent can exert its –R or +R effect only when it is at ortho or para position. Therefore, the acid-strengthening or acid-weakening effect of a substituent is very much pronounced when it is present at otho or para position but not at the meta position. The acidic strength of nitro substituted phenols, for example, increases in the following order: phenol < m-nitrophenol < o-nitrophenol < p-nitrophenol. A nitro (—NO2) group present in ortho- or para–position is capable of withdrawing electrons from the negatively charged oxygen atom of the phenoxide ion by its –I and –R effects. It, therefore, stabilizes the conjugate base by dispersing the negative charge. In fact, there is a relatively stable structure (P) in which the negative charge is placed on the highly electronegative oxygen atom of the —NO2 group and its contribution makes the hybrid stable. A similar resonance structure

1.90

Organic Chemistry—A Modern Approach

(Q) makes the o-nitrophenoxide ion considerably stable. Hence, o- and p-nitrophenoxide ions are relatively more stable than phenoxide ion. In fact, the relative stabilization of the phenoxide ion with respect to the undissociated phenol is more effective with these nitrophenols compared to phenol and for this reason, o- and p-nitrophenols are more acidic than phenol.

The m-nitrophenoxide ion is stabilized only by the –I effect of the –NO2 group and this is because the nitro group is unable to delocalize the negative charge on oxygen due to lack of proper conjugation. For this reason, m-nitrophenol is a stronger acid than phenol, but weaker acid as compared to its other isomers.

In the o-isomer, there occurs intermolecular hydrogen bonding (chelation) which, in fact, disfavours proton release to some extent. For this reason, the o-isomer is less acidic than the p-isomer, even though the —I effect of the relatively close —NO2 group is much stronger than the former.

1. Acidic character of active methylene compounds An active methyl compound contains a —CH2— or —CHR— group flanked by two strong electron-attracting (—I and —R) groups. Acetylacetone (CH3COCH2COCH3), diethyl malonate (EtO2C – CH2—CO2Et), ethyl acetoacetate (CH3COCH2CO2Et), ethyl cyanoacetate (EtO2C — CH2 — CN), dinitromethane

Structure, Bonding and Proper es of Organic Molecules

1.91

(O2N — CH2 — NO2), etc. are the most important active methylene compounds. The acidic methylene hydrogens of these compounds can be easily abstracted by base because the corresponding conjugate bases are well stabilized by resonance. However, their acidity differs due to the presence of different electron-withdrawing groups. For example, acetylacetone is a stronger acid than ethyl acetoacetate. A resonance argument can be used to explain the difference in acidity between these two active methylene compounds. Acetylacetone and ethyl acetoacetate ionize in the presence of a base (:B) as follows:

Charge delocalization occurs in both the conjugate bases and also the same number of resonance structures can be written for both of them. The resonance structures of acetylacetonate anion are all significant contributors. On the other hand, two of the three resonance structures of acetoacetate anion (the first and second) contribute significantly to the hybrid. The contribution of the last structure to the hybrid is, in fact, very small because delocalization of electrons within the ester group itself lowers the ability of the C == O group to disperse the negative charge.

Because of this, delocalization of charge is more effective in acetylacetonate anion than in ethyl acetoacetate anion, i.e., the former conjugate base is more stabilized by resonance than the latter conjugate base. The difference is stability between the neutral molecule and the conjugate base is, therefore, greater in the case of ethyl acetoacetate (CH3COCH2CO2Et) than in the case of acetylacetone (CH3COCH2COCH3) and hence the equilibrium involved in the case of aceylacetone is relatively more favourable than that involved in the case of ethyl acetoacetate. That is, acetylacetone is more acidic than ethyl acetoacetate.

1.92

Organic Chemistry—A Modern Approach

(2) Acidic character of imides Imides are often sufficiently acidic to form alkali metal salts. The acidity of an imide, e.g., phthalimide, is due to —I and —R effects of the two C == O groups. The unshared pair of electrons on nitrogen is sufficiently delocalized and as a consequence, it becomes considerably positively polarized. Because of this, the polarity of the N—H bond increases and hence the tendency of N—H bond fission to release proton increases. For this, imides exhibit acidic character. The acidity of phthalimide can also be explained by considering phthalimide-phthalimide anion equilibria. Both phthalimide and its conjugate base can be represented as a hybrid of three resonance structures such as follows:

In phthalimde, the unshared pair of electrons on nitrogen is delocalized by the two adjacent C == O groups. However, this resonance is less effective and less stabilizing because it involves charge separation. On the other hand, resonance is more effective and more stabilizing in phthalimide anion because there occurs no charge separation. Resonance thus lowers the energy difference between the imide and its conjugate base. Because of this, the ionization becomes much favourable, i.e., the compound exhibits considerable acidity and dissolves readily in alkali metal hydroxide (e.g., NaOH) solution to form salts.

(3) Acidic Character of Nitroform [CH(NO2)3], Chloroform (CHCl3) and Fluoroform (CHF3) Nitroform is more acidic than chloroform which in turn is more acidic than fluroform. These observations can be well explained by resonance. Any thing that stabilizes a conjugate base A①: makes the starting acid H—A more acidic. The conjugate base of nitroform, i.e., @

C(NO2 )3 , is stabilized by —I and —R effect (p-p p bonding) of the three —NO2 groups.

Structure, Bonding and Proper es of Organic Molecules

1.93

@

The conjugate base of chloroform, i.e., CCl3 , on the other hand, is stabilized by —I effect and d-orbital resonance (p-d p bonding) involving three Cl atoms. Because of longer bond length and the difference in size between the 2p and 3d orbitals, the p-d p bonding is far less significant than p-p p bonding. Also, a —NO2 group exerts stronger —I effect than a Cl @

@

atom. For these reasons, C(NO2 )3 is very much stable than CCl3 and therefore, nitroform is a much stronger acid than chloroform.

@

Fluorine has no d orbital. So, unlike CCl3 there is no question of resonance stabilization @ @

of CF3 . It is@only stabilized by the —I effect of fluorine. Thus, CCl3 is relatively more stable than CF3 and for this reason, chloroform (CHCl3) is more acidic than fluoroform, even though fluorine is more electronegative than chlorine.

1.6.6.3 Basicity of Organic Compounds (1) Basic character of arylamines (e.g., aniline, C6H5N¨H2) The basicity of nitrogenous bases depends on the availability of the unshared pair of electrons on the N atom. Any factor that increases the electron density on the N atom increases an amine’s basicity and any factor that decreases the electron density on the N atom decreases an amine’s basicity. Aniline is a weak base and its basicity is much weaker than an aliphatic amine like ethylamine (C2H5NH2). This can be explained by resonance. Aniline can be represented as a resonance hybrid of the following five resonance structures:

1.94

Organic Chemistry—A Modern Approach

The unshared pair of electrons on the N atom of aniline is involved in resonance interaction with the ring p electrons. As a result of this electron delocalization, the N atom acquires a partial positive charge. Consequently, aniline exhibits little tendency to take up a proton. So, aniline is a very weak base. The decrease in basicity is partly due to the –I effect of the aryl group. In the case of aliphatic amine like C2H5NH2, on the other hand, similar delocalization of electrons by resonance is not possible. Naturally, the electron density on the N atom is not reduced. In fact, due to +I effect of the alkyl group (–C2H5), the electron density on the N atom is somewhat increased. As a consequence, nitrogen can easily donate its unshared electron pair to a proton (H+) to combine with it. Thus, aniline (an arylamine) is a weaker base than ethylamine (an alkylamine).

Apart from this, the relative basicity can also be explained by considering aniline-anilinium ion and amine-ammonium ion equilibrium systems. In fact, the difference in stability between a free amine and its conjugate acid is to be considered. The higher the difference, the lesser the basicity of the amine.

Aniline is a resonance hybrid of five resonance structure (I–V) of which two are Kekule structures (I & V). In anilinium ion, the lone pair of electrons on the N atom is localized in the N—H bond and so only two Kekule structures (VI and VII) can be drawn for its hybrid. Therefore, more effective electron delocalization occurs in aniline as compared to anilinium ion and consequently, aniline is more resonance-stabilized with respect to the

Structure, Bonding and Proper es of Organic Molecules

1.95

anilinium ion (in the following energy diagram b > a). As a result of this, the difference in stability between the base and its conjugate acid is increased and protonation of aniline ≈

is disfavoured. The conjugate acid of ethyl amine, i.e., CH 3CH 2 NH 3 , is stabilized only by the weak +I effect of the —C2H5 group. None of the amine and its conjugate acid can be stabilized by resonance. For this reason, the difference in energy content between the conjugate acid and the base is relatively small and actually it is smaller than the difference in energy content between aniline and anilinium ion (DE2 > DE1). The energy required for protonation of aniline is, therefore, greater than the energy required for protonation on for ethylamine and hence aniline (an arylamine) is less basic than ethylamine (an alkylamine).

(2) Effect of ring substituents on the basic strength of arylamines The effect of a ring substituent on the basic strength of arylamines depends on whether the substituent is electron-attracting or electron-donating and also on its position with respect to the —NH2 group in the ring. An electron-donating group, like —CH3, —OCH3, etc., increases the electron density of the ring and thereby prevents the unshared electron pair on nitrogen of the —NH2 group, to some extent, from entering into resonance with the ring p electrons. As a result, the unshared electron pair on nitrogen becomes more available for coordination with a proton. Therefore, such a group increases the basic strength of aniline. On the other hand, an electron–attracting group, like —X, —NO2, —COR, etc., favours the resonance interaction of the NH 2 group with the ring and thereby make the lone pair on the N atom less available for coordination with a proton. Therefore, such a group lowers the basic strength of aniline. The effect of a ring substituent on the basicity of anilines can also be explained in terms of the difference in stability, i.e., the energy content, between the base and its conjugate acid,

1.96

Organic Chemistry—A Modern Approach

i.e., the anilinium ion. An electron-releasing group tends to spread the positive charge of the anilinium ion and thus stabilizes the conjugate acid but not the base. As a result, the difference in stability between the conjugate acid and the base decreases, i.e., the energy required for protonation decreases and consequently, the protonation equilibrium becomes favoured. Thus, an electron-releasing group increases the base strength of the aniline. An electron-withdrawing group causes intensification of the positive charge of the conjugate acid and thus destabilizes the conjugate acid but not the base. As a result, the difference in energy content between the conjugate acid and the base increases, i.e., the energy required for protonation increases and consequently, the protonation equilibrium becomes disfavoured. Therefore, an electron-withdrawing group lowers the basic strength of the aniline.

A group can exert its +R and —R effects from ortho and para positions and due to that the base-strengthening or base-weakening effect is most pronounced if the group is present at these positions. For example, p-nitroaniline is less basic than m-nitroaniline. This can be explained as follows. A nitro group para to the —NH2 group withdraws the nonbonded electron pair on N of the —NH2 group by its —R and —I effects and thereby makes the lone pair less available for protonation. An m-NO2 group, on the other hand, cannot enter into resonance with the —NO2 group because it is not in proper conjugation with the unshared pair of electron on nitrogen. Only a weak electron-attracting inductive effect (—I) operates. The availability of the unshared pair is, therefore, somewhat better in this case and so, m-nitroaniline is more basic than its para isomer.

Structure, Bonding and Proper es of Organic Molecules

1.97

O Ê ˆ || Á ˜ (3) Basic character amides Ë R — C — NH2 ¯ Because of electron-attracting inductive and resonance effect (—I and —R) of the C == O group, the lone pair of electron on amide nitrogen is very much less available for taking up a proton. In an amine, on the other hand, the alkyl group pushes electrons towards nitrogen by its +I effect makes nitrogen’s electron pair readily available for sharing with a proton. Amides are, therefore, less basic than amines.

NH || (4) Basic character of guanidine H2N — C — NH2 That guanidine is an extremely strong base can be well explained by resonance. Both the neutral molecule and the conjugate acid can be represented as a resonance hybrid of three resonance structures.

1.98

Organic Chemistry—A Modern Approach

The neutral molecule contains two charge-separated structures which contribute less to the resonance hybrid. Electron delocalization is, therefore, less effective in the free base and so, it is not much resonance-stabilized. The conjugate acid, on the other hand, involves no charge separation; only the positive charge is delocalized. Also, the canonicals are all exactly equivalent. So, there occurs extremely effective charge delocalization and as a consequence, the conjugate acid is highly stabilized by resonance with respect to the neutral molecule. The difference in stability (i.e., the energy content) between the conjugate acid and the base is actually very small and so, protonation of the base required very small amount of energy. For this reason, guanidine behaves as a very strong base.

1.

Designate the following pairs as structural isomers or resonance structures and give your reasoning. (b)

O ==

(a)

and CH3—O—N==O

CH3—N

O

(d)

O

==

(c)

OH

CH3—C—CH3 and CH3—C==CH2

(e) CH3CH2CH==CHCH3 and

CH3CH2CH2CH==CH2

Solution Two resonance structures have the same placement of atom but a different arrangement of nonbonding and p electrons. However, two structural isomers differ in the arrangement of both atoms and electrons. Therefore, the two structures in each of (a) and (c) are two resonance structures, while the two structures in each of (b), (d) and (e) are structural isomers. 2. In the following set of resonance structures, label the major and minor contributors and give reasons for such labelling:

(a)

(b)

Structure, Bonding and Proper es of Organic Molecules

(c)

1.99

–

O CH3CH2—C

´ O—H

I

O CH3CH2—C

+

O—H

II

(d)

(e)

–

+

+

–

CH2—CH==CH—CH 2 ´ CH 2 ==CH—CH—CH 2 I

II

(f)

(g)

(h) (i)

(j)

Solution (a) Resonance structures with a greater number of covalent bond are more stable and more contributing than those with a fewer number of covalent bonds. Therefore, the structure I with ten covalent bonds is the major contributor and the structure II with nine covalent bonds is the minor contributor. (b) In structure I, octets of all the atoms are filled up while in structure II, octet of positive carbon is not filled up. Therefore, the structure I being more stable, is the major contributor and structure II is the minor contributor. (c) When two structures have the same number of covalent bonds, then the uncharged structure is more stable and more contributing than the charge separated

1.100

(d)

(e)

(f)

(g)

(h)

(i)

(j)

3.

Organic Chemistry—A Modern Approach

structure. Therefore, the structure I is the major and the structure II is the minor contributor. Resonance structures with negative charge on a more electronegative atom is more stable and more contributing. Therefore, the structure II with a negative charge on more electronegative oxygen atom is the major contributor and the structure I with a negative charge on less electronegative carbon atom is the minor contributor. Resonance structures in which unlike charges are on the adjacent atom are more stable and more contributing than those in which the charges are not adjacent. Therefore, the structure II is the major and the structure I is the minor contributor. Since the structure II constitutes a stable aromatic system [(4n +2)p, n = 1], therefore, it is more stable than the structure I. Therefore, the structure II is the major contributor and the structure I is the minor contributor. The structure I (a 2° carbocation) is relatively more stable than the structure II (a 1° carbocation) and therefore, the former structure is the major contributor and the latter structure is the minor contributor. Resonance structures with like charges on adjacent atoms are less stable and less contributing than those in which the charges are no adjacent. Therefore, the structure II is the major and the structure I is the minor contributor. Less electronegative nitrogen accommodates the positive charge better than more electronegative oxygen. Therefore, the structure II is the major contributor and the structure I is the minor contributor. Since the structure II constitutes a stable aromatic system [(4n + 2)p, n = 1], therefore, it is more stable than the structure I. Hence, the structure II is the major contributor and the structure I is the minor contributor. Arrange the resonance structures in each group in order of increasing contribution to the resonance hybrid: (a)

:O:@ :O:@ :O: | | || ≈ CH3 —C—OC2 H5 ´ CH3 — C —OC2 H5 ´ CH3 — C == OC2 H5

(b)

(C2 H5 )2 C == N — NH2 ´ (C2 H5 )2 C — N== N H2 ´ (C2 H5 )2 C — N —NH2

(c)

:O:@ :O:@ :O≈ :O: | ≈ | | || H—C—OH ´ H —C == OH ´ H — C —OH ´ H — C — OH

I

III

≈ II

≈

@

≈

II

I

I

II

III

≈ III

≈

@

@

@

IV ≈

(d)

CH2 == CH— Cl: ´ CH2 — CH == C l: ´ CH2 — CH== Cl:@

(e)

CH2 == N == N: ´ CH2 — N== N: ´ CH2 — N ∫∫ N ´ CH2 — N== N:

II

I

≈

I

@

≈

III

@

II

@

≈

III

≈

@

IV

Structure, Bonding and Proper es of Organic Molecules

Solution

4.

1.101

(a)

II, III, I æææææææææææÆ contribution increases

(b)

III, II, I æææææææææææÆ contribution increases

(c)

IV, III, II, I æææææææææææÆ contribution increases

(d)

III, II, I ææææææææææ contribution increasesÆ

(e)

IV, II, III, I ææææææææææ contribution increasesÆ

Draw the resonance structure and the resonance hybrid of phenoxide ion (C6H5—O①) and arrange then in order of decreasing stability. Give your reasoning. What does the resonance hybrid illustrate?

Solution The resonance structures and the resonance hybrid of the phenoxide ion my be drawn as follows:

The resonance structures II, III and IV have the negative charge on a carbon, a less electronegative element than oxygen. As a result, structures II–IV are less stable than structures I and II which have the negative charge on oxygen. Moreover, resonance structures I and V have intact aromatic rings, whereas structures II–IV do not. This, too, make structures II–IV less stable than I and V. A resonance hybrid is more stable, i.e., it has lower energy than any of its contributing structures. Therefore, the order of decreasing stability is VI > I = V > II = IV ª III. The resonance hybrid of the phenoxide ion illustrates that its negative charge is dispersed over four atoms — three C atoms and one O atom. 5. Explain why (a) the two oxygen–oxygen bonds in ozone (O3) have the same length and (b) the two nitrogen–nitrogen bonds in the azide ion (N3①) have the same length. Solution (a) Ozone (O3) can be represented as a hybrid of the following two resonance structures:

Since the two resonance structures are equivalent, they make equal contribution to the hybrid. Therefore, the two oxygen–oxygen bonds have the same order and so, the bonds have the same length.

1.102

(b)

Organic Chemistry—A Modern Approach

The azide ion (N3①) can be represented as a hybrid of the following three resonance structures: +

2–

N∫∫ N—N

6.

´

–

+

–

N==N==N ´

2–

+

N—N∫∫ N

I II III Each of the nitrogen–nitrogen bonds is triple in one structure, double in one structure and single in one structure. Hence, their bond order is the same and for this reason, they have the same length. @ The carbon–carbon bond length in CO23 (1.31 pm) is greater than that ① found in CH3COO (127 pm). Explain this observation.

Solution The CO23@ ion can be represented as a hybrid of three equivalent resonance structures. In this ion, two negative charges are evenly distributed over three oxygen

atoms and so, each oxygen atom acquires

2 3

@

charge. On the other hand, the HCOO①

ion is resonance hybrid of two equivalent canonical and in it, one negative charge is evenly distributed over two oxygen atoms, i.e., each O atom carries

1 2

@

charge. Since the

charge accumulation is greater in oxygen of CO23@ than in oxygen of CH3COO①, therefore carbon–oxygen bond order in CO23@ is less than that in CH3COO① and because of this, the carbon–oxygen bond length in CO23@ is greater than that in CH3COO①.

7.

Explain why acetic acid is more acidic than phenol.

Solution Any thing that stabilizes a conjugate base A:① makes the starting acid H—A more acidic. The acetate ion (the conjugate base of acetic acid) is a resonance hybrid of two equivalent resonance structures, both of which places the negative charge on a more electronegative oxygen atom. Although the phenoxide ion (the conjugate base of phenol) is resonance hybrid of five resonance structures, the negative charge is accommodated by three less electronegative carbon atoms and only one more electronegative oxygen atom. Furthermore, the two resonance structures of acetate ion are equivalent. Therefore, resonance stabilizes phenoxide ion but not as much as it stabilizes acetate ion. It is for this reason acetic acid is more acidic than phenol.

Structure, Bonding and Proper es of Organic Molecules

8.

1.103

Explain why picric acid is nearly as acidic as mineral acids.

Solution 2,4,6-Trinitrophenol is known as picric acid. Due to cumulative —I and —R effects of the three electron attracting —NO2 groups placed in para and in two ortho positions, 2,4,6-trinitrophenoxide ion, the conjugate base of picric acid, is very much stabilized with respect to the unionized phenol. Because of this, 2,4,6-trinitrophenol or picric acid behaves as a very strong acid and in fact, it is nearly as acidic as mineral acids.

9.

Acetic acid (CH3COOH) can act as a weak base in the presence of a strong acid like H2SO4. Which of the two oxygen atoms is more basic and why?

Solution In the presence of H2SO4, a proton can be added either to the carbonyl oxygen or to the hydroxyl oxygen of the carboxyl group. The two protonation equilibria are as follows: !

:O: || CH3 — C —OH + H2SO4

:OH

|| CH3 — C — OH + HSO4@

:O: || CH3 — C — OH + H2SO4

:O: || ! CH3 — C — O H2 + HSO@4

I

II

The conjugate acid I is a resonance hybrid of two equivalent canonicals and therefore, it is considerably stabilized by resonance. ´ CH3—C

CH3—C

OH

==

==

+

OH OH

+

OH

I

1.104

Organic Chemistry—A Modern Approach

On the other hand, the conjugate acid II is not at all stabilized by resonance. The positive charge is localized on the oxygen atom. Therefore, II is less stable than I and the first equilibrium in more favourable than the second. That is, the carbonyl oxygen is more basic than the hydroxyl oxygen. 10. Explain the following observations: (a) The central C—C bond in 1,3-butadiene (CH2 == CH—CH == CH2) is shorter in length than the C—C bond in ethane (CH3 —CH3). (b) The heat of hydrogenation of 1,3-cyclohexadiene in less than that of 1,4-cyclohexadiene (c)

(e) (f) (g) (h) (i)

(j)

(

The doubly bonded N atom of imidazole the other N atom.

(d)

(

(

(

N N H

(

( is relatively more basic than

The a-hydrogens in propanal (CH3CH2CHO) are much more acidic than its b-hydrogens. 1,2,3-Triphenylcyclopropane is a very much stable cyclopropane derivative. The Lewis acid character of boron trihalides decreases in the order: BI3 > BBr3 > BCl3 > BF3. Tri-p-nitrophenylamine is not at all basic. p-Nitrotoluene is more acidic than toluene. In the following compound, Ha is more acidic than Hb which in turn is more acidic than Hc:

Sulphonic acids (RSO3H) are very strong acids (pKa ª –7).

Solution (a) As the percent of s-character increases, the size of the hybrid orbital decreases and consequently, the length of the bond formed by overlapping two such orbitals decreases. Hence the C-2—C-3 bond in 1,3-butadiene, CH2 == CH—CH == CH2 (a C 2 — C 2 bond) is shorter than the C — C bond in ethane (a C 3 — C 3 bond). sp

sp

sp

sp

This can also be explained by resonance. 1,3-Butadiene can be represented by three resonance structures:

Structure, Bonding and Proper es of Organic Molecules

(b)

1.105

The two charge separated structures II and III contain a double bond between C-2 and C-3. Therefore, the hybrid must have a partial double bond there and for this reason, the central C—C bond is shorter than the C—C bond in ethane. 1,3-cyclohexadiene is a conjugated diene while 1, 4-cyclohexadiene is an isolated diene. Because a conjugated diene has overlapping p orbitals on four adjacent atoms, its p electrons are delocalized over four atoms. This electron delocalization cannot take place in an isolated diene. Therefore, 1,3-cyclohexadiene is stabilized by resonance.

Since on hydrogenation both the dienes produce the same cycloalkane, the more stable diene has the smaller heat of hydrogenation. Therefore, 1,3-cyclohexadiene has the smaller heat of hydrogenation as compared to 1,4-cyclohexadiene.

(c)

The more the corresponding conjugate acid is stable, the more the N atom is basic. No resonace structure can be written for the conjugate acid obtained on protonation of the singly bonded N atom of imidazole, i.e., this conjugate acid is not stabilized by resonance. Furthermore, on protonation aromaticity of the system is lost.

1.106

Organic Chemistry—A Modern Approach

On the other hand, two equivalent resonance structures can be written for the conjugate base obtained on protonation of the doubly bonded nitrogen, i.e., the corresponding conjugate acid is well stabilized by resonance. Also, on protonation aromaticity of the system is not lost.

It thus follows that the doubly bonded N atom which coordinates with a proton to from a stable conjugate acid is relatively more basic than the other N atom. (d) The anion obtained by the loss of an a-H atom of propanal (CH3CH2CHO) is stabilized by resonance. :O:@ H :O: @ a| | || ≈ b CH3 — C H — CHO + B [CH3 — CH — C — H ´ CH3 — CH == C — H] + BH Propanal Resonance-stabilized conjugate base Owing to an intervening saturated carbon atom, the anion obtained by the loss of a B — H atom is not stabilized by resonance. H @ | ! a b CH2 —CH2 — CHO + B CH2 — CH2 — CHO + BH Conjugate base Propanal (not resonance-stabilized)

(e)

Therefore, the difference in energy between the anion (the conjugate base) and the neutral molecule (the acid) is much greater in the second case than in the first case. Hence, a-hydrogens are much more acidic than b-hydrogens in the aldehyde molecule. 1,2,3-triphenylcyclopropane is highly stabilized by resonance because there occurs extensive electron delocalization due to the presence of a cyclopropane ring which behaves in some respect like a double bond and makes the system a perfect conjugated one.

Structure, Bonding and Proper es of Organic Molecules

(f)

1.107

This order of relative Lewis acid strength of boron trihalides, which is just the reverse of what may be expected on the basis of the electronegativities of the halogen atoms, can well be explained on the basis of the tendency of the halogen atom to donate its lone pair of electrons to the boron atom through pp–pp back bonding. Since the vacant 2p-orbital of B and the 2p-orbital of F atom containing a lone pair of electrons are equal in size, therefore, the tendency of F atom to donate the unshared pair by pp–pp back bonding is maximum. In fact, BF3 can well be represented as a resonance hybrid of four resonance structures. As a result of resonance involving pp–pp back bonding, the electron density on the boron atom increases effectively and so its strength as a Lewis acid decreases considerably.

As the size of the halogen atom increases on going from Cl to I, the extent of overlapping between the 2p orbital of boron and a large p orbital of halogen (3p of Cl, 4p of Br and 5p of I) decreases. As a consequence, the electron deficiency of boron increases and thus, the Lewis acid strength increases on going from BF3 to BI3. (g)

In tri-p-nitrophenylamine

the lone pair of electrons on the

N atom is extensively delocalized over the three p-nitrophenyl groups. Because of this, the unshared pair is not at all available for coordinating with a proton. Hence, by ordinary standard the compound has no basic strength.

1.108

Organic Chemistry—A Modern Approach

(h)

Due to the presence of an electron-withdrawing —NO2 group para to the —CH3 group, the conjugate base of p-nitrotoluene is stabilized not only by the —I effect of the —NO2 group but also by its strong —R effect. On the other hand, the conjugate base of toluene is only less effectively stabilized by resonance involving the benzene ring. Therefore, the difference in energy between benzyl anion (the conjugate base) and toluene is much greater than the difference in energy between p-nitorbenzyl anion (the conjugate base) and p-nitrotoluene. Thus, the second equilibrium is more favoured than the first and so, p-nitrotoluene is more acidic than toluene.

(i)

Anything that stabilizes a conjugate base A : makes the starting acid H—A more acidic. The conjugate base obtained by the loss of Ha proton is stabilized by resonance involving the benzene ring.

@

The conjugate base obtained by the loss of Hb proton is also stabilized by resonance involving the benzene ring but it is somewhat destabilized by the +I effect of the adjacent —CH3 group. And, the conjugate base obtained by the loss of Hc proton

Structure, Bonding and Proper es of Organic Molecules

1.109

is not at all stabilized by resonance due to the presence of intervening methylene group.

Therefore, the stability of the conjugate bases decreases in the order:

Hence, Ha is more acidic than Hb which in turn is more acidic than Hc. Sulphonic acids (RSO3H) are very strong acids because their conjugate bases are highly stabilized by resonance involving three equivalent resonance structures containing a negative charge on the highly electronegative oxygen atom.

R—S—O—H + B

–

–

O

R—S—O ´ R—S==O ´ R—S==O

O

O

O

O

–

Sulphonate anion (highly stabilized by resonance)

Sulphonic acid

11.

O

==

O

== ==

== ==

O

==

(j)

What do you mean by push-pull or captodative ethylene? Explain why they have a very low barrier to rotation as compared to simple alkenes.

Solution The compound containing two electron-withdrawing groups on one doubly bonded carbon and two electron-releasing groups on the other doubly bonded carbon is called a push-pull or captodative ethylene. For example:

:N∫∫ C :N∫∫ C

S Me C==C N Me2

1.110

Organic Chemistry—A Modern Approach

The contribution of the diionic resonance structure such as shown below decrease the double bond character of this ethylene derivative and allows easier rotation. For this reason, such compounds have a much lower barrier to rotation compared to a simple alkene.

12.

The bond ‘a’ of the following cross-conjugated compound is shorter than the double bond of 1,3-butadiene (CH2 == CH— CH == CH2) while the bond ‘b’ of this compound is longer than the double bond of 1,3-butadiene. Explain a

CH2 == CH — C — CH == CH2 ||b CH2 3-methylene-1,4-pentadiene Solution 3–Methylene–1, 4–pentadiene (a cross-conjugated alkene) can be represented as a hybrid of the following five resonance structures: @

!

[CH2 == CH — C — CH == CH2 ´ CH2 — CH == C — CH == CH2 ´ CH2 — CH == C — CH == CH2 || | | @ CH2 ..CH2 ≈CH2 ≈

@

CH2 == CH — C == CH — CH2 ´ CH2 == CH — C == CH — CH2 ] | | @ CH ≈CH2 2 Resonance in 3-methylene-1,4-pentadiene

1,3-Butadiene can be represented as a hybrid of the following three resonance structures. ≈

@

@

≈

[CH2 == CH — CH == CH2 ´ CH2 — CH == CH — CH2 ´ CH2 — CH == CH — CH2 ] Resonance in 1,3-butadiene The bond ‘a’ in the cross-conjugated compound is double in three of the five resonance structures while the double bond in 1,3-butadiene is double is one of the three resonance structures. Therefore, the bond ‘a’ of the cross-conjugated compound possesses more double bond character than the double bond of 1,3-butadiene and hence the bond ‘a’ is shorter than the double bond of 1,3-butadiene. On the other hand, the bond ‘b’ is double in one of the five resonance structures. Therefore, the bond ‘b’ of the cross-conjugated compound

Structure, Bonding and Proper es of Organic Molecules

1.111

possesses less double bond character as compared to the double bond of 1,3-butadiene. Hence, the bond ‘b’ is longer than the double bond of 1,3-butadiene. 13. When 1 mole of benzene is allowed to react with 1 mole of hydrogen under high temperature and pressure and in the presence of an active catalyst, a mixture of 2/3 mole of benzene and 1/3 mole of cyclohexane is isolated. The intermediated products such as 1,3-cyclohexadiene and cyclohexene are not isolated. Account for these observations. Solution

Because of aromatic stability, benzene is much more stable than 1,3-cyclohexadiene and for this reason, the first reaction is endothermic and unfavourable. Cyclohexene is thermodynamically more stable than 1,3-cyclohexadiene and for this, the second reaction is exothermic and occurs rapidly. Cyclohexane is much more stable than cyclohexene and the third reaction is somewhat more exothermic and occurs more readily. From the enthalpy values it becomes clear that a hydrogen molecule reacts with a benzene molecule under harsh conditions to yield 1,3-cyclohexadiene. The second molecule of H2 does not add to a second molecule of benzene, rather it adds to a more reactive molecule of 1,3-cyclohexadiene to yield cyclohexene and the third molecule of H2 does not add to 1,3-cyclohexadience rather it adds to cyclohexene (a thermodynamically more favourable reaction) to yield cyclohexane. Thus, the intermediate products are not isolated. In this hydrogenation process, 3 moles of hydrogen reacts with 1 mole of benzene or 1 mole 1 of hydrogen reacts with 1/3 mole of benzene. As a result of this, mole of cyclohexane is 3 Ê 1ˆ obtained and Á 1 - ˜ or 2/3 mole of benzene remains unreacted. Therefore, when 1 mole of Ë 3¯ hydrogen is allowed to react with 1 mole of benzene, a mixture of 1/3 mole of cyclohexane and 2/3 mole of benzene is obtained.

1.112

1.

Organic Chemistry—A Modern Approach

Explain the following observations: (a)

(

Imidazole :N

( (b)

(

NH is approximately 100 times more basic than pyridine

(

N:

(c)

4–Aminopyridine has a larger dipole moment (4.4 D) than 4–cyanopyridine (1.6 D). The C—O bond length in an ester is shorter than that in an anhydride.

(d)

In the homologous series of compounds

(e)

values of n are stronger bases. I is less basic than II:

(f) (g) (h) (i)

NH

, those with smaller

(CH2)n

Benzene always reacts with O3 to give triozoride but not mono- or diozonide. A substituent such as –NO2 bonded at ortho or para position has a greater effect on acidity of a phenol than a benzoic acid. The doubly bonded N atom in guanidine is more basic than the singly bonded N atom. Indan-2-one

is more acidic than its acyclic model C6H5CH2COCH3. NH2

(j)

Amidines (R—C==NH) are stronger bases than saturated amines.

(k)

CH2(SCH3)2 is more acidic than CH2(OCH3)2 @

[Hint: The conjugate base of the thioacetal, i.e., C H(SCH 3 )2 , is stabilized by (l)

d orbital resonance involving two S atoms (dp-pp back bonding).] The compound I will have higher electron density at the marked (*) carbon atom than the compound II:

Structure, Bonding and Proper es of Organic Molecules

1.113

(m) p-Chlorophenol is more acidic than p-fluorophenol, even though fluorine is more electronegative than chlorine. [Hint: The corresponding p-chlorophenate ion is more stabilized by the extended resonance involving vacant 3d orbital of chlorine atom. In the case of p-fluorophenate, this is not possible because of the absence of the d orbital in fluorine atom.]

2.

(n)

–COOH has –I effect but –COO@ has +I effect.

(o)

p-Nitroaniline is less basic than m-nitroaniline.

(p)

Benzylamine (PhCH 2NH 2 ) is more basic than aniline (Ph NH 2 ) .

(q)

Benzyl alcohol is less acidic than phenol (PhOH).

(r)

p-Nitrophenol is more acidic than m-nitrophenol.

(a) Which of the two nitrogen atoms of the following compound undergo protonation and why?

(b)

Where would protonation occur (at ‘N’ or ‘O’) in the following molecule? Give reasons. NH2 O==C NH2

3.

Arrange the following compounds in order of increasing basicity. Give reasons for your answer.

4.

Arrange the following acids in order of increasing acidity and explain the order: 4-Hydroxybenzoic acid, Benzoic acid, Salicylic acid

5.

Which one between 1,3,5- and 1,3,6-heptatriene is more stable and why?

6.

The methyl hydrogens of g-picoline CH3— CH3 hydrogens of b–picoline N Explain.

( (

7.

(

(

N are more acidic than the methyl

Although 2-methoxyacetic acid (CH3OCH2COOH) is a stronger acid than acetic acid (CH3COOH), p-methoxybenzoic acid than benzoic acid

—COOH. Explain.

is a weaker acid

1.114

8. 9.

Organic Chemistry—A Modern Approach

Explain why peroxy-acids (R COOOH) are weaker acids than carboxylic acids (RCOOH) which are in turn weaker acids than sulphonic acids (RSO3H). Arrange the anions in each group in order of increasing basicity and give your reasoning. O

(a)

–

O

–

,

O

–

(b)

, NO2

10.

–

–

–

C6H5CH2, C6H5O , C6H5NH

Arrange the compounds in each group in order of increasing acidity and explain the order: (a) NO2 OH

(b)

OH ,

O2N

11.

OH , O2N

Br

When acetic acid (CH3COOH), labelled at its OH oxygen with 18O isotope, is treated with aqueous base, and then the solution is acidified, two products having the 18O labelled at different locations are formed. Explain this observation. 18 O O O || || || 1.NaOH 18 CH3 — C—18 OH ææææ ! Æ CH 3 — C — OH + CH 3 — C— OH 2.H3O

Which compound has the greater electron density on its oxygen atom and why? O O ==

(a)

==

12.

—NH—C—CH3

(b)

or

—NH—C—CH3

Which compound has the greater electron density on its nitrogen atom and why? or N H

13.

N H

Arrange the following compounds in order of increasing acidity of the indicated hydrogen and give your reasoning: O O O O || || || || CH3 — C — CH2 — C — CH3 , CH3 — C — CH2 — CH2 — CH2 — C — CH3, I

II

≠

O O || || CH3 — C — CH2 — CH2 — C — CH3 II

≠

≠

Structure, Bonding and Proper es of Organic Molecules

14.

1.115

In each of the following pairs, which species is the stronger base and why? (a)

CH 3COO@ or CH 3CH 2O@

(b)

NH2 NH | || CH3 — CH — CH3 or CH3 — C — NH2

(c)

(CH3 )2 C — CHO or (CH3 )2 C — CH == CH2

@

@

(d) 15.

16.

Explain why N, N-di-(2,4,6-trinitrophenyl) amine is about 100 times more acidic than acetic acid. [Hint: The conjugate base of N, N-di-(2,4,6-trinitrophyenyl) amine is highly stabilized by resonance.] Arrange the following diazonium ions in increasing order of stability with reason: (a)

+

—N∫∫N

O2N—

+

(b)

(c)

—N∫∫N

17.

Arrange the following dienes in order of increasing stability: 1,4-pentadiene, 1,3-pentadiene, 1,2-pentadiene [Hint: The order of increasing stability: 1,2-pentadiene (CH3CH2CH==C==CH2) < 1, 4-pentadiene (CH2 == CH — CH2 — CH == CH2) < 1, 3-pentadiene (CH3CH == CH — CH == CH2)]

18.

Which compound of the following pairs will have higher electron density at the marked (*) carbon atom and why? Cl

(a)

F and

(b)

19.

List the following compounds in order of increasing acidity and explain the order:

20.

Which one is more acidic and why? O

S O

or

S

1.116

Organic Chemistry—A Modern Approach

21.

DBN and DBU are two strong sterically hindered nitrogen bases useful for E2 reactions. Explain why they are strong bases.

22.

The basicity constants of N, N-dimethylaniline and pyridine are almost the same, while 4-(N, N-dinmethylamino) pyridine is considerably more basic than either.

23.

24. 25.

26.

Which one of the two nitrogens of 4-(N, N-dimethylamino)pyridine is more basic and why? Suggest an explanation for its enhanced basicity as compared with pyridine and N, N-dimethylaniline. O O In R—C C == O and C — O bond distances are different but in R—C – they O OH are same. Give your reasoning. Tricyanomethane [HC(CN)3] is as strong as mineral acids. Explain. [Hint: The conjugate base is extensively stabilized by resonance.] The dipole moment of 2-butenal (CH3CH == CHCHO) is higher than that of butanal. Explain.

Guanidine

NH2 | (HN== C — NH2 )

is a much stronger base than acetamidine

CH3 | (HN == C — NH2 ) . Explain. 27.

p-Fluorobenzoic acid is less acidic than p-chlorobenzoic acid. Explain.

28.

p-Nitrophenol is more acidic than m-nitrophenol which in turn is more acidic than phenol. Explain. Explain why phthalimide dissolves in alkali.

29. 30.

Explain why chloroform is more acidic than fluoroform, even though fluorine is more electronegative than chlorine.

Structure, Bonding and Proper es of Organic Molecules

1.117

O

31.

Ph—C

–

is more stabilized by resonance compared to

. Explain.

O

≈

32.

Ph NH 2 is more stabilized by resonance compared to Ph N H3 . Explain.

33.

Account for this increase in acidity (loss of proton): N N H

1.7

N H

O

N H

N H

O

N H

O

HYPERCONJUGATION

When a carbon containing at least one hydrogen atom is attached to a multiple bond such as C == C, C == C, C == O, C == N, etc. The s-electron of the C—H bond is involved in delocalization with the p-electrons of the unsaturated system, i.e., there occurs a s–p conjugation. This type of conjugation may also take place when a carbon containing at least one H-atom is attached to a carbon containing a partially filled or vacant p orbital. This special type of resonance or conjugation giving stability to the species concerned (molecule, free radical or carbonation) is called hyperconjugation. For example: (1) Hyperconjugation in propene (CH3CH = CH2) may be shown as follows:

(2)

≈

Hyperconjugation in ethyl cation (CH 3 C H 2 ) may be shown as follows:

1.118

(3)

Organic Chemistry—A Modern Approach

Hyperconjugation in ethyl radical (CH 3 CH 2 ) may be shown as follows:

Electron displacement towards the p-MO or the vacant or partially filled p-AO caused by hyperconjugation is called hyperconjugative effect. Since in the contributing form no bond exists between C and H, this phenomenon is often termed as ‘no bond resonance’. Although one C—H bond is shown to be broken in each hyperconjugation structures, H≈ never becomes free from the rest of the molecule nor does it change its position in the molecule. For effective hyperconjugatation to take place, the p-orbital concerned and the a C—H bond, i.e., the sp3–s orbital must remain in the same plane. Orbital representations of hyperconjugation in propane or in ethyl cation may be shown as follows:

The stability of a molecule, ion or free radical increases due to hyperconjugation. However, this stability is less than that contributed, if any, by resonance. After the names of the scientists who advanced this theory, hyperconjugation is also called Baker–Nathan effect. The electron-releasing power of alkyl groups attached to unsaturated systems or electrondeficient carbon atoms actually depends on the number of a–H atoms. Methyl (—CH3) group having three a–H atoms has the highest hyperconjugative effect while this effect is non existent with the tert-butyl (—CMe3) group having no a–H atom. So, the electronreleasing power of various alkyl groups when attached to a double bond (or an electron deficient-carbon) follows the order: —CH

—CH2CH3

—CH(CH3)2

—C(CH3)3

This order is exactly the reverse of the order of +I-effect of the alkyl group.

Structure, Bonding and Proper es of Organic Molecules

1.7.1

1.119

Sacrificial and Isovalent Hyperconjugation

Sacrificial hyperconjugation: Each of the hyperconjugative forms involved in hyperconjugation of an alkene, for example, propene (CH3 — CH == CH2), bears isolated charges and has one less real bond than the main uncharged form. This types of hyperconjugation involving ‘sacrifice’ of a bond is termed as sacrificial hyperconjugation. Isovalent hyperconjugation: Each of the hyperconjugative forms involved in hyperconjugation of a free radical (e.g., ethyl radical, CH 3CH 2 ) and a carbocation (e.g., ≈

ethyl cation, CH 3 C H 2 ) displays no more charge separation than the main form and has the same number of covalent bonds as the main form. This type of hyperconjugation involving no ‘sacrifice’ of a bond is termed as isovalent hyperconjugation.

1.7.2 1.7.2.1

Effect of Hyperconjugation on the Physical and Chemical Properties of Molecules and on the Stabilities of Intermediates Heat of combustion and relative stabilities of alkenes

Stabilities of alkenes can easily be explained by hyperconjugation. Greater the number of a-hydrogen atom (H-atom present on the carbon atom attached directly to a double bonded carbon), i.e., greater the number of hyperconjugative structure, higher will be the stability of the alkene due to hyperconjugation and less will be the heat of combustion. For example, the heat of combustion of isobutene, (CH3)2C == CH2 (646.1 kcal/mol) is lower than that of 1-butene, CH3CH2CH == CH2 (649.8 kcal/mol), even though the two isomers have the same number of C—C, C == C and C—H bonds. 1-Butene (CH3CH2CH == CH2) contains two hyper conjugable a-hydrogen atoms while isobutene [(CH3)2C == CH2] contains six such hydrogens. Thus, 1–butane can be represented as a hybrid of three no bond resonance or hyperconjugation structures, whereas isobutene can be represented as hybrid of seven no bond resonance or hyperconjugation structures.

1.120

Organic Chemistry—A Modern Approach

Delocalization of s-electrons into the adjacent p-bond (s-p conjugation) is, therefore, more effective in isobutene than in 1–butene and so isobutene is thermodynamically more stable than 1-butene. Therefore, the heat of combustion of isobutene is less than that of 1-butene.

1.7.2.2

Directive influence of alkyl groups

The ortho- and para-directive influence of alkyl groups can be explained by hyperconjugation. For example, the directive influence of —CH3 group in toluene can be explained on the basis of hyperconjugation. Toluene can be represented as a hybrid of the following hyper conjugative structures.

It is clear from these structures that hyperconjugation increases the electron density at o- and p-positions and hence the electropohilic aromatic substitution reactions in toluene occur mainly at these two positions. Alkyl groups are, therefore, o, p-directing.

1.7.2.3

Bond length

Shortening of carbon–carbon single bond and lengthening of carbon–carbon double bond can be explained by hyperconjugation. For example, in propene (CH3 — CH == CH2) the C—C bond length is 1.488 Å as compared to 1.543 Å in ethane (CH3 — CH3) and the C == C bond length is 1.353 Å as compared to 1.334 Å in ethylene (CH2 == CH2). Propene can be represented as a hybrid of the following hyperconjugation structures.

The C — C bond in propene is single in one structure but double in three structures while the C == C bond in propene is double in one structure but single in three structures. Thus, the carbon–carbon single bond possesses some double bond character and the carbon– carbon double bond possesses some single bond character. For this reason, the C — C bond in propene is somewhat shorter in length than the C — C bond in ethane and the C == C bond in propene is somewhat longer in length than the C == C bond in ethylene.

Structure, Bonding and Proper es of Organic Molecules

1.7.2.4

1.121

Dipole moment

Hyperconjugation may influence the polarity of compounds. For example, the higher dipole moment of acetaldehyde (2.721) as compared to formaldehyde (2.27D) is due to hyperconjugation.

1.7.2.5

Relative stabilities of carbocations

Due to hyperconjugation, the C—H bonding electron pair is attracted towards the positively charged C atom of the carbocation. This helps in dispersing the positive charge in different parts of the alkyl group, i.e., charge delocalization resulting in stability of the carbocation occurs. As the number of a–H atom increases, the number of no bond resonance structures of cabocation increases. As a result, the extent of charge delocalization and consequent stabilization increases. In tert-butyl cation, there are nine hyperconjugable H atoms. Hence, nine equivalent hyperconjugation structures can be written for the cation: H≈ H È ˘ | | Í ˙ ≈ ÍH — C — C (CH 3 )2 ´ H — C ==C(CH3 )2 ´ eight more structures ˙ | Í ˙ | Í ˙ H H Í ˙ Hyperconjugation in tert-butyl cation ÍÎ ˙˚ ≈

≈

Six and three such hyperconjugation structures can be written for (CH3 )2 CH and CH3 CH2 , ≈

respectively, and none can be written for C H 3 . Hence, the increasing order of hyperconjugation stability of methyl, ethyl, isopropyl and tert-butyl cation is

1.7.2.6

Relative stabilities of free radicals

Because of hyperconjugation, the odd electron of a free radical undergoes delocalization for which it becomes stabilized. With increase in the number of a-H atom, the number of no bond resonance structures of free radical increases. As result, delocalization of the

1.122

Organic Chemistry—A Modern Approach

odd electron takes place to a greater extent and stability of free radical increases. Nine hyperconjugation structures can be written for tert-butyl cation: H H È ˘ | Í ˙ ÍH — C — C (CH ) ´ H — C — C (CH ) ´ eight more structures˙ 3 2 3 2 Í ˙ | | Í ˙ H H Í ˙ ÍÎ ˙˚ Hyperconjugation in tert-butyl radical Six and three such hyperconjugation structures can be written for isopropyl radical (CH3CHCH3 ) and ethyl radical (CH3CH2 ) , respectively, and none can be written for methyl radical (CH 3 ). Therefore, the increasing order of hyperconjugation stability is

1.

Which one between 2–methylbut -2-ene and 2-methylbut-1-ene has higher heat of hydrogenation and why?

Solution 2-methylbut-2-ene contains nine hypeconjugable a-H atoms. So, this molecule is highly stabilized by hyperconjugation and because of this, it has relatively lower heat of hydrogenation. On the other hand, 2-methylbut-1-ene contains only five hyperconjugable a-H atoms. So, this molecule is relatively less stabilized by hyperconjugation and because of this, it has relatively higher heat of hydrogenation.

2.

The C — C bond in acetaldehyde (CH3CHO) is shorter than that in ethane, while the C — C bond in trifluoroacetaldehyde (CF3CHO) is essentially the same as that in ethane. Explain.

Solution Acetaldehyde molecule contains three a-H atoms. These hydrogens is involved in hyperconjugation with the double bond of the carbonyl group. As a result, the carbon–carbon

Structure, Bonding and Proper es of Organic Molecules

1.123

bond in acetaldehyde possesses some double bond character. Thus, the carbon–carbon bond in acetaldehyde is shorter than that in ethane (CH3 — CH3). Trifluoroacetaldehyde (CF3CHO) contains no a-H atoms. So, hyperconjugation is not possible in CF3CHO. Thus, the C — C bond in CF3CHO is essentially the same as that in ethane.

3.

Acid-catalyzed dehydration of 2-methylpent-2-ol, CH3CH2CH2C(OH)(CH3)2 gives 2-methylpent-2-ene, CH2CH2CH == C(CH3)2, as the major product and 2-methylpent-1-ene, CH3CH2CH2(CH3)C == CH2, as the minor product. Explain this observation.

Solution The acid-catalyzed dehydration of 2-methylpent-2-ol (a 3° alcohol) occurs through the steps as follows:

Since the dehydration process is reversible, therefore, the more stable alkene will be formed predominantly. The stabilization through hyperconjugation is greater for 2-methylpent-2ene (which offers eight hyperconjugable a–H atoms) than for 2-methylpent-1-ene (which offers only five hyperconjugable a-H atoms) and because of this, 2-methylpent-2-ene is obtained as the major product.

1.124

4.

Organic Chemistry—A Modern Approach

Arrange the following isomeric alkenes in order of increasing heat of combustion and explain the order:

(CH 3 )2 C == C(CH 3 )2 ,CH 2 == CHCH 2CH 2CH 3 ,CH 3CH == CH CH(CH 3 )2 ,CH 3CH == C(CH 3 )CH 2CH 3 I II III IV

Solution The stability of an alkene is determined by the number of a-H atoms present in the molecule, i.e., by the number of no bond resonance structures involved in hyperconjugation. The greater the number of structures, the higher the stability of the alkene. Now, the number of a-H atoms in the alkene I, II, III and IV are 12, 2, 4 and 8 respectively. Thus, the stability of the alkenes increases in the order: II < III < IV < I. Since the heat of combustion increases with decrease in stability of the alkene, therefore, the order of increasing heat of combustion is I < IV < III < II.

1.

2. 3.

4.

5.

The carbon–carbon single bond in methylacetylene (CH3 — C ∫∫ CH) is approximately 1.46 Å in length, much less than the value (1.54 Å) found in saturated hydrocarbons. Explain. Heat of hydrogenation of ethene is greater than that of propene. Explain. Alkyl groups attached to benzene ring release electrons in the order: CH3 —> CH3CH2 — > (CH3)2CH — > (CH3)3C —. This is quite opposite to that expected on the basis of inductive electron release (+I effect). Explain this anomaly. What three alkenes yield 2-methylbutane on catalytic hydrogenation ? Match each alkene with its correct heat of hydrogenation. Heat of hydrogenation in kJ mol–1: 112, 118, 126 Which compound of the following pair will have higher electron density at the marked (*) carbon atom ? Explain.

Structure, Bonding and Proper es of Organic Molecules

1.8

1.125

STERIC EFFECT

Steric effect is an effect on the relative reaction rates caused by the space-filling properties of those parts of a molecule attached at or near the reaction site, i.e., steric effect is an effect on the rates of chemical reactions caused by van der Waals repulsions. The most common steric effect, however, is the classical steric hindrance. When the sheer bulk of groups at or near a reacting site of a molecule hinders or retards a reaction, it is called steric hindrance, i.e., a steric effect which decreases the reactivity of a compound is called steric hindrance. In an SN2 reaction, a backside approach by the nucleophile to a distance within the bonding range at the carbon atom bearing the leaving group is required. For this reason, bulky substituents on or near the carbon atom have a dramatic inhibiting effect. They cause the energy of the transition state to be increased and consequently, they increase the activation energy of the reaction for which the reaction becomes slow. Methyl halides react most readily in SN2 reaction because only three small hydrogen atoms interfere with the approaching nucleophile. Neopentyl and tert-butyl halides are least reactive because bulky groups present a strong hindrance to the approaching nucleophile. In fact, tertiary substrates do not react by an SN2 mechanism.

Steric hindrance increases Steric strain is another kind of steric effect. If the constituent atoms or groups of a molecule or ion owing to their bulky nature require more space than what is available for them, i.e., when they are forced too close to one another, then mechanical interference amongst the groups or atoms takes place and the molecule or ion is then said to be under steric strain. Steric strain makes the species unstable, i.e., its energy increases. For example, owing to

1.126

Organic Chemistry—A Modern Approach

great bulk of the tert-butyl group and small available space for it, 1,2,3-tri-tert-butylbenzene is under severe strain and so it is very difficult to prepare. Gauche conformation of butane are higher in energy (less stable) than the anti conformation because of steric strain.

Cyclic compounds twist and bend in order to have a structure that minimizes the three different kinds of strain that may destabilize a cyclic compound. These are (i) Angle strain: It is the strain induced in a molecule when the bond angles are different from the desired tetrahedral bond angle of 109.5°. (ii) Torsional strain: It is caused by repulsion of the bonding electrons of one substituent with the bonding electrons of a nearby susbstituent. (iii) Steric strain.

1.8.1 Properties of Molecules Influenced by Steric Effect 1.8.1.1

Acidic character of compounds

(1) Acidic character of aromatic acids In order to achieve maximum overlap of p-orbitals, which is necessary for delocalization, it is essential that the skeleton should be planar in the conjugated system. In other words, any structural feature that destroys this coplanarity of the conjugated system inhibits resonance. This inhibition is referred to as steric inhibition of resonance (SIR). The higher acidity of 2,4,6-trimethylbenzoic acid with respect to p-toluic acid can be explained in terms of steric inhibition of resonance (SIR). p-Toluic acid can be represented as hybrid of the following resonance structures:

Structure, Bonding and Proper es of Organic Molecules

1.127

The carboxyl group withdraws electrons from the ring by its —R effect and as a consequence, the lone pair on hydroxyl oxygen is prevented to a greater extent from entering into resonance with the C == O group. As a result, the OH oxygen becomes less positive by entering into resonance with the C == O group and proton release becomes disfavoured. For effective resonance to take place the carboxyl group must be planar or nearly planar with the aromatic ring. In 2,4,6-trimethyl benzoic acid, the planar conformation required for electron delocalization experiences steric strain. To avoid this strain, the carboxyl group is forced out of the plane by the two ortho-methyl groups. As a result, the resonance interaction of the carboxyl group with the ring decreases and consequently, the resonance interaction of the unshared electron pair on OH oxygen with C == O increases. Because of this, the hydroxyl oxygen becomes relatively more positive and proton release becomes favoured.

Steric inhibition to solvation often plays an important role in determining the acidity of aromatic acids. For example, benzoic acid (pKa = 5.05) is found to be more acidic than 2,6di-tert-butylbenzoic acid (pKa = 6.25) in a mixture of ethanol and water. Because of steric inhibition of resonance, the latter acid is expected to be more acidic than the former acid. But an opposite result is obtained when H2O-EtOH is taken as a solvent and that is due to steric inhibition to salvation. Solvation (stabilization by solvent) in less effective for 2,6-ditert-butylbenzoate ion than for the benzoate ion and this is because the negative charge on the former ion is somewhat shielded from the solvent molecules by surrounding tert-butyl groups. Therefore, the difference in stability between the conjugate base and the acid is less in the case of benzoic acid than in the case of 2,6-di-tert-butylbenzoic acid. Therefore, the former equilibrium is relatively more favourable than the latter and so, benzoic acid is found to be more acidic than 2,6-di-tert-butybenzoic acid in H2O — EtOH.

1.128

Organic Chemistry—A Modern Approach

(2) Acidic character of phenols 2,6-Dimethyl-4-nitrophenol is more acidic than 3,5dimethyl-4-ntrophenol. This observation can be explained on the basis of steric inhibition of resonance (SIR). In the latter compound, the oxygen atoms of the p-nitro group becomes involved in steric interaction with the two adjacent methyl groups and as a consequence, they are forced out of the plane of the ring. Hence, the — NO2 group cannot enter into resonance with the — OH group. As a result, the oxygen of — OH becomes less positive and proton release is less favoured. In 2,6-dimethyl-4-nitrophenol, on the other hand, the —NO2 group enters into resonance with the — OH group. As a result , the — OH oxygen becomes considerably positive and proton release in highly favoured. Hence, 2,6-dimethyl4-nitrophenol is more acidic than 3,5-dimethyl-4-nitrophenol.

1.8.1.2

Basic character of compounds

(1) Basic character anilines N,N,2,6-Tetramethylaniline is a stronger base than N,N-3,5-tetramethylaniline. This can be explained by steric inhibition of resonance. In N,N,3,5-tetramethylaniline, the unshared pair of electrons on nitrogen is well delocalized by resonance with the benzene ring because the smaller ortho H atoms do not interfere sterically with the methyl groups attached to nitrogen and so, the —NMe2 group is not

Structure, Bonding and Proper es of Organic Molecules

1.129

forced out the plane. The unshared pair of electrons on nitrogen is, therefore, not well available for coordinating with a proton. Hence, it behaves as a weaker base.

On the other hand, in N,N-2,6-tetramethylaniline, the unshared pair of electrons on nitrogen cannot enter into resonance with the ring because the relatively bulky ortho methyl groups interfere sterically with the methyl group attached to nitrogen and as a consequence, the - N Me2 group is forced out of the plane (conformation A). The unshared pair of electrons is now localized on the N atom and is, therefore, N,N,2,6-tetramethylaniline is stronger base than N,N,3,5-tetramethylaniline.

That 3,5-dimethyl-4-nitroaniline is stronger as a base than 2,6-dimethyl-4-nitroaniline can also be explained on the basis of steric inhibition of resonance (SIR). In 3,5-dimethyl4-nitroaniline, the oxygens of the p-NO2 group is involved in steric interaction with two adjacent methyl groups and as a consequence, they are forced out of the plane of the ring. Hence, the —NO2 group cannot enter into resonance with the —NH2 group and the unshared pair of electrons on nitrogen is well available for coordinating with a proton. On the other hand, in 2,6-dimethyl-4-nitroaniline, the —NO2 group enters into resonance with the —N H 2 group because the smaller H atoms of the —NH2 group do not interfere sterically with the two o-methyl groups. Therefore, the unshared electron pair on nitrogen is not well available for coordinating with a proton. This explains why 3,5-dimethyl-4nitroaniline is stronger as a base than 2,6-dimethyl-4-nitroaniline.

1.130

Organic Chemistry—A Modern Approach

Hydrogen bonding sometime plays an important role in making electron delocalization much more effective. 2,4,6-trinitroaniline is a very much weaker base than 2,4,6-trinitro-N, N-dimethylaniline, even though aniline and N, N-dimethylaniline have almost same basic strength. That 2,4,6-trinitro-N,N-dimethylaniline is a much stronger base than 2,4,6trinitroaniline can be well explained in terms of steric inhibition of resonance. In the former compound, the oxygens of the two o-NO2 groups is involved in steric interaction with the two N-methyl groups and as a consequence, the two —CH3 groups are forced out of the plane of the ring. Hence, the —NMe2 group cannot enter into resonance with the —NO2 groups, i.e., the —NO2 groups cannot exert their —R effect to reduce the electron density on nitrogen. They only exert their —I effects to reduce the availability of the unshared pair. The unshared pair of electrons is, in fact, localized on the nitrogen atom and is, therefore, well available for coordinating with a proton. On the other hand, in the latter compound, the —NH 2 group enters into resonance with the three —NO2 groups because the smaller H atoms of the —NH2 group do not interfere sterically with the two o-NO2 groups. Again, there occurs intramolecular hydrogen bonding between the oxygens of the o-NO2 groups and the hydrogens of the —NH2 group. This help to hold the two —NO2 groups in the planar conformation required for maximum resonance interaction. Therefore, due to very powerful —R and —I effects of the three —NO2 groups, the availability of the unshared pair of electrons on nitrogen is markedly reduced. Because of this, 2,4,6-trinitroaniline behaves as a very much weaker base than 2,4,6-trinitro-N,Ndimethylaniline. The contribution of the two electron-releasing (+I) —CH3 groups on nitrogen in increasing the basic strength of the latter compound is actually very small.

Structure, Bonding and Proper es of Organic Molecules

1.131

(2) Basic character of tertiary amines (R3N) As the alkyl groups of tertiary amines (R 3 N) become gradually more bulkier, the additional strain due to crowding is somewhat reduced by increasing bond angles. As a consequence, the bonding orbitals acquire less p character, while the orbital containing the lone pair acquires more p character. This type of modification is actually possible when there is an unshared pair of electrons in one of the hybrid orbitals. When the amine coordinates with a proton, the bonding orbitals are forced towards tetrahedral (sp3) shape with reduction of bond angles and increased steric strain among the alkyl groups. This type of steric strain present in ‘back’ of the amine and away from the entering H! is called B strain (back strain). The bulkier the alkyl groups, the more B strain there is and weaker is the base. Another factor that partially reduces basicity of the 3° amine is the increasing steric hindrance to solvation of the conjugate ≈

acid, R 3 NH .

1.8.1.3

Bond length

The difference in length of a similar type of bonds can be explained in terms of steric inhibition of resonance. For example, the difference in the C—N bond length o- and p-nitro groups in picryl iodide can be explained as follows. O O I + + a N N – – O O (Picryl iodide) b

O

+

N

O

–

The large iodine atom in picryl iodide forces the oxygens of the o-nitro groups out of the plane of the ring but it has no such effect on the p-nitro group. Therefore, only the p-nitro group enters into resonance with the ring. As a result, the carbon–nitrogen bond in the p-nitro group acquires some double bond character and hence, the C—N bond length ‘a’ is longer (145 pm) than ‘b’ which is only 135 pm.

1.132

1.8.1.4

Organic Chemistry—A Modern Approach

Dipole moment

Steric inhibition of resonance is sometime responsible for reducing the dipole moment of compounds. For example, the dipole moment of p-nitroaniline is higher than the dipole moment of 2,3,5,6-tetramethyl-4-nitroaniline. In the latter compound, the oxygen of the p–NO2 group is involved in steric interaction with the two adjacent methyl groups and as a consequence, the group is forced out of the plane of the ring. Hence, the —NO2 group cannot withdraw electrons from the —NH2 group by exerting its —R effect. As a result, the extent of charge separation becomes less and the magnitude of dipole moment also becomes less. Since steric inhibition of resonance is absent in p-nitroaniline, there occurs greater charge separation and for this, its dipole moment is much higher.

1.8.2

Proton Sponges

To become free from steric strain, some organic bases possess greater affinity for protons, i.e., behave as much stronger bases. These compounds are known as proton sponges. For example, 1,8-bis (diethylamino)-2,7-dimethoxynaphthalene (I) is severely strained because the lone pairs of two N atoms are forced to be near each other. On protonation the stain is relieved because one lone pair coordinates with proton which subsequently forms a hydrogen bond to the other lone pair. The much higher basicities of 4,5-bis (dimethylamino) fluorene (II) and 3,5-bis(dimethylamino)phenanthrene (III) can similarly be explained. Compounds like I, II and III are called proton sponges.

Structure, Bonding and Proper es of Organic Molecules

1.8.3

1.133

Face Strain or F-Strain

When two atoms each containing three large groups approach towards each other to form a covalent bond, a steric strain develops at the time of bond formation. This is what is called face strain or F-strain. Because of such strain, the formation of adduct does not always take place smoothly. For example, tripropylamine gives no detectable adduct with trimethylborane, while quinuclidine gives a very stable adduct (both the amines have comparable base strength). The difference of these two amines can be explained in terms of F-strain. In tripopylamine È(CH 3CH 2CH 2 )3 N,˘ at least two of the three freely rotating Î ˚ propyl groups remain folded towards, and partly covering the unshared pair of electrons of nitrogen. Since this conformation causes F-strain, the approach of trimethylborane towards nitrogen is inhibited. Tripropylamine thus gives no detectable adduct with trimethylborane.

On the other hand, in quinuclidine, the substituents or nitrogen are held back by the ring system so that they cannot involve in steric interaction with trimethylborane molecule. For this reason, quinuclidine forms a very stable adduct.

1.134

1.8.4

Organic Chemistry—A Modern Approach

Steric Acceleration and Steric Retardation

Steric acceleration: Substrates, which are very much compressed, always try to avoid their steric strain. A reaction will be speeded up if the strain of the substrate is relieved in attaining the transition state of the rate-determining step of the reaction. This phenomenon is termed as steric acceleration. The rate of ionization and hence the rate of solvolysis of a substrate in which there is B-strain is, therefore, expected to be much higher. Tri-isopropylmethyl chloride, for example, undergoes solvolysis (SN1) at a much faster rate than that expected on the basis of stabilization of the carbocation through +I and hyperconjugative effect.

Due to sp3 angle of 109.5°, the bulky alkyl groups of this alkyl chloride are pushed together. As a result, the compound experiences large steric strain in the ‘back’ of the molecule (B-strain or back strain). On ionization, i.e., on going from the starting halide (with a tetrahedral disposition of four groups about the Csp3 atom) to the carbocation (with a planer disposition of only three group about the Csp2 atom), the bond angle increases from 109.5° to 120°. As a result, the steric strain, i.e., the B-strain resulting from the nonbonded interactions amongst the alkyl groups, is reduced because space between two alkyl groups increases. Consequently, the ionization (the rate-determining step) becomes favourable and the alkyl halide undergoes solvolysis at a faster rate than expected. Steric retardation: The sheer bulk of groups or atoms on the reacting part of the substrate may slow down or even stop a reaction. This phenomenon is called steric retardation or steric hindrance. Tertiary substrates do not undergo SN2 reaction because of steric hindrance. For example, when tert-butyl bromide is treated with KI in acetone solution, practically no tert-butyl iodide is obtained because the bulky methyl groups present a strong hindrance to the approaching nucleophile I①.

Structure, Bonding and Proper es of Organic Molecules

1.8.5

1.135

Bredt’s Rule

A double bond (C== C or C== N) cannot be introduced at the bridgehead position in bridged bicyclic compounds with small rings. This is what is called Bredt’s rule. For example, 1-chlorobicyclo [2.2.1]heptane (1) does not undergo base-promoted dehydrochlorination to yield bicyclo[2.2.1]hept-1-ene (II)

The p-orbitals of a bridgehead double bond are not coplanar and actually these are at right angles to each other. Because of this, significant orbital overlap is not possible and hence the formation of a double bond is not possible. To bring the orbitals in one plane (a condition required for maximum overlap) this regid molecule is to be distorted. However, such distortion resulting in severe strain in this molecule is not energetically permissible. A number of properties of compounds can be explained in terms of Bredt’s rule. These are as follows:

(1) Basicity of compounds That the compound I is more basic than the compound II can be explained by Bredt’s rule.

Since the Bredt’s rule is violated in the resonance structures of I, they have practically no contribution to the hybrid. Hence, the unshared pair of electrons on nitrogen is not delocalized with the ring p electrons and is, therefore, more available for coordinating with a proton.

1.136

Organic Chemistry—A Modern Approach

Since II is not a bridged bicyclic compound, therefore, the resonance structures with no bridgehead double bond are considerably stable and have significant contribution to the hybrid. Hence, the unshared pair of electrons on nitrogen is well delocalized with the ring and is not much available for coordinating with a proton. Thus, the compound I is more basic than the compound II.

(2) Acidity of compounds That the compound I is more acidic and readily soluble in alkali as compared to the compound II can be explained in terms of Bredt’s rule.

O

I

O

O

II

O

The acidity of a b-dicarbonyl compound and its solubility in aqueous alkali depend on the stability of the enolate ion (the conjugate base). The more the anion is stable the more the diketone is acidic and the more it dissolves in alkali. The diketone I and II react with alkali as follows:

Three resonance structures can be written for the conjugate base of I. Two of them are equivalent and relatively more stable because the negative charge is accommodated by the highly electronegative oxygen atom. Therefore, these two structures have significant contribution to the hybrid. Thus, the negative charge on carbon is highly delocalized by resonance involving the two adjacent C== O groups and consequently, the conjugate base is well stabilized. Because of this, the diketone I considerably acidic and dissolves readily in aqueous alkali. Two of the three resonance structures of the conjugate base of II are very unstable because they have double bond at the bridgehead position, i.e., they violate Bredt’s rule and hence, they have, in fact, no contribution to the hybrid. The conjugate base of II is, therefore, not

Structure, Bonding and Proper es of Organic Molecules

1.137

stabilized by resonance. Because of this, the bicyclic dicarbonyl compound II is relatively less acidic and reacts with aqueous alkali with much reluctance, i.e., its solubility in aqueous alkali is actually very poor.

(3) Decarboxylation of b-keto acids When b-keto acids are heated, they undergo ready decarboxylation via a six-membered cyclic transition state to produce initially an enol which readily tautomerizes to the more stable ketone. For example:

However, the following bridged b-keto acid does not undergo decarboxylation even when heated to 300°C.

This observation can be explained by Bredt’s rule. This b-keto acid is expected to decarboxylate through the formation of a transition state with the partial double bond at the bridgehead position and thus, through an enol in which the double bond is placed at the bridgehead position. Both the transition state and the enol are very much less stable because they violate Bredt’s rule. Because of this, the bicyclic b-keto acid resists decarboxylation even when heated to 300°C

1.138

Organic Chemistry—A Modern Approach

(4) Formation of Grignard reagent from vicinal dibromide When a vicinal or 1,2-dibromide is allowed to react with Mg in dry ether, it leads to the formation of an alkene along with MgBr2. Thus, the vicinal dibromide I smoothly forms the corresponding alkene when treated with Mg in dry ether.

However, when the bridged bicyclic vicinal dibromide II is treated with Mg in dry ether elimination of MgBr2 to form the corresponding alkene does not take place because the alkene and the transition state leading to it violates Bredt’s rules and as a consequence, the corresponding Grignard reagent is obtained.

1.

Explain why o-tert-butylbenzoic acid is more acidic than the para-isomer.

Solution The higher acidity of o-tert-butylbenzoic acid as compared to its para-isomer can be explained in terms of steric inhibition of resonance (SIR). p-tert-Butylbenzoic acid can be represented as a resonance hybrid of the following contributing forms:

The carboxyl group by its –R effect withdraws electrons from the ring. The lone pair of electrons on hydroxyl oxygen is thus prevented to a greater extent from entering into resonance with the C== O group. As a result, oxygen becomes less positive and proton

Structure, Bonding and Proper es of Organic Molecules

1.139

release becomes disfavoured. The essential requirement for effective electron delocalization is that the carboxyl group must be planar or nearly planar with the aromatic ring. In o-tert-butylbenzoic acid, the planar conformation needed for delocalization experiences steric strain. To avoid this strain, the carboxyl group is forced out of the plane by the bulky tert-butyl group. As a consequence, the resonance interaction of the carboxyl group with the ring decreases and the resonance interaction of C== O with the unshared electron pair on OH oxygen increases. The hydroxyl oxygen, therefore, becomes relatively more positive and this causes a more facile proton release.

2.

In the following triptycene derivative, rotation of the aryl group around the O-aryl bond is not completely free — Why? Me

CMe3

O

H

H

Me

Solution A complete 360° rotation of the aryl group around the O-aryl bond requires the aryl group to pass over three rotational barriers; one of which is the C — CMe3 bond and the other two are the top C — H bond of the other two rings. For this reason, the triptycene derivative, rotation of the aryl group around the O-aryl bond is not completely free. 3. It is possible to prepare cis and trans isomers of 5-amino – 2,4,6-triiodo – N, N, N¢, N¢-tetramethylisophthalamide. Explain. Solution It is possible to prepare cis and trans isomers of 5-amino-2,4,6-triiodo-N, N, N¢, N¢-tetramethylisophthalamide because each —CONMe2 group being flanked by two bulky iodine atoms cannot rotate freely due to steric strain. This is, in fact, an example of cistrans isomerism resulting from restricted rotation about single bonds.

1.140

4.

Organic Chemistry—A Modern Approach

Explain why the compound I is more acidic than the compound II.

Solution The conjugate base of I, i.e., the carbanion derived from the compound I is well stabilized by resonance because all the three aromatic rings are tightly held with each other and hence coplanar. Also, the carbanion constitutes an aromatic system [(4n + 2) p electron, where n = 1]. On the other hand, the benzene rings in the conjugate base of II, i.e., the carbanion derived from II, prefer to exist in different planes due to steric reason (steric interaction between ortho H atoms on adjacent rings) and so, it is not well stabilized by resonance. Because of greater stability of the conjugate base, the compound I is more acidic than the compound II.

Structure, Bonding and Proper es of Organic Molecules

5.

1.141

The amide I is more basic than the isomeric amide II—Why?

Solution Nitrogen’s electron pair in amide becomes involved in resonance interaction with the adjacent C== O group. That is, the C== O group withdraws the unshared pair on nitrogen by is strong –R effect. In these two amides, this resonance interaction is not equally well. Two resonance structures can be written for each of the two isomeric bicyclic amides I and II.

1.142

Organic Chemistry—A Modern Approach

The dipolar resonance structure of I with a double bond at the bridgehead position is very unstable because it violates Bredt’s rule. So it has practically so contribution to the hybrid of I. Hence, electron delocalization does not take place in this amide and consequently, the unshared pair of electrons on nitrogen is well available for coordinating with a proton. On the other hand, the dipolar resonance structure of II with a double bond not in a bridgehead position is quite stable and so contribute significantly to the hybrid of II. Hence, effective electron delocalization occurs in II and the lone pair on nitrogen is poorly available for sharing with a proton. Thus, the amide I is more basic than the amide II. 6. The following b-hydroxyketone does not undergo dehydration when heated with NaOH solution — Why?

Solution Base-catalyzed dehydration b-hydroxyketones takes place by E1CB mechanism which involves abstraction of a proton from the a-carbon by base to form an enolate ion in the first step followed by loss of OH① ion in the second step. When this bicyclic b-hydroxyketone is heated in the presence of NaOH solution, an enolate is first formed but the second step, i.e., the loss of OH① ion does not take place because this leads to the formation of a double bond at the bridgehead position which is not feasible sterically (violates Bredt’s rule). This explains why the given b-hydroxyketone does not undergo dehydration when heated with NaOH solution.

1. 2. 3.

The length of C—C bond connecting the ring in 2,2¢,6,6¢-tetramethylbiphenyl is longer than that in biphenyl. Explain. Hydroxide is a stronger nucleophile than tert-butoxide, despite its lower base strength – Why? 2,6-Di-tert-butylpyridine is a specific proton scavenger. Explain. [Hint: Because of steric reason (F-strain), the compound does not act as a

Structure, Bonding and Proper es of Organic Molecules

1.143

nucleophile, but it can easily take up a small proton because the space between the two bulky tert-butyl groups is enough for the entry of a proton. So this pyridine derivative is a specific proton scavenger.

Rank the following compounds in order of decreasing basicity and give your reasoning. ==

4.

NMe

N

O

5.

O

N

O

Me I II III Solvolysis of the alkyl chloride A in aqueous ethanol proceeds about 600 times faster than the alkyl chloride B. Account for this observation.

CH3 | (Me3C CH 2 )2C Cl A

Me3C Cl B

[Hint: Steric acceleration] 6.

7.

It is possible to prepare cis and trans isomers of 1,8-di-o-tolylnaphthalene – Why?

[Hint: Free rotation of the o-tolyl groups about the single bond is not possible due to steric reason.] Quino[7,8–h]quinoline (I) is a much stronger base than quinoline (II). Account for the fact. N

N N I

8.

II

Out of the following two diols (A) and (B), one forms intermolecular hydrogen bonds while the other forms intermolecular hydrogen bonds. Identify them and give your reasoning.

1.144

Organic Chemistry—A Modern Approach

CMe3 HO

CMe3 HO

OH

OH

A

B

[Hint:

In the stable conformations of these two diols, the bulky —CMe3 group is always equatorial.] 9.

2,6-Di-tert-butylpyridine is a weaker base than pyridine — Why? [Hint: This can be explained in terms of entropy effect caused by steric effect. The conjugate acid of di-tert-butylpyridine is less stable than the conjugate acid of nonsterically hindered pyridine because in the former conjugate acid, the bulky tertbutyl groups restrict rotation in water molecule which forms hydrogen bond with it and thereby lowers the entropy and decrease its stability.

10.

The heat of hydrogenation of cis-2-petnene is higher (28.6 kcal/mol) than that of trans-2-pentene (27.6 kcal/mol) — Why ? The bicyclic lactam A undergoes hydrolysis 107 times faster than the monocyclic lactam B. Explain.

11.

12.

1,3-Cyclohexadione readily undergoes base or acid-catalyzed deuterium exchange, whereas the following diketone does not — Why ?

Structure, Bonding and Proper es of Organic Molecules

13.

1.145

[Hint: Enolate formation is not possible because electron delocalization violates Bredt’s rule.] Predict the increasing order of equilibrium constants of the reaction of the following bases with Me3B and explain your answer.

III, II, I Æ (because F-strain decreases)] [Hint: ææææææ K eq increases

14.

Predict with proper reasoning which member in each of the following pairs behaves as a stronger base towards BMe3. (a)

Et2NH and

(b)

N

and

15.

Compare C—N bond length (a vs a¢) and (b vs b¢) in the following compounds and justify.

16.

The following bicyclo trisulfone (I) readily dissolves in aqueous sodium bicarbonate solution, while the triketone (II) does not — Why?

17.

[Hint: Compounds containing S atoms seem to defy Bredt’s rule. The conjugate base of I is highly stabilized by resonance, while the conjugate base of II is not at all stabilized by resonance.] Predict whether any change will take place or not. Justify your answer.

1.146

Organic Chemistry—A Modern Approach

Which one of the following two compounds is more acidic and why?

19.

[Hint: The enolate of II is resonance stabilized, but the enolate of I is not. Therefore, II is relatively more acidic.] The following compound being an amide reacts more or less like a ketone — Why? N O ==

18.

H3 C

20. 21.

The following b-keto acid does not decarboxylate even when heated to 500°C. Explain. O ==

22.

CH3 CH3 [Hint: It is an extreme case of a twisted amide. The overlap of the lone pair of electrons on nitrogen with the p-system of the carbonyl group is highly prevented. So, this compound reacts more or less like a ketone.] N, N-Dimethylation triples the basicity of aniline but increases the basicity of 2,4,6-trinitroaniline by 40,000 fold. Explain. Are the following two structures resonance forms? Explain why or why not ?

COOH

23.

The following quinclidinone forms an oxime — Why?

N O

Structure, Bonding and Proper es of Organic Molecules

24.

25.

1.147

The bridged bicyclic b-keto acids I, II and III undergo ready decarboxylation on heating while the b-keto acids IV and V do not — Why?

[Hint: A bicyclic ring system [a, b, c] with bridgehead double bond is sterically feasible when (a + b + c), i.e., the sum of the number of atoms in the bridges or the S number is seven or more, S ≥ 7. The numerical value S is 7 for I and II, 9 for III and 5 for IV and V.] Which one of the following two carboxylic acids is more acidic and why? COOH H

H COOH

I II [Hint: The conjugate base (i.e., the corresponding carboxylate ion) of the cisisomer I is not well solvated due to steric crowding caused by C–3 and C–5 axial hydrogens. On the other hand, the conjugate base of the trans-isomer II is well solvated because the equatorial —COOH group is free from any steric crowding.]

1.9

INTERMOLECULAR FORCES

The forces that hold molecules together may be called intermolecular forces. There are two kinds of intermolecular forces: (1) dipole–dipole interactions and (2) van der Waals forces.

1.9.1

Dipole–Dipole Interactions

The attraction of the positive end of one polar molecule for the negative end of a second polar molecule is called dipole–dipole interaction. The polar hydrogen fluoride molecules, for example, are held together by dipole–dipole interactions that operate either head to tail (I) or laterally (II).

1.148

Organic Chemistry—A Modern Approach

This attractive force is effective over a very short distance and its strength is about 2 kcal/mol. As a result of dipole–dipole interaction, polar molecules are generally held to each other more strongly as compared to nonpolar molecules of comparable molecular weight. The presence of dipole–dipole interaction in a compound is reflected in its physical properties.

1.9.1.1

Hydrogen bonding

It is a specifically strong kind of dipole–dipole attraction that operates between a hydrogen atom that is covalently bonded to an atom of high electronegativity and smaller size such as F, O, and N and a second such atom. The energy of such a bond varies from 5 to10 kcal/mol. The covalent bond between a hydrogen atom and a highly electronegative element is considerably polar because the electron cloud is greatly distorted towards the electronegative atom. The partial positive charge developed on the thinly shielded hydrogen nucleus is strongly attracted by the partial negative charge on the electronnegative atom of a second molecule (same or different) and leads to the formation of what is called a hydrogen bond. The dipolar HF molecules, for example, are held together by hydrogen bonding (represented by dotted lines) as shown below:

1.9.1.2

Types of hydrogen bonding

There are tow types of hydrogen bonding:

(1) Intermolecular hydrogen bonding When H-bonding occurs between different molecules of same or different compounds, it is called intermolecular hydrogen bonding. For example, methanol (CH3OH) molecule contains one highly electronegative oxygen atom bonded to H atom. Thus, the oxygen atom becomes partially negatively charged and the H atom becomes partially positively charged. The negative end of one molecule attracts the positive end of the other to form hydrogen bond. In this a way, a large number of methanol molecules become associated through hydrogen bonding to form a macromolecule which can be represented as (CH3OH)n.

Structure, Bonding and Proper es of Organic Molecules

1.149

(2) Intramolecular hydrogen bonding When hydrogen bonding takes place within the same molecule, it is called intramolecular hydrogen bonding. This type of hydrogen bonding is also known as chelation as it results in formation of a ring. It is favoured when a six or five membered ring is formed. It is normally found in disubstituted benzene compounds in which the substituents are attached to adjacent carbon atoms, i.e., they are ortho to each other. For example, o-nitrophenol possesses intramolecular hydrogen bonding.

Hydrogen bonds markedly influence the physical properties like melting points, boiling points and solubilities of compounds. Acidity of certain acids can also be influenced by intramolecular hydrogen bonding.

1.9.2

van der Waals Forces

Since nonpolar compounds can also solidify, therefore, there must be some forces acting between the molecules of a nonpolar compound. Such attractive forces are called van der Waals forces. Such attractions are also referred to as instantaneous dipole-induced dipole attractions. In nonpolar molecules (i.e., methane, CCl4, etc.), the centre of positive charge density coincides with the centre of negative charge density and so the molecules possess no net dipole moment. However, due to random movement of electrons around the nucleus, a momentary distortion of their distribution may occur. This results in momentary loss of electrical symmetry and formation of a momentary dipole in the molecule. This instantaneous dipole induces a dipole in a neighbouring molecule. These dipoles then attract each other and as a result, the molecules become held together. The van der Waals forces act only between the portions of different molecules that are in close contact, that is, between the surfaces of molecules. Therefore, the size of van der Waals forces depends on the area of contact between the molecules. The greater the area of contact, the stronger the van der Waals forces and the greater the amount of energy needed to overcome these forces. Since the surface areas of heavier molecules are usually much greater, therefore, intermolecular van der Waals attractive forces are also much larger. Another factor which effects the strength of van der Waals forces is polarizability. Polarizability indicates the measure of how the electron cloud around an atom can be distorted. The larger the atom, the more loosely it holds the outermost electrons and the more they can be distorted. The more polarizable the atom, the stronger the van der Waals attractive forces. The strength of van der Waals forces (also called London forces) is only about 1 kcal/mol and this is significant only at extremely short distances. These forces are, in fact, the weakest of all kinds of intermolecular forces and these help us to understand the effect of molecular size and shape on physical properties.

1.150

Organic Chemistry—A Modern Approach

(1) Boiling Point (bp) The boiling point of a compound is the temperature at which the liquid form of the compound becomes a gas. In other words, it is the temperature at which the vapour pressure of the compound equals the atmospheric pressure. In order for a compound to vaporize, the forces that hold the molecules close to each other must be overcome. Therefore, the boiling point of a compound depends on the strength of the attractive forces between the individual molecules in the more ordered liquid state. The stronger the intermolecular attractive forces, the higher the boiling point and this is because a lot of energy is required to pull the molecules away from each other. (2) Melting point The melting point in the temperature at which a solid is converted into its liquid phase. In melting, energy is required to overcome the attractive forces in the more ordered crystalline state. The stronger the intermolecular forces, the higher the melting point. Symmetry also plays an important role in determining the melting point of compounds containing the same functional group and similar molecular masses. A compact symmetrical molecule which packs well into a crystalline lattice has a much higher melting point. The more the compound is symmetrical, the higher is the melting point. (3) Solubility Solubility is the extent to which a compound (called solute) dissolves in a liquid (called solvent). The condition for dissolution of a solute in a solvent is that the solute–solvent interactions must be equal or larger in magnitude than the combined solute–solute and solvent–solvent interactions. That is, the old attractive forces must be replaced by new ones. Compounds generally dissolve in solvents possessing similar kinds of intermolecular forces. Polar compounds dissolve in polar solvents while nonpolar or weakly polar compounds dissolve in nonpolar or weakly polar solvents. The general rule that explains solubility on the basis of polarity of molecules is that ‘like dissolves like’. Table 1.1 Summary of types of intermolecular forces Type of force

Relative strength

Exhibited by

Example

van der Waals forces

weak

all molecules

CH3CH2CH2CH2CH3 CH3CH2CH2CH2CHO CH3CH2CH2CH2CH2OH

Dipole–dipole interactions

moderate

molecules with a net dipole

CH3CH2CH2CH2CHO CH3CH2CH2CH2CH2OH

Hydrogen bonding

strong

molecules with an O–H, N–H or H–F bond

CH3CH2CH2CH2CH2OH CH3CH2CH2NH2

1.9.3 1.9.3.1

Effect of Intermolecular Forces on Different Properties of Compounds Boiling points

(1) Boiling points of alcohols Alcohols have much higher boiling points than those of ethers and alkanes of comparable molecular masses. For example, n-butyl alcohol,

Structure, Bonding and Proper es of Organic Molecules

1.151

CH3CH2CH2CH2OH (molecular mass 74) boils at 118°C, whereas the boiling points of diethyl ether, CH3CH2OCH2CH3 (molecular mass 74) and n-pentane, CH3(CH2)3CH3 (molecular mass 72) are 35°C and 36°C respectively. Due to comparable molecular masses, the compounds should have similar volatility and so they are expected have comparable boiling points. But the boiling points of alcohols are much higher. Hydrogen bonding, the strongest intermolecular force is responsible for such behaviour of alcohols. Since in alcohol molecules a hydrogen atom remains bonded to a highly electronegative and relatively small oxygen atom, therefore, they remain associated through intermolecular hydrogen bonding.

Hydrogen bonding does not take place in ethers and in alkanes because hydrogen in them is attached to less electronegative carbon atoms. In the case of ethers, relatively weak dipole–dipole interactions operate along with very weak van der Waals forces and in the case of alkanes only very weak van der Waals forces operate. Since at the boiling temperature a considerable amount of thermal energy is required to separate the molecules by breaking numerous H-bonds, therefore, alcohols boil at a much higher temperature as compared to ethers and alkanes. The boiling points of isomeric alcohols decrease with increase in branching of the carbon chain. For example, the boiling points of n-butyl alcohol, CH3CH2CH2CH2OH, isobutyl alcohol, (CH3)2CH CH2OH, sec-butyl alcohol, CH3CH2CHOHCH3 and tert-butyl alcohol, (CH3)3COH, are 118°C, 108°C, 100°C and 83°C, respectively. As branching increases, alcohol molecules tend to become spherical, thereby resulting in a decreased surface area available for van der Waals interactions. Steric hindrance also affects the extent of hydrogen bonding. As the number of alkyl groups around the carbon bearing the –OH group increases, the extent of hydrogen bonding decreases due to steric crowding.

(2) Boiling points of carboxylic acids The boiling points of carboxylic acids are higher than those of alcohols of comparable molecular masses. For example, the boiling point of formic acid (100.5°C) is higher than that of ethyl alcohol (78.5°C), even though their molecular masses are the same (46). Both formic acid and ethyl alcohol contain polar O — H bond and remain associated through hydrogen bonding. Because of resonance, the polarity of the O — H bond present in the —COOH group of formic acid is much higher than that of the O — H bond in ethyl alcohol. Furthermore, the negative end of the polar carbonyl group (—C == O) is also involved in hydrogen bonding.

1.152

Organic Chemistry—A Modern Approach

Therefore, the hydrogen bond in formic acid is much stronger than that in ethyl alcohol and for this reason, the boiling point of formic acid is higher than that of ethyl alcohol.

(3) Boiling points of nitrophenols O-Nitrophenol has a much lower boiling point than its meta- and para-isomers. The —NO2 and —OH groups in o-nitrophenol molecule are attached to the adjacent ring carbons. Due to close proximity of these groups, intramolecular H-bonding giving rise to a stable six-membered ring (chelation) is possible in this isomer. Consequently, the —OH group is no longer available to form molecular association by intermolecular hydrogen bonding and so, the molecules exist as single unit (monomer). However, intermolecular hydrogen bonding is not possible in the meta- and para-isomers because in these two compounds the —OH and —NO2 groups are held farther apart. They form molecular association by intermolecular hydrogen bonding. Therefore, it requires a considerable amount of energy to separate the molecules by breaking hydrogen bonds. Hence, these isomers boil at a much higher temperature than o-nitrophenol.

Structure, Bonding and Proper es of Organic Molecules

1.153

(4) Boiling point of nitroalkanes (R—NO2) Nitroalkane molecules are highly polar (m = 3.5 – 4.0 D) and relatively strong dipole–dipole interactions hold the molecules together. On the other hand, weak van der Waals forces hold the nonpolar hydrocarbon molecules together. Since the dipole–dipole interactions (dissociation energy 8.373 kJ/mol) are stronger than the van der Waals forces (dissociation energy 4.186 kJ/mol), therefore, it requires more thermal energy to separate the nitroalkane molecules and hence nitroalkanes possess higher boiling points than alkanes of comparable molecular masses. (5) Boiling point of HF That the strength of the F–H---F H-bond (41.8 kJ mol) is the highest can be demonstrated by the following observation. Hydrogen fluoride (HF) and ethyl fluoride (C2H5F) are equally polar and the latter is heavier than the former. Yet the boiling point of HF (19.34°C) is much higher than that of C2H5F (–37.7°C). Since the compounds are equally polar, the strengths of dipole–dipole interactions involved in these two compounds are almost the same and hence the compounds are expected to boil at nearly the same temperature. But due to large molecular mass, C2H5F is expected to boil at higher temperature than HF. In fact, HF has a boiling point 57° higher than that of C2H5F. This difference in boiling point can be explained in terms of hydrogen bonding. Since in hydrogen fluoride, a hydrogen atom is bonded to the highest electronegative fluorine atom, the molecules remain associated through strong intermolecular hydrogen bonding.

Hydrogen bonding in C2H5F is not possible because there is no hydrogen atom bonded to electronegative F atom. The molecules are held together by relatively weak dipole–dipole attractions. Since it requires more energy to cleave the hydrogen bonds, therefore, HF boils at a much higher temperature than the heavier C2H5F.

(6) Boiling points of alkanes The boiling point of isomeric alkanes decreases with increase in branching. For example, the boiling point of n-pentane, CH3(CH2)3CH3 is 36.1°C, whereas the boiling point of isopentane, (CH3)2CH CH2CH3 and neopentane, (CH3)4C are

1.154

Organic Chemistry—A Modern Approach

27.9°C and 9.5°C, respectively. Because the strength of van der Waals forces depends on the area of contact between molecules, branching in a compound lowers its boiling point because it reduces the area of contact. A branched hydrocarbon molecule has a more compact, nearly spherical shape. If we think of the unbranched alkane pentane as a cigar and branched alkane neopentane as a tennis ball, we can see that branching decreases the area of contact between molecules. Two cigars make contact over a greater area than do two tennis balls. Thus, among the isomeric alkanes, the more highly branched one will have lower boiling point and the unbranched one will have a higher boiling point. It thus follows that n-pentane with no branching of the carbon chain boils at 36.1°C while isopentane with one branch-chain and neopentane with two branch chain boil at 27.9 and 9.5°C, respectively.

1.9.3.2

Melting points

(1) Melting points of isomeric 1,2-dibromoethylenes The cis-isomer of 1,2-dibromoethylene melts at a lower temperature than the trans-isomer. This can be explained in terms of intracrystalline forces. The more symmetrical a compound, the better it fits into a crystal lattice and hence the stronger the intracrystalline forces and the higher the melting point. The cis-isomer of 1,2-dibromoethylene is less symmetrical than the trans-isomer. So, the cis-isomer fits into a crystal lattice less tightly as compared to the more symmetrical transisomer. Thus, the cis-isomer has lower melting point than the trans-isomer.

Structure, Bonding and Proper es of Organic Molecules

1.155

(2) Melting points of isomeric xylenes o-Xylene melts at a much lower temperature than p-xylene. The highly symmetrical p-xylene fits better into a crystal lattice than the less symmetrical o-xylene. So, the melting point of the p-isomer having stronger intracrystalline forces is considerably higher than that of the o-isomer

(3) Melting points of carboxylic acids The melting point of a carboxylic acid containing even number of carbon atoms is higher than the acids having odd number of carbon atoms lying immediately below or above it in the series. This can be explained as follows. Carboxylic acids containing even number of carbon atoms have carboxyl and terminal methyl groups on the opposite sides of the zig-zag carbon chain and hence they fit better in the crystal lattice thereby increasing intermolecular attractive forces resulting in higher melting point. On the other hand, carboxylic acids containing odd number of carbon atoms have the carboxyl and the terminal —CH3 groups on the same side of the zig-zag carbon chain. Consequently, these molecules being less symmetrical fit less tightly in the crystal lattice. As a result, the magnitude of intermolecular forces of attraction becomes relatively less in this case and hence these acids have lower melting points.

1.9.3.3

Conformation Because of intramolecular hydrogen bonding some com-

pounds prefer to exist in a particular conformation. For example, 5-hydroxy-1,3-dioxane prefers to exist in the conformation with OH axial. In this conformation, the OH hydrogen is involved in the formation of hydrogen bond with the ring oxygen atom and so it becomes relatively more stable.

Similarly, because of intramolecular hydrogen bonding, cis-1,4-cyclohexanediol exists preferably in the twist-boat conformation.

1.156

1.9.3.4

Organic Chemistry—A Modern Approach

Solubilities

(1) Solubility of alcohols The solubility of alcohols in water decreases as the molecular mass increases, i.e., as the carbon chain becomes longer and among isomeric alcohols the solubility increases with increase in branching of carbon chain. An alcohol becomes soluble in water due to the formation of hydrogen bond with water molecules. In alcohols, the hydrocarbon part, i.e., the alkyl group (—R) is nonpolar and hydrophobic (water avoiding) while the hydroxyl (—OH) group is highly polar and hydrophilic (water seeking). In lower alcohols (up to four carbons), the polar and hydrophilic —OH group constitutes a considerably large part of the molecule and as its characteristic effect predominates over the effect of the nonpolar and hydrophobic alkyl group, these alcohols are soluble in water by forming H-bonds. With increase in molecular mass of alcohols, the size of the water insoluble hydrocarbon part increases and since the hydrophilic —OH group then contributes a very small portion of the molecule, the alcohol becomes progressively less water-soluble. Also, for a nonpolar hydrocarbon chain to be accommodated by water, the water molecules have to form a more ordered ‘ice-like’ structure around the chain, and for this, the entropy change is unfavourable. Thus, solubility decreases as the hydrocarbon chain gets larger. For example, methyl alcohol CH3OH, with a small hydrophobic part, is highly soluble in water while n-undecyl alcohol, with a large hydrophobic part is completely insoluble in water.

The solubility of alcohols also depends on the structure of the alkyl group. Since branching decreases the relative volume of the hydrophobic hydrocarbon portion, therefore, the

Structure, Bonding and Proper es of Organic Molecules

1.157

alcohols become progressively more water soluble with increase in branching. For example, tert-butyl alcohol (CH3)3COH is much more soluble in water than n-butyl alcohol, CH3CH2CH2CH2OH.

(2) Solubility of amines Low-molecular-mass amines are soluble in water because amines can form hydrogen bonds with water. Comparing amines with the same number of carbon atoms, primary amines are more soluble than secondary amines because primary amines have two hydrogens that can engage in hydrogen bonding. Tertiary amines, on the other hand, have unshared pair of electrons that can accept hydrogen bonds but do not have hydrogen to donate for hydrogen bonds. Therefore, tertiary amines are less soluble in water than secondary amines with the same number of carbon atoms. Amines are more water soluble than alcohols with the same number of carbons. Due to formation of H-bonds with water molecules, both amines and alcohols are soluble in water. However, the hydrogen bonds of the type involved in the case of amines are stronger than those involved in the case of alcohols because amine nitrogen is a better hydrogen bond acceptor (base) than oxygen in alcohol. It is the stronger hydrogen bonding for which amines are more soluble in water than alcohols.

(3) Solubility of nitrophenols o-Nitrophenol has much lower solubility in water than its meta- and para-isomers. Due to the formation of intramolecular hydrogen bond, the molecules of o-nitrophenol cannot form hydrogen bonds with water molecules and so it is less soluble in water. Intramolecular hydrogen bonding is not possible in the metaand para-isomers because in these two compounds, the —OH and —NO2 groups are held farther apart. For these reasons, they can form hydrogen bonds with water and becomes considerably soluble in water.

1.9.3.5

Acidic character of organic compounds

(1) Acidity of diastereoiosmeric dicarboxylic acids Maleic acid, cis —HOOC—CH== CH—COOH (pKa1 = 1.92, pKa2 = 6.04) is stronger for the first ionization but weaker for the second ionization than the isomeric fumaric acid, trans– HOOC—CH== CH—COOH (pKa1= 3.03, pKa2 = 4.44). In maleic acid, the two carboxyl (—COOH) groups are placed on the same side of the double bond and so, the maleate monoanion is stabilized by intramolecular hydrogen bonding. In fumaric acid, on the other hand, the two —COOH groups are placed on the different sides of the double bond

1.158

Organic Chemistry—A Modern Approach

and consequently, intramolecular hydrogen bonding is not possible for stabilization of the fumarate monoanion. Because of greater stability of the monoanion, maleic acid has a greater tendency to ionize than fumaric acid, i.e., for the first ionization it is a stronger acid than fumaric acid. Due to stabilization of the maleate monoanion by intramolecular hydrogen bonding, it is very difficult to remove a proton from it. Proton release is also disfavoured because the field effect generated by the —COO① group on the nearby —COOH group is free and there is no significant field effect caused by the —COO① group. On the other hand, in fumarate monoanion, the —COOH group is free and there is no significant field effect caused by the —COO① group. So, it easily ionizes to form the fumarate dianion, i.e., for second ionization fumaric acid is stronger than maleic acid.

The compound I is more acidic than the compound II can also be explained in terms of hydrogen bonding and field effect.

Each of these two diastereoisomeric acids (I and II) is locked in that particular chair conformation in which the bulky —C(CD3)3 group is placed in an equatorial position. In

Structure, Bonding and Proper es of Organic Molecules

1.159

such conformation, the —COO① and —COOH groups are both axial in I and equatorial in II. Because of close proximity of these two groups, intramolecular hydrogen bonding occurs in II, but not in I in which they are held farther apart. Therefore, ionization of II is relatively more disfavoured compared to I. Proton release from II is further disfavoured because of field effect generated by the —COO① group, on the —COOH group. No significant field effect operates in I. Hence, the compound I is more acidic than the compound II.

(2) Acidity of hydroxybenzoic acids Hydroxybenzoic acids have two acidic functional groups; one is —COOH and the other is —OH.

1.160

Organic Chemistry—A Modern Approach

The first ionization (i.e., dissociation of carboxyl proton) of o-hydroxybenzoic acid (salicylic acid) is more favourable than that of the p-hydroxybenzoic acid, whereas the second ionization (i.e., dissociation of hydroxyl proton) of the latter acid is more favourable than that of the former acid. This observation can be explained by hydrogen bonding and field effect. Since the —OH and the —COOH groups in o-hydroxybenzoic acid (salicylic acid) are placed in adjacent ring carbons, therefore, the conjugate base, i.e., the monoanion, is stabilized by intramolecular hydrogen bonding leading to the formation of a stable sixmembered ring (chelation). On the other hand, the two groups in the p-isomer are held farther apart and no such intramolecular H-bonding is possible. o-Hydroxybenzoate ion is, therefore, more stable than the p-hydroxybenzoate ion and because of this, ionization of the —COOH group of o-hydroxbenzoic acid is relatively more favourable as compared to that of p-hydroxybenzoic acid. Since the o-hydroxybenzoate ion is stabilized by internal H-bonding, it is difficult to remove the —OH proton from this negatively charged cyclic system, i.e., the second ionization does not take place easily. Also, the ionization is disfavoured because of the field effect generated by —COO① group on the nearby —OH group. In p-hydroxybenzoate ion, on the other hand, the —OH group is free and so the second ionization occurs easily in this case.

Structure, Bonding and Proper es of Organic Molecules

1.161

[Although intramolecular H-bonding occurs in undissociated o-hydroxybenzoic acid, it is less effective than its anion because a negative charge on oxygen leads to a stronger H-bonding.]

1.

How can o-nitrophenol be separated from its m- or p-isomer ? Give reasons.

Solution A compound can be steam-distilled if it has an appreciable vapour pressure at the boiling point of water. Due to intramolecular H-bonding, o-nitrophenol is a low-boiling compound and it has appreciable vapour pressure at 100°C. So it can be separated from its m- or p-isomer by steam distillation. 2. Explain why in the vapour state acetic acid (CH3COOH) has molecular weight of 120. Solution In the vapour phase, acetic acid (CH3COOH) exists as dimmers in which a pair of acid molecules are held together by intermolecular hydrogen bonds. Because of this, acetic acid in the vapour state has a molecular mass of 120.

3.

The compound II is more basic than the compound I. Explain.

Solution The conjugate acid (monoprotonated from) of the cis-diamine II is stabilized by intramolecular hydrogen bonding while the conjugate acid of the trans-diamine I is not stabilized by intramolecular hydrogen bonding because the groups are held further apart. For this reason, the diamine II is relatively more basic than the diamine I.

1.162

4.

Organic Chemistry—A Modern Approach

Nonpolar ethane (bp – 88.2°C) boils at a temperature higher than methane (bp – 162°C) at a pressure of 1 atm — Why?

Solution Heavier molecules require greater thermal energy in order to acquire velocities sufficiently great to escape the surface of the liquid and since the surface areas of heavier molecules are usually much greater, therefore, intramolecular van der Waals attractions are also much larger. Ethane (CH3CH3) is heavier than methane (CH4) and also its surface area is greater than methane. For these reasons, ethane boils at a temperature higher than methane. 5. The fluorocarbon C5F12 has a slightly lower boiling point than pentane (C5H12), even though it has a far higher molecular mass. Explain. Solution Since the molecular mass of C5F12 is much higher than C5H12 and also its size is somewhat larger than C5H12 (F is slightly larger than H), therefore, C5F12 is expected to boil at a temperature higher than C5H12. This can be explained in terms of relative polarizability of the electrons of the atoms involved. The highly electronegative fluorine atoms show a very low polarizability because their electrons are very tightly held with the nucleus. Because of this, C5F12 molecules are held together by very small van der Waals forces and in fact, it is much smaller than that operates in C5H12. For this reason, C5H12 has a slightly lower boiling point than pentane (C5H12), even though it has a far higher molecular weight. 6. Arrange the following compounds in order of increasing boiling point and explain the order:

Solution To predict relative boiling points, we have to consider the differences in (i) hydrogen bonding and (ii) molecular mass and surface area. 2,2-Dimethylpropane (neopentane) is the lightest among these compounds and it has compact spherical shape for which the van der Waals attractions are minimum. For this reason, 2,2-dimethylpropane is the lowest-boiling compound. The other four compounds have similar molecular masses. Neither hexane nor 2,3-dimethylbutane is hydrogen bonded, so they will be the next higher in boiling points. Because 2-3-dimethylbutane is more highly branched and has a smaller surface area than hexane, the former will be lower boiling than the latter. The

Structure, Bonding and Proper es of Organic Molecules

1.163

two remaining compounds are alcohols and they remain associated with intermolecular hydrogen bonding. Therefore, their boiling points are higher than the three hydrocarbons. Now, 1-pentanol has more surface area for van der Waals forces to operate and hence its boiling point is higher than the relatively compact 2-methyl-2-butanol. Therefore, the increasing order of boiling point is: 2,2-dimethylpropane < 2,3-dimethylbutane < hexane >2-methyl-2-butanol < 1-pentanol. 7.

The cis-isomer of 1,2-dibromethylene boils at a higher temperature than its trans-isomer — Why?

Solution The trans-isomer of 1,2-dibromoethylene is nonpolar because the two C — Br bond moments orienting in opposite directions cancel each other. So relatively weak van der Waals forces hold the molecules together. The cis-isomer, on the other hand, is polar because the two Br atoms lie on the same side of the double bond and the two C — Br bond moments do not cancel each other. A net moment resulting from two individual bond moments operates. As a result, there operates relatively stronger dipole–dipole attractions amongst the molecule along with van der Waals forces. Since more energy is needed to separate the molecules by overcoming these forces, the cis-isomer boils at a higher temperature than the trans-isomer.

8.

o-Xylene boils at a higher temperature than p-xylene — Why?

Solution o-Xylene molecules are weakly polar. So there operates weak dipole–dipole attractions amongst the molecules in addition to the van der Waals forces. On the other hand, p-xylene molecules are nonpolar and so only weak van der Waals forces hold the molecules together. Because of this, o-xylene boils at a higher temperature than p-xylene, but the difference in boiling points is actually very small.

1.164

9.

Organic Chemistry—A Modern Approach

The melting point of neopentane (Me4C) is much higher than that of isopentane [CH3CH2CH(CH3)2]. Explain.

Solution The compact symmetrical molecules of neopentane (MC34C) pack well into a crystal lattice while isopentane [CH3CH2CH(CH3)2], which has a —CH3 group dagling from a four-carbon chain, does not. Therefore, intracrystalline forces are much stronger in neopentane compared to isopentane. Consequently, neopentane has a much higher melting point than isopentane. 10. Methyl fluoride (CH3F) is more soluble in water than methyl chloride (CH3Cl) — Why? Solution Methyl fluoride (CH3F) containing highly electronegative and small-sized F atom can form hydrogen bond with water but methyl chloride does not. For this reason, methyl fluoride is more soluble in water than methyl chloride. O H F—CH3 CH3—F H

11.

tert-Butyl alcohol is more soluble in water than n-butyl alcohol. Explain.

Solution Alcohols containing branched alkyl groups are more water soluble than nonbranched alkyl groups with the same number of carbon atoms because branching minimizes the contact surface of the nonpolar portion of the molecule and thereby decreases the van der Waals forces of attraction. It is for this reason, the relatively more compact tert-butyl alcohol is more soluble in water than n-butyl alcohol.

CH3 | H3C — C — OH | CH3 tert-Butyl alcohol (bp 83∞ C) 12.

CH3CH2CH2CH2OH n-Butyl alcohol (bp 118∞ C)

Butane (CH3CH2CH2CH3) is soluble in carbon tetrachloride but not in water — Why ?

Solution The molecules of both butane and carbon tetrachloride (CCl4) are nonpolar and the molecules of each compound are held to each other by weak van der Waals interaction. Since ‘like dissolves like’ is a rule for dissolution of a solute in a solvent and the dissolution process involves a large increase in entropy, i.e., thermodynamically favourable, butane dissolves in carbon tetrachloride. The highly polar water molecules are held together by strong dipole–dipole interactions and hydrogen bonds. There could be only very weak attractive forces (dipole-induced dipole attractions) between butane molecules on the one hand and water molecules on the other hand. Since solute–solvent interactions cannot outweigh or equal to the sum of solute–solute interactions (van der Waals forces) and solvent–solvent interactions (hydrogen bonding forces), butane does not dissolve in water.

Structure, Bonding and Proper es of Organic Molecules

13.

≈

1.165

@

Ammonium chloride (N H 4 C l) is insoluble in the nonpolar solvent ≈

@

carbon tetrachloride while tetramethylammonium chloride (Me4 N C l) is appreciably soluble in this solvent. Explain. ≈

Solution In nonpolar solvent CCl4, stabilization of N H 4 and Cl① ions by solvation does

not take place and because of this, ammonium chloride is insoluble in CCl4. Tetramethyl≈

@

ammonium chloride, Me4 NCl, on the other hand, is appreciably soluble in CCl4 because the cation in which the positive nitrogen is surrounded by four methyl groups presents a large nonpolar hydrocarbon surface to this solvent and becomes involved with it by van der Waals forces of attraction. 14.

Explain why glycerol (HOCH2CHOH CH2OH) is a very viscous liquid.

Solution The degree of molecular association through intermolecular hydrogen bonding is much higher in glycerol containing three —OH groups. For this reason, this triol is a very viscous liquid.

15.

The melting point of NaCl (801°C) is very much higher than that of CCl4 (–24°C) — Why ?

Solution The crystal units in NaCl are cations and anions, which are held together by strong electrostatic forces of attraction. Hence it requires a large amount of energy to separate the ions. On the other hand, the crystal units in CCl4 are nonpolar molecules, which are held together by weak van der Waals forces. Therefore, these molecules can be separated by applying a small amount of thermal energy. This explains why the melting point of NaCl is very much higher than that CCl4.

16.

In which of the following solvents would cyclohexane have the lowest solubility: diethyl ether, 1-pentanol, hexane and ethanol ?

Solution Cyclohexane is a nonpolar compound which is expected to be more soluble in a nonpolar solvent like hexane or a weakly polar solvent diethyl ether (‘like dissolves like’) than in an H-bonding and polar solvent like 1-pentanol and ethanol. Again, the nonpolar hydrocarbon part of ethanol (C2H5OH) is smaller than that in 1-pentanol (CH3CH2CH2CH2CH2OH). Therefore, cyclohexane is least soluble in ethanol.

17.

Salicylic acid, o-HOC6H4COOH, is a stronger acid than o-CH3OC6H4COOH. Explain why?

Solution Because of the ortho-effect of the bulkier —OCH3 group, o-CH3OC6H4COOH is expected to be more acidic. However, the reverse is true, and this is because, the conjugate base of salicylic acid is stabilized by internal H-bonding (a more dominating factor).

1.166

18.

Organic Chemistry—A Modern Approach

2,6-dihydroxy-4-methylbenzoic acid is a much stronger acid than 2-hydroxy-4-methylbenzoic acid. Explain.

Solution The conjugate base of 2,6-dihydroxy-4-methylbenzoic acid is highly stabilized by intramolecular H-bonding (chelation) involving both the o-hydroxyl groups and for this reason, it is a much stronger acid than 2-hydroxy-4-methylbenzoic acid, the conjugate base of which is stabilized by intramolecular H-bonding involving only one —OH group.

19.

Arrange the following compounds in order of increasing boiling point and explain your answer.

Structure, Bonding and Proper es of Organic Molecules

CH3CH2OCH2CH3 Diethyl ether

1.167

CH3CH2CHCH3 | OH sec-Butyl alcohol

CH3CH2CH2 CH2CH3 Pentane

Solution Pentane has no polar groups, so its molecules are held together only by weak van der Walls forces. Diethyl ether has the polar ether group, so it can have dipole–dipole interactions in addition to van der Waals forces. 2-Butanol is a polar bent molecule, so it can have dipole–dipole interactions in addition to van der Waals forces. Because it has an O—H bond 2-butanol molecules are held together by intermolecular hydrogen bonds. Since hydrogen bonding is stronger than dipole–dipole attractions which in turn stronger than van der Waals forces, therefore, the boiling point of these three compounds increases in the order: pentane < diethyl ether < sec-butyl alcohol.

1.

Which compound in each of the following pairs would have the higher boiling point? Explain your answer. (a) HOCH2CH2OH or CH3CH2CH2OH (b)

(c)

(d)

(e)

(f)

2.

Arrange the following compounds in order of increasing boiling point and explain the order:

3.

Which compound in each pair has the higher melting point ? (a)

(b)

1.168

Organic Chemistry—A Modern Approach

(c) 4.

5. 6.

7.

8.

9.

Which compound in each pair is more soluble in water and why? (a)

(b)

(c)

(d)

Symmetry affects the melting point of a compound but not the boiling point — Why? The boiling point of ethanol (78°C) is more than 100°C higher than that of dimethyl ether (–25°C) while ethylmethylamine has a boiling point (37°C) only 34° higher than that of triethylamine (3.5°C). Explain. 1-Butanol (bp 118°C) has a much higher boiling point than its isomer ethoxyethane (bp 35°C). However, both of them show same solubility (8 g pre 100 g) in water. Account for these observations. Comment on the solubility of the following compounds in water and in organic solvents (such as CCl4): NaCl, CH3CH2CH2CH3, CH3CH2CH2OH, CH3(CH2)10OH What types of intermolecular forces are present in each of the following compounds? (a)

10. 11.

12.

13.

(d)

(b)

(c) (CH3)3N

(d) CH2 == CHBr (e) CH3CH2CH2COOH (f) CH3CH2C ∫∫ C CH2CH3 o-Hydroxybenzaldehyde has much lower boiling point and much lower water solubility than its m- and p-isomers. Explain. Which of the following compounds exhibit chelation: (a) methyl salicylate, (b) o-iodophenol, (c) o-fluorophenol, (d) o-cyanophenol, and (e) o-cresol? [Hint: Chelation occurs only in (a) and (c). In (d), although the N atom of —CN is electronegative, the linearity of the —CN group places the N atom too far away from the —OH group to form an H-bond.] C6H6 2CH 3COOH DS = +20 e.u.; despite Acietic acid dimmer, (CH 3COOH)2 the favourable entropy of dissociation, the equilibrium lies to the left. Explain why? [Hint: Strength of intramolecular H-bonding in the dimer is more than enough to comprensate for the loss of freedom in the dimer.] 2-methylpyrrolidine boils at temperature higher than the boiling point of pyrrolidine — Why.

Structure, Bonding and Proper es of Organic Molecules

1.169

14.

Between the anti- and syn-isomers of pyridine-2-carboxaldoxime the antiform predominates in the isomeric distribution of the compound. Explain this observations. [Hint: The anti-form gets extra stability due to intramolecular hydrogen bonding (chelation). No such chelating effect is observed in the syn-isomer. For this reason, the anti-form predominates in the isomeric distribution of the compound.

15. 16.

Resorcinol has higher boiling point than 2-nitroresorcinol — Why? 8-hydroxyquinoline can be separated from 4-hydroxyquinoline by steam distillation. Account for this observation. [Hint: Intramolecular H-bonding occurs in 8-hydroxyquinoline but not in 4-hydroxyquinoline.

17.

Draw the H-bonding arrangements in CH3OH — H2O and CH3NH2 — HCHO systems. [

18.

19. 20.

]

Why is the mp of sulphanilic acid so high? [Hint: Sulphanilic acid exists as a salt (a dipolar ion or zwitter ion): – H3N— —SO3 . Because of this, sulphanilic acid has a much higher melting point.] Explain why H2O has a higher boiling point than CH3OH (65°C), NH3 (–33°C) and HF (20°C). Arrange the following compounds in order of increasing boiling point. Explain your answer.

1.170

21. 22. 23. 24.

25.

26.

Organic Chemistry—A Modern Approach

Explain why 1-pentanol has solubility of 2.7 g per 100 ml of water, where as ethanol is completely miscible in water. Why does one expect the cis-isomer of an alkene to have a higher boiling point than the trans-isomer? Alcohols with fewer than four carbons are soluble in water, but alcohols with more than four carbons are insoluble in water — Why? The boiling point of propylamine (bp 49°C) is higher than that of ethylmethylamine (bp 37°C) which in turn is higher than that of triethylamine (bp 3.5 °C). Explain. [Hint: Trimethylamine (MC3N) has no N — H bond and so it cannot form hydrogen bonds with each other. Ethylmethylamine (CH3 CH2NH CH3) has one N — H bond and as it remains associated through intermolecular hydrogen bonding. Propylamine (CH3CH2CH2NH2) with two N — H bonds is more extensively hydrogen bonded. This explains their boiling points] There are four amides with the molecular formula C3H7 NO. Write their structures. One of these amides has melting and boiling point that is substantially lower than that of the other three. Identify this amide and explain your answer. [Hint: CH3CH2CONH2, CH3CONHCH3, HCONHCH2CH3 and HCON (CH3)2. The last one has a melting and boiling point that is substantially lower than that of the other three because it does not have a hydrogen that is covalently bonded to nitrogen and, therefore, its molecules cannot form hydrogen bonds to each other. The other molecules all have a hydrogen covalently bonded to nitrogen, and therefore, hydrogen-bond formation is possible.] Which of the following compounds is expected to volatilize easily and why? OH NH CH N I

NH CH N II

[Hint: Due to intramolecular hydrogen bonding, I is expected to volatilize easily.]

1.10

REACTIVE INTERMEDIATES

A chemical reaction involves conversion of a molecule into a new molecule. The new molecule, i.e., the product, has different arrangement of atoms compared to the starting molecule, i.e., the reactant. A redistribution of electrons also occurs during this change. In most of the organic reactions, conversation of reactants into products takes place through some specific steps. The mechanism of an organic reaction is a detailed description of these steps. It not only acquient us the number of steps involved in the reaction, but also informs us regarding the sequence of breaking old bonds and making new bonds. Study of reaction mechanism is of immense importance in organic chemistry as thousands of apparently different reactions occur through a limited number of common steps.

Structure, Bonding and Proper es of Organic Molecules

1.171

Organic reactions usually involve fission of weaker covalent bonds and formation of stronger ones, so that a relatively stable molecule is formed from a less stable molecule. Breaking of bonds requires energy while formation of bonds involves release of energy. A covalent bond is represented by a dash (—) and the transfer of electrons is shown by using arrow signs. Curved arrow signs containing two barbs ( ) indicate the shifting of a pair of electrons while the transfer of a single electron is indicated by curved arrow signs containing one barb ( ) or fishhook arrow [it is to be noted that the symbol ( ) is incorrect]. Fission or cleavage of covalent bonds can take place in two ways depending on the nature of the bond involved, the nature of the attacking agent and the conditions of the reaction. (1) Homolytic fission or homolysis: It a covalent bond in a molecule undergoes fission in such a way that each of the two bonded atoms gets one electron of the shared pair, it is called homolytic fission or homolysis. This type of bond cleavage results in formation of neutral species called free radicals, or often simply radicals, i.e., a radical is a reactive intermediate with a single unpaired electron. Homolytic fission is usually favoured by conditions such as nonpolar nature of the bond, high temperature, presence of high energy (uv) radiations or presence of radical initiators such as peroxides. The homolytic fission of a bond A—B leading to the formation of free radicals A and B (each containing odd electrons), may be shown as follows:

e.g.,

(2)

Homolytic bond cleavage requires less energy than heterolytic bond cleavage. Heterolytic fission or heterolysis: When a covalent bond breaks in such a way that the pair of electrons stays with the more electronegative atom, such a fission is called heterolytic fission or heterolysis. This type of fission results in formation of ionic (cationic and anionic) intermediates. If in the molecule A–B, B is more electronegative than A, the heterolytic fission of the bond leading to the formation ≈

@

of the cation A and the anion B: may be shown as follows:

e.g.,

1.172

Organic Chemistry—A Modern Approach

This type of bond cleavage resulting in the formation of charged species, i.e., ions, is favoured by conditions such as polar nature of the covalent bond and the presence of polar solvent. If the fragments of a heterolytic fission are carbon species, then the cation is called carbocation and the anion is called carbanion. Both of them are unstable reactive intermediates. Reactive intermediates: Under the influence of attacking reagent, most of the organic compounds (substrates) undergo either hemolytic or heterolytic fission of a bond to form certain short lived and highly reactive (hence cannot be normally isolated) chemical species which are called reactive intermediates or reaction intermediates. Some common examples of reactive intermediates are carbocations, carbanions, free radicals, carbenes, nitrenes, arynes, etc. A one-step reaction is called a concerted reaction and in such a reaction, no matter how many bonds are broken or formed, a starting material is converted directly into a product, i.e., a concerted reaction involves no reactive intermediate. A stepwise reaction involves more than one step. The substrate is first converted into an unstable intermediate, which then goes on to form the product. Concerted reaction: Substrate Æ Product Stepwise reaction: Substrate Æ Reactive intermediate Æ Product An understanding of the structure and properties of these intermediates (molecules, ions or radicals) is very much important in understanding organic reaction mechanism. Matters related to various intermediates are discussed below:

(1) Carbocations Carbocations are a group of reactive intermediates having positively charged carbon atom bearing only six electrons. These are represented by the symbol R!. For example, ≈

≈

≈

CH3 , CH3CH2 , (CH3 )2 CH, etc. Because of having a strong tendency to complete the octet of the electron-deficient carbon, carbocations are highly reactive species. Generation: Carbocations are generated by heterolytic fission of a bond to carbon in which the leaving group is removed along with its shared pair of electrons.

The principal ways of generating carbocations are as follows: (i) Heterolysis of C — L bond (L = halogen, OTs, OBs, etc.) of neutral substrates by the influence of polar solvents like H2O, EtOH, etc. leads to the formation of carbocations. For example:

Structure, Bonding and Proper es of Organic Molecules

1.173

(ii)

Deamination of aliphatic amines by nitrous acid leads to the formation of carbocation. For example:

(iii)

Protonation of alcohols followed by loss of H2O leads to the formation of carbocations. For example: (CH3)3C—OH

(iv)

H

+

H2SO4

+

+

(CH3)3C—OH2

(CH3)3C + H2O

Protonation of alkenes leads to the formation of carbocations. For example:

(v)

Carbocations can be generated by the action of Lewis acid such as AlCl3 on alkyl halides. For example:

(vi)

Carbocations can be obtained by treating alkanes with FSO3H — SbF5 called super acid. FSO3H — SbF5 is, in fact, an extremely strong acid which donates H! even to an alkane in order to form a carbocation by extracting a hydride ion (H①). H2 is liberated in this reaction. For example:

(vii)

A more stable carbocation can be obtained when a relatively less stable carbocation abstracts H①. For example: H +

Ph3C +

H

+ –

H-shift

Ph3CH + Cycloheptatrienyl cation (aromatic)

Nomenclature: In naming a carbocation, the word ‘cation’ is added to the name of the ≈

≈

≈

alkyl or aralkyl group. For example, CH 3 ,(CH 3 )2 CH, and C6 H5 CH 2 are named as methyl cation, isopropyl cation and benzyl cation, respectively. Classification: Carbocations are classified as primary (1°), secondary (2°) and tertiary (3°) on the basis of the number of carbon atoms (one, two or three) directly bonded to

1.174

Organic Chemistry—A Modern Approach ≈

the positively charged carbon atom. Fro example, ethyl cation (CH 3 CH 2 ) is a primary, ≈ ≈˘ È ≈ isopropyl cation (CH 3CHCH 3 ) is secondary and tert-butyl cation Í(CH 3 )3 C˙ is a tertiary Î ˚ ≈ carbocation. Methyl cation (CH 3 ) with one carbon atom is a special case.

Structure: The positively charged carbon atom of a carbocation is sp2-hybridized. Therefore, the shape of a carbocation is trigonal planar and the bond angle is 120°. The three sp2 orbitals are utilized in making bonds to three substituents. The vacant p orbital is perpendicular to the plane of sp2 hybridized orbitals.

Stability: Any factor which tends to delocalize the positive charge must increase the stability of carbocations while any factor which tends to localize or intensify the positive charge must decrease the stability of carbocation.

The stability of alkyl carbocations decreases in the following order:

Carbocations are stabilized mainly by +I, +R and hperconjugation effects. The relative stability of carbocation can be easily assessed by determining the heterolytic R — H Æ R! + H① dissociation energies in the gas phase. The lower the value of energy, the greater the stability of the carbocation. The bond dissociation energies D (R+ — H–) where R = alkyl, aralkyl or aryl are given in the following table.

Structure, Bonding and Proper es of Organic Molecules

1.175

Heterolytic R — H Æ R+ + H– dissociation energies in gas phase D(R+ + H–) in kcal/mol

Ion ≈

CH 3

314.6

C6 H5≈

294

Stability increases

≈

CH2 == CH

287

C2 H5≈

276.7 ≈

CH == CH — CH2 ≈

(CH 3 )2 CH

256 249.2 246

≈

C6 H5 CH 2 ≈

(CH3 )3 C

238 231.9

Factors which determine the stability of carbocations: (a) Inductive effect: An ion is stabilized by a factor that disperses its charge. The electron-donating inductive effect (+I) exerted by an alkyl group attached to the positive carbon of a carbocation neutralizes the charge partially. As a result, the charge becomes dispersed over the alkyl groups and the system becomes stabilized. For example, the methyl group in ethyl cation stabilizes the system through its +I effect.

(b)

The stabilities of carbocations increase with increasing the number of electronreleasing alkyl groups attached to the positive carbon. Hyperconjugation effect: An alkyl group may also delocalize the positive charge of a carbocation by the hyperconjugation effect. The charge becomes dispersed over the a–H atoms and consequently, the system becomes stabilized. Hyperconjugation in ethyl cation, for example, occurs as follows:

1.176

Organic Chemistry—A Modern Approach

H≈ H H H È | ≈ | | Í ÍH — C — CH 2 ´ H — C == CH 2 ´ H≈ C == CH 2 ´ H — C == CH 2 | | | Í H≈ H H ÍÎ H Hyperconjugation in ethyl cation

˘ ˙ ˙ ˙ ˙˚

As the number of a-hydrogens, i.e., the number of hyperconjugation structures, increases, the stability of carbocations increases. Therefore, the stability of methyl substituted carbocations increases in the following order: ≈

≈

(least stable) C H 3 CH 3 C H 2 (3 a - H)

(c)

≈

≈

(CH)2 C H (CH 3 )3 C (most stable) (6 a - H)

(9 a - H)

Resonance effect: Resonance is the most important factor influencing the stability of carbocations. When there is a double bond a to the positive carbon of a carbocation, effective charge delocalization with consequent stabilization occurs. Allyl and benzyl cations, for instance, are found to be highly stabilized by resonance.

A nonbonding lone pair on a heteroatom which is directly attached to the electrondeficient C stabilizes a carbocation more than any other interaction. This resonance satisfies the octet of the cationic carbon, even though there is developed a positive ≈

charge over the heteroatom. For example, CH 3OCH 2 is a very stable carbocation in spite of its primary nature. In fact, the methoxymethyl cation is obtained as a ≈

stable solid, CH 3OCH 2 SbF6@ .

(d)

Steric effect: Stability of tertiary carbocation increases due to steric effect. For ≈˘ È example, the alkyl groups of triisopropyl cation Í(Me2CH)3 C˙ having planar Î ˚ arrangement with a bond angle of 120° are far apart from each other and so, there

Structure, Bonding and Proper es of Organic Molecules

1.177

is less steric interaction among them. However, when this carbocation undergoes nucleophilic attack, i.e., when a change of hybridization of the central carbon atom from sp2 (trigonal) to sp3 (tetrahedral) occurs, the bulky isopropyl groups are pushed together. This results in a steric strain (B strain) in the product molecule. Because of this, carbocation is not much willing to react with a nucleophile, that is, its stability is enhanced due to steric reason.

(e)

Constituting an aromatic system: The vacant p orbital of a carbocation may be involved in constituting a cyclic and planar (4n + 2)p electron system, where n = 0, 1, 2 …. etc., i.e., a carbocation may be stabilized by constituting an aromatic system. In fact, it acquires maximum stability by this process. Cycloheptatrienyl cation, for example, is unusually stable because it is a planar 6p electron system and aromatic.

1.10.1

Nonclassical Carbocation

A carbocation in which the positive charge is delocalized by a double or triple bond that is not in the allylic position or by a single bond is called a nonclassical carbocation. 7-Norbornenyl and norbornyl cations are two examples of nonclassical carbocation.

1.178

Organic Chemistry—A Modern Approach

In a carbocation, if there is one carbon atom between the positively charged carbon atom and the double bond, it is called a homoallylic carbocation. 7-Norbornenyl cation is also an example of homoallylic carbocation.

1.10.2

Carbonium Ion and Carbenium Ion or Carbocation

There are intermediates in which a carbon bears a positive charge, but the formal covalency of the carbon atom is five rather than three. The simplest example is the methanonium ion, ≈

CH5 . Such pentacoordinated positive ions are called carbonium ions. The intermediates in which there is positive charge at a carbon atom which is trivalent are called carbenium ≈

≈

ions or carbocations. For example, methyl cation (CH 3 ) and ethyl cation (CH 3 CH 2 ), etc. ≈

[N.B. The methanonium ion CH5 has a three-centre two-electron bond. It is not known whether this ion is a transition state or a true intermediate, but an ion CH5≈ has been detected by mass spectroscopy.

(2) Carbanions Carbanions are a group of reactive intermediates carrying a negative charge on carbon @

atom possessing eight electrons in its valence shell. For example, CH 3 (methyl carbanion) @

and CH 3 CH 2 (ethyl carbanion), etc. They are represented by symbol R①. Their reactivity is due to the presence of formal negative charge on the carbon atom. Generation: Carbanions are generated by heterolytic fission of a bond to carbon in which the bonding electron pair remains with the carbon atom.

Structure, Bonding and Proper es of Organic Molecules

1.179

The principal methods of generating carbanions are as follows: (i) By an abstraction of an acidic hydrogen alpha to an electron-withdrawing group: Compounds containing acidic hydrogen attached to carbon alpha to —NO2, —CN or C == O groups produce carbanions when treated with a strong base. For example:

(ii)

By abstraction of hydrogen from terminal alkynes using a strong base: Terminal alkynes being acidic produce carbanion when treated with strong bases. For example: –

–

H2N: + H—C∫∫ C—CH3

(iii)

C∫∫ C—CH3 + NH3

By metal halogen exchange: When organic halogen compounds are treated with strongly electropositive metals like Li, Na, etc. in an inert solvent, carbanions are obtained in the form of organometallic compounds. For example: ≈

@

ether CH 3 — Br + 2Li æææ Æ CH 3 L i + LiBr @

≈

ether Ph3C — Cl + 2Na æææ Æ PH 3 C Na + NaCl @

≈

ether Ph — Br + 2Li æææ Æ Ph Li + LiBr

1.180

Organic Chemistry—A Modern Approach

(iv)

By decomposition of carboxylate ions: When metal carboxylates are heated, they undergo decarboxylation to yield carbanions. For example:

(v)

By addition of an anion to multiple bond containing electron-withdrawing groups: Carbanions are obtained when an alkene containing electron-attracting group undergoes nucleophilic attack by an anion. For example:

(vi)

By addition of electrons to an unsaturated system: Unsaturated compounds may accept electrons from electropositive metals to generate carbanions. For example, when cyclooctatraene is treated with metallic sodium, it is converted to cyclooctatetraenyl dianion which is a stable aromatic system containing (4n + 2)p electrons, where n = 2.

Nomenclature: In naming a carbanion, the word ‘anion’ is added to the name of the @

@

@

alkyl or aralkyl group. For example, CH 3 , (CH 3 )2 CH and C6 H5 CH 2 are named as methyl anion, isopropyl anion and benzyl anion, respectively. Classification: Carbanions are classified as primary (1°), secondary (2°) and tertiary (3°) on the basis of the number of carbon atoms (one, two or three) directly bonded to @

the negatively charged carbon atoms. For example, ethyl anion (CH 3 CH 2 ) is a primary, @

≈

isopropyl anion (Me2 CH) is a secondary and tert-butyl anion Me3 C is a tertiary carbanion. @

Methyl anion (CH 3 ) with one carbon atom is a special case.

Structure, Bonding and Proper es of Organic Molecules

1.181

Structure: In simple carbanions, the negatively charged central carbon atom is sp3 hybridized; it is surrounded by three bonding electron pairs and one unshared pair of electrons occupying an sp3 orbital. Therefore, a carbanion is expected to have the tetrahedral shape. However, the shape is not exactly that of a tetrahedron and in fact, it is found to have the pyramidal shape just like ammonia. Since the repulsion between the unshared pair and any bonding pair is greater than the repulsion between any two bonding pairs, therefore, the angle between two bonding pairs (i.e., between two sp3-s bonds) is slightly less than the normal tetrahedral value of 109.5° and for this reason, a carbanion appears to be shaped like a pyramid with the negative carbon at the apex and the three groups at the corners of a triangular base.

The central carbon atoms in resonance-stabilized carbanions are, however, sp2 hybridized and hence they are planar. This is due to the fact that planarity is an essential criterion for resonance to occur. Allyl anion, for example, is a planar carbanion.

The negative carbon atom of the following conjugated anion is, however, not resonancestabilized because according to the Bredt’s rule a double bond at the bridgehead position cannot be formed in bridged bicyclic compounds with small rings and for this reason, this carbon is sp3-hybridized. Its shape is pyramidal.

1.182

Organic Chemistry—A Modern Approach

Stability: Any factor that tends to delocalize the negative charge must increase the stability of carbanions while any factor that tends to localize or intensify the negative charge must decrease the stability of carbanions.

[W = Electron-withdrawing group; it disperses the negative charge and thus, stabilizes the carbanion]

[D = Electron-donating group; it intensifies the negative charge and hence destabilizes the carbanion] @

@

The stability of carbanions follows the order: Methyl anion (CH 3 ) > R CH 2 (primary or 1°) @

@

> R 2 CH (secondary or 2°) > R 3 C (tertiary or 3°) because an alkyl group destabilizes a carbanion. Carbanions are stabilized mainly by –I and –R effects. Functional groups in the a position stabilize carbanions in the following order: —NO2 > —COR > —COOR > —CN ~ —CONH2 > —X > —H. The structural features responsible for the increased stability of carbanions are as follows: (a) s character of the anionic carbon atom: An s orbital being closer to the nucleus than the p orbital in a given main quantum level possesses lower energy. An electron pair in an orbital having large s character is, therefore, more tightly held by the nucleus and hence of lower energy than an electron pair in an orbital having relatively small s character. Thus, a carbanion in which the anionic carbon is sphybridized (50 percent s character) is more stable than a carbanion in which the anionic carbon is sp2 hybridized (33.33 percent s character), which in turn is more stable than a carbanion in which the negative carbon is sp3-hybridized (25 percent s character). Therefore, the order of decreasing stability of carbanions is

(b)

Inductive electron withdrawal: Substituents possessing electron-withdrawing inductive effects (–I) stabilize a carbanion by dispersing or delocalizing the negative charge. Examples of some carbanions experiencing such stabilizing effect are as follows:

Structure, Bonding and Proper es of Organic Molecules

1.183

(c)

Resonance effect: If there is a double or triple bond conjugated with the anionic carbon, the carbanion is stabilized by delocalization of the negative charge with the p orbitals of the multiple bond. Thus, allylic and benzylic carbanions and carbanions attached to the groups containing polarized multiple bond such as —NO2, —C ∫∫N, C ∫∫ O, etc. are well stabilized by resonance. For example:

(d)

Formation of an aromatic system: A carbanion becomes highly stabilized if it is involved in constituting an aromatic system [a planar (4l + 2) p electron system, where n = 0,1,2, …, etc.]. Cyclopentadienyl anion, for example, is unusually stable because it is a close loop of 6p electron system and hence aromatic.

@

Optical inactivity of asymmetric carbanions like CH 3CH 2 CHCH 3 : Asymmetric carbanions are found to be optically inactive and cannot be separated into two enantiomers. This can be explained in terms of umbrella effect. There

1.184

Organic Chemistry—A Modern Approach

occurs rapid interconversion between two enantiomeric pyramidal carbanions (called umbrella effect) through a low energy planar transition state and because of this, asymmetric carbanions exist as racemic modifications and are found to be optically inactive. The pyramidal inversion is so rapid that such carbanions cannot be separated into two enantiomers, i.e., cannot be resolved.

[N.B. Due to the same basic reason, a chiral tertiary amine such as CH 3CH 2 N H CH 3 is found to be optically inactive and cannot be separated into two enantiomers.]

(3) Free radicals An atom or a group of atoms containing an odd (unpaired) electron is called a free radical or simply a radical. Although free radicals are electrically neutral species, they are, due to the presence of an odd electron, weakly attracted by the magnetic field and hence are paramagnetic in nature. They are very much reactive. Some examples of free radical are methyl radical (C H 3 ), ethyl radical (C H C H 2 ), phenyl radical (P h) , etc. They are intermediates in a group of reactions called radical reactions. Generation: Homolytic cleavage of a covalent bond leads to the formation of free radicals. Homolysis of bonds to carbon leads to the formation of free radicals containing an odd electron on carbon; these are collectively called alkyl free radicals or alkyl radicals or simply free radicals.

Structure, Bonding and Proper es of Organic Molecules

1.185

Free radicals can be generated by the following three principal methods: (a) Thermolysis: When compounds containing weak bonds are strongly heated, they undergo hemolytic cleavage of a covalent bond to form free radicals. For example: (i) Tetramethyl lead decomposes to form lead and methyl radicals when heated strongly at 600°C. 600∞ C (CH 3 )4 Pb æææÆ Pb + 4CH 3

(ii)

The O–O bond of a peroxide or a peracid has a strength of only 36 kcal/mol and so it undergoes cleavage to form radicals at room temperature. For example:

(iii)

The formation of radicals is often facilitated by the simultaneous release of very stable molecule like N2. Thermolysis of azo compounds, for example, leads to the formation of free radicals by the expulsion of N2 molecule. For example:

(b)

Photolysis: Absorption of visible or ultraviolet light provides a molecule with sufficient energy to cleave covalent bonds and thus, photochemical dissociation leading to the formation of radicals may take place. Acetone, for example, undergoes photochemical decomposition in the vapour phase by light having a wavelength of 320 nm to give CO and methyl radical (CH3 ).

Azoalkanes or diacyl peroxides also undergo photolytic cleavage to form radicals. For example:

(c) (i)

Redox reactions: One-electron transfer reactions are frequently employed to produce radicals. For example: In Kolbe electrolysis, a reaction that has considerable synthetic importance, the carboxylate ion gives up one electron to the anode and becomes oxidized to a radical.

1.186

Organic Chemistry—A Modern Approach

The resulting radical then dissociates to give an alkyl radical which dimerizes to yield an alkene.

(ii)

Inorganic ions such as Fe2+, Cu+, etc. capable of changing their valence state by accepting or losing a single electron are used for the generation of radicals. These, in fact, act as a reducing agent in such reactions. For example

Fenton’s reagent behaves as an oxidant by forming another radical. For example:

A stable phenoxide radical is obtained when Fe3+ ion acts as an oxidizing agent by accepting an electron from a phenoxide ion. For example:

Nomenclature: In naming a free radical containing an odd electron on carbon, the word ‘radical’ is added to the name of the alkyl or aralkyl group. For example, CH 3 , (CH 3 )2 CH and C6 H5CH 2 are named as methyl radical, isopropyl radical and benzyl radical, respectively. Classification: Free radicals are classified as primary (1°), secondary (2°) and tertiary (3°) according to the number of carbon atoms (one, two or three) directly attached to the carbon atom bearing the unpaired electron. For example, ethyl radical (CH3CH2) is a primary, isopropyl radical (Me2CH) is a secondary and tert-butyl radical (Me3C) is a tertiary radical. Structure: Simple alkyl radicals may have two possible shapes. They may have a planar triangular configuration (the carbon containing the unpaired electron is sp2-hybridized) and the unpaired electron is accommodated in a p orbital or a rapidly inverting pyramidal configuration (the central carbon is sp3-hybridized and the unpaired electron is in an sp3 orbital).

Structure, Bonding and Proper es of Organic Molecules

1.187

Available evidence indicates that simple alkyl radicals prefer a planar or near-planar shape, even though the energy difference between a planar and a pyramidal free radical is not large. Free radicals in which the central carbon is connected to atoms of high electronegativity, e.g., CF3, prefer to exist in pyramidal shape. The deviation from planarity increases with increase in electronegativity. The shape of CF3 can be well explained by valence shell electron pair repulsion theory (VSEPR). Due to the strong –I effect of fluorine, the electron density around the central carbon in CF3 is reduced and as a result, the unpaired electron–bond pair repulsion becomes greater than bond pair–bond pair repulsion. As a consequence, the angular distance between the p orbital containing the odd electron and the C — F bonding orbital is increased to give pyramidal shape of the radical. When a bridgehead carbon of bicyclic system bears the unpair electron, a planar trigonal configuration is completely prevented by the geometric requirements and a pyramidal configuration becomes mandatory. For example, the bridgehead radicals I and II are pyramidal. It is to be noted that bridgehead radicals are less stable than the corresponding open-chain radicals.

Like carbanions, the central carbon atoms of resonance-stabilized free radicals are sp2 hybridized and hence they are also planar. Allyl radical, for example, is planar.

Stability: Free radicals are highly unstable and reactive species because they have a strong tendency to gain an electron, i.e., to share with some other atom or group to have a complete octet. The stability of alkyl radicals follows the order: R3C (tertiary or 3°) > R2CH (secondary or 2°) > RCH2 (primary or 1°) > CH3 (methyl radical).

1.188

Organic Chemistry—A Modern Approach

The factors responsible for the stability of free radicals are as follows: (a) Hyperconjugation: Free radicals are stabilized by hyperconjugation (s–p conjugation) involving a–H atoms.

As the number of a–H atom increases, hyperconjugation becomes progressively more effective and consequently, the radicals become progressively more stabilized due to more effective electron delocalization. The relative stability of simple alkyl radicals is found to follow the order: (most stable) R3C (tertiary) > R2CH (secondary) > RCH2 (primary) > CH3 (methyl) (least stable). For example, tert-butyl radical (CH3)3C (with nine hyperconjugable a–H atoms) is more stable than isopropyl

(b)

(c)

radical, (CH3)2CH (with six hyperconjugable a–H atoms) which in turn is more stable than ethyl radical, CH3CH2 (with only three hyperconjugable a–H atoms). The methyl radical, CH3, is the least stable radical because the unpair electron is not at all delocalized due to lack of a–H atoms. Resonance: Resonance is a major and important factor influencing the stability of free radical. When the carbon atom bearing the odd electron is a to a double bond or a benzene ring, effective delocalization of the unpaired electron with the p orbital system with consequent stabilization occurs. Allyl and benzyl radicals, for example, are found to be particularly stable because of resonance.

Steric effect: Steric effect sometimes becomes very much significant in stabilizing tertiary radicals. There occurs considerable relief of steric strain (B strain) when an sp2-hybridized tertiary radical is formed from an sp3-hybridized substrate and this is because the repulsion between the bulky alkyl groups is relieved to a certain extent by an increase in bond angles from 109.5° to about 120°. As a consequence, the radical becomes much reluctant to react further, i.e., its stability is enhanced due to steric reason.

Structure, Bonding and Proper es of Organic Molecules

1.189

[Steric inhibition of resonance often decreases radical stability. For example, the resonance stability of triphenylmethyl radical (Ph3C) is not so as expected and this is because due to steric hindrance between ortho H-atoms of the adjacent rings, the three benzene rings cannot be all in the same plane for effective delocalization to take place. In fact, it exists in propeller-shaped conformation in which the three rings being aligned at about 30° out of the common plane.

In fact, the major reason for the greater stability of Ph3C is its greater reluctance towards dimerization because dimerization will introduce a considerable amount of steric strain due to close proximity of the bulky phenyl groups in hexaphenylethane (Ph3C — C Ph3), the expected dimer. Thus, it is not resonance, but steric hindrance to dimerization is the major cause for the stability of triphenylmethane radical. Detection: A very simple method to know whether a particular reaction proceeds through a radical intermediate or not is to study the kinetics of the reaction (i.e., the reaction rate) in the presence and in the absence of inhibitors such as hydroquinone. If the reaction proceeds through radical intermediate, its rate is expected to be seriously affected by the presence of the inhibitor. Electron spin resonance (esr) spectroscopic studies offer the most useful method for detecting radicals. Bond dissociation energy (a measure of free radical stability): The bond dissociation energy is the energy needed to cleave a covalent bond homolytically.

Since bond breaking requires energy, therefore, bond dissociation energies are always positive numbers and homolysis is always endothermic. The shorter the bond, the higher the bond dissociation energy. Again, shorter bonds are stronger bonds. Bond dissociation energies (DH° values) of R—H bonds provide a measure of relative inherent stability of free radicals R . The following table lists such values. The higher the DH° values, the less stable the radical.

1.190

Organic Chemistry—A Modern Approach

DH° values of some R—H bonds R Ph

Bond dissociation energy, DH° (Kcal/mol) 111

CF3

107

CH 2 == CH

106

Stability increases

106

CH3

105

CH3CH2

100

(CH3 )3CCH2

100

CH3CH2 CH2

100

Cl3C

96

(CH3 )2CH

96

(CH3 )3C

95.8 95.5

C6 H5CH2

88

HCO

87

CH 2 == CH — CH 2

86

The bond dissociation energy for a benzylic hydrogen is less than that of a methane hydrogen which can be explained by resonance.

Structure, Bonding and Proper es of Organic Molecules

1.191

From the enthalpy values, it becomes clear that 17 kcal/mol less energy is needed to form the benzyl radical from toluene than to form methyl radical from methane. This difference in energy requirement can be explained on the basis of the difference in energy content between the radical and its precursor. Both toluene and benzyl radical can be represented as resonance hybrids.

Since toluene is a resonance hybrid of only two resonance structures (I and II) and the benzyl radical is a resonance hybrid of five resonance structures (III–VIII), therefore, the benzyl radical is stabilized by resonance to a greater extent than toluene. As a result of this, the difference in energy content between benzyl radical and toluene becomes relatively small. Both methane and methyl radical, on the other hand, can be represented satisfactorily by a single (localized) structure.

None of these species are resonance-stabilized and thus, the difference in energy content between the methyl radical and methane is relatively large. Because of this, it requires higher energy to form methyl radical from methane than to form benzyl radical from toluene. This becomes clear from the following energy diagrams.

The bond dissociation energy of CH3 — H is equal to the hypothetical but greater than the actual bond dissociation energy of PhCH2 — H (DE1 > DE2).

1.192

Organic Chemistry—A Modern Approach

1.10.2.1

Radical inhibitors or ‘scavengers’

Compounds that destroy reactive radicals by creating unreactive stable radicals or neutral molecules with only paired electrons and thereby prevent reactions that take place by mechanisms involving radicals are called radical inhibitors. Phenol is an example of the first type, and nitric oxide (NO) is an example of the second type.

Oxygen molecule (O2) can also function as a highly efficient radical inhibitor. This is because it has two unpaired electrons and behaves as a diradical (O — O) . By combining with highly reactive radical intermediates (R) , it converted into very much less reactive peroxy radicals (R — O — O) that cannot carry out the chain reaction.

1.10.2.2 (a)

(b)

Electrophilic and nucleophilic radicals

Electrophilic radicals: The radicals that have a tendency to abstract electron-rich hydrogen atoms and which behave like electrophiles in that the rate is increased by the presence of electron-releasing groups in the substrate are called electrophilic radicals. For example, an electron-releasing group at the para-position of toluene increases the rate of hydrogen abstraction by bromine while an electron-attracting group decreases it. Therefore, B r is an example of electrophilic radical. Nucleophilic radicals: The radicals that have a tendency to abstract electronpoor hydrogen atoms and that behave like nucleophiles that which the rate is increased by the presence of electron-attracting groups in the substrate are called nucleophilic radicals. For example, an electron-attracting group at the paraposition of toluene increases the rate of hydrogen abstraction by tert-butyl radical while an electron releasing group decreases it. Therefore, (CH3)C is an example of nucleophilic radical.

So far, we have discussed points 1, 2 and 3 in previous sections, however, the below mentioned points 4, 5, 6 and 7 are discussed in detail in Volume-II of this book. (4) Carbenes [see Chapter 5 of Vol.-II] (5) Ylides [see Chapter 6 of Vol.-II] (6) Benzyne [see Chapter 7 of Vol.-II] (7) Nitrenes [see Chapter 8 of Vol.-II]

Structure, Bonding and Proper es of Organic Molecules

1.10.2.3

1.193

Electrophiles and nucleophiles

(a) Electrophiles: The reactive sites of some molecules and ions (i) are capable of accepting more electrons or (ii) are the d+ end of polar bonds. These electron-deficient sites are electrophilic and the species possessing such sites are ‘electron loving’ and hence ≈

called electrophiles. BF3 is an example of neutral electrophile and Me3C is an example of ionic electrophile. (b) Nucleophiles: The reactive sites of some molecules or ions have higher density of electrons because the site (i) may have an unshared pair of electrons or (ii) may be the d– end of a polar bond or (iii) may have C ∫∫ C p electrons. Such electron-rich sites are nucleophilic and the species possessing such sites are ‘nucleus loving’ and hence called nucleophiles. N H 3 is an example of neutral nucleophile and I@ is an example of ionic nucleophile. Bond formation: Like bond fission, bond formation occurs in two different ways. Two free radicals can combine, in which each donates one electron, to form a bond. Again, two ions with unlike charges combine to form a bond; the negatively charged ion donates both the electrons. Bond formation is always an exothermic process.

1.

≈

Triphenylmethyl fluoroborate (Ph 3C BF4@ ) can be stored indefinitely as a stable ionic solid — Why? ≈

Solution The triphenylmethyl cation (Ph 3C) is exceptionally stabilized by resonance involving three phenyl groups and also the BF4① ion is nonnucleophilic in nature. For ≈

these reasons, triphenylmethyl fluoroborate (Ph 3C BF4@ ) can be stored indefinitely as a stable ionic solid. 2. Acetylene, unlike ethylene, does not dissolve in concentrate H2SO4 — Why? Solution An unsaturated hydrocarbon is expected to dissolve in conc. H2SO4 by forming a bisulphate salt of a carbocation. The more the s character in the positively charged carbon atom, the less stable is the carbocation and the less likely it is to be formed, i.e., less likely the compound is to be dissolved in conc. H2SO4. Protonation of ethylene (CH2 == CH2) ≈

produces the relatively more stable alkyl carbocation salt, CH3CH2 HSO@4 (the positive

1.194

Organic Chemistry—A Modern Approach

carbon is sp2-hybridized and being relatively less electronegative it accommodates the positive charge better). On the other hand, protonation of acetylene is expected to produce ≈

the relatively less stable vinyl carbocation salt, CH2 == CH HSO4@ (the positive carbon is sp-hybridized and being more electronegative it is less willing to accommodate the positive charge). For this reason, ethylene dissolves in conc. H2SO4 but acetylene does not.

3.

An alkyl cation can survive long enough when it is generated from an alkyl fluoride (RF) in the presence of Lewis super acid antimony pentafluoride (SbF5) dissolved in liquid SO2 — Why?

Solution It is possible for a simple alkyl cation to survive long enough if its counter ion being very weakly nucleophilic cannot combine with it. When an alkyl fluoride is treated with SbF5 dissolved in SO2 at low temperature, SbF6@ is obtained as the counter ion.

SbF6@ , the conjugate base of the extremely strong acid HSbF6(pKa = –25), lacks basicity as well as nucleophilicity. So, in the presence of SbF6@ , an alkyl cation can survive long enough. 4.

Explain why cyclohexyl cation , but cyclohexyl anion

is more stable than phenyl cation is less stable than phenyl anion

.

Solution The stability of carbocations decreases with increase in the s character, i.e., the electronegativity of the cationic carbon atom. The positive charge in cyclohexyl cation is on a less electronegative sp2-hybridized (33.33 percent s character) carbon atom while the positive charge in phenyl cation is on a more electronegative sp-hybridized (50 percent s character) carbon atom. Therefore, the positive carbon in phenyl cation is less capable of accommodating the positive charge compared to the positive carbon in cyclohexyl cation. Because of this, cyclohexyl cation is relatively more stable than phenyl cation. On the other hand, the stability of carbanions increases with increase in the s character, i.e., the electronegativity of the anionic carbon atom. In cyclohexyl anion, the negative charge is on a less electronegative sp3-hybridized (25 percent s character) carbon atom while in phenyl anion, the negative charge is on a more electronegative sp2-hybridized (33.33 percent s character) carbon atom. The more the carbon is electronegative, the better it accommodates the negative charge. Cyclohexyl anion is, therefore, less stable than phenyl anion.

Structure, Bonding and Proper es of Organic Molecules

1.195

5.

Identify A, B and C and explain these observations. Solution The strong acid HBF4 converts triphenylcarbinol, Ph3COH, to the salt of the very ≈

stable carbocation, Ph3 C (triphenylmethyl cation). Its stability is due to extensive charge delocalization involving three benzene rings. Because of such electron delocalization, the carbocation absorbs radiation of wavelength 480–490 nm in the visible range and as a ≈

consequence, it appears in orange colour. Therefore, A is Ph3C BF4@ . +

Ph3C—OH + HBF4

–

+

Ph3C—OH2BF4

–

Ph3C BF4 + A Triphenylmethyl fluoroborate (a bright orange salt)

H2O

≈

Addition of H2O or EtOH converts Ph3 C to Ph3COH or Ph3COEt and as a consequence, the colour of the salt is discharged.

6.

When a primary aliphatic amine is allowed to react with nitrous acid (i.e., NaNO2/HCl), it is converted into an alcohol. For example: H NO 2 (CH 3 )2 CH NH 2 ææææææ Æ (CH 3 )2 CH OH (NaNO 2 / HCl) Isopropylamine Isopropyl alcohol (a) Give the mechanism of the reaction ≈

(b) Explain why benzenediazonium ion (Ph N 2 ) is more stable than ≈

isopropyldiazonium ion (Me 2CH N 2 ).

1.196

Organic Chemistry—A Modern Approach ≈

(c) Predict the product obtained when an acidic solution of Ph N 2 Cl @ is warmed. Solution (a) Isopropylamine reacts with HNO2 (i.e., NaNO2/HCl) to form isopropyl alcohol as follows:

≈

(b)

Nucleophilic attack by isopropylamine on the nitrosonium ion (NO) obtained from nitrous acid produces an N-nitrosoamine which, in turn, tautomerizes to a diazohydroxide. The diazohydroxide then undergoes protonation followed by loss of water to form a diazonium ion. The diazonium ion being very unstable dissociates extremely readily to form isopropyl cation which reacts with water to yield isopropyl alcohol. The instability of diazonium ion in the absence of any stabilizing structural feature is due mainly to the extreme effectiveness of N2 as a leaving group. Benzenediazonium ion is stable if the diazotization reaction is carried out below ≈

5°C. The stability of benzenediazonium ion (as compared to Me2 CH ) is attributed to resonance as follows:

≈

The increased stability of Ph — N2 is also due to greater difficulty of forming ≈

relatively less stable Ph! as compared to Me2 C H . (c)

Benzenediazonium ion dissociates on warming and forms N2 and phenyl cation which subsequently undergoes nucleophilic attack by water to give phenol, after proton loss.

Structure, Bonding and Proper es of Organic Molecules

1.197

7. (a) Explain why carbocation (not all) undergoes rearrangement. (b) What are the important nodes of rearrangement of carbocation and why are they called 1,2-shifts? (c) Give examples of three reactions in which different types of 1,2-shift occur. Solution (a) Carbocations undergo rearrangement because they become stable as a result of rearrangement. For example, a 1° carbocation may rearrange to a more stable 2° or 3° carbocation and a 2° carbocation sometimes rearranges to a more stable 3° carbocation. (b) A less stable carbocation can rearrange to a more stable carbocation by shift of a hydrogen atom or an alkyl group or a ring bond. These rearrangements are called 1,2-shifts because they involve migration of an alkyl group or a hydrogen atom or a ring bond from one carbon to an adjacent carbon atom. The migrating group moves with the two bonding electrons. (c) (i) Electrophilic addition of HBr to 3-methyl-1-butene produces 2-bromo-2methylbutane as the major product and 2-bromo-3-methylbutane as the minor product. However, according to the Markownikoff’s rule 2-bromo-3-methylbutane is expected to be the major product.

CH3 | CH3 CH CH == CH2 + HBr Æ 3-Methyl-1-butene

CH3 CH3 | | CH3 CH CHCH3 + CH3 CCH2CH3 | | Br Br 2-Bromo-3-methylbutane 2-Bromo-2-methylbutane (major product) (minor product)

This observation can be explained by the fact that the initially formed 2° carbocation rearranges to a more stable 3° carbocation by a 1,2-hydride shift. As a result of this carbocation rearrangement, two alkyl bromides are formed — one from attack of the nucleophile (Br①) on the unrearranged carbocation and the other from attack on the rearranged carbocation. The bromide corresponding to the more stable carbocation is obtained as the major product. The reaction occurs as follows:

1.198

Organic Chemistry—A Modern Approach

(ii)

Similarly, electrophilic addition of HCl to 3,3-dimethyl-1-butene leads to the formation of 2-chloro-2,3-dimethybutane as the major product and 3-chloro-2,2dimethylbutane as the minor product which, in fact, expected to be the major product according to the Markownikoff’s rule.

(iii)

Carbocation rearrangements can also take place by ring expansion involving the shift of a ring bond. When 1-methyl-1-vinyl-cyclobutane, for example, is treated with HBr, 1-bromo-1,2-dimethylcyclopentane is obtained. The initially formed 2° carbocation undergoes ring expansion by a 1,2- shift to form a more stable 3° carbocation. The process is also favoured by the release of angle strain (a fourmembered ring is converted into a five-membered ring).

Structure, Bonding and Proper es of Organic Molecules

8.

1.199

Explain why carbocations undergo rearrangement while free radicals do not.

Solution The central carbon atoms of a carbocation and a free radical are sp2-hybridized. The unhybridized p-orbital is vacant in carbocations but in free radicals it has one electron. In carbocation rearrangements, the two electrons of the migrating s-bond can easily move into the vacant p-orbital of positive carbon since in accordance with Pauli exclusion principle, an orbital at maximum can accommodate two electrons. For this reason, less stable carbocations rearrange to yield more stable carbocations. On the other hand, in free radicals, the two electrons of the migrating s-bond cannot move into the p-orbital which already has one electron because by doing so the p-orbital will have three electrons which is not in accordance with the Pauli exclusion principle. If the rearrangement occurs at all, the third electron has to go to the antibonding molecular orbital making the rearrangement energetically unfavourable. Therefore, a carbocation undergoes rearrangement because it is energetically favourable but a free radical does not undergo rearrangement because the process is energetically unfavourable. 9. Rank the following carbocatioss in the order of increasing stability and explain the order:

Solution The 3° carbocation I is stabilized by +I effect of the three ring bonds. However, its stabilization by hyper conjugation is not possible because the formation of a double bond at the bridgehead position is not possible according to Bredt’s rule. Also, this carbocation suffers from angle strain because the angle between the bonds is somewhat less than the sp2 bond angle of 120°. The 3° carbocation II is more stable than I because it is stabilized by the +I effect of three methyl groups directly attached to the positive carbon and hyperconjugation effect involving nine a-H atoms. The carbocation III is the most stable one because it is not only stabilized by the inductive and hyperconjugation effect of two methyl groups but also stabilized by resonance involving the adjacent double bond.

Therefore, the order of increasing stability of these carbocations is

10.

Which one of the following two carbocations is more stable? Give your reasoning. ≈

≈

(CH 3 )3 C; (CD3 )3 C

1.200

Organic Chemistry—A Modern Approach

Solution Each of these two 3° carbocations is stabilized by +I effect of the three alkyl groups. However, the +I effect of —CH3 group is more stronger than the +I effect of —CD3 ≈

≈

group (D is more electronegative than H). Hence, (CH 3 )3 C is more stable than (CD3 )3 C from inductive point of view. Again, both the ions are stabilized by hyperconjugation as follows:

Since the C — D bond is stronger than the C — H bond, therefore, the hyperconjugative ≈

forms of (CD3 )3 C are relatively less stable and contribute less to the hybride. Hypercon≈

≈

jugation is, therefore, less effective in (CD3 )3 C compared to (CH 3 )3 C . Hence, also from ≈

≈

hyperconjugation point of view, (CH 3 )3 C is relatively more stable than (CD3 )3 C. 11. Which of the following carbocations would you expect to undergo rearrangement and why? (a)

(b) C6H5CH == CHCH2CH2≈

(c)

(d)

(e)

(f) (CH 3 )2CH CHCH 3

(g)

(h)

≈

(i)

≈

(CH 3 )2CHCH CHCH 3

(j)

Solution (a) This carbocation does not undergo rearrangement because rearrangement would yield other 2° carbocations of nearly equal stability. That is, there is no energetic advantage in these processes.

Structure, Bonding and Proper es of Organic Molecules

1.201

(b)

This carbocation will undergo rearrangement because a 1,2-hydride shift will convert this primary carbocation into a more stable allylic carbocation which is stabilized by resonance.

(c)

This 1° carbocation will rearrange because a 1,2-hydride shift will convert it into a more stable 3° carbocation.

(d)

This secondary carbocation will rearrange because a 1,2-hydride shift will convert it into a more stable tertiary carbocation.

(e)

This 2° carbocation will rearrange to give another 2° carbocation because the resulting carbocation is stabilized by resonance involving the cyclopropane ring.

(f)

This 2° carbocation will undergo rearrangement because a 1,2-hydride shift will convert it into a more stable 3° carbocation.

1.202

Organic Chemistry—A Modern Approach

(g)

This 2° carbocation will undergo rearrangement because a 1,2 methyl shift will convert it into a more stable 3° carbocation.

(h)

This 1° carbocation will undergo rearrangement because a 1,2-bond shift will convert it into a more stable 2° carbocation. 1,2-Hydride shift leading to the formation of a 3° carbocation does not take place because the resulting carbocation is expected to suffer from a good amount of angle strain (120°–90° = 30°).

(i)

This 2° carbocation will not undergo rearrangement because a 1,2-hydride shift will convert it into an equally stable 2° carbocation. Therefore, there is no energetic advantage in the process.

(j)

This 2° carbocation will rearrange because a 1,2-bond shift will convert it into a more stable 3° carbocation.

12.

Arrange the following carbocations in order of increasing stability and explain the order:

Structure, Bonding and Proper es of Organic Molecules

1.203

Solution A carbocation in which the cationic carbon is directly attached to a cyclopropyl group is especially stable because of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p orbital of the cationic carbon. Hence, III is considerably stabilized by resonance.

The stability of the carbocation increases with each additional cyclopropyl group. Therefore, the carbocation I bearing two such groups is more stable than III bearing only one such group. Again, a cyclopropyl group stabilizes an adjacent positive charge even better than a phenyl group. Hence, IV is less stable than III. II is less stable than IV because unlike IV, this 1° carbocation is not resonance-stabilized. Therefore, the order of increasing stability of these carbocations is

13.

Rank the following carbocations in order of decreasing stability and give your reasoning:

Solution In p-methoxybenzyl cation (III), the unshared pair of electrons on the oxygen atom of the —OMe group is in proper conjugation with the vacant p orbital of the positive carbon. Because of this, this carbocation is more stabilized by resonance as compared to the benzyl cation (I).

In p-nitrobenzyl cation(II), the electron-attracting —NO2 group intensifies the positive charge of the cationic carbon by placing a positive charge on the carbon adjacent to the cationic carbon by its –R and –I effects and thereby makes this carbocation considerably unstable. Hence II is less stable than I

1.204

Organic Chemistry—A Modern Approach

Therefore, the order of decreasing stability of these carbocations is as follows: +

CH3O—

14.

—CH2 (III)

+

O2N—

—CH2 (I)

+

—CH2 (II)

≈

Explain why the carbocation F3 C is more stable than the carbocation ≈

F3C — CH 2 . ≈

Solution In F3C — C H 2 , the electron-attracting (–I) –CF3 group withdraws electrons of the C — C bond towards itself and thereby destabilizes the carbocations by intensifying the ≈

positive charge. On the other hand, in the carbocation F3 C , the unshared pair of electrons on each of the three F atoms overlaps with the empty p orbital of the cationic carbon and thereby stabilizes the carbocation by delocalizing the positive charge. Furthermore, the p orbitals overlap is much more effective because the orbitals of fluorine and carbon are of comparable size. In fact, the resonance stabilization outweighs the destabilization caused ≈ by the —I effect of fluorine. This explains why carbocation F3 C is relatively more stable ≈

than the carbocation F3C — C H 2 .

15.

Rank the following carbocation in order of decreasing stability and explain the order:

Solution The decreasing order of stability of these carbocations is I > II > III > IV.

In the carbocation IV, the positive charge is stabilized by the inductive (+I) and hyperconjugation effect of one methyl group while one methyl and one ethyl group are involved in stabilizing the charge in III. Again, an sp-hybridized doubly bonded positive

Structure, Bonding and Proper es of Organic Molecules

1.205

carbon is more electronegative than an sp2-hybridized singly bonded positive carbon and hence the former carbon is less capable of accommodating a positive charge than the latter. Because of these, the vinylic cation IV is less stable than the carbocation III. The charge in the allylic cation II is highly stabilized by resonance and because of this, it is more stable than the carbocation III where only weak inductive and hyperconjugation effects are operative.

In the carbocation I (a cycloprophenyl cation derivative), the positive charge is involved in constituting an aromating system [(4n + 2)p, where n = 0]. Hence, it is more stable than the carbocation II. Therefore, the decreasing order of stability of these carbocations is as given above. 16.

Triptycenyl cation is less stable than trityl cation. Explain.

Solution In trityl or triphenylmethyl cation, the positive charge is well stabilized by resonance involving three phenyl groups. However, in triptycenyl cation, the empty p orbital is, in fact, perpendicular to the p orbitals of the three benzene rings and since bond rotation is not possible in this rigid system; therefore, they can never become parallel for conjugation to take place. Therefore, the positive charge in triptycenyl cation is not at all stabilized by charge delocalization. This cation is, therefore, less stable than trityl cation.

17.

Identity A and B and explain these observations.

1.206

Organic Chemistry—A Modern Approach

Solution Sodium amalgam reduces triphenylmethyl chloride to give the salt of the very @

stable carbanion, Ph3 C (triphenylmethyl anion). Its stability is due to extensive charge delocalization into three benzene rings.

@

Because of extensive electron delocalization, the carbanion Ph3 C absorbs radiation of wavelength 490–500 nm and as a consequence, it appears red. The strongly basic carbanion takes up a proton from H2O or EtOH to form colourless Ph3CH which does not absorb radiation in the visible range. 18.

Rank the following carbanions in order of increasing stability and explain the order:

Solution The order of increasing stability of these carbanions is

The external orbitals (i.e., the orbitals directed towards outside bonds) in cyclopropane (the conjugate acid of I) have larger (33%) s character, i.e., they are approximately sp2-orbitals. Because of this, the unshared pair of electrons on the anionic carbon in cyclopropyl anion (I) is more tightly held with the carbon nucleus than the unshared pair in ethyl anion (IV) that occupies an sp3 orbital (25% s character). Consequently, the former anion is more stable than the latter. In the vinylic anion II, the unshared pair of electrons occupies an sp2orbital (33.33%) s character and hence this anion is somewhat more stable than cyclopropyl anion (I). In the allylic anion III, the negative charge is delocalized by resonance with the adjacent double bond and so it is relatively more stable than the vinylic anion II. 19.

Arrange the following carbanions in order of increasing stability and give your reasoning: @

@

CH 3 CH NO2

@

CH 3 CHCOCH 2CH 3

I

@

CH 3 CH CO2Et CH 3 CH C6 H5

II

III

IV

Solution The order of increasing stability of these carbanions is @

@

@

@

C6 H5 C HCH 3 (iv) < CH3 CH CO2 Et (III)
Structure, Bonding and Proper es of Organic Molecules

1.207

All these carbanions are stabilized by resonance, but the anions conjugated with a carbon– oxygen or carbon–nitrogen double bond are relatively more stabilized than the benzylic anion IV and this is because the more electronegative oxygen atom accommodates the negative charge better than the less electronegative carbon atom. Therefore, the benzylic anion IV is the least stable one.

@

The C == O group in CH 3 C H CO2Et (III) is less effective in stabilizing the carbanion than @

the C == O group in CH 3 C H COC2H5 (II) and this is because unlike keto carbonyl, the ester carbonyl is involved in resonance interaction with the unshared pair of electrons on the oxygen atom of the –OEt group. Therefore, the anion II is more stable than III.

All the groups (except methyl) attached to the anionic carbon exert their –I effects in addition to their resonance effects. However, the –I effect exerted by the —NO2 group is the strongest one because the N atom in —NO2 bears a positive charge. Because of this, the NO2 group is more effective in stabilizing the carbanion than the > C == O group. Hence, the carbanion I is the most stable one. 20.

Classify each of the following species as electrophile and nucleophile and give your reasoning. @

≈

(a)

C H3

(b)

(CH 3 )3 C

(c)

:I:@

(d)

:Cl≈

(e)

NH 3

(f)

BF3

(g)

Cl 2 C:

(h)

: PPh 3

(i)

CH3CH == CHCH3

(j)

SiF4

(k)

CH 3 O:@

Solution Electrophiles: (b), (d), (f), (g) and (j).

The central atoms (C, B and C) of (b), (f) and (g) have six electrons and can accept an electron pair from a nucleophile to complete the more desirable octet. Therefore, these are

1.208

Organic Chemistry—A Modern Approach

electrophiles. For the same reason, the cation (d) is an electrophile. The Si atom in (j) can acquire more than eight electrons by utilizing its d orbitals. Hence, it is also an electrophile. Nucleophiles: (a), (c), (e), (h), (i) and (k). The anions (a), (c) and (k) and the neutral molecules (e) and (h) have unshared pair of electrons. Therefore, these are nucleophiles. The alkene (i) has two available electrons in a p bond. Hence, it is also a nucleophile. 21. Give example of a very stable salt which can be obtained by combining a very stable carbanion with a very stable carbocation. Solution An example of salt obtained by combining a very stable carbanion with a very stable carbocation is as follows:

The cation (a cyclopropenyl cation) is stable because it is an aromatic system. The anion is stabilized by resonance involving contributing structures containing aromatic cyclopentadienyl anion systems.

22.

Show how hydroquinone can act as a radical inhibitor.

Solution When hydroquinone traps a radical (R) , it forms semiquinone. Semiquinone is stabilized by resonance and is, therefore, less reactive than other radicals. Furthermore, semiquinone traps another radical and itself becomes converted into quinone (a compound whose electrons are all paired). The reaction occurs as follows:

Structure, Bonding and Proper es of Organic Molecules

23.

1.209

Give two examples of radical inhibitors that are present in biological system.

Solution The two radical inhibitors present in the biological system are vitamin C (ascorbic acid) and vitamin E (a-tocopherol). Like hydroquinone, they form relatively stable radicals. CH2OH H OH O O

H

O

HO OH Vitamin C (ascorbic acid)

24.

Vitamin E (a-tocopherol)

Rank the following radicals in order of increasing stability and give your reasoning: i

(CH3 )3 C, I

i

CH2 — CH == CH2 , II

i

CH3CH2 , III

i

(C6 H5 )2CH IV

Solution The stability of a free radical depends on the extent of delocalization of the unpaired electron by resonance or hyperconjugation. The unpaired electron in ethyl radical (III) is stabilized by hyperconjugation involving only three a-hydrogen atoms and so it is the least stable among the four alkyl radicals. The unpaired electron in tert-butyl radical (I) containing nine hyperconjugable a-H atoms is well delocalized by hyperconjugation. Hence, its stability is higher than that of ethyl radical (III).

1.210

Organic Chemistry—A Modern Approach

The unpaired electron on carbon in allyl radical (II) is delocalized by resonance interaction with the p-electrons of the double bond and the delocalization is more effective because the two resonance structures are equivalent. Since the stability due to hyperconjugation is less than that caused by resonance, therefore, the allyl radical (II) is more stable than the tert-butyl radical (I)

In diphenylmethyl radical (IV), the unpaired electron is extensively delocalized over the p orbital system of the two benzene rings and so, delocalization is much more effective in this case. Because of this, it is relatively more stable than the allyl radical (II).

Therefore, the order of increasing stability of these radicals is CH 3CH 2 (III) < (CH 3 )C (I) < CH 2 — CH == CH 2 (II) < (C6 H5 )2 CH (IV) 25.

Arrange the following compounds in order of increasing rate of thermal decomposition to yield nitrogen and give your reasoning:

CH 3 — N == N — CH 3 I

C6 H5 CH 2 — N == N — CH 2C6 H5 II

CH 3CH 2 — N == N — CH 2CH 3 III

Solution The rate of thermal decomposition of an azo compound is much higher if it produces radicals that have considerable stability. Dimethyldiazene (I), dibenzyldiazene (II) and diethyldiazene (III) decompose to give methyl, benzyl and ethyl radicals, respectively. Benzyl radical is highly stabilized by resonance and it is, therefore, more stable than ethyl radical which is stabilized by hyperconjugation involving three a-H atoms (a less important stabilizing effect). The methyl radical (CH 3 ) is the least stable because it is not stabilized either by hyperconjugation or by resonance. Because of this, dibenzyldiazene decomposes at a rate faster than diethyldiazene which in turn decomposes at a faster rate than dimethyldiazene.

Structure, Bonding and Proper es of Organic Molecules

1.211

26.

Identify A and B and explain these observations. Solution Triphenyl methyl chloride (Ph3CCl) reacts with Ag or Zn to give very stable (stabilized by resonance) triphenyl methyl radical which gives a yellow solution in benzene. The radical remains in equilibrium with its dimer. When the solution is evaporated, the dimer is obtained as the residue (a white solid). Therefore, A is Ph3C and B is its dimer.

The formation of the dimer may be shown as follows:

Triphenylmethyl radical reacts with O2 to give colourless ditriphenylmethyl peroxide or with I2 to give colourless triphenyl iodomethane. Therefore, C is Ph3COOCPh3 and D is Ph3C — I.

1.212

27.

Organic Chemistry—A Modern Approach

Explain why the halogenation of alkenes in the presence of tetraethyl lead (Et4Pb) occurs at a temperature lower than the temperature required when halogenation is carried out in its absence.

Solution Halogenation of alkenes is normally carried out between 525 and 675K and this is because homolysis of the X — X bond occurs to form halogen free radical (X) at this

temperature range. Tetraethyl lead decomposes at a much lower temperature of 423 K to produce ethyl radical (CH 3CH 2 ) which subsequently reacts with halogen molecules to form halogen free radicals.

Since in the presence of tetraethyl lead, halogen radicals are produced at a lower temperature, therefore, halogenation also takes place at a lower temperature.

1.

Arrange the following carbocations in order of decreasing stability and explain the order: ≈

≈

≈

≈

≈

≈

(CH 3 )3 C, (CH 3 )2 CH, C6 H5 CH 2 , CH 3 , CH 3 CH 2 , (C6 H5 )2 CH 2.

Rank the following carbocations in order of increasing stability and give reasons: ≈

≈

≈

(a)

CH 3 CH ClCH 2CH 2 C H 2 , CH 3 CH 2 CH 2 CH ClC H 2 , CH 3 CH 2 CH ClCH 2 C H 2

(b)

CH3 CH2CHCH2 , CH3CHCH2 CH2 , CH3 CH2 CH2 CH | | | OCH3 OCH3 OCH3

≈

≈

≈

Structure, Bonding and Proper es of Organic Molecules

3.

1.213

Which one is more stable and why? ≈

≈

(a)

CH 3 NH C H 2 or CH 3 OC H 2

(b)

(c)

CH3 OCH == CH — CH2 or CH2 == C — CH2 | :OCH3

≈

≈

(d)

(e)

(f)

4.

[Hint]: (b) Consider ‘steric inhibition of resonance’ (f) Consider ‘Bredt’s rule.’] Arrange the following ions in order of increasing stability. Give reasons. (a)

(b)

(c)

≈

CH3 CHCH3 , I

≈

≈

CH2 == CH — CH2 , CH3 C H2 CH2 II

III

(d)

5.

Which one is more stable and why? (a)

(c)

(b) O O @ || || CH3 CH — N — C — Ph or CH2 — N — CH2 — C — Ph | | C 2H5 C2H5 @

@

@

(d) CCl3 or CF3

1.214

Organic Chemistry—A Modern Approach

(e) 6.

(f)

Arrange the following carbanions in order of decreasing stability and give your reasoning: @

@

Ph3 C , 7.

,

@

@

(CH 3 )3 C, (CH 3 )2 C H, C H 3

Predict the state of hybridization of the anionic carbon in each of the following carbanions: @

(a)

CH 3 , C H 2

(d)

CH 3 CO C H COCH 3

(g)

O2 N C H 2

(b)

@

(c)

Ph C H 2

@

CH 3 CO C H 2

@

@

(j)

(e)

(f) @

(h)

Ph CH2 CH2

(k)

CCl3

@

@

(i)

HC ∫∫ C

(l)

C F3

@

8.

Bridgehead carbocations are not much stable yet 1-tris homobarrelyl cation (I) have been prepared in super acid solution at –78°C — Why?

9.

AIBN is widely used as a free radical initiator. Write appropriate equation to show how AIBN can produce free radicals. CN CN | | [Hint: AIBN is azobisisobutyronitrile, Me2 C — N == N — C Me2 ]

10.

How can chlorine atoms or radicals be generated using an azo compound? [Hint:

Structure, Bonding and Proper es of Organic Molecules

11.

1.215

Arrange the following free radicals in order of decreasing stability:

What is the expected common reason for the same stability order between carbocations and free radicals? 12.

13.

Explain why the radical I undergoes ready dimerization while the radical II does not.

[Hint: The planar radical I has much less steric hindrance to dimerization than the nonplanar (due to steric reason) radical II.] Explain why acetylene (HC ∫∫ CH) is more acidic than ethane (CH3CH3), even though the C — H bond in acetylene has a higher bond dissociation energy than the C — H bond in ethane. [Hint: The bond dissociation energy is a measure of how easily the bond breaks homolytically

while acidity is a measure of how

easily the bond breaks heterolytically 14.

)]

Rank the following free radicals in order of increasing stability and give your reasoning: C6 H5 — C H — CH 3 , C6 H5 — CH — C H == CH 2 , C6 H5 CH 2 CH 2 C6 H5 CH 2 I

15.

16.

II

III

IV

Explain the following observations: (a) CH 3 is planar but CF3 is pyramidal. (b) (c)

Me3C — H bond is weaker than CH3 — H bond. Toluene undergoes free radical bromination at a rate faster than ethane.

(d)

Monobromination of (S)-3-methylhexane yields a racemic product.

Rank the following free radicals in order of increasing stability and give your reasoning: Benzyl, Vinyl, Methyl, Allyl

1.216

Organic Chemistry—A Modern Approach

17.

Mention the structure for A in the following reaction and classify the carbons in it by type.

18.

Electrolysis of an aqueous solution of CH3CH2COONa produces butane, ethane, ethylene and ethyl propanoate suggest a suitable mechanism to account for the formation of these compounds. [Hint:

19.

Which C—H bond in each of the following compounds is expected to be broken most readily during radical halogenation. Give your reasoning.

20.

Predict the major product obtained when 3-ethylpentane is heated with Br2. [Hint: The more substituted the carbon atom, the weaker the C—H bond. Therefore, the major bromination product of 3-ethylpentane will be obtained by the cleavage of the 3° C—H bond (its weakest C—H bond), i.e., the product is (CH3CH2)3CBr (3-bromo-3-ethylpentane).] The azo compound I decomposes 20 times faster than the azo compound II. Suggest a reason for this observation.

21.

Structure, Bonding and Proper es of Organic Molecules

22.

23.

24.

25.

Rank the following azo compounds in order of their expected rate of thermal decomposition to produce N2. Give your reasoning. (a) CH3—N== N—CH3 (b) (c)

PhCh2—N == N–CH2Ph

(e)

(CH3)3C—N== N—C (CH3)3

(d)

Which of the following bicyclic alkenes would you expect to undergo allylic bromination with N-bromosuccinimide (NBS)? Give your reasoning.

[Hint: A bridgehead radical is not much stable.] What is called a ‘captodative radical’? Give an example. [Hint: When a free radical is stabilized by the presence at the radical centre of both, an electron-donating and an electron-withdrawing group, that radical is called a ‘captodative radical’. For example:

Predict the major product of each of the following reactions and provide mechanism that accounts for its formation:

(a)

26.

1.217

;

(b)

The phenoxy radical I exists not as the dimer but as such both in solution and in the solid state. Explain this observation.

1.218

27.

Organic Chemistry—A Modern Approach

[Hint: The two bulky-CMe3 groups in the two ortho positions of one radical come in steric interaction (F-strain) with the two-CMe3 groups of another approaching radical and as a result, their combination to form a dimer is prevented.] The radical anion of benzophenone (Ph2C== O) is more stable than that of cyclohexanone

. Explain.

[Hint: The unpaired electron in the radical anion of benzo-phenone (Ph2 C — O@ ) is highly delocalized over the p orbital system of the two benzene rings as well as over the C == O system. But the unpaired electron in the radical anion of cyclohexanone is only slightly delocalized (only two resonance structures can be written).

] 28. 29. 30.

31.

1.11

A benzylic hydrogen is 6.4 × 10 times more reactive in bromination than a hydrogen atom in ethane. Explain. a-Naphthoxyl radical is more stable than b-naphthoxyl radical. Explain. A benzene solution of 2,4,6-tri-tert-butylphenoxy radical is decolorized in atmospheric oxygen in 30 min, whereas 2,6-di-tert-butyl-4-phenylphenoxy radical requires 8 hr. Explain. There is no enormous energy barrier against the formation of a bridgehead radical as there is against the formation of a carbocation. Explain. [Hint: Unlike carbocation a radical may have pyramidal shape.]

TAUTOMERISM

Tautomerism is a phenomenon in which two structural isomers undergo rapid interconversion and exists in dynamic equilibrium in the liquid state or in solution under normal laboratory conditions. The two forms or isomers which exist in dynamic equilibrium with each other are called tautomers of tautomerides. Toutomerism may be classified into cationotropy and anionotropy depending upon whether atoms or groups of atoms shift as cations or anions. A great majority of cases of cationotropy, however, is prototropy in which the migrating cation is a proton. The most important and classical example of prototropy is the keto-enol tautomerism and this can be represented as follows:

Structure, Bonding and Proper es of Organic Molecules :O:

|| | — C— C— | H Keto form

ææ Æ ¨æ æ

1.219

OH | | — C == C — Enol form

Some other examples of prototropy are listed below: (a) Nitro-acenitro tautomerism:

| (b) Amido-imidol tautomerism: —NH — C == O Amido form

| ææ Æ —N == C— OH ¨æ æ Imidol form

| | | | ææ Æ — C == C— NH2 (c) Imine-enamine tautomerism: — C H — C == NH ¨æ æ Imine form Enamine form | (d) Nitroso-oximino tautomerism: —C H — N == O Nitroso form

| ææ Æ —C == N —OH ¨æ æ Oximino form

H | (e) Azo-hydrazono tautomerism: —N == N — C — | Azo form

ææ Æ —NH — N == C— ¨æ æ | Hydrazono form

The essential structural need for tautomerism (prototropy) to be exhibited by compounds like carbonyl, nitro, nitroso, etc. is that the compound must contain at least one a-hydrogen atom. It thus follows that tautomerism is not possible in benzaldehyde (C6H5CHO), benzophenone (C6H5COC6H5), 2,2-dimethylpropanal (Me3CCHO), 2-methyl-2-nitro propane (Me3CNO2), etc. but possible in acetone (CH3CCH3), propanal (CH3CH2CHO), acetophenone (C6H5COCH3), nitroethane (CH3CH2NO2), etc.

(1)

Keto-enol tautomerism

Ethyl acetoacetate is the best studied classical example of keto-enol tautomerism: OH O O O || || | || ææ Æ CH 3 — C == CH — C — OC2H5 CH 3 — C — CH 2 — C — OC2H5 ¨æ æ Keto form of EAA Enol form of EAA

1.220

Organic Chemistry—A Modern Approach

The presence of the keto form is supported by the fact that the compound forms a bisulphite addition product with sodium bisulphite, a cyanohydrin with hydrogen cyanide, an oxime with hydroxylamine and a phenylhydrazone with phenylhydrazine.

The presence of the enol form in the equilibrium mixture is proved by the fact that ethyl acetoacetate gives an intense reddish violet colouration with ferric chloride solution | (this is characteristic of compounds containing the grouping: C == C — OH), it rapidly decolourizes bromine and reacts with diazomethane to form an ether.

Therefore, ethyl acetoacetate is an equilibrium mixture of two distinct substances, the keto form and the enol form. A convenient method of preparing pure samples of keto and enol forms was devised by Meyer, who placed the equilibrium mixture in a quartz (since glass has a catalytic effect on interconversion) distilling flask and separated the enol form, which is more volatile than the keto form, by distillation.

Structure, Bonding and Proper es of Organic Molecules

1.221

The estimation of the enol content of keto-enol mixture is based on the fact that the enol forms absorb bromine rapidly, whereas the keto form does not. Thus, if an equilibrium keto-enol mixture is titrated rapidly with bromine, at the end point (when the colour of Br2 is no longer discharged rapidly) all the enol form would have been converted to the bromo compound. The a-bromo compound is then reduced by adding HI, and the liberated iodine is titrated with thiosulphate. R R R | | | Br2 HI R — CH == C — OH æææ Æ R — CH — C == O æææ Æ R — CH — C == O + I2 2 fast | enol form Br

1.11.1

Mechanism of Keto-enol Tautomerism

Tautomerization, the process of converting one tautomeric form into another, is catalyzed by both acids and bases.

(1) Acid-catalyzed Tautomerism In an acidic solution, the carbonyl oxygen of the keto tautomer undergoes protonation in the first step. In the second step (the rate-determining step), water removes a proton from the a-carbon to form the enol.

(2) Base-catalyzed Tautomerism In a basic solution, the hydroxide ion abstracts a proton from the a-carbon atom of the keto tautomer to form a resonance-stabilized enolate ion in the first step (the rate-determining step). In the second step, oxygen of the enolate ion undergoes protonation to form the enol tautomer.

Protonation on the a-carbon reforms the keto tautomer. In addition to its mechanistic importance, keto-enol tautomerism affects the stereochemistry of carbonyl compounds. An a-hydrogen atom may be lost and regained through keto-enol tautomerism. Such a hydrogen is said to be enolizable. If an asymmetric carbon atom

1.222

Organic Chemistry—A Modern Approach

contains an enolizable hydrogen atom, a trace of acid or base allows that carbon to invert its configuration through the intermediate formation of an enol or enolate ion. A racemic mixture is the result. An optically active carbonyl compound will, therefore, be found to have lost its optical activity if it is left standing in the presence of acid or base. For example:

1.11.2

Difference between Resonance and Tautomerism Resonance

Tautomerism

1. Resonance involves delocalization of p electrons 1. Tautomerism involves movement of both electrons and atoms, and hence tautomers differ and hence the resonance structures (canonicals both in the position of electrons and atoms. For or contributing forms) differ only in the position example: of electrons. The position of atoms remains unchanged. For example:

2. Various resonance structures are only imaginary, 2. Tautomers are real molecules, which actually and do not exist. The real molecule is a hybrid of exist and can be isolated under suitable condithese structures. tions. 3. The resonance structures are not in dynamic 3. Tautomers exist in dynamic equilibrium with equilibrium as they are not real structures. each other as they are actual compounds. 4. Resonance leads to stabilization of the molecule. 4. Tautomerism does not lead to stabilization of the The actual molecule (i.e., the hybrid) is more molecule. stable than any resonance structure. 5. Resonance structures are shown by double 5. Tautomeric structures are shown by using equiheaded arrow (´). librium arrows ( or )

Structure, Bonding and Proper es of Organic Molecules

1.11.3

1.223

Position of the Tautomeric Equilibrium

Both the tautomers are not equally stable. The keto form is about 50 kJ/mol more stable than the enol form largely because a C == O bond is much stronger than a C == C bond and for this reason, under ordinary conditions mono-ketones do not exhibit much tendency to enolize. For example, an aqueous solution of acetone exists as an equilibrium mixture of more than 99.9% keto form and less than 0.1% enol form. O OH || | ææ Æ CH2 == C — CH3 Acetone: CH3 — C — CH3 ¨æ æ enol fom Keto form (<0.1%) (> 99.9%) The fraction of the enol form is considerably greater for a b-diketone because the enol tautomer is stabilized by intramolecular hydrogen bonding and by conjugation of the carbon–carbon double bond with the second C == O group.

The aforesaid difference in thermodynamic stabilities of the two tautomers is not the only factor that determines the position of equilibrium between them. The position of the equilibrium is significantly affected by other important factors such as follows:

(1) Entropy factor Although acyclic monoketones show negligible tendency to enolize, cyclomonoketones (e.g., cyclohexanone) enolize appreciably and this is because introduction of a C == C bond in enolization decreases the conformational freedom of the acyclic monoketone but no such extra restriction is imposed in the case of cyclic monoketones.

(2) Steric factor The enols of dicarbonyl compounds are generally stabilized by intermolecular hydrogen bonding. Therefore, any additional substituents on the C-atoms in the ring adversely affects ring formation by H-bonding due to steric strain. As a result, the enol content decreases.

(3) Nature of the solvent In the enol form of dicarbonyl compounds, the intramolecular hydrogen bond is quite strong and it is not considerably affected by solvents like water and ethanol. But the Keto form is well stabilized by these solvents due to hydrogen bonding. Therefore, such protic solvents tend to reduce the enol content. However, the nonpolar solvents like hexane, benzene, etc., which are unable to form hydrogen bond, tend to increase the enol content.

1.224

Organic Chemistry—A Modern Approach

The extent of enolization in some compounds may be explained as follows:

(a) Acetone: Acetone (CH3COCH3) exists almost exclusively in the keto form. OH O || | ææ Æ CH3 — C — CH3 ¨æ CH — C == CH2 æ 3 Keto form Enol form (> 99%) (1.5 ¥ 10-4 %) | | This keto-enol tautomerization actually involves inter conversion of a — CH — C == O | | | group and a — C == C(OH) –group. The sum of bond energies of the — CH — C == O group | is 347.9 kcal/mol and that of the — C == C(OH) –group is 336.0 kcal/mol. C — H (98.9 kcal/mol) + C — C (83.0 kcal/mol) + C == 0 (166.0 kcal/mol) = 347.9 kcal/mol C == C (145.0 kcal/mol) + C — 0 (80.0 kcal/mol) + O — H (111.0 kcal/mol) = 336.0 kcal/mol | The enthalpy of the formation of the groups CH — C == O and C== C(OH)– are, therefore, 347.9 and 336.0 kcal/mol, respectively. Thus, the keto tautomer is thermodynamically more stable than the enol tautomer by 11.9 kcal/mol and for this reason, at equilibrium acetone exists almost exclusively in the keto form.

(b) Pentane-2,4-dione or acetylacetone: The enol content of pentane-2, 4-dione or acetylacetone (CH3COCH2COCH3) at equilibrium is very high (76%).

Because of extended conjugation, the enol form of pentane-2,4-dione is stabilized by resonance. Also, the enol is significantly stabilized by intramolecular hydrogen bonding. For these reasons, the enol content of pentane-2,4-dione at equilibrium is very high.

(c) Ethyl acetoacetate: The enol content of ethyl acetoacetate (CH3COCH2CO2Et) at equilibrium is very small (8 percent).

Structure, Bonding and Proper es of Organic Molecules

1.225

Since the ester carbonyl group is involved in resonance interaction with — OR oxygen ( ) it is less effective than the ketone carbonyl group in stabilizing the enol by resonance. Also, due to lack of resonance, the O — H hydrogen is not appreciably positive and intramolecular hydrogen bond is relatively weak. For these reasons, the amount of enol present at equilibrium is very small in ethyl acetoacetate.

(d) 3,4-Hexanedione: 3,4-Hexanedione (CH3CH2COCOCH2CH3) exists almost exclusively in the keto form.Conformation plays a dominant role in deciding the extent of enolization in the a-diketone. In this compound, the two C == O groups can rotate around the single bond. Therefore, to avoid repulsive interaction between two negatively polarized oxygen atoms, the two C == O groups take up an anti-conformation in which the two electronegative oxygen atoms are as far as possible from each other and in which the carbonyl dipoles are opposed. As a consequence, the formation of intramolecularly hydrogen bonded as well as conjugated enol through the less stable syn-conformation is not at all favoured. The equilibrium, therefore, lies almost exclusively over in favour of the keto form.

(e) 1,2-Cyclohexanedione 1, 2-Cyclohexanedione exists almost exclusively in the enol form. 1,2-Cyclohexanedione:

In this diketone, the two carbonyl groups are syn oriented. The energy of the molecule d+

d-

rises due to repulsion between these two polar >C == O groups. Its energy cannot be lowered by rotating the C == O dipoles, i.e., by orienting them anti to each other because such conformation would give an impossibly strained ring. However, the energy content of the molecule can be lowered by the formation of an enol because enolization eliminates one of these two polar groups. Also, the enol is stabilized by intramolecular hydrogen bonding and conjugative effect. For these reasons, the equilibrium lies essentially completely over in favour of the enol form.

1.226

Organic Chemistry—A Modern Approach

(f) Cyclohexa-2, 4-dien-1-one: This unsaturated ketone exists totally in its enol form which is known as phenol.

Phenol is an aromatic compound and it is, therefore, stabilized by the resonance energy of the benzene ring. In fact, phenol is thermodynamically much more stable than its keto form cyclohexa-2,4-dien-1-one (DH ª 20 kcal/mol) and for this reason, the equilibrium lies exclusively over in favour of the enol form, i.e., phenol.

(g) (Me3CO)3CH: This triketone exists exclusively in the keto form. This is due to the fact that the enol form, which suffers from severe steric strain, is energetically very much less stable than the keto form.

(h) 2,2-Dimesitylethanal: The enol content of 2,2-dimesityl-ethanal at equilibrium is very high.

Structure, Bonding and Proper es of Organic Molecules

1.227

In the enol form of this aldehyde, the two very large mesityl groups are about 120° apart, but in the keto form they are about 109.5° apart. So, unlike in the enol form, they remain much closer together in the keto form and become involved in severe steric interaction. Therefore, the keto form is thermodynamically less stable than the enol form and as a consequence, the percentage of the enol form at equilibrium is very high.

(i) 1,2-Diphenyl-1-ethanone: The enol content of this ketone at equilibrium is very high.

Enolization, by creating a double bond, links up the two phenyl groups so as to form a completely conjugated system which is highly stabilized by resonance. For this reason, the enol content of PhCOCH2Ph at equilibrium is very high.

(j) 1,3,5-Trihydroxybenzene: 1,3,5-Trihydroxybezene shows more ketonic activity than 1,3-dihydroxybenzene which in turn shows more ketonic activity than phenol. Phenol:

1,3-Dihydroxybenzene or resorcinol:

1,3,5-Trihydroxybenzene or phloroglucinol:

As the number of — OH groups in a benzene ring increases, the difference in stability between the enol and the keto form progressively decreases because the aromatic stability of only one ring becomes progressively less able to make the enol form very stable. As a

1.228

Organic Chemistry—A Modern Approach

consequence, the keto form is progressively more favoured. And, for this reason, 1,3,5trihydroxybenzene shows more ketonic activity than 1,3-dehydroxybenzene which in turn shows more ketonic activity than phenol.

1.11.4

Ring-chain Tautomerism

The type of tautomerism in which one tautomer is acyclic and the other is cyclic is denoted as ring-chain tautomerism. These ring and chain forms of tautomers are possible where one functional group of a bifunctional acyclic molecule reacts with other and forms a cyclic system. Sugars like D-glucose, D-mannose, etc. exhibit ring-chain tautomerism. In these cases, the open chain aldehydic form is converted into the cyclic hemiacetal form by intramolecular reaction between the alcoholic–OH group and the —CHO group.

1.11.5

Valence Tautomerism

Tautomerism in which there occurs change in interatomic distances through the formation of new bonds by redistribution of valence electrons within a molecule without the migration of any atom from the rest in any intermediate step is termed as valence tautomerism. For example, cyclooctatetraene and bicyclooctatriene represent a pair of valence tautomers.

1.

When (R)-3-phenyl-3-butanone is dissolved in ethanol containing HCl, the optical rotation of the solution gradually drops to zero. Explain why this is so.

Solution The asymmetric carbon atom of (R)-3-phenyl-2-butanone contains an enolizable hydrogen atom. When this compound is dissolved in ethanol containing hydrochloric acid,

Structure, Bonding and Proper es of Organic Molecules

1.229

it undergoes acid-catalyzed racemization through the intermediate formation of an achiral enol. Because of racemization, the optical rotation of the solution gradually drops to zero. The process may be shown as follows:

2.

Explain why the enol content of 2,6-bicyclo[2.2.1]heptanedione is very small, even though it is a b-diketone.

Solution Although the bridgehead enol is stabilized by resonance involving the C == O group, the compound actually shows no tendency to form a bridgehead enol and this is because it is impossible for this rigid bicyclic system to place a double bond at the bridgehead position (Bredts’s rule). The compound, therefore, enlizes in another direction. Since the resulting enol is not stabilized by resonance involving C == O, the enol content of this diketone at equilibrium is very small indeed.

1.230

3.

Organic Chemistry—A Modern Approach

The following ketone does not undergo base-catalyzed exchange in D2O, even though it has an a-hydrogen. Explain this observation.

Solution The ketone does not react with alkali to form the corresponding enolate ion because the enolate ion is not stabilized by resonance which due to the presence of an unstable structure containing a double bond at the bridgehead position is practically insignificant. For this reason, this bicyclic ketone does not undergo base-catalyzed exchange in D2O through intermediate formation of an enolate ion.

4.

1,3,5-Trihydroxybenzene reacts with hydroxylamine (NH2OH) to give a trioxime. However, 1,2,3-and 1,2,4-trihydroxybenzenes do not react with NH2OH. Account for these observations.

Solution 1,2,3- and 1,2,4-Trihydroxybenzenes do not react with NH2OH because the triketo tautomers of these two trihydroxybenzenes are destabilized by the adjacent C == O groups (repulsive destabilization caused by similar charges) and have an insufficient concentration to react with NH2OH. The triketo tautomer of 1,3,5-trihydroxybenzene suffers from no such destabilization because the C == O groups are not adjacent. For this reason, the concentration of the triketo tautomer is sufficient to react with NH2OH to form a trioxime.

Structure, Bonding and Proper es of Organic Molecules

5.

1.231

2-Ethyl-3-methylcyclobuta-2-en-1-one exists exclusively in the keto form. Explain.

Solution The enol form of 2-ethyl-3-methylcyclobuta-2-en-1-one is very unstable because it is an antiaromatic system (a planar closed loop of 4np electron system, where n = 1). For this reason, the compound exists exclusively in the keto form.

2-Ethyl-3-methylcyclobuta-2en-1-one:

6.

The enol form of 9-formylfluorene is favoured at equilibrium — Why?

Solution The enol form of 9-formylfluorene is favoured at equilibrium because the enol form is considerably stabilized by resonance involving two benzene rings and one aromatic cyclopentadienyl system [(4n + 2)p electrons, where n = 1].

1.232

7.

Organic Chemistry—A Modern Approach

Explain why the following compound exists exclusively in the keto form:

Solution The enol form of the this cyclic diketone is a close loop of 10 p electrons [(4n + 2) p, where n = 2]. So, the enol is expected to be an aromatic system of much higher stability.

Although the enol has no angle strain, the hydrogens at C–5 and C–10 interfere sterically with each other and force the molecule out of planarity. However, to achieve aromatic stability, the system should adapt a planar shape. So, the enol does not possess aromatic stability and, therefore, the keto form is much more stable than the enol form and the diketone exists exclusively in the keto form.

8.

Explain why cis-5-tert-butyl-2-methlcyclohexanone undergoes isomerization to yield the trans-isomer in the presence of a base.

Solution In the cis-isomer of 5-tert-butyl-2-methyl-cyclohexanone, the bulky–CMe3 group locks the molecule in that particular conformation in which this group is placed in an equatorial position and the methyl group is placed in an axial position. However, this conformation with the — CH3 group axial suffers from syn-axial interactions between the —CH3 group and H’s at C-4 and C-6. On the other hand, there is no such steric strain in the trans-isomer because the —CH3 group is equatorial. Again, the —CH3 group is too far to experience any appreciable steric interaction with C == O and also eclipsing of —CH3 and C == O is electronically favourable. The trans-isomer is, therefore, relatively

Structure, Bonding and Proper es of Organic Molecules

1.233

more stable than the cis-isomer and because of this, when treated with a base, the cisisomer undergoes isomerization to yield the trans-isomer through the formation of an intermediate enolate ion.

9.

The enol content of 3,3,3-trifluoropropanal is larger than that of propanal. Explain.

Solution The enol of 3,3,3-trfluoropropanal is stabilized by intramolecular hydrogen bonding involving a stable six-membered ring (chelation) while the enol of propanal (CH3CH2CHO) is not similarly stabilized. Because of this, the enol content of the former aldehyde is larger than that of the latter.

10.

At equilibrium the enol content of 3,5-heptanedione (CH3CH2COCH2COCH2CH3) is larger than that of 4-methyl-3,5-heptanedione [CH3CH2COCH(CH3)COCH2CH3]. Account for this observation.

Solution In the enol form of 4-methyl-3,5-heptanedione, the —CH3 group at C–4 becomes involved in steric interaction with the adjacent —C2H5 groups. This steric strain also makes the internal hydrogen bonding weak. No such steric strain is involved in the enol

1.234

Organic Chemistry—A Modern Approach

form of 3,5-heptanedione because hydrogen is relatively smaller in size. So, the former enol is relatively less stable than the latter enol and hence at equilibrium the percentage of the former enol is less than the percentage of the latter enol.

11.

The enol content of 2,4-pentanedione or acetylacetone at equilibrium is only 15 percent in water, 58 percent in acetonitrile (CH3CN) and 92 percent in hexane (C6H14). Account for these observations.

Solution The keto form of a b -diketone like 2,4-pentanedione is more polar than the enol form because intramolecular hydrogen bonding and resonance involving the other carbonyl group results in a decrease in polar character of the enol. Therefore, the enol content is expected to increase with decrease in solvent polarity. Hexane (CH3CH2CH2CH2CH2CH3) with a dielectric constant 1.9 is a nonpolar solvent, acetonitrile (CH3CN) with a dielectric constant 38 is an aprotic polar solvent and water with a dielectric constant 79 is a protic and much polar solvent. So, the enol content in hexane is greater than in acetonitrile and the enol content in acetonitrile is greater than in water. The percentage of enol drops dramatically in water because water forms hydrogen bonds with C == O’s, thereby inhibiting the intramolecular hydrogen bonding that helps stabilize the enol.

12.

Account for the given trend in enol content for each of the following reactive methylene compounds: EtO2C — CH2 — CO2Et(<1 percent); CH3COCH2CO2Et(7.7 percent); CH3COCH2COCH3 (76 percent); PhCOCH2COCH3 (89 percent)

Solution Cross-conjugation (delocalization of a pair of nonbonding electrons on the oxygen of the –OR group of the ester onto the carbonyl oxygen) retards the ability of the

Structure, Bonding and Proper es of Organic Molecules

1.235

C = O of the ester group to help stabilize the enol. It thus follows that the enol content will decrease with increase in the number of ester group. Therefore, diethyl malonate (I) with two ester groups has a lower enol content than ethyl accetoacetate (II) with one ester group which in turn has a lower enol content than acetylacetone (III) containing no ester group. The enol content of 1-phenyl-1,3-butanedione (IV) is somewhat higher than that of acetylacetone (III) because the –Ph group stabilizes the enol by extending the conjugation for p-bond delocalization.

13.

Which one of the following two carbonyl compounds will enolize and why?

Solution The compound I will not enolize because its enol being a flat ring containing 4n(n = 1)p electrons is an antiaromatic system and hence very unstable. On the other hand, the compound II will enolize readily because its enol being a flat ring containing a close loop of (4n + 2)p electrons, where n = 0, is an an aromatic system and hence very stable.

14.

When cis-1-decalone is treated with sodium ethoxide in ethanol, it undergoes epimerization to yield trans-1-decalone. Explain this observation.

1.236

Organic Chemistry—A Modern Approach

Solution This epimerization occurs because trans-1-decalone in which no bulky groups are axial is relatively more stable than cis-1-decalone in which a large group is axial.

1.

2. 3.

Which compound in each of the following pairs has the greater enol content and why? (a)

(b)

(c) CH3CH2COCH3 or CF3CH2COCH3

(d) Ph CH2 COCH3 or CH3CH2COCH3

(e)

(f)

Compare the extents of enol content of 2,4-pentanedione (CH3COCH2COCH3) in water and benzene. Explain why the compound A exists exclusively in the enol form while the compound B exists exclusively in the keto form

Structure, Bonding and Proper es of Organic Molecules

4.

Label the following pairs of structures as tautomers or resonance structures: (a)

5.

1.237

≈

≈

CH3CH2CHOCH3 and CH3CH2CH = OCH3

(b)

(c)

(d)

O OH || | (e) C6 H5 — C — NH2 and C6 H5 —C == NH

Write down the structures of all possible enol tautomers of acetylacetone (CH3COCH2COCH3) and justify their relative stabilities. Write down the s-cis and s-trans forms of one of the possible tautomers. [Hint: The possible enol tautomers of acetylacetone are as follows:

The enol II is less stable than the enol I because in it the enol double bond is not conjugated with C == O and hence it is not stabilized by resonance. The s-cis and s-trans conformations of I are as follows:

6. 7.

8.

State and explain which of CH3COCH2COCH3 and CH3COCHPhCOCH3 has got higher enol content in tetrahydrofuran. Which of the following has higher enol content? Give reasons for your answer.

Which of the following pair of compounds would predominantly remain in the enol form? Give your reasoning. (a)

CH3COCOCH3 and CH3COCH2COCH3

(b)

1.238

9.

Organic Chemistry—A Modern Approach

Arrange the following compounds in order of increasing enol content. Give reasons. CH 3COCH 2 COCH 3 ; CH 3 COCH 2CH 2COCH 3 ; CH 3COCH 2CO2Et I

10. 11.

12.

II

III

[Hint: II < III < I] Ethyl acetoacetale exists in the enol form much more in hexane than in water — Why? Explain why the following two structures are tautomers but not resonance forms: CH 3 — C == CH 2 and CH 3 — C — CH 3 || | O OH Draw the enol tautomers for each of the following compounds. For those compounds that have more than one enol tautomer, indicate which one is more stable.

(a) (d)

(b)

(c)

—CH2COCH3 (e) CH3CH2COCH2COCH2CH3

(f) CH3COCH2CO2Et

[Hint: (a)

(b) OH

(c)

CH3

(more stable);

(d)

(e)

OH OH | | CH 3 CH 2COCH == CCH 2CH 3 (more stable); CH 3CH == CCH 2COCH 2CH 3 (less stable)

(f)

OH OH | | CH 3 — C == CH — CO2Et (more stable) CH 2 == C CH 2CO2Et (less stable) ]

13.

Would optically active ketones like those given below undergo acid- or basecatalyzed racemization? Explain your answer.

Structure, Bonding and Proper es of Organic Molecules

1.239

[Hint: In the ketone I, these is no hydrogen attached to its a-carbon atom (which is a chirality centre) and hence enol formation involving the chirality centre is not possible. In the ketone II, the chirality centre is a b- carbon atom and hence it remains unaffected by enol formation. Therefore, none of these two ketones undergo acid- or base-catalyzed racemization.] 14.

Explain the following observation: CH3 O CH3 O | || | || @ OD /D2O H5C2 — CH — C — C6 H5 æææææ ≈ Æ H 5C2 — CD — C — C6 H 5 or D3O

(±)

15.

(±)

Write down the resonance structures and tautomers of the following compound: O HN O

NH N H

O

[Hint: Resonance structures:

Tautomers:

16.

Explain the position of the following equilibrium:

[Hint: Ring aromaticity usually stabilizes the phenolic tautomer. However, in this case the reverse is true. Since anthrone has two distinct intact aromatic rings in addition to the C == O, it is relatively more stable than anthranol, which suffers a loss of aromaticity per ring because its three rings are fused.]

1.240

17.

Organic Chemistry—A Modern Approach

Draw a stepwise mechanism for each of the following reactions: –

(a)

(b)

OH

OH H2O

== O

18.

Draw the enol intermediate and the ketone product formed in the following reaction:

19. 20.

Draw all of the enol forms of 2-butanone. Which of them is the least stable and why? Give a curved-arrow mechanism for the exchange of a-hydrogens for deuterium:

21.

(a)

(b)

1.12

Draw the structure of the conjugate base of each of the following compounds. What is the relationship between the two conjugate bases? O OH H H I Which compound is more acidic and why?

II

AROMATICITY

Cyclic compounds like benzene and those which resemble benzene in chemical behaviour have unusually large resonance energies and hence possess unusual stability. These compounds are called aromatic compounds and the quality which renders these compounds especially stable is referred to as aromaticity.

1.12.1

Criteria for Aromaticity

For a compound to be classified as aromatic, it must fulfil the following criteria: (1) The compound must have an uninterrupted cyclic cloud of p electron above and below the plane of the molecule (often called a p cloud). To form a cyclic p cloud, the structure of the molecule must be cyclic. To form an uninterrupted p cloud, each atom in the ring must have a p orbital, i.e., the molecule must be completely conjugated. And, to form a p cloud, each p orbital must be able to overlap with the p orbitals on either side of it, i.e., the molecule must be planar. (2) An aromatic compound must have an odd number of pairs of p electrons, i.e., the uninterrupted p cloud must contain (4n + 2) delocalized p electrons, where n = 0, 1, 2, etc., (this is what is called Hückel’s rule of the 4n + 2 rule) and therefore, all bonding molecular orbitals (and HOMOs) must be completely filled.

Structure, Bonding and Proper es of Organic Molecules

(3)

Delocalization of the p electrons over the ring must lower the electronic energy of the species, i.e., the species must be stabilized by delocalization. Therefore, planar, cyclic and completely conjugated compounds containing (4n + 2) p electrons, where n = 0, 1, 2, 3 and so on (i.e., rings containing 2, 6, 10, 14,…, etc., p electrons) are unusually stabilized by electron delocalization and are said to be aromatic. Benzene is the best known example of compounds possessing aromatic character, since this molecule is a planar, cyclic and completely conjugated system containing six p electrons [(4n + 2)p electrons, where n = 1]. All of its bonding molecular orbitals (and HOMOs) are completely filled.

Some other examples of aromatic systems: (1)

(2)

(3)

1.241

1.242

Organic Chemistry—A Modern Approach

(4)

1.12.2

Antiaromatic Compounds

Planar, cyclic and completely conjugated compounds containing 4np electrons, where n = 1, 2, 3, ... etc., are especially unstable (destabilized by delocalization). These are called antiaromatic compounds. Cyclobutadiene or [4]annulene is an example of antiaromatic compound, since this molecule is a planar, cyclic and completely conjugated one containing four p electrons (i.e., 4np electrons, where n = 1). It is so highly reactive that it can only be prepared at extremely low temperatures.

Some other examples of antiaromatic systems: (1)

Cyclopropenyl anion. A 4np electron system, where n = 1. The system is destabilized by resonance. In fact, the unconjugated conjugated

(2)

is more stable than

.

Cyclopentadienyl cation. A 4np electron system, where n = 1. The system is destabilized by electron delocalization. In fact, the unconjugated stable than conjugated

(3)

is more

.

Cycloheptatrienyl anion (planar). A 4np electron system where n = 2. The system is destabilized by resonance. The unconjugated conjugated

1.12.3

is more stable than

.

Nonaromatic Compounds

Compounds that lack one (or more) of the criteria fulfilled by aromatic or antiaromatic compounds are called nonaromatic compounds or simply the compounds which are neither aromatic or antiaromatic are called nonaromatic compounds. 1,3,5-Cycloheptatriene, for example, is a nonaromatic compound because the ring is not completely conjugated.

Structure, Bonding and Proper es of Organic Molecules

1.243

A (4n + 2)p electron system that has a continuous ring of p orbitals may also be monaromatic it overlapping of orbitals is inhibited due to nonplanarity.

1.12.4

Classification of Compounds as Aromatic, Antiaromatic and Nonaromatic by Comparing their Stabilities with that of the Corresponding Open Chain Compounds

If the ring is more stable (lower p-electron energy) than the open chain counterpart, the compound is classified as being aromatic. Benzene, for example, is aromatic because it is more stable than 1,3,5-hexatriene.

If the ring is less stable (higher p-electron energy) than the open-chain counterpart, the compound is classified as being antiaromatic. Cyclobutadiene, for example, is antiaromatic because it is less stable than 1,3-butadiene.

Cyclobutadiene (less stable: antiaromatic)

1,3-Butadiene (more stable)

And, if the ring and its open-chain counterpart have the same stability (same p-electron energy), the compound is classified as being nonaromatic. For example, 1,3-cyclohexadiene is similar in stability to cis, cis-2, 4-hexadiene.

1.12.5

Modern Definition of Aromaticity

The modern definition of aromaticity is the ability to sustain an induced ring current by a planar, cyclic and conjugated system with (4n + 2) delocalized p electrons, where n = 0, 1, 2, 3, ... etc. An applied magnetic field perpendicular to a conjugated and planar

1.244

Organic Chemistry—A Modern Approach

ring containing (4n + 2)p electrons will cause the p electrons to circulate around the ring and thereby generating their own magnetic field. This circulation is called induced ring current.

1.12.6

Molecular Orbital Energy Diagram of Some Ions and Molecules

The molecular orbital energy diagram of cyclopropenyl anion (I), cyclopropenyl cation (II), cyclobutadiene (III), cyclopentadienyl anion (IV), cyclopentadienyl cation (V), benzene (VI), cycloheptatrienyl cation (VII) and cyclooctatetraenyl dianion (VIII) may be given as follows: ≠ ≠

——

—

—

——

— Antibonding

——

——

——

——

—— Nonbonding

≠ ≠ ≠ E

≠Ø ≠Ø

≠ ≠

≠Ø ≠Ø

≠Ø ≠Ø

≠Ø

≠Ø

≠Ø

≠Ø

≠Ø ≠Ø ≠Ø ≠Ø

≠Ø

≠Ø

≠Ø

Bonding

≠Ø

–

–

–

+

I

II

III

IV

–

+

V

VI

VIII

The solution of Schrodinger wave equation for the p electron system of any cyclic, planar and conjugated polyene reveals that energies of the molecular orbitals in such systems have a very simple characteristic pattern. In such systems, there is always one orbital of lowest energy which is followed by degenerate pairs of orbitals (i.e., orbitals possessing same energy) in the order of increasing energy. The first pair of electrons is filled in the lowest energy molecular orbital and then the rest of the electrons go to degenerate pairs of orbitals according to Hund’s rule. In a 4np electron system, the orbital of lowest energy will be filled by two electrons. After filling the remaining orbitals according to Hund’s rule,

Structure, Bonding and Proper es of Organic Molecules

1.245

there will always be two singly occupied degenerate orbitals in these systems. Because of such diradical structure, these systems are very unstable and possess higher energy than their open chain analogies or the hypothetical cyclic structures where delocalization does not occur, i.e., these systems are antiaromatic. In a (4n + 2)p electron system, the lowest orbital will be filled by a pair of electron and each of the degenerate pairs of orbitals will be filled by four electrons. Thus, in such systems all the occupied bonding orbitals are doubly filled. This is an energetically favourable configuration. Because of this closed shell filling of orbitals with electrons, such systems are unusually stable, i.e., aromatic. It thus follows that a compound is aromatic if all bonding MOs ( and HOMOs) are completely filled and no p electrons occupy antibonding MOs. On the other hand, a compound is antiaromatic if it has electrons in antibonding molecular orbitals or if it has half-filled bonding or nonbonding molecular orbitals.

1.12.7

Use of Inscribed Polygon Method to Determine the Relative Energies of p Molecular Orbitals for Cyclic Planar and Completely Conjugated Compounds and to Classify Them as Aromatic and Antiaromatic

In the inscribed polygon method, a regular polygon is drawn inside a circle with its vertices touching the circle and one of the vertices pointing down. The total number of corners is equal to the number of p molecular orbitals. The points where the corners of the polygon touch the circle correspond to the energy levels of the p molecular orbitals of the system. A line is then drawn horizontally through the centre of the circle and the molecular orbitals are labelled as bonding, nonbonding and antibonding. Corners below the horizontal diameter are bonding (p), those above are antibonding (p*) and those on the diameter are nonbonding (pn). It is assumed that all the systems are planar. The method may be applied to 3–, 4–, 5–, 6–, 7–, and 8-carbon systems and from the character of the p molecular orbitals the systems can be classified as aromatic and antiaromatic as follows:

1.246

Organic Chemistry—A Modern Approach

Structure, Bonding and Proper es of Organic Molecules

1.12.8

1.247

Classification of Some Molecules and Ions as Aromatic, Antiaromatic and Nonaromatic

(a)

The compound is aromatic because it has a continuous ring of overlapping p orbitals (boron contains a vacant p orbital) and it has a 6p electron [(4n + 2), p, where, n = 1] electron system.

(b)

The compound is nonaromatic because it does not have a continuous ring of overlapping p orbitals (there is a saturated carbon containing no p orbital), even though it contains 6p electrons.

(c)

Since boron contains a vacant p orbital, therefore, the compound has a continuous ring of overlapping p orbitals. However, it has 4p (4np, n = 1) electrons. Therefore, the compound is antiaromatic,

B H

(d)

Oxygen is sp2-hybridized and hence the compound has a continuous ring of overlapping p orbitals. One unshared pair of electrons on oxygen is used to complete the aromatic sextet, i.e., it has a 6p[(4n + 2)p, n = 1] electron system. Therefore, the compound is aromatic.

1.248

Organic Chemistry—A Modern Approach

(e)

Due to the formation of dative p bond between boron and oxygen, the molecule possesses a continuous ring of overlapping p orbitals and the system contains 6p[4n + 2)p, where n = 1) electrons. Therefore, this planar molecule is aromatic.

(f)

The molecule contains 10p [(4n + 2)p, n = 2] electrons. However, the molecule is nonplanar because the two hydrogens pointing to the interior of the ring interfere sterically and for this reason effective overlapping of p orbitals does not take place. Hence the molecule is nonaromatic.

(g)

The compound is aromatic because it is a planar close loop of 10p [4n + 2)p, n = 2] electrons. Two interfering H’s are removed to form a bridge.

(h)

The ion is aromatic because it has a continuous ring of overlapping p orbitals and has a 6p[(4n + 2) p, n = 1] electron system. Pyridine takes ≈

up a proton to form its conjugate acid, the pyridinium ion (C6 H5 N H) . The additional proton has no effect on the electrons of the aromatic

Structure, Bonding and Proper es of Organic Molecules

1.249

sextet. It simply bonds to the nonbonding pair of electrons present in an sp2 orbital of nitrogen.

(i)

It is a planar, cyclic and completely conjugated system containing 6p [(4n + 2)p, n = 1] electrons and therefore, it is aromatic. The uncharged nitrogen atom is sp2 hybridized, with a lone pair of electrons in the p orbital. This p orbital overlaps with the adjacent p orbitals to form a continuous ring.

(j)

The sp3-hybridized N atom is bonded to four atoms (two carbon atoms and two hydrogen atoms) and hence it contains no unhybridized p orbital. Therefore, the ion does not have a continuous ring of overlapping p orbitals and so, it is nonaromatic.

+

N H

(k)

H

N The compound is aromatic because it is a planar, cyclic and completely N conjugated 10p [(4n + 2)p, n = 2] electron system. The unshared electron pairs on

Ns do not interfere to make the system nonplanar. Ph

Ph +

(l)

The ion is aromatic because it is a planar, cyclic and completely conjugated system containing 2p[4n + 2)p, n = 0] electrons.

+

Ph

Ph

Ph

Ph + +

Ph

Ph

Ph Ph

´ Ph

´ Ph

+

+

Ph +

Ph

+

+

Ph

Ph ∫

´

Ph ++

+

Ph Ph

Ph

Ph

Stabilized by electron delocalization

Ph

Ph

Ph Aromatic

1.250

Organic Chemistry—A Modern Approach

Ph

–

–

Ph

(m) Ph

Ph

(n)

CH3 ==

N

The ion is aromatic because it is a planar, cyclic and completely conjugated system containing 6p[(4n + 2) p, n = 1] electrons.

The molecule can be represented as a resonance hybrid. The resonance structure II has a major contribution because it is a combination of two aromatic ring systems: cyclopentadienyl anion and pyridinium ion. Therefore, the molecule is aromatic.

== I

(o)

(p)

+

´

–

II

N

+

´ etc. ∫

N

–

Aromatic

The molecule can be represented as a resonance hybrid. The resonance structures II and III have equal and major contribution because in these two structures all the three rings are aromatic: cyclopentadienyl anion, benzene and cycloheptatrienyl cation. Therefore, the molecule is aromatic.

O +

(q)

N

CH3

CH3

CH3

Since the more stable and more contributing resonance structure II of this ion is a planar, cyclic and completely conjugated system containing 6 p [(4n +2)p, n = 1] electrons, the ion is aromatic.

The compound is antiaromatic because it is a planar, cyclic and completely conjugated system containing 8p (4np, n = 2) electrons.

Structure, Bonding and Proper es of Organic Molecules

1.251

(r)

Although the species is planar and it has a continuous ring of overlapping p orbitals, it contains 5p electrons. Therefore, it is nonaromatic.

(s)

Although the ion contains 6p [(4n + 2)p, n = 1] electrons, it does not have a continuous ring of overlapping p orbitals (one sp3-hybridized carbon has no p orbital). Therefore, this ion is nonaromatic.

1.12.9

Homoaromatic Compounds

A homoaromatic compound is a compound that contains one or more sp3-hybridized carbon atoms in an otherwise conjugated cycle [a(4n + 2)p electron system]. When cyclcooctatetraene is dissolved in concentrated sulphuric acid, a proton adds to one of the double bonds to form the homotropylium ion. In this species, an aromatic sextet [(4n + 2)p electrons, where n = 1] is spread over seven carbons, as in tropylium ion. The eighth carbon is an sp3-hybridized one and cannot take part in aromaticity. This ion is an example of homoaromatic compound. In order for the p orbitals to overlap most effectively so as to close a loop, the sp3 carbon is forced to lie almost vertically above the plane of the aromatic atoms.

1.12.10 Some Chemical and Physical Consequences of Aromaticity (1)

Cyclopentadiene (Ka = 10–15) is much more acidic than cycloheptatriene (Ka = 10–45). The cyclopentadienyl anion, the conjugate base of cyclopentadiene, can be represented as a hybrid of five equivalent resonance structures while the cycloheptatrienyl anion, the conjugate base of cycloheptatriene, can be represented as a hybrid of seven equivalent resonance structures.

1.252

(2)

Organic Chemistry—A Modern Approach

Although a greater number of resonance structures can be drawn for the cycloheptatrienyl anion, it is less stable than the cyclopentadienyl anion and this is because the former anion containing 8p (4np, n = 2) electrons is antiaromatic (destabilized by resonance) while the latter anion containing 6p [(4n + 2)p, n = 1] electrons is aromatic (highly stabilized by resonance). Due to the formation of a relatively stable conjugate base, cyclopentadiene is a much stronger acid than cycloheptatriene. In hydrogen-exchange reactions, the nitrile A loses its proton at a very much slower rate than the nitrile B.

Loss of proton from the compound A leads to the formation of the antiaromatic (4np electron, where n = 1) and hence very unstable cyclopropenyl anion while loss of proton from the compound B leads to the formation of a relatively stable ordinary carbanion. For this reason, the compound A loses proton at a very slower rate than the compound B in hydrogen exchange reactions.

==

Croconic acid is almost strong an acid as H2SO4. O O O

==

HO

==

(3)

OH

Croconic acid

Structure, Bonding and Proper es of Organic Molecules

1.253

The dianionic conjugate base of croconic acid is aromatic and much stable because it is a planar, cyclic and completely conjugated system containing 2p [4n + 2)p, n = 0] electrons and for this reason, the compound is almost as strong as acid such as H2SO4.

(4)

The carbonyl oxygen of 4-pyrone is more basic than the ring oxygen and the compound does not form an oxime or phenylhydrazone. Protonation of the carbonyl oxygen of 4-pyrone leads to the formation of a stable aromatic system [a(4n + 2)p electron system, where n = 1] while protonation of the ring oxygen produces a nonaromation less stable ion with a localized positive charge. For this reason, the carbonyl oxygen is more basic than the ring oxygen of 4-pyrone.

1.254

Organic Chemistry—A Modern Approach

Because of aromatic character of 4-pyrone, its carbonyl reactivity (positive character of the carbonyl carbon) reduces and for this reaction, the compound does not form an oxime or phenylhydrazone.

(5)

The amine B is more basic than the amine A.

The unshared pair of electrons on nitrogen in the amine A is well delocalized because the charge-separated structures constitute a stable aromatic system [cyclopentadienyl anion with (4n + 2)p electrons, where n = 1]. On the other hand, the unshared electron pair on nitrogen in the amine B is not at all delocalized by resonance because the charge separated structures constitute an unstable antiaromatic system (cycloheptatrienyl anion containing 4np electrons, where n = 2). Therefore, the availability of the lone pair of electrons on nitrogen in B is greater than that in A. Hence B is more basic than A.

Structure, Bonding and Proper es of Organic Molecules

(6)

1.255

Pyrrole is a much weaker base than pyridine.

In pyrrole, the unshared pair of electrons on nitrogen is used to complete the aromatic sextet [a(4n + 2)p electron system, where n = 1] and so these are not readily available for coordinating with a proton.

In pyridine, on the other hand, the nonbonding electrons on nitrogen are not used to complete the aromatic sextet and so, these are readily available for coordinating with a proton.

Pyrrole is, therefore, a much weaker base than pyridine.

1.256

==

The hydrocarbon A is much more acidic than the hydrocarbon B. ==

(7)

Organic Chemistry—A Modern Approach

==

A B The conjugate base of B is resonance-stabilized because the unshared electron pair on the negative carbon is in proper conjugation with the three double bonds. The conjugate base of A is also resonance-stabilized. However, because of constituting an aromatic system (a close loop of 6p electrons), it is much more stable than the conjugate base of B. For this reason, the hydrocarbon A is much more acidic than the hydrocarbon B.

(8)

When the compound I is treated with silver perchlorate in propanoic acid, it undergoes solvolysis while the compound II does not. I

I

I II The reaction proceeds through the rate-determining formation of an intermediate carbocation. The carbocation expected to be obtained from I in relatively stable 2° carbocation while the carbocation expected to be obtained from II is a very much unstable carbocation (cyclpentadienyl cation) which is antiaromatic [a planar

Structure, Bonding and Proper es of Organic Molecules

1.257

conjugated system containing 4p (4np, where n = 1) electron]. In fact, this carbocation is not at all formed. For this reason, the compound I undergoes solvolysis when treated with silver perchlorate in propanoic acid while the compound II does not.

(9)

Cyclooctatetraene (COT) is nonplanar (actually it is tub-shaped) while its dianion is planar. In the planar conformation, cyclooctatetraene (COT) suffers from angle strain (a regular octagon has a angle of 135° while sp2 angles are most stable at 120°). Also, the molecule containing 8p (4np, n = 2) electrons is antiaromatic in planar form and hence destabilized by delocalization. To avoid the destabilizing antiaromaticity and strain, the molecule assumes a nonplanar shape (like a tub) which is now nonaromatic.

Cyclooctatetraenyl dianion is a 10 p [(4n + 2)p, n = 2] electron system and is aromatic. The delocalization energy of the system is, therefore, large enough to overcome the angle strain in it and hence the ion exists in the stable planar state.

(10)

The compound B is more stable than the compound A.

In [10] annulene (A), the C–1 and C–6 hydrogens interfere sterically with each other and force the molecule out of planarity and for this reason, conjugation is inhibited. So, the compound is not aromatic, even though it is a 10p [(4n +2)p, n = 2] electron system. On the other hand, in 1,6-methano[10]annulene (B), the internal H’s in [10]annulene are replaced by a methylene bridge above the molecule, permitting it to be flat. So, there occurs extensive electron delocalization in this 10p electron system and therefore, the compound is aromatic. Hence, the compound B is more stable than the compound A.

1.258

(11)

Organic Chemistry—A Modern Approach

The following fulvalenes are significantly polar:

Pentafulvanlene containing 8p electrons can be represented as a hybrid of a nonpolar resonance structure and a dipolar resonance structure which, in fact is a hybrid of several charge-separated structures. The dipolar form which consists of two aromatic ions (a cyclopentadienyl anion and a cyclopropenyl cation) is considerably stable and has significant contribution to the hybrid. For this reason, pentafulvalene is significantly polar.

The dipolar form of sesquifulvalene also containing two aromatic rings (cyclopentadienyl anion and cycloheptatrienyl cation) is considerably stable and has significant contribution to the hybrid. Therefore, this compound is also significantly polar.

Structure, Bonding and Proper es of Organic Molecules

==

Tropone is more polar than cycloheptanone.

==

(12)

1.259

O Tropone

O Cycloheptanone

In tropone, the carbonyl oxygen atom being highly electronegative takes up the p electrons of the carbonyl group to form a dipolar structure which, in fact, is a hybrid of several charge-separated structures that constitute an aromatic system (a cycloheptatrienyl cation system). The dipolar structure is, therefore, considerably stable and has large contribution to the hybrid. For this reason, tropone is significantly polar.

Tropone:

The contribution of the charge-separated structure of cycloheptanone to its hybride is not much significant because it is not much stable. For this reason, it is relatively less polar than tropone. ´

==

Cycloheptanone:

O

(13)

+ –

O (less stable)

Azulene is polar and it undergoes electrophilic substitution at C–1 of the five-membered ring.

Azulene, a 10p electron system, can be represented as a resonance hybrid of two nonpolar structures (I and II) and a dipolar structure (III) which, in fact, is a hybrid of several charge-separated structures.

1.260

Organic Chemistry—A Modern Approach

In structure III, both the rings possess a closed shell of six p electrons. The sevenmembered ring is an aromatic cycloheptatrienyl cation system while the fivemembered ring is an aromatic cyclopentadienyl anion system. For this reason, the structure is considerably stable. Since this stable dipolar structure has significant contribution to the hybrid, azulene possesses dipole moment, i.e., it is a polar molecule. The five-membered ring being negatively polarized can accommodate a positive charge while the seven-membered ring being positively polarized cannot accommodate a positive charge. Because of this, it is the five-membered ring which undergoes electrophilic substitution. Again, the substitution occurs at 1-position because the 1-attack leads to a stable intermediate in which the seven-membered ring is aromatic (a cycloheptatrienyl cation system).

(14)

The rotational energy barrier around the C — N bond of the compound B is higher than the compound A.

In the compound A, the unshared pair of electrons on nitrogen is involved in maintaining an aromatic system of 6p electrons in the pyrrole ring and as a consequence, its resonance interaction with the m-tolyl ring is actually very small. Because of this, the C — N bond possesses a very small amount of double bond character (very small contribution of the resonance structures with carbon–nitrogen double bond).

Structure, Bonding and Proper es of Organic Molecules

1.261

Since the unshared pair of electrons on nitrogen in compound B is not used to complete an aromatic sextet, it is highly delocalized with the m-tolyl ring. As a consequence, the C — N bond possesses considerable double bond character (significant contribution of the resonance structures with carbon–nitrogen double bond). Because of much greater double bond character of the carbon–nitrogen bond, the rotational energy barrier of the compound B is higher than that of the compound A.

(15)

Dehydro [14]annulene is aromatic while [14]annulene is a nonaromatic compound.

Fourteen is a Hückel number [(4n + 2)p, where n = 3]. Therefore, [14]annulene is expected to be aromatic. However, the H’s that point to the interior of the ring interfere sterically with each other and force the molecule out of planarity. Because of this, it behaves like a nonaromatic compound. In dehydro[14]annulene, on the other hand, two of the interfering Hs are removed to form a triple bond. As a result, steric strain is released and the molecule becomes planar. One of the p bond of the triple bond becomes involved in conjugation. The other p bond simply exists as a localized bond because this p orbital remains at right angles to the extended p system. Therefore, it is a 14p [(4n + 2)p, n = 3] electron system and hence it is aromatic.

1.262

(16)

Organic Chemistry—A Modern Approach

The following two [10]annulenes would be expected to be aromatic on the basis of electron count. However, they are not aromatic.

The [10]annulene with all cis double bonds would, if it were planar, have considerable angle strain because the internal bond angles would be 144° instead of 120°. As a consequence, any stability this isomer gained by becoming planar in order to become aromatic would be more than offset by the destabilizing effect of the increased angle strain. Hence this isomer of [10]annulene is not aromatic. Because of the same large angle strain, the cis-trans–cis-cis-cis-isomer of [10]annulene is prevented from being aromatic.

1.

3-Chlorocyclopropene reacts with the Lewis acid SbCl5 to form a salt while 3-chlorocyclopropane does not. Explain these observations.

Solution Due to the formation of aromatic cyclopropenyl cation, 3-chlorocyclopropene reacts with SbCl5 to form the corresponding salt. Cyclopropyl cation is not an aromatic system. Furthermore, it is much destabilized due to angle strain. For this reason, the corresponding salt of chlorocyclopropane is not obtained.

Structure, Bonding and Proper es of Organic Molecules

2.

1.263

Using the theory of aromaticity, explain the finding that I and II are different compounds, but III and IV are identical. D

D

D

D D

D I

D

D II

III

IV

Solution Since cyclobutadiene is an antiaromatic compound (destabilized by resonance), it contains localized double bonds and because of this, I and II are different compounds (these are actually two isomers of dideuterocyclobutadiene). Since benzene is an aromatic compound (stabilized by electron delocalization), it contains delocalized double bonds. Therefore, III and IV are identical compounds (these are actually two resonance structures of 1,2-dideuterobenzene).

3.

The reaction (i) takes place readily to form a pale yellow precipitate of silver bromide but the reaction (ii) does not take place at all. Explain these observations. AgNO3 (i) Cycloheptatrienyl bromide ææææ Æ A nitrate salt + AgBr Ø (Pale yellow)

AgNO3 (ii) Cyclopentadienyl bromide ææææ Æ No reaction

Solution The reaction (i) takes place readily to form a pale yellow precipitate of AgBr because the substrate cycloheptatrienyl bromide exists in the ionic form, i.e., as a salt. This is because cycloheptatrienyl cation or tropylium ion containing 6p [(4n +2)p, where n = 1] electrons is an aromatic species and very much stable. The reaction (ii), on the other hand, does not take place because it leads to the formation of the antiaromatic and hence very unstable cyclopentadienyl cation (a 4np electron system, where n = 1).

(i)

1.264

Organic Chemistry—A Modern Approach

(ii)

4.

The hydrocarbon A (pKa ª 14) is much more acidic than the hydrocarbon B (pKa ª 22). Explain.

Solution In compound A, an aromatic and dipolar azulene system is, in fact, fused with a cyclopentadiene system through the cycloheptatrienyl cation system. On the other hand, in compound B, an azulene system is fused with a cyclopentadiene system through the cyclopentadienyl anion system. Both of the two compounds react with a base to form an aromatic cyclopentadienyl anion system. In the conjugate base of A, the two cyclopentadienyl anion systems are separated by a cycloheptatrienyl cation system, while in the conjugate base of B, the two cyclopentadienyl anion systems are adjacent. Therefore, due to the presence of like charges on adjacent rings, the conjugate base of B is relatively less stable than the conjugate base of A in which two negative rings are separated by a positive ring and for this reason, A is more acidic than B.

Structure, Bonding and Proper es of Organic Molecules

5.

1.265

Explain the following observation:

Solution When cyclopentadiene containing radioactive carbon (14C) at C–5 is treated with potassium hydride (KH), the potassium salt of cyclopentadienyl anion (a stable aromatic system containing a close loop of 6p electrons) is obtained. Because of electron delocalization, all carbons in this anion, except for the isotope, are equivalent.

In fact, the isotope makes no detectable difference in the relative importance of the resonance structures; except for the position of isolope, all canonicals are equivalent. Therefore, cyclopentadienyl anion undergoes protonation by water at each carbon with equal facility (20 percent). Protonation at C–1 and C–4 (i.e., from III and IV) gives the same labelled cyclopentadiene and protonation C–2 and C–3 (i.e., from V and II) gives the same labelled cyclopentadiene. Hence the yield of each of these labelled cyclopentadiene is 2 × 20 percent, i.e., 40 percent. These two remain mixed with 20 percent of the original cyclopentadiene labelled at C–5. 6.

Although there are eight free electrons in pyridine and phenyl anion (C6H5①), they are aromatic. Explain.

Solution The unshared pairs of electrons of pyridine or phenyl anion are located in an sp2 orbital which is coplanar with the sigma skeleton of the ring but perpendicular to its p electron cloud. Therefore, the unshared electron pair in each case cannot be delocalized through

1.266

Organic Chemistry—A Modern Approach

resonance interaction with the p electron cloud of the aromatic nucleus. Therefore, each of these species (planar) is a closed loop of six [(4n + 2)p, n = 1]p electron and hence aromatic.

7.

Unlike an alkyl halide cycloheptatrienyl bromide is insoluble in nonpolar solvents but is readily soluble in water. Explain.

Solution Cycloheptatrienyl bromide is not aromatic in the covalent form because it has an sp3-hybridized carbon, so it does not have an uninterrupted ring of p orbital-bearing atoms. In the ionic form, the cycloheptatrienyl cation (also called the tropylium cation) is aromatic because it is a planar cyclic ion having an uninterrupted ring of p orbital-bearing sp2-hybridized atoms and it has 6p [(4n + 2)p, n = 1] electrons. The stability associated with the aromatic cation causes this alkyl halide to exist in the ionic form. Being ionic it is insoluble in nonpolar solvents but readily soluble in highly polar solvent water (dielectric constant = 79).

8.

Predict the product of the following reaction and explain:

Solution The strong base butyllithium removes two protons from the compound and this acid-base reaction leads to the formation of the 10p electron pentalene dianion (an aromatic ion).

Structure, Bonding and Proper es of Organic Molecules

1.267

1.

The following molecule shows very high dipole moment. Explain the reason.

2.

[10]Annulene contains Hückel number of electrons but is not considered as an aromatic compound. Explain.

3.

Which one of the following compounds is more polar and why ? ==O

==O

4.

I II Tropolone does not form 2,4–dinitrophenylhydrazone derivative. Explain.

[Hint: Tropolone

5.

possesses aromatic character [a(4n + 2)p electron

system, where n = 1) and because of this, it shows a lack of ketonic properties like formation of 2,4-dinitrophenylhydrazone, etc.] Arrange the following carbocations in order of their increasing stability and explain the order: +

+

6.

+

==

==

I II III Squaric acid is almost strong as acid as H2SO4. Explain. O OH

OH

O

Squaric acid

7.

Explain why it would be incorrect to write resonance structures as shown below: ´

8.

Rank the indicated C — C bonds in order of increasing bond length, and explain why you choose this order. a c d b

1.268

Organic Chemistry—A Modern Approach

Ph

9.

Compare the C == O bond lengths in Me2C == O and

== O Ph

10.

Apply Hückel’s rule to explain why azulene is more stable than pentalene and heptalene.

Azulene

Pentalene Heptalene

11.

Discuss the differences in aromatic behaviour between pyrrole and furan.

12.

Coronene is an aromatic hydrocarbon, even though it contains 24p electrons. Explain.

Coronene

[Hint: It consists of two aromatic systems. An inner aromatic ring containing 6p electrons [(4n + 2)p, n = 1] and an outer aromatic ring containing 18 p electrons [(4n + 2)p, n = 4].] 13.

Explain why borazine (also called inorganic benzene) is a very stable compound. H B

H N

H N

B

B N

H

H

H Borazine

14.

The compound A is more acidic than the compound B. Explain this observation.

A

H

H

COC2H5

COC2H5

Structure, Bonding and Proper es of Organic Molecules

15. 16.

1.269

Explain why benzene reacts with ozone to give triozonide but not mono- or diozonide. Account for the following observations: ==

C2H5 + PhLi

C2H5 + C2H5 Li Ph

–

C2H5

==

C2H5

No reaction

The barrier for rotation about the central double bond in the following compound is very low (only about 14 kcal/mol). Explain. O == == CH — C CH3

==

17.

+ PhLi

C2H5

18.

[Hint: Because of the stable (both rings are aromatic) dipolar resonance structure, the central double bond possesses considerable single bond character.] Which one of the following two molecules is more polar and why?

19. 20.

Which species is the smallest aromatic substance? Compare the basicities of the following compounds:

21.

[Hint: III > I > II] Explain the following order of acidity:

22.

Classify the following molecules and ions as aromatic, antiaromatic, or nonaromatic. Give your reasoning.

(a)

1.270

Organic Chemistry—A Modern Approach

,

(b) –

–

,

(c)

+

,

,

O

+

O

O

(d)

(e)

(f)

23.

Although cyclopropene is an extremely unstable compound, it is possible to prepare its cation. Give reasons.

24.

Comment on the relative stabilities of the following species: (a)

(b)

+2

(c)

–2

25.

Compare the acidities of the compounds I and II and justify:

26.

Classify each nitrogen atom as strong basic and weak basic, according to the availability of its lone pair of electrons. N (a) (b) N H Compare the dipole moments of the following compounds with reasons:

27.

[Hint: Azulene can be represented as a combination of two aromatic systems such as cyclopentadienyl anion and cycloheptatrienyl cation. Therefore, the first compound in which the electron-withdrawing —CN group is attached to the negatively polarized five-membered ring and the electron-releasing –OMe group is attached to the positively polarized seven-membered ring is relatively more polar, i.e., possesses higher dipole moment than the second compound in which the electron-attracting group is attached to the positively polarized ring and electrondonating group is attached to the negatively polarized ring.

Structure, Bonding and Proper es of Organic Molecules

+

–

1.271

+

CN

MeO

Less polar

Arrange the following compounds in order of increasing acidity and explain.

==

==

==

[Hint: III < II < I] Predict which ones are likely to be aromatic, and explain why they are aromatic. – + + O (a) (b) (c) (d) (e) O N H O O O O NH2 H NH N N N (f) (g) (h) (i) O N N O H O ==

29.

OMe

CN More polar

28.

–

==

==

30.

Which of the two ‘C == O’ bonds is shorter than the other? Give reasons.

31.

Which of the following two bromo compounds would undergo silver ion-assisted hydrolysis at a faster rate? (a)

32.

33.

(b)

Br

Br Label the following compounds as aromatic, homoaromatic and antiaromatic: – + H (a) (b) (c) (If planar) – H Complete the following reactions:

(a)

(b)

hn CH2N2

A

Br2

B

D

C

1.272

Organic Chemistry—A Modern Approach

[Hint: 34.

The compound given below is a somewhat stronger base than ammonia. Which nitrogen do you think is protonated when it is treated with an acid and why? Write the structure of its conjugate acid.

35.

Explain why the following compounds are aromatic.

–

(a)

36.

S

(b)

(c)

(d)

[Hint: (a) Sulphur uses its acant d orbital to make a cyclic conjugated system containing (4n + 2)p electrons, where n = 2. (b) One of the p bonds of the acetylenic function and the central p bond of the 1,2,3-triene system are perpendicularly oriented with the conjugated cyclic p system and do not involve in electron delocalization. The conjugated system thus contains Hückel number of electrons. Also, the system is planar because the H’s pointing to the interior of the ring do not interfere sterically with each other.] Explain why fulvene possesses considerable dipole moment (m = 1.5 D). [Hint: Due to greater contribution of the stable dipolar resonance structure in which the ring is an aromatic cyclopentadienyl anion moiety possesses considerable dipole moment.]

37.

38.

When 3-chlorocyclopropene is treated with AgBF4, AgCl precipitates. The organic product can be obtained as a crystalline material, soluble in polar solvents such as nitromethane but insoluble in hexane. Explain these observations. Thymine is one of the heterocyclic bases found in DNA. Do you expect thymine to be aromatic? Explain.

Structure, Bonding and Proper es of Organic Molecules

1.273

39.

Predict the relative pKa values of cyclopentadiene and cycloheptatriene.

40.

When this compound ionizes, is Cl① or Cl≈ formed? Give your reasoning. Cl

41.

Classify each species as aromatic, antiaromatic or nonaromatic. (a)

(d)

(b)

(c)

N H

N H

(e)

(f) N

42. 43.

Predict the relative pKa values of cyclopropene and cyclopropane. Give your reasoning. Which is more soluble in water, 3-bromocyclopropene or bromocyclopropane? Explain. [Hint: 3-Bromocyclopropene.]

44.

Pyrene is aromatic, even though it contains 16 p electrons. Explain.

Pyrene

45.

Draw the p molecular orbitals of benzene and place them in an energy level diagram. Explain why it is extraordinarily stable. [Hint: The p molecular orbitals of benzene are as follows:

1.274

The energy of molecular orbitals increase with increase in the number of node. The six electrons completely fill the three bonding MOs with two electrons in each. This accounts for the extraordinary stability of benzene.] Explain why the following compound shows a reduced reactivity towards acetylation. O ==

46.

Organic Chemistry—A Modern Approach

N

47.

48.

49.

H Unlike quinoline and isoquinoline, which are of comparable stability, indole and isoindole are quite different from each other. Which one is more stable and why?

[Hint: Indole is more stable than isoindole because the six-memebred ring in indole corresponds to benzene while the six-membered ring in isoindole does not have the same pattern of bonds as benzene.] What does a comparison of the heats of combustion of benzene (781 kcal/mol), cyclooctatetraene (1086 kcal/mol), [16]annulene (2182 kcal/mol) and [18]annulene (2346 kcal/mol) reveal? One of the two dipolar resonance structures I and II is expected to contribute appreciably to the resonance hybrid of calicene. Which of these two resonance structures is more reasonable and why?

Structure, Bonding and Proper es of Organic Molecules

1.275

==

+

+

51.

I II Calicene Which one of the following two compounds should be stabilized by resonance to a greater extent? Give your reasoning.

Imidazole is a stronger base than pyrrole. Explain. N N H Imidazole

Compare the dipole moments of the following compounds with reasons: O O OH == (a) (b) and CH and ==

52.

N H Pyrrole

==

50.

I II I II [Hint: (b) The compound II has higher dipole moment. The ionic forms of both the compounds have an aromatic system (cycloheptatrienyl catation). But in the case of II, it is further stabilized by intramolecular hydrogen bonding.

53.

Which of the following molecules is likely to be planar and which nonplanar? Explain your answer.

1.276

1.13

Organic Chemistry—A Modern Approach

THERMODYNAMICS, ENERGY DIAGRAMS AND KINETICS OF ORGANIC REACTIONS

Let us consider the reaction A + B — C Æ A — B + C. Does the reaction practically occur? It occurs, to what extent and how fast? We cannot answer these questions so easily, unless we know properly the thermodynamics and kinetics of the reaction. The thermodynamic factors tell us about the feasibility of the reaction, i.e., whether the reaction will take place at all or not, while the kinetic factors tell us about the rate of the reaction, i.e., how fast the reaction will take place.

1.13.1

Thermodynamics

Thermodynamics describes the properties of a system at equilibrium. The relative concentrations of reactants and products at equilibrium can be expressed numerically by the equilibrium constant, Keq, of the reaction. For example, when starting materials A and B react to give products C and D, the equilibrium constant may be given by the following expression. ææ ÆC+D A + B ¨æ æ K eq =

[products] [C][D] = [reactants] [A][B]

The value of Keq tells us about the position of equilibrium, i.e., it expresses whether the reactants or products are more stable and therefore predominate once equilibrium has been reached. If the products are more stable (have a lower free energy) than the reactants, there will be higher concentration of products than reactants at equilibrium and Keq will be greater than 1, i.e., the reaction is favoured as written from left to right. On the other hand, if the reactants are more stable than the products, there will be a higher concentration of reactants than products at equilibrium and Keq will be less than 1, i.e., the reverse reaction is favoured as written from right to left. For a reaction to be useful, the equilibrium must favour the products and Keq > 1. A reaction at equilibrium can be described by several thermodynamic parameters. The difference between the free energy of the products and that of the reactants at equilibrium under standard conditions is called the Gibbs standard free energy change (DG°). The symbol° indicates that the reaction takes place under standard conditions, i.e., all species at 1M, temperature at 25°C and pressure at 1 atm. DG∞ =

∞ ∞ Gproducts — Greactants (free energy of (free energy of the reactants) the products)

It becomes clear from this equation that DG° will be negative if the products have a lower free energy (or more stable) than the reactants and are favoured at equilibrium. Since the reaction will release more energy to the system than it will consume from the system, it will be an exergonic reaction. If the products have a higher free energy (or less stable)

Structure, Bonding and Proper es of Organic Molecules

1.277

than the reactants, i.e., if the reactants are favoured at equilibrium, DG° will be positive. Since the reaction will consume more energy from the system than it will release into the system, it will be an endergonic reaction. These two terms should not be confused with terms exothermic and endothermic (discussed later). Whether reactants or products are favoured at equilibrium can, therefore, be indicated either by the equilibrium constant (Keq) or by the change in free energy (DG°). DG° is related to Keq by the following equation: DG∞ = – RT ln K eq = – 2.303 RT log K eq where R is the gas constant (1.986 kcal/mol/K or 8.314 × 10–3 kJ/mol/K) and T is the absolute temperature in Kelvin (K). The equation can be used to determine the relationship between the equilibrium constant and the free energy change between reactants and products. When Keq > 1, i.e., when log Keq is positive, DG° is negative. Thus, for reactions with a negative DG°, the formation of products at equilibrium is favourable. When Keq < 1, i.e., when log Keq is negative, DG° is positive. Thus, for a reaction with a positive DG°, the formation of products at equilibrium is unfavourable. Since DG° depends on the logarithm of Keq, a small difference in DG° gives rise to a large difference in the relative concentrations of products and reactants. For example, a difference in energy of only ~ 4.184 kJmol–1 or ~1 kcal mole–1 means that there is 10 times as much of the more stable species at equilibrium. A difference in energy of ~16.592 kJ mol–1 or ~ 4.2 kcal mol–1 means that there is essentially only one species (99.9%), either reactant or product, at equilibrium.

1.13.1.1

Enthalpy and entropy

The free energy change (DG°) has two components. These are enthalpy change (DH°) and entropy change (DS°). These three thermodynamic quantities are related as follows: DG∞ = DH ∞ - T DS∞ (at a fixed temperature T) The enthalpy change (DH°) is the heat liberated or the heat consumed during the course of a reaction. Heat is liberated when bonds are formed and heat is consumed when bonds are broken. If the bonds that are formed in a reaction are stronger than the bonds that are broken, energy will be released and DH° will be negative. A reaction with a negative DH° is called an exothermic reaction. If the bonds that are formed are weaker than those that are broken, energy will be consumed and DH° will be positive. A reaction with a positive DH° is called an endothermic reaction. DH° = (Energy of the bonds broken during the course of reaction) – (Energy of the bonds formed during the course of the reaction) For an exothermic reaction (DH° is negative) DG° is always negative at all temperatures if DS° is positive, i.e., if the number of species increases on going from reactants to products. For an exothermic reaction (DH° is negative) if the number of species decreases on going

1.278

Organic Chemistry—A Modern Approach

from reactants to products, i.e., if DS° is negative, DG° is negative only when the term TDS° is numerically less than DH°, i.e., at low temperature. If DS° is positive for an endthermic reaction (DH° is positive), then the reaction becomes feasible (DG° is negative) only at high temperature. On the other hand, if DS° is negative for an endothermic reaction (DH° is positive), it is not feasible at all. Therefore, the feasibility of a reaction is based on the thermodynamic parameter, DG° and is temperature dependent. It is relatively easy to calculate the values of DH° which is the largest factor in the driving force in most organic reactions. If the reaction involves only a small change in entropy, the TDS term will be small and the value of DH° will be very close to the value of DG°. However, we must be cautious in making this approximation because many organic reactions have relatively small changes in enthalpy, larger changes in entropy or occur at high temperatures and so have significant TDS° terms.

1.13.2

Energy Diagram

A schematic representation of the energy changes that take place as reactants are converted into products is called an energy diagram or a reaction coordinate diagram. From an energy diagram we can know how readily a reaction proceeds, how many step are involved in a reaction and how the energies of the reactants, products and intermediates compare. In an energy diagram, energy is plotted on the y-axis and the progress of the reaction is plotted on the x-axis. As the reactants are converted into products, the reaction passes through a maximum energy state which is known as transition state. The reactants attain the transition state by acquiring the energy supplied by collisions. The transition state is a hypothetical arrangement of atoms possessing a definite shape and charge distribution. Its structure is somewhere in between the structures of the reactant and product. Any bond that is partially cleaved or formed is drawn with a dashed line. Any atom that loses or gains a charge contains a partial charge in the transition state. Transition states are | ). Because of high energy and often drawn in brackets, with a superscript double dagger (= very short life time, it can never be isolated or trapped. @

Let us consider a concerted reaction between the molecule B—C and the anion A : to form @ products A — B and C: . Since the reaction occurs in a single step, the bond between A and B is formed as the bond between B and C is broken. Let us also assume that the reaction is exergonic, i.e., the products are lower in free energy than the reactants.

The anion A①: attacks the molecule B — C from the side remote from C and along the bonding line of B — C. This mode of attack requires minimum energy because the electrostatic repulsion between the entering and leaving groups is minimum. As the bond A — B begins to form, the bond B — C begins to break. Since the energy released by bond formation (which lags behind) is less than that required for bond cleavage, the energy of the system increases on going from reactants to the transition state which is represented

Structure, Bonding and Proper es of Organic Molecules d-

1.279

d-

by A B C (B is partially bonded to A and C). The energy of the system then begins to decrease on the going from transition state to products and reaches a minimum when C is completely displaced by A. The energy diagram of this reaction is given below. Since DG° values are used, y-axis is free energy.

In this exergonic reaction, the energy difference between the reactants and products is DG°. The energy difference between the transition state and the reactants is called the | free energy of activation, symbolized by DG=. DG° relates to the equilibrium constant of the | = reaction, while DG relates to the rate of the reaction. If a reaction involves only a small change in entropy the TDS° term will be small and the value of DH° will be close to the value of DG°. In that sense the above energy diagram may also be represented as follows. Since DH° values are used, the y-axis is here potential energy. The energy difference between the transition state and the reactants is called energy of activation, symbolized by Ea. The reaction is now called an exothermic one (DH° = –ve).

1.280

Organic Chemistry—A Modern Approach

The two variables, DG=| and DG° (or Ea and DH∞), are independent of each other. Two | reactions can have identical values of DG° (or DH°) but very different DG= (or Ea) values. A multistep reaction proceeds through transient chemical species which unlike transition state possesses some degree of stability and can be isolated and detected in many cases. These are known as intermediates. Examples are carbocations, carbanions, free radicals, etc. If the same exergonic (or exothermic) reaction occurs by a two-step pathway and bond breaking occurs before bond making, it can be represented as follows: Step 1: Cleavage of the B — C bond

Step 2: Formation of the A — B bond Each step has its own energy barrier, with a transition state at the energy maximum. Step 1 is endergonic (or endothermic) because energy is required to break the B — C bond, making DG° (or DH∞) a positive value and placing the products at higher energy than reactants. In the transition state, the B — C bond is partially broken. Step 2, on the other hand, is exergonic (or exothermic) because energy is released due to formation of the A — B bond, making DG° (or DH°) a negative value and placing the products at lower energy than the reactants. In the transition state the A — B bond is partially formed. The species B! is a reactive intermediate which is formed in step 1 and consumed in step 2. The two transition states are separated by an energy minimum, at which the reaction intermediate B! is located. Since the step 1 has the higher energy transition state (DG1|= > DG2|= or Ea1 > Ea2 ), it is the slower, that is, the rate-determining step of the reaction. The energy diagram of the reaction can be represented as follows:

Structure, Bonding and Proper es of Organic Molecules

1.13.3

1.281

Kinetics

From a knowledge of DG° (or DH°) we cannot predict how fast a reaction occurs, i.e., it does not tell us any thing about the energy barrier or the energy hill that must be climbed for the reactants to be converted into products. The higher the energy barrier, the slower the reaction. Kinetics is the field of chemistry that describes the reaction rate and the factors | that affect them. The energy barrier of a reaction, indicated by DGG=, is called the free energy of activation. It is the difference between the free energy of the transition state and the free energy of the reactants. DG=| = (free energy of the transition state) – (free energy of the reactants) The rate of a chemical reaction is determined by measuring the decrease in the concentrations of the reactants or the increase in the concentrations of the products over time. The rate of a reaction depends on the following three factors: (i) the number of collisions that take place between the reacting molecules in a given period of time, (ii) the fraction of the collisions that occur with sufficient energy to get the reacting molecules over the energy hill and (iii) the fraction of the collisions that occur with proper orientation. A rate law or rate equation is an equation that shows the relationship between the rate of a reaction and the concentration of the reactants. A rate law can be determined experimentally, and it depends on the mechanism of the reaction. The two important terms of a rate law are the rate constant k and the concentration of the reactants. All reactant concentrations may not appear in the rate equation. The rate law or rate equation of a reaction may be represented as rate = k [reactants]

1.282

Organic Chemistry—A Modern Approach

The rate constant k (a fundamental characteristic of a reaction) is a complex mathematical term that takes into account the dependence of the rate of a reaction on temperature and the experimental energy of activation, Ea, and this is because the Arrhenius equation relates the rate constant of a reaction to Ea and the temperature at which the reaction is carried out. The Arrhenius equation: k = Ae– Ea / RT The rate constant k and DG=| (or Ea) are inversely related. A high DG=| (or Ea) corresponds to a small k. Fast reactions have large rate constants, while slow reactions have small rate constants. A rate equation contains concentration terms for all reactants in a onestep (concerted) reaction, while a rate equation contains concentration terms for only the reactants involved in the rate-determining step in a multistep reaction. For the following one-step reaction, both reactants appear in the transition state of the rate-determining step (the only step here) of the reaction. This type of reaction involving two reactants in the transition state is said to be bimolecular. A:

+

B—C Æ

A—B + C:

The order of a reaction is the summation of the exponents of the concentration terms appearing in the rate equation. The rate equation of the above reaction is rate = k [A①:] [B — C] In this rate equation, there are two concentration terms, each with an exponent of one. Therefore, the sum of the exponents is two and it is a second-order reaction. Since the rate of the reaction depends on the concentration of both reactants, doubling the concentration of either A①: or B —C doubles the rate of the reaction. If the concentrations of both A①: and B — C are doubled, the rate of the reaction will be quadruple. However, the situation is different if the same reaction occurs in two steps such as follows:

In multistep reactions, only the concentrations of the reactants involved in the ratedetermining step appear in the rate equation. The rate of this reaction, therefore, depends only on the concentration of B — C only, because only B — C is involved in the ratedetermining step. rate = k [B — C] Since there is only one concentration term (raised to the first power), it is a first-order reaction. Also, it is an unimolecular reaction because only one reactant is involved in the transition state of the rate-determining step. Since the rate of the reaction depends on the concentration of only B — C, doubling its concentration doubles the reaction rate, but doubling the concentration of A①: has no effect on the reaction rate. In fact, neither its concentration nor its identity affects the reaction rate.

Structure, Bonding and Proper es of Organic Molecules

1.283

The rate constant of a reaction should not be confused with the rate of a reaction. The rate constant informs us how easy it is to reach the transition state, i.e., to cross the energy barrier. Low energy barriers (low DG=| ) are associated with large rate constant while high energy barriers (high DG=| ) have small rate constants. The rate of a reaction is a measure of the amount of reactant disappeared or the amount of product formed per unit of time. The reaction rate is the product of the rate constant and the concentration of the reactant(s). Thus reaction rates depend on concentration, while rate constants are independent of concentration. Therefore, when two reactions are compared to see which one occurs more easily and at a faster rate, their rate constants and not their concentration-dependent rates of reaction are to be compared. Again, the free energy of activation, DG=| , should not be confused with the experimental energy of activation, Ea. The free energy of activation has both enthalpy and entropy components (DG=| = DH=| – TDS=| ), whereas the experimental energy of activation has only an enthalpy component (Ea = DH=| + RT). Ea is an approximate energy barrier to a reaction, while DG=| is a true energy barrier to a reaction and this is because some reactions proceed by a change in enthalpy, some by a change in entropy, but most by a change in both entropy and enthalpy. Without going to detail suffice it to say that the two quantities DG=| and Ea are closely related and that both measure the difference in energy between the reactants and the transition state.

1.13.4 Catalysis Some reactions proceed faster when certain substances are added in small amounts to the reaction. These substances are called catalysts. A catalyst may be described as a substance that increases the rate of a reaction, i.e., speeds up a reaction without appearing in the stoichiometric equation of the reaction (i.e., by remaining unaltered in mass and chemical composition). A catalyst operates in such a way that it combines with one of the reagents so as a provide an alternative pathway for the reaction and is regenerated in a later stage of the reaction. The position of equilibrium of the reaction remains unaffected because the catalyst increases the rate of the forward as well as the reverse reaction to the same extent, i.e., a catalyst does not change the amount of reactant and product at equilibrium (DH° of the reaction remains unchanged). In fact, a catalyst helps to achieve equilibrium quickly. The energy of activation (Ea) is related to the rate constant (k) as follows: k = A. e– Ea / RT The rate of the reaction can thus be accelerated either by increasing the value of A or by decreasing the values of Ea. Since the value of A is usually constant, therefore, in the presence of a catalyst a reaction follows a course of lower Ea. This is shown in the following energy diagram. The uncatalyzed reaction (solid line) has the transition state denoted by TS1, and the catalyzed reaction (dotted line) by TS2. A catalyst thus speeds up the reaction by lowering the activation energy of both the forward reaction as well as the backward reaction.

1.284

1.13.5

Organic Chemistry—A Modern Approach

Hammond Postulate

According to the Hammond postulate, the transition state of a reaction resembles the structure of the species (reactant or product) to which it is closer in energy. In exothermic reactions, the transition state is closer in energy to the reactants. Therefore, the structure of the transition state will more closely resemble the structure of the reactant than that of the product. In endothermic reactions, the transition state is closer in energy to the products. Therefore, the structure of the transition state will more closely resemble the structure of the product.

That chlorination is less selective than bromination can be well understood by the help of Hammond postulate. The first propagation step, i.e., abstraction of a hydrogen atom by a bromine or chlorine radical, is the rate-determining step, in bromination or chlorination. This step is endothermic for bromination but exothermic for chlorination. According to the Hammond postulate, the product-like transition state for endothermic bromination has a great deal of radical character on the carbon atom. Therefore, the energy difference

Structure, Bonding and Proper es of Organic Molecules

1.285

of the two transition states forming the 1° and 2° radicals reflects most of the energy difference of the radical products. On the other hand, the reactant-like transition state for exothermic chlorination has a little radical character on the carbon atom. So, the energy difference of the two transition states forming 1° and 2° radicals reflects only a small part (about one-third) of the energy difference of the radical products. Consequently, the transition states leading to the formation of 1° or 2° radical for bromination have a larger energy difference than those for chlorination, even though the difference in energy of the products is the same in both reactions. Chlorination of propane is, therefore, less selective (product ratio 60:40) than bromination of propane (product ratio 97:3).

1.286

Organic Chemistry—A Modern Approach hn CH3CH2CH3 + Br2 ææææ Æ CH3CH2CH2 Br + CH3CHBrCH3 + HBr 125° C Propane 1-Bromopropane (3%) 2-Bromopropane (97%) hn CH3CH2CH3 + Cl2 ææææ Æ CH3CH2CH2Cl + CH3CHClCH3 + HCl 125° C Propane 1-cholorpropane (40%) 2-chloropropane (60%)

1.13.6

Kinetic Control versus Thermodynamic Control of a Chemical Reaction

When a reaction produces more than one product, the product that is formed most readily is called the kinetically controlled product or simply kinetic product. Reactions that produce the kinetic product as the major product are said to be kinetically controlled reactions. The most stable product is called the thermodynamically controlled product or simply thermodynamic product. Reactions that lead to the formation of the thermodynamic product as the major product are said to be thermodynamically controlled reactions. Example: Electrophilic addition of HBr to 1,3-butadiene is an example of a reaction in which the kinetic product and the thermodynamic product are different. When the reaction is carried out at low temperature (–80°C), the major product (80%) is formed by 1,2-addition and the minor product (20%) is formed by 1,4-addition. -80∞ C CH2 == CH — CH == CH2 + H Br ææææ Æ CH3CH BrCH == CH2 + CH3CH == CH CH2 Br 1,3-Butadiene 1,2 addition product 1,4-addition product (80%) (20%)

At lower temperature, the relative amounts of the products are determined by the relative rates at which the two additions occur. The 1,2-addition occurs at a faster rate (TS is stable and Ea is lower because the +ve charge is on a secondary allylic carbocation) and so, the 1,2-addition product is the major product. In fact, at lower temperature, there is enough energy for the reactants to overcome the energy barrier of the first step of the reaction to form the carbocation, and enough energy for the carbocation to form the two addition products. However, there is not enough energy for the reverse reaction to take place. When the reaction is carried out at higher temperature (40°C), the major product (80%) is formed by 1,4-addtion and the minor product (20%) is formed by 1,2-addition. 40∞ C CH2 == CH — CH == CH2 + H Br æææ Æ CH3CH BrCH == CH2 + CH3CH == CH CH2 Br 1,3-Butadiene 1,2-addition product 1,4-addition product (20%) (80%) At higher temperature, the relative amounts of the products are determined by the position of an equilibrium. The 1,4–addition product is thermodynamically more stable (stabilized more by hyperconjugation), so it is the major product. In fact, at higher temperature (40°C), there is enough energy for the reactants to get over the energy barrier to form the products and enough energy for one or more of the products to go back to the intermediate carbocation. Since the product can interconvert, the relative amounts of the two products at

Structure, Bonding and Proper es of Organic Molecules

1.287

equilibrium will depend on their relative stabilities. The thermodynamic product reverses less readily because it has higher energy barrier (larger Ea) to the common intermediate. As a result, it gradually becomes the predominant product in the product mixture. The reaction will be more clearly understood if we examine the energy diagram given below.

1.

How can the thermodynamic stability of the product compared with the reactants be predicted from the DG° value of the reaction?

Solution If DG° of the reaction is negative, the product is thermodynamically more stable compared with the reactant and if DG° is positive, the product is thermodynamically less stable compared with the reactant. 2. Explain why the standard free energy change (DG°) of a reaction can often be approximated by the change in bonding energy, i.e., the enthalpy change (DH°) of the reaction.

1.288

Organic Chemistry—A Modern Approach

Solution If a reaction involves only a small change in entropy and it is not carried out at high temperature, the TDS° term is small compared to the enthalpy term (DH°) and it can be neglected. For this reason, the standard free energy change (DG°) of a reaction can often be approximated by the change in bonding energy, i.e., the enthalpy change (DH°) of the reaction. 3. Draw the energy diagram of (i) an exothermic reaction that takes place at room temperature; (ii) an exothermic reaction that cannot take place without adding energy above that provided by the external thermal conditions; (iii) an endothermic reaction having small experimental energy of activation and (iv) an endothermic reaction having large experimental energy of activation. Solution

(a)

(b)

Structure, Bonding and Proper es of Organic Molecules

1.289

(c)

(d)

4.

Mention the factors on which the rate of reaction depends.

Solution The rate of a reaction depends on the following three factors: (i) The number of collisions taking place between the reacting molecules in a given period of time. The greater the number collisions, the faster the reaction. (ii) The fraction of the collisions that occurs with sufficient energy to get the reactant molecules over the energy ‘hill’ or energy barrier. If Ea is small, more collisions will lead to reaction than if Ea is large. (iii) The fraction of collisions that occur with proper orientation. For example, in the reaction of propene with HBr, the reaction will take place only if the molecules collide with hydrogen of HBr approaching the p bond of propene. If collision occurs with hydrogen approaching the methyl group of propene (CH3CH== CH2), no reaction will take place regardless of the energy of the collision. 5. Explain why the rate of a reaction increases on increasing the concentration of the reactants and on increasing the temperature at which the reaction is carried out.

1.290

Organic Chemistry—A Modern Approach

Solution Increasing the concentration of the reactants increases the reaction rate because it increases the number of collisions that occur in a given period of time. The rate of reaction also increases on increasing the reaction temperature because an increased temperature increases the frequency of collisions and the number of collisions that have sufficient energy to get the reactant molecules over the energy barrier. 6. What do you mean by the kinetic stability of a compound ? Solution A compound is said to be kinetically stable if it is not very reactive, i.e., if DG=| is large.

7.

Write down the Arrhenius equation and from this predict how the experimental activation energy (Ea) and temperature affect the rate constant of a reaction.

Solution The Arrhenius equation: k = A. e– Ea /RT where k is the rate constant, Ea is the

experimental energy of activation, R is the gas constant (1.986 × 10–3 kcal mol–1 K–1 or 8.314 × 10–3 kJ mol–1 K–1), T is the absolute temperature (K) and A is the frequency factor. The frequency factor accounts for the fraction of collisions that occur with the appropriate orientation for the reaction to take place. The factor, e– Ea / RT , accounts for the fraction of collisions having minimum energy (Ea) needed to react. Taking logarithms on both sides, the Arrhenius equation becomes ln k = ln A - Ea /RT From this equation, it becomes clear that the rate constant decreases on increasing the experimental activation energy (Ea) and increases on increasing the temperature of the reaction. 8. What is called free energy of activation (DG=| )? How does it govern the rate of a reaction? Solution The free energy of activation (DG=| ) is the difference between the free energy of the transition state and the free energy of the reactants.

According to the transition state theory, the reactants are in equilibrium with transition state and the equilibrium constant K=| is related to DG=| by the equation: DG=| = –2.303 RT log K=| This theory further tells that all the transition states collapse into the products at the same rate. Therefore, the rate constant of the reaction depends only on the position of equilibrium between the reactants and the transition state, that is, the rate constant depends on the value of the equilibrium constant, K=|. A lower value of DG=| indicates a higher rate constant, i.e., a faster reaction. Conversely, a higher value of DG=| means a slower reaction. 9. What are intermediates? How does transition states differ from intermediates?

Structure, Bonding and Proper es of Organic Molecules

1.291

Solution Multistep reactions proceed through transient chemical species which possess some degree of stability and can be isolated and detected in many cases. These are known as intermediates. Carbocations, carbanions, free radicals, arynes, carbenes, etc., are examples of intermediates.

The transition state, on the other hand, are hypothetical arrangement of atoms possessing a definite shape and charge distribution. Because of high energy and very short lifetime, they can never be isolated or trapped. 10. The energy diagram of a reaction is as follows:

Answer the following questions by considering the energy diagram: (a) Mention the number of steps involved in this reaction. (b) Label DH°, Ea and transition states for each step and labels the DH∞ of the reaction. (c) At which point on the graph the reactive intermediate is located. (d) Which is the rate-determining step of the reaction ? (e) Predict whether the overall reaction is exothermic or endothermic. Solution (a) The reaction involves two steps. (b)

1.292

Organic Chemistry—A Modern Approach

(c) (d) (e) 11.

The two transition states are separated by an energy minimum, at which the intermediate is located. Since Ea1 > Ea2 , the first step is slower than the second step and hence, it is the rate-determining step of the reaction. Since DH° of the overall reaction is +ve, the overall reaction is endothermic.

Draw an energy diagram for the following reaction in which C is the most stable and B is the least stable of the three species and the transition state for the conversion of A Æ B is more stable than the transition state for the conversion of B Æ C. A (a) (b) (c) (d) (e)

k1

B

k -1

k2 k -2

C

How many intermediates are involved in this reaction ? How many transition states are there in the overall process? Identify the faster step in the forward and reverse directions. Which out of the four steps is the fastest one? Mention the rate-determining steps in the forward and reverse processes.

Solution The energy diagram of the reaction is as follows: TS2

Potential energy

TS1

(a) (b) (c) (d) (e) 12.

B A C

Progress of the reaction The reaction involves only one intermediate (B). There are two transition states in the overall process. A Æ B is the faster step in the forward direction and B Æ A is the faster step in the reverse direction. Out of the four steps B Æ A is the fastest one. B Æ C is the slower step, i.e., the rate-determining step in the forward process and C Æ B is the rate-determining step in the reverse process. Write a rate equation for each of the following reactions:

(a)

ii

ii

: @ Æ CH 3 CH = CH 2 + ii Br:@ + EtOH (CH 3 )2 CHBr + Et O ii ii

Structure, Bonding and Proper es of Organic Molecules

1.293 ≈

@

slow OH (b) (CH 3 )3 CBr æææ Æ Br @ + (CH 3 )3 C æææÆ (CH 3 )3 COH fast

Solution (a)

13.

rate = k [(CH3 )2 CHBr][EtO@ ]

(b)

rate = k [(CH3)3CBr]

In which two cases the entropy changes of reactions are very important?

Solution Entropy changes are important in the following two cases: (i) When the number of molecules of reactant differs from the number of molecules of product in the balanced chemical equation. (ii) When an acyclic molecule undergoes cyclisation, or cyclic molecule is converted to an acyclic one. 14. For which reaction DS∞ is expected to be more significant and why? B + C or A + B C+D (a) A (b) A B or A + B C Solution (a)

15.

DS∞ is more significant for the first reaction because the number of particles increases, i.e., disorder increases on going from starting material to products. (b) DS∞ is more significant for the second reaction because the number of particles decreases, i.e., disorder decreases, on going from starting material to product. What do you mean by each of the following values of Keq, DG∞, DH∞ and DS∞? (a) Keq > 1 (b) DG∞ < 0 (c) DH∞ < 0 (d) DS∞ > 0

Solution (a) (b)

(c) (d) 16.

Keq > 1 means that more products than reactants are present at equilibrium. DG∞ < 0 means that the free energy of the product is lower than the free energy of the reactants. DH∞ < 0 means that bonds in the products are stronger than the bonds in the reactants. DS∞ > 0 means that the products are more disordered than the reactants.

Predict whether the value of DS∞ for the dissociation of Cl2 is positive, i.e., favourable or negative, i.e., unfavourable. Give an idea about the sign of the value DG° of this reaction. hn Cl 2 ææ æ Æ 2Cl.

Solution Two isolated chlorine atoms have much more freedom of motion than a single Cl2 molecule. Thus, the disorder of the system increases, and therefore, the change in entropy is positive. However, the positive (unfavourable) value of DH° of this reaction is much larger than the positive (favourable) value of TDS°. Since DG° = DH° – TDS°, the value of DG° of this reaction is positive, i.e., the reaction is unfavourable.

1.294

17.

Organic Chemistry—A Modern Approach

Using bond dissociation energies predict whether the following reaction is endothermic or exothermic. [Bond dissociation energies: Cl — Cl: 58 kcal/mol; CH3 — H: + 104 kcal/mol; CH3 — Cl: –84 kcal/mol and H — Cl: –103 kcal/mol]

Solution

Bonds broken

DH ∞ (per mole)

Cl — Cl CH 3 — H

+ 58 kcal + 104 kcal

Total :

+ 162 kcal

Bonds formed

DH ∞ (per mole)

H — Cl CH3 — Cl

- 103 kcal - 84 kcal

Total :

- 187 kcal

DH∞ (The overall enthalpy change) = Sum of DH∞ of bonds broken + (–) sum of DH° of bonds formed = + 162 + (–187) kcal/mol = –25 kcal/mol Since the value of DH∞ is negative, the reaction is an exothermic one. 18. The experimental energy of activation (Ea) of the following reaction is +4 kcal/mol and the enthalpy change (DH∞) is +1 kcal/mol. Draw an energy diagram for this reaction. What are the activation energy and heat of reaction (DH∞) for the reverse reaction. CH + Cl. Æ .CH + HCl 4

3

Solution The energy diagram of this reaction is as follows:

The activation energy for the reverse reaction is +3 kcal/mol and DH° for the reverse reaction is –1 kcal/mole. 19. Explain why the tert-butyl cation is formed at a rate faster than isobutyl cation when 2-methylpropene is subjected to react with HBr. Give an energy diagram.

Structure, Bonding and Proper es of Organic Molecules

1.295

Solution

In between the two resulting carbocation tert-butyl cation (a tertiary or 3° carbocation) is more stable than the isobutyl cation (a primary of 1° carbocation). Since the formation of a carbocation is an endothermic reaction, the transition state resembles the carbocation and it possesses a significant amount of positive charge on a carbon atom. The same factors that stabilize the positively charged product, i.e., carbocation, will stabilize the partially positive charged transition state. Therefore, the transition state leading to the formation of the tert-butyl cation is more stable than the transition state leading to the formation of isobutyl cation and so, the tert-butyl cation (lower Ea) is formed at a faster rate than isobutyl cation (higher Ea).

1.

Using the bond dissociation energies, calculate the DH° for the addition of HBr to ethene. [DH°: p-bond of ethane = 61 kcal/mol; H — Br = 87 kcal/mol, C — H = 101 kcal/mol and C — Br = 69 kcal/mol] Do you expect the reaction to exergonic or endergonic?

1.296

2.

3.

4.

5. 6.

Organic Chemistry—A Modern Approach

Draw an energy diagram for a reaction in which the product is both thermodynamically and kinetically unstable and for a reaction in which the product is thermodynamically unstable and kinetically stable. The DG° for the conversion of ‘axial’ phenylcyclohexane to ‘equatorial’ phenylcyclohexane at 25°C is –2.9 kcal/mol. Calculate the percentage of phenylcyclohexane molecules that have the phenyl substituent in the equatorial position. State whether you expect the entropy change (DS°) to be negative, positive or approximately zero for each of the following reactions: (a) A Æ B + C (b) A + B Æ C (c) A + B Æ C + D What is the value of DG° for a reaction where Keq = 1 and where Keq = 10? Assuming DS° ª 0, what change in DH° is required to produce 100-fold increase in the Keq? (a) Which Keq corresponds to a negative value of DG°, Keq = 100 or Keq = 0.01? (b)

Which Keq corresponds to a lower value of DG°, Keq = 10–5 or Keq = 10–3?

7.

Considering the following values predict whether the reactant or product is lower in energy? (a) Keq = 102 (b) DG° = 3.0 kcal (c) Keq = 10–3 (d) DG° = –3.5 kcal

8.

Neglecting entropy and considering each of the following values predict whether the reactant or product is favoured at equilibrium: (a) DH° = 30 kcal/mol, (b) DH° = –20 kcal/mol, (c) DS° = 2 k cal/mol, (d) DS° = –2 kcal/mol.

9.

Which value indicates a faster reaction: (a) Ea = 15 kcal or Ea = 3 kcal; (b) a reaction temperature of 5°C or a reaction temperature of 30°C; (c) Keq = 10 or Keq = 150; (d) DH° = –4 kcal/mol or DH° = 4 kcal/mol? Consider the following energy diagram for the conversion of M Æ S and answer the following questions: P

Potential energy

10.

N R

O Q

M

S Progress of the reaction

Structure, Bonding and Proper es of Organic Molecules

1.297

(a) (b) (c) (d)

11. 12.

Mention the number of steps present in the reaction? Which point on the graph corresponds to reactive intermediates? Which point on the graph corresponds to transition states? Label each step of the reaction and the overall reaction as endothermic or exothermic. A reaction will not take place at all if DH° is +ve and DS° is –ve — Why? For a reaction DH° = – 14 kcal/mol and DS° = 0.01 kcal/K/mol, calculate DG° and Keq at 20°C and at 100°C. How does DG° and Keq change as T increases?

13.

Which of the following factors affect the rate of a reaction? (a) DH°, (b) DG° (c) temperature, (d) DS°, (e) Ea (f) concentration, (g) Keq, (h) catalysts (i) k

14.

In esterification reaction, a carboxylic acid is subjected to react with an alcohol in the presence of sulphuric acid to form an ester with less of water. An example of intermolecular esterification reaction and an example of intramolecular esterification reaction are given below. Explain why Keq is different for these two apparently similar reactions.

15.

At 298K the enthalpy change, DH° for the ionization of trichloroacetic acid (Cl3C COOH) is +6.3 kJ.mol–1 and the entropy change, DS°, is +0.0084 kJ mol–1K–1. Calculate the pKa value of trichloroacetic acid. [Hint: DG∞ = DH ∞ - T DS°= 6.3 kJ mol -1 - (298 K)(0.0084 kJ mol -1K -1 ) = 3.8 kJ mol -1 –

– DG∞ = log K eq = log K a = -pK a 2.303 RT

Therefore 16.

pK a =

3.8 kJ mol -1 DG∞ = = 0.66] 2.303 RT (2.303) (0.008314 kJ mol-1 K -1 ) (298K)

Label each product in the following reaction as a 1,2- or 1,4-product, and decide which is the kinetically controlled product and which is the thermodynamically controlled product. H3C H3C H3C + HBr + Br Br

1.298

1.14

Organic Chemistry—A Modern Approach

METHODS OF DETERMINING MECHANISMS OF REACTIONS

A mechanism is the actual process by which a reaction takes place. There is no specific way to know how exactly a reaction proceeds. There are a number of commonly used methods for determining mechanism. In most cases, one method is insufficient, and in most cases, the problem is approached from several directions. The techniques for determining mechanisms of reactions include: (i) identification of products (ii) stereochemical evidence, (iii) isotope labelling, (iv) determination of the presence of intermediates along with spectroscopic, and crystallographic analysis, (v) crossover experiments and (vi) kinetic evidence.

(1) Identification of Products

Any mechanism proposed for a reaction must account for all the products obtained and for their relative proportions, including the by-products (if any). A proposed mechanism cannot be correct if it fails to predict the products in approximately the observed proportions. For example: (1)

hn Any mechanism for chlorination of methane (CH4 + Cl2 ææ æ Æ CH3Cl + HCl), that fails to account for the formation of a small amount of ethane (CH3CH3) cannot be correct. Since the following mechanism shows the formation of ethane, it is the actual mechanism of chlorination.

(2)

One of the oximes of 2-bromo-5-nitro acetophenone readily undergoes ring closure to benzisoxazole, thus establishing its structure as I. The other isomer II does not undergo ring closure because —OH and —Br are not close to each other. However, it readily gives the Beckmann rearrangement product, the amide III. The identification of III thus leads to the conclusion that the rearrangement involves a trans migration of the substituted phenyl group. CH3 O2N

C==

I

N

Br OH

O2N OH

–

CH3 N O A benzisoxazole

Structure, Bonding and Proper es of Organic Molecules

1.299

Therefore, the mechanism of Beckmann rearrangement is as follows:

(2) Stereochemical Evidence The stereochemical course of a reaction, which can be obtained from a knowledge of the stereochemistry of the products, is often helpful in determining the mechanism of a reaction. For example, bromination of trans-2-butene yields meso-2,3-dibrombutane and not (±)-2, 3-dibromobutane.

The formation of optically inactive meso-product indicates that the two bromine atoms add to the double bond from the opposite sides, i.e., it is a case of trans-addition. This trans-mode of addition rules out the possibility of one-step (concerted) mechanism because the atoms in a Br2 molecule are very close and unable to add simultaneously in an anti fashion. Hence, the addition is a two-step process such as follows:

1.300

Organic Chemistry—A Modern Approach

(3) Isotopic Labelling The mechanism of a reaction can be ascertained by carrying out the reaction with isotopically labelled compound and then identifying the position of the isotope in the products. The mechanism of the following reaction, for example, can be ascertained by isotopic labelling experiment.

Direct substitution of chlorine with an amide ion (NH 2@ ) appears to be an attractive possibility, but experiments performed with 1-14C-chlorobenzene provided a sample of aniline, about half of which was labelled in the 1-position, and remaining half in the 2-position. A two-step mechanism involving the formation of an intermediate known as benzyne is consistent with the experimental observations.

The mechanism of Cannizzaro reaction, which involves transfer of hydrogen from one molecule to another, has been investigated by the use of deuterated benzaldehyde (PhCDO), instead of benzaldehyde. O || @ + 50% NaOH 2PhCDO æææææÆ Ph — C — O Na + PhCD2OH

Structure, Bonding and Proper es of Organic Molecules

1.301

Since the deuterium is exclusively located on the carbon atom of benzyl alcohol bearing the hydroxyl group, the possibility of exchange of hydrogen with the hydrogen atoms in the solvent (H2O) is ruled out. Thus, there occurs a direct transfer of hydrogen (or deuterium) between the carbonyl carbons of two aldehyde molecules. A plausible mechanism for Cannizzaro reaction can thus be written as follows:

(4) Determination of the Presence of Intermediates A lot of organic reactions proceed through the formation of one or more intermediates. A knowledge of the structures of that intermediates (isolated, detected by spectroscopic methods or trapped) gives useful information about the mechanistic course of a reaction. (a) Isolation of intermediates: Sometimes an intermediate can be isolated from a reaction mixture by stopping the reaction after a short time or by use of very mild conditions. For example: (1) the formation of cyclobutanone from a ketene and diazomethane (CH2N2) at low temperature has been known for a long time. It has been demonstrated by 14C labelled diazomethane that cyclopropanone is the most likely intermediate. Recently, direct evidence for its formation has been provided by preparing cyclpropanone and then allowing it to react with CH2N2 to yield cyclobutanone.

(2)

In Hofmann rearrangement in which amides are converted into amines by Br2 NaOH and alkali (RCO NH 2 + Br2 ææææ Æ RNH 2 ), three intermediates, e.g., an N-bro@

moamide (RCONHBr), its anion (RCO N Br) and an isocyanate (R — N == C == O), have been isolated. When these intermediates are subjected to normal reaction conditions, they lead to the formation of usual reaction products at a rate not slower than the rate of the overall reaction. Therefore, the mechanistic pathway to be sujected for this reaction is that which accounts for the formation of these intermediates. The mechanism of the reaction is as follows:

1.302

Organic Chemistry—A Modern Approach

(b)

Detection of intermediates: An intermediate, which cannot be isolated, can often be detected by spectroscopic studies, such as IR, NMR, ESR or other spectra. The detection of nitronium ion (NO≈ 2 ) by Raman spectra, for example, was regarded as a strong evidence that this is an intermediate in the nitration of benzene.

(c)

Trapping of intermediates: An intermediate can often be trapped by running the reaction in the presence of a compound with which the intermediate reacts. The compound obtained as a result of this is isolated and from a knowledge of its structure the structure of the intermediate is predicted. For example, arynes react with dienes to give adducts (a Diels-Alder reaction). In any reaction where benzyne is a suspected intermediate, the addition of anthracene (which acts as a diene) to the reaction mixture and the detection of triptycene (the Diels-Alder adduct) indicate that the reaction involves benzyne as an intermediate.

(5) Crossover Experiments Intramolecular rearrangements are those in which a group migrates from one position to another without detaching itself completely from the molecule. On the other hand, the rearrangements in which the group becomes completely free during its migration are known as intermolecular rearrangements. Whether a particular rearrangement is intra- or intermolecular can be known by crossover experiments. In such experiments, the rearrangement is carried out with a mixture of two structurally related reactants and the products are then identified. If the reactant X—Y—Z is allowed to rearrange in the presence of a structurally related reactant A—Y—B (where X is closely related to A and Z is related to B), a mixture of four or a mixture of two rearrangement products is expected to be obtained.

Structure, Bonding and Proper es of Organic Molecules

1.303

Y—Z—X + Y—B—A Normal products

+ Y—Z—A + Y—B—X

[X—Y—Z + A—Y—B]

Crossover products

Y—Z—X + Y—B—A Normal products

It only Y—Z—X and Y—B—A (the normal intramolecular products) are obtained, the process is intramolecular, but if Y—Z—A and Y—B—X (the crossover products) are obtained along with the normal products, the process is intermolecular. The crossover experiment can be used, for example, to establish that the pinacol–pinacolone rearrangement is intramolecular and the Fries rearrangement is intermolecular. The pinacol–pinacolone rearrangement involves conversion of a pinacol (a 1, 2-diol) to a pinacolone in the presence of moderately concentrated H2SO4. For example: ≈

H Me2 C —CMe2 ææææ Æ Me3C — C — Me + H2O H2SO4 | | || OH OH O When two structurally related pinacols are allowed to rearrange in the same solution no crossover products are obtained. For example

1.304

Organic Chemistry—A Modern Approach

This observation suggests that this rearrangement is an intramolecular process. The Fries rearrangement involves conversion of a phenyl ester into a hydroxyl-ketone in the presence of the Lewis acid AlCl3 COCH3 1. only AlCl3

OH COCH3

2. H3O +

C2H5

C2H5

When this rearrangement is carried out with a mixture of two structurally related phenyl esters, two crossover products along with two normal products are obtained. For example:

Formation of crossover products indicates that this rearrangement is an intermolecular one.

(6) Kinetic evidence The order of a reaction (determined by kinetic methods) gives us an important information about which reactants and how many of them take part in the rate-determining step of the reaction. The benzidine rearrangement, for example, has been found to follow a thirdorder kinetics (second-order in [H+] and first-order in [PhNHNHPh]).

Structure, Bonding and Proper es of Organic Molecules

1.305

From the rate law it can be suggested that the reaction involves rate-determining protonation of hydrazobenzene followed by rapid rearrangement of the hydrazonium cation to benzidine. + 2H + slow

NH—NH

+

+

NH2—NH2

fast

N2 N

NH2

An indirect method is quite helpful in formulating the most plausible mechanism of a reaction. In this process, the rate law of a reaction is determined experimentally. Subsequently, various mechanisms are postulated intuitively for the reaction. The rate equations are then derived for each of these individually postulated mechanisms. The calculations, which may sometimes be quite complex, can be simplified by the steady-state assumption. Any derived rate law that corresponds to the experimentally observed rate law is then selected. The mechanism corresponding to the law is then suggested to be the most plausible mechanism for the reactions. For example, the mechanism for the hydrolysis of tert-butyl chloride in aqueous acetone can be established by this procedure. H2O (CH 3 )3CCl ææææ Æ (CH 3 )3 COH + HCl acetone

The reaction is found to be first-order and obeys the following rate law: d[(CH 3CCl] = k[(CH 3 )3CCl] dt The rate law suggests that the rate-determining step of the reaction involves only the concentration of tert-butyl chloride. The reaction sequence can then be written as -

Step 1: Step 2: If we apply the steady-state hypothesis assuming the tert-butyl carbocation in a steadystate concentration, we can arrive at a rate law similar to the experimental rate law written above. Hence, the proposed mechanism involving rate-determining cleavage of the C — Cl bond is the most plausible one.

1.14.1

Kinetic Isotope Effects

Changes in reaction rates brought about by isotopic substitution are known as kinetic isotope effects. They provide important information about the mechanism of the reaction. (1) Primary kinetic isotope effect: A large difference in the rate of a reaction is observed when one of the atoms of a bond that breaks in the rate-determining step is replaced by its isotope (the rate of a reaction for a compound containing a heavy isotope is slower than that of the compound with a lighter isotope). This is what is called primary kinetic isotope effect. If a C — H bond cleaves in the rate-determining step of a reaction, substitution of D for H results in decrease in the reaction rate and that is due to the fact that a C — D

1.306

Organic Chemistry—A Modern Approach

bond is stronger than a C — H bond. For example, the rate of dehydrobromination of 2-bromopropane (CH3CHBrCH3) on treatment with NaOEt in EtOH has been found to become slow when the hydrogens in the methyl group of the alkyl halide are replaced by deuteriums. A large kinetic isotope effect (kH/kD = 6.7) is observed. This observation, therefore, suggests that the rate-determining step of the reaction involves cleavage of the C — H (or C — D) bond. The reaction thus proceeds through the E2 mechanistic pathway and not through the E1 mechanistic pathway. E2 pathway:

E1 pathway:

(2) Secondary kinetic isotope effect: There are many reactions in which substitution of hydrogen by deuterium in a bond which is not even broken in the rate-determining step may still bring about a measurable change in the reaction rate by isotopic substitution. These are known as secondary kinetic isotope effects. The rate of solvolysis of tert-butyl chloride in 60% aqueous ethanol is found to be faster than the rate of solvolysis of its nonadeuterio analog, i.e., (CD3)3CCl. The value of kH/kD was found to be 2.32. C H OH/ H O

2 5 2 (CH3 )3Cl ææææææ Æ (CH3 )3COC2 H5 + (CH3 )3 COH kH tert-Butyl chloride

C H OH/ H O

2 5 2 (CD3 )3CCl ææææææ Æ (CD3 )3 COC2H5 + (CD3 )3 COH kD tert-Butyl chloride-D9

This secondary kinetic isotope effect indicates that the reaction proceeds through the ratedetermining formation of a 3° carbocation which is differently stabilized by hyperconjugation involving — CH3 and — CD3 groups (because of greater strength, the C— D bond is not a good hyperconjugative participant when compared to C—H). The reaction, therefore, proceeds through SN1 pathway. SN1 pathway:

Structure, Bonding and Proper es of Organic Molecules

1.

1.307

a-Methylallyl chloride and crotyl chloride yield a mixture of ethyl a-methylallyl ether and ethyl crotyl ether when treated with ethanolic sodium ethoxide under SN1 conditions. Predict the mechanism of these two reactions from the knowledge of the products obtained.

Solution Since each of the isomeric allylic chlorides gives the same mixture of isomeric ethers, the two substitution reactions proceed through a common intermediate carbocation which is a hybrid of two resonance structures.

2.

How would you establish by stereochemical studies that the acid-catalyzed hydrolysis of the following ester involves alkyl-oxygen bond cleavage?

Solution This dextrorotatory ester of acetic acid undergoes hydrolysis to give racemic alcohol.

Since racemization occurs, therefore, an intermediate 3° carbocation must have been formed by alkyl-oxygen bond cleavage. This may be shown as follows:

1.308

3.

Organic Chemistry—A Modern Approach

The base-promoted hydrolysis of an ester involves acyl-oxygen bond cleavage:

Establish this mechanistic course by a labelling experiment. Solution When ethyl propanoate labelled with 18O in the ethyl-type oxygen of the ester is subjected to hydrolysis with aqueous NaOH, all the 18O shows up in the ethanol that is

produced. None of the 18O appears in the propanoate ion (CH 3CH 2COO@ ) This labelling result is completely consistent with the mechanism given above, i.e., the hydrolysis involves acyl-oxygen bond cleavage.

4.

How can you establish that the alkaline hydrolysis of chloroform involves dichlorocarbene (:CCl2) as the reactive intermediate?

Structure, Bonding and Proper es of Organic Molecules

1.309

Solution That the alkaline hydrolysis of choloroform involves dichlorocarbene (:CCl2) as the reactive intermediate can be proved by trapping this intermediate.

When cis-2-butene is added to the reaction mixture, a cyclopropane derivative is obtained. This observation suggests that the reaction proceeds through the formation of dichlorocarbene (:CCl2).

5.

3-Pentanol undergoes chromic acid oxidation to yield 3-pentanone. Give the mechanism of the reaction. OH O | || Na2Cr2O7 /H2SO4 CH 3CH 2 CH CH 2 CH 3 æææææææ Æ CH 3CH 2 — C — CH 2CH 3 3- Pentanol

3- Pentanone

Which step of this reaction is rate-determining? Give evidence in favour of your answer. Solution When sodium dichromate (Na2Cr2O7) is dissolved in a mixture of sulphuric acid and water, chromic acid (H2CrO4) is obtained.

O || ≈ Na 2Cr2O7 + 2H 2SO4 + H 2O Æ 2HO — Cr — OH + 2Na + 2HSO@4 || O The oxidation of 3-pentanol by chromic acid proceeds through the steps as follows: Step 1:

1.310

Organic Chemistry—A Modern Approach

Step 2: The chromate ester undergoes E2 elimination to give 3-pentanone. Water acts as a base and chromium readily accepts an electron pair and is thus reduced from Cr (VI) to Cr (IV).

The decomposition of the chromate ester (step 2) is the rate-determining step of the reaction. When the rate of oxidation of 3-pentanol to 3-pentanone is compared with the rate of oxidation of its deuterated derivative (Et2CDOH), an isotope effect is observed (kH/kD = 6.6). This observation suggests that step 2, i.e., the decomposition of the chromate ester (an E2 elimination reaction) involving C—H bond cleavage is the rate-determining or ratelimiting step of the reaction. 6. What is solvent isotope effect? Mention the factors responsible for such effect. Solution Reaction rates often change when the solvent used is isotopically labelled, i.e., when the solvent is changed from H2O to D2O or from ROH to ROD. This kind of isotope effect is called solvent isotope effect.

This isotope effect may be due to any of the following three factors or a combination of all of them: (1) If the solvent is a reactant and an O — H bond of the solvent is cleaved in the rate-determining step, there will be a primary isotope effect. There may also be a secondary effect (caused by the O — D bonds that are not breaking) if the molecules involved are D2O or D3O≈ . (2) Labelling of substrate molecules may take place by rapid hydrogen exchange, and then the resulting labelled molecule may become cleaved in the rate-determining step. (3) The nature and extent of solvent–solute interactions may be different in the deuterated and nondeuterated solvents. This may change the energy of the transition state, and hence the activation energy (Ea). These are secondary isotope effects. 7. Base-promoted hydrolysis of methyl mesitoate occurs through attack on the alcohol carbon instead of the acyl carbon:

Structure, Bonding and Proper es of Organic Molecules

Me

1.311

O

Me

C

C O—CH3

Me

Me

O :O + CH3OH

–

+ :OH

Me

Me

Suggest an experiment with labelled compounds that would confirm this mode of attack. Solution When the above reaction is carried out in weakly alkaline solution in water enriched with 18O, it resulted in the formation of methanol containing 18O label.

This observation, therefore, confirms the given mode of attack. 8.

Explain the following observation:

Solution The mechanism of the reaction in which phenoxide group acts as a neighbouring group explains the scrambling of 14C label in the products.

9.

Account for the following differences in the primary kinetic isotope effect: kH/kD ª 7; kH/kT ª 10 – 20, where T is tritium.

Solution Isotope effects are due solely to the mass differences, which are manifested in the vibrational frequencies of the C — H, C—D, and C — T bonds. The greater the mass of the isotope whose bond is broken in the rate-determining step, the greater the activation energy of the bond breaking and so, the greater is the primary kinetic isotope effect. Since the mass of T is greater than the mass of D, therefore, kH/kT > kH/kD.

1.312

1.

Organic Chemistry—A Modern Approach

Salts of aliphatic or aromatic carboxylic acids can be converted to the corresponding nitriles by heating with BrCN. 250 – 300∞C RCOO@ + Br CN æææææ Æ RCN + CO2 + Br @

By isotopic labelling prove that the CN group in the product does not come from the CN in BrCN. *

*

[Hint: R COO@ + Br CN ææÆ R CN + CO2 + Br @ (* =

2.

3.

4. 5.

14

C)

Using water enriched in 18O how will you prove that the hydration equilibrium OH is operative in the case of acetone? C == O + H2O C OH In the oxidation of Ph2CHOH and Ph2CDOH by alkaline permanganate, the kH/kD ratio was found to be 6.6. From this kinetic isotope effect, suggest an acceptable mechanism for this reaction. How the general mechanism of electrophilic aromatic substitution can be established by the kinetic isotope effect? The relative rates of chromic acid oxidation of isotopically labelled isopropyl alcohol are as follows: relative rates: CH3CHOHCH3 CH3CHODCH3 CH3CDOHCH3 D3CCHOHCD3

(

(

1.0

1.0

0.16

1.0

Comment on how these observed relative rates relate to the mechanism of oxidation. 6.

18

Oxidation of Ph2C = 18O by m-chloroperbenzoic acid yields only Ph COO Ph , i.e., there is no scrambling of the 18O label in the product ester. 18

m-ClC H COOH

18

6 4 Ph2C == O æææææææ Æ Ph COOPh

7.

Suggest a mechanism for this reaction. Where would you expect to find the labelled oxygen if you carry out an acidcatalyzed hydrolysis of methyl benzoate in 18O-labelled water? [Hint: The labelled oxygen atom should appear in the carboxyl group of the acid.]

Structure, Bonding and Proper es of Organic Molecules

8.

1.313

Predict the product and give the mechanism of the following reaction:

[Hint: The oxirane ring undergoes SN2 attack at the less crowded labelled carbon to give an alkoxide. The alkoxide then undergoes intramolecular SN2 displacement to give the product.

9.

10.

When chlorobenzene is treated with sodamide in liquid ammonia in the presence of furan, a 1,4-addtion product (I) is obtained. Explain this observation

Explain the results of the following crossover experiment:

2

PRINCIPLES OF STEREOCHEMISTRY

CHAPTER

Chapter Outline Introduction 2.1

Projection Formulas of Stereoisomers 2.1.1 Flying-Wedge Projection Formula 2.1.2 Fischer Projection Formula 2.1.3 Sawhorse Projection Formula 2.1.4 Newman Projection Formula 2.1.5 Interconversion of Projection Formulas

2.2

Symmetry Elements 2.2.1 Simple Axis of Symmetry or Rotational Axis of Symmetry (Cn) 2.2.2 Plane of Symmetry (s) 2.2.3 Centre of Symmetry (i) 2.2.4 Alternating Axis of Symmetry (Sn) 2.2.5 Symmetric, Asymmetric and Dissymmetric Molecules

2.3

Isomerism: Constitutional Isomers and Stereoisomers 2.3.1 Constitutional Isomers 2.3.2 Stereoisomers 2.3.3 Enantiomers and Diastereoisomers

2.4

Molecular Chirality 2.4.1 Chiral and Achiral Molecules 2.4.2 Source of Chirality 2.4.3 Stereocentre or Stereogenic Centre 2.4.4 Meso Compounds

2.5

Configuration and Configurational Nomenclature 2.5.1 D, L-System of Configurational Designation 2.5.2 Specification of Configuration: The R, S-System 2.5.3 Erythro and Threo Nomenclature of Compounds with Two Adjacent Chiral Centres 2.5.4 The E-Z System of Designating Alkene Diastereoisomers 2.5.5 R, S and E, Z Assignment in the Same Molecule (Geometric Enantiomerism) 2.5.6 Syn-anti Nomenclature for Aldols 2.5.7 Number of Stereoisomers for Compounds with Chiral Centres 2.5.8 Chirotopic and Achirotopic Atom in a Molecule 2.5.9 Prostereoisomerism and Topicity

2.6

Optical Activity of Chiral Compounds 2.6.1 Plane-polarized Light 2.6.2 Optical activity 2.6.3 Polarimeter 2.6.4 Specific Rotation 2.6.5 The Necessary and Sufficient Condition (the ultimate criterion) for Optical Activity

Organic Chemistry—A Modern Approach

2.2

2.6.6 2.6.7 2.6.8 2.6.9 2.7

Racemic Modification Enantiomeric Excess(ee) or Optical Purity (op) Racemization Resolution of Racemic Modification

Conformation of Acyclic Organic Molecules 2.7.1 Conformations of Ethane (CH3—CH3) 2.7.2 Conformations of Propane (CH3CH2CH3)

2.7.3 2.7.4 2.7.5 2.7.6 2.7.7

Conformations of Butane (CH3CH2CH2CH3) Conformations of Chloroethane (CH3CH2Cl) Conformations of 1,2-dichloroethane (ClCH2CH2Cl) Conformations of Some Typical Acyclic Molecules Invertomerism

INTRODUCTION The nature of the functional groups does not solely determine the properties of organic compounds. The difference in orientation of the atoms or groups in space also responsible in determining molecular properties. If molecules having the same molecular formula are superimposable, then these are called homomers, and those that are not superimposable are called isomers. Isomers having different atomic connectivities are called constitutional isomers, and isomers having same atomic connectivities, but they differ in orientation of their atoms in space are called stereoisomers. This phenomenon is known as stereoisomerism. The part of chemistry that deals with the properties of stereoisomers and the chemical effects of stereoisomerism is called stereochemistry. In fact, stereochemistry is the study of the three-dimensional structure of molecules. It is not possible to understand organic chemistry, biochemistry or biology without the knowledge of stereochemistry. Biological systems are very much selective, and they often discriminate between molecules with very small stereochemical differences. Difference in spatial orientation seems apparently to be unimportant, but stereoisomers often process remarkably different physical, chemical and biological properties. For example, the cis and trans-isomers of butenedioic acid are a special type of stereoisomer called cis- trans or geometric isomers. Both compounds have the same structure (HOOC— CH==CH—COOH), but they differ in how these atoms and groups are arranged in space. The cis-isomer in which the two –COOH groups are on the same side of the double bond is called maleic acid, and the trans-isomer in which the two –COOH groups are in opposite sides of the double bond is called fumaric acid. Fumaric acid is an essential metabolic intermediate in both animals and plants, whereas maleic acid is toxic and irritating to tissues.

Principles of Stereochemistry

2.3

O H H

C C

C C

O OH OH

H

cis-isomer HO

stereoisomer

O

C

C C

C

OH

trans-isomer

H

O

Maleic acid (mp 138°C, a toxic and irritating substance)

Fumaric acid (mp 287°C, an essential metabolite)

The discovery of stereochemistry is, in fact, one of the most important breakthroughs in the structural theory or organic chemistry. In this chapter, we will study the principles of stereochemistry and in the following chapters, we will see how stereochemistry plays a major role in the properties and reactions of organic compounds. Stereochemistry is now an overgrowing subject that helps one to understand and explain the chemical, physical and biological properties of molecules. However, the subject is so vast that it is not possible to cover it in one or two chapters. An attempt has been made in this chapter to discuss the basic principles of organic stereochemistry.

2.1

PROJECTION FORMULAS OF STEREOISOMERS

It is inconvenient to represent three-dimensional (3D) molecules in two-dimensional (2D) surface of paper or blackboard. So several 2D-formulas have been developed to represent the molecules having 3D-structures and these are known as projection formulas. The important projection formulas that are used to represent organic molecules are (i) Flyingwedge projection formula, (ii) Fischer projection formula, (iii) Sawhorse projection formula, and (iv) Newman projection formula.

2.1.1

Flying-Wedge Projection Formula

This is a three-dimensional representation of a three-dimensional molecule on a twodimensional surface and is generally used for compounds containing chiral carbon (often marked * ), i.e. a carbon which is attached to four different substituents, such as in alanine, *

CH3 CH(NH2)CO2H. In this representation, the ordinary lines represent bonds in the plane of the paper, a solid wedge ( ) represents a bond above the plane of the paper and a hashed wedge ( ) represents a bond below the plane of the paper. The Flying-wedge projection formula of (R)-alanine, for example, can be shown as follows: CO2H

bonds in the plane of the paper

C H3C

NH2 H

(R)-Alanine

bond below the plane of the paper bond above the plane of the paper

Organic Chemistry—A Modern Approach

2.4

Compounds containing two or more chiral carbons may be expressed similarly. The following figure represents the Flying-wedge formula (2S, 3S)-tartaric acid.

2.1.2

Fischer Projection Formula

Although the Flying-wedge formula represents the actual three-dimensional structure of a molecule, it is not convenient to draw this formula especially for a molecule containing a number of chirality centres. In that case, three-dimensional chiral molecules are represented with two-dimensional formulas which are called Fischer projection formulas. In this representation, the chiral carbon lies in the plane of the paper and the four bonds are shown by two vertical lines representing bonds projecting behind the plane of the paper (i.e., pointing away from the viewer) and two horizontal lines representing bonds projecting above the plane of the paper (i.e. pointing towards the viewer). The point of intersection of these lines represents the chiral carbon. The Fischer projection, therefore, looks like a cross. To represent the 3D-structure of a molecule, for example, (S)-alanine, in a Fischer projection, the asymmetric or chiral carbon is viewed in such a way that the two of the bonds to the carbon are vertical and pointing away from the viewer, and two are horizontal and pointing towards the viewer. When this view is projected on the plane of the paper, the Fischer projection of the molecule is obtained.

∫

S

S

Fischer projection is very much useful for representing molecules containing two or more chiral carbons that are the part of a carbon chain. In such cases, formula for Fischer projection of molecules is written with the main carbon chain extending from top to bottom (usually according to IUPAC numbering), i.e., along the vertical line with all groups eclipsed. For example, the Fischer projection formula for (2R, 3R)-3,4-dichlorohexane may be written as follows:

Principles of Stereochemistry

CH3CH2

2.5

CH2CH3

H Cl

C

CH2CH3

Cl

C

H

Cl

H

H

C

Cl

H

Cl

90° rotation

C

Cl H

CH3CH2

CH2CH3

CH2CH3 Fischer projection of (2R, 3R)-3,4dichlorohexane

3D-representation of (2R,3R)-3,4-dichlorohexane

A 3D-structure can be written in Fischer projection directly as follows: hashed wedge to right horizontal line COOH C H3C 3Drepresentation of (S)-alanine

H NH2

upper full line to upper vertical line 3D to Fischer

COOH

H2 N

H CH3

solid wedge to left horizontal line

Fischer projection of (S)-alanine

lower full line to lower vertical line

The 3D-molecule is to be viewed between the solid and the hashed wedges. Then the solid wedge bond (on the left hand side of the viewer) is written as the left horizontal line in the Fischer projection, the hashed wedge bond (on the right hand side of the viewer) is written as the right horizontal line, the upper full line bond as the upper vertical line and the lower full line bond as the lower vertical line. Similarly, if the solid and hashed wedges are on the left side of the 3D-figure, then the corresponding Fischer projection can be written as follows:

Organic Chemistry—A Modern Approach

2.6

CO2H

CO2H 3D to Fischer

C

H2N

CH3

H

H

H 2N CH3

Fischer projection Fischer projection of (S)-alanine of (S)-alanine

3D-representation of (S)-alanine

General rules for exchanging ligands or rotation of Fischer structures (a) An even number of interchanges result in no change in the configuration of a molecule, i.e., an identical structure is obtained.

2

3 Allowed

(i)

2 3

R

(exchanging ligands across the horizontal bonds and the vertical bonds)

2

R

(ii)

(b)

An odd number of interchanges convert the molecule into its enantiomers (non superimposable mirror image). Therefore, this operation is not allowed. For example, (R)-lactic acid becomes its enantiomer (S)-lactic acid when only one exchange is made.

Principles of Stereochemistry

2.7

2

2 Forbidden

(i)

(interchanging ligands across the horizontal bonds) 3

3

(ii)

(iii)

(c)

180° horizontal or vertical rotation of a Fischer formula outside of the plane of the paper changes the configuration of the molecule, i.e. converts it into its enantiomer. Therefore, this operation is not allowed. For example:

(i)

Organic Chemistry—A Modern Approach

2.8

(ii)

R

(d)

S

90° or 270° rotation of a Fischer formula in the plane of the paper is not allowed because this operation does not obey the convension ‘horizontal lines forward, vertical lines back’ and hence, the configuration of the molecule is changed, i.e., the molecule is converted into its enantiomer. For example: not identical but enantiomers CO2H H

(e)

OH OH

Forbidden (90° rotation in the plane of the paper)

HO2C

CH3

CH3

H

( )-Lactic acid

( )-Lactic acid

180° rotation of a Fischer projection formula in the plane of the paper is permissible because this operation follows the convension ‘horizontal lines forward, vertical lines back’ and hence, there occurs no change in configuration, i.e., an identical molecule is obtained. identical CO2H

(i)

H

CH3 Br

Allowed (180° rotation in the plane of the paper)

Br

H

CH3

CO2H

(R)-2-Bromopropanoic acid

(R)-2-Bromopropanoic acid

Principles of Stereochemistry

2.9

Cl

(ii)

C2H5

H

C2H5

H

Cl

Allowed (180° rotation in the plane of the paper)

Cl

H

H5C2

H

C2H5

Cl

(3R, 4R)-3, 4-Dichlorohexane

(f)

(3R, 4R)-3, 4-Dichlorohexane

Clockwise or counterclockwise rotation of any three groups is allowed because it leads to the same molecule. 3

3 Allowed

Allowed

(clockwise rotation of F, Cl and Br)

(clockwise rotation of CH3, F, Cl)

3

(i) A

A

A

3

A C2H5

OH

(ii)

C2H5

H

H

Br

Allowed (clockwise rotation of C2H5, OH and H)

C2H5 (3S, 4R)-4-Bromo3-hexanol

C2H5

H

OH

H

Br C2H5

Allowed

(anticlockwise rotation of Br, H and C2H5)

H

OH

Br

C2H5 H

Organic Chemistry—A Modern Approach

2.10

2.1.3

Sawhorse Projection Formula

When it is required to know the spatial relation between two adjacent C-atoms, both chiral or achiral, Sawhorse representation is quite convenient. In this projection, the bond between the two key C-atoms is joined by a diagonal line (slightly elongated) in the plane of the paper. The remaining bonds are shown by smaller lines projected above and below the plane of the paper, and is a free rotation around the diagonal line either 1

2

3

4

clockwise or anticlockwise. For example, 2,3-dibromobutane (CH3 CHBr CHBr CH3 ) can be represented by the following Sawhorse projections when C-2 and C-3 carbon atoms are taken as the two key carbon atoms. 180° rotation of the front carbon around the diagonal line Br

H CH Br

back carbon Br front carbon

H

CH H

CH

I (2R, 3S)-2, 3-Dibromobutane (eclipsed conformation)

Configuration remains unchanged

Br

H CH

II (2R, 3S)-2, 3-Dibromobutane (staggered conformation) In conformation I, two Br and two H atoms are above the plane of the paper and two –CH groups are below the plane of the paper. In II, one of each groups above the plane and one of each group below the plane.

Both I and II represent two conformations of (2R, 3S)-2,3-dibromo-butane. The conformation II in which the groups attached to C-2 and C-3 are oriented as far away from each other as possible is known as the staggered conformation and the conformation I in which the groups attached to C-2 and C-3 are nearest is known as the eclipsed conformation. Transformation of one Sawhorse projection into another can be effected simply by rotating one of the two

Principles of Stereochemistry

2.11

key carbon atoms around the diagonal line either clockwise or counterclockwise. However, the configuration of that carbon (if chiral) remains unchanged. It is to be noted that the conformation with the dihedral angle (q) = 0° is called the eclipsed conformation and the conformation with the dihedral angle (q) = 60° is called the staggered conformation.

2.1.4

Newman Projection Formula

This is also a perspective formula which is drawn from Sawhorse projection formula by viewing the molecule along the bond joining the two key C-atoms of the molecule, both chiral or achiral. The key C-atoms are represented as superimposed circles. Only one circle is drawn the centre of which represents the front C-atom, and the circumference of the circle represents the rear C-atom. The three s-bonds of the front carbon are represented by three small lines drawn from the centre of the circle. The angle between any two bonds is 120°. The three s-bonds of the back carbon are drawn from the circumference of the circle. The Newman projection of n-butane, for example, can be shown as follows:

n

n

n

n

Organic Chemistry—A Modern Approach

2.12

2.1.5

Interconversion of Projection Formulas

(a) (i) Fischer projection to Flying wedge projection This transformation may be carried out as follows:

A

R

B

The above figure demonstrates the conversion of the Fischer projection of (R)-mandelic acid to its Flying-wedge projections (A and B). This conversion involves placing the vertical bonds in the Fischer projection in the plane of the paper and the horizontal bonds above and below the plane of the paper. If the lower vertical bond in Fischer projection is bent on the right side with respect to the vertical line (as in projection A), the group on the right side horizontal bond in Fischer projection is to be placed above the plane of the paper (to be represented by solid wedge), and the group on the left side is to be placed below the plane of the paper (to be represented by hashed wedge). If the lower vertical bond in Fischer projection is bent on the left side with respect to the vertical line (as in projection B), the group on the left horizontal bond in the Fischer projection is to be placed above the plane of the paper and the group on the right horizontal bond is to be placed below the plane of the paper.

Principles of Stereochemistry

2.13

(ii) Flying-wedge projection to Fischer projection The reverse method is to be followed to convert a Flying-wedge representation to Fischer projection. For example: bent on the left with respect to the vertical line

COOH COOH C

H C

H N

H NH

CH vertical line

Flying-wedge projection of ( )-alanine

H

Fischer projection of ( )-alanine

In (S)-alanine molecules, the —COOH and —CH3 groups are in the plane of the paper. Since the —CH3 group is bent on the left side with respect to the vertical line, the —NH2 group, which is above the plane in the Flying-wedge structure, is placed on the left horizontal bond in the Fischer projection and the H atom, which is below the plane, is put on the right horizontal bond.

(b) (i) Fischer projection to Sawhorse projection This conversion in a molecule, for example in (2R, 3R)-2,3-dibromobutanoic acid, may be shown as follows:

R

R

The Fischer projection is to be transformed into that eclipsed form of Sawhorse projection in which the horizontal groups of the back and front carbon remain above the plane of the paper. In this conversion the last chirality centre (C-3 in this example) in the Fischer projection (counting from the top) is considered as the front carbon in sawhorse projection

(ii) Sawhorse projection to Fischer projection For this conversion, the Sawhorse projection of the molecule is to be first transformed into that eclipsed conformation in which two groups of the front carbon and two groups of the back carbon remain above the plane of the paper. The groups above the plane occupy the horizontal position (front carbon becomes the last chiral centre) and the groups below the plane become the two vertical groups. The Sawhorse projection of meso-2,3-dichlorobutane, for example, can be converted into the Fischer projection as follows:

Organic Chemistry—A Modern Approach

2.14 clockwise 60° rotation of front carbon

3 3

2

2

3

3 3

3

3

3

(c) (i) Sawhorse projection to Newman projection The Sawhorse projection (staggered and eclipsed form) of (2S, 3R)-3-bromobutan-2-ol, for example, may be converted into the Newman projection (staggered and eclipsed form) as follows. The front and back carbon in Sawhorse projection becomes the front and back carbon, respectively, in the Newman projection. Groups attached to each key carbon are placed accordingly.

S R

Principles of Stereochemistry

2.15

(ii) Newmann projection to Sawhorse projection The reverse method is to be followed to convert a Newman projection to a Sawhorse projection. The bond between front and back carbons is drawn diagonally and the groups attached to them are placed as follows: Br Cl

Br

H H Cl CH3 CH3

Newman projection of (2 , 3 )-2-bromo-3chlorobutane (eclipsed)

CH3 Br

H

H

Cl

CH3 (Staggered)

H CH3 H

CH3 Sawhorse projection (eclipsed)

Br CH3

H

Cl Cl H Sawhorse projection (staggered)

(d) Conversion of a Flying-wedge projection to Newman projection through Fischer and Sawhorse projection This process can be illustrated by taking active tartaric acid as an example:

Organic Chemistry—A Modern Approach

2.16

H HO

OH

H

OH H

H

HOOC

H

OH

COOH Newman projection (staggered)

1.

OH

HO

H

H CO2H Fischer projection

Sawhorse projection (eclipsed)

COOH OH

H

CO2H

CO2H

CO2H COOH Newman projection (eclipsed)

H

CO2H H

HO

OH

OH

CO2H

Sawhorse projection (staggered)

Convert the following Flying-wedge projections to Fischer projections: 3 3

(b)

(a)

2

2

2

Cl (d) F

(c)

Br

C CH3

Br

H

H Cl

C

(e)

C C

H 3C HO

CH3 H

(f) HO H

CH2

CH2

C

C

H3C CO2H

H Br

Principles of Stereochemistry

2.17

Solution

H

CH3

H C

(a)

HO2C

HO

CH3

CO2H OH 3 3 2

(b) 2 2

2

CHO

(c)

H

CHO H

Br

C

Br CH3

CH3

F

Cl

(d)

F

Br

C

90° rotation (in-plane)

H3C

F

CH3

H3 C

Cl

C

Cl Br

Br Viewing direction

Br

H

Cl

C

(e)

H

H Br

C

CH3

CH2

CH2

C

C CO2H

H

Cl C CH3

H3C

HO2C

HO

CH3

H3C

H Br

CH3

OH H

CH2

CO2H OH CH2

CH2

CH2

C

C H 3C

H H H

Br HO Cl

H

HO

(f)

OH H C

C

C

H 3C

H

H Br

H

Br

H CH3

Organic Chemistry—A Modern Approach

2.18

2.

Convert the following Flying-wedge projections to Fischer projections: H

O

Cl

O

(a)

(b)

CH3 H

(d)

C

(e) H

H OH

(c) H

F Br CH2 6

Br

C

NH2 H

CH3

Solution

O

(a)

1

C

O H

CH3

(b)

CH3

3 5 4

H H

(c)

2 3

Cl H

1 6

1

2

3 4

H H

5

Br

2 1 6

4

Cl Br

5 4

5

(d)

3

4 3

6 7

1 2

8

HO H OH

2 1

5

H

6 8

7

NH2 1

(e)

CH2 6 H

C2

H

1

NH2 (CH2)6

(CH2)6 C

1

C

NH2

H

2

H

CH3

3.

H

C

H3C

2

H

CH3

Convert the following Fischer projections to Flying-wedge projections: CO2H

(a) H

OH CH3

CH3

(b) Cl

OH Br

F

(c)

CH3 H

Principles of Stereochemistry

2.19

CO2H

(d) H

CH3

Br

H

(e)

(f) Cl

OH CH3 C2H5

(g) H

CHO H (h) HO H H

OH (CH2)6

H

Cl

OH H OH OH

(i)

CH2OH

CH3 Solution

CO2H

CO2H

(a)

H

OH

C

(b)

C

H HO

Br Cl

F F OH

OH

(c)

or

CH3

Br

Cl

CO2H

OH H

H3C

CH3 CH3

CH3

C

CH3

H

H Br CO2H

(d)

H

Br

H

OH

HO2C

C C

H3C

CH3

CH3

CH3

C

(e) Cl

Cl

H

(f)

H 2N H

H OH

H OH

H 2N H

NH2 OH H

H

OH H

C CH3

H H

Cl Br

Organic Chemistry—A Modern Approach

2.20

2

5 2

5

(g) 2 2

2 6

3

3

H CHO

(h)

H HO H H

OH

HO H

OH H OH OH

C CHO

C

H

C

CH OH C

HO

CH OH

OH

H Cl H H

(i)

H

Cl Br

Br H

4.

Draw a Fischer projection for each of the following compounds: *

(a) CH3 C HOHCOOH *

*

(b)

CH3 C H(NH2)CO2H

*

*

(c) CH3 C HBr C HClC2H5 (d) *

*

CH3 C H(OH) C H(OH)C2H5

*

(e) CH3CH2 C HFCH2 C HOHCH3 Solution

CO2H

(a)

H

(b) H

OH CH3

NH2 CH3

CH3

(d)

CH3

CO2H

H

OH

H

OH

(e) H H

C2H5

CH2CH3 F CH2 OH CH3

(c) H

Br

H

Cl C2H5

[Other Fischer projections, which may be the enantiomer or the diastereoisomer, may also be written.]

Principles of Stereochemistry

5.

2.21

Convert the following Fischer projections to Sawhorse projections: CO2H

(a) H

CH3

OH

H

(b) H

Br

Cl

H CH3

CHO

CHO Br

(e) H

H

Cl

H

(c) H

OH

H

OH

NH2

CH3

(d) H

CO2H

CO2H

(f)

H

Br

Cl

H

NO2

CH2OH Solution

CO H

(a)

OH

H

H

OH

H

Br

H

H Br CO H

(eclipsed) 1

3

2

2

3 3

3 2

3

4

3

3 1

3

3

2

2

2

2

2

2

(c)

3

4

1 CHO

H

3 4

H Br Cl

CH2OH

H

2

2

2

2

H

3

2

3

2

CO H

(staggered)

2

(b)

(d)

Br

H C

CH

CH

OH

H

2

Br

H H

3

Cl

CHO

CH2OH (eclipsed)

3

HOH2C

Cl

2

Br

CHO

(staggered)

Organic Chemistry—A Modern Approach

2.22

CHO

(e)

H

H H

H

H H

NO2 CHO

NO2

CHO NO2 (staggered)

(eclipsed)

(f)

H

Br

Cl

H

H Cl

H

Br

Br

H Cl H

(eclipsed)

(staggered)

6.

Convert the following Sawhorse projections to Newman projections: H

(a) HO

OH

H CO2H

CH2Br

H

Br

(b) CHO

H

Cl

CH2OH NH2 H

(c) H

Br

H CH3

(d) H

Cl

OH CO2H

Ph

H2N

Cl CH3

(e) Br

H

(f) H Br

OH CH3

Br CH3

Principles of Stereochemistry

2.23

Solution

H

(a)

HO

OH

H HO

H CO2H

OH H CO2H CH2OH

CH2OH

(b)

H

BrH2C

Br CHO

H

H

CH2Br Br

H

Cl

Cl

CHO NH NH

H

(c)

CH

H OH

H

OH

H

CH CO H

CO H Br

H

(d)

H

H2N

Ph Cl

Ph CH3

(e) Br

Br Cl

H H

Cl

OH

H2N

Cl

Br

OH CH3

H

(f)

H Br CH3

7.

Br CH3

H H

Br Br CH3 CH3

Draw meso-2,3-dibromobutane as indicated below: Flying-wedge (staggered form) Æ Fischer Æ Sawhorse Æ Newman

Organic Chemistry—A Modern Approach

2.24

Solution 1 1

3 2

3

3 2 1 3

3

2

3 3

4

3

4 3

3

4

3

2 3

3

3 3 3

3 3 3

3

8.

Convert the following Fischer projections into Newman projections: (a)

H Cl

(c) H H

CO2H D Br

CHO

(b) F H

CH3 NO2

CH2OH

CH2Cl

Ph

CH2Br Br Cl

Cl Br CO2H

(d) H H

CO2Cl

Principles of Stereochemistry

2.25

Solution

(a)

H Cl

CO2H D Br

H Cl

CO2H CH2OH

CH2OH CHO

(b)

F H

F H

CH3 NO2

Ph H H

CH3 NO2 CHO CH2Cl

CH2Cl

(c)

D Br

H H

Cl Br

Cl Br Ph CO2H

CO2H

(d)

1.

2.

Draw Fischer projection of any stereoisomer of each of the following compounds: (a) Lactic acid (b) Alanine (c) Atrolactic acid (d) Mandelic acid (e) Glyceraldehyde (f) Tartaric acid (g) Amphetamine (h) Malic acid (i) 2-Bromo-3-chlorobutane (j) 2-Fluoro-3-iodobutanoic acid (k) 2,3,4-Trichloropentane (l) 2-Bromo-4-chloropent-3-one Convert the following Flying-wedge projections into Fischer projections: CN

NH2

(a)

H H3C

C CO2H

(b) H

C Ph

OH

Organic Chemistry—A Modern Approach

2.26

C2H5 H3C

CO2H

(d) H

C

(c) Ph

OH

C

OH Ph Cl

(e)

Ph

C

3

(f)

Br

2

H

NH2

HO2C

(g)

C

H

C

(h)

H

H C

H HO

OH

OH

3.

H3C

CO2H

C CH3

Convert the following Fischer projections into Sawhorse projections (eclipsed and staggered): CO2H Br CH3 D F Br (a) F (b) H (c) H3C F H H3 C Br Cl H CH3

C2H5

F Ph

Br

(e) Et Cl

(d)

CO2H CO2H

H H (f) 3C

CH3 H Ph

CH3

2

CHO

(g)

4.

H H

Cl Br

(h) H H

NH2 OH CH2Cl

Convert the following Sawhorse projections into Newman projections: CHO H

D H

(a)

(b) H

Cl CH3 Br

(c) H

Br Cl

Ph

Principles of Stereochemistry

2.27

Br

Ph CH3 CH3 Br

(d)

H

CH2F

Cl

5.

HO

H2 N

(e)

Br

Br

H

Ph

Convert the following Fischer projections into Newman projections: CH3

(a)

H H

Cl Br

(b)

CO2H OH H CO2H NH2 Cl Br CH3

H HO

CH3 CHO

(c)

6.

(f) Br

H

H HO

OH H CH2OH

(d) Ph Ph

Convert the following Flying-wedge projections into Newman projections through Fischer and Sawhorse projections: H OH H3C CH3 H3C (a)

C

H

C H

Br H Br

Ph H3C F

(e)

H HO

C

H C2H5

H

(d) H C 3 Ph

Ph C

HO

(f) HO2C CH3 Cl

C

CO2H

C Ph

C C

HO2C

C

H CHO

H C

H D H

C

HO

Cl

C

(g)

C

(b)

H

C

(c)

D

H Cl H C2H5

Organic Chemistry—A Modern Approach

2.28

7.

Identify the equivalent Fischer projections in each case: CO H

(a)

(b)

H

CH

OH ; H

H H

CO H OH ; HO

CH

CH ; HO

CH I

CO H II

H III

CO2H Cl H ; Br H CH3 I

CO2H CO2H H Cl ; ; Br H3C H Br Cl II III

OH CO H ; H

CH

H IV

Cl

CO H V

Cl

CO2H Cl ; H

H Br

HO2C H3C

H H Br V

CH3 IV 3

3

(c) 3 3

8.

3

Identify the equivalent Sawhorse projections: H

Cl

HO C

; Br Br CO H

H

H

CH

CH

CH

I

II

Cl

H

CO H

Br

Br CO H ;

; H C

H

Cl

Cl CH

H

H III

IV CO H Cl Br

H CH

H V

9.

Identify the equivalent projections in each case:

(a)

CO2H H H Br ; H F H CH3 I

CO2H

F Br

Br ; H

Br

H F ; H

CH3 CO2H

CO2H CH3

II

III

CH3 F

H ;

CH3 Br

H

H

F CO2H

IV

V

;

Principles of Stereochemistry

2.29

(b)

2.2

SYMMETRY ELEMENTS

Organic molecules, because of directional property of covalent bonds, assume definite geometric shape. On the basis of the geometric distribution of atoms and groups they can be classified as symmetric molecules and nonsymmetric molecules. Four fundamental elements of symmetry are encountered in organic molecules and these are: (a) simple axis of symmetry or rotational axis of symmetry (Cn), (b) plane of symmetry (s), (c) centre of symmetry (i), and (d) alternating axis of symmetry (Sn).

2.2.1

Simple Axis of Symmetry or Rotational Axis of Symmetry (Cn)

When rotation of a molecule by an angle of 360°/n around an imaginary axis passing through the molecule results in an equivalent structure or orientation that is superimposable with the original one, that axis, denoted by the symbol Cn (Latin word Circulate), is called the n-fold simple axis of symmetry. Iodoform, for example, has a three-fold (n = 3) simple axis of symmetry. When the molecule is rotated by an angle of 120° (360°/3) about the axis passing through the C—H bond, a structure indistinguishable from the original is obtained. indistinguishable

H C

I

I

I

120° rotation about the C3 axis

H C I

I I

120°

C3 axis in iodoform molecule

C3

Some other examples: 1. Deuterium oxide (D2O): The molecule has a two-fold simple or rotational axis of symmetry (C2) which bisects the D—O—D bond angle. indistinguishable O Da

Db 180°

C2

180° rotation about the C2 axis

O Db

Da

Organic Chemistry—A Modern Approach

2.30

2.

cis-1,2-Dideuterioethene: The molecule has a C2 axis that passes through the molecular plane and the mid-point of the C==C bond. indistinguishable

H D

C C

H

H

D

D

C C

H D

180°

C2

3.

trans-1,2-dideuterioethene: The molecule has a C2 axis that is perpendicular to the molecular plane and passes through the mid-point of the C==C bond. indistinguishable

H D

H

C C

180° rotation about the C2 axis

D

H D

C C

H D

180°

C2

4.

Difluoromethane (CH2F2): The molecule has a C2 axis that passes through the two H and two F atoms bisecting H—C—H and F—C—F bond angles. indistinguishable F

F

C

H H

F

180° rotation about the C2 axis

C

H H

F

C2

5.

Carbontetrachloride (CCl4): The molecule has four C3 axes each of which is perpendicular to the plane of three Cl atoms and passes through the fourth Cl atom and the carbon atom. Also, the molecule has three C2 axes each of which bisects two pairs of Cl—C—Cl bond angles.

Principles of Stereochemistry

2.31

Cl

Clc

C

Cl

120°

C C3

Cl

Cl

Cld

Clb

Cla

C

C3

120°

Clb

180°

Cld

Clc

C2

120° 120°

Clc C 180°

Cla

C2

Cla

Cld Clb

180°

C2

C3

C3

6.

Copper acetylide (Cu – C ∫ C – Cu): The molecule is linear. It has a C axis coincident with the internuclear axis since rotation around it by any angle gives an indistinguishable structure. The molecule also has an infinite number of C2 axes perpendicular to the centre of the C axis.

Cu

C

C

Cu

C

180°

C2

7.

180°

C2

cis-1,3-Dibromocyclobutane: The molecule has a C2 axis that passes vertically through the centre of the molecule. indistinguishable

Br 3 H

1

4

H

Br

2

1

H

H

H

H

Br

H

180° rotation about the C2 axis

H 4 H

H

2 Br

H

3 H

180°

C2

8.

trans-1,3-Dibromocyclobutane: The molecule has a C2 axis that passes through C–2 and C–4.

Organic Chemistry—A Modern Approach

2.32

indistinguishable Br 3

1

4

H

H

H

2

H

Br

H

180° rotation about the C2 axis

H

H

4

H

2

1

H

H 3

C2

180°

9.

Boron trifluoride (BF3): Trigonal planar BF3 molecule has one C3 axis that passes through the B atom and perpendicular to the molecular plane. Also, it has three C2 axes each of which passes through the B–F bond and bisects the angle between two other B–F bonds.

H

H

Br

F

F

180°

F

B

C2

F

B

F

10.

180°

C2

120°

Br

F

180°

C2 C3 Hexadeuteriobenzene (C6D6): The molecule has a C6 axis passing through the centre of the regular hexagon. It has six C2 axes. Three of them pass through the mid-point of two opposite bonds and the other three pass through the two opposite corners of the molecule. The molecule has also a C2 and a C3 axis collinear with C6 axis, i.e., C6, C2 and C3 are on one and the same axis. C6 is the principal axis of the molecule. D

180°

D

D

C2

D

D D

D

D

180°

D

D

180°

180°

C2

C2

D

D 180°

C2

D

C2

D

D 180°

C2

D D

D 60°

C6(C3, C2)

Three relevalent topics (i) Principal axis: If a molecule is found to possess Cn axis with different values of n, then the Cn axis having maximum value of n (fold or order) is called the principal axis. For example, C6 in the principal axis of C6D6 or C6H6 molecule. When a molecule has several Cn axes with the same value of n, then the principal axis is the one that passes through maximum number of atoms of the molecule.

Principles of Stereochemistry

(ii)

(iii)

2.2.2

2.33

Trivial axis: Rotation through an angle of 360° about any axis through any molecule gives back the identical structure. C1 axis is, therefore, known as the trivial axis. It is not included in the category of Cn axis. Indistinguishable and identical structures: The terms indistinguishable and identical have different meaning in the present context. The former refers to any equivalent structure arrived at by exchanging similar atoms or groups while the latter refers strictly to the original.

Plane of Symmetry (s)

An imaginary plane that bisects a molecule into two equal halves that are mirror image of each other is called the plane of symmetry. This is indicated by s (sigma, originated from the German word Spiegel which means mirror). The plane of symmetry is also called a mirror plane. For example, the eclipsed conformation of meso-2, 3-dibromobutane has a plane of symmetry. This may be shown as follows:

3

3

3

3

meso

It is to be noted that every planar molecule has a plane of symmetry coinciding with the molecular plane. The molecular plane of H – I, for example, is the s-plane. H — I Æ Molecular plane (s) 1.

2-Bromo-2-chloropropane: The molecule has a plane of symmetry passing through Br, C and Cl atoms. plane of symmetry Br

Br C

CH3 CH3 Flying-wedge projection Cl

H3C

CH3 Cl

Fischer projection

Organic Chemistry—A Modern Approach

2.34

2.

cis-1,3-Dichlorocyclobutane: The molecule has two s-planes passing through C-1 and C-3 and through C-2 and C-4. Cl

H

4

H

H

Cl 1

2

H

3.

s-plane passing through C-2 and C-4

H

3

s-plane passing through C-1 and C-3

H

trans-1,3-Dichlorocyclobutane: The molecule has a plane of symmetry which passes through C-1 and C-3. H

H

4

3

Cl

H

H

Cl 1

2

H

4.

s-plane passing through C-1 and C-3

H

1-chloro-3-methyl-1,2-butadiene: The molecule has a plane of symmetry passing through Cl, H and the three double bonded carbons. CH3

Cl C

C

CH3

H

5.

s-plane

C

Water (H2O): The molecule has two s-planes. One is the molecular plane and the other is perpendicular plane bisecting the O atom. perpendicular plane (s) O H

6.

H

molecular plane (s)

(2R,4S)-2,4-Dibromopentane-1,5-diol: The molecule has a plane of symmetry which passes through C-3, H and the –OH group (the designations R and S have been discussed later in this chapter).

Principles of Stereochemistry

2.35

CH OH

HOH C

CH OH

C s H Br

C R

C H

OH

H Br

Br

H

Br

H

Br

H

s-plane

CH OH Fischer projection

3D-representation or Flying-wedge projection

sh, sv and sd The symbol s is usually found to carry three subscripts indicating the position of the symmetry plane relative to the principal axis (Cn). (i) sh: It represents the s-plane perpendicular to the principal axis (h stands for horizontal). (ii) sv: It represents the s-plane containing the principal axis (v stands for vertical). (iii) sd: It represents the s-plane which contains the principal axis and bisects the angle between the two C2 axes (d stands for diagonal). Tetrachloroethene, for example, has both sh and sv and tetrabromoallene, for example, has two sd. sh Cl 2

sd

sd

Br

Cl C

sv

C

2

Cl

Cl

2

C

C

C

Br 2 Tetrabromoallene

Tetrachloroethylene

Br

principal axis

Br

The presence of sh, sv and sd in a single molecule, for example, in square plane XeF4 molecule may be shown diagrammatically as follows: C sd

sv sh F

F Xe

F

C

C F

principal axis Square palanar XeF (Xenon tetrafluoride) molecule

Organic Chemistry—A Modern Approach

2.36

2.2.3

Centre of Symmetry (i)

The centre of symmetry or a point of symmetry or an inversion centre (i) is a point within a molecule such that if an atom or a group is joined to it by a straight line and then extended an equal distance towards the opposite direction, it meets an identical atom or group. It is symbolized as i because inversion of all the atoms or groups in the molecule through this point provides a structure which is indistinguishable from the original. For example, trans-1,3-dimethylcyclobutane has a centre of symmetry.

Some other examples. 1. trans-1,2-Dichloroethene: i

2.

meso-Tartaric acid:

2 2

3.

trans-(1R,3S)-Di-sec-butylyclobutane:

S

R

Principles of Stereochemistry

2.2.4

2.37

Alternating Axis of Symmetry (Sn)

If a molecule is rotated by an angle 360°/n around an axis and then reflected on a plane perpendicular to that axis, a structure indistinguishable from the original is obtained, the molecule is said to have an ‘n’-fold alternating axis of symmetry (Sn). According to the mode of operation it may also be called rotation–reflection symmetry. For example, a-truxillic acid has a two-fold alternating axis of symmetry. S2

S2 180°

4

2 2

3

3

2

1

1

2

180° rotation about S2

ind

4

2

isti

2

ngu ish

a

abl

e

2

2

Some other examples 1. 1,3-Dichloro-2,4-difluorocyclobutane (I): The molecule has a two-fold alternating axis of symmetry. Cl F

3

H

180°

Cl I

4

H

H

180° rotation about S2

4

F ind

1

Cl

Cl 3

H

F

isti

ngu ish

abl

e

H mirror

H

H H

2

H

Cl F

2.

F

H

H

1

2

H

S2

F Cl I

5l5-Azaspiro[4.4]-2,3,7,8-tetramethylnona-5-ylium ion: The molecule has a four-fold alternating axis of symmetry.

Organic Chemistry—A Modern Approach

2.38

S4-axis

Me H

90°

H Me

2

H N

1

H

8

90° rotation about S4 6

5

9

Me

3 4

Me

7

H Me

Me

ind ist

ing

H Me N H H

Me

mirror

H

uis ha b

le

Me

Me H N

H Me

Me H

It is to be noted that a plane of symmetry (s is equivalent to one-fold alternating axis of symmetry (S1) and a centre of symmetry (i) is equivalent to two-fold alternating axis of symmetry (S2).

2.2.5

Symmetric, Asymmetric and Dissymmetric Molecules

A molecule is said to be symmetric if at least one of the symmetry elements s, i or Sn is present in it. Molecules in which there is no element of symmetry, i.e., Cn (except C1 which is present in all objects) and Sn axes (i.e. s, i and other Sn axes) are absent are called asymmetric molecules. On the other hand, molecules, in which Sn axis is absent, but Cn axis may be present or absent are known as dissymmetric molecules. Therefore, all asymmetric molecules are dissymmetric, but not all dissymmetric molecules are asymmetric. For example, trans-1,2-dibromocyclopropane is a dissymmetric molecule and not an asymmetric molecule because it has a C2 axis. However, the alanine molecule is both an asymmetric and dissymmetric one because both Sn and Cn are absent in it. H Br H

Br

CO2H 180°

C2

trans-1, 2-Dibromocyclopropane (a dissymmetric molecule)

H

C NH2

CH3 Alanine (an asymmetric or a dissymmetric molecule)

Principles of Stereochemistry

1.

2.39

Indicate the simple axis of symmetry present in the following molecules or ions: (b) trans-1,2-Dimethylcyclobutane (a) Ammonia (NH3) (c) CO32@ (d) 1,3-Dibromoallene (e) cis-(1R, 3R)-Di-sec-butylcyclobutane (f) trans-(1R, 3R)-Di-sec-butylcyclobutane (g) Chair form of dodecadeuterocyclohexane (h) Boat form of dodecadeuterocyclohexane (i) Anthracene Me

(k)

(j)

N

(l)

Me Br

(m)

Br

Cl Br

Cl

(p)

Me Cl

(n)

(o)

Cl

Cl (q) Tetramethylallene

Cl (r) trans-1,2-Dichlorocyclopropane

H HOOC

(s)

NH 2

OH

(t) H

COOH OH

Solution (a) Ammonia ( NH 3 ): The shape of the ammonia molecule is pyramidal. It has a C3 axis that passes through N atom and the perpendicular to the plane of three H atoms.

N H

H H 120°

C3

Organic Chemistry—A Modern Approach

2.40

(b)

trans-1,2-Dimethylcyclobutane: This molecule has a C2 axis that passes through the midway of C-1—C-2 and C-3—C-4 bonds.

H3C

H

H

CH3

C2

(c)

CO32 (carbonate ion): The planar carbonate ion has one C3 axis that passes through the C atom and is perpendicular to the plane of the species. It has three C2 axes, each of which passes through the C atom and one O atom and bisects the angle between two other C O bonds. 2–

2–

O3

O3

2–

2– O3

C

C

C2

2– 3 O

2– 3 O

C2

120°

C2

C3

(d)

180°

O3

1,3-Dibromoallene (BrCH=C=CHBr): This molecule has only one C2 axis that passes through the central sp-hybrizided carbon. It can be clearly demonstrated by its Newman projection. Br H

Br C

C

180°

H

180°

Br

Br

H

C

H 2 2

(e)

cis-(1R, 3R)-Di-1-methylpropylcyclobutane: This molecule has a C2 axis that passes vertically through the centre of the molecule. H

H

Et

H

Me C

C

4

3

H H

1

H

2

H 180°

C2

H

Me Et

Principles of Stereochemistry

(f)

2.41

trans-(1R, 3R)-Di-1-methylpropylcyclobutane: This molecule has a C2 axis which passes through C-2 and C-4. H

H

C

4

H

3

H H

1

H

2

H

C

Me

Me Et

H Et

C2

(g)

Dodecadeuterocyclohexane (chair form): The chair conformation of cyclohexane has a C3 axis passing vertically through the centre and three C2 axes passing through the centre and the mid-points of two opposite bonds. D D = Carbon atoms below D D D D the plane of the paper 180° D D 180° = Carbon atoms above D D C D 2 the plane of the paper C2 180° D 120°

C2

C3

(h)

Dodecadeuterocyclohexane (boat form): The boat conformation of cyclohexane has a C2 axis passing vertically through the centre of the boat. D D

D D

D D

D

D D

D D

D

180°

C2

(i)

Anthracene: Anthracene molecule has two C2 axes one of which passes through C-9 and C-10 and the other passes through the mid-points of C-2—C-3 and C-6—C-7 bonds. 7

8

9

5

10 4 180°

6

C2

1

2 180° 3

C2

Organic Chemistry—A Modern Approach

2.42

(j)

The molecule has three C2 axes passing through the mid-point of C-2—C-3, C-5—C-6 and C-7—C-8 bond and one C3 axis passing through C-2 and C-4 carbon atoms. 8 7 4 5

3

180°

C2

180° 6

1 120°

C2

2 180°

C3 C2

(k)

The triphenylene molecule has three C2 axes passing through the mid-point of C-2—C-3, C-6—C-7 and C-10—C-11 bonds. It has also a C3 axis passing vertically through the centre of the middle ring. 10 11

9 180°

C2

8

12

7 6 5

4 180° 3

1 2

C2

(l)

120°

180°

C3

C2

The molecule has only one C3 axis passing through nitrogen and the carbon opposite to it (C-4). 8

Me 7

4 5

3

N

6

Me

Me C3

(m)

2

120°

The molecule has three C2 axes each of which pass through Br—C and the opposite C—H bonds. It has also a C3 axes passing vertically through the centre of the ring. Br

Br Br Br

Br C2

180°

180°

C2

180°

Br

C2

120°

C3

Principles of Stereochemistry

(n)

2.43

The molecule has two C2 axes passing through two Cl atoms and the mid-points of C-2—C-3 and C-5—C-6 bonds. It has also a third C2 axis passing vertically through the centre of the ring. Cl 6

1 2

5

4

180°

Cl

C2

3

Cl

Cl

180° 180°

C2

(o)

C2

The molecule has only one C2 axis passing through C-2 and C-5. Cl 1 2

6 180° 5

4

3

Cl

C2

(p)

The molecule has only one C2 axis passing through the mid-points of C-1–C-2 and C-4—C-5 bonds. 5

6 1

Cl

2

Cl

C2

4 3

(q)

Tetramethylallene (Me2C=C=CMe2): The molecule has three C2 axes. One of them passes through the C atoms and each of the rest two axes passes diagonally through the central sp-carbon atom. H

C2

180°

180°

H 3C

C

C

C

CH3

CH3

H3C

CH3 C

C

CH3 H3C

18°

CH3

18°

C2

(r)

C

CH3

H3C 180°

C2

CH3

180°

C2

C2

trans-1,2-Dichlorocyclopropane: The molecule has only one C2 axis passing through C-3 and the mid-point of C-1—C-2. H

3

1

Cl 2

H

Cl 180°

C2

Organic Chemistry—A Modern Approach

2.44

(s)

Amide ion ( NH 2 ): The ion has a C2 axis that bisects the H—N—H bond.

H

N

–

H 180°

C2

(t)

This anti-staggered conformation of the active tartaric acid molecule has a C2 axis that passes through the mid-point of C-2–C-3 bond. indistinguishable

H

HO2C H

3

2

OH

CO2H OH

180°

CO2H 180° rotation about the C2 axis

H

2

H

3

OH

CO2H OH

C2

2.

Indicate the element(s) of symmetry (other than Cn) present in each of the following molecules: (a) cis-1,2-Dichlorocyclopropane (b) a-Truxillic acid CH3

H H Et C C C C H C H Br H H O

Et

(c)

H N

(e) H3C C O

Br

H

(d) CH3

C

C CH3

D

N H

(f)

H

D O

H CH3

C NH

(h)

H CH2OH H

CH2OH

NH C

(g) H3C

Br

H

H

C

D H 3C

C

C Cl

D CH3

H

O H

H Cl

(i)

(j) CH3

Cl H

CO2H H

Principles of Stereochemistry

2.45

CH3

CH3 H3C

O

O

(k)

(l)

O

(m) H3 C

CH3

H3C H

H

H

(n)

Me Me

C

H3 C

C

CH3

H

H

(o)

Et

O

(p)

Et

O

H

H

Cl

(q) Br H

O

C

O

H

O

(r) Me

H Br

Me

Cl

H Me

(s)

H H

+

(t)

Me

N H Me

Me H

H3C C H CH3

C

CH3 H CH3

Cl

(u) Br H

H Cl

O

H Solution

(a)

cis-1,2-Dichlorocyclopropane: The molecule has a plane of symmetry passing through C-3 and the mid-point of C-1—C-2 bond. Cl 1

H Cl 2

H

H 3

H

plane of symmetry (s)

Organic Chemistry—A Modern Approach

2.46

(b)

a-Truxillic acid: The molecule has a centre of symmetry. HO2C

H Centre of symmetry (i)

H

Ph H

(c)

Ph

H

CO2H

The molecule has a plane of symmetry passing through Br, C and H atoms. s-plane H

Et H

H

C C

C C

C

Et H

Br H

(d)

The molecule has a plane of symmetry passing through the two Br atoms, two doubly bonded carbons and the mid-point of a ring bond. CH3 H H3C

Br

H

(e)

The molecule has a centre of symmetry. It has also an S2 axis which is equivalent to (i). Centre of symmetry (i)

H

O

H N

C

C

N

H3C

(f)

plane of symmetry (s)

Br

C

CH3

O

H H

The molecule has a plane of symmetry passing through the mid-points of C-2—C-3, C-5—C-6 and C-7—C-8 bonds 8 7 5 6

plane of symmetry (s)

D 3

D

2

H

H CH2OH CH2OH

Principles of Stereochemistry

(g)

2.47

The molecule has a centre of symmetry. H N

s-plane

O C

H 3C

(h)

C

N H

C

D CH3

centre of symmetry O It has also a plane of symmetry which is the molecular plane and an S2 axis which is equivalent to (i). The molecule has a plane of symmetry passing through Cl, C and H atoms. H H D

C C

H3C

(i)

CH3

H

Cl

plane of symmetry (s)

The molecule has a centre of symmetry and a plane of symmetry passing through C-1, C-4, two H and two Cl atoms. plane of symmetry (s) H H

H

Cl

Cl

H

H H

H Cl

(j)

H

H centre of symmetry (i)

H

Cl

H H

The molecule has a plane of symmetry passing through C-1 and C-4. H 4

H3C

plane of symmetry (s) 1

(k)

CO2H H

The molecule has a centre of symmetry and it also has an S2 axis. H 3C

O O

H O

H3C CH3

O

H

CH3 centre of symmetry (i)

Organic Chemistry—A Modern Approach

2.48

(l)

The molecule has a plane of symmetry passing through C-2 and C-5. H

CH3

1

H3C

5

6

H

plane of symmetry (s)

3

(m)

CH3

2

H3C

4

The molecule has a plane of symmetry passing through C-1 and C-4. O CH3

CH3 O H3 C

(n)

O

plane of symmetry (s)

O

O O

1

H3C

H 4

CH3

H

CH3

H

The molecule has a plane of symmetry passing through H—C— C —O—C—H || system.

O s

(o)

The molecule has a plane of symmetry passing through the mid-points of C-2–C-3 and C-5–C-6 bonds and C-7 including the —CH3 group and H atom. s

(p)

The molecule has a centre of symmetry and an S2 axis. There is also a plane of symmetry passing through the two methylene carbons and two opposite ring bonds. H

H centre of symmetry (i)

plane of symmetry (s) H

H

Principles of Stereochemistry

(q)

2.49

The molecule has a centre of symmetry and so also it has an S2 axis. H

Cl Br

H

H

(r)

Br

H

Cl centre of symmetry (i)

The molecule has a plane of symmetry passing through C=O and the mid-point of C-3–C-4 bond. Me

O Me

Me 4

O

H Me

3

4

H

(s)

3

plane of symmetry (s)

The molecule has a four-fold alternating axis of symmetry. H

Me

Me

H

N Me H

H Me

H Me

Me H 90° rotation about the S4 axis

N Me

H

90°

Me

H S4

ind

ist

ing

uis

ha

ble

mirror

H

Me H

Me

N Me H

(t)

H Me

The molecule has a plane of symmetry passing through two phenonium ion ring carbons and two cyclopropane ring carbons.

Organic Chemistry—A Modern Approach

2.50

plane of symmetry (s)

+ CH3

H3 C H H3C

(u)

C

C

H CH3

The molecule has a plane of symmetry passing through C==O and C–3 carbon including Br and H. Cl Br

H

plane of symmetry (s)

O

Cl

H

H

3.

Which of the following molecules have plane of symmetry? H

H

(a)

(d)

HO H

(b) Br H

CO2H OH CO2H

HO H

CHO H OH

OHC

OH

H CH3 Br

(c) OHC HO

CH3

HO

H

CHO

H CH3

CH3

(e)

OH H

Br Cl

H Cl

H3C

Br

(f) H

Br C2H5

H CH3 H

H

(g)

(i)

H3C

C C

C C

Br H CO2H HO H H OH H HO

OH H CO2H

H

CH3

CH3

(h)

C

H D

CO2H

CH3

C CH3

H D

Principles of Stereochemistry

2.51

Solution

H

(a)

CO2H

HO H

CO2H OH

H H

CO2H

H

Br H

CH3 Br

Br Br

CH3 CH3

CH3

(c)

plane of symmetry (s)

CO2H

H

(b)

OH OH

plane of symmetry (s)

H

The molecule has no plane of symmetry (s).

(d) s

CH3

(e)

Br Cl H3 C

CH3 H Cl

Br Cl

H Cl

Br

Br

H

H

(f)

plane of symmetry (s)

CH3

The molecule has no plane of symmetry. CH3 H H

(g) H3C

C C

C

CH3

C H

H

Br

H H3C

Br H

C

C

C H

C CH3

CH3

H plane of symmetry (s)

Organic Chemistry—A Modern Approach

2.52

CH3

(h)

C

H H

CO2H

CH3

C CH3

H D plane of symmetry (s)

CO2H

(i)

HO H

H OH

H HO

OH H

plane of symmetry (s)

CO2H

4.

Draw various conformations of n-butane in Newman projection considering rotation about the C-2—C-3 bond and indicate the conformation that has a plane or centre of symmetry.

Solution The various conformations of n-butane obtained by rotating the front carbon about the C-2—C-3 bond may be shown as follows:

Principles of Stereochemistry

2.53

The conformation I has a plane of symmetry. The conformations II, III, V and VI have neither a plane nor a centre of symmetry. The conformation IV has a centre of symmetry which exists in the mid-point of C-2—C-3 bond. 5.

Identify the elements of symmetry (if any) in the following molecules as shown below. F H

Cl

(b)

(a)

H Me

(d)

Cl

C

C

C

C

H

Cl

Cl H

Br

(e) CHCl3

Me

H

Br Cl

H H

H C

(f)

(c)

C

C

H

(g) CH4

H

Solution (a) It has no element of symmetry. (b) It has only one plane of symmetry, i.e., the molecular plane. (c) It is a cumulene with odd number of double bonds and so, the molecule is planar. It has one C2-axis perpendicular to the molecular plane and passing through the centre of the molecule, a sh bisecting the C2 and a centre of symmetry (i).

180° H

Cl C

C

C

H

C Cl

H

Cl sh

C H

C

C

C

Cl centre of symmetry (i)

C

(d) (e)

It has a centre of symmetry. It has one C3 axis and three sv planes. The C3 axis and one sb plane is shown in the following figure.

Organic Chemistry—A Modern Approach

2.54

H Cl Cl

sv

C Cl 120°

C3

(f)

It has three C2 axes and two sd planes (i.e., planes that contain the principal axis and bisect the angle between the two C2 axes. C2 H

C

H

sd

C C2

C2 C H

(g) 6.

H

sd

It has three C2 axes, four C3 axes, three S4 axes, and six sd planes. Identify the elements of symmetry (Cn and s) of o-, m- and p-dichlorobenzenes.

Solution o-Dichlorobenzene: It has a C2 axis passing through C-1—C-2 and C-4—C-5 bonds. Also, it has two sv planes.

Cl

180°

Cl

sv (molecular plane)

sv

C2

m-Dichlorobenzene: It has a C2 axis passing through C-2 and C-5. Also, it has two sv planes. s

Cl

Cl

180°

C2

s (molecular plane)

Principles of Stereochemistry

2.55

p-Dichlorobenzene: It has three C2 axes. Also, it has two sv and one sh planes. sv sv (molecular plane)

Cl

sh

C2

Cl

Cl

180° 180°

Cl

C2 180°

C2

7.

Identify which of the following objects possess at least one plane of symmetry: (a) an empty spool, (b) a hand, (c) a cup with DAD written opposite the handle, (d) a cup with MOM written opposite the handle, (e) a spoon, (f) a six-pointed star-like cake, (g) a bicycle, (h) a spectacle, (i) a sock.

Solution The objects with at least one plane of symmetry are (a), (d), (e), (f), (h) and (i).

8.

Draw a square planar molecule with the formula X2Y2A (A is the central atom) which has a centre of symmetry and at least two planes of symmetry. How many planes of symmetry are actually present in the structural formula you have drawn?

Solution The structural formula of the molecule (X2Y2A) having a centre of symmetry and at least two planes of symmetry is as follows:

Centre of symmetry (i) X

Y A

Y

X

s (molecular plane) s-plane passing through Y-A-Y

s-plane passing through X-A-X

There are three planes of symmetry in the molecule. They are: (i) the plane through X–A–X and perpendicular to the molecular plane, (ii) the plane through Y–A–Y and perpendicular to the molecular plane and (iii) the plane that cuts through all five atoms of the molecule, i.e., the molecular plane.

Organic Chemistry—A Modern Approach

2.56

1.

Indicate the simple axis of symmetry present in each of the following molecules/ ions: ≈

(a)

(c) CH3COO@

(b) CMe3

Cyclohexane

O

(d)

BBr3

(e)

O

N

(f) N N O

≈

(g)

H3O

O

Et

F

Et

F

(h)

H

H

Br

(i)

Br H Br

Br

(j)

(k)

(m)

(n)

H H

≈

(l)

NH4

F C

C

F O

OCH3

(o)

(p)

CH3 C

H3C

H O

(q)

C

D

C

D

O

H

(r)

(s)

(t)

(v)

(Basketane)

O Cl

(u) 2.

Cl

O Indicate the element(s) of symmetry (other than Cn present in each of the following molecules: (a) 3-bromopentane (b) 1-chloro-3-methyl-1,2-butadiene (c) quinuclidine (d) 4-bromopiperidine (e) trans-2,5-dimethyl-1,4-dioxan (f) meso-3,4-dibromohexane (g) cis-1,3-dibromocyclohexane (h) 8-iodospiro [4,5] decane

Principles of Stereochemistry

(i) (k) (l) (n) (o) (p) (q)

2.57

trans-1, 4-dimethylcyclohexane (j) trans-diethyldiketopiperazine 1,1-dichlorocyclopropane cis-cyclobutane-1, 2-dicarboxylic acid (m) cis-1,3-dibromo-cyclopentane cis-2,3-dimethylcyclohexane-r-1-ol cis-1 4-ditrichloromethyl-cyclohexane cis-3, cis-5-dimethylcyclohexane-r-1-ol trans-3, trans-5-dimethylcyclohexane-r-1-ol (r) 1,3-dioxan Cl

(s)

cubane

Cl C

(t)

C

C

C

F

F

CH3

Br

C CH3

(u)

Br

H

(v)

Br

H

Br H

H

,

Br

H Cl

Br

Cl

(w) Br

CH3

H Br

Cl

Cl

N C Br (z) H H3 C CH3 H Br Cl Br Which of the following molecules may have a plane of symmetry? C

(y)

3.

CH3 C C C

(x)

H

H

(a)

HO

CH3

H

(b) Cl

CH2CH3

H

OH

Cl CH2CH3

CH3 H

(c)

HO2C

CCl3 OH

HO

H

HO

CO2H H

(d)

Cl

H

H

Cl

Cl3C

Cl H

Organic Chemistry—A Modern Approach

2.58

Et

(e)

CH3

I

(f) H

H C O

Et

Cl CH(CH3)2

I H Et H H

(g) Et

C C

C

H

(h) CH3

C

CHO

Et H CH3

C

C

H CH2OH CH3

Cl

H

H

2

(i)

(j) H

B N

N

B H 2

4.

5. 6.

Et

H

B N

H

H

Identify which of the following objects possess at least one plane of symmetry: (a) a chair, (b) a glove, (c) a nail, (d) a cup, (e) a nose, (f) a shoe, (g) a tree, (h) a sevenpointed star, (i) a plate, (j) a book, (k) a gas cylinder. Identify which of the following objects possess a centre of symmetry: (a) five-pointed star, (b) a dumb-bell, (c) a carom board (d) an eight-point star, (e) a tennis ball. Out of twenty six letters in English language how many of them are symmetric and how many of them are non-symmetric (considering them two-dimensional)? [Hint: F, G, J, L, P, Q and R are non-symmetric and remaining nineteen letters are symmetric.]

2.3 ISOMERISM: CONSTITUTIONAL ISOMERS AND STEREOISOMERS 2.3.1 Constitutional Isomers Isomers are different compounds that have the same molecular formula. Molecules having the same molecular formula but different molecular constitution, i.e., different atom-to-atom bonding sequence or atomic connectivity are called constitutional isomers or structural isomers and the phenomenon is called constitutional or structural isomerism. Examples of constitutional isomers are as follows:

Principles of Stereochemistry

2.59

Molecular formula

Constitutional isomers

C4H10

CH3 —CH 2 —CH 2 —CH3 Butane

C3H8O

CH 3 —CH 2 —CH 2 —OH 1-Propanol

2.3.2

and

CH3CH 2OH 1-Propanol

C2H6O

CH3 | CH3 —CH — CH3 2-Methylpropane

and

and

OH | CH3 —CH — CH3 2-propanol

CH 3 —O—CH3 Dimethylether

(chain isomers)

(position isomers)

(functional group isomers)

Stereoisomers

Isomers having similar molecular constitution but differing in the spatial arrangement of the various groups or atoms about a rigid part of the molecule (e.g., an asymmetric carbon atom, a double bond, etc.), i.e., having different configurations are called stereoisomers and the phenomenon is called stereoisomerism. Some examples of stereoisomers are given as follows. CO2H CO2H (i)

HO

C

and H

CH3

(R)-Lactic acid

(ii)

H3C

H

C

OH CH3

H

(two optical isomers)

(S)-Lactic acid H3C

H (two cis-trans isomers)

H3C

(iii)

H

cis-2-Butene CO2H HO2C

H

CH3

trans-2-Butene HO2C

H

(two cis-trans isomers) H H cis-Cyclopropane-1, 2-dicarboxylic acid

2.3.3

CO2H H trans-Cyclopropane-1, 2-dicarboxylic acid

Enantiomers and Diastereoisomers

Stereoisomers can be classified into two categories: (i) those that are enantiomers of each other, and (ii) those that are diastereoisomers of each other.

Organic Chemistry—A Modern Approach

2.60

2.3.3.1

Enantiomers

Stereoisomers whose molecules are related to each other as non-superimposable mirror images are called enantiomers. For example, alanine, CH3CH(NH2)CO2H, has two stereoisomers having mirror image relationship, but one is not superimposable on the other. To superimpose an object on its mirror image means to align all parts of the object with its mirror image. With molecules, this means aligning all atoms and bonds. Therefore, they represent a pair of enantiomers.

Since enantiomers are usually optically active, i.e., they turn the plane of polarized light to an equal degree but in opposite directions, they are also called optical isomers or optical antipodes.

2.3.3.2

Diastereoisomers

Stereoisomers whose molecules are not mirror images of each other are called diastereoisomers or diastereomers. The simplest example of diastereoisomers may be cited in the isomers of 2-butene.

By examining the structural formulas for cis- and trans-3-butene, we see that they have the same molecular formula (C4H8) and the same connectivity (both compounds have two central carbon atoms joined by a double bond and both compounds have one methyl group and one H atom attached to each carbon atom). But their groups/atoms have different arrangement in space (around the regid C==C bond) that is not interconvertible from one to another (because of large barrier to rotation of the C==C bond), making them stereoisomers. They are different compounds, as revealed from their physical and chemical properties. Furthermore, they are stereoisomers that are not mirror images of each other.

Principles of Stereochemistry

2.61

Therefore, they are diastereoisomers and not enantiomers. In fact, they represent a pair of cis-trans isomers and since they are not optically active, they are not included in the class of optical isomers. Cis- and trans-isomers of cycloalkanes provide another example of stereoisomers that are diastereoisomers. For example: diastereoisomers Et Et

Et H

H

H

H

-1,2-Diethylcyclopentane (C8H16)

Et

-1,2-Diethylcyclopentane (C8H16)

These two compounds have the same molecular formula (C8H16), the same atom-to-atom bonding sequence or atomic connectivity, but different arrangement of their atoms/groups in space, i.e., different configurations. In one compound, both ethyl groups are bonded to the same face of the ring, while in the other compound, the two ethyl groups are bonded to opposite faces of the ring. Furthermore, interconversion of the positions of the ethyl groups cannot be possible by conformational change. Therefore, these two compounds are stereoisomers and since they are not mirror images of each other, they can be further classified as diastereoisomers.

[Geometric or cis-trans isomerism Isomers having same molecular constitution, i.e., the same atom-to-atom bonding sequences or atomic connectivity, but differing in spatial arrangement of the groups or atoms owing to the restricted rotation about a double bond (C==C, C==N, etc.) or about a ring are called geometric or cis-trans isomers and phenomenon of existence of such isomers is called geometric or cis-trans isomerism. For example, 1,2-dibromoethene exists in two stereoisomeric forms (I and II) as shown below, one being different from the other in the spatial arrangement of the two bromine atoms and the two H atoms. The isomer I with the similar groups on the same side of the double bond is called the cis-isomer and the isomer II with similar groups on opposite sides of the double bond is called the trans-isomer. Similarly, III is the cis-1,2-dimethylcyclopropane and IV is the trans-1,2-dimethylcyclopropane.

Organic Chemistry—A Modern Approach

2.62

The condition for an alkene to exhibit geometric or cis-trans isomerism, i.e., to exist as cis- and trans- stereoisomers, is that each of the doubly bonded carbon atoms must be attached to two different atoms or groups. Therefore, an alkene Cab=Ccd will exist in two stereoisomeric forms if a π b and c π d. It does not matter whether a and/or b are the same as c and/or d. Similar condition is applicable for cyclic systems for showing geometric or cis-trans isomerism. In case of polyenes, the number of cis-trans isomers depends on the number of double bonds in the molecule as well as the terminal substituents. If the general formula of a polyene is R—(CH==CH)n—R¢ (with different terminal substituents), the number of stereoisomers is 2n and if the general formula is R—(CH==CH)n—R (with identical terminal substituents), the number of stereoisomers is (2n–1 + 2p–1) where p = n/2, when n is even, and p = (n + 1)/2 when n is odd. For example, four cis-trans isomers are possible for octa-2,4-diene (with different terminal groups) and three cis-trans isomers are possible for octa-3,5-diene (with identical terminal groups). 1

CH3

CH3

2

3

H

C

C

4

5

C

C

H

6

7

H

8

C

C

CH2CH2CH3

C

H CH2CH2CH3

(2Z, 4E)-Octa-2, 4-diene 1

H C

C

H

C

H H (2E, 4Z)-Octa-2,4-diene

H

C

H 3C

C

CH3

H

CH2CH2CH3

C

C

CH2CH2CH3 H (2E, 4E)-Oct-2, 4-diene H

H

H

C C H H (2Z, 4Z)-Octa-2, 4-diene

2

CH3CH2

3

4

C

C

H

H

H 5

6

C

C

H 7

8

C

H

C

CH2CH3 C

H

C

C

H H (3Z, 5Z)-Octa-3,5-diene

H

H

CH2CH3

C

CH3CH2

CH2CH3 H (3E, 5E)-Octa-3,5-diene CH3CH2

H C

H C

C

H

CH3CH2

C

C C H CH2CH3 (3Z, 5E)-Octa-3,5-diene

H

(3E, 5Z)-Octa-3,5-diene Identical

There are some other molecules (acyclic systems containing no C==C bond) that can exist as diastereoisomers. For example, active tartaric acid and meso-tartaric acid represent a pair of diastereoisomers.

Principles of Stereochemistry

2.63

diastereoisomers

CO2H H

CO2H H

OH

H CO2H Active tartaric acid

OH

OH CO2H meso-Tartaric acid

HO

H

Acylic diastereoisomers are possible if a compound contains more than one asymmetric or chiral carbon.

2.3.3.3

Determination of the stereochemical relationships between molecules

A series of questions may be framed with yes or no answers in such a way that the relationship between two compounds may be traced through no stereochemical relationship to a diastereoisomeric relationship. They are not isomers

No

Yes

Do the molecules have the same molecular formula?

No

They are homomers, i.e., identical molecules which are superimposable

No

They are constitutional or structural isomers

No

Are they superimposable?

They are isomers

Yes

Do they have the same molecular constitution, i.e., atom-to-atom bonding sequence or connectivity?

They are stereoisomers Are they non superimposable mirror-images? Yes

They are enantiomers

No

They are diastereoisomers

Organic Chemistry—A Modern Approach

2.64

2.3.3.4

Chemical and physical properties of enantiomers and diastereoisomers

The atoms in two enantiomers have exactly the same spatial arrangements with respect to distance and dihedral angles (interactions). Two enantiomers are, therefore, isomeric to each other. Because of this, enantiomers exhibit the same reactivity (Eact values are the same) towards achiral reagents, solvents, catalysts and conditions. However, two enantiomers react at different rates, i.e., show different reactivity towards chiral reagents, solvents, catalysts and conditions. The transition states developed from the individual enantiomer and the chiral reactant are not enantiomeric. They are, in fact, diastereoisomeric and hence possess different energy of activation (DG=|) due to different enthalpies of activation (DH =|). Because of this, the two enantiomers react with a chiral reagent at different rates. The two enantiomers have identical physical properties like melting points, boiling points, solubilities, densities, refractive indices, dipole moments, etc. because they are isometric to each other. However, they differ in their behaviour towards plane-polarized light (rotation in two opposite directions but in the same magnitude) and this is because a plane-polarized light is composed of two dissymmetric components, a right-handed and a left-handed circularly polarized light beam and thus constitutes a chiral environment. The atoms in two diastereoisomers have different spatial arrangements of atoms with respect to distance and dihedral angles (interaction) and are, therefore, anisometric relative to one another. For this reason, neither any two diastereoisomers nor their transition states have the same energies (Eact values are different) and as a consequence, they react at different rates with both chiral and achiral reagents. Because any two diastereoisomers are anisometric to each other, they also differ in physical properties like melting points, boiling points, refractive indices, dipole moments, specific rotations, etc. They may have the same or opposite sign of specific rotation or some may be optically inactive (do not rotate in plane-polarized light). They can be conveniently separated from each other by fractional distillation or fractional crystallization because their boiling points and solubilities are different. Again due to differences in molecular shape and polarity, they undergo differential adsorption on an adsorbent and can be separated easily by chromatography.

1.

How are the compounds in each pair related to of each other? Are they homomers (i.e., identical), enantiomers, diastereoisomers constitutional isomers or not isomers of each other? CH3

H3C

(a)

C

C

Br

H Br

H

H Br

H3C and

C H Br

C CH3

Principles of Stereochemistry

(b)

Cl C Cl

(c)

H C D3 C

2.65

H Cl and C H H

C

Cl H

C

CD3 H and C D3C H

C

H CD3

C

Br Cl and

(d)

Br Cl 3

(e) 3

(f)

and HO

(g)

CH3 and

H

H

Br Cl

C I

C

CH2Br

Cl

(h)

CH2OH

H3 C

C

and

Br

H

C

H

Br I C2H5

C2H5

(i)

N

H

Ph

and

H

and

H

Br Br

(k)

Ph

H

CH(CH3)2

(j)

N

(CH3)2CH

H Br Br H

H3C

OCH3

CH3O

CH3

and CH3O

(l)

H Me

CH3 S

and O

Me H

H3 C S O

OCH3

Organic Chemistry—A Modern Approach

2.66

O

O

P

(m)

and D OMe

EtO

P EtO

OMe D

O

O

(n)

Cl

and

Cl

Solution (a) Homomers:

H H 3C H Br

2

3

C

C

H 3C

CH3

Br H 1st structure

180° rotation of C-3

H Br

C

H Br

H3C

CH3

H Br

C

1st structure

C

Br

C

CH3

2nd structure

Superimposable and therefore, identical

(b) (c) (d) (e) (f) (g) (h)

Constitutional isomers: (1,1-dichloroethene and 1,2-dichloroethene) Diastereoisomers: (they are geometric or cis-trans isomers) Constitutional isomers: (1-bromo-2-chloronaphthalene and 2-bromo-1chloronaphthalene) Constitutional isomers: (1,1-Dimethylcyclopentane and cycloheptane) Not isomers of each other: (naphthalene and decaline) Constitutional isomers: (1-bromo-2-propanol and 2-bromo-1-propanol) Enantiomers: Mirror plane Cl C I

Cl Br H

Br I

Cl

C

C H

1st isomer

H

Br I

2nd isomer

1st isomer nonsuperimposable mirror images

Principles of Stereochemistry

(i)

2.67

Enantiomers: Mirror plane C H

C H

N H Ph CH(CH )

Ph

N

C H (CH ) CH H

CH(CH ) H

1st isomer

1st isomer

N

Ph

2nd isomer nonsuperimposable mirror images

(j) (k)

Diastereoisomers: (they are geometric or cis-trans isomers) Homomers:

H C

OCH

CH

CH O

CH O CH 1st structure

CH

CH O

H C OCH 1st structure

H C OCH 2nd structure

superimposable and therefore, identical

(l)

Enantiomers: Mirror plane S H O Me 1st isomer

180° rotation about the axis

S

Me H 1st isomer O

Me H

S O

2nd isomer

nonsuperimposable mirror images

Organic Chemistry—A Modern Approach

2.68

(m)

Enantiomers: Mirror plane O P EtO

O 180° rotation about the axis

D OMe

P

MeO D

1st isomer

O P EtO

OEt

1st isomer

OMe D

2nd isomer

nonsuperimposable mirror images

(n) 2.

Enantiomers: (nonsuperimposable mirror images) Label the following pairs of structures as homomers, constitutional isomers, diastereoisomers or enantiomers: H F F H Cl

(a)

(b)

F H H F Cl

and

and

Cl

Cl Cl

(c)

Cl I

O

(d)

Cl

and

Cl I

O and

I

I CH3

(e)

C

C

and

C

H

CH3

Br H

H

H

CH3

Br

H C

C

CH3 H

Br and

C

H3C

H

CH3

(f)

H

H

H3 C H

H

H Br

CH3

Principles of Stereochemistry

2.69

5 2

2

5

(g)

CH3

and

(h) CH3 Br

(i)

(j)

Br F

F

and

F

F Br H

Et

H

H

Et and H

H

Br H

Et

H

H Et H OH

H3C

(k)

H

H HO

C CH3

and

H

C

H 3C

H

CH3

H Br

(l)

H

Br C C C C

H

Br and

H

H C C C C

Br

Solution (a) Enantiomers:

Mirror plane H F F H 1st isomer

180° rotation about the axis

F H H F

F H H F 1st isomer

2nd isomer

nonsuperimposable mirror images

Organic Chemistry—A Modern Approach

2.70

(b) (c)

Diastereoisomers: (they are geometric or cis-trans dichlorocyclobutane and trans-1,2-dichlorocyclobutane) Homomers:

isomers:

cis-1,2-

Cl Cl

Cl

180° rotation about the axis

Cl

Cl

1st structure

Cl 2nd structure

1st structure

superimposable and therefore, identical

(d) (e)

Diastereoisomers: (they are geometric or cis-trans isomers) Enantiomers: Mirror plane

CH3

H C

C

C

H

H 90° rotation about the axis

CH3

H C

C

C

CH3

1st isomer

H

H C

CH3

C

C

CH3

1st isomer

CH3 2nd isomer

non-superimposable mirror images

(f)

Homomers: H

CH3 Br H

H

H

CH3

Br

H

Br 180° rotation about the axis

H3C H

H

H Br

1st structure

CH3

H

Br CH3 H

H

H Br

2nd structure

1st structure superimposable and therefore, identical

CH3

Principles of Stereochemistry

(g)

2.71

Homomers: H 5 C2

H

H

C2H5

H

C2H5

180° rotation about the axis

1st structure

2nd structure

1st structure superimposable and therefore, identical

(h)

Enantiomers:

nonsuperimposable mirror images

(i)

Homomers: Br

Br F F

180° rotation about the axis

F

F

Br

F

F

Br

Br

Br

1st structure

1st structure

2nd structure

superimposable and therefore, identical

(j)

Homomers: [1st structure on 180° rotation in the plane of the paper gives the second structure.]

Organic Chemistry—A Modern Approach

2.72

(k)

Homomers: H

HO H3C

H

H HO

180° rotation about the axis

C CH3

H

C

CH3

H

CH3

1st structure

H 1st structure

H

OH

H

C

CH3

H

CH3

2nd structure

superimposable and therefore, identical

(l)

Diastereoisomers: (cumulated trienes with two different terminal groups exist as cis-trans or geometric isomers)

3.

Tell whether the compounds of each pair are enantiomers, diastereoisomers, constitutional isomers or homomers. CO2H

(a) H

OH

OH

and HO2C

CH3 CH2OH

(b) H H

H CH2OH

OH OH

and

HO

H

HO

H

CH2Br CO2H

(c)

H HO

NH2 H

CH2Br CH2OH H and

H

and

H

OH

H

OH

CH2CH3 CO2H

(e)

H

H CO2H CHO

H OH

OH

H2 N

CH2OH CHO

(d) HO

CH3

CH2CH3 CO2H

Br and

Br

H

Br

H

H

Br

Br

H

H

Br

Br

H

H

Br

CO2H

CO2H

Principles of Stereochemistry

2.73

H

CHO

HO

(f)

CHO

HO

H

HO

H

H

and

H

OH

H

OH

HO

OH

H

H

OH

CH2OH

CH2OH

Solution

(a)

Enantiomers: Mirror plane CO2H H

OH

OH H3C

CH3 1st isomer

OH CO2H

HO2C

CH3

H

H

1st isomer

2nd isomer

nonsuperimposable mirror images

(b) (c)

Enantiomers: (nonsuperimposable mirror images) Enantiomers: Mirror plane COOH H HO

NH2 H CH2OH

1st isomer

180° rotation in the plane of the paper

CH2OH H H2 N

OH H

CH2OH HO H

COOH 1st isomer

H NH2 COOH

2nd isomer

nonsuperimposable mirror images

Organic Chemistry—A Modern Approach

2.74

(d)

Diastereoisomers: CHO HO H

CHO H

H

OH

OH

H

OH

CH2CH3 1st isomer

CH2CH3 2nd isomer

stereoisomers which are not mirror images of each other

(e)

Enantiomers: (nonsuperimposable mirror images)

(f)

Diastereoisomers: H HO HO HO H

CHO CHO H H OH

H HO HO H

CH2OH

CHO OH H H OH

CH2OH 1st isomer

1st isomer

H H HO H

OH OH H OH CH2OH 2nd isomer

stereoisomers which are not mirror images of each other

4.

Label the following pairs of structures as homomers, constitutional isomers, enantiomers or diastereoisomers: C2H5

(a) H

Cl

Cl

C2H5

H

and H

C2H5

Cl

Cl

H C2H5

Principles of Stereochemistry

2.75

(b) H3C H3C

H Br

H3 C H3C

and

Br H

Br H

(c)

F

I

H Br H

F

C2H5

I

and F

(d) H

F

CH3

H CHO

C2H5

OH

CH3

CH2OH OH

H and

H

(e)

H

OH CH2OH H

H

H

HO

CHO

H

H

H

H

H

H

H

and Cl

H

H

H

H

CH3 H

(f)

Cl

F

H

Cl H3C F

and

C2H5

C2H5

CH3

(g) HO H

H H OH

HO

OH

CH3

H

Cl

CH3 Cl Br CH3

CH3

and

CH3

(h) H H

H

H

and H

CH3 Br

H CH3

Cl

Organic Chemistry—A Modern Approach

2.76

Solution (a) Homomers:

C H

C H

H H

C H

Cl

Cl

Cl

Cl

C H

C H H

Cl

H

Cl

1st structure

H

C H

H

2nd structure

1st structure superimposable and therefore, identical

(b)

Enantiomers: Mirror plane H3C H3C

H Br

180° rotation about the axis

Br H

Br

H Br 1st structure

H

1st isomer

(c) (d)

CH3 CH3

H3 C H3C

Br H

H Br 2nd structure

nonsuperimposable mirror images

Constitutional isomers: [2,3-Difluoro-2-iodopentane and 2,3-difluoro-3-iodo-pentane] Diastereoisomers: 3

2 180° rotation about the axis

2

2

Principles of Stereochemistry

(e) H Cl

2.77

Homomers: H

H

H

H

H H

180° rotation about the axis passing through the centre of the molecule

1st structure

H H

H

H

H H 1st structure

H

H

Cl

H

H

H

H

H

H Cl

2nd structure

superimposable and therefore, identical

(f) (g)

Constitutional isomers: [3-chloro-2-fluoropentane and 2-chloro-3-fluoropentane] Diastereoisomers: 3

3

3

3

3

180° rotation of back carbon

3

3 3

(h)

Enantiomers: Mirror plane H

CH H H

Cl Br CH

1st isomer

H

Br

Cl CH

CH 1st isomer

H H C

H

Cl

Cl CH

Br

H

CH

H CH

Br

1st isomer

2nd isomer

non-superimposable mirror images

Organic Chemistry—A Modern Approach

2.78

5.

How is each compound related to the sugar D-erythrose? Is it an enantiomer, diastereoisomer, or homomer?

2

H CHO

HOCH2

(a)

C

(b)

C

HO

OH

C

C

HO

H

H

OH

OHC

CH2OH

H

H OH

OHC

2

(c)

(d)

C

C

H HO

CH OH

Solution

(a)

HOH C C

CHO

OHC

OH

H

C

HO H

H OH

OHC C

(b) HO

C

HO

C

C H

H OH

HO H

CHO C

H OH nonsuperimposable mirror-image of D-erythrose, i.e., enantiomer HOH C

OH H C CH OH

OH identical with D-erythrose, i.e., homomer

C

CH OH H

(c)

C

H

OHC

CH OH

Principles of Stereochemistry

2.79

(d)

This stereoisomer is not the mirror image of D-erythrose, i.e. it is a diastereoisomer of D-erythrose.

1.

Label the following pairs of structures as enantiomers, diastereoisomers, constitutional isomers or homomers. H H Br H and (a) Br H Br Br (b) 2

2

3

Br

(c)

and Br CH3

(d)

CH3

N

Ph

N

and

D

CMe3 and

(e)

(f)

H

H

C

C CH3 and H3C Br Br

(g)

and CH3 H

CH3

Br

(h)

Br

H

F H

CH3

and H3C F O

(i)

CMe3

Ph

D

H CH3 CH3

CH3 and O

CH3

Organic Chemistry—A Modern Approach

2.80

Br

(j)

Br and

Br

Br Cl

Cl and

(k)

Cl

Cl Br

Br

Br

and

(l) 2.

Br

Br Br Br Br How is compound I related to compounds II to IV? Choose from diastereoisomers, enantiomers, constitutional isomers or homomers (i.e., identical molecules). OH

OH OH

OH OH I

3.

II

III

IV

How each pair of the following compounds related: (a) A and B; A and C; (c) A and D; (d) C and D? Choose from homomers, enantiomers or diastereoisomers. CHO

CHO

H

CH2OH

H

Br

4.

II

Br H

CHO H

Br

Br CH2OH

A

B

CHO

H

Br

H

Br

H CH2OH

Br

C

CH2OH H Br D

How are the compounds in each pair related to each other? Are they homomers, enantiomers, diastereoisomers, constitutional isomers or not at all isomers of each other? Me Br Br and

(a)

Me Me and

(b)

H

Me Me

Me

H

Principles of Stereochemistry

2.81

Cl

Cl

(c)

CH2CH3

and CH3

H

H

(d)

Cl

H

H

Cl

Cl Br

H

(e)

Cl

and

C C C Br

C C C

and H

H

H

and

C

C

Et

Et

Br H

Et

H

(f)

H

Br

Et

H

3 3

(g)

3

3

Br

(h)

and Br H H

H

(i)

C H

H C

C C

H

C and

H

C

H C

H

C

D

D

D

C C H

O C

(j)

C

and

O

O

O

OH

H

(k)

N

H

H3C

CH3

F

(l)

F C C C

H

N

and F and

C H

OH H

C C C C H

F

H C

D

C H

Organic Chemistry—A Modern Approach

2.82

Me

H D

H

Me

O and

(m)

D O

5.

Label the following pairs of structures as homomers, constitutional isomers, enantiomers or diastereoisomers: CHO H

(a)

CH3

H2N

and

H

H

CH2OH NH2

H

CH3

CH2OH

(b)

H

COOH OH

H

OH

CHO

and

HO

COOH H

HO

H

COOH Br

COOH Cl

(c) F

Cl and F

Br

I CO2H

(d)

H

OH

H

Cl

H

OH

I CO2H and

H

OH

Cl

H

H

OH

CO2H

CO2H

CH3

(e)

H3C

Br H

HO

H

and H

OH Br

(f)

H

CH3 CH3

Principles of Stereochemistry

2.83

F

(g)

H

Me H

H and

Me

Cl Me

H

Cl

Ph

H

CH3

Ph

(h)

Br and

Cl H

Me

F

H3 C

H

H

Br

Cl Br

(i)

H H

Br and

Br

Br

H

H O

O H

(j) H

CH3

H

and

CH3 H3 C

C H3C

C

H

C

C

CH3 H

(l)

H

H and

C

Br

C CH3

Br

Br

and

Br

H

6.

H

H

H3C

(k)

CH3

H

H

Predict the product (A and B) of the following reaction and mention how they are related to each other.

2 3

3 3 2

Organic Chemistry—A Modern Approach

2.84

2.4 2.4.1

MOLECULAR CHIRALITY Chiral and Achiral Molecules

A molecule is called a chiral molecule if its mirror image is not superimposable on it. On the other hand, if the mirror image of a molecule is superimposable on the original, the molecule is called an achiral molecule. Enantiomers only occur with compounds whose molecules are chiral. A chiral molecule and its mirror image are called pair of enantiomers giving rise to a type of isomerism known as enantiomerism. The word chiral comes from the Greek word cheir, meaning ‘hand’. Chiral objects are said to possess ‘handedness’. The term chiral is used to describe molecules of enantiomers because they are related to each other in the same way that a left hand is related to a right hand. Alanine, for example, is chiral, whereas glycine is achiral. enantiomers Mirror plane CO2H CO2H H2 N

C

H

H

C

CH3

NH2 CH3

Mirror images of alanine. They are not superimposable and therefore are chiral and enantiomeric.

H 2N

identical Mirror plane CO2H

CO2H

C

C

H H

H

NH2

H

Mirror images of glycine. They are superimposable and both of these two structures represent the same, achiral molecule.

Similarly, the cis- and trans-isomers of 1,2-dibromoethene are both achiral because each isomer is superimposable on its mirror image, as illustrated by the following formulas. Mirror plane Br

Br

C C

Mirror plane

H

H

H

H

C C

Mirror images of cis1,2-dibromoethene

Br

Br

Br

H

C C

H

H

Br

Br

C C

Br H

Mirror images of trans1,2-dibromoethene

The mirror images of the cis-isomer are superimposable on each other (one structure on 180° rotation gives the other structure) and therefore, the cis formulas both represent the same, achiral molecule. This analysis is true also for the trans-isomer.

Principles of Stereochemistry

2.85

Chirality The property of any molecule of being nonsuperimposable on its mirror image is called chirality. Chiral centre or chirality centre or asymmetric centre A chiral centre or chirality centre or asymmetric centre is an atom bonded tetrahedrally to four different atoms or groups (also called ligands). A chiral centre is usually a C atom, but may also be N, P, S, Si, etc. A chiral centre is usually indicated with an asterisk (*).

2.4.2

Source of Chirality

The source of chirality may be as follows:

1. The presence of chiral centre A molecule must be chiral if it contains a chiral centre * or chirality centre or asymmetric centre. Thus, the chirality of 2-butanol, CH3CH(OH) CH2CH3, for example, is due to the presence of a chiral centre that bears four nonidentical ligands: CH3, H, OH and CH2CH3. 2-Butanol, thus exists in two enantiomeric forms. It is to be noted that if a molecule contains more than one chirality centre, it may or may not be chiral. For example, D-erythrose containing two chiral centres is chiral, whereas meso-tartaric acid containing two chiral centres is achiral.

It thus follows that the presence of more than one chiral centres is not always sufficient to induce chirality in a molecule. A molecule may even be chiral without having any chiral centres and this may happen when a molecule contains a chiral axis or a chiral plane.

2. The presence of a chiral axis This is an important source of chirality which may be observed in some allenes, alkylidenecycloalkanes, spiranes, adamantanes, biphenyls, etc. if they are properly substituted. These constitutionally nonplanar compounds show enantiomerism if they are devoid of any Sn axis. Elongated tetrahedron approach can be applied to explain the principle of axial chirality. When a chiral centre is replaced by a linear grouping, e.g., C—C or C==C==C, the tetrahedron becomes elongated, i.e., extended

Organic Chemistry—A Modern Approach

2.86

along the axis of the grouping. Such an elongated tetrahedron having lesser symmetry than a regular tetrahedron will be chiral if only the two lingands at each end of the axis are different, i.e., the minimum condition for chirality of an allene, abC==C==Cab, for example, is that ligand a π b. Thus, the structure IA, which represents an elongated tetrahedron (a desymmetrised tetrahedron of the type Caabb), becomes chiral and have enantiometric relationship with its mirror image (IB). The axis along which the tetrahedron is elongated (shown by dotted line) is called the chiral axis or the stereoaxis and the molecular chirality of this type is called axial chirality.

3

3

3

3

Orbital structure of an allene The central carbon atom in an allene is sp-hybridized (linear) and the two-terminal carbon atoms are sp2-hybridized (trigonal). The sp-hybridized central carbon atom uses its two mutually perpendicular p orbitals to form two p bonds with two outer carbon atoms. Therefore, the two p bonds must also be perpendicular. As a consequence, the terminal ligands lie in mutually perpendicular planes. This arrangement places the four ligands in the positions of an elongated tetrahedron.

Principles of Stereochemistry

2.87

p-bond a sp2

a C

C

sp2

C

orbital picture of an allene

b b

sp

The two mutually perpendicular planes of the terminal lingand can be shown , for example, by a molecule of 1,3-dibromoprop-1,2-diene. orthogonal planes

H

H C Br

C

C Br

1,3-Dibromoprop-1,2-diene

Atropisomerim If the two ortho positions of a biphenyl are occupied by sufficiently large groups, free rotation about the single bond joining the two benzene rings (the pivotal bond) is no longer possible because the planar conformations are destabilized due to steric repulsion. As a result, the two ring planes remain approximately perpendicular to each other. Provided each ring has no vertical plane of symmetry (i.e., dissymmetrically substituted), a nonplanar combination of two such phenyl groups would give rise to axial chirality. For example, 6, 6¢-dinitrodiphenyl-2, 2¢-dicarboxylic acid having a chiral axis exists as two enantiomers. Mirror plane chiral axis chiral axis NO2 HO2C CO2H NO2 CO2H NO2

NO2 CO2H

Enantiomers of 6,6¢-dinitrodiphenyl-2,2¢-dicarboxylic acid (atropisomers)

This type of stereoisomerism due to restricted rotation around a single bond is known as atropisomerism and the isomers are called atropisomers.

Organic Chemistry—A Modern Approach

2.88

3. The presence of chiral planes: Various cyclic molecules like paracycloplanes, ansa compounds and trans-cycloalkenes are chiral due to presence of chiral planes. For example: Mirror plane (CH )

(i)

(CH )

H C

CH O

H C

O

CH O

O

Cl

Cl

Enantiomers of an

compound ( = 8)

Mirror plane

(ii)

H2 C

CH2

H2 C

CH2

H2 C

CH2

H2 C

CH2

HO2C

CO2H R

S

enantiomers of paracyclophane containing two aromatic rings Mirror plane

(iii)

H

H C

C

C H

R

C

H S

enantiomers of trans-cyclooctene

2.4.3

Stereocentre or Stereogenic Centre

If two ligands attached to an atom of a stereioisomer are interchanged spatially to produce a new stereoisomer, the atom is called a stereocentre. For example, C-2 of 2-butanol * (CH3CHOHCH2CH3) is a stereocentre as well as a chiral centre. But the C-2 and the

Principles of Stereochemistry

2.89

C-3 carbons of cis- and trans-1, 2-dichloroethene are stereocentres but not chiral centre because interchange of H and Cl atoms on any of these C-atoms gives a new stereoisomer (a diastereoisomer) which is not an enantiomer. Therefore, all stereocentres are not chiral centres but all chiral centres are stereocentres. The stereocentre of 2-butanol is a tetrahedral stereocentre, whereas that of cis- and trans- dichloroethene is a trigonal planar stereocentre. chiral centre and stereogenic centre

OH 2C

H

3

4

CH2CH3 CH3 1

–CH3 and –CH2CH3 groups are interchanged spatially

chiral centre and stereogenic centre

OH C

CH3 CH2CH3

H

enantiomers of 2-butanol stereogenic centre but not chiral centre H Cl

1

C

2

C

H

Cl -isomer

H and Cl atoms of C2 are interchanged spatially

H Cl

1

C

2

C

Cl

H -isomer

Diastereoisomers of 1,2-dichloroethene

2.4.4 Meso Compounds A meso-compound is an achiral compound containing chirality centres. Because they are achiral, they are optically inactive (does not rotate the plane of polarized light). Mesocompounds may be acylic as well as cyclic stereoisomers. A meso-compound is achiral because at least one of its conformer has either a plain or a centre of symmetry. For example, the following stereoisomer of 2,3-dibromobutane (CH3CHBrCHBrCH3) is a meso-compound because it is achiral (a plain of symmetry is present in the fully eclipsed form and a centre of symmetry is present in the anti-staggered form) even though it contains two asymmetric centres or chiral centres.

Organic Chemistry—A Modern Approach

2.90

Plane of symmetry

Centre of symmetry H3C Br H

C

C

H Br

Br

H C

CH3

C

H Br CH3

H3C

(fully-eclipsed conformation)

2,3-Dibromobutane (anti-staggered conformation)

Identical

Chiral centre

H

Mirror plane Br Br

H

Br

CH3

Chiral centre

CH3 H

Br

CH3

H CH3

superimposable on each other after 180° rotation in the plane of paper

cis-1,3-Dichlorocyclohexane is an example of a cyclic meso compound. It contains two chiral centres (C-1 and C-3) and it has a plane of symmetry passing through C-2 and C-5 for which it is achiral. plane of symmetry (s)

H Cl 2

1

H

5

6

Cl 3 4

In fact, a meso compound is possible if a molecule of a compound with two or more asymmetric atoms or chiral centres can be divided into structurally identical halves. For example: HO2CCH(OH)

CH(OH)CO2H

structurally identical halves

CH3CH2CHCl

CHBr CHClCH2CH3

structurally identical halves

Principles of Stereochemistry

1.

2.91

Identify the chiral centre(s), if any in each of the following compounds and indicate them with an asterisk (*): (a) CH3CHClCH2CH2CH3 (b) CH3CH2CH=CH—CHBrCH(CH3)2 ≈

(c) CH3 CH2 N H(CH3) CH2 CH(CH3)2 O ≠ (d) CH3CH2CH2 N (CH3)CH2CH(CH3)CH2CH3 (e) CD3 SiCl (CH3) CH(CH3)2 ≈

(f)

CH3CH2CH2CH2 P H (CH3) CH(CH3)2 D

(g)

D

(h)

O CH3

O

D

(i)

D H3C Br

H3C Br

16

Br

(j)

(k)

CH3

(l)

2

O

18

D18O CH3

(m)

(n)

(o)

D

D

H O H

H H

(p)

C

C

H3C

C

C

CHCl

CH3

D

CH3 H

OH O

CH3

O

(q) Solution *

*

(a)

CH3 C HClCH2CH2CH3

(b) CH3CH2CH=CH— C HBrCH(CH3)2

(c)

CH3 CH2NH (CH3) CH2 CH(CH3)2

(d)

CH3CH2CH2 N (CH3)CH2 C H(CH3)CH2CH3 Ø O

*≈

*

*

Organic Chemistry—A Modern Approach

2.92

(e) (g)

D

(h)

*≈

*

CD3 S iCl (CH3) CH(CH3)2 (f) CH3CH2CH2CH2P (CH3) CH(CH3)2 O CH There is no chiral centre in the molecule (two of the substituents at H O C-2 are the same). 2

D

1 3

5

H3C

4

There is no chiral centre in the molecule (two of the substituents at C-4 are the same).

Br D

(i) *

H 3C H *

D Br H

Br

*

CH3 16

O

*

(k)

CH2

(l) *

(m)

*

(j)

*

*S

O

18 O H H There is no chiral centre in the molecule (cubane). Each of the eight carbons is bonded to three identical substituents.

CH3 *

(n) H

(o) There is no chiral centre in the molecule.

*

O CH3

3

(p)

*

(q)

OH O

*

O

CH3

3

2.

Classify the following objects as to whether they are chiral or achiral. Give your reasoning. (a) scissors, (b) knife, (c) shoe, (d) child’s block, (e) cricket bat, (f) a sixpointed star, (g) hand (h) foot (i) a p orbital (j) a spiral staircase (k) helix, (l) spool of thread, (m) spoon, (n) a golf club, (o) a tennis racket.

Solution (a) chiral, (b) achiral, (c) chiral, (d) achiral, (e) achiral, (f) achiral, (g) chiral, (h) chiral, (i) achiral, (j) chiral, (k) chiral, (l) chiral, (m) achiral, (n) chiral, (o) achiral.

3.

Write names and structures of the two lowest molecular-weight alkanes (with no isotopic atom) that are chiral.

Principles of Stereochemistry

2.93

Solution The two lowest molecular-weight chiral alkanes (with no isotopic atom) are: *

(i) 3-methylhexane [CH3CH2 C H(CH3)CH2CH2CH3] and 2,3-dimethylpentane [CH3CH2 *

C H(CH3)CH(CH3)2]. Each of them contains one chiral centre. 4.

In each of the following pairs of compounds one is chiral and the other is achiral. Identity them and give your reasoning. CH3

CH3

(a)

H Br

Br H

and

H H

Br Br

CH3

CH3

(b) Br CH2 CHBr CH2CH3 and Br2 CH CH2CH2 CH3

and

(c)

Cl Cl H Br

F

(d) H

H

H

H

and

F

Br

Br

F H

H

H

H

Br

F H

H Cl

Cl

(e)

Cl

and

Cl

H H Br

(f)

(g)

H

Br and

O

CH3CH2

H O H

O

H

O H

and

H

O O

CH3CH2

CH3CH2

O O

CH2CH3

Organic Chemistry—A Modern Approach

2.94

(h) H2C C CHCl

and

ClCH C CHCl

O

(i)

Cl

Cl

and

OH

OH

OH

Cl

CH3

(j)

HO H H

CH3 H OH OH

and

CH3

H H H

OH OH OH CH3

Solution

CH3

(a)

H Br

Br H

:

Chiral (It has no plane, centre and other Sn axis of symmetry.)

:

Achiral (It has a plane of symmetry.)

BrCH2 C HBrCH2CH3

:

Chiral (It contains one chiral C-atom.)

Br2CHCH2CH2CH3

:

Achiral (It has a plane of symmetry.)

:

Achiral (It has a plane of symmetry passing through C-1, C-4, C-7 and C-8.)

:

Chiral (It has no plane, centre and other Sn axis of symmetry.)

CH3 s plane

CH3 H H

Br Br CH3

(b)

*

(c) Cl

Cl H

Principles of Stereochemistry

Br

F

(d)

H

H

H

H

:

Achiral (It has a centre of symmetry.)

:

Chiral (It has no plane, centre and other Sn axis of symmetry.)

:

Achiral (It has a plane of symmetry passing through C-1 and C-4 including two Cl and two H atoms.)

:

Chiral (It has no plane, centre and other Sn axis of symmetry.)

:

Chiral (It contains one chiral C-atom.)

:

Achiral (It has a plane of symmetry.)

:

Achiral (It has a centre of symmetry.)

:

Chiral (It has no plane, centre and other Sn axis of symmetry.)

F

Br

Br

F H

2.95

H

H

H

Br

F

H 4

Cl

(e)

Cl 1

H H Cl

Cl

H Br

(f)

*

H

Br H

HO

O CH2CH3

(g) CH3CH2 O

CH3CH2 O H O

OH

O CH3CH2 OH

Organic Chemistry—A Modern Approach

2.96

H

(h)

H2C C CHCl H H

H C C C

:

Achiral (It has a plane of symmetry.)

:

Chiral (It has no plane, centre and other Sn axis of symmetry.)

:

Achiral (It has a plane of symmetry passing through C-1 and C-4.)

:

Chiral (It has no plane, centre and other Sn axis of symmetry).

:

Chiral (It has no plane, centre and other Sn axis of symmetry.)

:

Achiral (It has a plane of symmetry.)

Cl

H C C C Cl

Cl 1

(i)

4

Cl

OH

H

Cl

3

(j) 3

3 s-plane

3

5.

*

Compare the two enantiomers of 2-butanol (CH3 CHOHCH2CH3) with respect to: (a) melting point, (b) boiling point, (c) relative density, (d) refractive index, (e) specific rotation, (f) solubility in 200 g of water, (g) rate of reaction with HBr, (h) adsorption on alumina (Al2O3), (i) rate of reaction with an enantiomer of CH3CH2CH(CH3) COOH to form an ester, (j) specification as R or S.

Principles of Stereochemistry

2.97

Solution They have equal but opposite specific rotation, opposite R/S specification or designation and different rate of esterification with an enantiomer of 2-methylbutanoic acid. The different rate of esterification is due to formation of two diastereoisomeric transition states of different energies, i.e., the Eact values are different. All other properties are the same.

6.

What is the necessary and sufficient condition for enantiomerism?

Solution Chirality (i.e., nonsuperimposability on mirror image) is the necessary and sufficient condition for enantiomerism.

7.

What do you mean by epimers? Give examples.

Solution Two diastereoisomers which are different in the configuration of a single chiral centre are called epimers. For example, D-ribose and D-xylose (two carbohydrates) are two epimers because they differ only in the configuration about C-3.

CHO OH OH OH

H H H

H HO H

CH OH D-Ribose

8.

CHO OH H OH CH OH D-Xylose

Draw the eight constitutional isomers having molecular formula C5H11Br. Label the isomers as chiral and achiral.

Solution The isomeric alkyl bromides having molecular formula C5H11Br are as follows: *

CH3CH2CH2CH2CH2Br,

CH3CH2CH2 C HBrCH3,

CH3CH2CHBrCH2CH3,

I

II

III

CH3CH2 C H(CH3)CH2Br,

(CH3)2CHCH2CH2OH,

CH3CH2CBr(CH3)2,

IV

V

VI

and

(CH3)3CCH2Br

*

*

(CH3)2CH C HBrCH3 VII

VIII

Among these eight alkyl bromides, only three II, IV and VII are chiral because each of them contains a chiral carbon atom. Other five alcohols (I, III, V, VI, and VIII) are achiral because each of them has a plane of symmetry (s). 9.

Which pairs of the salts shown below are expected to have identical solubilities in methanol? Explain.

Organic Chemistry—A Modern Approach

2.98

6

5

6

5

3 –

–

6

5

3

6

5

6

5

3

3

6

6

5

6

5

5 –

–

Solution

Mirror plane C6H5 H

Et

C6H5 Et

NH3

C6H5

NH3

COO H

–

C6H5 H

H Et

Et COO

I

C6H5 Et

–

H COO

NH3 H

C6H5 –

I

Et IV

enantiomers Mirror plane C6H5

H

NH3

C6H5

H

NH3

–

H3N

H

Et H

Et COO

III

–

C6H5

H OOC Et

–

Et C6H5

Et

C6H5

COO H C6H5

III

Et II

enantiomers

Since the salts I and IV are enantiomeric, therefore, they have identical solubilities in methanol. Similarly, III and II have identical solubilities in methanol.

Principles of Stereochemistry

10.

2.99

Identify the stereocentres in each of the following compounds and predict whether each compound is chiral or not. Et

(a) Et

Et

H

(b)

Et

H Et

H

H

H Et

Et

(c)

Et

(d)

H

H Et

Et Et

Et

H

Solution (a) The compound is chiral and it contains four stereocentres.

stereocentres Et Et

H

stereocentre

Et H

(b)

H

The compound is chiral and it contains three stereocentres. stereocentres Et H Et

(c)

H The compound is chiral and it contains one stereocentre. Et

stereocentre

H Et

(d)

Et The compound is achiral and it contains two stereocentres. Et stereocentres

Et

H Et

H

Et

2.100

11.

Organic Chemistry—A Modern Approach

Give an example of compound that contains a tetrahedral stereogenic centre and a trigonal planar stereogenic centre. How can you verify that each of these carbons is a stereogenic centre?

Solution The following alkylidenecycloalkane contains one tetrahedral stereogenic centre and two trigonal planar stereogenic centre.

H

C

stereocentre Et

Et H

stereocentres

Whether any one of these carbons is a stereogenic centre or not can be verified by interchanging any two groups bound to it. This interchange may generate the other enantiomer. For illustration, the two ring bonds at the middle stereogenic centre are interchanged. This results in the formation of an enantiomer. bond cleavage H Et

C

Et H

enantiomers

H

180°

Et

H Et (positions of the groups are interchanged by rotating 180°)

Et

H

H

Et

180° rotation about the axis

Et

Et C H

H

ring is reformed

Et

Et

H

H

A group that interchanges at each of the other two stereogenic centres also gives enantiomers. 12.

Which of the following compounds has a stereoisomer that is a meso compound? (a) Cyclohexane-1, 4-dicarboxylic acid (b) 4,5-Dipropyloctane (c) 1,2,5-Tribromopentane (d) 2,3,4-Trihydroxybutanal (e) 1-Bromo-2-fluorocyclopentane (f) 1,3-Dichlorocyclopentane (g) 1,2-Dimethylcylobutane (h) 2,4-Dibromo-3-pentanol (i) 3,5-Dichloroheptane (j) 3-Heptene (k) 4-Methyl-2-pentyne

Solution A compound must have a stereoisomer that is a meso compound if it contains at least two chiral atoms which are bonded to the same four substituents.

Compounds (a), (b), (j) and (k) do not have a stereoisomer that is a meso compound because they do not have any chiral centre.

PRINCIPLES OF STEREOCHEMISTRY

Principles of Stereochemistry

2.101

CH2CH2CH3 CO2H

CH3CH2CH2CH CHCH2CH2CH3

HO2C

CH2CH2CH3 4,5-Dipropyloctane (b)

Cyclohexane-1,4dicarboxylic acid (a) CH3CH2CH CHCH2CH2CH3 3-Heptene (j)

CH3 C C CH(CH3)2 4-Methyl-2-pentyne (k)

The compound (c) does not have a stereoisomer that is a meso compound because it has only one chiral centre. Br | * BrCH2 —CHCH2CH2CH2Br 1,2,5-Tribromopentane Each of the compounds (d) and (e) contains two chiral atoms. But each of the chiral atoms is not bonded to the same four substituents. Therefore, they do not have a stereoisomer that is a meso compound.

Compounds (f), (h) and (i) have a stereoisomer that is a meso compound because each of them has two chiral atoms which are bonded to the same four substituents.

*

*

Cl Cl 1,3-Dichlorocyclopentane (f)

13.

Br

Br

*

*

CH3 CH

CH(OH) CH CH3

2,4-Dibromo-3-pentanol (h)

Identify the meso compounds and give your reasoning: H

(a)

CH3

H C

C

C C

H

CH3 H

Cl

Cl

Br

(b)

H Br

H

2.102

Organic Chemistry—A Modern Approach

H D3 C

(c)

C

C C

H5C2 H

H3C

H CD3

CH3 CH3

(d)

C2H5 F

CH3 O H

N Et H

(e)

(f)

Cl

H O

N Et H

CH3 H (g) H H

OH OH OH

H

(h)

COOH

HOOC

H

CH3 H3 C

(i)

HO2C H

CO2H

C

H Br

(j)

H

C H3C

H Br

Solution (a) It is a meso compound because it is achiral due to the presence of a plane of symmetry (s) and contains two chiral centres.

H

H C

CH3

C

C

C

H

H Cl

(b)

CH3

s-plane Cl

It is not a meso compound because it contains no chiral centre. However, it is achiral due to the presence of a s-plane and a point of symmetry. Br s-plane

H

H

Br

point of symmetry

Principles of Stereochemistry

(c)

2.103

It is a meso compound because it contains two chiral centres and is achiral due to the presence of a plane of symmetry. H D3 C

C

C

*

H5C2

*

C H

(d)

H CD3 C2H5

F

The compound has a plane of symmetry and so, it is achiral. However, it contains two chiral centres. Hence, it is a meso compound. 3

3 *

*

3 Me

Me

s

* *

3

(e)

It is not a meso compound because the molecule is chiral. * *

n

(f)

The compound is achiral because it has a centre of symmetry (i). However, it contains two chirality centres. Therefore, it is a meso compound. O H

N Et H H O

N Et H Centre of symmetry (i)

(g)

It is a meso compound because it is achiral due to the presence of a plane of symmetry (s) and it contains two chiral centres. CH3 H H H

*

*

OH OH OH

CH3

s-plane

2.104

(h)

Organic Chemistry—A Modern Approach

The molecule is chiral and so, it is not a meso compound. 2

Sn

2

(i)

The molecule is achiral due to the presence of a plane of symmetry (s) and also, the molecule contains two chirality centres. Therefore, it is a meso compound.

HO C * H

CO H * H

s-plane

(j)

It is a meso compound because it contains two chiral centres and is achiral due to the presence of a plain and a centre of symmetry.

s

1.

Identify the chirality centre(s), if any, in each of the following compounds and indicate them with an asterisk (*): CH3 (a) CH3CH2CH(CH3) (b) CH3

(c)

(d) CH3CH2O P(OCH3)NMe2 O

(e)

O

OH CH3

Me

Me

OMe

(f) Ph N CH2 CH2 CH CH2

Me

CH2Ph O

(g)

HO HO

OH OH OH

(h)

3

2

2

3

Principles of Stereochemistry

2.105

CH3

(i)

(j)

CH3CH2CH2Si CH2CH(CH3)2 Cl

O

(l) CH3CH2N(CH3)CH(CH3)2

(k) O Cl

(m) C6D6 CH C6H6

(n)

(o)

(p)

H3C

OH

3

O

Cl

(q)

3

(r)

Cl

H

Cl H

2.

3. 4.

H

Classify the following objects as to whether they are chiral or achiral and give your reasoning: (a) a bucket, (b) a brick, (c) an s orbital, (d) a double helix, (e) a tennis racket, (f) gloves, (g) a scarf tied around your neck, (h) a filled spool of thread, (i) an empty spool, (j) a five-pointed star, (k) a balance, (l) a boot, (m) a hammer, (m) a mug with HAT written opposite the handle, (n) a mug with TAT written opposite the handle. The positive carbon of a carbocation cannot be a chiral centre – Why? Write the structural formulas and names for the lowest molecular-height chiral: (a) alkene, (b) alkyne, (c) alcohol, (d) aldehyde, (e) ketone, (f) carboxylic acid, (g) amine and (h) ester. *

*

[Hint: (a) CH3CH2 C H(CH3)CH==CH2 (3-methylpent-1-ene), (b) CH3CH2 C H (CH3)CH∫∫CH (d)

*

(3-methylpent-1-yne),

CH3CH2 C H(CH3)CHO

(c)

(2-methylbutanal), *

*

CH3CH2 C HOHCH3 (e)

(2-butanol),

*

CH3CO C H(CH3)CH2CH3

(3-methyl-2-pentanone), (f) CH3CH2 C H(CH3)CO2H (2-methylbutanoic acid), (g)

2.106

Organic Chemistry—A Modern Approach

*

*

CH3CH2 C H(NH2)CH3 (2-aminobutane), (h) CH3CH2 C H(CH3)COOCH3 (methyl2-methylbutanoate)]. 5.

Locate the stereogenic centre(s), if any, in each of the following compounds: H HO CH3

(a)

C H

CH2Br

H CH

(d)

Me

H

Me

Cl

(c)

H

Cl

H

(e)

CH

H

Cl

(f) H

C C H

Br

H

OH

(b) HO

Br

CH3 C C C C

CH3 H

H CH3

(g) H3C

H CH3 CH3

(h)

C C C

(j) H

H

OH

(i) H

Br

Br C2H5

H

H

Cl

6.

CH3

Br

(k) Cl

H

Br

Indicate whether each compound is chiral. Identify the chiral carbons and stereocentres (if any) in each: Br

(a)

Et

CH Et C

C

Et

H

(b) Et

C

N Br

Et CH3CH2

(c)

CH2CH3

CH3

H

H H

Cl

(e)

CH2CH3

(d)

C C

(f)

H

H CH3

CH3

(g)

C CH3

Principles of Stereochemistry

2.107

H CH3

Br

(h)

Br

O

(i) H

(j)

H

O

Cl

7.

8.

9. 10.

11. 12.

H

Two stereoisomers of the compound (H2N)2 PtCl2 have different physical properties. From this observation predict whether the shape of the molecule is tetrahedral or square planar. Indicate whether each of the following statements is true or false. If false, explain why? (a) Constitutional or structural isomers may be chiral. (b) If a compound has an enantiomer it must be chiral. (c) Every chiral compound has a diastereoisomer. (d) Each molecule containing one or more chiral carbons is chiral. (e) Any molecule containing a stereocentre must be chiral. (f) Any molecule with stereogenic centre must have a stereoisomer. (g) Mirror-image molecules are in all cases enantiomers. (h) All chiral molecules have no plane of symmetry. (i) If a structure has no plane of symmetry it is chiral. (j) Some chiral compounds are optically inactive. Draw the structure of the chiral cyclic alkane (with no isotopic atom) of lowest molecular mass. Draw structures of all compounds having molecular formula C6H12Br2 that exist as meso compounds. Indicate how many meso compounds are possible for each structure. Neither cis-2-hexene nor trans-2-hexene (both achiral) a meso compound, why? Identify the meso compound and give your reasoning. H3C

(a)

Cl H

C

C C C

D

Et

H C

C C C

Et

CCl3 C

H

(b)

H Cl

Cl3C

(c)

Br

CH3

C

H D

(d) O

O

H Br

PRINCIPLES OF STEREOCHEMISTRY

2.108

Organic Chemistry—A Modern Approach

CH3

(e)

Br

Br

(f)

F3C

CF3

H

H

CH3 Cl

(g)

F3C

H

H

(h) Cl H

CH3

H Et

Ph Cl

(i)

Ph

(j)

H

H

(k)

CH3 H Cl

2.5

H

C

C

H

H

(CH2)3 H Et Cl

CH3

CH3 H (l) H

D Br

Br Br CH3

CONFIGURATION AND CONFIGURATIONAL NOMENCLATURE

The arrangement of atoms that characterizes a particular stereoisomer is called its configuration. Using the test of superimposability, we may conclude, for example, that there are two stereoisomeric 2-methyl-1-butanols. Their configurations are I and II. CH3 H

C I

enantiomers

CH2OH C2H5

HOH2C

2-Methyl-1-butanol

CH3 C

H

C2H5 II

In the laboratory samples of two compounds of formula CH3CH2CH(CH3)CH2OH, we find that one rotates the plane of polarized light to the right and the other to the left. These are taken in two bottles and labelled as (+) -2-methyl-2-butanol and (–) -2-methyl-2-butanol. Now, an important question arises, which configuration does each isomer have? Does the (+) isomer, say, have the configuration I or the configuration II ? How can it be known? Which structural formula, I or II, is to be drawn on the label of each bottle? That is to say,

Principles of Stereochemistry

2.109

how can the configuration of each isomer be assigned. For any optically active compound, the question of configuration could not be answered in an absolute sense until 1951. In that year, J.M. Bijvoet, the director of the Van’t Hoff Laboratory of the University of Utrecht in the Netherlands, reported by using a special technique of X-ray diffraction that he had determined the actual arrangement, i.e., absolute disposition in space of the atoms of the optically active sodium rubidium (+) -tartaric acid. Over the years, prior to 1951, the relationships between the configuration of (+) -tartaric acid and the configurations of hundreds of optically active compounds have been determined (through reaction of known stereochemistry, e.g. SN2, or reactions in which no bond to the chirality centre is broken). The configurations obtained by this way is called relative configurations. When the actual or absolute configuration of (+) -tartaric acid became known, the absolute configurations of other compounds also became known. (In the case of 2-methyl-1-butanol, for example, the (+) -isomer is known to have the absolute configuration I and the (–) -isomer is known to have the absolute configuration II.) Therefore, the absolute configuration of a chirality centre is the actual arrangement of groups in space at the chirality centre that distinguishes it from its mirror image and can be designed by R or S.

2.5.1

D, L-System of Configurational Designation

The relationship between the configurations of two chiral molecules is the basis of D, L-system of designation. When two chiral molecules can be interconverted chemically without breaking any bond to the chirality centre, they are said to have the same relative configuration independent of the direction of rotation of the plane polarized light. The oldest system of nomenclature of optical isomers containing chiral centres is the D, L-system, introduced by as early as 1890s, by German chemist Emil Fischer, who worked extensively with carbohydrates and established the relative configuration of (+) -glucose. Glyceraldehyde (HOCH2CH(OH)CHO), the simplest carbohydrate molecule, was chosen by Fischer as the standard for defining configurations of molecules with chirality centres. The (+) -enantiomer (that rotates the plane polarized light clockwise) was arbitrarily labelled as D-glyceraldehyde and the other enantiomer was L-glyceraldehyde. According to Fischer, the projection formulas of D- and L-glyceraldehydes are as follows: CHO H

OH CH2OH

D-(+)-Glyceraldehyde

CHO HO

H CH2OH

L-(–)-Glyceraldehyde

Any molecule containing a single asymmetric centre was assigned as D or L on the basis of a resemblance between the group on its chirality centre and those in glyceraldehyde, i.e., the enantiomer containing ‘similar’ groups in the same place, as D-glyceraldehyde,

2.110

Organic Chemistry—A Modern Approach

was designated as ‘D’ and its mirror image as ‘L’. Thus, the naturally occurring amino acid alanine, for example, was labelled as L-enantiomer. CO2H H2N

H

CH3 L-Alanine

In fact, L-alanine can be chemically correlated to L-glyceraldehyde. This generic nomenclature, however, did not work since in many cases, both of the enantiomers of a compound may be chemically correlated to the same glyceraldehyde. For example, both (+)- and (–)-lactic acid can be correlated to D-glyceraldehyde because the former can be obtained by reduction of —CHO to —CH3 and oxidation of —CH2OH to —COOH and the latter by oxidation of —CHO to —COOH and reduction of —CH2OH to —CH3. To circumvent this drawback, Rosanoff (1906) modified the system and suggested a projection nomenclature according to the following conventions. In this modified system of D, L-nomenclature, the molecule of the type RCHXR¢ is written in Fischer projection such that the carbon atoms of the longest chain R—C—R¢ are placed on the vertical line with C-1 (according to the IUPAC system of nomenclature) at the top. Then, if X (which is taken to be a negative group) is on the right horizontal bond, the molecule is given D designation and if it is on the left, the molecule is given L designation. R H

R X

R¢ D-enantiomer

X

H

R¢ L-enantiomer

Examples:

[It is to be noted that the D, L-symbols have no relation with the sign of rotation of the optically active compound.] When a chirality centre remains attached with a ring system, then the total number of carbon atoms including the ring carbons are to be considered for writing the main chain in Fischer projection and for this; a C6H5-group attached to a chirality centre is generally placed at the lower vertical bond in Fischer projection. For example:

Principles of Stereochemistry

2.111

CH2CH2CH3 H

CO2H

CH3

CH3

C6H5 D-2-Phenylpentane

CH2CH2CH3

OH

H3C

Ph D-Atrolactic acid

H

C6H5 L-2-Phenylpentane

When the chiral centre contains two different substituents of comparable electronegativity, then D- and L-symbols are separately used for each chiral centre to specify the configuration. This is also true when two different alkyl groups are present in horizontal bonds. For example: CH3 Br

CO2H

F

H3C

C2H5 2L-Bromo-2Dfluorobutane

C2H5

Ph 2D-Ethyl-2L-methyl-2-phenyl -butanoic acid

D, L-designation to compounds containing more than one chirality centre involves specifying each of the chirality centres by D, L-notations. For example: 1

CH3 2 HO H H 3 OH 4

5

H HO 6

CH2CH2CH3

Hexane-2L, 3D-diol

2.5.2

CO2H Br CH3

1

H

OH 2

Cl

Et

H

CO2H

5 3

H

CH3

Me

HH

OH

4

C2H5 2D-Bromo-3D-methyl -3L-hydroxypentanoic acid

2L-Chloro-3Dmethyl-1Dcyclopentanol

1L-Ethyl-3Dhydroxy-5L-methyl 1D-carboxylic acid

Specification of Configuration: The R, S-System

The two enantiomers of bromochloroiodomethane are the following: I H Cl

I C

C Br

H

Br Cl

I

II

If we name these two enantiomers using only the IUPAC system of nomenclature, both enantiomers will have the same name: bromochloroiodomethane. This is not desirable because each compound must have its own distinct name. Moreover, we must be able to write the structure of the compound from its name alone. Given the name bromochloroiodomethane, we could write either structure I or structure II.

2.112

Organic Chemistry—A Modern Approach

Three chemists, R.S. Chan (England), Sir Christopher Ingold (England) and V. Prelog (Switzerland), devised a system of nomenclature that, when added to the IUPAC system, solves both the problems. This system of configurational nomenclature is called the R, S-system or the Cahn-Ingold-Prelog (CIP) system and since this system of specifying configurations is independent of any reference compound, it is often termed Absolute Configuration assignment. In this system, configurations of stereoisomers are designated on the basis of individual chiral centres. According to this system, one enantiomer of bromochloroiodomethane should be designated (R)-bromochloroiodomethane and the other enantiomer should be designated (S)-bromochloroiodomethane. [(R) and (S) are from the Latin words rectus and sinister, meaning right and left, respectively.] These molecules are said to have opposite configurations.

2.5.2.1

Assigning (R) and (S) configurations

For assigning R or S designation to any chiral centre of a molecule, certain conventions are to be followed. These are given below: 1. The four different groups or atoms attached to each of the chiral centres are ranked in order of priority and a priority symbol 1, 2, 3, 4 or a, b, c, d is assigned to them based on some sequence rules such that the decreasing order of priority is 1 > 2 > 3 > 4 or a > b > c > d (1 or ‘a’ having the highest priority and 4 or ‘d’ the lowest). 2. The molecule is then oriented with the lowest priority group (4 or ‘d’) back (bonded by a hatched wedge), and the relative positions of the remaining three groups are visualized when a hypothetical path is traced from the first-priority group, through the second, to the third, i.e., 1 Æ 2 Æ 3 or a Æ b Æ c. If the path describes a clockwise motion, then the chiral centre is said to have R configuration and if the path describes a counterclockwise or anticlockwise motion, then the chiral centre is said to have S configuration. The designation (R) or (S) is written in italics within parentheses followed by a hyphen before the name of the compound (when necessary with appropriate locants). Æ

Æ

Æ

Æ

Principles of Stereochemistry

2.113

If in the figure, the group or atom of lowest priority is shown above the plane of the paper or in the plane of the paper, the molecule is to be rotated by 120° about the axis passing through the chiral carbon and a group in the plane of the paper, so as to place the lowest priority group back (bonded by hatched wedge). Then, the configuration of the chiral centre is determined by usual process. For example:

120° rotation

Æ

Æ S

120° rotation

Æ

Æ

S 120° 120° rotation

Æ

Æ R

The lowest priority group (4) may also be bonded by a hatched wedge by simply interchanging group 4 with the group that is bonded by a hatched wedge. However, in that case the actual configuration will be opposite to that determined.

Æ

Æ

2.114

Organic Chemistry—A Modern Approach

Example: The chiral centres of the two enantiomers of bromochloroiodomethane are designated as R or S as follows. The atoms bonded to the chiral centre are ranked in order of priority. The atomic number of atoms attached to the chiral centre determines the relative priorities. The higher the atomic number of the atom, the higher the priority (given in rule 1 for assigning priority). Therefore the priority order is: –I

–Br

1 2 highest

–Cl –H 3

4 lowest

priority decreases

Then, the chiral centre of each enantiomer is viewed from the opposite side of the lowest priority group (4 or ‘d’). In the enantiomer A, 1 Æ 2 Æ 3 traces a clockwise path. Hence, its configuration is R, i.e., it is (R)-bromochloroiodomethane. On the other hand, in the enantiomer B, 1 Æ 2 Æ 3 traces a counterclockwise path. Hence, its configuration is S, i.e. it is (S)-bromochloroiodomethane. 1 I 4 H Cl 3

1

viewer

C Br 2

3

2

1 Æ 2 Æ 3 traces a clockwise path. Hence the configuration is R, i.e. the IUPAC name of this enantiomer is (R)-bromochloroiodomethane.

Enantiomer A of bromochloroiodomethane

1 I viewer

C

H 4 Cl 3 Enantiomer B of bromochloroiodomethane Br 2

1 2

3

1 Æ 2 Æ 3 traces a counterclockwise path. Hence the configuration is S, i.e. the IUPAC name of this enantiomer is (S)-bromochloroiodomethane.

Principles of Stereochemistry

2.115

Sequence rules or CIP chirality rules for determining priority of ligands attached to a chiral centre. Rule 1. If the four atoms bonded to the chiral centre and are all different, priority depends on atomic number, with the atom of higher atomic number getting higher priority. An example has already been discussed. Rule 2. In cases where two of the attached atoms are isotopes of each other, higher atomic mass has higher priority over lower. For example, the priority order of the isotopes of hydrogen is T > D > H. However, —12CH2NH2 will get preference over —14CH2CH3 because the atomic number of the second atom (N) in the former group is higher than that of the second atom (C) in the latter group, even though the mass number of methylene carbon in the latter group is higher than that in the former. Again, —CH218OH has the higher priority than —CD216OH because 18O has highest atomic number as well as higher mass number, even though D is present in the latter group. Similarly, —CH2OH has priority higher than —CD2NH2. Rule 3. When the chirality centre is attached to two or more identical atoms, i.e., if the priority remains undecided, then the atomic numbers of next sets of atoms are to be considered and exploration is continued until a point of difference is reached. The priorities are then assigned at the first point of difference. For example, in 2-butanol 1

2

3

4

(CH3 CHOHCH2 CH3 ) , the oxygen atom gets highest priority (1) and H gets lowest priority (4) using Rule 1. The two remaining groups are —CH3 and —CH2CH3. The first point of difference is at the two carbons attached to the chiral centre. The carbon of the methyl group, i.e., C-1 is bonded to three H atoms, i.e., to (H, H, H), while the first carbon of the ethyl group, i.e., C-3 is bonded to one carbon and two H atoms, i.e., to (C, H, H). Since carbon has higher atomic number than hydrogen, ethyl (—C2H5) gets higher priority over methyl (—CH3). Rule 4. When multiple bonded groups like H C=O, —CH=CH2, —C∫∫N, etc. are found to H be attached with the chiral carbon, both atoms attached to the multiple bond are treated as if they are duplicated or triplicated single bonds. This pretreatment for assigning priority of groups containing multiple bonds is termed as ‘replication’. The replicated atoms are enclosed in parentheses in the expanded form of the group. Each replicated atom, except for H atom, is converted to single bond tetracovalency by adding the so-called ‘phantom atoms’. Phantom atoms are denoted by the cipher zero which are often shown as subscript. These are used to bring the valency up to four. Phantom atoms are imaginary atoms having an atomic number zero. A few examples of replications of multiple bonds are shown below:

2.116

Organic Chemistry—A Modern Approach

00

2 000

000

000

000

000

000

000

000

000 000

000

000

000

000

000 000 000

000

The priority sequence of the four groups —CHO, —CH==CH2, —C∫∫CH and —C6H5 may be determined as follows. The first atoms are connected, respectively, to (H, O, O), (H, C, C), (C, C, C) and (C, C, C). Therefore, —CHO gets highest priority and —CH==CH2 gets lowest priority because only one oxygen outranks three carbons and three carbons outrank two carbons and a hydrogen. To rank the remaining two groups, we have to proceed further along the chains. The phenyl group (—C6H5) has two of its (C, C, C) carbons attached to (C, C, H), while the third carbon is attached to (000) and is, therefore, gets priority over the —C∫∫CH group, which has one of its (C, C, C) carbons is connected to (C, C, H), while the other two are connected to (000)s. the carbons bonded to the chiral centre C (C, C, C) (C, C, H) (C, C, H) (000) two

CH

(C, C, C) (C, C, H) (000) one

for this –Ph group has higher priority than –C ∫ CH group

(000)

Principles of Stereochemistry

2.117

Rule 5. When two enantiomeric groups are found to be attached with the chiral centre, the group with R configuration gets priority over the group with S configuration. Again, (R, R) or (S, S) gets priority over (R, S) or (S, R) configuration. Rule 6. When the chiral centre is attached to two groups having cis and trans configuration, then cis gets priority over trans. In case of E and Z groups, Z gets priority over E. A list of common groups and atoms with increasing priority 1. 2. 3. 4. 5. 6. 7. 8. 9.

10. 11. 12. 13. 14. 15. 16. 17. 18.

H D CH3 CH2CH3 CH2CH=CH2 CH2C∫CH CH2C6H5 CHR2 CH=CH2

CH=CHR CR3 C(R) = CR2 C∫CH C6H5 C∫CR CH2OH CH(OH)R CR2OH

19. 20. 21. 22. 23. 24. 25.

CHO COR CONH2 COOH CO2R COX CX3 (X = halogen) 26. NH2

27. 28. 29. 30. 31. 32. 33. 34. 35.

NHR NR2 NHCOR NO NO2 OH OR OPh OCOR

36. 37. 38. 39. 40. 41. 42. 43. 44.

F SH SR SOR SO2R SO3H Cl Br I

H gets priority over (..) lone pair of electrons

R and S assignments in compounds with two or more stereogenic centres When a compound has more than one stereogenic centre, the R or S configuration must be assigned to each of them. For example, in the following stereoisomer of 2,3-dichloropentane, C-2 has the S configuration and C-3 has the R. So, the complete IUPAC name of the compound is (2S, 3R)-2, 3-dichloropentane. — According to the sequence rules, the priority order of groups around C-2 is —Cl CHClCH2CH3 —CH3 H and around C-3 in —Cl —CHClCH3 —CH2CH3 H. Configuration of C-2: 3

2

2

3

Æ

3

Æ

S.

Configuration of C-3: H3C H Cl

2

3

C 2

C

CH2CH3 3

1 Æ 2 Æ 3 traces a clockwise path. Hence, the configuration of C-3 is .

3

H4 Cl 1

2

1

2.118

Organic Chemistry—A Modern Approach

Assigning R, S-designation to chiral centres when a compound is represented by Fischer projection The procedure to determine the configuration of a compound drawn as a Fischer projection involves two steps: (i) the groups or atoms bonded to the chiral centre are to be ranked in order of priority and (ii) a semicircle is to be traced joining 1 Æ 2 Æ 3 (or a Æ b Æ c), ignoring 4 (or d), the lowest priority group. When the group with lowest priority (4 or d) is on a vertical bond in Fischer projection, the sequence gives the correct configurational descriptor. That is, if 1 Æ 2 Æ 3 (or a Æ b Æ c) traces a clockwise path, the configuration is R, but if the sequence traces counterclockwise path, the configuration is S. For example: (i) According to the sequence rules, the priority order of groups in 2-chloropropanoic *

acid (CH3 C HCl COOH) is —Cl

—COOH

2 COOH 3 H3C

Cl 1

H.

2 3

1 4

H 4

(ii)

—CH3

(the lowest priority group is on a vertical bond)

Since 1 Æ 2 Æ 3 traces a counterclockwise path, the configuration of this enantiomer of 2-chloropropanoic acid is S, i.e., the compound is (S)-2-chloropropanoic acid. According to the sequence rules, the priority order of groups in 2-butanol (CH3CH2CHOHCH3) is —OH —CH2CH3 —CH3 —H. 4 H 1 HO

4 CH2CH3 2

CH3 3

1

2 3

(the lowest priority group is on a vertical bond)

Since 1 Æ 2 Æ 3 traces a clockwise path, the configuration of this enantiomer of 2-butanol is R, i.e., the compound is (R)-2-butanol. If the group or atom with the lowest priority is on a horizontal bond, the answer we get from the direction of the arrow will be opposite of the correct answer. For example, if the arrow points clockwise, suggesting the R configuration, then the compound actually has the S configuration and if the arrow traces a counterclockwise path, suggesting the S configuration, then it actually has the R configuration. For example, in the following enantiomer of lactic acid, the group with the lowest priority is on a horizontal bond, so a clockwise arrow signifies the S configuration.

Principles of Stereochemistry

2.119

The configuration of a molecule drawn in Fischer projection with the lowest priority group is on a horizontal bond and may also be determined by bringing the lowest priority group into a vertical bond, either above or below, by changing the positions among three groups sequentially on the chiral centre without disturbing the position of the fourth or by exchanging two pairs of groups simultaneously. Now, if 1 Æ 2 Æ 3 traces a clockwise path, the configuration is R and if the arrow traces a counterclockwise path, the configuration is S. For example: 2 COOH 4H

3

NH2 1

4

1 2

CH3 3 Alanine

interchange of positions of 1, 4 and 3 sequentially

4 1 2

3 (lowest priority group is on a vertical bond)

Since 1 Æ 2 Æ 3 traces a clockwise path, the configuration of this enantiomer of alanine is R, i.e., the compound is (R)-alanine. This procedure of assigning R, S-designation can also be applied to Fischer projections with more than one chiral centre. For example: H H

CHO Br 3 Cl 2

CH3 2-Bromo-3-Chlorobutanal In 2-bromo-3-chlorobutanal, the priority order of groups attached to C-2 chiral centre is: —Br CHClCH3 —CHO —H and the priority order of groups linked to C-3 chiral centre is: —Cl —CHBrCHO —CH3 —H.

Configuration of C-2:

3

Since 1 Æ 2 Æ 3 traces a counterclockwise path, the configuration of C-2 is S.

2.120

Organic Chemistry—A Modern Approach

Configuration of C-3: 2 CHBrCHO 4H

Cl 1

2 4

1 3

CH3 3

interchange of positions of 1, 4 and 3 sequentially

2 1

3 4

(the lowest priority group is on a vertical bond)

Since the arrow from 1 Æ 2 Æ 3 points clockwise, the configuration of C-3 is R. Therefore, the complete stereochemical name of the compound is (2S, 3R)-2-bromo-3-chlorobutanal. A ‘very good’ (mnemonic) procedure for assigning absolute configuration to chiral centres is due to Epling (1982). The procedure, applicable to Fischer projections only involves two operations: (a) assigning priority symbols 1, 2, 3, 4 or a, b, c, d to each of the groups attached to the chiral centre and (b) tracing a semicircle joining 1 Æ 2 Æ 3 or a Æ b Æ c ignoring 4 or d, the lowest priority group. If 4 or d is found on either of the vertical bonds, the path traced gives the correct description, i.e., R for clockwise motion and S for counterclockwise motion. But if 4 or d remains on a horizontal bond, the counterclockwise path indicates R configuration and the clockwise path indicates S configuration, i.e., the descriptors arrived at from the sequence 1 Æ 2 Æ 3 or a Æ b Æ c have to be reversed. This procedure is very helpful for assigning absolute configuration to chiral centres in a molecule containing any number of chiral centres. For example: 2 3

4 2

3

Priority of groups at different chiral centres is as follows: C-2 : —OH C-3 : —Cl C-4 : —OH

—CClFCHOHCH2CH3 —F

—CHOHCHO

—CClFCHOHCHO

—CHO

H

—CHOHCH2CH3 —CH2CH3

—H

Principles of Stereochemistry

2.121

At C-2 and C-4, the lowest priority ligand (H) is on the horizontal bond and therefore, the absolute configuration of C-2 is S, because 1 Æ 2 Æ 3 or a Æ b Æ c traces a clockwise path and the absolute configuration of C-4 is R, because 1 Æ 2 Æ 3 or a Æ b Æ c traces a counterclockwise path. Since the lowest priority group, i.e., —CHOHCH2CH3 on C-3 is on a vertical bond and the sequence 1 Æ 2 Æ or a Æ b Æ c traces a clockwise path, the absolute configuration of C-3 is R. Therefore, the complete IUPAC name of the compound is (2S, 3R, 4R)-3-chloro-3-fluoro-2,4-dihydroxyhexanal. Ra, Sa-nomenclature of allenes For assigning Ra, Sa designations to allenes having chiral axis, the structure of the allene is to be regarded as an elongated tetrahedron and the molecule is to be viewed along the axis. It is immaterial from which end the allene is viewed. The four groups (two on each terminal sp2 carbon) are projected onto a plane right angles to the chiral axis. In the resulting Newman projection formula, the near groups are to be given priority over the far. For better visualization, the near groups are to be placed on a thick line (horizontal or vertical). In the chiral allene xyC = C = Cxy, if the group x has priority higher than y, then the horizontally placed (nearest to the viewer) x and y are to be numbered 1 and 2 and the vertically placed (rear) x and y are to be numbered 3 and 4, respectively. Then, if 1 Æ 2 Æ 3 traces a clockwise path, the configuration is Ra (as shown in the figure below) and if 1 Æ 2 Æ 3 arrow points counterclockwise, the configuration is Sa (‘a’ stands for ‘axial’). The same procedure is followed to assign R, S-designations to cumulenes having even number of double bonds (4, 6, 8, etc.).

Æ

Æ

It is to be noted that if in the projection formula the lowest priority group (4) remains in a horizontal bond, it is not necessary to transfer it to any vertical bond before tracing the path in the sequence 1 Æ 2 Æ 3.

2.122

Organic Chemistry—A Modern Approach

Example: 3 CO2H viewed from the side A

Viewer H A

CO2H C C C

CH2CH3 1

2H H 4

B

H

Viewer I Hexane-2,3-dienoic acid (CH3CH2– H– ; –COOH –H)

1 CO2H

CH3CH2

viewed from the side B

H 4

3 CH3CH2

H 2 1 Æ 2 Æ 3 traces a clockwise path. Hence, the configuration of the molecule is Ra. 2 viewed from the side A

A

2

2

3

B

2

3

2 viewed from the side B

Æ

3

2

Æ Sa

Ra, Sa-nomenclature of biphenyls Ra, Sa-descriptors to the atropisomers of biphenyl derivatives can also be assigned by the same method as in the cases of substituted allenes with chiral axis. In this method, the four carbon atoms C-2, C-6, C-2¢ and C-6¢, which correspond to the four vertices of

Principles of Stereochemistry

2.123

an elongated tetrahedron are to be taken into considerations for the application of the sequence rules according to CIP conventions. In the following biphenyl derivative, for example, the priority sequence in the left ring is C-2 C-6 (–NO2 –COOH) and in the right ring is C-2¢ C-6¢ (–NO2 –COOH). Therefore, the configuration is Sa when the molecule is viewed from either direction (A or B). 1 NO2 viewed from the side A along 1-1¢ bond

Viewer

5 A 4

(

3

CO2H 2

NO2 CO2H

2¢ 3¢

6

2

1

4¢

1¢

NO2 3

4 HO2C

B

6¢ 5¢

3 NO2

Viewer

HO2C NO2

)-6,6¢-Dinitrobiphenyl-2,2¢dicarboxylic acid

viewed from the side B along 1¢-1 bond

1 O2 N

CO2H 2

CO2H 4 The traced path from 1 Æ 2 Æ 3 is counterclockwise. Hence, the configuration is .

When C-2 and C-6 in a ring are attached to identical groups and C-3 and/or C-5 are also substituted, the priority order of the ortho carbons is then determined through exploration around the ring or side chain. Thus, in the following example, the priority order of the ortho carbons in the right ring is C-2¢ C-6¢ (–Me –H). 2 viewed from the side A

2 2

6¢

A 1

1¢

5¢ 4¢

B

2¢ 3¢ 2

2

viewed from the side B

2

Æ

Æ a

2.124

Organic Chemistry—A Modern Approach

2.5.3 Erythro and Threo Nomenclature of Compounds with Two Adjacent Chiral Centres The terms erythro and threo are commonly used to describe pairs of enantiomeric molecules of the type abxC—Caby. That enantiomeric pair is designated by the prefix erythro in which at least two pairs of similar groups eclipse each other and that enantiomeric pair is designated by the prefix threo in which only one pair of similar or like groups eclipse each other. Therefore, A and B are the erythro-forms because in these structures a eclipses a and b eclipses b and C and D are the threo-forms because in these structures a eclipses a and b eclipses b but not both a, a and b, b at the same time. Mirror plane a b

a C

a b

C

y

b

x

y b b

y

a a

y

x

b

y a

b a

x

a

a

C

x

b

a

b C

x

y

b b

a b

C

y

b

x

a

a C

a

b

x

a

y

y

a a

b b

C

x

b

x

b

a C

b

a a

Mirror plane

x

b y

a b

b a

x

y

x

b a

a b

x

y

x

x

x

a

b

b

a

a

b

b

a

a

b

b

a

b

a

a

b

y

y erythro enantiomers

y

y

threo enantiomers

4-Bromo-3-hexanol containing two chiral centres, for example, can exist in two pairs of enantiomeric forms. One pair is erythro-4-bromo-3-hexanol and the other pair is threo-4bromo-3-hexanol. This may be shown as follows:

Principles of Stereochemistry

2.125

Mirror plane CH CH Br

H H

OH CH CH

Mirror plane

Br

CH CH H

Br

HO

H

H

CH CH

(±)-erythro-4-Bromo-3-hexanol

CH CH H OH

H

CH CH Br

HO

CH CH

H CH CH

(±)-threo-4-Bromo-3-hexanol

An erythro-isomer is diastereoisomeric with a threo-isomer.

2.5.4

The E-Z System of Designating Alkene Diastereoisomers

In article 2.3.3, we learned the terms cis and trans to designate the stereochemistry of alkene diastereoisomers (cis-trans-isomers). The terms cis and trans are unambiguous only when used to designate the stereochemistry of alkenes of the type Cab==Cab or Cab==Cac. When, however, all the four groups attached to the two doubly bonded carbons are different, cis and trans prefixes cannot be applied at all. For example, it is not possible to designate two isomeric 1-chloro-1-fluorobut-1-enes (I and II) as cis and trans since no pairs of substituents are the same. C2H5 Cl C C F H I

H Cl C C F C2H5 II

To avoid this difficulty, E-Z system of nomenclature has been introduced. The E-Z system is used as follows: 1. The two groups attached to one doubly-bonded carbon are examined and relative priorities of the groups are determined according to the CIP sequence rules. 2. The operation is repeated at the other carbon atom. 3. The group of higher priority on one carbon atom is compared with the group of higher priority of the other carbon atom. If the two groups of higher priority are on the same side of double bond, the alkene is said to have the Z-configuration (Z from the German word Zusammen, meaning ‘together’). If the two groups of higher priority are on opposite sides of the double bond, the alkene is said to have E configuration (E from the German word entgegen, meaning ‘opposite’). These configurational designations are added to the name of the alkene as an italics prefix in parentheses.

2.126

Organic Chemistry—A Modern Approach

higher priority

higher priority

higher priority

lower priority

C C

C C

lower priority

lower priority

higher priority

lower priority

Z-configuration (groups of higher priority are on the same side of the double bond)

E-configuration (groups of higher priority are on the opposite sides of the double bond)

Example: higher priority

higher priority

F

I

F F

C C

H

higher priority

CH3

CH3

F

H CH3

C C

H

I higher priority

(Z)-1-Fluoro-2-iodopropane

(E)-1-Fluoro-2-iodopropane

Although in most cases, Z corresponds to conventional cis and E corresponds to trans, this may not always be the case. For example: F H

F Cl

Cl

F

F>H Cl > F

C C

H

C C F

(Z)-1-chloro-1, 2-difluoroethene [trans-1-chloro-1,2-difluoroethene]

(E)-1-chloro-1, 2-difluoroethene [cis-1-chloro-1,2-difluoroethene]

When cis-trans- isomers contain two or more double bonds, the configuration of each double bond is to be specified with appropriate locants before the designations E and Z. For example: H 1

3

4

C C

2

CH3CH2

H 5

6

C C

H

H 7

8

9

CH2CH2CH3

(3Z, 5E)-3,5-Nonadiene

When there is a choice in counting the carbon atoms in a molecule having both (Z) and (E) descriptors, Z gets preference over E. For example: 1

2

CH3CH2 H

3

4

C C H

H 5

7 6

C C

8

CH2CH3 H

(3Z, 5E)-3,5-Octadiene [Lower numbers are assigned to the (Z) double bond]

It is to be noted that E-Z descriptors are not used for disubstituted cycloalkanes.

Principles of Stereochemistry

2.127

2.5.5 R, S and E, Z Assignment in the Same Molecule (Geometric Enantiomerism) Geometric isomers are optically inactive. However, if they develop molecular chirality (with or without chiral centre), then they also exhibit optical isomerism (rotates plane polarized light). In such cases, the isomerism is known as geometric enantiomerism. If a molecule contains both a chiral centre or chiral axis and an appropriately substituted double bond (C = C, C = N) to give E, Z-isomers, then four stereoisomers are possible. According to the CIP system, they may be designated as (R, E), (S, E), (R, S) and (S, Z). For example: (i) 1-Phenylpent-2-en-1-ol (PhCHOHCH=CH CH2CH3), which contains a chiral centre and a suitably substituted double bond capable of giving cis-trans isomers, exists as four stereoisomeric forms as given below:

2

5

5 2

2

5

5 2

(A, B) and (C, D) are two pairs of enantiomers. But each of (A, C), (A, D), (B, C) and (B, D) represents a pair of diastereoisomers.

2.128

(ii)

Organic Chemistry—A Modern Approach

1,5-Dichloropent-1,2,4-triene (ClCH=C=CH–CH=CHCl) shows geometric enantiomerism because the allene moiety is chiral and the double bond is appropriately substituted to give cis-trans isomers. Its four possible stereoisomers are as follows: Mirror plane Cl C H

C R

C

H C

H

Cl

H

H E

C

C

H C

Cl

Cl

A (1 , 4 )-1,5-Dichloropent1, 2, 4-triene

C Z

C S

C H

H

B (1 , 4 )-1,5-Dichloropent1, 2, 4-triene

A pair of enantiomers Mirror plane Cl C H

C R

C

Cl C

H

Cl

H

H Z

C

H

C (1 , 4 )-1,5-Dichloropent1, 2, 4-triene

C

Cl C H

Z

C

H

C S

C H

D (1 , 4 )-1,5-Dichloropent1, 2, 4-triene

A pair of enantiomers

(A, B) and (C, D) are two pairs of enantiomers. But each of (A, C), (A, D), (B, C) and (B, D) represents a pair of diastereoisomers.

2.5.6 Syn-anti Nomenclature for Aldols Aldol condensation can create two asymmetric and also stereogenic centres in the products (i.e., aldols) depending on the nature of the reactants. Many aldol condensations are found to be diastereoselective.

Principles of Stereochemistry

2.129

R¢ O

O +

R¢

R≤

O +

R¢

R≤

R

Base

R

I

R≤

H

OH

OH

O

R

I¢ OH

O R¢

R≤

OH

O +

R¢

R≤ R

R II

II¢

Diastereoisomers of the type (I, I¢) are commonly referred to as syn or erythro and those of the type (II, II¢) as anti or threo.

2.5.7 Number of Stereoisomers for Compounds with Chiral Centres The number of stereoisomers (both chiral and achiral, i.e., meso compound) for a compound having chiral centres depends upon the number as well as the nature of chiral centres present in the compound. Case 1: If a molecule contains even or odd number of ‘n’ different chiral centres and the molecule cannot be divided into two mirror-image halves, the number of possible stereoisomer = 2n, all of which are chiral. For example: (a)

*

Lactic acid (CH3CHOHCOOH) has n = 1 and therefore, the number of stereoisomer is 21 = 2, which are enantiomers. Mirror plane COOH HO

CH3

COOH H3C

H

OH H

(R)-Lactic acid

(S)-Lactic acid

Enantiomers

(b)

*

*

2-Bromo-3-chloropentane (CH3 C HBr C HClCH2CH3) has two different chiral centres and hence, the number of stereoisomers is 22 = 4. I and II and III and IV are two pairs of enantiomers, while each of I and III, I and IV, II and III and II and IV represents a pair of diastereoisomer.

2.130

Organic Chemistry—A Modern Approach

CH3

Chiral Mirror plane

CH3

CH3

Chiral Mirror plane

CH3

H

Br

Br

H

H

Br

Br

H

H

Cl

Cl

H

Cl

H

H

Cl

C2H5

C2H5

(2S, 3R) I

C2H5

(2R, 3S) II

C2H5

(2S, 3S) III

(2R, 3R) IV

Enantiomers Enantiomers Case 2: If the molecule contains an even number (n) of chiral centres, and the molecule can be divided into two mirror-image halves in one of the possible conformations, it can have 2(n–1) chiral stereoisomers and 2(n–2) achiral stereoisomers, i.e., meso compounds. *

*

For example, tartaric acid (HOOC C HOH C HOHCOOH) has two similar chiral centres and can be divided into two mirror-image halves. Therefore, it has 2(2–1) = 21 = 2 chiral stereoisomers (V and VI) and 2(2–2)/2 = 2° = 1 achiral stereoisomer (VII). In fact, VII and its mirror image are identical structures (superimposable on each other by 180° rotation of the Fischer projection in the plane of the paper).

H

Chiral COOH Mirror plane HO OH

HO

H

COOH H

H

COOH

OH COOH

(2S, 3S) V

(2R, 3R) VI

s-plane

H H

identical and achiral COOH Mirror plane OH HO OH

HO

COOH (2R, 3S)

COOH H

s-plane

H COOH (2S, 3R)

VII

Enantiomers

meso compound (optically active) V and VI represent a pair of enantiomers (optically active) and VII represents an achiral (optically inactive) meso compound having a plane of symmetry (s).

Case 3: If a molecule contains an odd number (n) of chiral centres and the molecule can be divided into two mirror-image halves by a plane passing through the central chiral centre in one of the possible conformations, it can have 2(n–1)–2(n–1)/2 chiral (optically active) isomers and 2(n–1)/2 achiral (optically inactive) meso compounds. For example, 4-bromoheptane-3, *

*

*

5-diol (CH3CH2CCHOHCHBr C HOHCH2CH3) has three chiral centres and can be divided into two mirror-image halves by passing a plane through C-4 chiral centre. Therefore, it has 2(3–1)–2(3–1)/2 = 4 – 2 = 2 chiral isomers (VIII and IX) and 2(3–1)/2 = 21 = 2 achiral isomers (X and XI).

Principles of Stereochemistry

2.131

chiral

achiral

Mirror plane HO OH Br Br H H

C2H5 H H HO

C2H5

C2H5

H H s-plane H H OH H

C2H5 VIII

OH Br OH

Mirror plane

C2H5 H Br H

OH H s-plane OH

C2H5

C2H5

C2H5

IX

X

XI meso compounds

Enantiomers

VIII and IX represent a pair of enantiomers (optically active) and X and XI represent a pair of achiral (optically inactive) meso compounds.

2.5.8

Chirotopic and Achirotopic Atom in a Molecule

The site symmetry of atoms in molecules may be classified as chiral and achiral. A chirotopic atom is one whose site symmetry is chiral, i.e., which resides in a chiral environment. Chirality is an all-inclusive property (as it affects all parts of a chiral molecule) and so, all the atoms in a chiral molecule are chirotopic. For example, all the five atoms in bromochlorofluoroiodomethane (CFClBrI) are chirotopic. [The molecules bearing chirotopic atom or centre need not be as a whole chiral.] An achirotopic atom is one whose site symmetry is achiral, i.e., the atom lies on a plane of symmetry (s) or on a centre of symmetry or an alternating axis of symmetry (Sn) passes through it that intersects its reflection plane. For example, in chlorofluoromethane (CH2ClF), the carbon atom is achirotopic since a s-plane passes through it. These terms including pseudoasymmetric atom or centre can be well illustrated by 2, *

*

*

3, 4-trihydroxyglutaric acid (HOOC C HOH C HOH C HOHCOOH). This dicarboxylic acid containing three chiral centres exists as two chiral and two achiral isomers. enantiomers 1

COOH 2 3 4

H HO HO 5

OH H H

COOH I (chiral)

Mirror plane

COOH HO H H

H OH OH

COOH II (chiral)

s-plane

COOH H H H

OH OH OH

COOH III (achiral)

COOH s-plane H OH OH H H OH COOH IV (achiral)

2.132

Organic Chemistry—A Modern Approach

(R) C-2 HO

chirotopic and nonstereogenic

C-2 (S)

H

OH

H

(R) C-4 I [C-3 is achiral because C-2 and C-4 are both R]

C-4 (S)

(R) C-2 H

OH

C-2 (R)

HO

(S) C-4

II [C-3 is achiral because C-2 and C-4 are both S]

achirotopic and stereogenic

H C-4 (S)

III

IV

[C-3 is chiral and the configuration is (r)]

[C-3 is chiral and the configuration is (s)]

[priority order of the groups at C-3 is: –OH R S –H]

Though the isomers I and II are chiral, C-3 in both of them is achiral. However, C-3 in them is chirotopic, as it resides in a chiral environment. Again, C-3 in them is nonstereogenic since interchange of ligands (say –H and –OH) bonded to C-3 does not produce a new stereoisomer (in fact, interchange of ligands followed by 180° rotation in the plane of the paper produces the original). 180° rotation in the plane of the paper

interchange of H and –OH at C-3

I

I Identical

C-3 in III and IV is chiral. However, it is achirotopic because a s-plane passes through it. Again, C-3 in them is stereogenic because interchange of –H and –OH at this centre generates new stereoisomer (III generates IV and II generates III). A pseudoasymmetric carbon atom or centre is one which is attached to two enantiomorphic ligands having opposite chirality sense and two other ligands different from the previous ones. A pseudoasymmetric carbon or centre is stereogenic but chiral even though the molecule is achiral. In spite of the chiral centre as C-3, the molecules III and IV are achiral and their C-3 is designated as a pseudoasymmetric centre. The configuration of a pseudoasymmetric centre is designated by ‘r’ or ‘s’ (instead of R or S). Therefore, the complete IUPAC name of III is (2R, 3s, 4S)-2, 3, 4-trihydroxyglutaric acid and of IV is (2R, 3r, 4S)-2, 3, 4-trihydroxyglutaric acid.

Principles of Stereochemistry

H H H

2.133

1

COOH

1

2

OH

2

3

OH OH

4

H HO H

5COOH

R

2.5.9

s

COOH OH H OH

3 4

5COOH

S

R

r

S

Prostereoisomerism and Topicity

Ligands (atoms or groups in a molecule) are called homomorphic (Greek homos means same and morphe means form) if they are indistinguishable when considered in isolation. In the case of atoms, they must be the atoms of the same element, e.g., two Cl atoms, two H atoms, etc. and in the case of groups they must have the same constitution and configuration, e.g., two Et or two Ph groups or two secondary butyl groups of same configuration (R or S). In an intact molecule, if the homomorphic ligands are bonded to constitutionally different ligating centres, the ligands are called constitutionally heterotopic. For example, H’s at C-2 in heptane are constitutionally heterotopic compared to H’s at C-4. constitutionally heterotopic H’s 1CH

2 3 4 5 6 7 3— CH2— CH2— CH2— CH2— CH2— CH3

Heptane

When the spatial relation of two ligands with the rest of the molecule is different, the ligands are called stereoheterotopic. For example, the geminal H’s at C-2 (designated as HA and HB) of heptane are stereochemically not equivalent, i.e., stereoheterotopic (actually enantiotopic).

Topicity describes the relationships of two or more homomorphic ligands (or faces) which together constitute a set. Homomorphic atoms or groups can be classified as homotopic, enantiotopic or diastereotopic. Ligands, which are not homomorphic, cannot be related topically.

2.134

2.5.9.1

Organic Chemistry—A Modern Approach

Homotopic ligands

A pair of homomorphic ligands (atoms or groups) in a molecule are called homotopic if substitution (replacement) of each of them in turn by a ligand, which is not already attached to the ligating centre, gives identical product. For example, the three hydrogens of acetic acid (CH3COOH) are homotopic because replacement of any of these H’s by atom, one at a time, produces the same molecule of bromoacetic acid (BrCH2COOH).

H.V.Z. reaction

2.5.9.2

Enantiotopic ligands

A pair of ligands (atoms or group) in a molecule is called enantiotopic if replacement of one or the other of them by a different achiral ligand, which is not already attached to the ligating centre, gives rise to one or the other of a pair of two enantiomers. For example, the two methylene hydrogens (HA and HB) or propanal (CH3CH2CHO) are enantiotopic because replacements of these hydrogens in turn, say by F, gives rise to two enantiomers of 2-Fluoropropanal (CH3CHFCHO). enantiotopic H’s HA

HB C

OHC CH3 Propanal CHO replacement of HA by F

HA

HB

replacement of HB by F

CH3 Mirror plane CHO

CHO F

HB

HA

CH3

F CH3

Enantiomers

Principles of Stereochemistry

2.5.9.3

2.135

Diastereotopic ligands

Two ligands (atoms or groups) in a molecule are called diastereotopic if replacment of one or the other of them by a different ligand produces one or the other isomer of a set of diastereoisomers. For example, the two H atoms on the same carbon (C-1) of propene are diastereotopic because sequential replacements of these H’s, say by Br, result in the formation of a pair of diastereoisomers (two cis-trans or geometric isomers). disastereotopic hydrogens Br

H C

cis-1-Bromopropene

H

Br C

replacement of HB by Br

C CH3

H

HB C

replacement of HA by Br

C CH3

HA

C H3C

H

Propene

H

trans-1-Bromopropene

diastereoisomers

The two H atoms on C-2 in 3-bromobutanoic acid are also diastereotopic because sequential replacements of these hydrogens, say by Cl, give rise to a pair of diastereoisomers.

replacement of HA by Cl

A

3

R

B

replacement of HB by Cl

3

R

3

S

R

R

2.5.9.4

Prochiral and prostereogenic centres

A chiral centre is created when one of the pair of enatiotopic or diastereotopic ligands is replaced by a ligand different from the ligands already present on the chiral centre. Therefore, an achiral centre bearing enantiotopic or diastereotopic ligands is called a prochiral centre. For example, C-2 of acetic acid, C-3 of 2-bromobutane, etc. are prochiral centres. Again, if the two faces of an sp2-hybridized carbon are enantiotopic or diastereotopic, that carbon is also called a prochiral centre. For example, the carbonyl carbon of chloral and the carbonyl carbon of 3-hydroxy-2-butanone are prochiral centres.

2.136

Organic Chemistry—A Modern Approach

prochiral centre

H

CH3

S

C

C H CH2 CH3 Br (S)-2-Bromobutane

CH3 CO2H

H

prochiral centre

H Cl3C

chiral centre prochiral centre

C

CH3 H HO

O

C C

chiral centre prochiral centre CH3

O (S)-3-Hydroxy-2-butanone

Chloral

A prostereogenic centre is a centre which can be converted into a stereogenic centre by replacing one of the homomorphic ligands (atoms or groups) by a different ligand. A prostereogenic centre may or may not be prochiral. For example, the C-2 of bromoethene is prostereogenic because on replacement of HA or HB by Br it becomes converted into a stereogenic centre. However, C-2 is not prochiral because it cannot be converted into a chiral centre by replacing HA or HB.

A B

replacement of HA by Br

4

3

2

1

However, when one of the homomorphic H atoms on C-3 of 2-butanone (CH3 CH2 COCH3 ) is replaced by F atom, the centre C-3 becomes stereogenic as well as chiral. Therefore, in this case, C-3 is both prostereogenic and prochiral. prochiral and prostereogenic centre CH3 HA

chiral and stereogenic centre

CH3 HB

COCH3 2-Butanone

H (or H ) is replaced by F

F

H

COCH3 (R)-3-Fluorobutan-2-one

Principles of Stereochemistry

2.137

2.5.9.5 Pro-R and pro-S designation of enantiotopic and diastereotopic ligands (groups or atoms) Enantiotopic and diastereotopic pair of atoms or groups on a prochiral centre in a molecule is designated as pro-R and pro-S. If replacement of one of them by a nonequivalent ligand with higher priority than the other (without disturbing the priority order with respect to the remaining ligands) gives (R)-configuration of that chiral centre, that ligand is specified as pro-R and if (S)-configuration, then as pro-S. For example, replacement of HA in chloroiodomethane (CH2ClI) by D gives (R)-chlorodeutereoiodomethane and replacement of HB by D gives (S)-chlorodeutereoiodomethane. Therefore, HA is pro-R and HB is pro-S. pro-R

D C H

H A

C

I Cl

H

prochiral centre B

I Cl

2.5.9.6

C

I Cl (S)-Chlorodeutereoiodomethane D

pro-S

(R)-Chlorodeutereoiodomethane

H

Chloroiodomethane

Homotopic faces

The two faces of a planar group like C=O, C=C, etc., are said to be homotopic if addition of the same reagent to either face generates the same product. Thus, the two faces of formaldehyde (HCHO) are homotopic because addition of MeMgI to either face of it leads to the formation of ethanol. d-

d-

d+

d+

top face addition

H3O

bottom face addition H3O

2.5.9.7

Enantiotopic faces

The two faces of a planer n-orbital system like C = O, C = C, etc. are said to be enantiotopic, if attack of a reagent on one side of the p-bond produces one enantiomer, whereas attack on the opposite side leads to the other enantiomer. For example, the two faces of acetophenone

2.138

Organic Chemistry—A Modern Approach

(PhCOCH3) are enantiotopic because reaction with the achiral reagent LiAlH4 at one face gives a molecule of phenylmethylcarbinol, whereas reaction at the other face produces the enantiomeric phenylmethylcarbinol. Acetophenone H

H

top face attack

Me Ph

–

Me C O

3

bottom face attack

H (hydride ion)

OH enantiomers

Me Ph

–

2.5.9.8

S

Mirror plane

1. LiAlH4 2. H O≈

Ph

C

OH

C

R

H Phenylmethylcarbinol

Diastereotopic faces

The two faces of a double bond, i.e., a p-system like C=O, C=C, etc. are said to be diastereotopic, if attack of a reagent on one side of the p-bond leads to one diastereoisomer, whereas attack on the opposite side produces the other diastereoisomer. For example, the two faces of the carbonyl group of chiral 2-phenylpropanal are diastereotopic because reaction with MeMgI at one face affords a molecule which is diastereoisomeric with the molecule formed by reaction at the other face.

H

d+

CH3

MgI

C Ph H3C

top face addition

d–

C

d– H3 C

O H

1. ether 2. H3O≈

d+

MgI

(R)-3-Phenylpropanal

2.5.9.9

bottom face addition

H Ph H3C

Me C R

C S

H Ph H 3C

C R

C

OH H OH H

diastereoisomers

R

Me 3-Phenyl-2-butanol

Re and Si designation of enantiotopic and diastereotopic faces

The enantiotopic and diastereotopic faces can be specified by an extended use of CIP convention of assigning configurational designation based on priority sequence rules in two dimensions. If the three groups of the trigonal centre as arranged by the sequence rules have the order a b c, that face (when viewed from the top of the molecule) in which the groups in this sequence are clockwise (as in X) is Re face (from Latin rectus) and that face in which the groups in this sequence are counterclockwise (as in Y) is the Si face (from Latin Sinister).

Principles of Stereochemistry

2.139

c

b

C a

[The structure Y is actually the bottom face of structure X. This can be obtained by 180° out-of-plane rotation of structure X.]

C a

b

c

X Re face

Y Si face

For example, the face of the carbonyl group of benzaldehyde as shown in structure I is the Re face because the ligands (=O Ph H) trace a clockwise path, whereas the face as shown in structure II is the Si face because the ligands trace a counterclockwise path. 3 H 2 Ph

2 Ph C O 1

3 H

I

1Æ 2 Æ 3 clockwise Hence, it is Re face

C O 1 II

1Æ 2 Æ 3 counterclockwise Hence, it is Si face

Two opposite faces of benzaldehyde

The two carbonyl faces of (S)-3-chloropentanal are diastereotopic (due to presence of a chiral centre). The face shown in structure III is Re (1 Æ 2 Æ 3 clockwise) and the other face shown in structure IV is Si. A further subscript S (the chirality description of C-3) may be added so that the double-lettered subscripts ReS and SiS are in true sense diastereotopic. 3 C2H5 H C C 2 H S CH2 Cl III ReS face

2.5.9.10

C2H5 O 1

H Cl

C S

2 CH2 C

O 1

3 H IV SiS face (S)-3-Chloropentanal

Retention of configuration

If the spatial arrangement of bonds with respect to a chiral or asymmetric carbon in a chiral molecule remains the same before and after the reaction, retention of configuration occurs. In the conversion of the substrate XCabc into the product YCabc, if the hypothetical path traced by the ligands a, b and c (a Æ b Æ c) remains the same (clockwise as shown in

2.140

Organic Chemistry—A Modern Approach

the following figures) in both the substrate and the product, the reaction is said to have taken place with retention of configuration. XCabc and YCabc have the same relative configuration. c c C b

+Y – –X –

X

C b

a

(a Æ b Æ c traces a clockwise path)

Y

a

(a Æ b Æ c traces a clockwise path)

configuration retained

In general, during a reaction, if none of the bonds to the chiral centre is broken, the product has the same relative configuration of the ligands around to the chiral centre as that of the substrate. Such a reaction is said to proceed with retention of configuration. For example, when 2-methyl-butanol is heated with concentrated HCl, 1-chloro-2-methylbutane is obtained. In this reaction, no bond to the asymmetric carbon is broken. Therefore, the reaction must take place with retention of configuration. C2H5 H

C

CH2

C2H5 HCl D

CH2 OH

H

C

CH3

CH2Cl

+ H 2O

(+)-1-Chloro-2-methylbutane

(–)-2-Methyl-1-butanol

Configuration retained

Mechanism: C2H5 H

H Cl

C

CH3

2.5.9.11

C2H5

CH2 OH

H CH3

C2H5 + Cl –

C CH2 OH2

SN2

H CH3

C CH2Cl + H2O

Inversion of configuration

If the spatial arrangement of bonds with respect to an asymmetric or chiral carbon in a chiral compound suffers a change after the reaction, the compound is said to have undergone inversion of configuration.

Principles of Stereochemistry

2.141

In conversion of the substrate XCabc, into the product YCabc, if the hypothetical path traced by the ligands a, b and c (a Æ b Æ c) changes (clockwise in XCabc but counterclockwise in YCabc), the substrate is said to have undergone the reaction with inversion of configuration. c c + Y – – X –

Y

C

–

b

X

Y

C

b

+ X

–

a

a

(a Æ b Æ c traces a clockwise path)

(a Æ b Æ c traces a counterclockwise path)

Inversion of configuration

When 2-chlorobutane, for example, is subjected to alkaline hydrolysis under SN2 conditions, the product 2-butanol is obtained with inversion of configuration. H

HO

–

C

Cl

Et Me 2-Chlorobutane

NaOH/H2O SN2

H

HO

C

Et Me 2-Butanol

+ Cl

–

Inversion of configuration

1.

Write down the Fischer projections of the following compounds: (a) L-Glyceraldehyde (b) D-2-Hydroxybutanedioic acid (c) D-2-Phenylbutan-2-ol (d) L-2-Fluoro-2-phenylpropanoic acid (e) 2D-Ethyl-2L-methyl-2-phenylbutanoic acid (f) D-2-Butylmethylether (g) 2L-Chloro-2D-fluoropentane (h) 5D-Ethyl-3D-hydroxy-1L-methylcyclohexane-1D-carboxylic acid

Solution

COOH

CHO

(a)

HO

H CH2OH

L-Glyceraldehyde

(b)

H

OH CH2COOH

D-2-Hydroxybutanedioic acid

2.142

Organic Chemistry—A Modern Approach

COOH

C2H5

(c)

H 3C

F

(d)

OH

CH3 Ph

C6H5

L-2-Fluoro-2-phenylpropanoic acid CH3

D-2-Phenylbutan-2-ol COOH

(e)

CH3

H

(f)

C2H5

CH2CH3

Ph

D-2-Butylmethylether

2D-Ethyl-2L-methyl-2phenylbutanoic acid CH3

(g)

Cl

OCH3

Me

(h)

F

H

CH2CH2CH3

1

Et H

5

COOH OH

3

5D-Ethyl-3D-hydroxy1L-methylcyclohexane1D-carboxylic acid Assign R or S designations for each of the following compounds: CH2OH 2L-Chloro-2Dfluoropentane

2.

(a)

C* H CH3

(b)

Br

COOH *C

(c) (CH3)2CH

H Br

*

O

CH CH2 OH

C*

(d)

HC C O2 N

(f)

Me CH2COOH

CN–

CH(CH3)2 N*

(e) Ph

H CH3

S* CH3

*

Me

(g)

(h)

*

Me O

(i)

Br *

H

*

H

Br

(j)

C

H

*

Cl

C

Cl

*

H

Principles of Stereochemistry

2.143

O

(k)

C CH

*

H3 C

H NH2

(m)

*C

Ph

(n)

CD2OH CH2OH

HSH2C

*C

(l)

CH2Cl

H

H H 3C

H

C*

(p) CH(OCH3)CH2CH3

S S

H

H

S

HO H3C

R

OH

C

C

*

C

CH3 H

CH3

CH3

(q)

CH3

D

CH(OCH3)CH(CH3)2

(o)

*

C

Ph C CCH3 Cl OH * C

R

C

R

OH H HO H

Solution

(a)

Priority order of ligands: CH2OH

CH3 > H 1

1 CH2OH

The arrow (1 Æ 2 Æ 3) traces a clockwise path, suggesting the R configuration.

C

4 H

2

3

2

3 CH3

1 CH2OH

or, 4 H

CH3 3

1 Allowed interchange

3

2 4

2 (in Fischer)

2.144

(b)

Organic Chemistry—A Modern Approach

Priority order of ligands: Br

O || CH2— C —

3

2

or,

Br 1 3

3 1

2

4 2

(c)

H

The arrow from 1 Æ 2 Æ 3 traces a clockwise path, suggesting the R configuration

H 4

O

CH2CH2—

1 4

Priority order of ligands: OH

*

COOH

CH=CH2

CH(CH3)2

120° rotation 2 3 2

3 2

2

Æ

(d)

Priority order of ligands: Br

NO2

CN

Æ

C∫CH

Æ

Æ S

Principles of Stereochemistry

(e)

2.145

Priority order of ligands: Ph 2 CH(CH3)2 N

or,

1

3

2

2 4

Allowed interchange

1

3

1

4

Priority order of ligands: electrons

CH2COOH

CH3

4 120° rotation

CH2COOH 2 or,

4 3

Priority order of ligands: Cl

Cl H

1 Cl 4 H

or,

C

120° rotation

H

C 3

3

Cl H

C

Allowed interchange

1

3 4

H

H

Et

E

2 1

3

Et

2 4

1

C

4

2 1

3

2 H

C

Et H

Z

H C C 2 Et C

1

The arrow from 1 Æ 2 Æ 3 points counterclockwise, suggesting the S configuration

2 1

H

2

S

4 3

1 CH3

(◊ ◊) lone pair of

Me

2

S

3 Me

(g)

H

The arrow from 1 Æ 2 Æ 3 traces a clockwise path, suggesting the R configuration.

CH3 3

3

(f)

CH3

2

H 4

1 Ph

CH(CH3)2

3

1

The arrow from 1 Æ 2 Æ 3 points counterclockwise, suggesting the S configuration.

2.146

Organic Chemistry—A Modern Approach

C

(h)

Priority order of ligands at C-1 and C-4: C(Me) = C(Me) — 4 CH2CH2—

7

CH2CH

C-4: * 4

5 * 6

3 1 2

180° rotation

7

Æ

Æ

Æ

Allowed interchange

C-1: 1

2 3

Æ

6 4

5

Allowed interchange

C

(i)

Priority order of ligands at C-2 and C-6: Br

CH

C

CH2CH2—

H

C

Principles of Stereochemistry

2.147

8 7 4

5

Br

3

1

*6

2 *

H

Br

H 3

4H 1 Br

3 6

2

2

C

H 4 Br 1

3

3

C-2: 2

or,

4 1

2

1

3 1

4 Allowed interchange

4

3

2

4 1 or,

3

The arrow from 1 Æ 2 Æ 3 points counterclockwise, suggesting the S configuration of C-6.

C 2

1

2 3

3 4

1

Allowed interchange

1

2

(j)

1 2

3

C-6:

The arrow from 1 Æ 2 Æ 3 points clockwise, suggesting the R configuration of C-2.

Priority order of ligands at C-2 and C-3: Cl

2 4

O—

CHCl—

H

2.148

Organic Chemistry—A Modern Approach

∫∫

Æ

Æ R

Æ

Æ R

Principles of Stereochemistry

(k)

2.149

O || C —}

Priority order of ligands: CH2Cl O

CH2CH2—}

2 O ∫

CH Cl

3

* H

H 4 CH Cl 1

2 or,

H

2

∫

1

3

The arrow from 1 Æ 2 Æ 3 traces a clockwise path, suggesting the configuration.

2 3

4

1

1 4

3

(l)

Priority order of ligands: C∫CCH3

Ph

C∫CH

4 H3C

Ph 2

or,

1

2 3

(m)

120° rotation

Priority order of ligands: NH2

C 1

C∫CCH3 1

3

3

3

C∫CH 3 *C

CH3

4

∫

2

The arrow from 1 Æ 2 Æ 3 traces a clockwise path, suggesting the S configuration.

CH2SH

CD2OH

CH2OH

1

2

2.150

Organic Chemistry—A Modern Approach

∫

Æ

Æ R

Cl OH

Ph H D

Priority order of ligands at C-2: Cl > OH > CHDPh > CH3

*

* C—C 3

Priority order of ligands at C-3: C(OH)ClCH3 > Ph > D > H

2

CH3

Configuration of C-2: 120° rotation

∫ 2 3

∫

(n)

Æ

Æ R

Principles of Stereochemistry

2.151

Configuration of C-3: 2 Ph 4 H

2 1 C—C(OH)ClCH3 ∫

∫

C

4

3 D

2

4

2 3

Allowed interchange

3 4

Priority order of ligands: CH(OCH3) CH(CH3)2 1 CH(OCH3)CH(CH3)2 4 H

∫

4

Allowed interchange

2 H H S 2 1 R OH HO C 3 C 5 4 2 CH 3 *C 1CH H 4 3 CH3 3

∫∫

(p)

∫ 2

3

2

3 4

3

CH3

1

C

4

CH(OCH3)CH2CH3 2 1

or,

CH(OCH3)CH2CHCH3

1

C*

3 H3C

The arrow from 1 Æ 2 Æ 3 traces a clockwise path, suggesting the S configuration of C-3

1

1

(o)

1

3 2

or,

3

1

3

1 2

The arrow from 1 Æ 2 Æ 3 traces a clockwise path, suggesting the R configuration.

Priority of ligands at C-3: (R) CH(OH)CH3

(S) CH(OH) CH3

CH3

H

H

2.152

Organic Chemistry—A Modern Approach

The arrow from 1 Æ 2 Æ 3 traces an anticlockwise path, suggesting the S configuration.

∫ 1 4

C 2

1

∫ 2

3

3

1

1 or,

3

Allowed interchange

4

3 4

2

(q)

2

Priority order of ligands at the central chiral carbon: (R, R) 3 *

S

*

*

Æ

R

CH3

H

Æ

* *

R

R

∫∫

S

(S, S)

∫

3.

Assign priorities to the following pairs of groups according to the sequence rules: (a) —CH2CH2CH2CH3 and —CH2CH(CH3)2 (b) —CH2CH2Br and —C(CH3)3 C(CH ) CH(CH ) and —CH (c) —CH CH CH F CH CHF 12 14 (d) — CH2NH2 and — CH2CH3 (e) —Ph and —C∫∫C—CH2CH3 CH2CH3 (f)

C==CH2 and —C∫∫CH

Principles of Stereochemistry

2.153

(g) —CD216OH and —CH218OH (h)

and (CH3)2CH—

Solution (a) The priority sequence is: —CH2CH(CH3)2 —CH2CH2CH2CH3. The first point of difference is at the carbons b to the chiral centre. The b carbon b

a

(b)

b

a

of — CH2 CH(CH3 )2 and — CH2 CH2 CH2CH3 are attached to (C, C, H) and (C, H, therefore, —CH2CH(CH3)2 gets priority over H, H), respectively. Since, C —CH2CH2CH2CH3. The priority sequence is: —C(CH3)3 —CH2CH2Br. The first point of difference is at the two carbons attached to the chiral centre. The first carbon (Ca) in —C(CH3)3 is attached to (C, C, C) whereas the first carbon —CH2CH2Br is attached to (C, H, H). Since C H, (CH3)3C— gets priority over a

b

— CH2 CH2 Br , even though Cb bears an atom (Br) having higher atomic number. (c)

(d)

(e)

The priority sequence is: —CH

C(CH3)3 CH2CH2F

(I)

—CH

CH(CH3)2 CH2CHF2

(II)

The first point of difference is found at the branch carbon at Ca which is attached to (C, C, C) in the case of I, but (C, C, H) in the case of II. Since C H, I gets priority over II. The priority sequence is: —12CH2NH2 —14CH2CH3. Although the mass number of carbon in —14CH2CH3 is greater than that in —12CH2NH2, the latter group is preferred because the atomic number of the second atom (N, AN = 7) is higher than that of the former group (C, AN = 6). The priority sequence is: —C∫∫C—CH2CH3 —C6H5 (C)000 (C)000 —C ∫∫ C—CH2CH3

replication

C

C CH2CH3

(C)000 (C)000 (C)000 H C replication

C (C)000 C

(C)000 C H H C C

(C)000

H H (C)000 (C)000

In each of these two groups, the carbon bonded to the chirality centre is attached to (C, C, C). 1-Butenyl (—C ∫ C—CH2CH3) has two of its (C, C, C) carbons connected to (000), while the third is (C, C, C) and is thus preferred to phenyl (—C6H5), which has only one (0, 0, 0) and two (C, C, H)s.

2.154

Organic Chemistry—A Modern Approach

CH CH

(f)

The priority sequence is : —C ∫ CH

C==CH .

(C)000 (C)000

CH2CH3

—C ∫ CH

replication

C

C H ; —C

CH2

C2H5 H replication

(C)000 (C)000

C

C H

(C)000 (C)000

In each of these ligands, the first carbon is connected to (C, C, C). In ethynyl (—C∫∫CH), two of these carbons are connected to (0, 0, 0) and one is connected to CH2CH3 (C, C, H), whereas in 1-ethylethenyl —C==CH2 , one of these carbons is connected to

(g)

(0, 0, 0) and two are connected to (C, H, H). Since C > H, ethynyl gets priority over 1-ethylethenyl. The priority sequence is: —CH218OH —CD216OH This is due to the fact that in —CH218OH, 18O has higher atomic number as well as higher mass number.

(h)

The priority sequence is: Cyclopentyl

isopropyl (Me2CH–).

In each of these two groups, the first carbon is attached to (C, C, H). Both of these carbons in cyclopentyl are connected to (C, H, H), whereas in isopropyl, they are connected to (H, H, H). Since C > H, cyclopentyl gets priority over isopropyl. 4.

Give flying wedge or 3D-representations of the following compounds: (a) (S)-Alanine (b) (R)-2-Pentanol (c) (R)-3-Bromocyclopentanone (d) (2R, 3S)-3-Chloro-2-hexanol (e) (1S, 2R)-1-Ethyl-2-fluorocyclohexane

Solution

(a)

*

In alanine [CH3 CH(NH2)COOH], the priority order of ligands is: NH2(1) > COOH(2) > CH3(3) > H(4). The flying wedge structure (perspective formula) of (S)-alanine can be drawn by putting the lowest priority ligand (4) on the hatched wedge and other ligands on the remaining bonds such that the arrow from 1 Æ 2 Æ 3 traces a counterclockwise path. 3

2

S

S

Principles of Stereochemistry

(b)

2.155

*

In 2-pentanol [CH3 C H(OH)CH2CH2CH3], the priority order of ligands is: OH (1) > CH2CH2CH3(2) CH3 (3) H (4). The flying wedge structure (perspective formula) of (R)-2-pentanol can be drawn by putting the lowest priority ligand (4) on the hatched wedge and other ligands on the remaining three bonds such that the arrow from 1 Æ 2 Æ 3 traces a clockwise path. 3

3

2

2

O

(c)

In 3-bromocyclopentanone

, the priority order of ligands is: Br (1) Br

O || CH2— C — (2)

CH2CH2— (3)

H(4).

Therefore, the flying wedge or 3D-representation of (R)-3-bromocyclopentanone is:

(d)

*

*

In 3-chloro-2-hexanol (CH3 C HOH C HClCH2CH2CH3), the priority order of ligands at C-2 is: OH(1) CHClCH2CH2CH3 (2) CH3 (3) H(4) and at C-3 is: Cl (1) CH(OH)CH3 (2) CH2CH2CH3 (3) H(4). The flying wedge formula of 3-chloro-2-hexanol can be drawn by putting the lowest priority ligand (4) on the hatched wedges and other ligands (1 and 3) on the remaining bonds (2 becomes automatically fixed after putting 1) such that the arrow from 1 Æ 2 Æ 3 traces a clockwise path around C-2 and a counterclockwise path around C-3.

R

S

R 3

R

S

S 2

2

3

2.156

Organic Chemistry—A Modern Approach

Et H H 2 F 1-ethyl-2-fluorocyclohexane 1

(e)

The priority order of ligands at C-1: CHFCH2— (1) > CH2CH2CH2— (2) > Et(3) > H(4) and at C-2: F(1)> CH(Et) CH2— (2) > CH2CH2CH2— (3) > H(4).

The flying wedge formula of (1S, 2R)-1-ethyl-2-fluorocyclohexane is as follows: H 3

2 1

C 4H S 3 Et

2

C R

Et S

H 4

H

R

F F 1 (1S, 2R)-1-Ethyl-3-fluorocyclohexane

5.

Draw Fischer projections for the following molecules: (a) (S)-Lactic acid (b) (R)-2-Butanol (c) (2R, 3S)-3-Iodo-2-hexanol.

Solution * (a) In lactic acid (CH3CHOHCOOH), the priority order of ligands is: OH (1) COOH (2) CH3 (3) H (4). Now, the Fischer projection of (S)-lactic acid can be drawn by putting the lowest priority ligand (4) on any of the bonds on the vertical line and other ligands on the remaining bonds such that the arrow from 1 Æ 2 Æ 3 traces a counterclockwise path.

2 3

COOH 1

H C

OH

4 or,

2

H 1

4

H

3

(S)-Configuration

(S)-Lactic acid

(S)-Configuration

(b)

HOOC

OH CH

(S)-Lactic acid

*

In 2-butanol (CH3CHOHCH2CH3), the priority order of ligands is: OH (1) CH2CH3 (2)

CH3 (3)

H (4).

Now, the Fischer projection of (R)-2-butanol can be drawn by putting the lowest priority ligand (4) on any of the vertical bonds and other ligands on the remaining bonds such that the arrow from 1 Æ 2 Æ 3 traces a clockwise path.

Principles of Stereochemistry

2.157

2

3 3

2

3

3

R

(c)

R

R *

R

*

In 3-iodo-2-hexanol (CH3 C HOH C HI CH2CH2CH3), the priority order of ligands at C-2 is: OH (1) CHICH2CH2CH3 (2) CH3 (3) H (4) and at C-3 is: I (1) CH(OH)CH3 (2) CH2CH2CH3 (3) H (4). For each chirality centre, the chiral ligand (2) is automatically placed on the vertical bond (lower for C-2 and upper for C-3). The Fischer projection of (2S, 3R)3-iodo-2-hexanol can be drawn by putting the lowest priority ligands (4) on the vertical bonds and two other ligands (1 and 3) on the horizontal bonds such that the arrow from 1 Æ 2 Æ 3 traces a counterclockwise path around C-2 (i.e., 1 is on the left bond and 3 is on the right bond) and a clockwise path around C-3 (i.e., 1 is on the left bond and 3 is on the right bond). 4 H CH3 3

1 HO 2 1 I

CH2CH2CH3 3 H 4

(2 , 3 )-3-Iodo-2-hexanol

6.

Designate R or S description for each chiral centre of the following compounds using the so-called ‘very good’ procedure: Cl I

CH3

(a)

Cl EtO H OH

H

OH

I

H

OH

CH

H

(b) Br

F

H

Cl

C2H5

Br

CO2H

CH3

Br

F

(c)

OH CH3

2.158

Organic Chemistry—A Modern Approach

Solution

(a)

C-2: Since the lowest priority group (4) is on a vertical bond, clockwise motion of the arrow from 1 Æ 2 Æ 3 indicates R configuration.

4 CH3 1 Cl

CO2H 2

2

3 2 1 EtO

H 4

3

3 2 4H

4

C2H5 3

OH 1 (2R, 3S, 4S)

(b)

3 Cl 1

1 I

Br 2 4 3

2

1 Br

F 2 4 3

2 Cl

1 Br

I 1

3

4

4 3 CH3 4

F 2 (1S, 2S, 3R, 4R)

C-3: Since the lowest priority group (4) is on a horizontal bond, clockwise motion of the arrow from 1 Æ 2 Æ 3 indicates S configuration. C-4: Since the lowest priority group (4) is on a horizontal bond, clockwise motion of the arrow from 1 Æ 2 Æ 3 indicates S configuration. Therefore, it is the (2R, 3S, 4S)-enantiomer.

C-1: The lowest priority group (4) is on a vertical bond. Therefore, counterclockwise motion of the arrow from 1 Æ 2 Æ 3 indicates S configuration. C-2: The lowest priority group (4) is on a vertical bond. Therefore, counterclockwise motion of the arrow from 1 Æ 2 Æ 3 indicates S configuration. C-3: The lowest priority group (4) is on a vertical bond. Therefore, clockwise motion of the arrow from 1 Æ 2 Æ 3 indicates R configuration. C-4: The lowest priority group (4) is on a horizontal bond. Therefore, counterclockwise motion of the arrow from 1 Æ 2 Æ 3 indicates R configuration. Therefore, it is the (1S, 2S, 3R, 4R)-enantiomer.

Principles of Stereochemistry

(c)

2.159

C-2: The lowest priority ligand (4) is on a horizontal bond. Therefore, a clockwise motion of the arrow from 1 Æ 2 Æ 3 indicates S configuration.

3 CH3 4 H

* 2

H

3

OH 1

2

C-4: The lowest priority ligand (4) is on a horizontal bond. Therefore, a clockwise motion of the arrow from 1 Æ 2 Æ 3 indicates S configuration.

OH

2 *

1 HO

H 4

4

Since both C-2 and C-4 have S configuration, C-3 is achiral. Hence, it is the (2S, 4S) enantiomer of chiral pentane-2,3,4-triol.

CH3 3 (2 , 4 )

7.

Specify the configuration (E or Z) of each of the following compounds containing various types of double bonds: H3C

(a)

H C

DH2C

C CH3

H

(c)

H 3C

C

CH CH2

C

(g)

H

H C

(i) O D3 C

(k)

H

(f)

Ph

CH2COOH

C COOH Cl H

C

(h)

H

H C

C H

C

CH2I

H

C5H5

C

C

CH2F

C

CH3

H C

ClCH2

H

CH3 C

C

Ph

N

CH3 H5C2

BrCH2

(d)

CH(CH3)2 N

(e)

(b)

C

C COOH

C H OH

C

C2H5

N

18

CH3

2

(j)

CHO

16 2

COOH C

C

C

C

C

C CH3

Solution

(a)

3 2

CH2D 3

Higher priority ligands on the same side: Z configuration

CH3 ; CH3

H

2.160

Organic Chemistry—A Modern Approach

(b)

2

2

2

2

CH2Br

CH2Cl; CH2I

CH2F

Higher priority ligands on the same side: Z configuration H

(c)

higher priority

CH CH C

H C

C CH(CH )

CH3

H; CH = CH2

CH(CH3)2

higher priority

Higher priority ligands on different sides: E configuration. higher priority

Ph

(d)

H C

H

C COOH

higher priority

Ph

H; COOH

H

Higher priority ligands on different sides: E configuration (e)

N

N

3

CH3

(..) unshared pair of electrons

CH3 higher priority

higher priority

Higher priority ligands on the same side: Z configuration C-2==C-3 : COOH H; CH==CH — > H C-4==C-5 : CH==CHCOOH H; CCl==C(CH3)Ph H C-6==C-7 : Cl CH==CH — ; Ph CH3

higher priority

(f) 3

7

6

3 5

higher priority

2

4

higher priority

Higher priority ligands on the same side of C-2==C-3 bond: Z configuration Higher priority ligands on the same side of C-4==C-5 bond: Z configuration Higher priority ligands on different sides of C-6==C-7 bond: E configuration

Principles of Stereochemistry

2.161

higher priority C-3==C-4 : CH2COOH H; CH = CHC2H5 H C-5==C-6 : CH = CHCH2COOH 2 C2H5 H

(g) 5 2

6

5 4

higher priority

3

H;

higher priority

(h)

Higher priority ligands on the same side of C-3 == C-4 bond: Z configuration Higher priority ligands on different sides of C-5 == C-6 bond: E configuration higher C6H5 OH C6H5 C2H5; OH (..) unshared C N priority higher H5C2 pair of electrons priority Higher priority ligands on the same side : Z configuration. higher priority

(i) O

C

O

CH3

CHO

CH3 ; CH2C—

CH2CH2—

CHO higher priority

Higher priority ligands on different sides : E configuration. (j)

18 2 16

∫

2

18 2

16 2

Higher priority ligands on the same side: Z configuration (k)

higher priority

higher priority D3 C H

COOH > CH3 ; CD3 > H

COOH C

C

C

C

C

C CH3

Higher priority ligands on the same side : Z configuration (This compound containing five cumulated double bond is planar and gives rise to cis-trans isomerism.) 8.

Write down the structural formula for each of the following compounds: (a) (Z)-1-Bromo-2-methylbut-l-ene (b) (E)-2-Chloro-1-nitropent-1-ene (c) (Z)-2-Butene (d) (E)-But-2-enoic acid

2.162

Organic Chemistry—A Modern Approach

Solution

(a)

CH3 | 1 2 CH3 CH1 C == CHBr (1-Bromo-2-methylbut-1-ene) 4

4

Since CH2CH3 CH3 and Br H, the structure of (Z)-1-bromo-2-methylbut-1-ene is that in which the higher-ranked groups (CH2CH3 and Br) are on the same side of the double bond. Hence, the configuration is: CH3CH2

Br C

H3C

C H

(Z)-1-Bromo-2-methylbut-1-ene

Cl | 1 CH3 CH2 CH22C = CHNO2 (2-Chloro-1-nitropent-1-ene) 5

(b)

4

3

Since Cl CH2CH2CH3 and NO2 H, the structure of (E)-diastereoisomer of this compound is that in which the higher-priority groups (Cl and NO2) are on the opposite sides of the double bond. Hence, the configuration is: Cl

H C

C NO2

CH3CH2CH2

(E)-2-Chloro-1-nitropent-1-ene 4

(c)

3

2

1

CH3 CH=CHCH3 (2-Butene) Since CH3 H, the structure of (Z)-2-butene is that in which the higher priority groups (-CH3) are on the same side of the double bond. Hence the configuration is: H

H C

C CH3

H3 C

( )-2-Butene 4

(d)

3

2

1

CH3 CH=CHCOOH (but-2-enoic acid) Since CH3 H and COOH H, the structure of (E)-diastereoisomer of this compound is that in which the higher-ranked groups (CH3 and COOH) are on the opposite sides of the double bond. Hence, the configuration is: H3C H

H C

C COOH

( )-But-2-enoic acid

Principles of Stereochemistry

9.

2.163

Designate each of the following structures with erythro or threo prefix:

(a) H H

OH

COOH Cl Br

H3C

(b) H

H

F

COOH CH3 CHO H

Cl

(c) NO2

10.

C CH3

C2H5

CHO Solution

Br C

(d) Br H

H

H

(a) erythro-isomer (b) threo-isomer (c) erythro-isomer (d) threo-isomer

What will be the change in conductivity of boric acid if (R, R)-2-, 3-butanediol, (R, S)-2, 3-butanediol and (S, S)-2, 3-butanediol are separately treated with boric acid?

Solution The borate complex of a chiral isomer [(R, R) or (S, S)] of 2, 3-butanediol is more stable than that of the achiral (meso) (R, S)-isomer because in the former complex the bulky methyl groups are anti to each other (sterically more favoured), whereas in the latter complex the methyl groups are gauch to each other (sterically less favoured). Therefore, the conversion of the chiral diol in to the borate complex is relatively more favourable than for the achiral diol. The conductivity of boric acid increases with increase in concentration of the complex. Thus, chiral 2, 3-butanediol [(R, R), or (S, S)-enantiomers] enhances the conductivity of boric acid more than does the achiral isomer.

anti-conformer

CH3 H

OH

H

OH

2

CH3

CH3 H

O

3

O

H

O

B H≈

O

+ H3BO3 m CH3

(S, S)-2,3-Butanediol (chiral)

CH3

OH + 3H2O OH CH3

Borate complex (more stable)

2.164

Organic Chemistry—A Modern Approach

anti-conformer

CH3 H

CH3

OH

2

H

O

3

O

O

B H≈

O

+ H3BO3 m H

H

OH CH3

CH3

gauche-conformer

H

OH

2 H3 C

OH

(R, S)-2,3-Butanediol (achiral)

11.

+ 3H2O H CH3

H

H

H

O

3

O

H3C

O

B H≈

O

+ H3BO3 m CH3

H

Borate complex (more stable)

(R, R)-2,3-Butanediol (achiral) H

CH3

steric strain

H + 3H2O

CH3

CH3 CH3

Borate complex (less stable)

steric strain

Isomers of CH3CH2CH==CHCH2CH3 differ widely in chemical properties but those of CH3CH2CH==C==CHCH2CH3 do not – Why?

Solution Isomers of 3-hexene (CH3CH2CH==CHCH2CH3) are cis-trans isomers, i.e., they are diastereoisomers and because of this they differ widely in chemical properties. On the other hand, isomers of 3, 4-heptadiene (CH3CH2CH==C==CHCH2CH3) are enantiomers and because of this, they do not differ in chemical properties (except reaction with a chiral reagent involving diastereoisomeric transition states having unequal energies).

cis

trans

Principles of Stereochemistry

12.

2.165

Assign R/S descriptions to the asterisked chiral centre (*) in each of the following compounds: NO2 CH2OCH2CH3 CH2CH2OCH2CH3

OCH3 *

(a)

(b)

H C* Br H3CO

NO2

O

C2H5

(c)

H H H Cl H

Cl Cl Cl H Cl

*

C2H5 Solution

OCH3

OCH3 NO2

(a)

Since —NO2 is close to OCH3 in

, it gets priority over NO2

. Therefore, the priority order of the ligands is: NO2 Br

H NO2 OCH3

OCH3

NO OCH H Br

C

CH O

(b)

2 NO

1

3

The arrow from 1 Æ 2 Æ 3 traces a clockwise path suggesting the R configuration.

For the determination of absolute configuration, the other ring is to be written in the disconnection form (as given below), where the chiral carbon itself serves as a duplicated atom at the end of the expanded chains. The duplicated atoms are written within parentheses. Phantom atoms (designated by subscript 0) are used to bring the valency of duplicated atoms up to four.

2.166

Organic Chemistry—A Modern Approach

Æ

Æ S

(c)

In this compound, the chiral centres C-2, C-3, C-5 and C-6 have absolute configurations, S, S, S and R, respectively. According to the sequence rules, (S, S) gets priority over (S, R). The absolute configuration of the asterisked chiral centre is determined as follows: C2H5 H H H Cl H

S

Cl Cl Cl H Cl

S * S R

2 (S, S)

13.

Cl 1

4 H (S, R) 3

C2H5

2 Allowed interchange

3

1 4

The arrow from 1 Æ 2 Æ 3 traces a clockwise path, suggesting the R configuration.

Identify all the asymmetric carbon atom(s) in each of the following structures: (a) CH CH CH CH

(b) H3C—CH—COOH

CH CH

NH2

CH3

3

(c)

(d)

(e) CH3 CH2CH CHCH3

(f)

Cl Br CH3

(g) H 3C

O

CH3 CH3

(h)

H 3C O

Principles of Stereochemistry

2.167

Solution (a)

* CH3CH2CH2CH

(b)

CHCH3 * H3C—CH—COOH

(c) (d)

NH2 No asymmetric carbon CH3 OH *

* *

(e)

(h) 14.

*

CH3CH3 CHClCHBrCH3

(f) (g)

(* denotes an asymmetric carbon atom)

3

No asymmetric carbon * CH3 H 3C * O State whether each of the following molecules is achiral or chiral Br H H

(a)

≈

H C I

(c) H3C N CH2CH3 Cl

(b) H C H

F

–

CH2CH2CH3

Cl H 14

(d)

2

12

Cl

(e) O

2

H

(g)

(f) 3

H

3

Cl

(h)

Cl Solution (a) chiral (b) achiral (c) chiral (d) chiral (e) achiral (f) achiral (g) achiral (h) achiral

15.

Locate the indicated centres as stereogenic/nonstereogenic or chirotopic/ achirotopic: H

2

(a) 2

(b) Cl Cl

CH3 CH3 H

2.168

Organic Chemistry—A Modern Approach

Br

(c)

C

He

CH3

Cl

(d)

H

C

Et

(e) H Cl Cl

Br

Cl I

Cl H H CH3

(f)

H H H

CH3

CH3

CO2Et OH Cl OH

(g) H

Cl H

H

(h) H3C

H

C C == C

H

CH3

CO2Et C

(i)

H

CH3 CH2CH3

(j) H3C Et

H C == C

C

(k) H

H

Cl

C

H CH3

H

Solution

H

C==CH C==C

(a)

H C=CH

H

H

(b) Cl Cl

achirotopic and stereogenic Br

(c)

HC Et

chirotopic and stereogenic

C

(e)

H

H

Cl

(d)

Cl

H

Cl

H CH3

C Cl

chirotopic and stereogenic chirotopic and nonstereogenic chirotopic and stereogenic

achirotopic and nonstereogenic

Cl

H

CH3

CH3 CH3

achirotopic and stereogenic

I Br

CO2Et

(f)

H

OH

H

Cl

H

OH CO2Et

chirotopic and stereogenic chirotopic and nonstereogenic chirotopic and stereogenic

Principles of Stereochemistry

2.169

(g)

H

H

C

Cl H

CH3

chirotopic and stereogenic

CH3

H

chirotopic and stereogenic

C == C

H3C

(h)

H

CH3 chirotopic and stereogenic 3

(i)

2

achirotopic and stereogenic

chirotopic and stereogenic

(j) 3

H H3C Et

H C == C

C

H

H

chirotopic and nonstereogenic

(k) 3

16.

Label the indicated (Æ) centres of the following compounds as stereogenic/ nonstereogenic or chirotopic/achirotopic. Give your reasoning. Br

(a)

(b) Cl Cl

F H

Br

H

H

H

H

F Cl F

H

(c)

Solution (a) The molecule is chiral, therefore, the indicated centre is chirotopic. However, this centre is not stereogenic because interchange of positions of F and H does not produce a new stereoisomer.

2.170

Organic Chemistry—A Modern Approach

Br

H

Br interchange of the positions of F and H

F H

Br

H H

H

Br F

identical molecules because one on upside down produces the other

chirotopic and nonstereogenic

(b) (c)

H

Due to the same reason, the indicated centre is chirotopic and nonstereogenic. The indicated centre is achirotopic because a s-plane passes through it. It is stereogenic because interchange of positions of F and H produces a new stereoisomer. achirotopic and stereogenic Br

F

Br interchange of positions of F and H

Br H

H

Br H

H

17.

H

two different stereoisomers

F H

Indicate whether each of the following compounds is chiral. Identify the asymmetric carbons and stereocentres or stereogenic centres (if any) in each. Br H ≈ CH—CH2CH2CH3 C N (b) (a) H5C2 H3C C C H5C2 CH3 Br Cl Cl (c) (d) H 3 H

Solution (a) The compound is chiral because it contains one asymmetric carbon or chiral carbon.

Br H 5 C2 H5C2

C

C

*CH—CH2CH2CH3 CH3 Stereocentre

Principles of Stereochemistry

(b)

2.171

The compound is chiral because it contains one asymmetric carbon. ≈ 3

(c)

The compound is achiral because it has a s-plane. It contains no asymmetric carbon. However, it has four stereocentres. stereocentres

CD3

(d)

This molecule is chiral due to presence of a chiral axis. It contains no asymmetric carbon. However, it has three stereogenic centres. Cl

Cl H

H stereocentres

18.

Identify Ha and Hb in each of the following structures as homotopic, enantiotopic or diastereotopic ligands: COOH

Ha

(b) Ha Hb

Br

(a) H Br H

Hb

(c)

Hb C

OH C H

Br H

Cl

Cl Ha

Ha

COOH

H

(d)

OH

Cl

(e) Hb

Ha Cl

(f) Ha

Hb

F

Hb

H

2.172

Organic Chemistry—A Modern Approach

(g)

(h) Ha H

Cl

C Hb

CH3

H Ha HO

Hb

C

(i)

Br

H C

C CH3

Ha

CH3

COOH

(j)

Hb

Br

Ha

Ha

COOH

OH Hb H

(k) Ha H

(l) HO2C

H Hb

CHOHCH3 Hb

COOH COOH Solution (a) Homotopic, (b) Enantiotopic, (c) Diastereotopic, (d) Enantiotopic, (e) Enantiotopic, (f) Diastereotopic, (g) Enantiotopic, (h) Diastereotopic, (i) Enantiotopic, (j) Homotopic, (k) Homotopic, (l) Diastereotopic.

19.

Complete the equations and label each product with D or L. Comment. COOH III

NaOH low [OH–]

H

Br

NaN3

H2 Ni

I

II

CH3 Solution

COO3 H

OH CH3 III (D–)

COOH NaOH low [OH–]

Br

H

CH3 D–

COOH NaN3

H

N3

CH3 I (L–)

COOH H2 Ni

H 2N

H CH3 II (L–)

D-2-Bromopropanoic acid reacts with NaN3 (an SN2 reaction) with inversion of configuration. For this reason, the configuration of I becomes L. When I is reduced to II, no bond to the chiral centre is broken and retention of configuration occurs. As a result, the configuration of II becomes L. When D-2 bromopropanoic acid is treated with NaOH solution having low [OH–], a neighbouring group participation occurs and the hydrolysis proceeds with retention of configuration (a result of double inversion). Therefore, the configuration of III is also D.

Principles of Stereochemistry

2.173

Mechanism: O

HO COOH –

N3 + H

Br

COOH SN2

H

N3

CH3 D

O

; H

–

Br

H

O OH

O

–

C

SN2

H

O

CH3

–

CH3 D

C S N2 1st inversion

OH

Br

O

C

20.

+ Br

CH3 L

O

–

C

–

OH

H

2nd inversion

CH3

CH3 D

Find out the configuration of the final product when a hydride ion from a chiral source is allowed to attack on the re-face of acetophenone.

Solution Since 1 Æ 2 Æ 3 traces a clockwise path, the re-face of acetophenone is as given below. Now attack of H@ on the re-face leads to the formation of an alkoxide having S configuration.

re-face

4 H

H

3 H C –

C==O 1 2 Ph

21.

H C

C

O

–

∫

Ph 2

3 H C

Ph S configuration

O 1

–

Draw all the possible stereoisomers of EtCH == CH—CHBr—CH == CHEt. Mention their configurations in terms of R/S and E/Z and predict whether they are chiral or achiral.

Solution Four stereoisomers are possible. Two of them are achiral and two of them are chiral (exists as enantiomers). These are as follows:

H H Et C C C C H E C E H

Et

Br H (E, E)-isomer (achiral)

diastereoisomers

H H H C C C C Et Z C Z Et H

Br H (Z, Z)-isomer (achiral)

2.174

Organic Chemistry—A Modern Approach

H H Et C C C C Et Z R C E H

H H H C C C C H E SC Z Et

H

H

Br

Et

H

Br

enantiomers

(E, R, Z)-isomer (E, S, Z)-isomer (chiral) (chiral) Explain why camphor exists only as a pair of enantiomers, even though it has two asymmetric carbons. Designate the asymmetric carbons as R or S.

22.

H3 C

CH3 H

CH3 O Camphor Solution Since camphor contains two dissimilar asymmetric carbon atoms, it might be expected to have 22 or 4 stereoisomers. However, it exists only as a pair of enantiomers

because the diastereoisomers having a trans-oriented (CH3)2C not possible for this rigid structure. H3 C 5

R

7

S

1

6

Mirror plane

CH3 H 4

7

2

O

O

CH3 RH 4

3

3 2

CH3 I

H3 C

bridge is structurally

5

1 S

6

CH3 II

Enantiomers of camphor

Configuration of C-1 of I: H3C

CH3 2

3

H

2

1

1

3

CH3 4

2

4

1

1

C 3

O

2

2

4

120° rotation

4 1

3

1

3

The arrow from 1 Æ 2 Æ 3 traces a clockwise path, suggesting the R configuration.

Principles of Stereochemistry

2.175

Configuration of C-4 of I: 2

2 4

3

1

3

1

The arrow from 1 Æ 2 Æ 3 traces a counterclockwise path, suggesting the configuration.

The absolute configurations of C-1 and C-4 of II will be S and R, respectively. 23.

Tri-sec-butylmethane has four optically active stereoisomers. Write down the isomers in terms of R/S designation.

Solution Each sec-butyl group [CH3CH2(CH3)CH–] has a chiral C that can be R or S. Since all three groups are equivalent, the order of groups is immaterial. RSR is identical with SRR or RRS. The four isomers are

1. 2. 3.

Give the R/S designation of each of the stereoisomers of 1, 3-dibromo-2methylbutane. Write down all the possible stereoisomers of MeCH==CH–CH==N–OH and specify their configuration as E/Z. Give an example of a molecule with enantiomeric groups along with pro-R and pro-S hydrogen atoms on a propseudoasymmetric centre. [Hint: pro-S

4.

COOH R H OH H H H S OH

pro-R

] COOH Predict configurations of the products when (a) Pro-R-hydrogen of propanal is replaced by Br. (b) CN@ is allowed to attack on the Re-face of propanal.

2.176

5.

Organic Chemistry—A Modern Approach

Designate the marked (*) centres of the following compounds as stereogenic/ nonstereogenic and chirotopic/achirotopic. Give reasons. Br H Me Me CH3 H3C * * * * C C (a) (b) Et Et H Br Cl COOH H * * C C C (c) H (d) Cl * H H Cl CH Cl H COOH

6.

Assign the following compounds with R/S descriptor. Ph H

H

(a)

D3 C

C

CH2CH3

C

(b)

C

C

Me

(c) H

OH

H CH2OH

CH(OCH3)2

N∫C C

(d)

H

C

C CHO

H3C—C

Ph H 3C C

(e)

O COOH

CN

(f) H3C

NH2

C∫CH

Ph

Ph OHC

O

CH2OCMe3 C

(g) D3CH2

P

(h) CD2CH3

Me

CD2CH3

(i)

H3C

C

OH

CH2CH2CH3

Cl

C H

C

C∫CCH3 Ph OHC

Me

H

CH2Ph Ph CH3

Cl CH3

(j) Cl CH3 H

Cl

Principles of Stereochemistry

2.177

CH2OH

H Br

(k)

H2 N Cl H

(m)

C

CH2SH CH3

(n)

F

I

CD2OH

C

(l)

Br Br

H C

(o) Cl

7.

List the substituents in each of the following sets in order of decreasing priority: (a) –Br, –OH, –SH, –H (b) –CH2Br, –CH2Cl, –CH2OH, –CH3 (c) –CH=CH2, –CH2CH3, –Ph, –CH3 (d) –CH2CH2Cl, –CH(CH3)2, –Cl, –CH2CH2CH2Br (e) –CH(CH3)2, –C(CH3)3, –H, –CH==CH2 (f) –OH, –OPO3H2, –CHO, –H (g) –OCH3, –CH3, –NEt2, –H (h) –D, –H, –OH, –CHO (i) –NH2, –F, –CH3, –OCH3

8.

Shown below are Newman projection formulas for (R, R)-, (S, S)- and (R, S)-4, 5-dibromooctane. (a) Which is which? (b) Which formula represents an achiral meso compound? H H

CH2CH2CH3 Br Br

CH2CH2CH3 Br

H

H

Br

CH2CH2CH3 I

9.

Br Br

CH2CH2CH3 II

CH2CH2CH3 H H CH2CH2CH3 III

For the following molecule, draw its enantiomer as well as one of its diastereoisomers. Assign the R or S configuration at each chirality centre. H Cl F H5C2 H

H

2.178

Organic Chemistry—A Modern Approach

10.

Draw one chiral (active) and one achiral (meso) isomer of EtO2C(CHOH)3CO2Et in Fischer projection formula. Will the interchange of H and OH at C-3 of the chiral stereoisomer, you have drawn, lead to another stereoisomer? What will happen if H and OH are interchanged at C-3 of the achiral (meso) isomer you have drawn? Explain stating whether C-3 is a stereogenic centre in each case.

11.

There are four diethylcyclopropane isomers. Write 3D-formulas for these isomers. Which of the isomers are chiral?

12.

PhCH2CH N H3COO@ has two diastereotopic ligands. Label them as pro-R and pro-S.

13.

Identify the enantiotopic, diastereotopic and homotopic hydrogens (HA, HB) in each of the following compounds: CO2H

≈

(a)

A

HB

(b) HA HB

B

(c)

OH

3

CH3 HB

2

C == C == C

(d)

COOH

CH3

HA

(e) HB H

H

HA

(f)

CH3

H3C

2

A

COOH

CH3

B

HA Me

3

`

(g)

ClHB

(h)

A

A

(i)

O

Cl Me

B B 3

H

(j)

H H C

A

(k) 2

H

14.

B 2

Locate the stereogenic centre(s) in each of the following compounds. A molecule may have no stereogenic centre. (a)

CH3CH2CH2CHClCH2CH2CH3

(b)

CH3CH2CH2OCH(CH3)CH2CH3

(c)

CH3CH2CHDCH2CH2CH3

(d)

(CH3)2CHCH2CH(CH3)CH2CHClCH2CHBrCH3

Principles of Stereochemistry

2.179

(e)

(f) O 3

Br CH2

(g)

(h)

Et 3

15.

Identify the missing substituents (A and B) in each of the following compounds: A A C

(a)

C2H5

(b)

CH3 B (S)-2-Bromobutane

HO

B

COOH (R)-2-Hydroxypropanoic acid

16.

When (S)-2-methylbutan-1-ol is allowed to react with concentrated HBr, (S)-1bromo-2-methylbutane is obtained. Explain with reasons whether the reaction proceeds with inversion or retention of configuration.

17.

Redraw each of the following molecules as a Fischer projection and then assign R or S designations to each stereocentre: COOH 3

OHC

C2H5

HO

C2H5

(b)

(a) 3

H2N

(c)

18.

C

D3 C

OH H C

OH H

H3 C

(d) H

Cl C

C

COOH CH3 H F Indicate whether the hydrogens marked Ha and Hb in each of the following compounds are homotopic, enantiotopic or diastereotopic: Cl Ha

(b)

(a) Hb

Ha Ph Hb Ph

2.180

Organic Chemistry—A Modern Approach

Ha

Hb

H

H

C

(c)

C

CH3

C

Br

Ha C H

Hb

Ha Hb

C

H5C2

(f)

C2H5

C

C

D3 C

C H H

Cl

3 3

(g)

a

Ha

C C

(j) Ha–CH2 O

Ha

Hb

C2H5

S

(l) Ha

CMe3

H

OH CH3 O– ≈ S—C Hb

O

CMe3 Hb

H CMe3

H

(m)

Hb

H

Hb–CH2

H

Br

Hb

(i)

(k)

H CD3

C C

(h) H5C2

b

Ha

F

H

H C

CH3

H

H

Hb

C

H 3C

H

H

(e)

C

(d)

C

Ha

H

H

b a

(n)

Ha Me

19.

Write down the flying wedge projections of each of the following compounds having noted configuration and ligands bonded to the chirality centre: 14

(a) (b)

—Ph, — C ∫ CH, —H, —C ∫ CH (with R configuration) —I, —CH2NH2, —CH2CH3, —CH2Cl (with S configuration)

(c)

—OD, —18OD, —OMe, —OH (with R configuration)

(d) 20.

Me, —OH, —D, —CN (with S configuration)

Predict the number of diastereoisomers for the compound with molecular formula C2BrClFI. Write down their structures and label each of them as (E) or (Z).

Principles of Stereochemistry

Br F 21.

22.

23.

24.

2.181

[Hint: Six diastereoisomers are possible; these are: Cl Br Br , I Br , I Cl Cl , F Cl , F and Cl F Cl F I I I F Br I Br Give examples of the following: (a) A chirotopic as well as stereogenic centre (b) A chirotopic but nonstereogenic centre (c) A stereogenic but achirotopic centre A triol [CH3CH2(CHOH)3CH2CH3] has three dissimilar asymmetric carbon atoms. List the isomers in terms of R/S designation. R S R S R S R S È ˘ ÍHint: R S ; R S ; S R ; S R ; four pairs of enantiomers ˙ Í ˙ ÍÎ ˙˚ R S S R R S S R Draw all possible stereoisomers of EtCH(OH)CH==CHEt and designate them by R/S and E/Z notations. [Hint: (3R, 4E), (3S, 4E), (3R, 4Z), (3S, 4Z).] How many stereoisomers of the following compound would you expect (mathematically speaking) to have? Only two of them are actually isolated – Why? Predict the number of stereogenic centre in the compound. 3

25.

Which of the following structures represents the naturally occurring amino acid (2S, 3R)-threonine? H

H3C

C

HO H

C

H

H

≈

≈

NH3 COO

H3N

–

C

CH3

OOC

I

OH

C

–

II

H

CH3

≈

H3N

C

C

H

–

OH

OOC

III

≈

NH3

HO H

C

C

H3C

COO – H

IV

2.182

2.6

Organic Chemistry—A Modern Approach

OPTICAL ACTIVITY OF CHIRAL COMPOUNDS

Enantiomers share many of the same properties, including the same melting points, the same boiling points, the same indices of refraction, same solubilities in common solvents, same infrared spectra, same rate of reaction with achiral reagents, etc. In fact, all the physical and chemical properties of enantiomers are the same except those that arise due to different spatial arrangement of groups bonded to the asymmetric or chiral centre. One property that enantiomers do not share is the way they interact with the plane-polarized light resulting in its rotation in opposite directions.

2.6.1 Plane-polarized Light Light is an electromagnetic phenomenon. A beam of normal or ordinary light (composed of rays of different wavelengths) consists of an electric field and a magnetic field which are oscillating at right angles to each other in every plane perpendicular to the direction of propagation. The same is true for monochromatic light, i.e., light of single wavelength. These oscillations of the electric field can be made to occur in a single plane when a beam of monochromatic light is allowed to pass through a polarizer (a Nicol prism) that interacts with the electric field. Such light is called plane-polarized light.

2.6.2

Optical activity

In 1815, the physicist Jean-Baptiste Biot discovered that some naturally occurring organic compounds rotate the plane of polarization of plane-polarized light. Some of them rotate it clockwise and some rotate it counterclockwise. It has been proposed that the rotation of the plane of plane-polarized light is due to some asymmetry in the molecule. When the plane-polarized light is allowed to pass through an achiral compound in pure liquid form or through its solution in a suitable solvent, the light emerges from the solution with its plane of polarization unchanged, i.e., there occurs no rotation of the plane of polarization. However, when the plane-polarized light is allowed to pass through a solution of a chiral compound or through the pure liquid chiral compound, the light emerges with its plane of polarization rotated either clockwise or counterclockwise. If one enantiomer of the compound rotates it clockwise, its mirror image will rotate it in exactly the same amount counterclockwise. A compound that rotates the plane of polarization of plane-polarized light is said to be optically active. It thus follows that chiral compounds are optically active, while achiral compounds are optically inactive. If an optically active compound rotates the plane of polarization clockwise, then the compound is called dextrorotatory (Latin dexter, meaning right), which can be indicated in the name of the compound by the prefix (+). If it rotates the plane of polarization counterclockwise, then it is called levorotatory (latin laevus, meaning “left”), which can be indicated by (–). Sometimes d and l are used instead of (+) and (–).

Principles of Stereochemistry

2.183

The (+) and (–) signs are not to be confused with R and S. In fact, there is no correlation between R and S configuration of enantiomers and the direction [(+) or (–)] in which they rotate the plane of plane-polarized light. The (+) and (–) symbols indicate the direction in which an optically active compound rotates the plane of plane-polarized light, whereas R and S indicate the spatial arrangement of the groups about a chiral centre. Compounds with the R configuration may be (+) or (–). Similarly, compounds with S configuration may be (+) or (–). For example, (R)-(+)-2-methyl-1-butanol and (R)-(–)-1-chloro-2-methylbutane have the same configuration, that is, they have the same general arrangement of their groups about the chiral centre. However, they have an opposite effect on the direction of rotation of the plane of plane-polarized light. The first one is dextrorotatory (+) and the second one is levorotatory (–). 2

2

R

5

2

3

2

5

3

R

The configuration (R or S) of a compound can be ascertained by looking at its structure, but the only way to know whether a compound is dextrorotatory (+) or levorotatory (–) is to put the compound in a polarimeter, an instrument that measures the direction as well as the amount the plane of polarization of plane-polarized light is rotated.

2.6.3

Polarimeter

A polarimeter is an instrument by which the rotation of the plane of a plane-polarized light by an optically active compound is detected and measured. The main working parts of a polarimeter are: (i) a light source, (ii) a polarizer, (iii) a cell or tube for holding the optically active compound or its solution in the light beam, (iv) an analyser, and (v) a scale for measuring the angle (in degrees) that the plane of plane-polarized light has been rotated. A polarimeter uses a monochromatic (single-wavelength) light source (usually a light from a sodium arc which is called the sodium D-line; wavelength = 589 nm) which produces unpolarized light. This light passes through the polarizer to generate the plane-polarized light. The plane-polarized light then passes through the cell or tube and then through the analyzer and finally reaches the eye of an observer. Before taken to the compounds in the cell, the polarizer and the analyser are so adjusted that the light passes through them without loss of intensity, i.e., the maximum amount of light reaches our eye. The compound in pure liquid form or as a solution in a suitable solvent like water, ethanol, etc., is then poured into the cell. If the intensity of light is found to be reduced (due to rotation of the plane of the polarized light), the compound is said to be optically active. In that case, the analyser is rotated in either clockwise or counterclockwise direction to bring back the original intensity of light. The angle in degrees that the analyser needs to be rotated is called the observed rotation (a) of the optically active compound. The clockwise rotation of

2.184

Organic Chemistry—A Modern Approach

the plane of plane-polarized light is taken as positive (+), while counterclockwise rotation of the plane is taken as negative (–). Two enantiomers rotate the plane of plane-polarized light by exactly the same amount, i.e., to an equal degree but in opposite directions. The observed rotation depends on the number of optically active molecules that the light encounters in the sample, which in turn depends on the concentration of the sample and the length of the polarimeter tube. The observed rotation also depends on the temperature and the wavelength of the light used. A schematic drawing of a polarimeter is as follows: Direction of light propagation

Light source (Sodium-D-line)

Ordinary light

Polarizer

Direction of light propagation

Planepolarized light a

Observed rotation

Viewer Polarimeter tube containing a chiral compound

2.6.4

Planepolarized light (after clockwise rotation)

Analyser

Specific Rotation

The observed rotation (a) is found to be proportional to the concentration of the optically active compound and the length of the cell through which the plane-polarized light passes. a l◊c [a] = the proportionality constant and is called the specific rotation a = the observed rotation c = the concentration of the solution in grams per millilitre of solution (or density in gm/L for pure liquids) l = the length of the cell in decimetre (1 dm = 10 cm). a = [a ] ◊ l ◊ c

where,

or [a ] =

Principles of Stereochemistry

2.185

Thus, the specific rotation, [a], of a compound may be defined as the rotation in degrees brought about by a pure liquid or solution containing 1 g of optically active substance per mL of solution, placed in a 1 decimetre polarimeter tube. Since the specific rotation is independent of c and l, it is used as a standard measure of optical activity. The specific rotation also depends on the temperature and the wavelength of light that is employed and for this, [a] is conventionally reported with a subscript that indicates the wavelength of light, used and a superscript that indicates the temperature in °C. Therefore, a specific rotation reported as [a]D25 indicates that it has been determined at 25°C and the wavelength of light used is a yellow emission line in the spectrum of Na, called the sodium D line. The specific rotation of (+)-2-methyl-1-butanol, for example, might be given as follows: [a ]25 D = + 5.756∞ Although the unit of specific rotation is degree cm2g–1, it is usually expressed in degree only. When the specific rotation is measured in solvents other than water, the name of the solvent must be mentioned. Under a given set of conditions the specific rotation of any compound is constant and the value may be used as a criterion for identification of optically active compounds. Specific rotation is a physical property of a compound just as mp, bp, density, etc. M 100 where M indicates the molecular mass of the chiral compound. Thus, optical rotation of 1 mole of compound in a sample tube of 1 dm length is called its molecular rotation.

t t t , such that [M]D = [a ]D ¥ Optical rotation is also expressed by Molecular rotation, [M]D

A single experiment of measurement of rotation of any chiral compound cannot be used to conclude that the rotation is either dextro (+) or levo (–). For example, if the observed rotation of a compound is reported to be +50° one can conclude that it could be –130° (concentration is 1 M). One can find out the correct direction of rotation by changing the concentration of the solution. If the concentration is reduced by ten times, then the decrease in dextrorotation should be ten times of +50° and consequently the new value would be +5°. Similarly, the change in levorotation should be ten times less than that of –130° and the changed value would be –13°. Therefore, if the second reading is +5°, the rotation must be dextro (+) and if the second reading is –13°, the rotation must be levo (–). Thus, the reading taken at two different concentrations almost always determine the sign of a unequivocally.

Calculation of specific rotation The observed rotation of a chiral compound, for example, (–) glyceraldehyde at 25°C using sodium D line has been found to be –1.74° when its solution containing 2.0 g/10 mL is placed in a 1 dm polarimeter tube. Then its specific rotation: a -1.74 [a ]25 = = - 8.7∞ D = cl (1)(0.2)

2.186

Organic Chemistry—A Modern Approach

Doubling the concentration doubles the observed rotation because the number of molecules the plane of polarized light encounters doubles. Therefore, the observed rotation, a = –1.74° ¥ 2 = –3.48°. Doubling the length of the sample tube also doubles the observed rotation due to the same basic reason. However, the specific rotation, [a ]25 D , is a constant and is independent on concentration and the length of the cell because the value of a/c or a/r remains unchanged.

2.6.5 The Necessary and Sufficient Condition (the ultimate criterion) for Optical Activity The same chirality (nonsuperimposability on the mirror image) that gives rise to enantiomerism also is responsible for optical activity. The relationship between optical activity and chirality is absolute. Many thousands of cases have been found in accord with it. The ultimate criterion, then, for optical activity is chirality. This is both a necessary and a sufficient condition.

Optical activity: A close look All optically active compounds are chiral, but the reverse is not always true. The reasons are as follows: (i) Some chiral molecules are configurationally unstable. Due to rapid pyramidal inversion (umbrella effect) such molecules undergo racemization (formation of an equimolar mixture of two enantiomers) and becomes optically inactive [the (+)-enantiomer would rotate the plane of polarized light clockwise and the (–)-enantiomer would rotate the plane counterclockwise by exactly the same amount]. For example, ethylisopropylamine (EtNHCHMe2) is found to be optically inactive although the nitrogen atom has four different groups around it (a hydrogen, an ethyl group, an isopropyl group and an electron pair) and the geometry of the molecule is essentially tetrahedral, i.e., it is a chiral molecule. enantiomers Et

Et N

N

H Me2HC (Chiral)

H

CHMe2

Transition state (achiral)

Et N H CHMe2 (Chiral)

A racemic mixture of ethylisopropylamine

As the inversion takes place, the large lobe of the electron pair appears to push through the nitrogen to the other side. As this occurs, the three other groups move first into a plane containing the nitrogen, then to the other side.

Principles of Stereochemistry

(ii)

2.187

Because of extremely low differences in polarizabilities among the alkyl groups, some chiral molecules have a vanishingly small rotation and are found to be optically inactive. For example, the specific rotation of the chiral alkane ethyln-propyl-n-butyl-n-hexylmethane is too low and actually far below the limits of deflection by any existing polarimeter (calculated specific rotation (0.00001°) and so, each of its enantiomer is found to be optically inactive. CH2CH3 C CH3(CH2)3

CH3 (CH2)5CH3

(R)-Ethyl-n-propyl-n-butyl-n hexylmethane (chiral but exhibits no optical activity)

(iii)

2.6.6

A 50:50 mixture of enantiomers is optically inactive. If we are to observe optical activity, the material we are dealing with must contain an excess of one enantiomer so that the net optical rotation can be detected by the particular polarimeter at the hand.

Racemic Modification

An equimolar mixture of two enantiomers is called a racemic modification or a racemic mixture or a racemate. A racemic mixture is symbolized by the prefix (±) to the name of the compound. For example, racemic 2-butanol would be symbolized by (±)-2-butanol. A racemic modification is optically inactive because the rotation caused by a molecule of one isomer is exactly cancelled by an equal and opposite rotation caused by a molecule of its enantiomer. CH3

Mirror plane

CH3

C H5C2

C H OH

H

C2H5 OH

(±)-2-Butanol (optically inactive)

2.6.7

Enantiomeric Excess(ee) or Optical Purity (op)

Whether a particular sample of a compound consists of a single enantiomer, a racemic mixture, or a mixture of two enantiomers in unequal amounts can be determined by its

2.188

Organic Chemistry—A Modern Approach

observed specific rotation, which is the specific rotation of the sample. For example, if a sample of (S)-(+)-2-butanol is enantiomerically pure or has an enantiomeric excess of 100% (meaning only one enantiomer is present), it will have an observed specific rotation of +13.52° because its specific rotation is +13.52. If, however, the sample of 2-butanol is a racemic mixture, it will have an observed specific rotation of 0°. If the observed specific rotation is positive but less than +13.52°, the sample must be a mixture of two enantiomers containing more of the S enantiomer than the R enantiomer, because the S enantiomer is dextrorotatory. The enantiomeric excess (ee), also known as optical purity (op), can be defined as the specific rotation of a mixture of two enantiomers, expressed as a percentage of the specific rotation of a pure enantiomer. Observed specific rotation ¥ 100% Therefore, Enantiomeric excess = Specific rotation of the pure enantiomer For example, if a sample of 2-butanol has an observed rotation of +6.76°, then the enantiomeric excess is 50%. In other words, the excess of one of the enantiomers comprises 50% of the mixture. +6.76∞ ¥ 100 = 50% Enantiomeric excess = +13.52∞ When we say that the enantiomeric excess of this mixture is 50%, we mean that 50% of the mixture is excess S enantiomer and 50% is a racemic mixture. Half of the racemic mixture plus the amount of excess S enantiomer equals the amount of the S enantiomer present in the mixture. It thus follows that 75% of the mixture is the dextrorotatory S enantiomer [(50 ¥ 1/2) + 50] and 25% is the levorotatory R enantiomer. The enantiomeric excess may also be defined as the excess amount of one enantiomer in a mixture of enantiomers expressed as a percentage of the entire mixture and can be expressed as follows: moles of one enantiomer - moles of other enantiomer ¥ 100 Enantiomeric excess = total moles of both enantiomers For example, the enantiomeric excess of a mixture containing 10 g of (S)–(+)-2-butanol and 6 g of (R)-(–)-2-butanol is 25%. 10 6 4 ¥ 100 = 25% Enantiomeric excess = 74 74 ¥ 100 = 16 16 74

2.6.8

Racemization

Racemization is the process of producing a racemic modification starting from either of the pure enantiomer. A racemic modification is an equimolar mixture of two enantiomers, i.e., a mixture of two different molecular species. Since mixing causes increase in disorder,

Principles of Stereochemistry

2.189

DS of racemization is a positive quantity and therefore, DG in the expression DG° = DH° – TDS° is negative (assuming DH° constant). Because of this, racemization is a process which is thermodynamically favourable. For example, at 27°C (300 K), DG° change is –1.8 kJ/mol. Driving force for racemization is, in fact, entirely entropic. Various methods used for racemization are discussed below:

(a) Thermal racemization In this process of racemization, one of the four bonds to the chiral carbon is broken temporarily by applying heat. An achiral free radical is obtained by homolytic bond cleavage. The carbon radical then combines with the breakaway radical from either faces of the sp2 (planar) carbon to form an equimolar mixture of two enantiomers (a racemic modification). For example, (+)- or (–)- a-chloroethylbenzene undergoes racemization when distilled under normal pressure. Cl H3C

D

C

Cl

H

Ph ( )-(+)-a-Chloroethylbenzene

H3C

C

H + Cl

or Ph Cl Ph

An achiral free radical

D

C

C

H3 C Mirror plane

H Ph

H3C Ph C

H

Cl (±)-a-Chloroethylbenzene

H

H3 C ( )-(–)-a-Chloroethylbenzene

(b) (i) Acid-catalyzed racemization via an enol This process of racemization involves temporary separation of an acidic hydrogen atom bonded to a chiral centre. Chiral carbonyl *

compounds of the type RR ¢ CHCOR in which the chiral centre containing hydrogen is adjacent to the C=O group undergoes racemization in the presence of acid through the formation of a planar and achiral enol (intermediate). Recombination of the proton with the achiral enol molecule then occurs from either side of the double bond with equal facility to produce a racemic modification. For example, 1-phenyl-2-methyl-1-butanone undergoes racemization when dissolved in ethanol containing HCl. The solution gradually loses its optical activity.

2.190

Organic Chemistry—A Modern Approach

HO O

OH H *

Ph C CHCH2CH3

H OH

Ph C

CH3

Ph

OH

C CH2CH3 CH3

C C(CH3)CH2CH3

HO

(+) or (–)-1-Phenyl-2-methyl1-butanone (kclo form)

Ph

–

C C(CH3)CH2CH3 Enol form (achiral)

(a) 2 2 2

3

3

3 H O H

3

2

–

3

3 (b) 2 2

3 3

3

3 2

(ii) Base-catalyzed racemization via an enolate anion This process of racemization involves the formation of a planar (achiral) enolate anion formed by expulsion of a proton from the chirality centre by base. The enolate anion then recombines with a proton from water. This process occurs from either side of the enolate with equal facility to give a racemic modification. For example, 1-phenyl-2-methyl-1-butanone undergoes racemization when dissolved in ethanol containing NaOH.

Principles of Stereochemistry

O

2.191

H

O

–

Ph C

C CH CH

Ph

CH

O

–

C C(CH )CH CH

Ph

–

C C(CH )CH CH

Enolate anion (achiral)

(+)- or (–)-1-Phenyl-2methyl-1-butanone H C

Ph C O

(a)

CH2CH3 CH3

(a) HO H (–OH – )

Mirror plane

(±)-1-Phenyl-2methyl-2-butanone O Ph

C

C

(b)

CH2CH3 CH3

O

CH2CH3

–

Ph

C C

CH3

(b)

H

(c) By cation formation This process involves the formation of a carbocation by separating a group from the chirality centre with its bonding electrons leaving behind a planar (achiral) carbocation. The anion then attacks the intermediate carbocation from either side with equal facility to form both the enantiomers in equal amounts. This process actually occurs when a substrate which can form a stable carbocation (benzylic, allylic or tertiary) is treated with a Lewis acid (e.g., SbCl5, AlCl3, BF3 and ZnCl2). For example, when (+)- or (–)-a-chloroethylbenzene is heated with antimony pentachloride (SbCl5), (±)-a-chloroethylbenzene is obtained. 3

3

SbCl

– –

5

3

6

a

Cl Ph

C

CH3 H

5

Mirror plane

(±)-a-Chloroethylbenzene Ph C Cl

CH3 H

(a) Ph

CH3 H

(b) 5

C

2.192

Organic Chemistry—A Modern Approach

A mineral acid may also cause racemization. For example, when optically active (2R, 4S)or (2S, 4R)-camphene is treated with acid, racemization occurs through the formation of a carbocation which undergoes a methyl shift followed by loss of a proton.

CH3

CH3 CH3

1,2-methyl–shift

CH3

1,2-methyl–shift

CH3

CH3 +H

+H

–H

CH3 CH3

–H

CH2 180° rotation

Mirror plane * * (2R, 4S)

H2C

CH2 CH3 CH3

H3C H3C

* * (2S, 4R)

(±)-Camphene

2.6.9

Resolution of Racemic Modification

Resolution is the method of separation of a racemic modification into its pure enantiomers. The two enantiomers in a racemic mixture have identical chemical and physical properties except towards optically active reagents. So, they cannot be separated by usual techniques like fractional distillation, fractional crystallization (unless the solvents are optically active), chromatography (unless the absorbents are optically active), etc. The chemical method, which involves the formation of diastereoisomers, is the best method of resolution of a racemic modification. Since diastereoisomers, unlike enantiomers, have different physical properties like solubility, boiling point, absorption coefficient, etc., they can easily be separated by fractional crystallization (if they are solids) or by fractional distillation (if they are liquids). The separated diastereoisomers are then individually

Principles of Stereochemistry

2.193

treated with a suitable reagent to regenerate the pure enantiomer. Resolution of three types of compounds are discussed below.

1. Resolution of a racemic carboxylic acid The racemic modification of a carboxylic acid * (RCOOH) is first allowed to react with an optically active base (the resolving agent, e.g., methylamine, 1-phenylethylamine and alkaloids like brucine, strychnine, ephedrine, morphine, etc.) to form two diastereoisomeric salts. These salts are then separated by fractional crystallization. Each of the separated diastereoisomer is then treated with dilute mineral acid (e.g., dil. HCl) to regenerate the enantiomer. Finally, the resolving agent and the pure enantiomer of the acid are isolated. A schematic representation of the process is as follows: O *

( )

R

C

O

H

+ O *

( )

R

C

( )

N

( )

N

+ O

H

A racemic mixture of a carboxylic acid

An optically active amine (base) (the resolving agent) An acid-base reaction

O ( )

*

R

O

–

C

O

( )

NH

+

( )

*

R

C

–

O

( )

NH

Two diastereoisomeric salts having different physical properties separated by fractional crystallization

*

( ) RCOO ◊ ( )

( )

NH

Single diastereoisomer

*

RCOO ◊ ( )

NH

Single diastereoisomer

dil. HCl *

–

dil. HCl

( ) RCOOH + ( ) NH Cl Pure enantiomer Hydrochloride (organic layer) salt of amine (aqueous layer) –

*

( ) RCOOH + ( ) NH Cl Pure enantiomer Hydrochloride (organic layer) salt of amine (aqueous layer) –

2.194

Organic Chemistry—A Modern Approach

Example: Resolution of racemic mandelic acid may be shown as follows:

–

–

–

–

–

–

Principles of Stereochemistry

2.195

The organic layer is separated and from that pure enantiomer is isolated by removing the solvent. *

2. Resolution of a racemic base The racemic modification of an organic base (RNH2 ) is converted into two diastereoisomeric salts by the reaction of an optically active carboxylic acid (the resolving agent, e.g., camphoric acid, camphor-10-sulphonic acid, glutamic acid, tartaric acid, etc.). The salts are then separated by fractional crystallization. Each diastereoisomeric salts is then treated with dilute alkali to regenerate the enantiomeric amines, leaving the acid as its salt. A schematic representation of the process is as follows: (R) RNH + (S) RNH

+ 2(S) RCOOH An optically active carboxylic acid (the resolving agent)

A racemic mixture of an amine

[(R) RNH (S) RCOO + (S) –

RNH (S) RCOO ] –

Two diastereoisomeric salts

(R) RNH (S) RCOO

(R) RNH

+ (S)

Pure enantiomer (organic layer)

–

RCOONa + H O Sodium salt of the acid (aqueous layer)

(S) RNH (S) RCOO

(S)

RNH

+ (S)

Pure enantiomer (organic layer)

–

RCOONa + H O Sodium salt of the acid (aqueous layer)

Example: Resolution of racemic 1-phenylpropylamine by using (S)-(+)-mandelic acid as the resolving agent may be shown as follows:

2.196

Organic Chemistry—A Modern Approach

2

2

R

S

S An acid-base reaction

–

–

3

3

R S

S S Separated by fractional crystallization

–

–

3

3

R S

S S dil. NaOH followed by extraction with an organic solvent

dil. NaOH followed by extraction with an organic solvent –

–

2

2 2

R

2

S S

S

Principles of Stereochemistry

2.197

The organic layer is separated and from that the pure enantiomer is isolated by removing the solvent.

3. Resolution of a racemic alcohol (±)-Alcohols are usually resolved by converting them into their half-ester by treating them with phthalic or succinic anhydride. These halfesters are then resolved as typical active acids. Pure enantiomeric alcohols are finally regenerated either by hydrolysis with hot aqueous sodium hydroxide or by reduction with LiAlH4. Reduction is preferred when there is possibility of racemization by base. Example: Resolution of (±)-2-methyl-1-butanol may be shown as follows: C H HOH C

O

C H CH

+ H C

H (R)-2-Methyl-1butanol

C O C O Phthalic anhydride

CH OH +

H (S)-2-Methyl-1butanol

A racemic mixture C H

COOH

C H CH

C O H C

+

CH

H C

HOOC O

C

H

H O (R)-Half-ester

O (S)-Half-ester

A racemic mixture

–

COO BH

CH

C O H C O

– C H HB OOC

C H

H [(R), (–)]-salt

+

CH

H C

O

H

C

O [(S), (–)]-salt

A mixture of two diastereoisomeric salts

2.198

Organic Chemistry—A Modern Approach

– C H HB OOC

C H

–

COO BH

H C

CH

C O H C

CH

O

H O [(R), (–)]-salt

O [(S), (–)]-salt

C H

C H

COOH

–

C O CH H (R)-Half-ester (organic layer)

O

C

H

HOOC

–

CH O BHCl + H C CH + BHCl C (aqueous (aqueous H O layer) layer) (S)-Half-ester (organic layer)

The half-esters are separated and then treated individually as follows: 2 aq.NaOH/D 2 2

2

5

3

R 3 LiAlH4

R

5

2

2

5

2

3

2

R

2 aq.NaOH/D

5

3

2

S 2 3

5 2 LiAlH4

S

2

2 3 2

S

5 2

Principles of Stereochemistry

1.

2.199

Give an example of each of the following compounds: (a) an optically inactive compound containing chiral carbons, (b) an optically active compound containing no chiral carbon.

Solution (a) Meso-tartaric acid containing two chiral carbons is optically inactive.

COOH H H

OH OH

COOH meso-Tartaric acid

(b)

Pent-2-3-diene (CH3CH=C=CHCH3) is an optically active compound containing no chiral carbon.

2.

Write structural formulas for the simplest optically active (a) alkene, (b) alkyne, (c) alcohol, (d) aldehyde, (e) ketone, (f) carboxylic acid with no isotopic atom.

Solution *

*

(a) CH2=CH– C H(CH3)CH2CH3 (3-methylbut-1-ene), (b) CH∫∫C– C H(CH3)CH2CH3 *

*

(3-methylbut-1-yne), (c) CH3 C H(OH)CH2CH3 (2-butanol), (d) CH3CH2 C H(CH3) *

CHO (2-methylbutanal), (e) CH3CO C H(CH3)CH2CH3 (3-methylpent-2-one), (f) *

CH3CH2 C H(CH3)COOH (2-methylbutanoic acid). 3.

In which of the following ways should enantiomers differ? (a) melting point, (b) boiling point, (c) sign of specific rotation, magnitude of specific rotation, (e) absolute configuration, (f) solubility in water, (g) solubility in chloroform, (h) solubility in benzene, (i) solubility in (+)-2-chloropentane, (j) solubility in (–)-2-chloropentane, (k) solubility in (±)-2-chloropentane, (l) refractive index, (m) infrared spectrum, (n) NMR spectrum, (o) rate of reaction with an optically inactive reagent, (p) rate of reaction with an optically active reagent, (q) mass spectrum, (r) colour, (s) interaction with left circularly polarized light, and (t) toxicity to human beings.

Solution

4.

Enantiomers differ in (c), (e) (i), (j), (p), (s) and (t).

Can a tertiary amine of the type RR ¢R ¢¢ N or a carbanion of the type RR ¢R ¢¢ C: exist as a pair of enantiomer?

2.200

Organic Chemistry—A Modern Approach

Solution Theoretically it can, but not in practice because a rapid pyramidal inversion (umbrella effect) converts either enantiomer to a racemic mixture. The energy required for this inversion is very low at room temperature, thus racemization is unavoidable. The inversion process is so fast that the enantiomers cannot be separated or isolated.

enantiomers

enantiomers

R≤

R≤ N

N

R¢ R (chiral)

R¢

R T.S. (achiral)

R≤

R≤

R≤ –

N

C

R¢ ; R¢ R R (chiral) (chiral)

A racemic mixture

5.

C

R≤ –

–

C R¢ R (chiral)

R¢

R T.S. (achiral)

A racemic mixture

An optically active alkene A has the molecular formula C5H9Cl. On hydrogenation A is converted to the optically inactive compound B. Predict the structure of A and B.

Solution Since on hydrogenation the compound A becomes optically inactive, i.e., achiral, the unsaturated group and the other alkyl group attached to the chiral centre must contain the same number of carbon atoms and become identical after hydrogenation. Therefore, the unsaturated group must be –CH==CH2 and the saturated group which may be obtained on its hydrogenation is –CH2CH3. Hence, the compound A is 3-chloropent-1ene (any enantiomer) and the compound B is 3-chloropentane.

Cl H3CH2C

Cl

Cl CH CH2 or H2C CH

H R

CH2CH3

H2/Ni

CH3CH2

H S A 3-Chloropent-1-ene (optically active)

6.

CH2CH3 H s-plane

B 3-Chloropentane (optically inactive)

∞ Esterification of (+)-lactic acid, [a ]25 D = + 4∞ , with methanol give (–)-methyl ∞ lactate, [a ]25 D = –8°. Has the configuration changed?

Solution When lactic acid (CH3CHOHCOOH) undergoes esterification by reacting with MeOH in the presence of acid, none of the bonds to the chiral centre is broken. Therefore, although the sign of rotation is changed, the configuration remains unchanged.

Principles of Stereochemistry

2.201

COOH HO

COOMe

CH3

MeOH/H

HO

H (+)-Lactic acid

CH3

H (–)-Methyl actate

Configuration retained

7.

No necessary correlation exists between the R and S designations and the direction of rotation of plane-polarized light. Explain.

Solution The rotation of a compound, (+) or (–), is something that we measure in the polarimeter, depending on how the molecule interacts with light. On the other hand, the R and S designation is our own artificial way of describing how the atoms or groups are arranged in space around a chiral centre. In the laboratory, we can measure a rotation and see whether a particular substance is (+) or (–). On paper, we can determine whether a particular drawing is designated as R or S. But it is difficult to predict whether a structure with R-configuration will rotate the plane of polarized light clockwise or counterclockwise. Similarly, it is difficult to predict whether a (+) substance in a bottle has the R or S configuration.

8.

(–)-Epinephrine is biologically active (stimulating effect is notified when given to a patient) but (+)-epinephrine is not. Why?

Solution Enzymes in living systems are chiral, and they are capable of distinguishing between enantiomers. Usually, only one enantiomer of a pair fits properly into the chiral active site of an enzyme. Only the (–)-enantiomer of epinephrine fits into the active site of an enzyme and for this reason, (–)-epinephrine is biologically active.

9.

(S)-Lactic acid has the physical characteristics listed under the structure. COOH C

H OH (S)-Lactic acid 25° [a]D = + 3.8° mp = 53°C

H3C

(a) What is the melting point of (R)-lactic acid? (b) How does the melting point of a racemic mixture of (R)- and (S)-lactic acid compare to the melting point of (S)-lactic acid? (c) What is the specific rotation of (R)lactic acid, recorded under the same conditions as that of (S)-lactic acid? (d) What is the optical rotation of a racemic mixture of (R)- and (S)-lactic acid? (e) Label each of the following as optically active or inactive: (i) a

2.202

Organic Chemistry—A Modern Approach

solution of pure (S)-lactic acid, (ii) a solution of an equimolar mixture of (R)- and (S)-lactic acid and (iii) a solution that contains 70%(S)- and 30%(R)-lactic acid. Solution (a) 53°C. (b) The melting point of a racemic mixture of (R)- and (S)-lactic acid will be different from that of (S)-lactic acid. (c) The specific rotation of (R)-lactic acid is –3.8°. (d) The optical rotation of a racemic mixture of (R)- and (S)-lactic acid is zero. (e) (i) optically active, (ii) optically inactive, (iii) optically active (dextrorotatory).

10.

Calculate the specific rotation [a ]20 D of (S)-2-butanol if a sample of (S)-2butanol has an observed rotation of +1.03°. The measurement was made with a 1.0 M solution of (S)-2-butanol in a polarimeter tube that is 10 cm long.

Solution To calculate the specific rotation, the sample concentration in g/mL must be determined. Since the molecular mass of 2-butanol is 74.12, the solution contains 47.1 g/L or 0.0741 g/mL of 2-butanol. This is the value of C. The value of l is 10 cm or 1 dm.

Therefore, 11.

[a ]20 D =

a +1.03 = = + 13.9∞ cl (0.0741)(1)

A sample of 2-methyl-1-butanol has an observed specific rotation, [a ]25 D , equal to +1.151. (a) Calculate the enantiomeric excess of the sample. (b) What is the actual stereoisomeric composition of the mixture? (The specific rotation of the pure enantiomer is + 5.756°).

Solution

Observed specific rotation ¥ 100% Specific rotation of pure enantiomer +1.151 = ¥ 100 = 20.00% +5.756

(a)

Enantiomeric excess =

(b)

Of the total mixture, (100 – 20) or 80% consists of the racemic form, which contains an equal number of the two enantiomers. Therefore, half of this 80%, or 40%, is the (+) enantiomer and 40% is the (–) enantiomer. The other 20% of the mixture (the excess) is also the (+) enantiomer. Consequently, the mixture contains 60% (+) enantiomer and 40% (–) enantiomer.

12.

A solution obtained by mixing 20 mL of a 0.1 M solution of the R enantiomer of a compound and 60 mL of a 0.1 M solution of the S enantiomer was found to have an observed specific rotation of +4.8°. What is the specific rotation of each of the enantiomers?

Solution (20 mL ¥ 0.1 M), i.e., 2 mmol of the R-enantiomer is mixed with (60 mL ¥ 0.1 M), i.e., 6 mmol of the S enantiomer. Now, 2 mmol of the R enantiomer and 2 mmol of the S enantiomer will form 4 mmol of a racemic mixture. So, there will be 4 mmol of the S enantiomer as left-over. Since 4 out of 8 mmol is excess S enantiomer, therefore

Principles of Stereochemistry

2.203

the solution has 50% enantiomeric excess. Now, from the values of enantiomeric excess and the observed specific rotation, the specific rotation of the pure enantiomer can be calculated. Enantiomeric excess = 50% =

Observed specific rotation ¥ 100% Specific rotation of the pure enantiomer

+ 4.8∞ ¥ 100% Specific rotation of the pure enantiomer

+ 4.8∞ ¥ 100 = + 9.6∞ 50 Hence, the S enantiomer has a specific rotation of +9.6° and the R enantiomer has a specific rotation of –9.6°.

Therefore, specific rotation of the pure enantiomer =

13.

What is the enantiomeric excess for each of the following mixtures of enantiomers X and Y? (a) 90% X and 10% Y (b) 75% X and 25% Y

Solution Enantiomeric excess tells how much one enantiomer is present in excess of the racemic mixture. (a) Since the mixture contains 90% of X enantiomer and 10% of Y enantiomer, the enantiomeric excess is 90% – 10% = 80%. There is an 80% excess of X enantiomer over the racemic mixture. (b) Similarly, the enantiomeric excess in this mixture is (75% – 25%) = 50%. There is a 50% excess of X enantiomer over the racemic mixture.

14.

Which of the following are optically active? Et

Et

Et

(b)

(a)

(c)

Et

Et

Et

(d)

Br

Et

Et

(e) Et Et

(g)

Br

Br

Et

(f) Et

Br

Et

Cl

(h) Br

Et

Et

Solution The compounds (c), (e) and (h) are chiral (nonsuperimposable on mirror image) and no Sn axis is. So, these are optically active. The compounds (a), (b), (d), (f) and (g) have a plane of symmetry and they are superimposable on their mirror images, i.e., these compounds are achiral and optically inactive.

2.204

15.

Organic Chemistry—A Modern Approach

The following conformer (I) of perfluorobutane is chiral, then why is perfluorobutane optically active? CF3 F CF3 F

F

F I Solution Perfluorobutane has another conformation (II) which is a nonsuperimposable mirror image of the conformer (I). The two enantiomers may interconvert simply by rotation about the C-2—C-3 bond and the rate of interconversion is much higher at room temperature because rotational barriers are generally very small. Since there exists an equimolar mixture of two conformational enantiomers I and II, all the rotations produced by individual molecules are cancelled and prefluorobutane is found to be optically inactive. In fact, these two exist as a nonresolvable dl-pair (a racemic mixture). CF3 CF3 120° or 240° rotation F CF3 F3C F of the back carbon

F

F

F

F I

16.

F II

nonresolvable enantiomers

Calculate the concentration of a solution of the alkaloid coniine ([a ]25 D = + 16∞ ) with an observed rotation of +4.0° in 1.0 dm polarimeter tube.

Solution

Concentration of the solution = =

17.

F

Observed rotation Specific rotation ¥ length of tube + 4.0∞ = 0.25 g/mL +16∞ ¥ 1.0

Draw all the stereoisomers of (a) 1, 2-dimethylcyclopropane, (b) 1, 2-dimethylcyclobutane and (c) 1,3-dimethylcyclobutane and indicate the optically active and inactive isomers.

Solution s-plane

(a)

H3 C H

CH3 H

cis (meso) (optically inactive)

,

H3C

H

H trans

CH3

H

CH3

CH3 trans

(optically active individually)

H

Principles of Stereochemistry

2.205 s-plane

(b)

H3 C

CH3

H3C

H

H

H

H

H

trans

H3C

H CH3

H

cis (optically inactive)

18.

CH3

Centre of symmetry s-plane

H

CH3

H3C

(optically active individually)

CH3 H3 C

H

trans

cis (meso) (optically inactive)

(c)

H

trans (optically inactive)

The ketone (+)-(CH3)2CHCOCH(OH)CH2CH3 racemizes on treatment with alkali, whereas the isomeric ketone (+)-CH3CH2COCH2CH(OH)CH2CH3 does not. Explain this observation.

Solution For racemization to take place the hydrogen attached to the chiral carbon must be acidic and for this to happen the C==O group must be adjacent to the chiral carbon. The carbonyl function is attached to the chiral carbon in the ketone (CH3)2CHCOCH(OH) CH2CH3 but not in the isomeric ketone CH3CH2COCH2CH(OH)CH2CH3. Because of this, the former ketone racemizes on treatment with alkali, whereas the latter ketone does not.

19.

(+)-MeO—

Solution

-MeO

—CH(CH3)C2H5 racemizes in the presence of AlCl3. Explain.

AlCl3 (a Lewis acid) abstracts a hydride ion from the chiral carbon of (+) *

CH(CH3)C2H5 to form a planar (achiral) carbocation. The process

is facilitated by the resonance stabilization of the resulting carbocation by the p-methoxylphenyl group. A hydride ion from a second molecule of the compound then attacks the carbocation from either side with equal facility to form racemic or (±) -MeO

CH(CH3)C2H5.

2.206

Organic Chemistry—A Modern Approach

H –

MeO

C—C2H5 + AlCl5

H – Shift

CH3

–

MeO

C—C2H5 + AlHCl3

CH3 A resonance-stabilized carbocation (achiral) OMe

(a) C2H5 MeO

C CH3

+ H—C—C2H5 CH3

(b) MeO—

—C(CH3)C2H5 +

OM

20.

MeO—

—C(CH3)C2H5 +

e

C H

C2H5 CH3

—CH(CH3)C2H5

Mirror plane

(b)

C2H5 CH3

MeO

–

H – Shift

C

(±)MeO—

H

(a)

Explain whether the following compounds are resolvable or not: O H3CHC==C==CHCH3; PhCH2P(Me)Ph; PhN(Me)Et I

II

III

Solution The allene I and the phosphine oxide II are resolvable because they are chiral and configurationally stable. The amine III is not resolvable because this chiral amine is not configurationally stable due to rapid pyramidal inversion (umbrella effect).

21.

On heating two moles of (±)-lactic acid loses two moles of water to give two diastereoisomeric products of which one is resolvable and the other is not. Draw the structures of the diastereoisomeric products and indicate which is resolvable and why?

Principles of Stereochemistry

2.207

Solution When heated (±)-lactic acid (two molecules) reacts intermolecularly to give cisand trans-products (two diastereoisomers). The cis-isomer is optically active (it has no Sn axis) and resolvable, whereas the trans-isomer is optically inactive (it has a centre of symmetry).

D —2H2O 3

3

3

3

cis 3

D

3

3

cis

3

—2H2O

cis i 3

3

D —2H2O

3

3

trans meso

22.

Which of the following amines could in principle be used as a resolving agent for a racemic carboxylic acid? (±) – Ph—CH—NH CH I

(–) – Ph—CH—NH CH II

Ph(CH ) C—NH III

Solution A resolving agent must be a pure enantiomer. Therefore, II could in principle be used as a resolving agent for a racemic carboxylic acid.

23.

Enantiomers have identical reactivities with achiral reagents — Why?

Solution Enantiomers have identical free energies. The enantiomeric transition states have also identical free energies, as do the enantiomeric starting materials. Because relative reactivity is determined by the difference in free energies of transition state and

2.208

Organic Chemistry—A Modern Approach

starting material, and because of this difference in same for both enantiomers, enantiomers react at identical rates. 24.

A compound is found to be optically inactive. Predict the nature of the compound.

Solution It may be (i) an achiral compound with no stereogenic centre or with two or more stereogenic centres (meso), (ii) a compound whose molecules are chiral but not configurationally stable, (iii) a compound whose molecules are chiral but each enantiomer has immeasurably small optical activity, or (iv) a racemic mixture—an equal amount of two enantiomers.

1. 2. 3. 4.

Explain why an optically inactive product is obtained when (–)-3-methyl-1-pentene is subjected to catalytic hydrogenation. What is the necessary and sufficient condition for a molecule to show optical activity? The presence or absence of a chirality centre is no criterion for optical activity. Explain. Explain why the compound I can be resolved into enantiomers, but the compound II cannot. 2

5

2

5

–

2

2

2

2

2

5.

When an optically active compound A (molecular formula C9H16) is allowed to react with hydrogen in the presence of a catalyst, an optically inactive product is obtained. Which of the following structures for A is (are) consistent with all the data? Et

Et

(a)

Et

(b) Et

(c) Et

Et Et Et

(d)

Et

(e) Et

Et

Principles of Stereochemistry

6.

2.209

Which of the following compounds can be resolved into enantiomers at room temperature? Explain. CH2CH2CH3 N

(a)

2

5

2

C2H5

(b)

3 3

C2H5 2

3 2

CH2CH2CH3

Et

Et

N

(c)

(d)

N N

CH3 H

Et Et

7.

8.

Of the nine possible stereoisomers of 1, 2, 3, 4, 5, 6-hexachlorocyclohexane, only two can be isolated in optically active form under ordinary conditions. Write down the structures of these enantiomers. Explain why compound A has two optically active stereoisomers, but B and C exist as optically inactive single compounds. ≈

CH3CH2CH2 N H(CH3 )CH = CH2Cl@ ;

≈

(CH3 )2 N HCH = CH2Cl@

A

CH3NHCHCH2CH = CH2

B

C

9.

The observed rotation of a solution of compound (0.187 g/mL) is found to be –6.52° when placed in a polarimeter tube. What is the percent of each enantiomer in the solution if [a ]25 D of the compound is –39.0° and the length of the tube is 1 dm?

10.

A sample of (R)-(–)-lactic acid was found to have an enantiomeric excess of 74%. How much R enantiomer is present in the sample? Are the following compounds chiral and exhibit optical activity? Give your reasoning. (b) CH3CH==C==CHCH3 (a) (CH3)2C==C==C(CH3)2

11.

(c)

CH3CH==C==C==CHCH3 (d) (CH3)2C==C==C==C(CH3)2 F

F

Br

(e)

(f) Br

F

H N

Br

O

(g) Br

F

NH O

2.210

12.

Organic Chemistry—A Modern Approach

Draw all possible stereoisomers of the following compound and indicate which are optically active. HOCH2CH CH

CHCH2OH

OH OH OH

13. 14. 15.

Pure (R)-mandelic acid has a specific rotation of –154°. If a sample contains 60% of the R and 40% of the S enantiomer, what is the observed rotation of this solution. When butane is allowed to react with Br2/hn, a bromoalkane is obtained which can be resolved. Explain this observation. Predict the products of the following reaction and indicate whether each product is optically active or not. CH2 H /Ni

CH2

16. 17.

18. 19.

20.

A chiral sample gives a rotation that is close to 180°. How can one tell whether this rotation is +180° or –180°? A solution of 4.0 g of (+)-glyceraldehyde, HOCH2CHOHCHO, in 20 mL of water was placed in a 100-mm polarimeter tube. Using the sodium D line, a rotation of +1.74° ∞ was found at 25°C. Determine the specific rotation [a ]25 D of (+)-glyceraldehyde. Calculate the enantiomeric excess and the specific rotation of a mixture containing 6 g of (+)-2-methyl-1-butanol and 4 g of (–)-2-methyl-1-butanol. When (+)-2-iodoactane is reflexed with KI in acetone, it loses optical activity. Explain this observation. [Hint: Racemization occurs by a reversible SN2 reaction in which the nucleophile and the leaving group are identical.] Predict the stereochemical outcome of the following reactions: Br

21.

22. 23. 24.

Br

2 2 Æ (b) trans-2-Butene æææ Æ (a) cis-2-Butene æææ How can an amino acid be resolved? [Hint: A racemic amino acid can be resolved by initial conversion to its formyl derivative (–NH2 Æ NHCHO) in which the basic character of the –NH2 group is eliminated owing to the –I and –R effects of the C==O group. This derivative then behaves as an acid and resolved by using an optically active base. The basic resolving agent and the formyl group are finally removed by acid hydrolysis.] In a given solution, a compound shows optical rotation of +300°. How will you prove that it is dextrorotatory? Explain why (S)-3-phenylbutanone loses its optical activity in alkali medium. Predict the specific rotation of a mixture of 30% (–)-2-bromobutane and 70% ∞ (+)-2-bromobutane. [a ]25 D of the pure enantiomer is –23.13°.

Principles of Stereochemistry

25.

26.

27.

28.

2.7

2.211

The product of the first reaction can be resolved but the product of the second reaction cannot be resolved into two enantiomers. Explain. (i) PhCHO + HCN Æ; (ii) Me2CO + HCN Æ The diketone CH3COCH(CH3)COCH3 is allowed to react with HCN (2 moles) in the presence of a basic catalyst. Identify the products and give their Fischer projections. Discuss their behaviour towards plane-polarized light. Which one of the following two reactions gives a product which is optically inactive? Give your reasoning.

Tetra-sec-butylmethane has four optically active and one optically inactive stereoisomers. List them in terms of R/S designation.

CONFORMATION OF ACYCLIC ORGANIC MOLECULES

Stereoisomers that can be interconverted through rotation around single bonds (or twisting of bonds) are known as conformational isomers or rotational isomer or simply conformers or rotamers and the phenomenon is known as conformational isomerism. A molecule could have an infinite number of confomations. The rotation around a C—C single bond is not actually completely free. It is hindered by very small energy barrier of 16–25 kJ/mol due to very weak repulsive interaction between the electron clouds of eclipsed s-bonds. Since the energy barrier to this rotation is actually very small, the different conformations are rapidly interconvertible and cannot be isolated, except in a few cases at very low temperatures, or when they possess unusual structures like ortho-substituted biphenyls. A particular conformer having a considerable amount of internal strain (bond angle strain, torsional strain and steric strain) is less stable and hence less populated than that without having strain. The more stable conformer is more populated and therefore, it represents the properties of a molecule at ordinary temperature. At higher temperatures, the population of the less stable conformers increases and the properties of molecules are then represented by those isomers. The study of physical as well as chemical properties of a compound with respect to the relative stability as well as the population of its conformers is known as conformational analysis. Unlike readily interconvertible conformers of molecules like ethane, propane, etc., the conformers of ortho-substituted biphenyls are stable enough at a particular geometry and do not interconvert. For example, 6, 6¢-dinitro-diphenyl-2,2¢-dicarboxylic acid is

2.212

Organic Chemistry—A Modern Approach

not planar since this geometry is highly destabilized due to strong steric interaction between the substituents at ortho positions of the ring. In order to avoid this strain, the compound exists in two isomeric nonplanar forms (I) and (II) in which the two rings are orthogonal. These are actually conformers since they are related to the rotation around the C—C s-bond connecting the two aromatic rings. They do not interconvert at ordinary temperature (energy barrier 80–100 kJ/mol) as interconversion involves highly unstable planar structure as an intermediate. These conformational isomers are nonplanar and nonsymmetric and hence chiral. In fact, they are enantiomers and therefore, it is a case of conformational enantiomerism, commonly known as atropisomerism. Their interconversion may, however, be effected by applying heat and this results in racemization.

Bond rotation requires higher energy

2

2

2

2

¢

¢

Dihedral angle: The dihedral angle (q), which is usually symbolized by the Greek letter theta, is the angle between the X–C–C plane and the C–C–Y plane of X–C–C–Y unit in a molecule. In the Newman projection formula, it is the angle between the C–X bond on the front carbon atom and the C–Y bond on the back carbon atom. dihedral angle

q=

60°

q

Torsion angle is synonymous with the term dihedral angle. However, in contrast to dihedral angle, torsion angle has been given directional sense. It is positive (+) when measured in a clockwise direction and negative (–) when measured in a counterclockwise direction, starting from the front substituent and ending at the rear substituent. Therefore, torsion angle has a range of value from 0° to ±180°.

Principles of Stereochemistry

2.7.1

2.213

Conformations of Ethane (CH3—CH3)

A molecule of ethane contains a carbon–carbon single bond (s-bond) and each carbon atom is attached to three hydrogen atoms. The two –CH3 groups can rotate freely around the C—C bond axis. Rotation of one carbon keeping the other fixed results into an infinite number of spatial arrangements of hydrogen atoms attached to the rotating carbon atom with respect to the hydrogen atoms attached to the fixed carbon atom. These are conformational isomers or conformers or conformations of ethane. Thus, there are an infinite number of conformations of ethane. However, there are two extreme cases. The conformation in which the hydrogen atoms attached to two carbons are as close together as possible, i.e., in which the dihedral angle between two nearest C—H bonds of two –CH3 groups is zero is called the eclipsed conformation. This is so-called because the Newman projection shows the hydrogen atoms on the back carbon atom to be hidden (eclipsed) by those on the front carbon atom. The other extreme conformation in which the hydrogen atoms are as far apart as possible, i.e., the dihedral angle between two C—H bonds is 60° called the staggered conformation. Any other intermediate conformation is called a skew conformation.

H H

H H q = 0° H

H H

H H Eclipsed conformation

q = 0 – 60°

° 60 q=H H

H Staggered conformation

H

H

H H

H H H

Skew conformation

In the eclipsed conformation, the electron clouds in the eclipsed C—H bonds on adjacent carbon atoms repel each other. The repulsive destabilization caused due to eclipsing of bonds on adjacent carbons is called torsional strain. On the other hand, in the staggered conformation, the electron clouds in the C—H bonds are separated as much as possible and so, the staggered conformation is free from torsional strain. Because of this, the eclipsed conformation is less stable (about 2.86 kcal/mol or 12 kJ/mol higher in energy) than the staggered conformation. Since a molecule must rotate from one staggered conformation to another through the more energetic eclipsed conformation, the rotation about the C—C s bond in ethane is somewhat restricted, i.e., not completely free. There is an energy barrier of about 2.86 kcal/mol or 12 kJ/mol. This is called torsional energy. How the potential energy of ethane changes with dihedral angle or torsion angle as one CH3 group rotates relative to the other can be represented graphically in a potential energy diagram or torsional curve as shown below:

2.214

Organic Chemistry—A Modern Approach

I

60° 60° 60° 60° 60° 60° II III IV V VI I rotation rotation rotation rotation rotation rotation H H

H H H H

H H H H

H H H H

H H III

H H

H H V

Potential energy H

H H

H

H

H

H II

0°

60°

H H

H

H

H

H IV

120°

eclipsed conformations (energy maximum)

All three eclipsed conformations (I, III and V) are identical and the same is true for all three staggered conformations (II, IV & VI).

H H

H H I

12 kJ mol–1 (2.86 kcal mol–1)

I

H H

180°

H

Staggered conformations (energy minimum)

H H VI

240°

300°

360° = 0°

Dihedral angle

The staggered conformation is the most stable arrangement, so it is at an energy minimum. As the C—H bonds on one carbon are rotated relative to the C—H bonds on the other carbon, the energy increases as the C—H bonds come close to each other until a maximum is reached after 60° rotation to the eclipsed conformation. As rotation continues, the energy decreases until after 60° rotation, when the staggered conformation is reached once again. This energy barrier is not large enough to prevent rotation at room temperature as collision between the molecules supply sufficient kinetic energy to overcome this energy barrier. Therefore, at any given moment, unless the temperature is extremely low (–250°C), most ethane molecules will have enough energy to undergo rapid bond rotation from one staggered conformation to another, i.e., at any given moment, all ethane molecules do not exist in the more stable staggered conformation; rather a higher percentage of molecules is present in the staggered conformation than any other possible conformation. The steric contribution to torsional strain or Pitzer strain due to non bonded interaction between the vicinal H atoms is negligible since the internuclear distance (0.23 nm) is almost equal to twice the value of van der Waals radius of hydrogen (0.12 nm).

2.7.2

Conformations of Propane (CH3CH2CH3)

The torsional curve, i.e., the potential energy diagram for propane is very similar to that of ethane because both C-1—C-2 and C-2—C-3 bonds are equivalent. However, the energy

Principles of Stereochemistry

2.215

barrier between the eclipsed and staggered forms is slightly higher (14.2 kJ mol–1). This small difference in potential energy barrier between propane and ethane (14.2 – 12 = 2.2 kJ/mol) further shows that steric interaction has no significant role (at least in this case) in the origin of energy barrier. If it would be, the eclipsed propane would have much higher potential energy due to CH3/H interaction in place of H/H interaction. Potential energy changes of propane as function of dihedral angle or torsion angle may be shown by the following energy diagram: I

60° 60° 60° 60° 60° 60° II III IV V VI I rotation rotation rotation rotation rotation rotation H H3C

H H H H

H H

H CH3

H H

H H3C

H3C

H

H

H

0°

60°

H

H3C

H

H

CH3 IV

120°

Staggered conformations (energy minimum)

H

H CH3

H II

eclipsed conformations (energy maximum)

All three eclipsed conformations (I, III and V) are identical and the same is true for all three staggered conformations (II, IV & VI).

14.2 kJ mol–1 H

H H I

V

H H

H

H H

H H

III

I

Potential energy

H H

H H H VI

180°

240°

300°

360° = 0°

Dihedral angle

2.7.3

Conformations of Butane (CH3CH2CH2CH3)

Butane contains two kinds of C—C bonds. So, conformations likely to be generated depend on that the particular C—C bond around which C-atoms are made to rotate. C-2 C-3 bond

C-3 C-4 bond 1

CH3

2

CH2

3

CH2

4

CH3 C-1 C-2 bond

2.216

Organic Chemistry—A Modern Approach

(i) Rotation about C-3—C-4 bond Rotations around C-3—C-4 and C-1—C-2 bonds generate homomeric conformations. The case is very similar to the case of propane (methyl group is replaced by ethyl group) and so, the torsional curve or the potential energy diagram is very similar to that of propane. (ii) Rotation about the C-2—C-3 bond Clockwise rotation of C-2 (the front carbon in the following figure) of butane in 60° increments (until the original conformation comes back) results in six possible conformations: 4

H CH3

CH3 H

H 60° rotation

2

H

1

CH3

H

H H

H H3C

Anti I

CH3 60° rotation

H3 C H

60° rotation

H H

60° rotation

H3C

CH3 60° rotation

H CH3

H

H

Gauche III

Eclipsed II

H CH3

H

CH3

H H

H Gauche V

H

CH3

60° rotation

H H

H H

Totally eclipsed IV The order of the increasing stability of the conformations of butane is as follows: Eclipsed VI

IV < II=VI < III = V < I Potential energy changes of butane as a function of dihedral angle or torsion angle may be shown by the following energy diagram:

Principles of Stereochemistry

2.217

The staggered conformations (I, III and V) of butane are at energy minima and are the stable conformations of butane. The different staggered conformations have been given special names. The conformations with a dihedral angle of ±60° (or in Fig. 60° and 300°) between the two C—CH3 bonds are called gauche conformations (III and V) and the conformation in which the dihedral angle is 180° is called the anti-conformation (I) [gauche is French meaning “to turn aside” and anti is Greek for “opposite off”]. The anticonformation (I) does not have torsional strain because the groups are staggered. Again, the two —CH3 groups are located farthest from each other and so also no steric stain operates. The methyl groups in the gauche conformations (III and V) are close enough to each other and for this, the gauche conformations are destabilized by van der Waals repulsions between nonbonded hydrogens on the two —CH3 groups. This repulsion causes the gauche conformation to have approximately 0.9 kcal/mol (3.8 kJ/mol) more energy than anti-conformation in which such van der Waals repulsions are absent. Therefore, the anticonformation is more stable than the gauche conformation. The eclipsed conformations of butane are unstable for the same reason that the eclipsed conformations of ethane are unstable. The eclipsed conformations (II, IV and VI) represent energy maxima in the potential energy diagram. The eclipsed conformations II and VI not only have torsional strain, but also have van der Waals repulsions arising from the eclipsed methyl groups and hydrogen atoms. The eclipsed conformation IV has the greatest energy (least stable) of all because, in addition to torsional strain, there is a large van der Waals repulsive interaction because the hydrogens in the two eclipsed –CH3 groups are even closer than they are in the gauche conformation. A particular structure of gauche-form without having any symmetry property is chiral but the compound as a whole is not resolvable (and optically inactive) due to rapid interconversion between them (III and V). The same is true for the enantiomeric conformations II and VI. The most stable anti-conformation is the predominant conformation of butane. At room temperature, there are about twice as many molecules of butane in the anti-conformation as there are in the gauche conformation (66% anti and 34% gauche). These two conformations interconvert rapidly at room temperature. Because the eclipsed conformations are unstable, they do not exist to any measurable extent. Butane-gauche interaction: The steric interaction between the two methyl groups in the gauche conformations of butane (q = 60°) makes them less stable (0.9 kcal/mol higher in energy) than the anti conformation (q = 180°). This fundamental interaction is encountered in many organic molecules and is specifically called butane–gauche interaction.

2.218

Organic Chemistry—A Modern Approach

CH3 H

CH3

butane-gauche interaction (0.9 kcal/mol)

H

H H A gauche conformation of butane

2.7.4

Conformations of Chloroethane (CH3CH2Cl)

This case is very similar to the case of propane (methyl group is replaced by Cl atom) and so, the torsional curve or the potential energy diagram is very similar to that of propane. The barrier to rotation is about 15 kJ/mol.

2.7.5

Conformations of 1,2-dichloroethane (ClCH2CH2Cl)

Clockwise rotation of C-1 (the front carbon in the following figure) of 1,2-dichloroethane in 60° increments (until the original conformation comes back) results in six possible conformations:

The order of increasing stability of the conformations of 1,2-dichloroethane is as follows: I III = V II = IV = VI. The potential energy changes of 1,2-dichloroethane (in liquid state) as a function of dihedral angle may be shown by the following energy diagram:

Potential energy

Principles of Stereochemistry

2.219

I

I

III

II 0°

60°

V

IV 120° 180°

VI 240° 300°

360°= 0°

Dihedral angle

The difference in energy between the anti- and gauche-conformations of 1,2-dichloroethane in the liquid state is about zero (in the vapour phase, the anti-conformation is more stable by 1.2 kcal/mol.). Two opposing forces are to be considered to account for this observation, and these are the dipole–dipole repulsion and the van der Waals attraction between the two Cl atoms. In the liquid phase, the dipole–dipole repulsion becomes equal to the van der Waals attraction (in the vapour phase the dipole–dipole repulsion is stronger than van der Waals attraction). Due to the presence of a number of 1,2-dichloroethane molecules in the neighbourhood of a particular molecule, the intermolecular dipole–dipole interaction causes reduction of the intramolecular dipole–dipole interaction to such an extent that the van der Waals attraction just cancels the intramolecular dipolar interaction (repulsion) in the gauche conformation. As a consequence, the energy difference between the anti- and gauche-conformations is zero. The anti-conformation as well as the gauche conformations are the most stable insofar as the torsional strain or Pitzer strain and dipolar repulsions are concerned. The eclipsed conformations are less stable than the anti and gauche conformations mainly because of torsional strain and a very small H/Cl steric interaction. The fully eclipsed form is the least stable one because of torsional strain and large electrostatic repulsion between the two negatively polarized Cl atoms which are close to each other (q = 0). Also, there operates Cl/Cl steric interaction of about same size as that operates in butane (the van der Waals radius of Cl atom is about the same as that of a –CH3 group).

2.7.6

Conformations of Some Typical Acyclic Molecules

The gauche conformation of butane is less stable than the anti-conformation. However, molecules like HOCH2CH2X (where, X = halogen, OH, NH2, OR, NHR, NR2, etc.) are found to be more stable in their gauche conformations due to intramolecular hydrogen bonding as shown below:

2.220

2.7.7

Organic Chemistry—A Modern Approach

Invertomerism

Two conformers which interconvert by inversion at an atom having an unshared pair of electrons are known as invertomers and this phenomenon is known as invertomerism. (R)and (S)-ethylmethylamine are two invertomers which cannot be resolved. p orbital

H N

H

sp

H

N

Me Et R-(Ethylmethylamine)

Me

N Me Et (S)-Ethylmethylamine

Et T.S.

invertomers

If the nitrogen atom is contained in a small ring, for example, it is prevented from attaining the 120° bond angles that facilitate inversion. Such a compound has a higher activation energy (Eact) for inversion and for this reason, the enantiomers can be resolved. The following chiral aziridine derivative, for example, can be resolved into two enantiomers because it is configurationally stable. enantiomers H3C H3C

CH3 N CH3

(R)-1,2,2-Trimethylaziridine

H3C

N

CH3

(S)-1,2,2-Trimethylaziridine

Principles of Stereochemistry

2.221

Unlike N and C (second period elements), the activation energy for pyramidal inversion at S or P (third period elements) is much higher. For this reason, a sulfonium salt with three nonidentical substituents on S, a sulfoxide with two nonidentical substituents on S and a phosphine with three nonidentical substituents on P do not undergo pyramidal inversion at room temperature and can be resolved into enantiomers. R1 S R2 R3

O

R1

X

S X

–

–

O S

R2

R1 R2

R3

S

R1 R2

enantiomeric sulfoxides (configurationally stable and resolvable)

enantiomeric sulfonium salts (configurationally stable and resolvable) R1

R1 P

P

R2 R3

R2 R3

enantiomeric phosphines (configurationally stable and resolvable)

1.

Explain why the following two conformations (I and II) of 3,4-di (1-adamantyl)-2,2,5,5-tetramethylhexane can be isolated. CMe3 Me3C R

CMe3 R

H I

H

R R

H

CMe3

H

R =

(adamantyl group)

II

Solution Due to severe steric interaction between two bulky alkyl groups (–CMe3/–R) in the eclipsed conformation, the rotational energy barrier is much higher in this case and for this reason, these two conformations of 3,4-di(1-adamantyl)-2,2,5,5-tetramethylhexane can be isolated.

2.222

2.

Organic Chemistry—A Modern Approach

Write the most stable conformation of n-pentane (in Newman projection formula).

Solution The anti-conformation is the most stable conformation for rotation around the C-2—C-3 bond because this conformation is free from torsional and steric strain (the – C2H5 and –CH3 groups are anti to each other)

CH3 H

H

H

H C2H5

Anti conformation of n-pentane

3.

The potential energy barrier for the direct interconversion of the two gauche conformations of butane is 15 kJ mol–1 or 3.6 kcal mol–1, while the potential energy barrier for the direct interconversion of the two gauche conformations of 1,2-dichloroethane, ClCH2CH2Cl, is 38.9 kJ mol–1 or 9.3 kcal mol–1 even though the van der Waals radius of a covalently bound Cl atom is about the same as that of a methyl group (–CH3). Explain why the energy barrier in 1,2-dichloroethane is so much higher than that in butane.

Solution Due to very strong electrostatic repulsion between the two very close negatively polarized Cl atoms in the fully eclipsed conformation, the potential energy barrier for the direct interconversion of the two gauche conformations of 1,2-dichloroethane is much higher (38.9 – 15 = 13.9 kJ/mol) than butane.

electrostatic repulsion d–

d– d–

Cl

H

H

H H

4.

d–

Cl

Cl

Cl

Cl H

H H

H H (unstable)

d–

Cl

H

d–

H H

Use a Newman projection, about the indicated bond, to draw the most stable conformation for each compound: (a) 3-methylpentane about the C-2—C-3 bond and (b) 3,3-dimethylhexane about the C-3—C-4 bond.

Principles of Stereochemistry

2.223

Solution 4

3 2

1

(a)

2 3

3 2

3 4

5 2

3

2 3 3

1 3

(b)

5.

The torsional energy in propane (CH3CH2CH3) is 14 kJ/mol. If each H/H eclipsing interaction is worth 4.0 kJ/mol of destabilization, then how much is one H/CH3 eclipsing interaction worth in destabilization?

Solution

Therefore Therefore 6.

Two H/H interactions + One H/CH3 interaction = 14 kJ/mol 2 ¥ 4.0 kJ/mol + One H/CH3 interaction = 14 kJ/mol One H/CH3 interaction = (14 – 8) or 6 kJ/mol

Haloethanes (CH3CH2X, X = Cl, Br, I) have similar barriers to rotation (13.4 – 15.5 kJ/mol) despite the fact that the size of the halogen increases on going from Cl to I. Suggest an explanation.

Solution Since the size of halogen atom increases in the order: Cl < Br < I, the H/X eclipsing interaction is expected to increase. However, due to increase in size of halogen, the C—X bond length gradually increases and as a result, the distance between H and X increases which causes decrease in eclipsing interaction. For this reason, haloethanes have similar barriers to rotation despite the fact that the size of the halogen increases on going from Cl to I.

7.

Draw the two conformations of 1,3-butadiene. Which one of them is more stable? Give your reasoning.

Solution The two conformations of 1,3-butadiene (CH2 = CH – CH = CH2) resulting from rotation about the C – C single bond are s-cis and s-trans. The prefix s stands for single bond.

2.224

Organic Chemistry—A Modern Approach

H

H

C

H

C

H

H

rotation

C C

H H

180°

H

C

C

H

H

C C

H

H s-trans conformation (more stable)

s-cis conformation (less stable) 1,3-Butadiene

Because of steric interactions between two terminal H atoms, the s-cis conformation is thermodynamically less stable than the s-trans conformation. 8.

The intramolecular H-bonding in active butane-2,3-diol is relatively stronger than that in meso butan-2,3-diol. Explain.

Solution

stronger H-bond CH3 H

OH

H H CH3

H

HO

OH

HO

H HO

CH3 H3 C

CH3

CH3

OH H

H

CH3

H O H

H

O CH3

Active Butane-2, 3-diol weaker H-bond H CH3 H

OH

H

OH CH3

meso-Butane-2,3-diol

H H OH CH3 CH3

OH

H H

OH OH

H

O O H CH3

H CH3 H3C

CH3 butane–gauche interaction

Because of butane–gauche interaction, the intramolecular H-bonding in the meso form of butane-2,3-diol is much weaker. On the other hand, in the H-bonded staggered

Principles of Stereochemistry

2.225

conformation of active butane-2,3-diol, the two –CH3 groups are placed anti to each other. So, unlike the meso form, it does not suffer from any steric strain and for this reason, the H-bonding in this case is relatively much stronger. 9.

Considering rotation around the indicated C—C bond in each compound. Draw Newman projections for the most stable and least stable conformations. 1

(a)

2

3

4

5

6

1

CH 3 — CH 2CH 2CH 2CH 2CH 3

2

3

4

5

6

(b) CH 3CH 2CH 2 — CH 2CH 2CH 3

≠

≠

Solution

(CH2)4CH3

(a)

H

H

H

H

H (CH2)4CH3 ;

H Most stable conformation

H

H

H

H C2H5

Most stable conformation

10.

H H

Least stable conformation C2H5 H5C2

C2H5

(b)

H H

;

H H

H H

Least stable conformation

Arrange the staggered conformations of 2,3-dimethylbutane in order of decreasing energy.

Solution

11.

Which in between butane (CH3CH2CH2CH3) and 2-methylbutane [(CH3)2CHCH2CH3] has more torsional strain or Pitzer strain?

Solution 2-methylbutane has more torsional strain than butane. It has two staggered conformations with two adjacent methyl groups and one methyl group at a distance. The

2.226

Organic Chemistry—A Modern Approach

third staggered conformation has all the three methyl groups adjacent to each other, and has the highest energy. Only two methyl groups are to be considered in the case of butane. It has two staggered conformations, one with two methyl groups adjacent and the other with two methyl groups anti (the lowest energy conformation). The energy difference between the lowest and highest energy conformations (the barrier to rotation) is, therefore, greater in 2-methylbutane than in butane, i.e., 2-methylbutane has more torsional strain. 12.

Explain why: (a) the gauche conformation of 1-chloropropane is found to be more stable than its anti-form and (b) the eclipsed form of CH3CH2CHO predominates over the anti-form.

Solution (a) Due to electrostatic forces of attraction between the two electronically opposite groups (Cl and –CH3), the gauche conformation of 1-chloropropane is found to be more stable than its anti-form. d+

Me

Me

d–

H

H

H

Cl

H

H

H

H

anti form Cl (less stable)

(b)

gauche form

H (more stable)

Due to electrostatic forces of attraction between the electronegative oxygen atom, and the electron-deficient –CH3 group (an electron-releasing group), the eclipsed from of CH3CH2CHO predominates over the anti-form. d+

Me d– O

O H

H

anti form

eclipsed form Me H (less stable)

13.

H

H

H (more stable)

Explain why 2,3-di-tert-butyl-1,3-butadiene exists nearly exclusively in s-trans conformation.

Solution Because of severe steric stain involved in the s-cis conformation due to close proximity of very bulky tert-butyl groups (–CMe3), 2,3-di-tert-butyl-1,3-butadiene exists nearly exclusively in the more stable s-trans configuration.

Principles of Stereochemistry

2.227

severe steric interaction Me3C

C C

H2C

CH2

Me3C

CMe3

Me3C

C

CH2

CH2

s-cis conformation (much less stable)

s-trans conformation (more stable)

1.

C

Which conformation in each pair is more stable? Calculate the difference in energy between the two conformations (butane–gauche interaction = 0.9 kcal/mol and H/CH3 eclipsing interaction = 1.4 kcal/mol). (a)

(b)

2.

3.

The eclipsed conformation of ethyl chloride (CH3CH2Cl) is 15 kJ/mol less stable than the staggered conformation. How much is the H/Cl eclipsing interaction worth in destabilization? Label the sites of torsional and steric strain in each of the following conformations:

(a)

4.

(b)

(c)

The gauche conformation of ethylene glycol is more stable than the anticonformation. Offer an explanation.

2.228

5.

6.

Organic Chemistry—A Modern Approach

Draw Newman projections of all staggered and eclipsed conformations that result from rotation around the C-2—C-3 bond in (a) pentane and (b) 3-ethylpentane. Also, draw a graph of potential energy versus dihedral angle for rotation around this bond for each of these two molecules. Considering rotation around the indicated bond in each of the following compounds, draw Newman projections for the most stable and least stable conformations: (b) CF3CF2CF2 CF2CF2CF3 CD3 CD2CD2CD2CD3 ≠ ≠ The anti-conformation of butane is more stable than the gauche conformation by about 0.9 kcal/mol. (a) Calculate the equilibrium constant and (b) the relative amounts of the two conformations at 25°C. [Hint: DH = –900 cal @ DG = –RT Keq. \ ln Keq = 1.52 and Keq = 4.57 for gauche m anti. The ratio of anti/gauche is 4.6/1. This indicates that about 82% of the molecules exist in anti-conformation and 18% in the gauche conformation at any moment.] Draw the eclipsed conformations of 2,3-dimethylbutane (rotation around the C-2—C3 bond) in order of increasing energy. Explain why the rotation about the C—C s-bond in ethane (CH3—CH3) is not completely free? What do you mean by the term conformation? How many conformations of ethane are possible? Name and draw the conformations with the highest and lowest energies at room temperature. Mention the term used for low energy conformations. [Hint: the low energy conformations are usually called ‘conformers’.] Explain how the diagram of potential energy versus angle of rotation around the C-2—C-3 bond of 2,2,3,3-tetramethylbutane to differ from that for ethane, if at all. Considering only rotation about the bond shown, draw a potential energy versus dihedral angle graph for: (a) Me2CH – CHMe2 (b) Me2CH – CH2Me (c) Me3C – CMe3 Both calculations and experimental evidence indicate that the dihedral angle between the two –CH3 groups in the gauche conformation of butane is actually somewhat larger than 60°. How would you account for this observation? Which of the following structures represent conformers of (a)? (a)

7.

8. 9. 10.

11. 12.

13.

14.

(a)

(b)

(d)

(e)

(c)

Principles of Stereochemistry

15.

16.

17.

2.229

In between the two conformations (I and II) of acetylcholine which one is more stable and why?

[Hint: II is more stable due to electrostatic attraction between the positively charged quaternary ammonium nitrogen atom and the slightly negative oxygen atoms in the acetyl group.] Draw the preferred conformation(s) of the following compounds: (b) FCH2CH2OH (a) ClCH2CH2Br ≈ (c) CH3COOCH2CH2 S (CH3)2 (e) CH3CH2CHO (d) (E) – CH2==CH—CH==CH—CH==CH2 (g) OHCCH2F (h) OHCNMe2 (f) CH3COOCH3 Explain your choice of conformations. Predict the optical nature of Tröger’s base.

18.

The following conformation of hexane is chiral, then why hexane is found to be optically inactive?

19.

At room temperature, the anti-conformer of butane is more stable than the gauche conformer by about 0.9 kcal/mol of energy; this results in an equilibrium constant (Keq) of about 4.6 favouring the anti form. Would you expect any change to the value of the equilibrium constant as the temperature is increased? Give you reasoning. Draw the most stable conformer for each of the following using Newman projections: (a) 3-Methylhexane (rotation around C-3—C-4 bond) (b) 3-Methylpentane (rotation around C-2—C-3 bond) (c) 3,3-Dimethylhexane (rotation around C-3—C-4 bond)

20.

2.230

Organic Chemistry—A Modern Approach

Hint: (a)

(b)

(c)

21.

Which of the following conformations of 1-bromo-2-methylpropane is the most stable?

22.

For rotation around the C-3—C-4 bond of 2-methylhexane: (a) Draw the Newman projection of the least stable conformation. (b) Draw the Newman projection of the most stable conformation. (c) How many of the C—C bonds in the compound have staggered conformation that are all equally stable. The potential energy diagram for the rotation about the C-2—C-3 bond in (R)2-iodobutane is given below. Assume that the energies of the conformations are determined by steric effect alone. Draw Newman projections of the conformations corresponding to position I – IV on the figure. Why is the plot of potential energy against dihedral angle (q) unsymmetrical?

23.

3 CHAPTER

NUCLEOPHILIC SUBSTITUTION REACTIONS AT SATURATED CARBON ATOM Chapter Outline 3.2.9

Introduction 3.1

3.2

The SN2 Reaction 3.1.1 Example of SN2 Reaction 3.1.2 Kinetics of SN2 Reaction 3.1.3 Mechanism of SN2 Reaction 3.1.3 Stereochemistry of SN2 Reaction 3.1.5 Evidence in Favour of SN2 Mechanism 3.1.6 Factors Influencing SN2 Reaction Rate or SN2 Reactivity The SN1 Reaction 3.2.1 Example of SN1 Reaction 3.2.2 Kinetics of SN1 Reaction 3.2.3 Mechanism of SN1 Reaction 3.2.4 Stereochemistry of SN1 Reaction 3.2.5 Evidence in Favour of SN1 Mechanism 3.2.6 Factor Influencing SN1 Reaction Rate or SN1 Reactivity 3.2.7 Carbocation Rearrangements in SN1 Reactions 3.2.8 Comparison of the SN2 and SN1 Reactions

3.2.10 3.2.11 3.2.12 3.2.13

3.3

Summary of reactivity of Alkyl Halides in Nucleophilic Substitution Reaction Factors Favouring SN1 and SN2 Reactions SNi and SNi¢ Mechanisms SN1¢ Mechanism Isotope Effects and Salt Effects (Methods used to Distinguish Between SN1 and SN2 Type Reactions)

Neighbouring Group Participation (NGP) 3.3.1 Definition; 3.3.2 Mechanism of NGP; 3.3.3 Example of NGP; 3.3.4 Anchimeric Assistance; 3.3.5 Evidence for Participation by a Neighbouring Group 3.3.6 Various Cases of Neighbouring Group Participation

3.2

Organic Chemistry—A Modern Approach

INTRODUCTION Substitution reactions involving displacement of a nucleophile by another nucleophile are called nucleophilic substitution reactions. A nucleophilic substitution reaction may be depicted in the following way: *

Nu: +

Nuelophile

R — LG Substrate (LG is bonded to an sp3 - hybridized C atom)

ææ Æ R - Nu + Product

LG*:

Leaving group

*

In such a reaction, a nucleophile (Nu) displaces a leaving group (LG *:) in the molecule that undergoes substitution (known as substrate). The nucleophile is always a Lewis base and it may be negatively charged or neutral. The leaving group (also known as nucleofuge) is always a species that takes a pair of electrons (the bonding electrons) with it when it departs. The substrate is often an alkyl halide (R — X) and the leaving group is a halide �� :* . For example: anion : X �� �� * + CH Cl ææ �� *: :OH Æ CH 3OH + :Cl 3 �� �� Nucleophilic substitution reactions are among the most important and fundamental types of organic reactions because they make it possible to convert readily available alkyl halides into a wide variety of other compounds. OH

–

–

OR CN R—X Alkyl halide

– X

–

R¢COO –

SH SR¢ NH3

d-

(alcohols)

R—OR¢

(ethers)

R—CN

(nitriles)

–

–

d+

R—OH

R—OCOR¢ (esters) R—SH

(mercaptans)

R—SR¢

(sulphides)

R—NH2

(amines)

The polar C — X bond causes the alkyl halides to undergo substitution reactions. Since substitution at an sp3 hybridized carbon involves breaking of the s bond to the leaving group and formation of the s bond to the nucleophile, we must know the timing of these two events, i.e., whether bond breaking and bond making occur at the same time (one-step mechanism) or bond breaking occurs before bond making (two-step mechanism).

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.3

One-step mechanism: Two-step mechanism: The mechanism that predominates depends on (i) the structure of the alkyl halide, (ii) the strength and the structure of the nucleophile, (iii) the concentration of the nucleophile and (iv) the solvent in which the reaction is carried out.

3.1

THE SN2 REACTION

Example, kinetics, mechanism, stereochemistry, evidence in favour of the mechanism and factors influencing SN2 reactivity, i.e., the SN2 reaction rate.

3.1.1 Example of SN2 Reaction The alkaline hydrolysis of methyl chloride is a typical example of SN2 reaction. H2O �� : * + CH Cl æææ �� :* HO Æ CH 3OH + : Cl 3 60∞C �� �� Methyl chloride Methyl alcohol

3.1.2

Kinetics of SN2 Reaction

The rate of this reaction depends on the concentrations of the reagents. For example, doubling the concentration of either the nucleophile, i.e., OH① or the substrate, i.e., the alkyl halide doubles the rate. If the concentrations of both the reactants are doubled, the rate of the reaction quadruples. From this relationship between the rate of the reaction and the concentration of the reactants, we can write the rate law for the reaction as follows: rate = k [CH3Cl] [OH①] Because the rate of this reaction depends on the concentration of both the reactants, it is a second-order reaction. Therefore, the reaction is bimolecular, i.e., both methyl chloride and the hydroxide ion are involved in the transition state of the rate-determining step.

3.1.3

Mechanism of SN2 Reaction

The most straight forward explanation for the observed second-order kinetics is a concerted reaction, i.e., bond breaking and bond making occur simultaneously. The mechanism of this reaction is designated as SN2 (Substitution, Nucleophilic, bimolecular) because the two reactants (the substrate and the nucleophile) are involved in the transition state of the rate-determining step (the only step).

3.4

Organic Chemistry—A Modern Approach

In this substitution reaction, the hydroxide ion (OH①), the nucleophile, attacks the sp3 hybridized C atom in methyl chloride from the opposite side of bromine. This approach of the nucleophile requires lowest energy because this avoids electrostatic repulsions between the negatively charged entering and leaving groups. The backside attack is also more feasible sterically. Since it is a one-step (concerted) reaction, it involves no intermediate. The formation of the C — O bond and the cleavage of the C — Cl bond take place at the same time. The reaction proceeds through a single transition state in which the carbon atom is half-bonded to the three hydrogens. In the transition state, the hydroxide possesses a diminished negative charge because it starts sharing its electrons with carbon and chlorine hand also acquires a partial negative charge as it tends to depart with the bonding electrons. The carbon and the three attached hydrogens become coplanar (every bond angle is 120°). In the transition state, the carbon is pentacoordinated (fully bonded to three atoms and partially bonded to two) rather than tetrahedral and it is sp2 hybridized. The energy required to break the C — Cl bond is supplied by the formation of the C — O bond. The process can be represented as follows:

The energy profile diagram for the reaction may be shown as follows:

3.5

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.1.4

Stereochemistry of SN2 Reaction

When an SN2 reaction carried out with a chiral substrate, a product having configuration opposite to that of the substrate is obtained. For example, when (S)-(+)-2-chlorobutane is subjected to alkaline hydrolysis, (R)-(–)-2-butanol is obtained.

Therefore, an SN2 reaction results in inversion of configuration and since inversion occurs, it must be a case of backside displacement. The nucleophile attacks the sp3 hybridized carbon of the tetrahedral substrate from the rear and forces the carbon to achieve a planar (Csp2 ) configuration in the transition state. The plain containing the three groups attached to the central carbon is perpendicular to the remaining p orbital, which by means of its two *

lobes overlaps with the orbitals of the nucleophile and the leaving group (L G). When the leaving group is completely detached, the central carbon becomes once again tetrahedral, but the configuration is opposite to that of the substrate. Therefore, the backside attack causes the arrangement of the three nonreacting groups on the central carbon at which the substitution occurs to be turned inside out like an umbrella in a high wind. The process may be visualized as follows:

3.6

Organic Chemistry—A Modern Approach

Frontier Molecular orbital theory offers a good explanation for the backside attack in an SN2 reaction. In fact, the inversion of configuration can easily be explained by looking at the frontier orbitals, which will be the HOMO of the nucleophile and the LUMO of the electrophile (the C — X bond of the substrate molecule). The overlap is bonding when the nucleophile approaches the electrophile from the rear leading to inversion of configuration (i), but the approach from the front leading to retention configuration (ii), is both bonding and antibonding. Therefore, the former approach, i.e., the backside attack, is clearly preferred.

3.1.5 Evidence in Favour of SN2 Mechanism (a) Stereochemical evidence The SN2 reaction involves inversion of configuration which can be established by an experiment, carried out by Hughes and his co-workers. In the experiment the optically active 2-iodooctane is allowed to react with radioactive iodide ion (128 I* ) in dry acetone solvent. The alkyl iodide is found to lose its optical activity and gets to exchange its ordinary iodine for labelled iodine. The reaction follows a second-order kinetics (first order in the concentration of each reactant) and, therefore, it is an SN2 one. 128 I I | | C6 H13CHCH 3 + 128I* ææ Æ C6 H13 CHCH 3 + I* rate = k [C6 H13CHICH 3 ][128 I* ]

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.7

The rate of substitution, i.e., the rate of radioactive exchange, is determined by measuring the radioactivity of the alkyl iodide at certain intervals. If each nucleophilic attack of the iodide ion leads to inversion of configuration, a (+)-2-iodooctane molecule will be converted to a molecule of its mirror image, i.e., its enantiomer (–)-2-iodooctane, which will cancel the rotation of a second (+)-2-iodooctane molecule by forming a recemic mixture. The reaction of one molecule of alkyl iodide, i.e., each inversion, will thus result in cancellation of optical rotation of two molecules of alkyl halide. As a consequence, the rate of racemization is expected to be twice the rate of inversion. The rate of recemization is determined polarimatrically and it is found to be twice the rate of iodine exchange. The rate of inversion and of iodine exchange are, therefore, identical and it thus follows that every act of substitution much proceed with inversion of configuration.

(b) Kinetic evidence Usually SN2 reaction involving nucleophiles other than solvents exhibits the first-order kinetics with respect to both the substrate and the nuclephile. Therefore, SN2 reactions usually follow the second-order kinetics and hence these are bimolecular (involve rate-determining attack of the nuclephile on the substrate). However, the kinetic evidence is not enough for establishing the mechanistic pathway actually involved because several pathways may be suggested by considering these data.

3.1.6 3.1.6.1

Factors Influencing SN2 Reaction Rate or SN2 Reactivity The effect of substrate structure on rate

As the number of R groups on the carbon with the leaving group increases, the rate of SN2 reaction decreases.

3.8

Organic Chemistry—A Modern Approach

CH3—X Methyl

R CH2—X 1°

R2CH—X 2°

R3C—X 3°

SN2 reaction rate decreases

The SN2 reaction is very much less affected by the electrical effects of the groups at the reaction centre and this is because the central carbon does not become considerably more negative or positive in the transition state as compared to its initial state. Steric effects have profound influence on the rates of SN2 reactions. As a small H atom is replaced by larger alkyl groups, hindrance caused by bulky R groups make the attack by the nuceleophile from the back side more difficult, slowing the reaction rate. This may also be explained in terms of the energy of the transition state. In an SN2 pathway, the tetrahedral central carbon becomes surrounded by five substituents in the transition state (carbon is pentacoordinated). Thus, on going from the substrate to the transition state the steric crowding around the central carbon increase. As consequence, the SN2 transition state becomes progressively more crowded with increase in the number of alkyl groups at the reaction centre. As crowding increases, the transition state energy increases and thus, the reaction rate decreases. Hence, due to steric reason, the SN2 reactivity decrease progressively on going from methyl to tert-butyl halide. In fact, the steric crowding in the transition state of a tertiary alkyl halide is so large, i.e., the transition state is so unstable that it is unable to undergo the SN2 reaction.

The following observations demonstrate how various structural effects influence the rate (reactivity) of SN2 reactions:

(1) Increased branching at the b-carbon retards an SN2 reaction. For example, the rate of SN2 reaction of the following bromides with I① in acetone decreases as the following series is traversed: acetone �� : * R — Br + :��I : * ææææ Æ R — I + : Br �� ��

3.9

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

CH 3CH 2Br > CH 3CH 2CH 2Br > (CH 3 )2CHCH 2Br > (CH 3 )3CCH 2Br Ethyl bromide n–Propyl bromide Isobutyl bromide Neopentyl bromide relative rate: 1 0.082 0.036 0.000012 The differences in electron-releasing inductive effect (+I) of the methyl groups at the b-carbon through the two saturated carbon atoms are very small to show any detectable difference in the SN2 reaction rate. It is the steric effect which is responsible for the rate differences. As the number of CH3-groups at the b-position increases, it becomes progressively more different sterically for the nucleophile I① to attack on C 3 bearing sp

bromine form the backside and consequently, the SN2 reaction rate decreases on going from ethyl to neopentyl bromide. Conformational analysis gives us a clear picture. In conformation I, the backside of n-propyl carbon is seriously blocked, whereas in other two conformations (II and III) the situation is no worse than ethyl. As a consequence, n-propyl bromide undergoes SN2 reaction only slightly less readily than does ethyl bromide.

In the case of isobutyl bromide, there is only one conformation in which the nucleophile can avoid a blocking methyl group. Also, that conformation is highly congested and therefore, has relatively high energy. Consequently, isobutyl bromide is much less reactive than either ethyl or n-propyl bromides.

3.10

Organic Chemistry—A Modern Approach

In neopentyl bromide, there is no conformation in which the attacking nucleophile can avoid a blocking methyl group. Thus, except under very drastic condition neopentyl bromide is practically unreactive in SN2 reactions. Its unreactive nature is due partially to the fact that the leaving group experiences severe van der Waals repulsions with hydrogens of the methyl branches.

Hence, the order of relative SN2 rate of these alkyl bromides is CH3CHBr > CH3CH2CH2Br > (CH3)2CHCH2Br > (CH3)3CCH2Br

(2) Small bicyclic systems containing leaving group at the bridgehead carbon do not undergo SN2 reaction: For example, 1-bromobicyclo [2.2.1] heptane is extremely unreactive in an SN2 reaction. This is because the backside attack on the carbon bearing bromine is prevented sterically by its cage-like structure and also by the impossibility of assuming the required planar distribution of bonds to the bridgehead carbon in the transition state.

The matter can be well demonstreated by the fact that the following ether (I) undergoes cleavage by HBr to give the corresponding bicyclic alcohol and CH3Br.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.11

The reaction proceeds through the SN2 pathway involving nucleophilic attack by Br① on the methyl carbon of the protonated ether.

Because of the formation of an unstable bridgehead carbocation, the reaction does not proceed through the SN1 pathway to give methanol and 1-bromobicyclo[2.2.1]heptane.

(3) A double bond at the b-position helps to increase the rate of an SN2 reaction. For example, allyl bromide, CH2 == CH — CH2 Br, undergoes SN2 reaction slightly more readily than does ethyl bromide. This is because the unhybridized p orbital involved in the transition state interacts with the p orbital system of the allyl group and as a result, the transition state becomes much stable and the activation energy of the reaction becomes relatively low.

3.12

Organic Chemistry—A Modern Approach

(4) A carbonyl group at the b-position helps to increase the rate of an SN2 reaction. For example, chloroacetone (CH3COCH2Cl) undergoes SN2 reaction at a rate faster than does n-propyl chloride (CH3CH2CH2Cl). This is because the b-carbonyl group accepts some of the negative charge of the nucleophile and thereby facilitates the nucleophilic attack on carbon bearing the leaving group by stabilizing the transition state.

(5) Cyclopropyl substrates does not undergo SN2 reaction. For example, halocyclopropanes are unreactive towards SN2 reactions. This is due to increased angle strain in the system. In the transition state of an SN2 reaction, the central carbon becomes sp2 hybridized in which the normal bond angle is 120°. In a cyclopropyl system, the bond angle strain increases (109.5° – 60° = 49.5° to 120° – 60° = 60°) with change in hybridization (sp3 to sp2) of the central carbon atom. As a result, the energy of the transition state becomes too high to make halocyclopropanes unreactive towards SN2 reactions.

(6) Halocyclohexanes react slowly by the SN2 mechanism. The SN2 reaction of halocyclohexanes take place through the relatively stable equatorial conformation. Ring strain does not appear to be an important factor in this case. However, the axial hydrogen at C-3 and C-5 interfere sterically with attacking nucleophile and as a result, the transition state becomes relatively unstable. For this reason, halocyclohexanes react by the SN2 mechanism rater slowly.

3.13

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(7) In cyclohexane systems, the conformer with axial leaving group reacts more rapidly than the conformer with equatorial leaving group. For example, the cis-isomer of 4-tert-butyl*

≈

cyclohexyl bromide reacts with sodium thiophenoxide (Ph S Na) in aqueous ethanol at a much faster rate than its trans-isomer. This is because the axial approach of the nucleophile (specially bulky) for an equatorial leaving group is sterically more hindered due to the axial hydrogens at C-3 and C-5, relative to the carbon (C-1) undergoing substitution. Also, the ground state energy of the axial isomer is higher than the equatorial isomer, i.e., the energy of activation for the axial isomer is relatively low. In this case, the bulky tert-butyl group locks the cis-isomer of 4-tert-butylcyclohexyl bromide in that conformation which places bromine in the axial position and so, this isomer reacts with the bulky nucleophile thiophenoxide ion (Ph S①) by the SN2 mechanism at a much faster rate.

On the other hand, the bulky tert-butyl group locks the trans-isomer in that particular conformation which places bromine in the equatorial position and so, due to steric hindrance, displacement of bromine occurs at a much slower rate in this case.

3.14

Organic Chemistry—A Modern Approach

If the leaving and entering groups are the same in such a reaction, any difference in reaction rate must due to difference in the energy of starting materials. For example, axial cyclohexyl iodide reacts with radioactive iodide ion, 128I① (an SN2 reaction) at a rate faster than equatorial cyclohexyl iodide. Since the equatorial conformer of cyclohexyl iodide is more stable than the axial conformer, the equatorial conformer must have the lower ground state energy. Since the energy of activation reflects the difference in energy between the transition state and the starting material, the reaction involving the axial conformer must have lower Ea. For this reason, the axial conformer reacts at a rate faster than the equatorial conformer. 128I d –

H 128

+

I

–

S N2

d–

H

128

+

–

I

Transition state

equatorial cyclohexyl iodide

I

I

axial cyclohexyl iodide

I

I

H +128 I

128I

H

–

S N2

d–

H 128I d –

H –

128I

+ I

Transition state

(8) Vinylic and aryl halides are unreactive towards SN2 reactions. Displacement of an aryl or vinyl halogen by an SN2 mechanism is prevented because the p electron cloud of the benzene ring of an aryl halide or the double bond of a vinylic halide repeals the approach of a nucleophile from the backside. Again, the C — X bond of aryl or vinylic halides is unusually strong because the bond has considerable double bond character due to delocation of the unshared electron pair on halogen with the p electrons of the ring or the double bond. Consequently, bond breaking requires more energy. It is for these reasons, vinylic and aryl halides are unreactive towards SN2 reactions.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.15

The unreactive nature of phenyl halides can be demonstrated by the fact that ethyl phenyl ether reacts with HBr to give ethyl bromide and phenol instead of bromobenzene and ethyl alcohol. The conjugate acid of ethyl phenyl ether undergoes SN2 attack by Br① not on the ring carbon (the p electron cloud of the ring repeals the backside approach of the nucleophile) to give bromobenzene and ethanol but on the methylene carbon (a favourable site for SN2 attack) to yield ethyl bromide and phenol.

(9) Intramolecular SN2 reaction is possible if geometry allows backside attack. For example, when treated with alkali, trans-2-bromocyclopentanol produces cyclopentene oxide but its cis-isomer produces trans-cyclopentane-1,2-diol.

3.16

Organic Chemistry—A Modern Approach

Since in trans-2-bromocyclopentanol the hydroxyl group is placed stereochemically anti with respect to bromine, intramolecular backside displacement of Br① by oxyanion occurs to yield cyclopentene oxide.

In the cis-isomer, on the other hand, the hydroxyl group and bromine are on the same side of the ring. Therefore, backside displacement of bromine by the oxyanion does not take place. An ordinary SN2 reaction with inversion of configuration occurs to give transcyclopentane-1,2-diol.

3.17

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(10) Williamson synthesis of ethers involves an SN2 reaction between an alkyl halide and an alkoxide. Therefore, if there are two possible ways for the synthesis of a particular ether, that alkyl halide should be used which causes less steric hindrance during during backside attack. For example, tert-butyl ether can be expected to be prepared either by treating ethyl chloride with potassium tert-butoxide or by treating tert-butyl chloride with sodium ethoxide. SN 2 Æ (CH 3 )3C — O — C2H5 + KCl CH 3CH 2Cl + (CH 3 )3COK æææ Potassium Ethyl chloride tert- Butyl ethyl ether tert-butoxide or

S 2

N (CH 3 )3CCl + CH 3CH 2ONa æææ Æ (CH 3 )3C — O — CH 2CH 3 + NaCl tert- Butyl chloride sodium ethoxide tert-Butyl ethyl ether

In the latter combination of reactants, a tertiary alkyl halide is used and because of this, an SN2 reaction leading to the formation of tert-butyl ethyl ether does not take place due to steric reason. However, in the presence of a very strong base like EtO①, the alkyl halide undergoes facile b-elimination reaction to yield 2-methylpropene as the major product. Such as elimination reaction does not take place when ethyl chloride is allowed to react with potassium tert-butoxide. Being a primary substrate, eithyl chloride undergoes ready substitution by the SN2 pathway to form tert-butyl ethyl ether predominantly. This combination of reactants is, therefore, more preferred than the other.

3.18

Organic Chemistry—A Modern Approach

(11) SN2¢ mechanism: When steric factors prevent normal SN2 attack at an allylic carbon, the nucleophil attacks at the g-carbon, i.e., an allylic rearrangement occurs under SN2 conditions. This is what is called SN2¢ (SN2-prime) mechanism (bimolecular nucleophilic substitution with allylic rearrangement). For example, when a, a-dimethylallyl chloride is allowed to react with sodium thiphenoxide (a bulky is nucleophile) in ethanol, the rearranged substitution product is obtained is good yield.

3.1.6.2

The effect of solvent on rate

In general, SN2 reactions are very slightly affected by the polarity of the solvent. In a most common type of SN2 reaction, an anionic nucleophile attacks a neutral substrate: –

Nu + R—X Reactants (concentrated charge: more stabilized by solvation)

d–

Nu

d–

R X

Transition state (dispersed charge: less stabilized by solvation)

Nu—R + X

–

Products

Since the charge is dispersed in the transition state, therefore, stabilization by the solvent is somewhat greater for the reactants than for the transition state. As a consequence, a small decrease in reaction rate is observed in increasing solvent polarity. A marked decrease in reaction rate occurs by the change from an aprotic to a protic solvent. An aprotic polar solvent [e.g., N, N-dimethyl formamide (DMF) Me2NCHO, dimethyl sulphoxide (DMSO) Me2S=0, dimethylacetamide (DMA), Me2NCOMe, hexamethylphosphoramide

3.19

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(HMPA) (Me2N)3PO, N-methylpyrrolidone (NMP), N, N-dimethylpropyleneurea (DMPU), etc] preferentially solvates the cations. They perform this by orienting their negative ends around the cation and by donating an unshared electron pair to the vacant orbitals of the cations. However, because they cannot form hydrogen bonds and their positive centres are well shielded by steric effects from any interaction with anions, aprotic polar solvents do not solvate anions to any appreciable extent. Consequently, the anions become relatively free and highly reactive as a nucleophile. ==

SMe2

O O== SMe2

Me2S == O Sodium cyanide in DMSO:

+

Na

:CN O== SMe2

Me2S == O

==

O

(A relatively free or ‘naked’ cyanide ion)

SMe2 (A sodium ion solvated by molecules of the aprotic polar solvent DMSO)

On the other hand, a protic solvent solvates the anions effectively through hydrogen bonding and thereby reduces its reactivity abnormally. The cyanide ion (CN①) in sodium cyanide (NaCN), for example, is far more reactive as a nucleophile in DMSO than in the protic solvent aqueous methanol.

3.20

Organic Chemistry—A Modern Approach

Therefore, an aprotic polar solvent is much more effective for an SN2 reaction which particularly depends on the reactivity of the nucleophile.

Protic and aprotic solvents A protic solvent is one that consists of molecules having acidic protons (usually attached with oxygen or nitrogen atom), i.e., protic solvent has O — H or N — H group. These groups form hydrogen bonds to negatively charged nucleophiles. Water, alcohols (ROH) and carboxylic acids (RCOOH) are examples of protic solvents. Solvent that cannot form hydrogen bonds are called aprotic solvents. Acetone (CH3COCH3), ether (ROR) and benzene (C6H6) are examples of aprotic solvents.

Polar and apolar solvents A solvent which has a high dielectric constant (e), i.e., which effectively separates ions of opposite charges from one another, is called a polar solvent and a solvent which has a low dielectric constant is called an apolar solvent. In fact, if a solvent has a dielectric constant of about 15 or greater, it is considered to be polar. Water (e = 79), methanol (e = 33) and formic acid (e = 59) are examples of polar solvents, whereas hexane (e = 1.9), ether (e = 4.3) and acetic acid (e = 6) are examples of apolar solvents. [N.B. In organic chemistry, the word polar has a double usage. When we say that a molecule is polar, we simply mean that it has a significant dipole moment (m). However, when we say that solvent is polar, we mean that it has a high dielectric constant (e). Actually, molecular polarity or dipole moment is a property of individual molecules, but solvent polarity or dielectric constant is a property of many molecules acting together. The contrast between acetic acid and formic acid is particularly striking:

These two compounds contain identical function groups and have very similar structure and dipole moments. Both are polar solvent. Yet there is substantial difference in their dielectric constants and they differ substantially in their solvent properties. Formic acid is a polar solvent, but acetic acid is not. Although it is true that all polar solvent consists of polar molecules, the converse is not true.]

3.21

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Dielectric Constants (Œ) of Some Solvent at 25°C Solvent

Dielectric constant

Cyclohexane (C6H12)

Increasing solvent polarity

p-Dioxane (

)

2.01 2.20

Carbon tetrachloride (CCl4) Benzene (C6H6) Carbon disulphide (CS2) Diethyl ether (C2H5OC2H5) Chloroform (CHCl3) Chlorobenzene (C6H5Cl) Ethyl acetate (CH3COOC2C2H5) Acetic acid (CH3COOH) Tetrahydrofuran ( )

2.23 2.27 2.60 4.24 4.8 5.63 6.09 6.1 (20°C)

1,2-Dichloroethane (ClCH2CH2Cl) Sulphur dioxide (SO2) n-Propyl alcohol (CH3CH2CH2OH) Acetone (CH3COCH3) Ethanol (C2H5OH) Methanol (CH3OH) Nitromethane (CH3NO2) O Ê ˆ � Á Dimethylformamide H - C - N Me2 ˜ Á ˜ ÁË ˜¯

10.36 14.1 20.1 20.7 24.33 32.63 35.9

Acetonitrile (CH3CN)

37.0

Sulfone (

)

Dimethyl sulfoxide (Me2S=O) Formic acid (HCOOH) Water (H2O)

7.6

36.7

42.0 46.7 59.0 79.0

That marked increase in reaction rate occurs by the change from a protic solvent to an aprotic solvent can be shown by carrying out the following reaction in methanol and in dimethyl sulphoxide (Me2SO). * SN 2 N3 + CH3 — I æææ Æ N3 — CH3 + : ��I : * �� The rate of the reaction increases very much (109-fold) on transfer from methanol to * DMSO. The nucleophile N3 remains highly solvated through hydrogen bonding in the protic polar solvent methanol. However, in the aprotic polar solvent DMSO (Me2SO), it remains weakly solvated (only through ion-dipole interactions). Also, its counterion (the

3.22

Organic Chemistry—A Modern Approach

cationic part) is solvated by this solvent making the nucleophile much freer to react. Thus, *

N3 reacts as a very powerful nucleophile in DMSO than when it reacts in methanol. For this reaction, the rate of this reaction increases very much on transfer from methanol (e = 32.63) to DMSO (e = 46.7).

3.1.6.3

The effect of the nucleophile on rate

The nature of the nucleophile plays a vital role in determining the rate of SN2 reaction because the nucleophile is involved in the transition state of the rate-determing step (the only step here). The SN2 reaction rate increases with increasing the concentration and the nucleophilic power (strength) of the nucleophile. A good nucleophile is one that reacts rapidly in an SN2 reaction with a given substrate and a poor nucleophile is one that reacts slowly in an SN2 reaction with the same substrate under comparable reaction conditions. Every nucleophile has basic character and in general, nucleophilicity and basicity run in parallel. For reagents having the same attacking atom, a parallelism between them is observed. For example, in the following anions (the attacking atom is oxygen) decreasing nucleophilicity parallels decreasing basicity: �� * > HO: �� * > PhO: �� * > CH COO: �� * > HCOO: �� * CH 3 O: 3 �� �� �� �� �� These two also run in parallel if the attacking atoms are elements with the same period but belonging to different groups of the periodic table. For instance, in the following series, the nucleophilicity decreases with decreasing basicity: *

*

�� �� :* : �� : �� * > : F CH 3 > NH2 > OH �� �� Group-17 Group-16 Group-15 Group-14 *

�� * > :Cl �� :* �� 2 > : SH : PH �� Group-17 Group-15 Group-16

(Period-2)

(Period-3)

The parallelism between nucleophilicity and basicity, however, is not observed when the attacking atoms are of elements belonging to the same group of the periodic table. For instance, the sequence of nucleophilic reactivity for the halides: I① > Br① > Cl① > F① (in protic solvent) is opposite to the sequence of basicity: F① > Cl① > Br① > I①. This order of nucleophilicity tells us that the nucleophilic reactivity of halides increases with increasing the size of the halogen atom. In fact, polarizability and solvation are involved in determining the nucleophilic power of a nucleophile. The outermost electrons of larger and less electronegative atoms are less tightly held by the nucleus. Thus, they are more polarizable and readily available for making bonds to C 3 in the SN2 transition state. Therefore, sp

the polarizability and hence nucleophilicity increases on going from the smaller F① ion to the larger I① ion. Again, in a protic solvent, a nucleophile remains solveted by hydrogen bonding and to participate in the reaction it must be released from the solvation cage.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.23

Solvation energies of ions increase with increasing charge to size ratio. Hence, desolvation of a smaller ion, in which the charge is more concentrated, requires more energy compared to a larger one bearing the same charge. The larger ions are, therefore, better nucleophiles in a protic solvent. In aprotic polar solvents, the cations remain well slovated but the anions remain less solvated and relatively free. Under such conditions, a small anion having its negative charge spread out over a comparatively small volume is more energetic than a large anion and tends to be a better nucleophile. Nucleophilicity and basicity, therefore, run in parallel. For example, in dimethyl sulphoxide (Me2S=0), the order of nucleophilicity for the halide ions is the same as their relative basicity: F① > Cl① > Br① > I①. A negatively charged nucleophile is always a more reactive nucleophile than its conjugate acid. For example, OH① is a better nucleophile than H2O and MeO① is better than MeOH. The nucleophilicity of a reagent may also be influenced by the steric effect. For example, the bulky tert-butoxide ion is unable to attack the central carbon of R — X (except R == Me) smoothly, even though it is more basic than the ethoxide ion which is a better nucleophile.

An interesting example of steric effect is as follows. Quinuclidine is a better nucleophile than tirethylamine. Since the carbon atoms in quinuclidine are held back by its cagelike structure, therefore, the unshared electron pair on nitrogen is more exposed and its approach towards the C 3 atom from the backside is not sterically hindered. On the other sp hand, the ethyl groups in triethylamine hinder sterically its close approach towards carbon in an SN2 reaction. Quinuclidine is, therefore, a better nucleophile than triethylamine.

3.24

Organic Chemistry—A Modern Approach

Like basicity, nucleophilicity depends on the availability of the unshared electron pair on �� is a good nucleophile while the attacking atom. For example, trimethylamine, (CH3 )3 N, �� is completely nonnucleophilic. In trimethylamine, the tris(trifluoromethyl)amine, (CF ) N, 3 3 three electron-releasing (+I) methyl groups increase the electron density on nitrogen and �� thereby increase the availability of its unshared electron pair. For this reason, (CH3 )3 N is a good nucleophile. In tri(trifluoromethyl)amine, however, the powerfully electronwithdrawing (–I) — CF3 groups decrease the electron density on nitrogen considerably and �� thereby decrease the availability of the unshared pair makedly. For this reason, (CF3 )3 N is completely nonnucleophilic. H3C N

+ R

X

S N2

+

(CH3)3N—RX

H3C CH3 F3C N F3C

CF3

+ R

X

S N2

No reaction

–

3.25

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Basic difference between nucleophilicity and basicity Nucleophilicity

Basictiy

1. The nucleophilicity of a regent is its tendency 1. The basicity of a reagent is its tendency to form to form a bond with a carbon atom. a bond with a proton. 2. Nucleophilicity is a measure of how rapidly a 2 Basicity is measure of the position of equilibrium reagent attacks the substrate in a substitution in an acid-base reaction. reaction 3 Basicity is determined by the equilibrium 3. Nucleophilicity is determined by the reaction constant which depends on the overall energy rate which depends on the free energy of change (DG°) of the reaction. | activation (DG=) of the reaction.

Some relevant topics are discussed below:

(a) The order of basicity of halide ions The order of basicity of a halide ions is opposite to the order of their nucleophilicity. Basicity is a measure of the position of equilibrium in the following reaction and it depends on the overall energy change (DG°) of the reaction:

The equilibrium constant (Keq) of the reaction is related DG° by the following expression: DG° = – 2.303 RT log Keq Basicity increases as the value of Keq increases, i.e., the value of DG° increases. Again, the value of DG° decreases as the value of DH° decreases (DG° = DH° – DS°), i.e., the value of H — X bond strength increases. Hence, basicities of halides increase with increasing H — X bond strength. Proton is a hard acid. Base hardness of the halide ions increases in the order: I① < Br① < Cl① < F①. A hard base prefers to combine with a hard acid and this result in formation of a stronger bond. Therefore, the bond strength of halogen acids decreases in the order: H — F > H — Cl > H — Br H — I and so, basicity increases in the order: I① < Br① < Cl① < F①.

(b) Gas phase nucleophilicity of halide ions In the gas phase, the order of SN2 reactivity, i.e., nucleophilicity of the halide ions towards CH3Br is F① > Cl① > Br① > I①. In the gas phase, where solvation effects are absent, the smaller anion having its negative charge spread over a comparatively small volume is more energetic than the larger anion and tends to be a better nucleophile. Therefore, in the gas phase, the order of SN2 reactivity, i.e., the order of nucleophilicity of halide ions towards CH3Br is F① > Cl① > Br①> I①. (c) Role of ion pairing in determining nucleophilicity In the following reaction, the order of nucleophilicity for various halides is I① > Br① > Cl① when LiX is used, but the order is ≈ *

Cl① > Br①> I① when Bu 4 N X is used. acetone �� :* �� :* ææææ CH3 CH2CH2CH2OBs + : X Æ CH3CH2CH2 X + BsO SN2 �� ��

3.26

Organic Chemistry—A Modern Approach

In the weakly polar solvent acetone (CH3COCH3), there is considerable ion pairing and the electrostatic attraction between the concentrated charges on two smaller ions is strongest. Therefore, the small Li≈ ion from tight ion pair with a small halide ion makes it less reactive as a nucleophile. Again, as the size of the halide ion increases, the freeness of the halide ion increases. Therefore, in the presence of Li≈ ion, the order of nucleophilicity is the same as that observed in protic solvents, i.e., I① > Br① > Cl① and for much the same reason. The large quaternary ammonium ion, with its charge shielded by the bulky n-butyl groups, forms only a very loose ion pairs. The halide ions are then relatively free and the small ion is more energetic and more nucleophilic, i.e., their reactivity order is same as their basicity order, i.e., Cl① > Br① > I①.

(d) Hard and soft nucleophiles and electrophiles Pearson’s HSAB (Hard and Soft Acids and Bases) theory has also been introduced in the domain of nucleophiles and electrophiles. Hard bases are hard nucleophiles in which the attacking atom possesses high electronegativity, low polarizability and low reducing power. Soft bases, on the other hand, are soft nucleophiles in which the attacking atom has low electronegativity, high polarizability and high reducing power. So, I① ion is a soft nucleophile and OH① ion is a hard nucleophile. Proton and all proton donors are hard electrophiles while a saturated carbon (C 3 ) is a soft electrophile. A few hard and soft electrophiles and nucleophiles are sp

given in the following table. Hard and so electrophiles and nucleophiles Electrophiles (Acids) Hard ≈

≈

≈

≈

H , Li , Na , K , Mg ≈ AlH3, CO, RCO , SO3

2≈

3≈

, Cr

Soft 3+

, Fe , BF3, AlCl3,

≈

≈

≈

≈

≈

Saturated carbon, Cu , Ag , Hg , I , Br , I2, Br2 metal atoms, carbenes, trinitrobenzene, quinones. ≈

≈

Borderline: Fe2+ , Cu2+ , Zn2+ , B(CH ) , Pb2+ , NO, Me C, C H≈ 3 3 3 6 5 Nucleophiles (Bases) Hard

Soft

NH 3 , RNH 2 , N2H 4 , H 2O, OH * , O* , ROH,

H* , R * , C2 H4 , C6 H6 , CN* , SCN* , R 3 P,

* * * 2* RO* , R 2O, CH 3COO* , NO* 3 , F , Cl , SO 4 , R 2 S, RSH, RS

PO34* , ClO* 4 Borderline: PhNH2, N3①

① ① N NO2 , Br

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.27

There is an empirical rule which governs acid–base and nucleophile–electrophile interactions. The rule sates that hard acids prefer to bind with hard bases and soft acids with soft bases. This means that hard–hard as well as soft–soft interactions are strong, i.e., these lead to the formation of strong bonds but hard–soft interactions are weak and these lead to weak bonds. For example, H≈ (a hard acid) reacts faster with OH① (a hard base) than with I≈ (soft base). Along a period from left to right, the harness of nucleophiles increases and the softness decrease and down a group, the softness of nucleophiles increases and the harness decreases. Thus, the order of increasing hardness amongst the following site of nuceleophiles is (i)

@

@

�� :@ : H@ < : CN < : OH < : F ��

��

(ii)

I < Br < Cl < F

(iii)

C@ < N@ < O@ < F@

@

@

@

@

A nucleophile with negative charge on the attacking atom is harder than its conjugate @

�� * is harder than H O �� , :OH �� : and HS �� @: is harder acid. Therefore, :N H2 is harder than NH 3 2 �� �� �� . than H2 :S Furthermore, a positively charged carbon is a harder centre (electrophile) than a partially positively charged carbon, while the latter is a harder centre than a less partially positively charged carbon.

(e) Ambident nucleophiles An ambident nucleophile is a species which has two (or more) different types of mucleophilic centres through which it can attack the electrophilic centres. Cyanide, nitrite and enolate ions are the examples of ambident nuceleophiles (ambi in Latin for ‘both’; dent in Latin ‘teeth’)

The nucleophilic centres in a cyanide ion are C and N atoms since both bear a lone pair of electrons. The same is the case with nitrite ion where the two O atom and the N atom are nucleophilic centres. In an enolate ion, anionic O and C atoms are the nucleophilic centres.

3.28

Organic Chemistry—A Modern Approach

The following sets of reaction illustrate the behaviour of ambident nucleophiles in distribution of products in substitution reactions: (a)

≈

*

attack by CH 3Br + KCN ææææææ Æ CH 3 — C ∫∫ N: + CH 3 — N ∫∫ C: CN* through Ethanenitrile Ethaneisonitrile C atom (major product) (minor product) attack by

≈

*

CH3 Br + AgCN ææææææ Æ CH3 — N ∫∫ C: + CH3 — C ∫∫ N CN* through N atom (major product) (minor product) (b)

These observations can be explained in terms of hard–soft nucleophile–electrophile interaction as discussed before. In each case. The reaction involves either a hard–hard interaction or a soft–soft interaction. The hard and soft centres of each nucleophile are labelled here following the fact that the hardness increases along a period from left to right.

In the presence of Na≈ or K≈ ions, the C atoms of CH3Br, CH3I and CH3CH2I are the @

attack the electrophilic carbons through soft centres soft centres. So, : CN : and NO@ 2 (less electronegative and more polarizable) C and N, respectively, and we get methyl cyanide and nitromethane as major products. In the presence of Ag≈, which is generated from the reagent AgCN or AgNO2, alkyl halides become more carbocation-like in which the central carbons are hard centres (being appreciably positively polarized). So, the nucleophiles attack the positively polarized (or cationic) carbons through their hard centres

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.29

*

(N of : CN : and O of NO2*) and naturally the major products are alkyl isocyanides and

alkyl nitrites. In the presence of Na≈ or K≈ ion, the reactions are SN2 but in the presence Ag≈ ion, the reactions are SN2 with considerable SN1 character or pure SN1.

This can be explained simply as follows. In the presence of AgCN, the reaction proceeds by the SN2 mechanism with considerable SN1 character or by pure SN1 mechanism (in the case of 2° or 3° substrates) and this is because Ag≈ ion coordinates with iodine and causes the development of considerable positive charge on central carbon (or promotes fromation of a carbocation by precipitating AgI). Nucleophilic attack on carbon then takes place through the more electronegative and more electron dense nitrogen atom and as a result, RNC (an alkyl isocyanide) is obtained predominatly. However, in the presence of NaCN, the reaction proceeds by the SN2 mechanism because unlike silver ion, Na≈ ion cannot make the central carbon positively polarized (or promote ionization of the C — I bond). Since bond formation occurs in the transition state of the rate-determining step, the cyanide ion attacks R — I through the more polarizable and less electronegative, i.e., more nucleophilic carbon atom, and thus, leads to the formation of R — CN (an alkyl cyanide) predominantly.

(f) Nucleophilic catalysis Highly polarizable nucleophile like I① (a good entering group and a good leaving group) can often be used as a catalyst to promote an otherwise slow displacement reaction. This is known as nucleophlic catalysis. For example, methyl chloride undergoes hydrolysis at a much faster rate in the presence of sodium iodide. The hydrolysis of CH3Cl in an aqueous medium take place at a much slower rate because H2O is a weak nucleophile (because the molecule is neutral and the attacking atom is highly electronegative and less polarizable) and Cl① is not very good as a leaving group (because it is not a very weak base and the C — Cl bond is not very weak).

3.30

Organic Chemistry—A Modern Approach

H2O: + CH3—Cl

SN2 very slow

HO—CH3

+

HCl

(i)

Methyl Methyl chloride alcohol ① The iodide ion (I ) catalyzes the hydrolysis of methyl chloride. Due to large size and low electronegativity, the iodide ion is highly polarizable. Also, its solvation energy is realatively small because the charge to size ratio is small. Furthermore, it is a very good leaving group because it is a weaker base and the C — I bond is relatively weak. For this dual ability to attack and depart readily, I① is an effective catalyst in substitution reactions. In the presence of NaI, the hydrolysis of CH3Cl takes place by the following two steps. I

–

+

H2O +

–

CH3—Cl

SN2 fast

I—CH3 + Cl

CH3—I

SN2 fast

HO—CH3 + HI

(ii) (iii)

Each of the reaction (ii) and (iii) takes place at a rate faster than the reaction (i) because I① is a better nucleophile than H2O and is a better leaving group than Cl①. Thus, the iodide ion increases the rate of formation of ethanol by changing a relatively slow SN2 reaction into two relatively fast reactions. The overall hydrolysis, therefore, occurs rapidly.

(g) Phase-transfer catalysts A compound that catalyzes a reaction by transferring a reagent (usually an inorganic ion) into the phase in which it is needed is called a phasetransfer catalyst. In the following reaction, tetrabutylammonium hydrogen sulphate acts as a phase transfer catalyst. ≈

CH3 (CH2 )5 Br (in decane) 1-Bromohexane

Bu 4 N HSO4* ææææææææ Æ aqueous NaCN/105∞ C

CH3 (CH2 )5 CN Heptanenitrile

≈

A quaternary ammonium salt like Bu4 N HSO@ because of its nonpolar alkyl groups, is 4 , soluble in a nonpolar solvent (like dissolves like), but it is also soluble in water because of its charge. Therefore, it can act as a mediator between the two immiscible solvents. When ≈

@ Bu4 N HSO4 ion passes into the nonpolar layer, it must carry a counter ion with it. The counter ion may be either HSO4* or CN* . Because there is more CN① ion than HSO4* ion in the aqueous layer, the former will be the main accompanying ion. Once it is in the nonpolar decane layer, the CN① ion can react with the alkyl halide. The catalysis effected by this technique is called phase-transfer catalysis. Some HSO* ion is also transported 4 into the organic layer, but it remains unreactive because it is a very poor nucleophile. The

≈

* Bu 4 N ion then passes back into the aqueous layer carrying with it either HSO* 4 or Br .

The reaction continues with the phase–transfer catalyst shuttling back and forth between

3.31

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

the organic and aqueous phases. Phase transfer catalysis has been used successfully in a wide variety of organic reactions.

Two other quaternary ammonium slats used commonly as phase transfer catalysts are as follows: ≈

*

CH 3 (CH 3 )14 CH 2 N(CH 3 )3 HSO4 Hexadecyltrimethylammonium hydrogen sulphate

+

–

— CH2N (C2H5)3 HSO4 Benzyltriethylammonium hydrogen sulphate

Crown ethers: A group of large-ring polyethers (cyclic polymers of ethylene glycol), because of their three-dimensional crown shape, are called crown ethers. They are named as x-crown-y, where x is the total number of atoms in the ring and y is the number of oxygen atoms. For example, 18-crown-6, 15-crown-5, 12-crown-4, dicyclohexano-18-crown-6, etc.

Crown ethers act as phase-transfer catalysts. They are able to bind various metal ions through ion-dipole interactions involving the unshared pair of electrons on the oxygen atom. A crown ether coordinates effectively with a particular cation which fits well into the cavity of the crown. For example, 18-crown-6 (hole diameter 2.7 Å) coordinates very effectively with K≈ ions (diameter 2.66 Å) because the cavity size is correct and because the six oxygen atoms are ideally situated to donate their unshared electron pairs to the central ion in a Lewis acid-base complex.

3.32

Organic Chemistry—A Modern Approach

The resulting complex is lipophilic on the outside, and has a positive charge burried within the molecule. Because of lipophilicity (hydrocarbon like properties), it is soluble in organic solvent of low polarity. When it passes into organic solvent, it gets the anion with it. The anion is shielded from the positive charge on K≈ by the bulky size of the crown, thus forming only loose ion pairs and is highly reactive. In this way, the crown ether permits an inorganic salt like KCN to be dissolved in nonpolar or less polar solvents like benzene, acetonitrile, etc. and thereby makes some SN2 reactions successful by acting as a phase-transfer catalyst. KCN (insoluble in benzene), for example, reacts with chloroethane in benzene containing a catalytic amount of 18-crown-6 to form CH3CH2CN almost quantitatively. 18- crown -6 K ≈ CN* + CH3CH2 Cl æææææ Æ CH3CH2CN + K ≈ Cl* Benzene (SN 2) Ethanenitrile Chloroethane

Alkyl fluorides (R–F) cannot be prepared easily. 18-Crown-6 provides a good method for the synthesis of alkyl fluorides. Potassium fluoride (KF) is an ionic compound which is particularly insoluble in the aprotic nonpolar solvent benzene. The crown ether 18-crown-6 makes this salt soluble in benzene. Every K≈ ion that enters into the benzene layer (as a Lewis acid-base complex with the crown ether) brings a fluoride ion (F①) with it. These relatively unsolvated fluoride ions with a greatly enhance reactivity displace halide ions from alkyl halides quantitatively. For example: KF, benzene 90∞ C CH 3 (CH 2 )4 Br æææææææ Æ CH 3(CH 2 )7 F 18-crown-6 1-Bromoctane 1-Flurooctane (92%)

The relationship between a crown ether and the ion it binds is called a host–guest relationship.

3.1.6.4

The effect of the leaving group on rate

The rates of SN2 reactions are influenced by the nature of the leaving group because the process of expulsion of the leaving group occurs in the rate-determining step (the only step here). In most of the SN2 reactions, the leaving group begins to acquire a negative charge as the transition state is reached. Stabilization of this developing negative charge at the leaving group stabilizes the transition state, i.e., lowers its energy. This, in fact,

3.33

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

lowers the free energy of activation (DG=| ) and thereby increases the rate of the reaction. Because weak base stabilizes a negative charge effectively, these are good leaving groups. For instance, in the following series, the leaving-group ability (fugicity) increases with decreasing basicity: @

@

�� :@ < PhO �� :@ < NH �� l :@ < : Br �� �� �� : @ < : ��I : @ : NH : : �� :@ < : C 2 < CH3 O 3 < CN < F �� �� �� �� �� �� Also, an effective leaving group should be linked to the central C 3 atom with a relatively sp

weak bond. The C — X bond strength, for example, increases in the following order: Bond dissociation energy (kcal/mol) :

C — I < C — Br < C — Cl < C — F 53

69

81

109

Therefore, the leaving ability of halide ions decreases in the order I① > Br① > Cl① > F①. The following sulfonate ions are good leaving groups like halides because these are all conjugate bases of strong acids:

The following fluorinated alkanesulfonate ions are very good leaving group (‘super’ leaving groups) because these are conjugate bases of very strong acids (pKa ≈ < –10): :O: :O: :O: � � � �� :* �� :* �� :* F3C — S — O CF3 (CF2 )3 — S — O CF2CH 2 — S — O �� �� �� � � � :O: :O: :O: (OTf *) TrifluoromethaneNonafluorobutane2,2,2-Trifluoroethane sulfonate ion sulfonate ion sulfonate ion (Triflate ion) (Nonaflate ion) (Tresylate ion) Strongly basic ions (e.g., OH①) rarely act as leaving groups and very powerful bases such as hydride ion (H①) and alkanide ions (:R①) virtually never acts as leaving groups. An SN2 reaction proceeds in the direction that allows the stronger base (the nucleophile) to displace the weaker base (the leaving group). Therefore, reactions such as the following do not take place:

3.34

Organic Chemistry—A Modern Approach –

–

Nu: + CH3—H

CH3—Nu + H:

–

–

Nu: + CH3—CH3

CH3—Nu + :CH3

Some relevant topics are discussed below:

(a) SN2 reactions are usually irreversible For example, the hydrolysis of CH3Cl with aqueous NaOH is essentially irreversible. �� :@ + CH — Cl ææ Æ HO — CH 3 + Cl��:@ HO ¥æ ¨æ 3 �� �� We can correctly predict the direction of equilibrium by comparing the basicity of the nucleophile and the leaving group. Equilibrium favours the products of an SN2 reaction when the leaving group is a weaker base than the nucleophile. In this reaction, the basicity of the nucleophile (OH①) and the leaving group (Cl①) may be compared by comparing the pKa values of their conjugate acids. The stronger the conjugate acid, the weaker the base and the better the leaving group. Nucleophile OH (Stronger base)

Leaving group Cl– (Water base)

Conjugate acid

H2O (Weaker acid)

HCl (Stronger acid)

pKa

15.7

–7

①

Because Cl①, the leaving group, is a weaker base then OH①, the nucleophile, the reaction favours the product, i.e., the reaction is essentially irreversible. If the difference between the basicities of the nucleophile and the leaving group is not very large, the reaction will be reversible. In the reaction of ethyl bromide with iodide ion, Br① is the leaving group in the forward direction and I① is the leaving group in the backward direction. Because the pKa values of the conjugate acids of the two leaving groups are similar (pKa of HBr = –9; pKa of HI = –10), the reaction is reversible. CH3CH2 Br + K ≈ I* � CH3CH2I + K ≈ Br * A completely reversible reaction is as given below:

3.35

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

When (+)-2-iodobutane is allowed to react with iodide ion (I①) in acetone, instead of only (–)-2-iodobutane (expected to be obtained by inversion of configuration) an optically inactive equimolar mixtrue of (+)- and (–)-2-iodobutane is obtained. This observation suggests that the reaction is completely reversible. Each SN2 attack by I① on (+) -2-iodobutane leads to the formation of (–)-2-iodobutane, a product with opposite relative configuration. However, complete conversion of the (+)-enantiomer into the (–)-enantiomer does not take place because the (–)-enantiomer also undergoes SN2 attack by I① with equal facility. In fact, as the concentration of the (–)-enantiomer in the reaction mixture increases, its tendency to react with I① increases. A time comes when the mixture contains an equimolar amount of (+)- and (–)-2-iodobutane and since both the enantomers then react with I① at the same rate (identical Ea), the composition remains unchanged. Hence, (+)-2-iodobutane reacts with I① in acetone to yield optically inactive racemic or(±)-2-iodobutane instead only (–)-2-iodobutane. The energy diagram for this reversible reaction is as follows: T.S.

Free energy

Free energy of activation is the same for the forward and the backward reaction DG

(+)-2-Iodobutane + I–

(–)-2-Iodobutane + I–

Progress of the reaction

(b) A poor leaving group converts to a good one by coordination with an electrophilic species Hydroxide ion, alkoxide ions and other strong bases are poor leaving groups for SN2 reactions. For example, the — OH group of an alcohol is a poor leaving group because it would have to leave as hydroxide ion (OH ). ①

–

Br

+ CH3—OH

–

Br—CH3 + OH (strong base)

Ions that are strong bases are poor leaving groups: �� * : OH �� Hydroxide

�� * : OR �� alkoxide

*

�� H2 :N �� amide

Some neutral molecules like H2O, ROH, R2NH, etc. are good leaving groups. A neutral molecule often severs as the leaving group from a positively charged electrophile.

3.36

Organic Chemistry—A Modern Approach

For example, if an alcohol is placed in an acidic solution, the hydroxyl group becomes protonated. Water then serves as the leaving group. It is to be noted that the need to protonate the alcohol (requiring acid) limits the choice of nucleophiles to those few that are weak bases, such as bromide and iodide. A strongly basic nucleophile would become protonated in an acidic medium. SN2

There are two main reasons for smooth SN2 reaction: (a) Br① is now attacking a positively charged, an opposed to neutral species, and (b) the very weakly basic H2O is a very much better leaving group than the strongly basic OH① ion. The well-known use of HI is to cleave ethers. I①, generated in a strongly acidic solution, is a very good nucleophile. The strong acid initiates the reaction by protonation of the �� — group. For example, the cleavage of anisole by HI is as follows: —O �� SN2

1.

Does the rate of SN2 reaction depend on the amount of d+ on the attacked Csp3 of R–X? Explain. The rate of an SN2 reaction does not depend an the amount of d+ on the attacked

Solution

Csp3 of R — X. The amount of d+ on the Csp3 decreases with decreasing electronegativity of X in the order F > Cl > Br > I. Therefore, RF is expected to be the most reactive and RI is the least. However, the reactivity order is RI > RBr > RCl > RF. The reactivity actually depends on leaving ability of X①. 2. tert-Butylamine cannot be prepared by Gabriel synthesis — Why? Solution The Gabriel synthesis of an 1° amine (usually with 1° R) may be given as follows:

O

O

C NH C

KOH H2O

O

C NK C

R—X SN2

O

C NR C

NH2NH2 EtOH/D

O

O

O

Phthalimide

Potassium phthalimide

N-Alkylphthalimide

C C

NH NH

O + RNH2 Phthalimide hydrazide

3.37

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

To prepare tert-butylamine (Me3C — NH2) we have to use tert-butyl halide (Me3C — X) in step II. A tertiary substrate cannot be used in this step because it is an SN2 reaction which, due to steric hindrance, does not take place. It is for this reason, tert-butylamine cannot be prepared by Gabriel synthesis. O C NK

+

C—X

SN2

No reaction

C O Potassium phthalimide

3.

A tert-butyl halide

Explain the following observation

1. TsCl/Py 2. LiBr / acetone

CH2OH

CH2Br

Solution In the first step, the bicyclic alcohol undergoes tosylation by TsCl in the presence of pyridine and as a result, the — OH group becomes a good leaving group. Although the intermediate tosylate is a neopentyl type of substrate, displacement of OTs① by Br① occurs under SN2 conditions in the second step to form the corresponding bromide. This is because the groups on the 3° bridgehead carbon are tied back and do not interfere sterically with the incoming nucleophile Br①.

3.38

Organic Chemistry—A Modern Approach

4.

Cleavage of ordinary ethers does not take place in the presence of base. However, basic regents may cause ready opening of epoxides. Explain these observations.

Solution An ordinary ether (R — O — R) does not undergo cleavage under basic conditions because that SN2 reaction would result in expulsion of strongly basic alkoxide ion (RO①) which, in fact, is a very poor leaving group.

HO

+

R—OR

R—OH

An ether

A nucleophile

+

An alcohol

RO

An alkoxide (a strong base and a poor leaving group)

Epoxides, on the other hand, suffers from significant ring strain (about 25 kcal/mol) and the strain is released when the ring opens up. For this reason, epoxides undergo ringopening reactions (SN2) under basic conditions although the leaving group in this case also an alkoxide ion which actually remains within the same molecule. The strain is enough to compensate for the poor alkoxide leaving group. In fact, the ring opening has a lower energy of activation because an epoxide is 25 kcal/mol higher in energy than an ordinary ether. O +

HO A nucleophile

5.

C——C An epoxide (suffers from angle strain)

O SN2

—C——C— OH An alkoxide ion (a strong base)

Are SN2 reactions stereospecific and/or stereoselective?

Solution SN2 reactions are stereospecific because stereoisomeric reagents give stereochemically different products. Also, they are stereoselective because they form exclusively only one of a possible pair of enantiomers or one of the possible diastereoisomers. 6. Predict the product and suggest a mechanism of the following reaction: 14

H2C——CH——CH2Cl

NaOMe (1 mole) MeOH

O Solution The epoxide first undergoes SN2 attack on the less crowded labelled carbon of the ring. The resulting alkoxide then undergoes intramolecular SN2 reaction to form again an epoxide in which the labelled carbon is not a part of the ring. This labelling experiment proves that the reaction is not a simple SN2 displacement of Cl① by OMe①.

3.39

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Cl 14

Me O + CH2—CH—CH2Cl

SN2

14

MeO—CH2—CH—CH2

O

7.

SN2

14

MeO—CH2—CH—CH2

O

O

Ethers can be easily cleaved by acids like HBr and HI. However, HCl cannot be used for this purpose. Explain these observations.

Solution In the presence of strong acids like HBr or HI, ether undergoes protonation and as a result, the very poor leaving group RO① becomes converted into a very good neutral leaving group ROH. The protonated ether then undergoes ready SN2 attack at the less crowded alkyl group by the good nucleophiles Br① or I① to yield an alcohol and an alkyl halide. For example:

HCl cannot be used for ether cleavage because it is not as strong an acid such as HI or HBr to protonate ether sufficiently and also Cl① is a very poor nucleophile. 8. Ethers can be used safely as a solvent during Grignard synthesis — Why? Solution Being a very strong base the leaving group of ether (R—O—R), i.e., RO① is a very poor leaving group. Consequently, ethers are unreactive towards nucleophilic substitution and so, they cannot be cleaved by base, i.e., the carbanion R① (from RMgX) does not cause cleavage of ROR.

R¢—MgX + R—O—R

R¢—O—R + RO MgX

Grignard reagent (source of :R¢ )

9.

Give SN1 and SN2 mechanisms for the cleavage of ethers with HI.

Solution

SN1 mechanism: If the departure of the leaving group (an alcohol) creates a relatively stable carbocation (e.g., a 3° carbocation), i.e., if R in ROR¢ is 3°, an SN1 reaction occurs. The steps involved are as follows: Step 1:

Protonation of ether oxygen by HI. R—O—R¢

+

H—I

R—O —R¢ + I

3.40

Organic Chemistry—A Modern Approach

Step 2: Departure of the leaving group R—O —R¢

R + (3°)

R¢OH

Step 3: Nuclephilic attack by I① on the carbocation R

+

I

R—I

SN2 mechanism: If departure of the leaving group would create an unstable carbocation (e.g., methyl cation or a primary carbocation), the leaving group cannot depart. It has to be displaced by the halide ion by an SN2 process. The halide ion preferentially attack that alkyl group which is less sterically hindered. The steps involved are as follows: Step 1: Protonation of ether oxygen by HI R—O—R¢

+

R—O —R¢

H—I

+

I

Step 2: I

+

R—O —R¢

SN2

R—I

+

R¢OH

(R = methyl or 1°)

10.

Predict the product(s) and suggest a mechanism for each of the following reactions of ether cleavage: (a)

+ HBr (excess) O

(b)

PhOCH2CH2CH(CH3 )CH2OCH2CH3 + HBr (excess) ææÆ

(c)

3 R — O — R ¢ ææææ Æ 2. H O

1. BBr 2

(R == CH 3 or 1∞) Solution

(a)

H—Br

O

O

Br

SN2

OH

—Br

H—Br

Br —

Br

+

— Br

— Br OH2 SN2 —H2O

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.41

(b)

(c)

11.

Explain why methyl propyl ether reacts with HI to yield 1-propanol and methyl iodide while methyl tert-pentyl ether is cleaved by HI to yield methanol and tert-pentyl iodide.

Solution If the presence of HI, the conjugate acid of a dialkyl ether (R — O — R) undergoes nucleophilic attack by I① on the less crowded alkyl group. It thus follows that the conjugate acid of methyl propyl ether undergoes SN2 attack by I① on the methyl carbon to give

methyl iodide and 1-propanal.

3.42

Organic Chemistry—A Modern Approach

Step 1:

CH3—O—CH2CH2CH3 + H—I

CH3—O—CH2CH2CH3 + I H

Methyl propyl ether CH2CH3

Step 2:

I

C —O—C

+

CH3—I

SN2

+

CH3CH2CH2OH

Methyl iodide less crowded centre

1-Propanol

more crowded centre

If, however, the protonated ether can cleave to give a stable carbocation, the reaction takes place by the SN1 mechanism. The conjugate acid of methyl tert-pentyl ether, therefore, cleaves to give methanol and stable tert-pentyl cation which subsequently reacts with I① to give tert-pentyl iodide. Step 1:

CH3—O—C(CH3)2CH2CH3 + H—I

Step 2:

CH3—O:—C(CH3)2CH2CH3 H

SN1

CH3—O—C(CH3)2CH2CH3 + I H

CH3OH + C(CH3)2CH2CH3 Methanol

A 3° carbocation (stable)

CH3

Step 3:

I

+

C(CH3)2CH2CH3

CH2CH2C—I CH3 tert-pentyl iodide

12.

Arrange the following alkyl bromides in order or decreasing reactivity as a substrate in an SN2 reaction and explain the order. Br

Br Br

Br I

II

III

IV

Solution To assess the reactivity of these alkyl bromides, we have to examine the steric hindrance to an SN2 reaction at the carbon bearing the leaving group. In I, it is 3° and, therefore, three groups would hinder the approach of a nucleophile towards the carbon bearing bromine. So, this alkyl bromide would react so slowly that it seems to be

3.43

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

unreactive. In II, the carbon bearing the leaving group is 2°, therefore, two groups would hinder the approach of the nucleophile. In III and IV, the carbon bearing bromine is 1° and so, only one group would hinder the nucleophile’s approach. Therefore, II is expected to react at a rate faster than I but slower than III and IV. Now, III has two methyl groups on the b-carbon (the carbon adjacent to the one bearing bromine), which provide more steric hindrance to the approaching nucleophile, while IV has only one alkyl substituent on the b-carbon which provides less steric hindrance to the approaching nucleophile. Hence, IV reacts at a rate faster than III. The order of decreasing reactivity, therefore, is IV > III > II > I. 13. Explain why each of the following ethers cannot be prepared by the Williamson synthesis: (a)

Me3C — O — CMe3

(b)

Me3C — O — CHMe2

(c)

Et2CH — O — CHEt2

(d)

Ph — O — Ph

(e)

MeCH == CH — O — CH == CHMe, and

(f)

Me3CCH2OCH2CMe3

Solution The Williamson ether synthesis consists of an SN2 reaction of a sodium alkoxide with an alkyl halide, alkyl sulfonate or alkyl sulfate. Because of steric reason, methyl and primary substrates are the best. Secondary and tertiary substrates are not suitable because steric hindrance causes them to undergo E2 elimination rather than substitution. Neopentyl type substrates are also inert due to steric reason. Aryl and vinyl halides are inert because backside attack does not take place due to repulsive interaction caused by the p electron cloud, and the C — LG bond in them is shorter and stronger because it possesses considerable double bond character due to electron delocalization.

Therefore, the given ethers cannot be prepared by the Williamson synthesis because in (a) both alkyl groups are tertiary, in (b) one alkyl group is tertiary and the other is secondary, in (c) both alkyl groups are secondary, in (d) both groups are aryl, in (e) both groups are vinyl and in (f) both groups are neopentyl. 14.

Predict the product and suggest a mechanism for the following reaction:

Solution The reaction of cis-1-chloro-3-methylcyclopentane (in which the carbon that bears the leaving group is chiral) with hydroxide ion (OH①) is an SN2 reaction which proceeds with complete inversion of configuration. The product is trans-3-methylcyclopentanol. The reaction occurs as follows:

3.44

Organic Chemistry—A Modern Approach

Cl H3 C H

Cl + OH H

SN2

H3C H

H OH

OH

cis-1-Chloro-3-methyl cyclopentane

15.

H3C H

H

trans-3-Methylcyclopentanol

Transition state

A cyclic ether can be prepared by an intramolecular Williamson synthesis. Give the mechanism of the following reaction and explain why it is carried out at high dilution. Cl

CH2

CH2

CH2

CH3

C OH

CH3

Na

CH3

high dilution

CH3

O

Solution The mechanism of the reaction is as follows:

The reaction is carried out at high dilution because under such conditions the intermolecular distance becomes greater than the intramolecular reactive centres and as a consequence, the cyclic compound is formed instead of polymeric product (a likely result in concentrated solution) as given below. Cl

Cl O Na

O Na

SN2

Cl

O ONa

etc.

16.

18–Crown–6 may be prepared by treating a mixture of triethylene glycol and the corresponding dichloride with aqueous KOH. Suggest a mechanism for the reaction.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.45

Solution The reaction proceeds through the SN2 mechanism and involves two successive displacements with chloride ion being the leaving group. Even though, a large ring is being formed, the reaction needs not be carried out at high dilution. After the initial alkylation, i.e., formation of ether linkage at one end, the K≈ ion apparently acts as a ‘template’ to bring the two reacting ends of the long chain close together for rapid reaction.

The mechanism of the reaction is as follows:

17.

Predict the product in each of the following SN2 reactions and give your reasoning: (a) (b)

(c)

1

4

1

PhS Na

Br

2

EtOH 5

3

I

4

PhS Na

I

2

EtOH

1

Br

5

3

Br

3 2

5 4

I

PhS Na EtOH

Solution (a) The dibromide undergoes ready SN2 attack by the nucleophile PhS① at the sterically less crowded primary site C-1. Nucleophilic attack does not take place at the sterically more crowded tertiary site C-4. Hence, the halogen at C-1 is more reactive than the halogen at C-4 in an SN2 reaction and the product corresponding to such displacement obtained.

3.46

Organic Chemistry—A Modern Approach

PhS /EtOH

Br

SN2

PhS

Br

(b)

I

PhS / EtOH SN2

PhS

I

+

I

Both of the carbons bearing halogen (i.e., C-1 and C-5) are primary sites and hence, they are equally favourable sites for SN2 attack from steric point of view. However, since I① is a better leaving group than Br①, the halogen at C-5 is more reactive. Nucleophilic attack, therefore, takes place preferably at C-5 to yield the corresponding product predominantly.

Br

18.

Br

Both C-1 and C-5 are primary sites and both are expected to be equally favourable for SN2 attack from steric point of view. However, the SN2 attack at C-1 is more favourable because the corresponding transition state is relatively more stable (stabilized by the adjacent double bond by p orbital overlap). Therefore, the iodine atom at C-1 is more reactive than the iodine atom at C-5 in an SN2 reaction and the product corresponding to that of the reactive centre is obtained predominantly.

I

(c)

+ Br

I

PhS / EtOH SN2

Br

S Ph +

I

Since SN2 reactions proceed with inversion of configuration, an (R)– compound always produce and (S)–compound and an (S)–compound always produce an (R)–compound. Comment on this statement and give example in favour of your answer.

Solution This statement is not correct. In an SN2 reaction of a chiral substrate, the relative configuration changes but the absolute configuration may or may not change since the absolute configuration depends on the priority of the groups or atoms present on the chiral carbon. For example, (S)-1-chloro-1-iodoethane reacts with Br① to give (R)1-bromo-1-chloroethane, while (S)-1-bromo-1-iodoethane reacts with Cl① to give (S)-1bromo-1-chloroethane.

3.47

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

19.

Comment on the stereochemical course involved in each step of the following reaction sequences and give three-dimensional structures (with R or S designation) for the compounds I — XIX: CH3

(a)

NaN

CH I

H2 3 3 ææææ Æ I æææ Æ II æææ Æ III Acetone Pt

C

Ph

Br H (S)-1-Bromo-1-phenylethane CH3

(b) H

CH CH O Na

HI 3 2 ææææææ Æ IV æææ ÆV CH CH OH

C

3

2

Br CH2CH2CH3

(R)-2-Bromopentane

CH COOK

KOH/ H O

TsCl 3 2 ææææææ Æ VI æææææ Æ VII æææææ Æ VIII Pyridine, 25∞ C Acetone D

(c)

H

(d)

C HO

PBr

CH3 C2H5

H SO

1. LiAlH

NaCN 3 2 4 4 ææææ Æ IX ææææ Æ X ææææ Æ XI æææææ ≈ Æ XII Pyridine DMSO H O 2

2. H3O

(R)-2-Butanol OH

(e)

CH3(CH2)4

C H

CH3 (S)-2-Heptanol NaN

H / Pt

NaNO / ACOH

TsCl 3 2 2 ææææ Æ XIII ææææ Æ XIV ææææ Æ XV ææææææ Æ XVI + XVII Pyridine Acetone 0 - 5∞ C

3.48

Organic Chemistry—A Modern Approach

H

(f)

C

H3C

Dioxan æææææ Æ XVIII + XIX H O/ NaN

OBs

2

3

(CH2)5CH3 (S)-2-Octyl brosylate Solution (a) Step 1: Being a secondary substrate, (S)-1-chloro-1-phenylethane reacts ① with the powerful nucleophile N3 (azide ion) in the aprotic polar solvent acetone by the SN2 mechanism to yield (R)-1-azido-1-phenylethane (I) with inversion of configuration.

CH3

CH3

N3

NaN3 / acetone

C

Ph

Br

(inversion of configuration)

C :N==N==N

Ph + Br

H I (R)-1-Azido-1-phenylethane

H (S)-1-Bromo-1-phenylethane

Step 2: The azide I on catalytic hydrogenation produces the corresponding amine II. Since the bond between the chiral carbon and nitrogen is not cleaved in this reaction, the configuration of the chiral centre is retained. CH3

CH3 H2 / Pt

C

Ph

:N==N==N

C

(retention of configuration)

H

H2 N

Ph + N ≠ 2

H II (R)-1-Amino-1-phenylethane

I

Step 3: Since methylation of the amine II does not involve reaction at the chiral carbon, no change in configuration takes place in this step. CH3 CH3 CH3—I

C

Ph

H2N H II

SN2 (retention of configuration)

C CH3NH

Ph + HI

H III (R)-N-Methyl-1-phenyl-1-ethanamine

(b) Step 1: When (R)-2-bromopentane is treated with ethoxide ion in ethanol, it undergoes an SN2 reaction to yield (S)-2-ethoxypentane with inversion of configuration.

3.49

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

CH3

CH3

C2H5O

C2H5O /C2H5OH

C

H

Br H

(R)-2-Bromopentane

SN2 (inversion of configuration)

C C2H5O

H

+ Br

CH2CH2CH3 IV (S)-2-Ethoxypentane

Step 2: In this step, the ether IV undergoes cleavage by HI. The conjugate acid of the ether undergoes SN2 attack at the less crowded methylene carbon of the ethyl group. In the resulting alcohol V, the configuration is retained.

(c) Step 1: Since tosylation of this chiral alcohol does not involve reaction at the asymmetric carbon no change in configuration occurs in this step.

Step 2: Being a primary substrate the tosylate VI undergoes SN2 displacement of OTs① (a good leaving group) by CH3COO① in acetone to yield (R)-1,2-dideuterioethyl acetate (VII) with inversion of configuration.

3.50

Organic Chemistry—A Modern Approach

Step 3: The alkaline hydrolysis of the acetate VII is known not to involve cleavage of the bond joining the acetate group to the chirality centre. Therefore, the alcohol VIII must have the same configuration as the acetate VII.

(d) Step 1: (R)-2-Butanol reacts with phosphorous tribromide (PBr3) to yield a phosphite ester with retention of configuration. The phosphite ester being a secondary substrate with a good leaving group (phosphoryl group, P == O) undergoes SN2 attack by Br① (a good nucleophile to yield (S)-2-bromobutane (IX) having configuration opposite to that of the phosphite ester. H C H5C2

H + OH

CH3

H N

PBr2 Br

C H5C2

+

O CH3

H

–

P Br2 Br

H5C2

O CH3

–

+

C

PBr2

Br +

N H

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.51

Step 2: The bromide IX (a secondary substrate) reacts with CN① (a very powerful nucleophile) in the aprotic polar solvent DMSO by an SN2 mechanism to give (R)-1-methylbutanenitrile (X) with inversion of configuration.

Step 3: Hydrolysis of the nitrile X produce (R)-2-methylbutanoic acid (XI). Since the bond joining the — CN group to the chiral carbon is not cleaved during hydrolysis, the configuration of the chirality centre is preserved.

Step 4: The reduction of the carboxylic acid XI by LiAlH4 to yield the corresponding 1° alcohol does not involve reaction at the chiral carbon. Hence, the configuration of the chirality centre remains unchanged in this case also.

(e) Step 1: Since tosylation of the alcohol does not involve reaction at the chiral carbon, the tosylate XIII has the same configuration as the original alcohol.

3.52

Organic Chemistry—A Modern Approach

Step 2: The tosylate XIII undergoes SN2 attack by the powerful nucleophile N3① (azide ion) in the aprotic polar solvent acetone to yield (R)-2-azidoheptane (XIV) with inversion of configuration. H N3 C

H O O—S—

H3C

(CH2)4CH3 O

— CH3 Cl

NaN3 / acetone

C

CH3

:N∫∫N—N

SN2 (inversion of configuration)

+ OTs

(CH2)4CH3 XIV (R)-2-Aziodoheptane

XIII

Step 3: The azide XIV on catalytic hydrogenation produces (R)-2-Heptanamine (XV). Since hydrogenation does not involve cleavage of the C — N bond, the configuration remains unchanged. H H C :N∫∫N—N

H2 / Pt

CH3

(CH2)4CH3 XIV

(retention of configuration)

C H2N

CH3 (CH2)4CH3

XV (R)-2-Heptanamine

Step 4: When the aliphatic amine XV is treated with nitrous acid (obtained from NaNO2 plus HCl), an unstable diazonium salt is obtained which readily dissociates to form a carbocation. The carbocation undergoes nucleophilic attack by water from either side with equal facility to give, after proton loss, an equimolar mixture of (R)- and (S)-2-heptanol.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.53

(f) (S)-2-Octyl brosylate undergoes SN2 attack by the more nucleophilic solvent dioxane to give an intermediate oxonium ion. The unstable intermediate reacts readily with water in another SN2 process to from (S)-2-octanol (XVIII) with the same configuration as the starting brosylate. The azide ion reacts similarly with the oxonium ion to yield (S)-2azidooctane (XIX). In both the cases, the retention of configuration has been brought about by two successive inversions.

3.54

20.

Organic Chemistry—A Modern Approach

Arrange the following compounds in order of decreasing reactivity �� :* in an S 2 reaction carried out in methanol: towards CH 3 O N �� CH3Cl, CH3I, CH3OH, CH3F, 14CH3OSO2CF3, CH3Br

Solution The leaving groups of the substrates can be arranged in order of increasing leaving ability as follows: �� * �� :* < : Br: �� :* �� :* < : Cl �� * < : ��I :* < CF SO O : OH <:F 3 2 �� �� �� basic) �� �� �� (least basic) (most

A better leaving group makes the substrate more reactive is an SN2 reaction. Therefore, the given compound can be arranged in order of decreasing SN2 reactivity as follows: 14

CH3OSO2CF3 > CH3I > CH3 Br > CH3Cl > CH3 F > CH3OH (most reactive) (least reactive)

21.

Explain the observed effects of increasing solvent polarity (solvent ionizing power) on the rates of the following SN2 reactions: �� * Æ CH OH + : Cl �� :* (a small decrease in the reaction rate) (a) CH3Cl + : OH 3 �� �� ≈ ≈ �� �� (b) CH 3 — N(CH 3 )3 + H 2S: Æ CH 3 — SH 2 + N(CH (a large increase in the reaction 3 )3 �� rate) ≈

* �� :��: (a large increase in the reaction rate) CH 3I + N(CH 3 )3 Æ CH 3 — N(CH 3 )3 + I �� ≈ * �� �� (d) CH 3 — N(NH 3 )3 + :OH Æ CH 3OH + N(CH 3 )3 (a large decrease in the reaction �� rate) Draw a comparative energy profile diagram to show the effect of solvent on the rate of attack by OH① ion on CH3Cl.

(c)

Solution (a) This SN2 reaction proceeds through the transition state as follows:

HO

+

CH3—Cl

HO

CH3

Cl

HO—CH3 + Cl

T.S.

(b)

In this reaction, an initial charge is dispersed in the transition state. Therefore, increasing solvent polarity stabilizes the reactants (specifically the negatively charged nucleophile) more than the transition state. Because of this differential stabilization, the difference in energy between the transition state and the nucleophile increases, i.e., the energy of activation (Ea) increases. The net result is that there is a slight decrease in the reaction rate. This SN2 reaction proceeds through the transition state as follows:

3.55

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(c)

In this reaction also, an initial charge is dispersed in the transition state. Therefore, increasing solvent polarity stabilizes the reactants (specifically the positively charged substrate) more than the transition state. As a result of this, the difference in energy between the transition state and the substrate increases, i.e., the energy of activation (Ea) increases. The net result is that there is slight decrease in the reaction rate. The reaction proceeds through the transition state as follows: N(CH3)3 + CH3—I

(CH3)3N

CH3

I

(CH3)4 N + I

T.S.

(d)

As both the substrate and the nucleophile are neutral, a charge is expected to be developed and concentrated in the transition state. Solvation is, therefore, much greater for the transition state than for the reactants. Increasing solvent polarity stabilizes the transition state much more than the reactants and thereby decreases the energy of activation (Ea) considerably. The net result is a large increase in the reaction rate. The reaction proceeds through the transition state as follows:

Both of the reactants are charged and the initial charges are decreased in the transition state. Increasing solvent polarity, therefore, stabilizes the reactant much more than transition state and thereby increases the energy of activation (Ea) considerably. The net result is a large decrease in the reaction rate. A comparative energy diagram showing the effect of solvent on the rate of the alkaline hydrolysis of CH3Cl is as follows:

3.56

22.

Organic Chemistry—A Modern Approach

Which one would you expect to be the stronger nucleophile in ethanol and why? (a)

�� * or NH �� :NH 2 3

(b)

(c)

�� or Ph P �� Ph3N 3

(d)

(e)

�� CH 3NH 2 or CH 3 NH 3

(g)

≈

—O or

(f)

�� :* or tert-C H O �� :* n-C4 H9 O 4 9 �� �� * * �� �� : : CH 3O or CH 3S �� �� * �� �� :* : p-CH3OC6 H4 O or p-NO2C6 H4 O �� ��

—CH2O

Solution (a) Anions are always stronger nucleophiles than their conjugate acids. The anion �� �� * is, therefore, more nucleophilic than its conjugate acid NH :NH 3. 2

(b)

Because of steric hindrance, tert-butoxide ion is a weaker nucleophile than n-butoxide ion, even though the former ion is a stronger base. The carbon adjacent to the negative oxygen atom is relatively less crowded in n-butoxide than in tertbutoxide. Consequently, the approach of n-butoxide towards C 3 in an SN2 reaction sp

is more easier than the approach of the very bulky tert-butoxide and therefore, the activation energy of the reaction involving n-butoxide is relatively low. Hence, n-butoxide ion is a stronger nucleophile than tert-butoxide ion. CH3CH2CH2

H3C C——O

C——O H3C

H

CH3

H n-Butoxide ion (stronger nucleophile)

(c)

(d)

(e)

tert-Butoxide ion (weaker nucleophile)

Phosphorous is less electronegative and larger in size than nitrogen; so, it is �� is relatively less solvated than Ph N �� in the relatively more polarizable. Also, Ph 3 P 3 protic solvent. For these reasons, Ph3P (triphenylphosphine) is more nucleophilic than Ph3N (triphenylamine). Because of low electronegativity and larger size, sulphur is more polarizable than oxygen. Furthermore, a protic solvent solvates CH3O① (through H-bonding) better than CH3S①. The more polarizable and weakly solvated CH3S① is, therefore, a better nucleophile than CH3O①. �� Due to the presence of an unshared pair of electrons on nitrogen, CH3 NH can act 3 ≈

as a nucleophile. However, CH3 NH3 is nonnucleophilic because the unshared pair of electron on nitrogen is involved in the formation of a coordinate bond and is no longer available to exhibit nucleophilic activity.

3.57

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

CH3 NH2 + H

(f)

(g)

23.

CH3N

H

or CH3NH3

The stronger base is expected to be the better nucleophile because the attacking atom is the same in these two anions. In p-CH3OC6H4O①, the CH3O-group increases the electron density of the phenoxy oxygen by its +R effect (by inhibiting the interaction of its negative charge with the delocalized p electrons of the ring). However, in p-NO2C6H4O①, the nitro group reduces the electron density of the negative oxygen by its — R and — I effects. The more stable anion p-NO2C6H4O① is, therefore, less basic and hence, less nucleophilic than p-CH3OC6H4O①. In phenoxide ion (Ph — O①), the negative charge on oxygen is delocalized with the ring p electron. However, in the alkoxide of benzyl alcohol, i.e., in PhCH2O①, the negative charge cannot be delocalized with the ring p electrons due to the presence of saturated carbon between oxygen and the ring. Therefore, PhO① is less basic than PhCH2O① and hence, PhO① is less nucleophilic than PhCH2O①. Arrange the following ions/molecules in decreasing order of nucleophilicity in protic solvents. Give reasons. (a) (c)

*

�� : : CH* : �� * : 3 , OH , N H2 , H2 O �� �� HOO �� :* , : OH �� * H2O, �� �� ��

�� :* , : OH �� * , C H O �� * (b) CH3 CH2COO 2 5 : �� �� �� * * �� : , : OH �� �� :* , CH COO �� :* (d) CH3OH, PhO , CH3S 3 �� �� �� ��

Solution (a) Since the atoms across a period of the periodic table have approximately the same size, the basicity and nucleophilicity run in parallel (stronger bases are better

nucleophiles). Since the basicity of the given anions (the attacking atoms are *

: �� : �� * elements of Period 2) decreases in the order :CH * 3 > NH 2 > OH , the nucleophilicity �� decreases in the same order. Again, a negatively charged nucleophile is always a more reactive nucleophile than its conjugate acid. Thus, OH① is a better nucleophile than H2O. Hence, the decreasing order of nucleophilicity of the given species is @

(b)

�� : : �� : �� * :CH * 3 > NH 2 > OH > H 2O . �� For reagents having the same attacking atom, nucleophilicity and basicity run in parallel, i.e., stronger bases are better nucleophiles. A stable anion (the conjugate �� :* base of a strong acid) is less basic and is, therefore, less nucleophilic. In C2 H5 O �� ion, the negative charge on oxygen is intensified by the +I effect — C2H5 group; �� * ion. Because of charge this ion is, therefore, less stable and more basic than :OH �� * �� : ion is much �� * delocalization by resonance, the C2 H5COO more stable than :OH �� �� * �� and so, it is less basic and less nucleophilic than :OH . Hence, the nucleophilicity ��

3.58

(c)

(d)

Organic Chemistry—A Modern Approach

�� :* > :OH �� * > C H COO �� :* of these ions decreases in the order C2 H5 O 2 5 �� �� �� Although the attacking atom is the same, nucleophilicity and basicity do not run in parallel in this series. Bases are better nucleophiles than their conjugate acids. �� * ion is more nucleophilic than H O �� : . Again, when the attacking Therefore, :OH 2 �� atom is bonded to an atom containing an unshared electron pair, the nucleophilicity �� �� :* is more nucleophilic than of the species increases (a-effect). Therefore, HOO �� �� * , even though it is less basic. Hence, the ��nucleophilicity :OH of these species �� * * �� �� �� �� decreases in the order HOO : > HO : > H 2O:. �� �� �� �� :* Since the sulphur atom is larger and more polariable than the oxygen atom, CH3S �� is the strongest nuclephile among these five species. CH3OH, containing neutral oxygen atom, is the poorest nucleophile. The decreasing order of basicity of the three �� :* > PhO �� :* > CH COO �� :* (a other reagents having the same attacking atom is OH 3 �� �� �� carboxylic acid is more acidic than phenol which in turn is more acidic than water). �� :* > PhO �� :* > CH COO �� :*. Therefore, the decreasing order of nucleophilicity is OH 3 �� �� �� Thus, the given species can be listed in the order of decreasing nucleophilicity in a protic solvent as follows: �� :* �� :* > PhO �� :* > CH COO �� :* > �� CH3S > HO CH3OH 3 �� �� �� �� �� (most nuclephilic) (least nucleophilic)

24.

Imidazole increases the rate of hydrolysis of phenyl acetate. Explain this observation. O

O

C

C

CH3

OH + PhOH (slow reaction)

CH3

OPh + H2O O

O

C CH3

:N

OPh + H2O

Phenyl acetate

C

NH

Imidazole

CH3

OH + PhOH (fast reaction)

Acetic acid

Phenol

Solution Imidazole is a better nucleophile than water, so it reacts at a rate faster than water with the ester phenyl acetate. The resulting acyl imidazole is particularly reactive because the positively charged nitrogen makes imidazole a very good leaving group. Therefore, it undergoes hydrolysis much more rapidly than the ester. Because the formation of acyl imidazole and its subsequent hydrolysis are both faster than the hydrolysis of ester, imidazole increases the rate of the overall reaction. The mechanism of

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.59

the reaction is as follows:

   

25.

The rate of SN2 reaction CH3CH2Cl + OH① CH3CH2OH + Cl① is given by the following equation: rate = k [CH3CH2Cl] [OH①] What happens to the rate of the reaction under each of the following conditions? (a) [CH3CH2Cl] is halved and [OH①] is doubled. (b) [CH3CH2Cl] is tripled and [OH①] remains the same. (c) [CH3CH2Cl] is halved and [OH①] stays the same. (d) Both [CH3CH2Cl] and [OH①] are doubled.

Solution (a) The rate expression for the reaction involving half the concentration of CH3CH2Cl and double the concentration of hydroxide ion is

[CH 3CH 2Cl] ¥ 2 [OH * ] 2 = k[CH 3CH 2Cl][OH * ]

rate = k

(b)

(c)

Hence, there will be no change of the reaction rate. The rate expression for the reaction involving triple the concentration of CH3CH2Cl is rate = k. 3[CH3CH2Cl] [OH①] = 3 k[CH3CH2Cl] [OH①] Thus, the rate increases by a factor of three. The rate expression for the reaction involving half the concentration CH3CH2Cl is rate = k ◊

[CH 3CH 2Cl] ◊ [OH * ] 2

3.60

(d)

26.

Organic Chemistry—A Modern Approach

The rate thus becomes half of the original rate. The rate expression for the reaction involving double the concentration CH3CH2Cl as well as OH① is rate = k ∙ 2 [CH3CH2Cl] ∙ 2[OH①] = 4 k CH3CH2Cl] [OH①] Thus, doubling the concentration of both reactants increases the rate by a factor of four. The nucleophilicity of F① in a weakly polar solvent like acetone increases in the order: LiF < KF < CsF — Why?

Solution There is considerable ion pairing in a weakly polar solvent. As the alkali metal cation gets smaller, ion-pairing to F① gets stronger and a result, its nucleophilicity decreases. Since Cs≈ ion is the largest and Li≈ ion is the smallest cation, F① in CsF is the most nucleophilic and F① in LiF is the least nucleophilic, i.e., the nucleophilicity of F① increases in the order: LiF < KF < CsF.

27.

List the following ions in decreasing order of leaving ability. Give your reasoning. O H3C—

—S—O

,

—O

,

O2N—

—O

O Solution The better leaving groups are those that become either a relatively stable anion or a neutral molecule (i.e., an weaker base) when they depart. p-Toluensulfonate ion (OTs①) is more stabilized by resonance than p-nitrophenoxide ion, which in turn is more resonance–stabilized than the phenoxide ion. Therefore, the leaving ability of these anions decreases in the order: p-CH3C6H4SO3① > p-NO2C6H4O① > C6H5O①.

28.

Methylation of propaonate ion (CH3CH2COO①) can be carried out smoothly by treating it not with methyl iodide, CH3I, but with dimethyl sulfate, Me2SO4. Explain these observations.

Solution Methyl sulfate ion, CH3OSO3①, being a very weak base, acts as a much better leaving group than I①. Because of this, methylation of propanoate ion occurs smoothly when treated with Me2SO4 than that with CH3I.

3.61

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

29.

Suggest a mechanism for each of the following reactions (methylation by diazomethane): (a)

ether CH 3CH 2COOH + CH 2N2 æææ Æ CH 3CH 2COOCH 2 + N2 0∞ C Propanoic acid Diazomethane Methyl propanoate

OCH3

OH

(b)

+ Phenol

(c)

CH2N2

ether

+ N2

0°C

Diazomethane

CH3CH2OH + Ethanol

Anisole

Al(OEt)3 CH2 N2 ææææææ Æ (a Lewis acid)

CH3CH2OCH3 + N2 Ethyl methyl ether

Explain why methylation of the alcohol requires a Lewis acid. Solution

(a)

O O CH3CH2—C—O—H + :CH2—N∫∫N Propanoic acid

C ether 0°C

CH3CH2

O + CH3—N∫∫N —N2 SN2

O C CH3CH2 OCH3 Methyl propanoate

Diazomethane methylates is a compound only in the presence of a proton source. ≈

CH3 — N ∫∫ N is a very reactive methylating agent because of having an excellent leaving group, gaseous N2.

3.62

Organic Chemistry—A Modern Approach

(b)

(c)

Because of low acidity, alcohol cannot supply a proton to diazomethane. However, when alcohol forms a complex with the Lewis acid Al(OEt)3, its hydroxyl proton becomes considerably acidic and then methylation occurs smoothly. 30. For each of the following pairs of SN2 reaction indicates which takes place with the larger rate constant: (a)

(CH3 )2CHCH2 Br + OH* or CH3CH2 CH Br + OH* | CH3

(b)

�� �� :* CH3CH2 Br + EtOH or CH3CH2 Br + EtO �� �� �� :* or CH CH Cl + CH O* (both in ethanol) CH3CH2Cl + CH3S 3 2 3 �� * * CH3CH2Cl + : ��I : or CH3 Br + : ��I : �� �� �� �� I CH2CH2CH2 NH or I CH CH 2 2 2 CH2 CH2 CH2 NH2

(c) (d) (e) (f) (g) (h)

�� :* or PhBr + MeO �� :* Ph CH2 Br + MeO �� �� * �� �� * : CH3CH == CBrCH3 + OH or CH3CH == CHCH BrCH3 + : OH �� �� CH3CH2CH2 Br (1.0 M) + MeO* (1.0 M) or CH3CH2CH2 Br (1.0 M) + MeO* (2.0 M)

Solution (a) (CH3 )2CHCH2 Br + OH* (Because the substrate is 1∞ and there is only two substituents at the b-position, i.e., the reaction is strerically more feasible.) �� :* (The anion is more nucleophilic than its conjugate acid.) (b) CH3CH2 Br + EtO ��

3.63

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(c)

�� :* [The more polarizable (less electronegative and larger in CH3CH2Cl + CH3S �� size) sulphur atom is more nucleophilic than less the polarizable oxygen.]

(d)

① CH3 Br + I* (CH3Br is a relatively less crowded substrate and Br is a better leaving group compared to Cl①.) �� I(CH2 )5 NH 2 (The intramolecular SN2 reaction leads to the formation of relatively strain free six-membered ring.)

(e) (f)

PhCH2 Br + MeO* (Since the substrate is a benzylic halide, the unhybridized p

(g)

orbital involved in the transition state interacts with the p orbital system of the ring and as a result, the transition state is stabilized. Bromobenzene does not undergo backside attack due to repulsive interaction.) Br | CH3CH == CH — CH — CH3 + OH* (Since the substrate is an allylic halide, the unhybridized p orbital involved in the transition state interacts with the p orbital system of the double bond and as a result, the transition state is stabilized. CH3CH == CBrCH3 being a vinylic halide does not undergo SN2 attack from the backside due to repulsive interaction.)

(h)

�� :* (2.0 M) (The concentration of the nucleophile is CH3CH2CH2 Br(1.0 M) + MeO �� higher.)

31.

Explain why each of the following SN2 reaction is unsuccessful: �� :* �� :* + CH — O — CH (a) EtS CH3 — S — Et + CH3O 3 3 �� �� �� * (b) : CN* + CH3CH2 C(CH3 )2 Br CH3CH2 C(CH3 )2 CN + : Br: �� * * �� CH3CH2CH2OH + : H (c) :OH + CH3CH2CH3 �� * CH3I + : CH3* (d) : ��I : + CH3 — CH3 �� (e) (f) (g)

*

�� * + �� : OH HOCH2CH2CH2 CH 2 �� * �� �� * PhS: + CH3CH2CH == CH — Br CH3CH2CH == CH — SPh + : Br: �� �� ≈

�� CH3 NH 2 + CH3 CH2 OH2

≈

CH3 NH2CH2CH3 + H2O

Solution

(a) (b)

Since the very strong base CH3O① is a very poor leaving group, it cannot be displaced �� :* . by EtS �� of steric hindrance, this tertiary substrate does not undergo backside Because attack required in an SN2 reaction.

3.64

Organic Chemistry—A Modern Approach

(c)

The reaction does not take place because the hydride ion, :H① (a very strong base), is a very poor leaving group.

(d)

Since methyl anion, :CH3 , is a very strong base and hence, a very poor leaving

(e) (f)

(g)

*

* group, it cannot be displaced by : ��I : . �� The reaction does not a take place because a carbanion (a very powerful base) is a very poor leaving group. �� :* because the p electron This vinylic substrate does not undergo SN2 attack by PhS �� cloud of the double bond repels the nucleophile and inhibits its approach from the backside of the leaving group. ≈ �� Since CH3 NH from the 2 is more basic than CH3CH2OH, it will take up H *

conjugate acid of ethanol, i.e., CH3CH2 OH2 , and instead of an SN2 reaction, an acid-base reaction will take place. CH3NH2 + H—O—CH2CH3 H

32.

CH3NH3 + CH3CH2OH

Which SN2 reaction of each pair is expected to take place at a faster rate? Give your reasoning. (a)

(b)

(c)

≈

�� ææÆ CH CH CH NPh + : Br �� :* CH3CH2CH2 Br + Ph 3 N 3 2 2 3 �� ≈ �� ææÆ CH CH CH PPh + : Br �� :* CH3CH2CH2 Br + Ph 3 P 3 2 2 3 �� EtOH/ H2O �� :* �� :* æææææ Æ CH3CH2SCN + : Cl CH3CH2Cl + SCN �� EtOH/ H2O �� :* �� :* æææææ CH3CH2Cl + SCN Æ CH3CH2 NCS + : Cl �� H2 O CH3I + NaOH æææ Æ CH3OH + NaI H O

2 CH3I + CH3COONa æææ Æ CH3 COOCH3 + NaI Br SPh

(d)

+ PhS

+ Br

+ PhS

SPh + Br

Br

(e)

CH3OH �� :* CH3CH2CH2CH2 Br + N3* ææææ Æ CH3CH2CH2CH3 N3 + : Br �� CH3CN �� :* CH3CH2CH2CH2 Br + N3* ææææ Æ CH3CH2CH2CH3 N3 + : Br ��

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(f)

3.65

acetone �� :* CH3CH2 Br + CN* ææææ Æ CH3CH2CN + : Br �� acetone �� :* CH3CH2 Br + CN* ææææ Æ CH3CH2 NC + : Br ��

Solution

(a)

(b)

(c)

(d)

(e)

(g)

33.

The second reaction is expected to take place more rapidly because �� , is a stronger nucleophile than triethylamine, Ph N �� triphenylphosphine, Ph 3 P 3 (larger and less electronegative phosphorus atom is more polarizable than nitrogen atom). SCN① is an ambident nucleophile in which sulphur and nitrogen are two nucleophilic sites. Its less electronegative and larger sulphur atom is more polarizable (soft base) while its more electronegative and smaller nitrogen atom is less polarizable (hard base). Attack by SCN① on the carbon of CH3CH2Cl (soft acid) thus occurs at a faster rate through sulphur (favourable soft–soft interaction) than through nitrogen (unfavourable hard–soft interaction). Therefore, the first reaction will take place at a faster rate. Since OH① is a stronger base than CH3COO①, it is a better nucleophile and therefore, the first reaction is expected to take place at a faster rate than the second reaction. Since the SN2 transition state involving cyclopentyl bromide is stable (less angle strain) compared to that involving cyclobutyl bromide (more angle strain), the second reaction is expected to take place at a faster rate than the first reaction. The polar aprotic solvent CH3CN does not solvate N3① ion through hydrogen bonding while it solvates the cation and thereby makes the nucleophile much freer to react. The protic polar solvent CH3OH solvates N3① by forming hydrogen bonds and thereby lowers its reactivity. So, the second reaction is expected to take place at a faster rate. :CN① is an ambident nucleophile. Its more polarizable carbon atom is the soft centre while its less polarizable nitrogen atom is the hard centre. The C atom of — CH2Br is a soft centre. Since the soft–soft interaction takes place rapidly than the hard–soft interaction, the first reaction is expected to take place at a faster rate. 1–Bromobutane (CH3CH2CH2CH2Br) can be prepared by heating 1–butanol (CH3CH2CH2CH2OH) with a mixture of NaBr and H2SO4. However, the reaction fails in the absence of H2SO4. Explain.

Solution The very bad leaving group OH① (a stronger base) cannot be displaced directly by Br①. So, 1-bromobutane cannot be prepared from 1-butanol by treating it with NaBr. However, in the presence of H2SO4, the — OH group undergoes protonation and becomes ≈

converted to a — OH 2 group. Since H2O is a very much better leaving group (as it is a very weak base), it can be displaced readily by Br① to yield 1-bromobutane.

3.66

34.

Organic Chemistry—A Modern Approach

Apply Corey–House method to synthesize compound A using two suitable substrates having four carbon and three carbon units respectively. Give the possible routes. Which route will you choose and why? (CH 3 )3 CCH 2 CH 2 CH 3 (A)

Solution The two possible routes for the synthesis of the compound A by Corey–House method are as follows:

Route 1: CH3CH2CH2I

Li

CuI

CH2CH2CH2Li

(CH3CH2CH2)2CuLi (CH3)3CCl SN2

(CH3)3CCH2CH2CH3

Route 2: (CH3)3CI

Li

(CH3)3CLi

CuI

(a 3° substrate)

(CH3)3C 2 CuLi SN2

(CH3)3CCH2CH2CH3

CH3CH2CH2Br (a 1° substrate)

A

The SN2 reaction in step III of route 1 involves a tertiary substrate and so, because of steric hindrance, the reaction does not take place. That is, no desired product A will be obtained through this route. On the other hand, the SN2 reaction in step III of route 2 involves a primary substrate and so, because of small steric hindrance, the reaction takes place smoothly to give the desired compound A in good yield. 35. Predict the products A, B and C obtained in the following two routes: (CH ) CHBr

NaNH

Route 1:

3 2 2 CH 3C ∫∫ CH ææææ Æ A ææææææ ÆB liq. NH liq. NH

Route 2:

NaNH2 3 (CH 3 )2 CHC ∫∫ CH ææææ Æ C ææææ ÆB liq. NH liq. NH

3

3

CH Br

3

3

3.67

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Are both the routes equally efficient for the preparation of the compound B? Explain your answer. Solution The three compound A, B and C are as follows: *

*

≈ ≈ �� Na �� Na A = CH 3C=C ; B =CH 3C ∫∫ C— CH(CH 3 )3 ; C= (CH 3 )2CHC ∫∫ C

Both the routes are not equally efficient for the preparation of B. Route 2 is more efficient because its 2nd step (an SN2 reaction which is very susceptible to steric effect) involves a relatively unhindered substrate methyl bromide (CH3Br) while the 2nd step of Route 1 involves a sterically hindered secondary substrate isopropyl bromide (Me2CHBr). 36. Complete the equations (products with structures) and assuming all but one of the steps is SN2, label each of the product with D or L. COOH ?

NaOH

Br

H

NaN3

?

H2 Ni

?

CH3 Solution

O

37.

Complete the following: MeCH——CH2 + Me2CH MgBr

1. ether 2. H3O

Solution SN2 attack by the carbanion (from the Grignard reagent) takes place at the less crowded epoxy carbon (the methylene carbon) to form the corresponding halomagnesium alkoxide. This is subsequently converted to the corresponding alcohol by the addition of dilute mineral acid.

3.68

Organic Chemistry—A Modern Approach

O MeCH——CH2 + Me2CH—MgBr

OMgBr SN2 attack at the less hindered carbon of the epoxide

MeCH—CH2CH2Me2 H 3O

OH MeCHCH2CHMe2 4-Methyl-2-pentanol

38.

Give methods for the following conversions: (R)-2-Ethoxypentane (R)-2-Pentanol — (S)-2-Ethoxypentane

Solution

3.69

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

39.

Compare the yields of the following two routes: (a) (CH3)2CH ONa + PhCH2Cl (CH3)2CH—O—CH2Ph Benzyl isopropyl ether

(b) (CH3)2CHCl + PhCH2ONa

Solution The route (a) gives better yield of benzyl isopropyl ether because this SN2 reaction (Williamson synthesis) involving a less crowded primary alkyl halide is relatively more facile from steric point of view. 40. Explain why 2,2,5,5-tetramethyl-3-hexyne cannot be prepared by using an acetylide anion. Solution The alkyl halide and the acetylide anion required for the synthesis of 2,2,5,5*

≈ �� Na. tetramethyl-3-hexyne (Me3C — C ∫∫C — CMe3) are Me3CBr and Me3CC ∫∫ C However,

an SN2 reaction involving a tertiary alkyl halide does not take place because of steric hindrance caused by three alkyl groups.

41.

Predict the product of the following reaction and explain:

Solution The epoxide undergoes SN2 attack by the acetylide ion at the less crowded ring carbon to give an alkoxide. The reaction occurs with inversion of configuration around the stereogenic centre. The alkoxide on subsequent protonation by water produces the corresponding alcohol.

+ :C∫∫ CH

O

CH3

H

SN2 (inversion of configuration)

O

H

H—OH

—— CH3

OH

H + OH

———— C∫∫ CH

CH3

C∫∫ CH

3.70

42.

Organic Chemistry—A Modern Approach

How can the following transformation be carried out using a 2,4,6triphenylpyrylium salt? H H

D

C

C

D

I

NH2 CH3

CH3

Solution The amine is converted to a pyridinium salt by treating with 2,4,6– tiphenylpyrylium iodide and as a result of this, — NH2 is converted into a good leaving group. When the salt is heated, the counterion I① acts as a nucleophile to give the desired compound with inversion of configuration.

43.

Explain the following observations: CH2CH2Cl OH

OH

I

A CH2CH2OH Cl

II

O

EtOH

CH2CH2OH

OH

CH2CH2OH +

EtOH

B

C

Solution The compound I is first converted into the corresponding alkoxide which then undergoes an intramolecular SN2 reaction (the carbon bearing chlorine is primary) to yield the spirocyclic compound A.

3.71

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Similar intramolecular SN2 reaction leading to the formation of the same spirocyclic compound does not take place in the case of the compound II because the carbon bearing chlorine is tertiary and the backside attack on it is not sterically feasible. As a consequence, the compound II undergoes elimination (E2) reaction to yield two isomeric alkenes B and C.

44.

Predict the product of each of the following reactions: (a)

Me3C

1. CH3O

O

2. H2O

(b)

1. CH3O

Me3C

2. H2O

O

Solution Epoxide ring fused with six-membered ring opens in a trans diaxial fashion. Thus, the nucleophile attacks from the axial direction to give a leaving group that also axial. Because of this, only one product is formed in each of the given reactions, even though both ends of the epoxide are equally substituted.

(a)

3.72

Organic Chemistry—A Modern Approach

(b)

45.

Which of the following two isomeric halocyclohexane derivatives (I and II) would you expect to be more reactive in an SN2 reaction and why? CH3 H

CH3

H Br

H I

CH3 Br H

H

CH3

H

II

Solution The two relatively bulky methyl groups lock the isomer I in that conformation which places bromine in the equatorial position. Since the axial approach of a nucleophile for an equatorial leaving group is sterically hindered due to the axial hydrogens at C-3 and C-5, relative to the carbon (C-1) undergoing substitution, displacement of bromine occurs very slowly.

On the other hand, the two relatively bulky methyl groups lock the isomer II in that conformation which places bromine in the axial position. In this case, the approach of the nucleophile is not sterically hindered and also, the ground state energy of this isomer is higher (i.e., Ea is lower) than the isomer I. For these reasons, this isomer reacts with the nucleophile by the SN2 mechanism readily. The cyclohexane derivative II is, therefore, more reactive than I in an SN2 reaction.

3.73

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

46.

Potassium permanganate (KMnO4), a common oxidizing agent, is insoluble in a nonpolar solvent because it is ionic. How, then, can KMnO4, oxidize compounds that are soluble only in nonpolar solvents?

Solution KMnO4 can be dissolved in a nonpolar solvent like benzene if 18-crown-6 is added to the reaction mixture. The crown ether binds K≈ in its cavilty and the crown ether-potassium complex with a nonpolar hydrocarbon-like exterior dissolves in benzene. In order to maintain electrical neutrality, the permanganate ion (MnO4①) accompanies the complexed K≈ ion into the benzene layer, permitting MnO4① to act as an oxidizing agent in the nonpolar solvent.

47.

Predict the product of each of the following reactions and explain the stereochemical course involved: Br

(a) MeO +

Br

MeOH

D

D H Me

(b) CH(CO Et) + 2 2

O—C— H

— NO2

O

Solution

(a)

This SN2¢ reaction proceeds with syn stereochemistry as given below:

The C — Br bond is attacked by the p bond from the back and the p bond is attacked from the back by the nucleophile. The result is an overall syn attack of the nucleophile MeO① with respect to the leaving group.

3.74

Organic Chemistry—A Modern Approach

(b)

This SN2¢ reaction also proceeds with syn stereochemistry. H

H Me

Me

SN2¢

CH (CO2Et)2 + O—C— H

— NO2

(EtO2C)2CH H

O

O

C

O

+ NO2

1.

Account for the following observations: (a) A small amount of NaI catalyzes the following reaction: R¢ONa + RCl Æ R¢OR + NaCl (b)

2.

Neopentyl halides are notoriously slow in nucleophilic substitution reactions, whatever the experimental conditions. (c) The decreasing order of reactivity of Me3CCH2Br(I), ClCH2CH ==CH2(II), ClCH2CH2CH3 (III) and BrCH2CH2CH3 (IV) in the Williamson synthesis of ethers is II > IV > III > I. Indicate the product of each of the following reactions with plausible mechanism: CH2OH

(a)

3.

CH2N2 ether

COOH

(b)

OH OH [Hint: The more acidic functional group will undergo methylation by diazomethane.] (a) 2,6–Di–tert–butylpyridine acts as a base but not as a nucleophile – Why? Ê �� * ˆ (b) Azide (N3①) ion is weaker base than amide Á : N H2 ˜ ion – Why? Ë ¯

(c)

4. 5.

CH2N2 ether

Many nucleophiles are anions, but some anions are not nucleophiles. Explain why BF4① is not a nucleophile. How will you convert (S)-1-phenyl-2-propanol to (R)-1-phenyl-2-propanol? Rank the following molecules/ions in the order of decreasing nucleophilicity and give your reasoning: (a) CH3COO①, CH3CH2O①, CH3CH2S① in MeOH (b) I①, Br①, Cl① and F① in DMSO and in ethanol

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(c)

3.75

�� *, NH �� , NH �� NH �� , PhNH �� : NH 2 3 2 2 2 in MeOH

(d) 6.

7.

�� H O , HOO �� :* , H O �� :* in MeOH H 2O, 2 2 �� �� �� When PhCH2Br is added to a suspension of KF in benzene, no reaction occurs. However, when a catalytic amount of 18-crown-6 is added, PhCH2F can be isolated in good yields. If, however, LiF is substituted for KF, there is no reaction even in the presence of crown ether. [Hint: Li≈ ion cannot be trapped by 18-crown-6 because its cavity size is larger than the size of Li≈.] Which compound in each pair exhibits greater SN2 reactivity. Explain your reasoning. Br or CH2Br (a) Cl

(b) (c) 8.

9. 10.

11.

12.

13.

or Cl I or Cl or

Cl

(d) Cl Treatment of substituted anisoles with HI produces CH3I and phenols rather than CH3OH and iodobenzenes. Explain. [Hint: SN2 attack by I① ion cannot take place at the ring carbon of the protonated ether] A concentrated aqueous solution of HBr reacts with CH3CH2OH to give CH3CH2Br, but a concentrated aqueous solution of NaBr does not. Explain. Acids of the type R3CCOOH cannot be prepared by the cyanide synthesis using a tertiary alkyl halide — Why? [Hint: R3C — CN cannot be prepared by treating R3C — Cl with NaCN in DMSO (an SN2 reaction) because of steric reason.] A three-membered ring formation reaction (by and intramolecular SN2) occurs 102–103 times faster than a comparable noncyclic reaction. Explain. [Hint: Ready formation of a three-membered ring is the consequence of a more favourable entropy contribution to the free energy of the reaction.] What is the main difference is using RS① and RO① as nucleophiles in SN2 reaction? [Hint: Elimination with RS① (a much weaker base) is not a serious problem as with RO① (a much stronger base).] What do you mean by necleofugality? Explain with suitable examples. [Hint: A leaving group that carries away an electron pair is known as a nucleofuge. Hence, nucleofugality is the leaving ability of a group that comes away with the

3.76

14. 15.

Organic Chemistry—A Modern Approach

electron pair. Nucleofugality increases with decreasing basicity. For examples, CN①, Br①, I①, SH①, etc. are good nucleofuges and HO①, RO①, NH2①, etc. are very bad nucleofuges.] Compare the nucleophilicities and basicities of tBuO① and EtO①. Predict the product of the following reaction and suggest a plausible mechanism: HI (S)- Me(Et)CH — O — CH 3 ææ æ Æ

16.

[Hint: (S)-Me(Et)CHOH; the configuration is retained because no bond to the asymmetric carbon is broken.] Convert (S)-2–pentanol to (R)-2–pentanol.

17.

Cl① ion of Bu 4 NCl* in acetone is a better nucleophile than that of LiCl in the

18.

same solvent. Explain. Predict the major product obtained from each of the following reactions and account for its formation:

≈

HI(exces)

(a) O

(b)

HI(exces) O

CH3

(c)

HBr CF3 — O — CH 3 æææ Æ

(d)

O

(e)

HI CH 3CH == CH — O — CH 2CH 3 ææ æ Æ

(f)

CH3 O

CH2CH3

HI

HI

CH3

19.

An increase in solvent polarity will increase the rate of the SN2 reaction between ≈ �� and an alkyl halide, RX, to form an ammonium salt, R NX * . an amine, R 3 N, 4 Explain.

20.

Write the appropriate combination of alkyl halide and alkoxide for the preparation of the following ethers by Williamson ether synthesis: (a)

CH2CH3

(b)

O

CH2 == CH — CH2 — O — CH(CH3)2

21.

For the SN2 reaction: KCN + R — X Æ R — CN + KX, which solvent among methanol, acetone and DMSO would be most appropriate and why?

22.

Which would you expect to be the stronger nucleophile in protic solvent? (a)

23.

CH3O① or CH3COO①

(b)

H2O or H2S

①

PhSO3①

(c)

Me3P or Me3N

Arrange CH3COO , PhO and anions as leaving groups in the decreasing order if the pKa values of their conjugate acids are 4.5, 10 and 2.6, respectively. ①

3.77

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

24.

Mention the solvent characters of the following: H2O, CCl4, CH3OCH3, DMSO

25.

Which of the following statements are correct? �� * . �� :* is a stronger nucleophile than : OH (a) RO �� �� �� :* is a stronger nucleophile than : OH �� * . (b) RCOO �� �� �� :* is a stronger nucleophile than ROH. �� (c) RCOO �� �� �� :. �� :* is a weaker nucleophile than H O (d) RO 2 �� In which of the following substrates chlorine is least reactive and why?

26.

27.

28.

(a)

Methyl chloride

(b) Allyl chloride

(c)

Ethyl chloride

(d) Vinyl chloride

Account for the following reaction sequences and give three-dimensional structures for A — K (with R and S designation): CH I

(a)

NaH 3 ( R)-2- Butanol æææ Æ A æææ ÆI

(b)

KOH/ H2O T sC l 3 ( R)-1- Deuterioethanol ææææææ Æ C æææææ Æ D æææææ ÆE Pyridine, 25∞ C acetone D

(c)

Na 3 (S)-sec-Butyl alcohol æææ Æ F æææ ÆG

(d)

NaSH 3 3 4 (S)-1-Chloro-1-phenylethane ææææ Æ H æææ Æ I ææææ Æ J ææææÆ acetone acetone 2. H O

(a)

Which one of the following compounds will react faster in an SN2 reaction and why? Br Br

CH COOK

CH I

CH I

Me I

29.

1. LiAH 2

Me

(b)

NaN

Me

Me II

[Hint: I will react at a rate faster than II.] Is an SN2 reaction exothermic or endothermic. Give your reasoning.

Nucelophilicity and basicity order of NH3, NH2NH2 and NH2OH are reverse to each other. Explain. �� �� �� �� [Hint: The basicity order is NH In the case of NH2NH2,the 3 > NH2 NH2 > NH2 OH. basicity is partly decreased due to — I effect of the other N atom. Since oxygen is more electronegative than nitrogen, basicity of NH2OH is less than that of NH2NH2. In this case, the nucleophilicity does not run parallel with basicity. In both NH2NH2

3.78

30. 31.

Organic Chemistry—A Modern Approach

and NH2OH, the tendency of nucleophilic attack by the lone pair on nitrogen is enhanced by the lone pair of electrons on the adjacent atoms. The phenomenon is called ‘a–effect’. This effect is more pronounced in the case of NH2NH2 compared to NH2OH and this is because the lone pair on less electronegative nitrogen is more polarizable than the lone pair on more electronegative oxygen. Therefore, the order �� NH �� �� �� �� of nucleophilicity is NH 2 2 > NH 2OH > NH 3 .] �� F① in CsF is more nucleophilic than F① in LiF in an aprotic solvent like acetone. Give an explanation. Which SN2 reaction in each pair is faster and why? (a)

– MeOH I + OMe

(b)

Me2 CHCH 2 Cl + N3* ææ Æ

– DMSO I + OMe

Me3 CCH 2 Cl + N3* ææ Æ (c)

H O

Cl + EtO

(d)

2 CH3 Br + NaOH æææ Æ

H O

2 CH3 Br + CH3COONa æææ Æ

(e)

–

Cl + EtOH

EtOH Br + Ph3N EtOH Br + Ph3P

32.

Which of the following SN2 reactions will take place and which will not? Give reasons. (a) (b) (c) (d)

NH2

+ I

I

–

–

+ NH2

CH2CH2CH2I + CH3O* ææÆ CH3CH2CH2OCH3 + : ��I : �� OH – – F + + F OH

*

Br + CN

–

– CN + Br

(e) (f)

CN + I

–

I + CN

–

3.79

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

33.

Rank the alkyl bromides in each group in the order of increasing SN2 reactivity and give your reasoning: Br (a) , , Br Br Br (b)

Br ,

Br , Br

Br

(c)

Br

, Br

(d)

Me H

, H

(a) (b)

H

H Br ,

H

H

34.

,

Br H

Phloroglucinol can behave as an ambident nucleophile. Explain with reactions. Two isomeric SN2 products are obtained when sodium thiosulfate (Na2S2O3) is allowed to react with one equivalent of CH3I in methanol. [Hint: (b) Thiosulfate (S2O32- ) ion is an ambident nucleophile.]

35.

Arrange the following carbanions in the order of increasing nucleophilicity: ≈

*

*

*

*

�� , (CH ) C :, HC ∫∫ C : :CH 3 , :C ∫∫ N, CH 2 == CH — CH 2 3 3 [Hint: The increasing acidity order of the conjugate acids is (CH 3 )3CH CH 4 CH 2 == CH — CH 3 HC ∫∫ CH HCN] 36.

Predict the major product in each of the following reactions and indicate the mechanistic pathway involved therein: (a)

CH3CHONa + CH3CH2Br Æ ;

(b)

(CH3)2CHBr + CH3CH2ONa

[Hint: (a) Since the substrate is a primary halide, substitution (SN2) is highly favoured over elimination (E2). CH 3CH 2Br + CH 3CH 2ONa ææ Æ CH 3CH 2OCH 2CH 3 + CH 2 = CH 2 (10%) E2 (90%) SN2

3.80

Organic Chemistry—A Modern Approach

(b)

With this secondary halide, the elimination is favoured over substitution because of steric hindrance. (CH 3 )2 CHBr + CH 3CH 2ONa ææ Æ (CH 3 )2 CHOCH 2CH 3 + CH 2 = CH CH 3 (21%)SN2 (79%) E2]

37.

38.

�� �� Alcohols (ROH) are weaker acids than phenols (ArOH) but are stronger �� �� nucleophiles. Explain. [Hint: The conjugate base of a phenol (ArO①) is relatively more stable (stabilized by resonance) than the conjugate base of an alcohol (RO①). For this reason, phenols are more acidic than alcohols. On the other hand, phenols are weaker nucleophiles than alcohols because the unshared pair of electrons on oxygen is delocalized with the ring p electrons in the case of phenols and therefore, less available compared to that on oxygen of alcohols.] Predict the product and explain their formations: H

==

O

HO

S

O

+

–

PhCHO Bisulfite ion [Hint: Hard–hard and soft–soft interactions are to be considered.]

39.

Explain the order of reactivity when each of the following compounds reacts with methyl iodide (CH3I) to yield a methyl pyridinium iodide salt. ≈

Nu : + CH 3 — I ææ Æ Nu — CH 3 I* CH3 Nu:

Relative rate

N

N

N

2.3

1

0.5

CH3

CH3

N 0.04

CH3

N

C(CH3)3

0.0002

40.

The trans-isomer of 2–chlorocyclohexanol can be converted to cyclohexene oxide by base, but the cis-isomer cannot. Explain.

41.

The optical rotation of a solution of NaBr and (+)–2–bromopentane in acetone goes slowly to 0°. Explain this observation.

3.81

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

42.

Give the mechanism of the following reaction:

What is the advantage of using 2,6–lutidine instead of OH①? 43.

The rate of SN2 reaction decreases in the order CH3Cl > CHCl2 > CHCl3 > CCl4. Explain.

44.

Write a detailed mechanism for the following reaction: HBr(excess) D

O

2

+ H—Br

Hint: O

Br

+ Br

+

O H

–

SN2 –

SN2

Br

45.

Br—H

+

OH + Br

OH2

Suggest a mechanism for each of the following reactions: OH

(a) (b)

Me

Br

PBr3 SOCl2

OH

(c) 46.

+

Br

Me

PBr3

OH

Cl Br

Predict the major product from each of the following competition experiments: (a)

Cl Cl

(b) (c)

H

F

Br

H

NaI(1 mole) DMF

NaI(1 mole) acetone NaSPh(1 mole)

I

I

EtOH

3.82

47.

Organic Chemistry—A Modern Approach

Predict the product obtained when ethyl amine is allowed to react with excess of methyl iodide in a basic solution of K2CO3. [Hint: The final product of the reaction is a quaternary ammonium iodide. Its formation may be shown as follows: Et—NH2 + Me—I

SN2

+

–

Et NH2Me I

K2CO3

Et NH Me

+

Me—I

Et NHMe2I

SN2

–

K2CO3 +

–

EtNMe3 I

I—Me SN2

EtNMe2

48.

Ammonia reacts with an alkyl halide (R — X) to give a low yield of 1° amine. A much better yield of 1° amine is obtained from the reaction of an alkyl halide (1°) with azide ion (N3①) followed by catalytic hydrogenation. Explain. [Hint: Unlike an alkyl amine (R — NH2) an alkyl azide is not nucleophilic.]

49.

To maximize the amount of alkyl iodide, the reaction of an alkyl chloride with KI is generally carried out in acetone (MeCOMe). What is the role of the solvent in increasing the yield of alkyl iodide? [Hint: KI is soluble in acetone but KCl is not. As a result, the backward reaction cannot take place.] How will the rate of the following SN2 reactions change if the polarity of the solvent

50.

(protic polar) is increased?

51. 52.

(a)

CH3CH2CH2 Br + HO* ææÆ CH3CH2CH2OH + Br *

(b)

CH 3 S CH 2CH 3 + C2H5O* ææ Æ C2H5OCH 3 + CH 3SCH 2C H 3 | CH3

(c)

CH3CH2CH2I + NH3 ææÆ CH3CH2CH2 NH3I*

≈

≈

Would you expect CH3COO① ion to be more reactive as a nucleophile in an SN2 reaction carried out in MeOH or DMSO? Starting with (R)-2-bromobutane, outline the synthesis of each of the following compounds: (a)

SH | (S)-CH 3CHCH 2CH 3

(c)

C2H5 | (S)-CH 3CHOC2 H5

(b)

(S)-C H 3 CHOCOCH 3 | C2H5

3.83

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

54.

1–Bromobicyclo[2.2.1]heptane is extremely unreactive in an SN2 reaction. Explain. Explain why halocyclohexanes react slowly by the SN2 mechanism.

55.

The trans-isomer of 4-tert-butylcyclohexyl bromide reacts with sodium thiophenoxide

53.

*

≈ (Ph SNa) in aqueous ethanol at a much slower rate than its cis-isomer. Explain this observation.

56.

57.

What are the possible combination of reactants required for the preparation of isopropyl methyl ether by Williamson synthesis? Which combination of reactants is preferable and why? Explain nucleophilic catalysis with a suitable example.

58.

CH3Br reacts with AgCN to give CH3 — NC (an isocyanide), whereas it reacts with NaCN to give CH3 — CN (a cyanide). Explain.

59.

(a)

Predict the major product from each of the following reactions: 1. t-BuO (i) CH 3COCH 2COCH 3 æææææ Æ 2. CH I *

3

AgNO2 BrCH 2CO2 Me ææææ Æ

(ii)

NaNO2 BrCH 2CO2 Me ææææ Æ

CH N

2 2 CH 3COCH 2COCH 3 ææææ Æ

(b)

Account for the following reaction, making clear the role played by tosyl chloride (TsCl):

soft base

soft base Hint: (a)(i) CH3COCHCOCH3

t-BuO

O

–

CH3

–

CH3—C—CH—C—CH3 –

O

´

H

CH3—I

CH3COCHCOCH3 + I

O

CH3—C ==CH—C—CH3 O—H CH3COCH2COCH3 Keto

CH3C ==CHCOCH3 Enol

–

+

CH2—N2

O

–

hard base

hard acid +

CH3—C ==CHCOCH3 + CH3—N2 OCH3 CH3—C ==CHCOCH3

–

3.84

60.

Organic Chemistry—A Modern Approach

Give the products of the following SN2 reactions: (a) (b) (c) (d)

61.

(3S, 4R)-3-Bromo-4-methylhexane + CH3S① Æ (3S, 4S)-3-Bromo-4-methylhexane + CH3S① Æ (3R, 4R)-3-Bromo-4-methylhexane + CH3S① Æ (3R, 4S)-3-Bromo-4-methylhexane + CH3S① Æ

For each pair of the following compounds, state which one is the better SN2 substrate. (a) 2–Methyl-1-iodopropane or tert-butyl iodide (b) Cyclohexyl bromide or 1-bromo-1-methylcyclohexane (c) 2,2–Dimethyl-1-chlorobutane or 2-chlorobutane (d) Isopropyl iodide or 1-iodo-2,2-dimethylpropane

62.

Draw a perspective structure of a Fisher projection for the products of the following SN2 reactions: (a) (S)-2-Bromopentane + KCN Æ (b) trans-1-Chloro-3-methylcyclopentane +KOH Æ CH3 (c)

Br

H CD3

H

+ NaI

acetone

CH2CH3 Br

(d)

F

C H

63.

+ NaSH CH2CH3

Elucidate the structures of compounds I–IV: I(CH ) Cl

NaNH2 2 7 1-Bromo-8-Fluorooctane ææ Æ I(C10 H17 F) ææææ Æ II (C10H16FNa) æææææ Æ + sodium acetylide NaCN II (C17 H 30Cl F) ææææ Æ IV(C18 H 30 NF)

[Hint:

3.85

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

64. 65.

66.

67.

68.

Predict the product obtained when 1–bromopropane is treated separately with methanolic solution of KSCN and AgSCN, respectively. Give your reasoning. Give the major product in each of the following reactions and account for its formation: O * � Na salt + CH3I �� — C H ææææææ (a) CH 3 — C — N Æ 2 5 �� O * � Ag salt + CH3I �� — C H ææææææ (b) CH 3 — C — N Æ 2 5 �� When methyl bromide is dissolved in methanol and an equimolar amount of NaI is added, the concentration of iodide ion quickly decreases, and then slowly returns to its original value. Explain these observations. Give the structures, including stereochemistry, of compounds I and II in the following sequence of reactions:

The reaction of cyclopentyl bromide with sodium cyanide to give cyclopentyl cyanide proceeds faster if a small amount of NaI is added to the reaction mixture. H Br

69.

CN

Can you suggest a reasonable mechanism to explain the catalytic function of sodium iodide? Select the combination of alkyl bromide and potassium alkoxide that would be the most effective in the synthesis of the following ethers: (a)

70.

H

NaCN ethanol-water

OCH3

(b) CH3OC(CH3)3

(c)

(CH3)3CCH2OCH2CH3

Suggest a mechanism for the following reaction: Br NH

+

N

I

[Hint: I is an SN2¢ product.]

N

II

3.86

3.2

Organic Chemistry—A Modern Approach

THE SN1 REACTION

Example, kinetics, mechanism, stereochemistry, evidence in favour of the mechanism and the factors influencing SN1 reactivity, i.e., the SN1 reaction rate.

3.2.1 Example of SN1 Reaction The alkaline hydrolysis of tert-butyl chloride is a typical example of SN1 reaction. H2O �� * + �� * :OH (CH 3 )3 CCl ææææ Æ (CH 3 )3 COH + : Br: acetone �� �� tert-Butyl chloride tert-Butyl alcohol

3.2.2

Kinetics of SN1 Reaction

The rate of the reaction is found to depend on the concentration of only the alkyl halide. For example, doubling the concentration of only the alkyl halide doubles the rate of the reaction and changing the concentration of the nucleophile has no effect on the rate of the reaction. From this relationship between the rate of the reaction and the concentration of the alkyl halide, we can write the rate law for the reaction as follows: rate = k [(CH3)3CCl] Because the rate of the reaction depends on the concentration of only one reactant, it is a first-order reaction. Therefore, the reaction is unimolecular, i.e., only tert-butyl chloride is involved in the transition state of the rate determining step.

3.2.3

Mechanism of SN1 Reaction

The most straightforward explanation for the observed first-order kinetics is a two-step reaction in which bond breaking occurs before bond making. The mechanism of the reaction is designated as SN1 (Substitution, Nucleophilic, unimolecular) because only one reactant (the substrate) is involved in the transition state of the rate-determing step. The two steps through which the reaction proceeds may be shown as follows: CH3 CH3 Step 1:

CH3—C—Cl

slow

H3C

CH3

Step 2:

H3C H3C

+ Cl

C CH3 CH3

C—CH3

OH fast

CH3—C—OH CH3

In the first step, tert-butyl chloride undergoes heterolysis of the C — Cl bond to yield tertbutyl cation and a chloride ion (the leaving group). This is the rate-determining (slow) step of the reaction. The cleavage of the C — Cl bond is an endothermic process. In the second

3.87

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

step, nucleophilic attack of the hydroxide ion on the carbocation leads to the formation of tert-butyl alcohol. This is a Lewis acid-base reaction. The formation of C — OH bond is an exothermic process and hence, a fast process. The energy profile diagram for the reaction may be shown as follows:

3.2.4

Stereochemistry of SN1 Reaction

In the first step in an SN1 reaction, the tetrahedral substrate undergoes slow dissociation of C — LG bond to form a trigonal carbocation. The hybridization of the central carbon changes from sp3 to sp2. The three s bonds connected to the positive carbon lie in a single plane which is perpendicular to the vacant p orbital of C 2 . The nucleophile may attack sp

the planar carbocation from either side or face with equal facility. Thus, if the reaction is carried out with a chiral substrate (the leaving group is bonded to the chirality centre), a mixture of equal number of molecules having configuration identical and opposite to the substrate will be obtained, i.e., the product will be mixture of exactly equal amounts of two enantiomers. Hence, complete racemization is the expected stereochemical result of an SN1 reaction. The process may be shown as follows:

3.88

Organic Chemistry—A Modern Approach

However, in SN1 reactions, complete racemization is not frequently observed. The product of inverted configuration generally exceeds its enantiomer, i.e., the usual stereochemical result of an SN1 reaction is racemization plus a few percent of inversion. This becomes clear if we consider the process of ionization of the substrate in the first step. When the alkyl halide ionizes, an intimate or tight ion pair is formed first. The ion pair then undergoes complete separation to give a pair of separately solvated free ions. A free carbocation obtained in a solvolytic reaction finally becomes symmetrically solvated by the solvent molecules. Nucleophilic attack on the carbocation by the solvent molecules then occurs from either side with equal readiness and results in formation of a racemic product. Attack on the ion pair, on the other hand, does not take place at either face of the carbocation with equal ease and this is because the leaving group remaining in close proximity shields the front side from nucleophilic attack. As a result, nucleophilic attack by the solvent molecule occurs preferably from the backside of the unsymmetrically solvated carbocation and substitutions product of inverted configuration is obtained predominantly. The process may be shown as follows:

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.89

In fact, the degree of racemization depends on the stability (lifetime) of the intermediate carbocation. A more stable carbocation that gets enough time to become symmetrically solvated reacts to form racemic product predominantly. However, if the carbocation is not much stable (short lived), the nucleophilic attack occurs mainly through the unsymmetrically solvated carbocation to produce the inverted product predominantly.

3.90

Organic Chemistry—A Modern Approach

An example of complete racemization is as follows. When optically active (S)-3-bromo3-methylhexane is heated with aqueous acetone, 3-methyl-3-hexanol is obtained as a mixture of 50% (R) and 50% (S).

3.2.5

Evidence in Favour of SN1 Mechanism

(a) Kinetic evidence The rate law, rate = k[RX], shows that the rate of SN1 reaction depends only on the concentration of the alkyl halide. This means that we must be observing a reaction whose rate-determining transition state involves only the alkyl halide, RX. The rate-determining transition state, therefore, is unimolecular because it involves only one molecule. In fact, a rate law similar to the experimentally observed one can be derived by applying the assumption of steady state and considering the first step to be rate-determining. Therefore, the two-step mechanism involving the rate-determining cleavage of the C — X bond is the most plausible one. (b) Stereochemical evidence The SN1 reaction of an alkyl halide in which the halogen atom is bonded to a chirality centre from two stereoisomers: one with the same relative configuration as the starting alkyl halide, and the other with the inverted configuration. These two enantiomers are often obtained in 1:1 ratio. Formation of two enantiomers suggests that these reactions proceed through planar carbocation which undergoes nucleophilic attack at either face to give product molecules with inversion and retention of configuration. (c) Change of rate with structure When the methyl groups of tert-butyl chloride are successively replaced by hydrogens, the rate of the SN1 reaction decreases progressively. The stability of carbocations decrease progressively for 3° to 1°. This observation, therefore, suggests that the rate-determining step of an SN1 reaction involves formation of a carbocation.

3.91

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.2.6 3.2.6.1

Factors Influencing SN1 Reaction Rate or SN1 Reactivity The effect of substrate structure on rate

As the number of R groups on the carbon with leaving group increases, the rate of an SN1 reaction increases. CH 3 — X methyl

RCH 2 — X 1∞

R 2CH — X 2∞

R 3C — X 3∞

An electrical effect is particularly important for SN1 reactions. An SN1 reaction involves rate-determining formation of a carbocation. Since the transition state for the heterolysis of C — X bond closely resembles the intermediate carbocation in energy, any effect that stabilizes the carbocation will stabilize the transition state and thereby causes rate enhancement. An alkyl group stabilizes a carbocation by releasing electrons through inductive (+I) and hyperconjugation effects. Thus, the stability of carbocation increases with increase in the number of alkyl groups on the central carbon and therefore, the stability decreases in the following order:

A tertiary carbocation is more stable and is, therefore, easier to form than a secondary carbocation, which in turn is more stable and easier to form than a primary carbocation. Therefore, a tertiary alkyl halide is more reactive than a secondary alkyl halide, which in turn is more reactive than a primary alkyl halide. Since methyl cation is least stable, therefore, a methyl halide is least reactive. Thus, the reactivity order agrees with the observation that the SN1 reaction rate increases as the hydrogen of methyl halide are successively replaced by alkyl groups. It is to be noted that primary carbocation and methyl cation are so unstable that primary alkyl halides and methyl halides do not at all undergo SN1 reactions. The very slow reactions reported for primary and methyl halides are actually SN2 reactions.

3.92

Organic Chemistry—A Modern Approach

Rela ve rates of SN1 reac ons (hydrolysis) of several alkyl bromides Class of alkyl bromide

Example

relative rate

Tertiary (3°) Secondary (2°) Primary (1°) Methyl

Me3C — Br Me2CH — Br MeCH2 — Br CH3 — Br

1,200,000 11.6 1.00* 1.05*

*SN1 rates are actually zero. These small rates are observed as a result of SN2 reactions.

Steric effects are less important in an SN1 reaction because the angular distance between the alkyl groups increases on going from the substrate to carbocation. However, for SN1 reactions of alkyl halides having large alkyl groups, an enhancement of reaction rate (steric acceleration) in observed. The rate enhancement is due to some relief of steric compression between the R groups (B strain) on going from the tetrahedral substrate (bond angle 109.5°) to the planar trigonal carbocation (bond angle 120°). The solvolysis (a nucleophilic substitution in which the nucleophile is a molecule of the solvent) of the alkyl chloride A in aqueous ethanol, for instance, proceeds about 600 times faster than the alkyl chloride B. H3C

H3C C—Cl

Me3CCH2 CH2CMe3 A (much steric strain)

C—Cl

H3C

CH3 B (less steric strain)

The following observations demonstrate how various structural effects influence the rate (reactivity) of SN1 reactions: (1)

The rate of hydrolysis (solvolysis in water) of alkyl halide increases as the number of the phenyl groups on the carbon bonded to the leaving group increases. For example, the SN1 reaction rate of the three phenyl substituted chloromethanes increases in the order: PhCH2Cl < Ph2CHCl < Ph3CCl. Because ≈

of charge delocalization by resonance, benzyl cation (Ph CH2 ) derived from benzyl chloride (PhCH2Cl) is resonably stable. For this reason, benzyl chloride undergoes heterolysis of the C — Cl bond readily in water (a protic polar solvent), i.e., it undergoes ready hydrolysis by the SN1 mechanism.

3.93

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

CH2—Cl

CH2

CH2

CH2 ´

´

CH2 ´

CH2 ´

+ Cl

Benzyl cation (resonance-stabilized) As the number of phenyl groups on the carbon bonded to the leaving group increases, resonance stabilization of the resulting carbocation increases in the Benzyl chloride

≈

≈

≈

order: Ph CH2 < Ph 2 CH < Ph 3 C and consequently, the rate of hydrolysis increases in the order: PhCH2Cl < Ph2CHCl < Ph3CCl. (2)

Small bicyclic systems containing leaving group at the bridgehead carbon does not undergo SN1 reaction. For example, 1-bromobicycl[2.2.1]heptane is exceedingly unreactive towards nucleeophilic substitution by the SN1 mechanism. Because of extremely rigid cage-like structure, the bridgehead carbocation expected to be formed cannot assume its usual planar trigonal orientation of bonds and the angle between bonds is somewhat less than the sp2 bond angle of 120°. So, the carbocation suffers from angle strain. Again, it will not be stabilized by hyperconjugation because the formation of a double bond at the bridgehead position is not possible according to Bredt’s rule. For these reasons, ionization of 1-bromobicyclo[2.2.1]heptane leading to the formation of an unstable bridgehead carbocation does not take place, i.e., the compound is unreactive towards SN1 reaction.

(3)

A halide producing an aromatic carbocation reacts at rate faster than a halide producing a nonaromatic carbocation which in turn reacts at a rate faster than a halide producing an antiaromatic carbocation. The SN1 reactivity of the following bromo compounds is II < I < III. Br

Br

Br O

I

II

III

3.94

Organic Chemistry—A Modern Approach

The compound III on ionization produces a carbocation which is aromatic [a(4n + 2)p electron system, where n = 1]. The compound I on ionization produces a carbocation which is nonaromatic and the compound II on ionization produces a carbocation which is antiaromatic (a 4np electron system, where n = 1).

(4)

The aromatic carbocation IIIa is more stable than the nonaromatic carbocation Ia which in turn is more stable than the antiaromatic carbocation IIa. Because the SN1 reactivity depends on the stability of the intermediate carbocation obtained in the rate-determing step, the order of increasing SN1 reactivity of these bromo compounds is II < I < III. In fact, the compound II does not undergo SN1 reaction. SN1 reactivity depends on the angle strain in the cyclic carbocation obtained from a cyclic substrate. The more it will be unstable and less readily it will be formed. The order of SN1 reactivity of the following chlorocycloalkanes is I < II < III. Cl —Cl

—Cl

I II III An SN1 reaction proceeds through the rate-determining formation of a trigonal planar carbocation in which the angle between any two bonds is 120°. The normal bond angles of three-, four- and five-membered rings are of the order of 60°, 90° and 108°, respectively. Thus, there are deviations of 60°, 30° and 12°, respectively from the trigonal angle. The bond-angle strain in the resulting cyclic carbocation thus

increases in the order: cyclopentyl cation cyclopropyl cation

( )

(

)

< cyclobutyl cation

(

)

<

. Also, in a cyclopentane system, change of hybridization

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(5)

3.95

from sp3 to sp2 involves relief of four bond oppositions. Therefore, the increasing order of stability of these carbocations is cyclopropyl cation < cyclobutyl cation < cyclopentyl cation. Because the SN1 reactivity depends on the stability of the intermediate carbocation obtained in the rate-determining step, the reactivity order of these chlorocyloalkanes is I < II < III. Vinylic and aryl halides are unreactive towards SN1 reactions. The positively charged carbons of vinyl or aryl cations are sp hybridized and because sp carbon are more electronegative than sp2 carbons that carry the positive charge of alkyl carbocations, sp carbons are more resistant to becoming positively charged, i.e., sp hybridized positive carbons are relatively very unstable. The formation of vinylic and aryl cations by the heterolysis of the C — X bond is, therefore, disfavoured, i.e., an SN1 mechanism is prevented from operating. Again, the C — X bond, the aryl and vinylic halides are unusually short and strong because in bond is formed by sp2– p overlap and the bond has a considerable double bond character due to delocalization of the unshared electron pair on halogen with the ring p electrons or with the p electrons of the double bond. Consequently, bond breaking requires more energy. For these reasons, aryl and vinylic halides are also unreactive towards SN1 reactions.

3.96

(6)

Organic Chemistry—A Modern Approach

A primary substrate may even undergo SN1 reaction readity if the carbon going to a carbocation is bonded to a heteroatom containing unshared electron pair. For example, chloromethyl ethyl ether, ClCH2OC2H5, undergoes ready SN1 solvolysis reactions, even though it is a primary substrate. CH3OH �� — CH — Cl ææææ �� — CH OCH C2 H 5 — O Æ C2 H 5 — O 2 2 3 25∞ C �� �� The carbocation obtained from C2H5OCH2Cl on heterolysis of the C — Cl bond is stabilized by resonance. It is for this reason chloromethyl ethyl ether undergoes ready SN1 solvolysis. SN1

C2H5—O—CH2—Cl

C2H5—O—CH2

C2H5—O==CH2

+ Cl

A 1° carbocation (resonance-stabilized) C2H5OCH2OCH3

–H

C2H5—O—CH2—O—CH3

CH3OH

H

(7)

Competition experiments are those in which two reactants at the same concentration (or one reactant with two reactive sites) compete for a reagent. When there are two halogen atoms in a substrate, which will take part in an SN1 reaction depends on the carbocation expected to be formed by C — X bond cleavage. For example, the circled halogen atom in each of the following compound is more reactive than the other in an SN1 reaction.

Br Br

Br Br I

CH2CH3

II Cl

I

I

III CH2— Cl

CH3O Cl IV

CH2CH2CHClCH3 V

3.97

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Heterolysis of the C — Br bond in compound I does not actually take place because this leads to the formation of a very unstable 1° carbocation. On the other hand, heterolysis of C— Br bond leads to the formation of a very much stable 3° carbocation. The circled bromine atom is, therefore, more reactive than the other.

The C — Br bond in compound II does not actually undergo heterolysis because this leads to the formation of a very unstable 3° carbocation (a nonplanar bridgehead carbocation which is not stabilized by hyperconjugation). Heterolysis of the C— Br bond, on the other hand, leads to the formation for a very much stable 3° carbocation. Therefore, the circled halogen atom is more reactive than the other one.

SN 1

Br Br II   

CH2CH3

SN 1

Br ; + Br ; CH2CH3

A bridgehead carbocation (very unstable)

Br + Br

Br Br

CH2CH3 II

Br

CH2CH3

A 3° carbocation (very stable)

Heterolysis of the C — I bond in compound III leads to the formation of a less stable 2° carbocation while heterolysis of the C — I bond leads to the formation of a very stable allylic cation (resonance-stabilized). Therefore, the circled iodine atom is relatively more reactive than the other one is an SN1 reaction.

3.98

Organic Chemistry—A Modern Approach

   

Heterolysis of the C— Cl bond in compound IV leads to the formation of a very stable (resonance-stabilized) 2° carbocation while heterolysis of the C — Cl bond leads to the formation of a relatively less stable simple 2° carbocation. Therefore, the circled chlorine atom is relatively more reactive than the other one. Cl SN1

CH3O

CH3O

+ Cl

CH3O Cl

Cl

Cl

A 2° carbocation (more stable)

IV Cl

Cl SN1

CH3O

CH3O IV

Cl

+ Cl

A 2° carbocation (less stable)

Heterolysis of the C— Cl bond in compound V leads to the formation of a very stable (resonance-stabilized) benzylic cation while heterolysis of the C — Cl bond leads to the formation of a relatively less stable 2° carbocation.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.2.6.2

3.99

The Effect of Solvent on Rate

The rate of an SN1 reaction increases markedly with increase in the ionizing power of the solvent. In the rate-determining step of the reaction a carbocation and an anion are formed from a neutral substrate. Since stretching of the C — X bond starts before its cleavage, therefore, the transition state is relatively more polar than the substrate.

The ionizing power of a solvent actually depends on two factors: (i) its dielectric constant (e) and (ii) its ability to solvate ions. As the dielectric constant of the solvent or solvent polarity increases, the electrostatic force of attraction between the two incipient ions decreases. This facilitates the reaction by lowering the energy of the polar transition state. As the molecular polarity, i.e., the dipole moment (m) of the solvent increases, solvation (through dipole–dipole interactions) and consequent stabilization of both the substrate and the transition state increases. However, such stabilization is far greater for the more polar transition state than the less polar substrate. Because of this differential stabilization, the energy of activation for ionization of the substrate is considerably reduced and the rate is increased. This rate enhancement, however, depends on whether the solvent is protic or aprotic. Since the transition state closely resembles the ions in energy and geometry, any factor that stabilizes the ions also stabilizes the transition state. Therefore, to understand

3.100

Organic Chemistry—A Modern Approach

the role played by a particular solvent, its ability to solvate the separated ions must be considered. Protic polar solvents (e.g., methanol, ethanol, water, etc.), which effectively solvate both cations and anions, are found to be very much effective for SN1 reactions. In such a solvent, cation and anion solvation occur through donation of the unshared electron pairs to the vacant orbital of the cation and through the formation of ion-dipole bonds or hydrogen bonding, respectively. d–

H2O H2O

d–

O

OH2 R

H

OH2

H2O

d–

OH2

d+

d+

H d+

d– H O H

H O

H

d+ H O

X d+ d– H O H

H

d– Hd+ H O A solvated anion

A solvated cation

Polar aprotic solvents like CH3CN, Me2SO, Me2NCHO, etc. processing high dielectric constants are not suitable for SN1 reactions because they unable to solvate anions (a very important factor for ionization) through ion-dipole bonds or hydrogen bonding. Water is the most effective solvent for promoting ionization because its dielectric constant is much higher and it can stabilize the anion through hydrogen bonding. However, most organic compounds do not dissolve appreciably in water. They usually dissolve in alcohol, and quite often in mixed solvents. Methanol-water and ethanol-water are the two common mixed solvents for nucleophilic substitution reactions. The rate of an SN1 reaction increases when the percentage of more polar component of a mixed solvent increases. For example, the rate of solvolysis of tert-butyl bromide in methanol-water mixed solvent increases with increasing the percentage of water. As the percentage of the more polar solvent water increases, the polarity of the mixed solvent increases. Increasing the polarity of the mixed solvent increases the rate of solvolysis because the more polar transition state is relatively more stabilized than the less polar substrate resulting in a decrease of activation energy (Ea). Me3C—Br Substrate (less polar)

SN1

d+

Me3C

d–

Br

Transition state (more polar)

Me3C +

Br

3.101

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

The effect of the polarity of the solvent on the rate of reac on of tert-butyl bromide in an SN1 reac on Solvent

relative rate

100% H2O

1200

80% H2O + 20% EtOH

400

50% H2O + 50% EtOH

60

20% H2O + 80% EtOH

10

100% EtOH

1

If, however, the compound undergoing SN1 reaction is charged, increasing the polarity of the solvent will decrease the rate of the reaction because the more polar solvent will stabilize the full charge on the reactant to greater extent than it will stabilize the dispersed charge on the transition state.

3.2.6.3

The effect of the nucleophile on rate

The rate of an SN1 reaction is independent of the nature of the nucleophile because the nucleophile is not involved in the rate-limiting (slow) step of this two-step process. However, if a better nucleophile is added to an SN1 reaction mixture, the rate remains unchanged

3.102

Organic Chemistry—A Modern Approach

but the product containing the added nucleophile is obtained predominantly. In some SN1 reactions, the solvent is the nucleophile. These are called solvolysis. For example, when water reacts with alkyl halides it serves as both the nucleophile and the solvent.

3.2.6.4

The effect of the leaving group on rate

The rate of an SN1 reaction is influenced by the nature of the leaving group because expulsion of the leaving group from the substrate occurs in the rate-determing step of the reaction. The leaving group in an SN1 reaction begins to acquire a negative charge as the transition state is reached. Stabilization of the developing negative charge at the leaving group stabilizes the transition state, i.e., lowers its energy. This in fact, lowers the free energy of activation (DG=| ) and thereby increases the rate of the reaction. Because the weak bases stabilize a negative charge effectively, i.e., stabilize the transition state effectively, these are good leaving groups. For instance, an alkyl iodide in the most reactive and alkyl fluoride is the least reactive of the alkyl halides in the SN1 reaction and this is because their basicity increases in the order: I① < Br① < Cl① < F①. This order of leaving ability is also due to the fact that C — X bond strength increases gradually from C — I to C — F. Relative reactivity of alkyl halides is an SN1 reaction:

Some relevant topics are discussed below:

(a) Electrophilic catalysis The expulsion of a halogen atom from an alkyl halide is facilitated by the presence of Ag≈ ion in the reaction mixture. This is called electrophilic catalysis of an SN1 reaction. For example, when optically active 1-bromo-1-phenylethane is treated with acetic acid in the presence of CH3COOAg, a racemic mixture of two enantiomeric acetates is obtained. CH3 H

OCOCH3

OCOCH3 CH3COO Ag

C

CH3COOH

Br

Ph Optically active

H

+

C Ph

CH3

C H3C

Racemic mixture (optically inactive)

H Ph

3.103

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

The mechanism of the reaction is as follows:

(b) Alcohols and ethers can act as SN1 substrates in the presence of acid The hydroxide ion is a strong base and thus reaction like the following do not take place. R——OH A 2° or 3° alcohol

R

+

OH

A very bad leaving group

However, when an alcohol is dissolved in a strong acid, it can undergo C — O bond cleavage to form a carbocation. Because the acid protonates the — OH group of the alcohol, the leaving group no longer needs to be a hydroxide ion; it is now a molecule of H2O which is @

a much weaker base than an OH ion and a good leaving group. R——OH

H

R——OH2

R + H2O A good leaving group

When a chiral alcohol like (R)-2-butanol is allowed to stand in aqueous acid, it is found to have lost its optical activity.

3.104

Organic Chemistry—A Modern Approach

H C2H5

H

H H3O

C OH

CH3 (R)-2-Butanol

C2H5

+

C OH

CH3 (R)-2-Butanol

C HO

C2H5

CH3 (S)-2-Butanol

Racemic mixture (optically inactive)

In the presence of aqueous acid, (R)-2-butanol undergoes protonation. The conjugate acid thus obtained undergoes heterolysis of the C — O bond (an SN1 process) to form a 2° carbocation by loss of water (a good leaving group). Recombination of the carbocation with water, which occurs with equal readiness at either face of the trigonal planar ion, leads to the formation of an equimolar mixture of two enantiomeric 2-butnaol (a racemic mixture) which is optically inactive.

An ether containing a 3° alkyl group (e.g., Me3C — O — Me) can easily be cleaved by HBr. Protonation makes–OMe a good leaving group (MeOH) and the ether undergoes ready ≈

cleavage of Me3C — O bond leading to the formation of a stable 3° carbocation (Me3 C).

3.105

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.2.7

Carbocation Rearrangements in SN1 Reactions

The SN1 reaction involves a carbocation intermediate. This intermediate can rearrange, usually by a hydride (H①) shift or an alkyl (R①) shift, to give a more stable carbocation. For example, the product obtained when 2-bromo-3-methylbutane undergoes an SN1 reaction is different from the product obtained when it undergoes an SN2 reaction. When the reaction is carried out under condition favouring an SN1 reaction, the initially formed secondary (2°) carbocation undergoes a 1,2–hydride shift, rearranging to a relatively more stable tertiary (3°) carbocation. H 1,2-hydride shift

SN1

(CH3)2CHCHCH3 Br 2-Bromo-3methylbutane

(CH3)2C—CHCH3 A secondary carbocation (less stable)

SN2

(CH3)2CHCHCH3

OH

OH 3-Methyl-2-butanol

(CH3)2CCH2CH3 A tertiary carbocation (more stable)

H2O

OH2

(CH3)2C CH2CH3 —H

OH (CH3)2CHCH2CH3 2-Methyl-2-butanol

The product obtained when 3-bromo-2,2-dimethylbutane is allowed to react with a nucleophile also depends on the conditions under which the reaction is carried out. The 2° carbocation formed initially under SN1 conditions undergoes rearrangement involving a 1,2–methyl shift to give a relatively more stable 3° carbocation. Since a carbocation is not formed under SN2 conditions, therefore, the carbon skeleton remains unchanged.

When neopentyl bromide is boiled with ethanol, it gives only a rearranged substitution product. The product results from a methyl shift (migration of a methyl group together with its pair of electrons). Without rearrangement, ionization of neopentyl bromide would form a very unstable primary (1°) carbocation.

3.106

Organic Chemistry—A Modern Approach

(CH3)3C—CH2—Br Neopentyl bromide

(CH3)2C CH2 + Br A 1° carbocation (very much less stable and not formed)

The methyl shift occurs while bromide ion is leaving, so that only more stable 3° carbocation is formed. Attack by ethanol on the 3° carbocation gives the product.

Similarly, neopentyl chloride cannot be prepared from neopentyl alcohol by treating with HCl. tert-Amyl chloride is obtained as the only product in this reaction.

Neopentyl alcohol first undergoes protonation in the presence of acid to form its conjugate ≈

acid. The — OH group is converted into — OH 2 which acts as a very good neutral leaving group (weak base). However, expulsion of water does not lead to the formation of an unstable 1° carbocation. It undergoes rearrangement at the same time by a 1,2–methyl shift to form a stable 3° carbocation. Nucleophilic attack by Cl① on this carbocation leads to the formation of tert-amyl chloride.

3.107

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

[It is to be noted that although neopentyl is a primary (1°) alkyl group, because of steric hindrance, neopentyl alcohol reacts with HCl extremely slowly to give neopentyl chloride by an SN2 mechanism.

The following bicyclic compound (a 1° alcohol) undergoes neopentyl type rearrangements in the presence of HBr/ZnCl2 to a bromide.

HBr / ZnCl2

+ H2O

SN1

Br

CH2OH

Under this SN1 condition, the bridge methylene group migrates to bond with the incipient 1° carbocation that would be formed by the loss of water. The resulting 3° bridgehead carbocation then undergoes attack by the nucleophile Br① to yield a bromo compound.

2 HBr ZnBr2

2 CH2OH

ZnBr 24

2

1,2-bond shift – H2O

2 2

CH2—OH2

Br—ZnBr2—Br

+ ZnBr2

2 Br

3.108

Organic Chemistry—A Modern Approach

3.2.8 Comparison of the SN2 and SN1 Reactions SN2 reaction 1. A one-step reaction

SN1 reaction 1. A two-step reaction

2. A bimolecular rate-determining transition 2. A unimolecular rate-determining transition state state 3. There is no carbocation rearrangements.

3. Carbocation rearrangements may take place.

4. The product has inverted configuration compared 4. The products have both identical and inverted with the reactant. configuration compared with the reactant. 5. Reactivity order: methyl >1° > 2° > 3°. In fact, 3° 5. Reactivity order: 3° > 2° > 1°> methyl. In fact, substrates do not undergo SN2 reaction because 1° and methyl substrates do not undergo SN1 of severe steric hindrance. reaction because the corresponding carbocations are extremely unstable.

3.2.9

Summary of Reactivity of Alkyl Halide in Necleophilic Substitution Reactions Alkyl halide

Reaction they undergo

methyl and 1°

SN2 only

2°

SN1 and SN2

3°

SN1 only

vinylic and aryl

neither SN1 nor SN2

1° and 2° benzylic and 1° and 2° allylic

SN1 and SN2

3° benzylic and 3° allylic

SN1 only

3.2.10

Factors Favouring SN1 and SN2 Reactions

Factor

SN1

SN2

Substrate

CH3X > RCH2X (1°) R2CHX (2°) R3CX (3°) > R2CHX (2°) (The reaction involves a relatively (The reaction requires unhindered substrates) stable carbocation)

Nucleophile

The reaction is favoured by weak The reaction is favoured by strong nucleophiles nuclepophiles (usually neutral (usually bearing a net negative charge). The rate solvent molecules increases by the high concentration of the nucleophile. In protic solvents, within a group of the periodic table, rate μ polarizability of the attacking atom of the nucleophile and for the same attacking atom, rate μ basicity of the nucleophile. In aprotic solvents, rate μ basicity of the nucleophile.

Solvent

Polar protic (e.g., alcohol, water or a Polar aprotic (e.g., DMF, DMSO, acetone, etc.) mixture of alcohol and water).

Leaving group The weaker the base after the group departs the better the leaving group for both SN1 and SN2 reactions.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.109

Depending on the reaction conditions a 2° substrate R2CHX may react by the SN1 or SN2 mechanism. Strong nucleophiles favour the SN2 mechanism over the SN1 mechanism. For example, RO@ favour SN2 while ROH favours SN1 mechanism. Protic polar solvents favour the SN1 mechanism while aprotic polar solvents favour the SN2 mechanism. For example, water and CH3OH favour the SN1 mechanism while acetone and DMSO favour the SN2 mechanism.

3.2.11

SNi and SNi¢ Mechanisms

3.2.11.1

SNi mechanism

Some nucleophilic substitution reactions occurs with retention of configuration in contrast to SN1 or SN2, even though there is no neighbouring group participation (see article 3.3). The mechanism of such reactions is termed as SNi mechanism (Substitution Nuclephilc internal). For example, (R)-2-octanol reacts with thionyl chloride (SOCl2) by the SNi mechanism to yield (R)-2-chlorooctane with the same configuration as the alcohol.

Mechanism: The mechanism of the reaction involves the steps as follows: Step 1: When (R)-2-octanol is allowed to react with thionyl chloride, an alkyl chlorosulphite with retained configuration is obtained and this is because the bond between the chiral carbon and oxygen is not cleaved during this reaction.

Step 2: The intermediate chlorosulphite ester undergoes slow dissociation to give an intimate ion pair.

3.110

Organic Chemistry—A Modern Approach

n-C6H13 H3C

C6H13-n

Cl C—O— S ==O

slow D

C CH3

H

+

–

O S ==O Cl

H

Intimate ion pair in solvent cage

Step 3: The ion pair rapidly collapses to yield (R)-2-chlorooctane. @

�� @ : ion originated from :O �� : SOCl ion attacks the carbocation from The nucleophile :Cl �� the front side because it is unable to get to �� the rear. As a result, the configuration is retained.

The overall reaction between the chiral alcohol and thionyl chloride, therefore, takes place with retention of configuration.

Evidence in favour of ion pair The reaction involves the formation of an ion pair that can be proved by the fact that 3-methyl-2-butanol, when allowed to react with SOCl2, produces 3-chloro-2-methylbutane. This is possible only if a carbocation forms and subsequently rearranges to the more stable one. CH3

CH3

O

(CH3)2CH—CH—OH + SOCl2 Æ (CH3)2CH—CH—O—S—Cl slow

3-Methyl-2-butanol H

–

O

+

(CH3)2C—CH2CH3

–

Cl Cl fast –SO2

+

–

S ==O H-shift (CH3)2C—CH—CH3 O SOCl

(CH3)2C CH2CH3

2-Chloro-2-methylbutane

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.111

Reaction of (R)-2-octanol with SOCl2 in the presence of pyridine The reaction proceeds with inversion of configuration when carried out in the presence of pyridine. The base pyridine reacts with HCl generated during the formation of the chlorosulphite ester to yield pyridinium chloride salt. The alkyl chlorosulphite then undergoes nucleophilic attack by the Cl@ ion from the back side (SN2) and results in formation of the corresponding alkyl chloride with inverted configuration.

The reaction may also take place as follows:

3.112

3.2.11.2

Organic Chemistry—A Modern Approach

SNi¢ mechanism

When an allylic substrate undergoes SNi¢ reaction, the nucleophilic attack occurs at the g-carbon rather than at the a-carbon. This is what is called SNi¢ mechanism. When pent2-en-1-ol, for example, is treated with thionyl chloride, 3-chloropent-1-ene is obtained exclusively. Cl | ether CH3CH2CH == CHCH2OH + SOCl2 æææÆ CH3CH2 CHCH == CH2 Pent-2-en-1-ol 3-Chloropent-1-ene (100%) The reaction takes place as follows:

3.2.12

SN1¢ Mechanism

Under SN1 conditions, allylic substrates undergo nucleophilc substitutions to give a rearranged product in addition to the normal product. For example, CH3CH==CHCH2Cl on solvolysis with 0.8(N) NaOH at 25°C yields 60% of CH3CH==CH CH2OH (normal product) and 40% of CH3CHOHCH== CH2 (rearranged product). 0.8 (N) NaOH CH 3 CH == CH CH 2Cl ææææææ Æ CH 3CH == CHCH 2OH + CH 3CH OH CH == CH 2 25∞C 1-Chlorobut-2-ene But-2-en-1-ol (60%) But-1-en-3-ol (40%) (normal prroduct) (rearranged product)

The mechanistic pathway which results in formation of the rearranged product is known as the SN1¢ (SN1-prime) mechanism (unimolecular nucleophilic substitution with allylic rearrangement). The reaction proceeds through the formation of a delocalized allylic carbocation as follows:

3.113

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom +

CH3CH ==CHCH2—Cl

+

[CH3CH ==CH—CH2 ´ CH3CH —CH ==CH2 ] + Cl Allylic carbocation

1-chlorobut-2-ene OH

–

CH3CH ==CH—CH2OH But-2-en-1-ol (SN1 product)

3.2.13

–

OH

–

CH3CHOH CH ==CH2 But-1-en-3-ol (SN1¢ product)

Isotope Effects and Salt Effects (Methods Used to Distinguish between SN1 and SN2 Type Reactions)

Isotope effects a and b secondary isotope effects have been employed as useful probes to distinguish between SN1 and SN2 type reactions. In b secondary isotope effect, substitution of D for H b to the position of bond cleavage slows the reaction. The effects are greatest when the T.S. possess considerable carbocation character. A large value of kH/kD, therefore, suggests that an SN1 reaction takes place. The kH/kD value for the solvolysis of tert-butyl chloride-d9 [(CD3)3CCl] in 50% acetone at 25°C, for example, is 2.38. This observation indicates that the compound undergoes solvolysis by the SN1 mechanism. a Secondary isotope effect results from a replacement of H by D at the carbon bearing the leaving group. These effects are also correlated with carbocation character. Substitution reactions not involving carbocation intermediates, i.e., SN2 reactions, have an isotope effect near unity. However, reactions involving carbocation, i.e., SN1 reactions, have higher a isotope effects (which depend on the nature of the leaving group). For instance, the kH/ kD value for the solvolysis of Ph2CDCl in 80% aqueous acetone is 1.16. This observation suggests that the compound undergoes solvolysis by the SN1 mechanism. The kD/kD value for the solvolysis of CH3(CH2)5CD(CH3)OTs in ethanol is near unity. This suggests that the compound undergoes solvolysis by the SN2 mechanism.

Salt effects SN1 reaction rates increase when non-common ion salts (which increase the ionic strengh of the medium) are added to the reaction mixture. This is what is called salt effect. SN1 reaction rates are sufficiently increased when there are ions present that specifically help in pulling the leaving group from the substrate. Some especially important ions are Ag ! , Hg 2! and Hg 2! . H ! helps to pull off F from alkyl fluorides by forming H-bond). The non-common ion such as the azide ion (N3@ ) can be used as a means of distinguishing between SN1 and SN2 mechanisms. This is because the azide ion markedly accelerates the rate of reaction as it is a strong nucleophile and provides an additional mode of attack in

3.114

Organic Chemistry—A Modern Approach

the alkyl derivative. A large enhancement in the rate of reaction on addition of an azide salt (e.g., NaN3) has been considered as an indication of SN2 mechanism. A slight increase, on the other hand, points to an SN1 mechanism.

1.

Rank the following compounds in order of increasing reactivity:

Solution The compound II on ionization produces a carbocation which is aromatic [a(4n + 2)p electron system, where n = 1], the compound III on ionization produces a carbocation which is nonaromatic and compound I on ionization is expected to produce a carbocation which is antiaromatic (a 4np electron system, where n = 1)

An aromatic system is more stable than a nonaromatic system, which in turn is more stable than an antiaromatic system. Hence, an increasing order of stability of the resulting carbocations is: (least stable) Ia < IIIa < IIa (most stable). Because the SN1 reactivity of a compound depends on the stability of the intermediate carbocation obtained in the rate-determining step of the reaction, the order of SN1 reactivity of these compounds is

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.115

I < III
Because an SN2 reaction is very susceptible to steric hindrance at the site of substitution and because steric crowding increases on going from primary to tertiary substrate, 1-bromo3-methylbutane (a primary substrate) reacts by the SN2 mechanism at a faster rate than 2-bromo-3-methylbutane (a secondary substrate), which in turn reacts at a much faster rate than 2-bromo-2-methylbutane (a tertiary substrate). In fact, the SN2 reactivity of the tertiary substrate is zero. 3. Solvolysis of the alkyl chloride I in aqueous ethanol proceeds about 600 times faster than the alkyl chloride II. Explain. CH3 | (Me3CCH 2 )2 CCl Me3 CCl I II

3.116

Organic Chemistry—A Modern Approach

Solution Compressed reactants try to avoid their steric strain. If the strain is relieved in attaining the intermediate of the rate-determining step or the transition state leading to the formation of the intermediate, that reaction will be speeded up. This phenomenon is termed as steric acceleration and because of this, an alkyl halide containing bulky substituents undergoes solvolysis (SN1) at a faster rate than that expected on the basis of stabilization of the carbocation through inductive and hyperconjugative electron release. In the alkyl chloride I, two out of three alkyl groups are very bulky neopentyl groups (— CH2CMe3), while in the alkyl chloride II, all the three groups are less bulky methyl groups. Steric acceleration is, therefore, more important in the case of I than in the case of II and so, the solvolysis of the alkyl chloride I in aqueous ethanol proceeds at a much faster rate (600 times) than the alkyl chloride II.

4.

In the investigation of the mechanism of solvolysis (in 80% EtOH) of 2-chloro-2-methylbutane, the following results were obtained for the solvolysis of: (a) Me2CClCD2Me (kH/kD = 1.41), (b) (CD3)2 CClCH2Me (kH/kD = 1.78). Explain the observations.

Solution From the kH/kD values it becomes clear that the substitution of deuterium for six b hydrogens [in (b)] causes much retardation of reaction (solvolysis) rate than the substitution of deuterium for two b hydrogens [in(a)] and this is because the carbocation !

!

(CD3 )2C CH 2Me is less stable than the carbocation Me2 C CD2Me (the C — D bonds are somewhat stronger than the C — H bonds and thus result in less stabilization by hyperconjugation). 5. Draw the conformational structures of the product(s) expected to be obtained from each of the following reactions: H

Br

(a) D

CH3

H2O CH3OH

Cl ;

(b)

H D

H

–

I CH3OH

3.117

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Solution (a) This tertiary substrate undergoes solvoysis (SN1) in methanol-water mixed solvent. Either nucleophile (H2O or CH3OH) can attack the intermediate carbocation from either the top or bottom side to give four substitution products (two with retention and two with inversion of configuration). H

Br

D

CH3

OCH3

∫∫

Br H2O

CH3

CH3OH

CH3 –

Br +

CH3 (retention)

H D

CH3OH

CH3

(–H + )

H

H

(a)

(a)

(b)

(b) D

(inversion)

H

∫∫

D

OCH3

D

OH CH3

(a)

(a) CH3 +

(retention)

H D

H2O

CH3

+

(–H ) H

(b) D

(b)

OH (inversion)

H D

(b)

This secondary substrate undergoes SN2 reaction with the strong nucleophile I@ to yield the corresponding iodo compound. The reaction, therefore, proceeds only with inversion of configuration around the stereogenic centre.

6.

Formation of ethers from alcohols by intermolecular dehydration using concentrated H2SO4 may proceed either by the SN1 or by the SN2 mechanism. Explain.

Solution The formation of an ether by acid-catalyzed intermolecular dehydration of an alcohol is, in fact, a nucleophilic substitution reaction in which the conjugate acid of alcohol is the substrate, H2O is the leaving group and a second molecule of alcohol is the nucleophile. Because of steric reason, methanol and primary alcohols react by the SN2 mechanism while, because of electronic reason, secondary and tertiary alcohols

3.118

Organic Chemistry—A Modern Approach

react by the SN1 mechanism. Both of these mechanistic pathways may be outlined as follows taking ethanol (CH3CH2OH) as an example for the SN2 route and isopropyl alcohol (CH3CHOHCH3) as an example for the SN1 route. SN2 route: Step 1:

CH3CH2 OH + H

+

fast

+

CH3CH2 OH2

Step 2:

Step 3: SN1 route: fast

Step 1:

(CH3)2CH—OH + H

Step 2:

(CH3)2CH—OH2

Step 3:

(CH3)2CH—OH + (CH3)2CH

slow SN1

(CH3)2CH—OH2

(CH3)2CH + H2O slow

(CH3)2CH—O—CH(CH3)2 H

Step 4: 7.

(CH3)2CH—O—CH(CH3)2 + H2O

fast

(CH3)2CH—O—CH(CH3)2 + H3O:

H 2-Ethyl-2-methyloxirane reacts with acidic methanol to yield the 1° alcohol A as the major product while it reacts with basic methanol to yield the 3° alcohol B as the major product:

Explain these observations regarding the regiochemistry of epoxide ring opening in acidic and basic medium. Solution In the presence of acid, 2-ethyl-2-methyloxirane undergoes ring cleavage by a pathway that is partially SN1 and SN2. It is not a pure SN1 reaction because a carbocation intermediate is not fully formed. It is not a pure SN2 reaction because the leaving group

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.119

begins to depart before the compound is attacked by the nucleophile. In acid, the epoxide or the oxirane first undergoes protonation. In protonated epoxide, the positive charge is borne not only by the oxygen but also by the two ring carbons. The sharing of the positive charge can be represented by drawing the following three resonance structures:

The structure I is the conventional structure for the protonated epoxide while the structures II and III show that the epoxied carbons share part of the positive charge. The structure II (a very stable 3° carbocation) is more important (contributing) than the structure III (a very unstable 1° carbocation). Therefore, the tertiary carbon bears a larger part of the positive charge, and it is more strongly electrophilic. As a consequence, the more substituted carbon–oxygen bond becomes longer and weaker than the less substituted carbon–oxygen bond. Thus, the transition state for attack at the more substituted carbon is lower in energy than that for attack at the less substituted carbon (even though attack at this carbon is less hindered) and so, attack by weakly nucleophilic methanol (which is sensitive to the strength of the electrophile) takes place preferentially at the more electrophilic tertiary carbon to give predominantly the alcohol A.

3.120

Organic Chemistry—A Modern Approach

Under basic conditions, the cleavage of the unprotonated epoxide ring occurs purely by �� :@ attacks the less substituted and less the SN2 mechanism and so, the nucleophile CH 3O �� sterically hindered methylene carbon of the epoxide preferentially to give the tertiary alcohol B as the major product.

8.

Explain the following observations:

Solution In an anhydrous medium, the low polarity of the solvent (ether) favours the SN2 mechanism and the nucleophile I@ attacks the sterically less hindered methyl carbon of the protonated ether to yield methyl iodide and tert-butyl alcohol. H—I

(CH3)3C—O—CH3

tert-Butyl methyl ether

(CH3)3C—O—CH3 + I

SN2

H

(CH3)3COH + CH3I tert-butyl Methyl iodide alcohol

In an aqueous medium, the high polarity of the solvent (H2O) favours the SN1 mechanism and therefore, the protonated ether dissociates to give methanol and tert-butyl cation which then undergoes nucleophilic attack by I@ to yield tert-butyl iodide. (CH3)3C—O—CH3

H—I

tert-Butyl methyl ether

(CH3)3C—O—CH3 + I

SN1

(CH3)3C + CH3OH + I

H (CH3)3C—I + CH3OH tert-Butyl Methyl alcohol iodide

3.121

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

9.

When the following ester of acetic acid is allowed to undergo acidic hydrolysis, a racemic alcohol is obtained: CH2CH (CH3)2 C

H5C2

O—C—CH3

CH3

H3O

CH3COOH + H5C2

CH2CH (CH3)2

CH2CH (CH3)2

C

C

OH + HO

CH3

CH3

O

(optically active)

C2H5

Racemic mixture (optically inactive)

Explain this observation. Solution Because heterolysis of the C — O bond results in the formation of a stable 3° carbocation, acid-catalyzed hydrolysis of this ester occurs by alkyl-oxygen cleavage. The reaction thus follows the SN1 mechanism and leads to the formation of acetic acid and a racemic alcohol. The reaction takes place through the steps as follows:

Step 1: CH2CH (CH3)2

CH2CH (CH3)2

O H5C2

C

OH H5C2

O—C—CH3 + H—OH2

CH3

H5C2

CH2CH (CH3)2

OH

C CH3

Step 3:

O—C—CH3 + H2O

CH3 CH2CH (CH3)2

Step 2:

C

O

C

slow

CH3

C

SN1

H5C2 CH3

+ CH3 COOH

3.122

10.

Organic Chemistry—A Modern Approach

When the following alkyl bromides are subjected to hydrolysis in dilute aqueous ethanolic sodium hydroxide solution, the rates of the reactions showed the following order: (CH3)3CBr > CH3Br > CH3CH2Br > (CH3)2CHBr Provide an explanation for this order of reactivity.

Solution Two different mechanisms are involved in the hydrolysis of these alkyl bromides. (CH3)3CBr undergoes hydrolysis by the SN1 mechanism and so, its hydrolysis takes place most readily. The other three alkyl bromides undergo hydrolysis by the SN2 mechanism (in the presence of strong nucleophile OH@, the 2° substrate also follow the SN2 pathway predominantly). Since the activation energy of an SN2 reaction is normally much higher compared to an SN1 reaction, the hydrolysis of these three alkyl bromides takes place relatively slow. Since an SN2 reaction is very susceptible to steric hindrance, the order of their SN2 reactivity is CH3Br > CH2CH3Br > (CH3)2CHBr.

11.

One reaction in each pair takes place more readily. Identify the reaction and explain your answer. (a) Me3CCl + H 2O ææ Æ Me3COH + HCl Me3CBr + H 2O ææ Æ Me3COH + HBr Ph Ph | | (b) (CH 3 )2 C — Br + H 2O ææ Æ (CH 3 )2 C — OH +HBr (CH 3 )3C — Br + H 2O ææ Æ (CH 3 )C — OH + HBr (c) (CH3 )3CCl + H2O ææÆ (CH3 )3 COH + HCl (CH3 )3CCl + CH3OH ææÆ (CH3 )COCH3 + HCl Æ Me3COH + HBr (d) Me3CBr + H 2O ææ H2O �� @ Me3CBr + OH @ æææ Æ Me3COH + :Br: �� EtOH �� @ (e) Me CCl (1.0M) + EtO@ (1.0M) ææææ Æ Me3COEt + Cl: 3 �� EtOH �� @ Me3CCl (1.0M) + EtO@ (2.0M) ææææ Æ Me3COEt + Cl: ��

(f)

Me3CCl + H 2O ææ Æ Me3COH + HCl PhCl + H 2O ææ Æ PhOH + HCl

3.123

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Solution (a) The second reaction will take place more rapidly because Br@ is a better leaving group than Cl@.

(b)

!

Since the 3° carbocation (CH 3 )2 C Ph is relatively more stable (resonance-stabilized) !

than the 3° carbocation (CH 3 )3 C, the first reaction (solvolysis) is expected to take place at a faster rate than the second. (c)

SN1 reactions take place faster in more polar solvents. Since water is a much polar solvent (e = 79) than methanol (e = 32.63) therefore, the first reaction will take place more rapidly.

(d)

Since the nucleophile is not involved in the rate-determining step of an SN1 reaction, the rate is independent of the nature of the nucleophile. Therefore, the two reactions are expected to take place at the same rate.

(e)

SN1 reactions are independent of the concentration of the nucleophile because the nucleophile is not involved in the rate-determining step of the reaction. Therefore, the two reactions will take place at the same rate. However, the predominant process in this pair of reactions would be E2.

(f)

The first reaction will take place rapidly because the substrate is a tertiary halide. Phenyl halides are unreactive towards SN1 reaction because, due to repulsive interaction, backside attack is not possible. Which of the following tosylates would undergo solvolysis in 80% ethanol at a faster rate and why?

12.

OTs

H OTs

H

or H

H I (trans)

II (cis)

Solution Owing to steric interaction between the axial-OTs group and the axial hydrogen atoms at C-3 and C-5, the cis-isomer (II) has a higher ground state energy than the transisomer (I). Now, solvolysis of both of them involves a common carbocation intermediate. So, as a first approximation, the transition state may also be assumed to be common. Therefore, the activation energy (Ea) for the cis-isomer (higher ground state energy) will be less than that for the trans-isomer and because of this, the cis-isomer undergoes solvolysis at a rate faster than the trans-isomer.

3.124

13.

Organic Chemistry—A Modern Approach

Compare the reactivities of isopropyl chloride and benzyl chloride in nucleophilic substitution reaction under SN1 and SN2 conditions.

Solution Benzyl chloride (PhCH2Cl) is more reactive than isopropyl chloride (Me2CHCl) !

both in SN2 and SN1 reactions. An SN1 reaction is favoured because benzyl cation (Ph CH 2 ) !

is relatively more stable (resonance-stabilized) than isopropyl cation (MeCH) and an SN2 reaction is favoured because it is a primary substrate (Me2CHCl is a secondary substrate and sterically more hindered) and also, the transition state is stabilized by p overlap.

3.125

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

14.

Write the structure of the intermediate carbocation expected to be formed in the following reaction:

solvolysis

Br Solution

SN1

∫

+ Br

Br Norbornyl cation (a nonclassical carbocation)

15.

In each of the following dihalides, determine which halogen atom is relatively more reactive in SN1 reaction. Give your reasoning. Br O (a) (b) (c) CH2Br CH3O Br Br Solution (a) The bromine atom adjacent to the ring, i.e., the benzylic bromine atom exhibits greater reactivity because the carbocation obtained on its expulsion (the benzylic cation) is relatively more stable (resonance-stabilized) compared to the carbocation obtained on expulsion of the other bromine atom (a 2° carbocation which is not stabilized by resonance due to the presence of intervening saturated carbon). Br Br (a)

–Br

Br

(b)

CH3O

Br

CH3O

(a)

(b)

A 2° carbocation (less stable)

etc. OCH3

Br

CH3O

A benzylic carbocation (more stable)

Br

3.126

Organic Chemistry—A Modern Approach

(b)

The carbocation obtained on expulsion of the iodine atom attached to the ring is a very unstable antiaromatic system ( 4np electron system, where n = 1) while the carbocation obtained on expulsion of the iodine atom attached to the side chain is a stable (resonance-stabilized) 3° allylic carbocation. Therefore, the latter iodine atom exhibits greater SN1 reactivity. In fact, the former iodine atom is unreactive towards SN1 reaction.

(c)

The carbocation obtained on expulsion of the Br atom attached to the ring is a very stable aromatic system [a(4n + 2)p electron system, where n = 1] while the carbocation obtained on expulsion of the other Br atom is a relatively less stable allylic cation. Therefore, the former Br atom exhibits greater SN1 reactivity.

16.

Which one of the following two halides is more reactive in SN1 reaction and why? O I

CMe2Br

O

CMe2Br

II

Solution The halide I is more reactive than the halide II in SN1 reaction because the intermediate 3° carbocation expected to be formed from the halide I in step 1 (the ratedetermining step) of an SN1 reaction is relatively more stable (stabilized by resonance) than the 3° carbocation (not stabilized by resonance) expected to be formed from the halide II.

3.127

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

O

CMe2—Br

–Br

etc. O

I

O

O

CMe2

A 3° carbocation (more stable)

CMe2—Br II

17.

CMe2

–Br

O CMe2 A 3° carbocation (less stable)

Predict the major product obtained from each of the following reactions and account for its formation: O

O

CH3O

H / H2O

(a)

(b)

CH3OH

Solution (a) The reaction under acidic conditions proceeds to form the major product A as follows:

In the protonated epoxide, one of the C — O bonds starts breaking in that direction which places a partial positive charge on the more substituted carbon and this is because the substituted carbocation is relatively more stable. Therefore, the protonated epoxide undergoes nucleophilic attack at the more substituted carbon to form the 1,2-diol A (with the — CH2OH group axial and the — OH group equatorial) predominantly. The reaction in fact occurs by a pathway that is partially SN1 and partially SN2. (b)

The reaction under basic conditions proceeds to give the major product B as follows: OH

O

O CH3O SN2

CH2OCH3

H—OCH3

CH2OCH3 + CH3O B

3.128

18.

Organic Chemistry—A Modern Approach

When a nucleophile attacks an unprotected epoxide, the reaction is a pure SN2. �� @ attacks the less substituted and Therefore, in this case, the nucleophile CH 3O: �� less satirically hindered ring carbon to yield the alcohol B (with the — OH group axial and the — CH2OCH3 group equatorial) predominantly. Propose a mechanism for each of the following reactions: OH O H3O

(a)

OH OH HBr

CHCH3

(b)

Br

CH2

CH==CHCH3 CH2

H AgNO3

(c)

H2O / EtOH

OH

CH2I

OH

CH3

(d)

CH—Br

CH3

H2O

(e)

Solution

O

(a)

H—OH2

d+ H O d+

OH

OH H2O

OH —H

OH2

OH

3.129

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(b)

(c)

CH3

(d)

CH—Br

H2O SN 1

Br +

CHCH3

CH3

bond shift

CH3 H2O –H

(e)

CH3

H 3C CH3 CHCH3 Br

H2O SN1

(b)

CH3

Br +

(a)

H

CH3 I

C (a)

OH

H

(b)

CH3

II

CH3 H

3.130

19.

Organic Chemistry—A Modern Approach

The relative rates of solvolysis of the following three tertiary halides in 80% ethanol at 25°C are as follows: Br Br Br

relative rate: Explain this observation.

I

II

III

1

ª10–6

ª10–14

Solution The bridgehead carbocation expected to be obtained from the [2.2.2] bicyclic !

system II is relatively very unstable than the 3° carbocation, Me3 C expected to be obtained from I because, being nonplanar, it suffers from angle strain and also it is not stabilized by hyperconjugation (according to Bredt’s rule double bond cannot be formed at the bridgehead position). Again, the bridgehead carbocation expected to be obtained from the [2.2.1] bicyclic system III is relatively more unstable as compared to the carbocation obtained from II and this is because it is less planar (as one of the bridges contains one less carbon) and hence, suffers from greater angle strain. It is also not stabilized by hyperconjugation. Because of such order of stability of the intermediate carbocations, the halide I undergoes solvolysis very much faster than the halide II, which in turn undergoes solvolysis very much faster than the halide III.

(from I)

(from II)

(from III)

Stability increases

20.

Describe the Lucas test for distinguishing primary, secondary and tertiary alcohols and discuss the reactions involved. How do you account for the anomalous behaviour of allyl alcohol, CH2 ==CHCH2OH, and benzyl alcohol, PhCH2OH, in this test?

Solution Whether an alcohol is primary, secondary or tertiary can be determined by taking an advantage of the relative rates at which the three classes of alcohols react with concentrated HCl plus ZnCl2 (the Lucas reagent). This is what is called Lucas test. When the reagent is added to the alcohol, the mixture forms a single homogeneous phase and this is because the concentrated HCl solution is very polar, and the polar alcohol-zinc chloride complex dissolves in it.

When an alcohol reacts with the Lucas reagent, the relatively nonpolar and insoluble alkyl chloride is obtained and so, the solution turns cloudy or the chloride separates into a second phase. When the test is carried out at room temperature, the solution turns cloudy immediately if the alcohol is tertiary, in about five minutes if the alcohol is secondary, and remains clear if the alcohol is primary. ZnCl2, a strong Lewis acid, encourages an

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.131

SN1 pathway for the formation of the alkyl chlorides (R — Cl). Tertiary alcohols react almost instantaneously because they form relatively stable 3° carbocations. Secondary alcohols react in about 1–5 minutes because secondary carbocations are less stable than the tertiary ones. Primary carbocations are too unstable to be formed and so, primary alcohols do not react to form the corresponding chlorides at room temperature. However, the turbidity appears when the mixture is heated or kept for a long period. In fact, the primary alcohols then react by the SN2 mechanism which is slower than the SN1 reaction of secondary and tertiary alcohols. Reactions: Secondary and tertiary alcohols react with the Lucas reagent by the SN1 mechanism as follows:

Primary alcohols react with the Lucas reagent at higher temperature by the SN2 mechanism as follows:

3.132

Organic Chemistry—A Modern Approach

Because each of the allyl and benzyl alcohols produces a stable carbocation (resonancestabilized), they readily react with the Lucas reagent to make the solution turbid very quickly even though they are primary alcohols.

21.

When cis-2-methylcyclohexanol reacts with the Lucas reagent, the major product is 1-chloro-1-methylcyclohexane. Propose a reasonable mechanism to account for the formation of this product.

Solution This secondary alcohol reacts with the Lucas reagent by the SN1 mechanism as follows:

The initially formed 2° carbocation rearranges by hydride shift to a more stable 3° carbocation which then undergoes nucleophilic attack by Cl@ to give 1-chloro-1-methylcyclohexane.

3.133

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

22.

Predict the products of the following reaction and explain their formations: H CH2 ==CH—C

CH3

SOCl2

A+B

OH Solution

The allyl cation obtained from the chiral chlorosulphite ester undergoes attack at the g-carbon in two fashions to yield two geometric isomers A and B. 23. Predict the product and suggest a mechanism of the following reaction:

3.134

Organic Chemistry—A Modern Approach

Solution The reaction takes place by an SNi mechanism i.e., the reaction proceeds with complete retention of configuration to yield (R)-a-chlorethylbenzene. The steps involved are as follows: Cl

Ph C—O

Me H

+ H

C

Ph O

–HCl

Ph

C—O—C—Cl

Me

Cl

O slow

(R)-a-Phenylethanol Phosgene

C

+

An alkyl chloroformate (retention of configuration)

C ==O Cl

Me

H

–

O

H

Intimate ion pair in solvent cage

Ph Me

C——Cl + CO2 H

(R)-a-Chloroethylbenzene (retention of configuration)

(R)-a-Phenylethanol reacts with phosgene to form an alkyl chloroformate with retained configuration. This ester undergoes slow dissociation to give an intimate ion pair. The ion pair rapidly collapses to form (R)-a-chloroethylbenzene. The nucleophile Cl@ ion originated @

from OCOCl ion attacks the carbocation from the front side because it is unable to get to the rear. This results in retention of configuration . It thus follows that the overall reaction between the chiral alcohol and carbonyl chloride or phosgene occurs with retention of configuration. 24. Predict the major product expected to be obtained when 1-methyl-1,2epoxycyclopentane is treated with (a) sodium ethoxide in ethanol and (b) H2SO4 in ethanol. Solution (a) In this case, an SN2 reaction takes place. EtO@ (a strong nucleophile) attacks the less hindered secondary carbon to give (E)-2-ethoxy-1-methylcyclopentanol. +

O H3C

–

Na OEt/EtOH SN 2 H

O

H3C

–

H OEt

H—OEt

OH H

+ EtO

–

CH3 OEt

1-Methyl-1,2-epoxycyclopentane

(b)

Under acidic conditions, the weakly nucleophilic alcohol attacks the more electrophilic tertiary carbon atom of the protonated epoxide. In this case, therefore,

3.135

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

nucleophilic attack occurs from the backside (an SN2 characteristic) at the more substituted carbon (an SN1 characteristic). H

+

H—OH2

O

O

d+

Et OH SN2 + SN1

d+

H3C

H

H3C

H

CHHO 3 +

H2O

H

EtO

+ CHHO + H3O 3

OEt H

H

1.

2.

Arrange the following compounds in the order of increasing reactivity towards aqueous alkali and give your reasoning: CH3(CH2)3Br, CH3CH2CH== CHBr, CH2== CHCH2CH2Br, CH3CH== CHCH2Br Explain the following observations: (a)

(b) (c)

Triphenylmethyl bromide and 1-bromotryptecene are found to differ in their rate of solvolysis under parallel conditions by a factor of 1:10–23.

CH3 | 2-Chloromethoxypropane (CH 3CHOCH 2Cl) undergoes solvolysis at a rate faster than 1-chloro-2-ethoxyethane (CH3CH2OCH2CH2Cl). The hydrolysis of ethyl vinyl ether in dilute aqueous acid takes place 103 times faster than the hydrolysis of diethyl ether. [Hint: Ethyl vinyl ether (C2H5OCH==CH2) undergoes hydration to form initially a hemiacetal. The hemiacetal then undergoes hydrolysis (by the SN1 mechanism) with greater facility because the resulting carbocation is resonance-stabilized.]

(d) EtO

CH2Cl

MeOH Solvolysis

EtO

CH2OMe

(very fast reaction)

O2N

CH2Cl

MeOH Solvolysis

O2N

CH2OMe

(very slow reaction)

3.136

Organic Chemistry—A Modern Approach

(e)

* Cl (* ∫

14

*

CH3COOAg

* O COCH3 +

C)

O COCH3

(33.33%)

(66.66%)

[Hint: Cyclopropenyl cation is a hybrid of three equivalent resonance +

structures:

*+ ´

* having equal contribution (33.33%) to

*´ +

the hybrid.] (f) The relative rate of solvolysis of the following allylic chlorides in formic acid containing a small amount of water is as follows: Relative rate: CH3 CH3 CH3 | | | CH 2 == CH — CH 2Cl CH 2 == C — CH 2Cl CH == CH CH 2Cl CH 2 == CH — CHCl I

1.0

3.

II

III

0.5

IV

3550

5670

Explain the following order of decreasing SN1 reactivity: Cl S

Cl—CH CH2CH3

Cl CH CH2CH3

CH CH2CH3

N

N

CH CH2CH3 Cl

4.

Which reaction in each of the following pairs will take place more rapidly and why? (a)

H2O Me3CCl æææ Æ Me3COH + HCl H2O (Me3C)3Cl æææ Æ (Me3C)3 COH + HCl

(b)

(c)

OH@ �� @ CH 3CH 2 CH 2 CH 2 Cl æææ Æ CH 3CH 2CH 2CH 2OH + Cl: H2O �� OH@ @ �� CH 3CH 2 OCH 2 Cl æææ Æ CH 3CH 2OCH 2OH + Cl: H2O ��

3.137

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Which mechanism, SN1 or SN2, is favourable for reactions with each of the following substrates? Explain.

5.

(a)

CH2Cl

MeO

(b)

O2N

(c)

CH2Cl

MeOCH2Cl

[Hint: SN1 for (a) and (c); SN2 for (b)] Which one of the following two bromo compounds would undergo silver ion-assisted hydrolysis at a faster rate? Explain your answer:

6.

On ozonolysis, a 14C labelled allyl chloride yields formaldehyde with 100 percent 14 C. Ozonolysis of the allyl acetate obtained by treating that allyl chloride with CH3COOAg in acetic acid yields formaldehyde with 50 percent 14C. Explain these observations. [Hint:

7.

14

CH COOAg

3 CH 2 == CHCH 2Cl ææææææ Æ 14CH 2 == CH — CH 2OCOCH 3 + CH 2 == CH 14CH 2 OCOCH 3 ] AcOH SN 1

Explain why 2-chloro-2-methylpropane undergoes solvolysis more slowly in 90% D2O–10% dioxane than in 90% H2O–10% dioxane solution. Account for the following observations: (a) Alcohols (ROH) react with CH3I in the presence of Ag≈ ion to form methyl ethers (ROCH3). However, the reaction fails in the absence of silver ion. (b) When tert-butyl alcohol is treated with equimolar mixture of HCl and HBr, tert-butyl bromide is obtained as the major product and tert-butyl chloride is obtained as the minor product. (c) tert-Butyl chloride undergoes solvolysis much faster than 2-chloro-1,l,1trifluoro-2-methylpropane. (d) Tetrahydrofuran can solvate a positively charged species better than diethyl ether can. (e) The ether A cleaves much faster than the ether B with concentrated HBr:

8. 9.

HBr H5C2 O CH CH H5C2 2 3 A

HBr

+ CH3CH2OH ; Br

OCH2CH3 B

+ CH3CH2Br OH

3.138

10.

Organic Chemistry—A Modern Approach

Write structures for the alcohols that you would expect from the following reaction: I CH3 (CH3)3C

11.

H2O

Explain the following observations: Cl

(a)

NaCN (0.01M) EtOH

CN (Major product)

NaCN (0.01M) EtOH

(b) Cl

OEt (Major product)

[Hint: SN1 reactions show only a slight tendency to discriminate between the weak and strong nucleophiles, whereas SN2 reactions show a marked tendency to discriminate.] 12.

13.

When 1-chloro-3-methylbut-2-ene or 3-chloro-3-methylbut-1-ene is treated with Ag2O in water, the following mixture of the two alcohols is obtained along with AgCl.

Suggest a mechanism that accounts for the formation of these products. What might explain the relative proportions of the two alkenes that are formed? Under which of the following reaction conditions (S)-2-chlorobutane would form the most (S)-2-butanol: (i) OH@ in 50% H2O / 50% EtOH or (ii) OH@ in 100% EtOH. Give your reasoning.

3.139

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

14.

Which of the following carbocation would you expect to rearrange and why? !

(a) (CH3 )2CHCHCH3 !

(d) CH3CH2 CHCH3 15.

+

(b)

CH3

(e)

CH2

(c)

+

C2H5 !

(f)

CH2 == CH — CH2 — CHMe3

Explain each of the following observations: (a) SN1 type of reactions are extremely uncommon in the gas phase. [Hint: solvation effect is absent.] (b) Hexane is not a common solvent for either SN1 or SN2 reaction. (c) (CH3)3CBr reacts at the same rate with F@ and H2O in substitution reaction, even though F@ has a net negative change but H2O is neutral. (d) When treated with silver perchlorate in propanoic acid, iodocyclopentane (I) undergoes solvolysis, whereas 1-iodocyclopenta-2,4-diene (II) does not. I I

16.

+

I II

Give the mechanism and the stereochemical course involved in each of the following reactions: (a)

SOCl2 trans-5-Methyl-2-cyclohexenol ææææ Æ

t-Bu

(b)

Me HO

[Hint: (a)

C——C∫∫ C CH3

SOCl2

3.140

Organic Chemistry—A Modern Approach

(b)

17.

18.

Account for the stereochemical course involved in each step of the following reaction sequences and give three-dimensional structures (with R or S designation) of the compounds A-G: SOCl

CH COOK

(a)

KOH/H2O TsCl /Pyridine Na I 2 3 (R)-1-Phenylethanol æææ Æ A æææææ Æ B ææææÆ C ææææææ Æ D ææææ ÆE acetone D 25° acetone

(b)

BsCl/Py (S)-2-Octanol ææææÆ F

ether

��

Dioxane G H 2O

Predict the product(s) of the following reaction and explain the mechanism: HCl/EtOH (CH 3 )3CCH 2C(CH 3 )Cl æææææ Æ

19.

Carry out the following transformation and give the mechanism involved: Me

20.

OH

Me

Cl

When 1-chloromethyl-4-methyl-1,3-cyclopentadiene is solvolyzed in aqueous acetone, three alcohols are obtained which are isomeric to each other. Give structures of the alcohols and arrange them in increasing order of their yield. Give your reasoning.

3.141

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

21.

22.

Explain why the following alkyl chloride does not undergo a substitution reaction regardless of the conditions under which the reaction is carried out.

Cl The compound I undergoes SN1 reaction very much slowly, but the isomeric compound II reacts readily, even though it is a primary alkyl halide. Explain. Me CH2Cl

Cl

23.

I II Predict the major product and explain its formation in each of the following reactions: –

(a) H

24.

MeO MeOH

O

(b)

Me

MeOH H+

O Me

H

State with reasons whether the following reactions will take place either by SN1 or by the SN2 mechanism: O

(a) HO

O

Me

O

OMe –

(b)

Me Br

O O

N3

H

+

Me + MeOH

O O

Me N3

–

O + Br O

[Hint: (a) The reaction proceeds by the SN1 mechanism because the carbocation intermediate is highly stabilized by resonance involving two oxygen atoms.

(b)

This is the only known example of SN2 reaction at a tertiary carbon and this occurs because an SN1 reaction involving the formation of a carbocation next to a carbonyl group cannot take place and also the carbonyl group accelerates the

3.142

Organic Chemistry—A Modern Approach

SN2 reaction very much by stabilizing the transition state (the unhybridized p orbital involved in the T.S interacts with the p orbital system of the carbonyl group.). Furthermore, the azide ion is an excellent nucleophile which, being sharp and marrow, is not affected much by steric hindrance.

25.

In each of the following compounds, determine which of the two halogens will be more reactive in the SN1 reaction. Explain. Cl

(a) O

26.

(c)

—CH2Br

Br—

Cl

Cl Explain the observed effects of increasing solvent ionizing power on the rates of the following reactions:

(a)

S 1

!

N Æ Ph C H + Br @ (a large increase in the reaction rate) Ph2CHBr æææ !

S 1

!

N Æ Ph C H + S(CH ) (a small decrease in the reaction Ph CH 2 S (CH 2 )3 æææ 2 3 2 rate) Explain the following observations:

(b)

27.

Cl CH2

(b)

Ph C H3C

Ph C

Cl H

(R)-(–)-1-Chloro-1phenylethane

H3C

Ph Cl

H

(R)-(–)-1-Phenyl-1ethanol

+

HO

C H

H3C

(S)-(–)-1-Phenyl-1ethanol

H2O æææ Æ

41%

59%

40% H2O æææææ Æ 60% acetone

47%

53%

20% H2O æææææ Æ 80% acetone

49%

51%

[Hint: The carbocation that is originally formed is not completely free to react with the nucleophile water on both sides. One side of the carbocation is shielded by the leaving group Cl@ for some time. Since the concentration of water molecules in pure water is very high, some of the reactions take place while the carbocation

3.143

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

is still shielded by Cl@ ion in the front. As the percentage of water in the mixed solvent decreases, an increasing number of carbocations can survive long enough without reacting with water that the ions can diffuse apart. Reaction with water can then take place on both sides of the carbocation to yield gradually an increased amount of racemized product.] 28.

tert-Alkyl fluorides (e.g., Me3C — F) are unreactive in solvolysis unless a strong acid is present. Explain. �� @ ion is a poor leaving group (F@ ion is the strongest base among the [Hint: :F: �� halide ions and the C — F bond is the strongest among the four carbon-halogen bonds) and because of this, tert-alkyl fluorides are unreactive in SN1 solvolysis reactions. However, in the presence of a strong acid, fluorine becomes strongly hydrogen bonded to a proton because it is most electronegative. The leaving group is then a stable H — F molecule rather than F@ ion and so, solvolysis takes place smoothly. This is an example of electrophilic catalysis.]

29.

Bridged alcohol as shown below cannot be chlorinated by SOCl2. Explain. SOCl2

OH

Cl

30.

The acetolysis of compound I gives a mixture of two products II and III. Give structures of II and III, and explain their formations.

31.

Which of the two alkyl chlorides in each set will undergo hydrolysis at a faster rate in an SN1 process? Give your reasoning. (a) (Me3C)3CCl,Me3CCl (b)

32.

Propose a mechanism for the following reaction: CH3 CH3 CH3 H2SO4 CH3 CH3

CH3

HO

33.

Unsymmetrical ethers are generally not prepared by heating two alcohols with H2SO4 — Why? However, when tert-butyl alcohol is heated in methanol containing H2SO4, a good yield of tert-butyl methyl ether results. Explain this observation by means of reaction mechanism. [Hint: Unsymmetrical ethers cannot be prepared by acid-catalyzed dehydration of two different alcohols because the reaction leads to the formation of all the three

3.144

Organic Chemistry—A Modern Approach

possible ethers (one unsymmetrical and two symmetrical ethers) which are not easy to separate. However, when a 3° alcohol is used with an 1° (or 2°) alcohol, a stable 3° carbocation generates readily and reacts with the 1° (or 2°) alcohol to give the desired mixed ether. In fact, there is no scope of formation of symmetrical ether corresponding to the 1° (or 2°) alcohol. Also, because of steric strain, the 3° alcohol is much more reluctant to attack the 3° carbocation to form the corresponding ether. It thus follows that a good yield of tert-butyl methyl ether results when tertbutyl alcohol is heated in methanol containing H2SO4. Me3C—OH

H

Me3C—OH2

—H2O

Me3C

CH3OH

Me3C—O—CH3 H

H2O

Me3COCH3 + H3O

34.

Explain why 2-cyclopropyl-2-propanol reacts with HCl to give 2-chloro-2cyclopropylpropane instead of 1-chloro-2, 2-dimethylcyclobutane. [Hint: The initially formed carbocation

—CMe2 is stabilized by resonance

involving the cyclopyl group and so, it does not rearrange to give finally a cyclobutane derivative.] 35.

Generally nucleophilic substitution reaction is difficult at the bridgehead position, but reaction given below proceeds — Why?

AgBF4 PhCl

O

+ CO2 + AgCl + HF

O C Cl

Cl

[Hint: Certain nucleophilic substitution reactions that normally involve carbocation (SN1) can take place at norbornyl bridgeheads if the leaving group used is of the type that cannot function as a nucleophile and thus come back once it has expelled. Since the expelled — OCOCl group (as CO2 and AgCl) cannot function as a nucleophile, PhCl acts as a nucleophile to give the product.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

36.

3.145

Select the reaction conditions that would allow you to carry out each of the following stereospecific transformations: O (R)-1,2-Propanediol

(a)

H CH3 O (S)-1,2-Propanediol

(b)

H CH3

[Hint: (a) The epoxide is to be cleaved by NaOH in ethanol-water. An SN2 reaction will take place at the less substituted carbon to form (R)-1-2propanediol.

(b)

37.

The epoxide is to be cleaved in the presence of acid. An SN2 reaction with considerable SN1 character will take place at the more substituted carbon to form (S)-1,2-propanediol.

When (R)-(+)-2-phenyl-2-butanol is allowed to stand in methanol containing a few drops of sulphuric acid, racemic 2-methoxy-2-phenylbutane is formed. Suggest a reasonable mechanism for this reaction.

3.146

3.3

Organic Chemistry—A Modern Approach

NEIGHBOURING GROUP PARTICIPATION (NGP)

3.3.1 Definition A suitably located group (with a pair of electrons to offer) close to the site of the reaction in a substrate is called a neighbouring group. In some substitution reactions, a neighbouring group participates temporarily in the reaction to control both the rate and the stereochemistry of the reaction. Such involvement of a group within the same molecule is known as neighbouring group participation (NGP). The enhancement of the rate of a reaction involving neighbouring group participation is called anchimeric assistance (Greek anchi + meros, meaning neighbouring parts).

3.3.2

Mechanism of NGP

A reaction involving neighbouring group participation consists of two SN2 substitutions. In the first step, the neighbouring group situated in the proper position for backside attack pushes out the leaving group. Conformation plays an important role in such participation. In the second step, the external nucleophile displaces the neighbouring group by another backside attack and the neighbouring group goes back to its original position. The net result of two consecutive SN2 substitutions involving inversion of configuration is retention of configuration at the reacting carbon.

Step 1:

Step 2:

The intermediates in most neighbouring group mechanisms are not symmetrical and it is therefore, not possible to get not a simple substitution product but a rearrangement product. This will happen if Nu@ attacks not the carbon atom form which LG left, but the one to which Z was originally attached.

3.3.3

Example of NGP

The displacement of chlorine form threo-3-chloro-2-butanol by the influence of dilute MeONa is an example of a reaction that involves neighbouring group participation. The product is formed with retention of configuration and the rate of the reaction does not

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.147

depend on the concentration of the nucleophile MeO@. The reaction is believed to proceed through the steps as follows:

3.3.4

Anchimeric Assistance

When the reacting groups are in different molecules, they must find each other in solution by random diffusion. It is relatively improbable that the two groups will diffuse together and collide in just the right way for a reaction to take place. However, when the two groups are present in the same molecule, they are already together; the only prerequisite for the reaction is that the C — Z: bond bends towards the backside of the carbon in the C — LG bond. Thus, the intramolecular attack of the neighbouring group on the reaction centre has a greater probability of occurring. A higher probability of reaction results in a lower standard free energy of activation. This means that this reaction occurring with greater probability takes place at a larger rate. Again, a reaction between the substrate and the external nucleophile :Nu@ involves a large decrease in entropy of activation | (DS==) because the reactants are far less free in the transition state than before. However, �� involves a much smaller loss of DS=|= because the the reaction of the neighbouring group -Z reactants are not very much free in the transition state than before. For this reason, attack by the neighbouring group (rate-determining step) takes place | faster (lower DG== ) than by the external nucleophile and as a result, a large enhancement of substitution rate occurs.

3.148

Organic Chemistry—A Modern Approach

3.3.5 Evidence for Participation by a Neighbouring Group The acceleration in the rate of a reaction is a good evidence for participation by a neighbouring group. Other evidence may be obtained from the stereochemical results of the reaction. Reactions involving neighbouring group participation are generally stereospecific. Usually such reactions yield products with retention of configuration as opposed to the normal racemization and inversion of the reaction centre.

3.3.6 Various Cases of Neighbouring Group Participation A neighbouring group may be any group containing an atom like sulphur, nitrogen, oxygen or bromine or a p bond or even a s bond. A successful NGP requires antiperiplaner geometry of the leaving group and the neighbouring group so that the stereoelectronic requirement for intramolecular SN2 reaction is fulfilled. Let us see different cases of neighbouring group participation.

3.3.6.1 (1)

Heteroatoms as neighbouring groups

There is a remarkable difference in the reaction rates of the following two substitution reactions, which are superficially very similar: (i)

100∞C CH3 (CH2 )4 CH2Cl ææææ Æ CH3 (CH2 )4 CH2OH + HCl 1 H2O 1-Chlrohexane 1-Hexanol

100∞C �� �� CH3CH2 SCH ÆCH3CH2 SCH 2 CH2 — Cl ææææ 2 CH2 — OH + HCl 3200 H 2O �� �� 1-Chloro-3-thiapentane 3-Thia-1-pentanol (b -chloroethyl ethyl sulfide) (b -Hydroxydiethyl sulfide) At first sight, both reactions appear to be simple SN2 reaction in which chloride ion is displaced as a leaving group by water. In fact, 1-chlorohexane reacts by the following mechanism:

(ii)

The presence of sulphur in the alkyl halide molecule should have little effect on the rate of the SN2 reaction, because the SN2 reaction is not very sensitive to the electronegativities of substituent groups. Yet the reaction of 1-chloro-3-thiapentane is thousands of times faster than the reaction of 1-chlorohexane with water. The rate of the second reaction is unusually large because a special mechanism facilitates the reaction, a mechanism not available to 1-chlorohexane. In the reaction of sulfide, it is the S atom (a powerful nucleophile that can participate

3.149

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

as a neighbouring group) that displaces the leaving chloride. This intramolecular reaction is much faster than the intermolecular displacement of chlorine by water either in the alkyl chloride or in the chlorosulphide. CH2CH3

CH2CH3 :S

S: CH2——CH2

Cl

SN2

CH2——CH2

(r.d.step)

An episulfonium salt

Cl

The episulfonium ion that results from this internal nucleophilic substitution reaction is very reactive because it contains a strained three-membered ring and a good leaving group. This intermediate is opened up by water in a second, intermolecular displacement step to give the product. CH2CH3 :S

H

CH2——CH2 + H2O

(2)

SN2

CH3CH2 S CH2CH2—OH

Cl

CH3CH2 S CH2CH2OH + HCl 3-Thia-1-pentanol

When (S)-2-bromopropanoate ion is treated with concentrated solution of sodium hydroxide, (R)-2-hydroxypropanoate is obtained. However, when the same reaction is carried out with a low concentration of hydroxide ion in the presence of Ag2O (where Ag ≈ ion acts as a Lewis acid), (S)-2-hydroxypropanoate ion is obtained. conc.NaOH CH 3CH Br COO@ æææææ Æ CH 3CH OH COO@ solution (S)-2-Bromopropanoate ion (R)-2-Hydroxypropanoate ion

dil.NaOH CH 3CH Br COO@ æææææ Æ CH 3CH OH COO@ solution (S)-2-Bromopropanoate ion (S)-2-Hydroxy propanoate ion Ag2O

In the presence of concentrated alkali, (S)-2-bromopropanoate ion undergoes normal SN2 reaction to yield (R)-2-hydroxypropanoate ion. The reaction proceeds with inversion of configuration at the chiral carbon.

3.150

Organic Chemistry—A Modern Approach

However, in the presence of Ag2O and a low concentration of OH@ ion, the same reaction occurs with retention of configuration at the chiral carbon. In this case, the reaction proceeds by the neighbouring group mechanism which consists of two inversion steps. In the first step, an oxygen of the carboxylate group attacks the stereocentre from the backside and displaces bromide ion to form an a-lactone (a cyclic ester) with inversion of configuration. The silver ion here acts as an electrophilic catalyst and aids the expulsion of bromine. In the second step, the highly strained three-membered lactone ring undergoes SN2 attack by H2O from the side as that of the expelled Br@ to produce (S)-2-hydroxypropanoate ion with inversion of configuration. Therefore, the net result of two inversions in two successive steps is an overall retention of configuration.

(3)

The fact that acetolysis of both 4-methoxy-1-pentyl brosylate (I) and 5-methoxy-2pentyl brosylate (II) gave the same mixture of products (III and IV) is an evidence for participation by a neighbouring group.

3.151

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

It is the ether oxygen that acts as a neighbouring group in the acetolysis of these @

two isomeric brosylates. The sulfonate ion (OBs) is an excellent leaving group and a five-membered ring is relatively unstained, so the intramolecular displacement of p-bromobenzenesulfonate to give a cyclic oxonium ion takes place in both the acetolysis reactions. The cyclic oxonium ion is opened by acetic acid in two ways to give a substitution product (III) and a rearrangement product (IV).

(4)

Since the more substituted carbon of the oxonium ion is more positive, nucleophilic attack by CH3COOH occurs preferably at that carbon to give the major product (IV). Optically active threo-3-bromo-2-butanol reacts with HBr to give (±)-2,3dibromobutane. This observation suggests that the reaction proceeds not by the ordinary SN2 or SN1 mechanism, but by the neighbouring-group mechanism in which bromine acts as the neighbouring group. A meso-dibromide is expected to

3.152

Organic Chemistry—A Modern Approach

be obtained if threo-3-bromo-2-butanel reacts with HBr by the SN2 mechanism. However, if the reaction proceeds by the SN1 mechanism, a mixture of meso and active products is expected.

Because the reaction produces only (±)-2,3-dibromobutane, it is not an SN2 or SN1 reaction. The formation of only racemic 2,3-dibromobutane can be explained by a neighbouring group mechanism. The suitably oriented bromine atom on the adjacent carbon acts as a neighbouring group and pushes out H2O (SN2) from the conjugate acid of the alcohol to form a bromonium ion intermediate. This intermediate then �� :@ on C-2 and C-3 equally well to produce undergoes nucleophilic attack by :Br �� optically inactive racemic or (±)-2,3-dibromobutane.

3.153

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

(5)

In a 1,2-disubstituted cyclohexane system, neighbouring group participation occurs only when there is a trans diaxial arrangement of the departing and participating groups and this is because antiperiplanar geometry is provided only in such conformation. Also, such conformation is very suitable for backside attack required for an internal SN2 . :G H

H

:G

H ∫

G: X H trans diequatorial conformation (not a suitable geometry for backside attack)

X

X

trans diaxial conformation (a suitable geometry for backside attack)

For example, the trans-isomer of 2-bromocyclohexanol, which can attain a diaxial geometry necessary for backside attack, undergoes intramolecular SN2 reaction (displacement of Br@ ion by the oxyanion) to yield cyclohexene oxide (an epoxide).

3.154

Organic Chemistry—A Modern Approach

The cis-isomer, on the other hand, cannot attain the antiperiplanar geometry necessary for backside attack and so, it undergoes a much slower reaction in which migration of a hydride ion occurs to form cyclohexanone.

Among the following, two isomeric 1,2-bromohydrins I reacts with alkali to form an epoxide but II does not. H

H OH

Me

Br H I

OH Me

Br H II

These are trans-decalin systems and so, these are locked in these two conformations. i.e., they cannot flip to any other conformation.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

3.155

The bromohydrin I will form an epoxide on treatment with alkali because the nucleophile (—O@) and the leaving group (—Br) are suitably placed (antiperiplanar arrangement) for a backside attack required for an internal SN2 reaction.

Since flipping is not possible, II cannot take up the conformation in which — OH and — Br are diaxial. For this reason, II cannot react with alkali to form an epoxide. (6)

When active trans-2-acetoxyclohexyl tosylate (I) is treated with potassium acetate (AcOK) in acetic acid (AcOH), (±)-trans-1,2-cyclohexanediol diacetate is obtained. However, when active cis-2-acetoxycyclohexyl tosylate (II) is treated similarly, active trans-1,2-cyclohexanediol diacetate is obtained.

3.156

Organic Chemistry—A Modern Approach

An acetoxy group, containing oxygen with unshared electrons, can function as a neighbouring group. In the diaxial form of the trans-isomer (I), the acetoxy (— OCOCH3) group is suitably placed for backside attack (antiperiplanar �� @ to form an acetoxonium arrangement) and so, it pushes out the tosylate anion :OTs �� ion (step 1) and, in doing this, inverts the configuration at the carbon under attack. The acetoxonium ion undergoes ring flip to form its mirror image. These two enantiomeric intermediates then undergoes SN2 attack by the acetate ion (step 2) again with inversion of configuration at the carbon under attack and as a result, racemic trans-1,2-cyclohexanediol diacetate is obtained.

3.157

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

This cis-isomer (II), on the other hand, cannot assume the diaxial conformation required for backside attack by the acetoxy group and therefore, neighbouring group participation is absent in this case. Thus, this isomer reacts by the normal SN2 mechanism to yield active trans-1,2-cyclohexanediol diacetate and the reaction is, therefore, much slower than for the trans-isomer (I) to which anchimeric assistance is available.

3.3.6.2 (1)

p bond as neighbouring group

If an olefinic C == C bond at Cg is stereochemically anti to the leaving group, it may act as a neighbouring group under SN1 conditions. Thus, acetolysis of anti-7norbornenyl tosylate (I) is 1011 times faster than that of its saturated analog (II) and proceeds with retention of configuration. OTs

H

I

OTs

H

II

Since the double bond in anti-7-norbornenyl tosylate (I) is situated in an especially favourable position for backside attack on the carbon bearing the leaving group, it acts as a very effective neighbouring group in the acetolysis of I and assists in the departure of OTs@. For this reason, I undergoes acetolysis at a rate much faster than II where there is no anchimeric assistance is available. The resulting nonclassical or bridged carbocation then reacts with AcOH forming the side opposite to the C == C group to from the corresponding acetate with retention of configuration.

3.158

(2)

Organic Chemistry—A Modern Approach

The C== C bond of an aryl ring at Cb with respect to the leaving group can act as a neighbouring group. For example, optically active isomer of 3-phenyl-2-butyl tosylate gives a racemic mixture of the corresponding acetates. H 3

H

2

H

1

CH3

C——C OTs

H3C Optically active threo -3-phynyl-2-butyl tosylate

3

CH3COOH

H H3C

2

H CH3 +

C——C OAc (2R, 3S)

H3C

2

3

C——C

AcO

H CH3

(2S, 3R)

Racemic mixture of 3-phynyl-2-butyl acetate (optically inactive)

The formation of these two enantiomeric acetates can only be explained if the phenyl group in the b position can function as a neighbouring group. Simple SN2 or SN1 reaction cannot explain their formations. The reaction takes place as follows:

3.159

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

On the other hand, optically active erythro-3-phenyl-2-butyl tosylate undergoes acetolysis to give optically active 3-phenyl-2-butyl acetate instead of racemic mixture. H

H CH3

C——C

H3C

CH3

CH3COOH

OTs H

Optically active erythro-3-phenyl-2-butyl tosylate (2R, 3R)

C——C

H3C

OAc H

Optically active-3-phenyl-2-butyl tosylate (2R, 3R)

3.160

Organic Chemistry—A Modern Approach

The reaction takes place as follows:

(3)

Since in the achiral or meso-phenonium ion the two methyl groups are cis while in the chiral or active phenonium they are trans, therefore, the latter phenonium ion is thermodynamically more stable (due to steric reason) than the former and for this, the erythro-isomer undergoes acetolysis at a rate somewhat faster than the threo-isomer. It is known that the properties of a cyclopropane ring are in some ways similar to those of a double bond. Therefore, it is not surprising that a suitably placed cyclopropyl ring can also be a neighbouring group. For example, acetolysis of endoanti-tricyclo[3.2.1.02,4]octan-8-yl-p-nitrobenzoate (A) is very much (~ 1014 times) faster than that of the p-nitrobenzoate of 7-hydroxybicyclo[2.2.1]heptane (B).

3.161

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

O H

O—C—

O —NO2

H

O—C—

—NO2

A B In the compound A, the suitably placed cyclopropyl ring acts as a very effective neighbouring group (even more effective than a double bond) and assists in the departure of p-nitrobenzoate ion (p-NO2C6H4COO@). Therefore, this compound undergoes acetolysis at a rate very much faster than that of the compound B where no such anchimeric assistance is available. The resulting nonclassical carbocation then undergoes nucleophilic attack by CH3COOH from the side of the expelled leaving group to form the corresponding acetate with retention of configuration.

3.3.6.3 C — C s bond as a neighbouring group Solvolysis in acetic acid of optically active exo-2-norbornyl brosylate produces a racemic mixture of the two exo acetates.

3.162

Organic Chemistry—A Modern Approach

AcOH

OBs H exo-2-Norbornyl brosylate (active)

OAc H

AcO H

(±)-exo-2-Norbornyl acetate

The brosylate undergoes s bond assisted ionization (the C-1 — C-6 s bond acts as a neighbouring group) to give a nonclassical carbocation. This intermediate carbocation then undergoes nucleophilic attack by CH3COOH with equal facility on the two equivalent carbons to give equimolar mixture of two enantiomeric exo-2-norbornyl acetates.

Acetolysis of the exo-isomer is about 350 times faster than its endo-isomer and this because the solvolysis of the endo-isomer is not assisted by the C-1 — C-6 bond because it is not in a favourable position for the backside attack.

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

1.

3.163

Account for the following observation of scrambling: CF COOH

3 Ph14CH 2CH 2OTs æææææ Æ PhCH 214CH 2OCOCF3 + Ph14CH 2CH 2OCOCF3 D

Solution A phenyl group at the b-position can function as a neighbouring group. The reaction occurs to give a mixture containing equal amounts of 1–14C–2-phenylethyl trifluoro acetate (I) and 2–14C–2-phenylethyltrifluoro acetate (II) as follows:

3.164

Organic Chemistry—A Modern Approach

In the first step, the phenyl group pushes out the tosylate anion (an SN2) to form a bridged ion, called a phenonium ion. In the second step, the phenonium ion undergoes nucleophilic attack by fluoroacetic acid at either of two methylene carbons with equal facility to form a mixture of I and II 2.

Explain why b-(p-hydroxyphenyl)ethyl bromide reacts with MeO@/MeOH about 106 times faster than b-(p-methoxyphenyl) ethyl bromide.

Solution In b-(p-methoxyphenyl)ethyl bromide, the — OMe group at the para-position makes the benzene ring a better neighbouring group because it stabilizes the arenium ion intermediate by its +R effect. However, the resulting intermediate is positively charged and relatively less stable.

OMe

OMe MeO MeOH

— Br slow

CH2—CH2—Br b-(p-Methoxyphenyl)ethyl bromide

OMe

CH2

CH2

CH2CH2OMe

An arenium ion intermediate (less stable)

When b-(p-hydroxyphenyl)ethyl bromide is treated with MeO@/MeOH, it is first converted into the corresponding phenoxide ion. The phenoxide ion is a much better neighbouring group because the resulting bridged dienone intermediate is neutral and relatively more stable. Because the neighbouring group effect or anchimeric assistance in b-(phydroxyphenyl)ethyl bromide is much more pronounced than in b-(p-methoxyphenyl) ethyl bromide, the former bromide reacts with MeO@/MeOH at a very much faster rate than the latter bromide.

3.165

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

O

OAC EtOH

3.

Me C

OEt OBs What is the significance of the reaction? O

Solution Formation of such a product indicates that the — OAc group acts as a neighbouring group in the displacement of OBs@.

4.

Give mechanisms of the following reactions:

(a)

N Et

D H2O

Cl N Et O

(b)

CF3COOH CH3C∫∫ C CH2CH2OTs H2O

C—CH3 + CH3

Solution

(a)

O

3.166

Organic Chemistry—A Modern Approach

(b)

5.

Both cyclopropylmethyl and cyclobutyl chlorides undergo hydrolysis at abnormally high rate to give cyclopropylmethyl alcohol, cyclobutyl alcohol and but-3-en-1-ol. Explain these observations.

Solution Cyclopropylmethyl and cyclobutyl chlorides undergo solvolysis very rapidly due to participation of the C — C s bond of the ring. It has been suggested that a common nonclassical intermediate is involved in both the processes and for this, both of these substrates give the three common products.

3.167

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

6.

When trans-3-bromocyclohexane carboxylic acid is treated with sodium hydroxide solution, a g-lactone is obtained. However, when its cis-isomer is similarly treated, no lactone is formed. Explain these observations.

Solution In conformation II of trans-3-bromocyclohexanecarboxylate anion, the — COO① group is suitably placed to displace bromide ion by a backside attack and because of this, an intramolecular SN2 reaction occurs to form a g–lactone.

H Br

O

NaOH

C

H2O

O C

H

–

–

Br

O C

Br H

H

H OH trans-3-Bromocyclohexanecarboxylic acid

–

+

O Na

O H

H

Ring flip

O

II

–

Br SN2

I

O O

C

H H A g -lactone

The backside attack is, however, not possible in either conformation of the cis-isomer because the — COO① group is not properly placed to displace bromide.

3.168

7.

Organic Chemistry—A Modern Approach

Predict the product of the following reaction and explain the mechanistic course involved: H HO CH3

C——C H3C

–

OH H2O

Cl

H

(optically active) Solution This optically active molecule of 3-chloro-2-butanol reacts with alkali to form optically inactive meso-2, 3-butanediol. In the first step, the oxyanion formed from the hydroxyl group pushes out chloride ion (an intramolecular SN2 to form an epoxide. In the next step, the epoxide undergoes substitution by external hydroxide ion in an intermolecular process. Ring opening occurs with equal probability at either carbon atom of the epoxide. The optically inactive meso-2, 3-butanediol is obtained from two stereospecific inversion steps.

8.

Propose a mechanism for the following reaction: OH

OH Me

NaOH

HBr

Me

Me

Cl

Br Me

+

Me

Br

Me

OH

3.169

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Solution The reaction proceeds through an intramolecular SN2 followed by an intermolecular SN2 to give these two isomeric products.

OH

H

H

Me Me ∫∫

OH Me Cl

HO

Me

Me

– –

Me Cl

O

H Me

H H

Cl

–

O

Cl Me

H In the boat form, the oxyanion is properly placed to displace chloride by a backside attack.

H

– Cl

–

Intermolecular SN2

+

O

O Me

H

Br–H

H

Me –

Br

H

Intermolecular Br SN2

OH

OH Me

H Me

Br

H

Me

H

Me

OH

Br

H

Me

Me

Me

H

H

Me Br

HO H ∫∫

∫∫

Br

OH

Br

OH

Me Me

–

Intermolecular SN2

H

Me

Me

Me

Me

Br

H

3.170

9.

Organic Chemistry—A Modern Approach

Explain why the following two alcohols react with HCl to yield the same alkyl chloride.

CH3 CH3 Cl | | | HCl HCl �� — CH — CH — OH æææ �� — CH — CH — OH + H O EtS Æ EtS — CH2 — CH — CH3 ¨ææ æ EtS 2 2 2 �� ��

Solution Both of these two alcohols produce the same episulfonium salt by an intramolecular SN2 reaction. In the episulfonium ion, the substituted carbon is more positive and so, like a protonated epoxide, the more substituted carbon undergoes nucleophilic attack by weak uncleophile Cl① to give the common product.

10.

Explain the formation of exo-2-norbornyl acetate from the following reaction:

OTs

CH3COOH OAc H

Solution Displacement of the leaving group by the double bond of the five-membered ring gives the 2-norbornyl cation.

3.171

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

Opening by acetic acid in an intermolecular SN2 reaction leads to the formation of the exo acetate.

OH +

SN2

+ O ==C

O—H

+

O ==C

CH3

OAc

CH3

H

11.

–H

+

H

Propose a mechanism for each of the following reactions: OH

H O H3O

(a)

(b)

HO—

—(CH2)4OBs

+

OH

(CH3)3COK (CH3)3COH

O Ph C

NHCOPh

(c) H

N

CH3 H

C——C

C3H5OH

OTs

H3COOC

O

C——C

H

KOAC

==

CH3

H

H3COOC

Solution

(a) +

O

+

OH

H—OH2

SN2

OH +

+

—H

H

H2O

OH OH

∫

OH HO

3.172

Organic Chemistry—A Modern Approach

(b)

(c)

12.

How do you account for the following relative rates of acetolysis of p–substituted cyclohexyl brosylates? In which case there is evidence of neighbouring group participation?

Z

Relative rates cis

trans

Cl Br I H

1.6 1.5

5.9 1250 2.2 ¥ 108 1.2 ¥ 104

Solution In solvolysis with SN1 character, halogen atom at b-position could exert two opposing effects. By its electron-attracting inductive effect, it could tend to slow down the reaction by intensifying the positive charge developing on the carbon atom. Through anchimeric assistance, it could tend to speed up to reaction. Only in the trans isomers –X and –OBs can take up the diaxial conformation required for anchimeric assistance. :Z

H H OBs

3.173

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

From the given data it becomes clear that there is no anchimeric assistance by cis-Cl or trans-Cl, or by cis-Br; there is only strong deactivation expected from the — I effect. The trans-Br compound is much more reactive than the cis-Br compound. This indicates that there is anchimeric assistance although not strong enough to offset the inductive effect completely. The trans-I compound provides powerful anchimeric assistance, more than offsetting any inductive (— I) effect and so, in this case solvolysis occurs at a rate 18000 times as fast as for the unsubstituted brosylate. Hence, the ability of halogens to give nucleophilic assistance falls in the order I > Br > Cl. 13.

Suggest a mechanism for each of the following reactions: (a)

EtO2C EtO2C

(b)

CH(CH2)2Br

NaOEt EtOH

CO2Et CO2Et

KOH Et 2N — CH 2 CHCH 3 æææ Æ Et 2N — CH — CH 2OH H2O | | Cl CH3

Solution (a) The base OEt① abstracts the acidic hydrogen to form the corresponding conjugate base. An intramolecular SN2 reaction then takes place to form a cyclopropane derivative. Strain in the three-membered ring is offset by the relatively favourable contribution of entropy to the activation energy.

H (EtO2C)2 C CH2CH2Br

OEt

–Br

(EtO2C)2 C—CH2—CH2—Br Intramolecular SN2

CH2 (EtO2C)2C CH2

(b)

Nitrogen pushes out chloride in an intramolecular SN2 process to form a threememebered ring of aziridinium ion. Attack by the strong nucleophile OH① then takes place at the less hindered position and as a result, the ring is opened up to give the product.

3.174

Organic Chemistry—A Modern Approach

14.

Explain the following observations: OCH3

Cl

OCH3

Cl

Cl

C O

C Cl

O

I

C OCH3

O

III

Cl II

Solution The reaction takes as follows:

Due to favourable geometry in the diaxial conformation of I, and nucleophilic substitution on the carbonyl group occurs to form a lactone containing positive charge on oxygen. An intermolecular SN2 reaction then takes place with inversion of configuration at the carbon under attack. The compound II does not react because in none of the conformations the — OCH3 group is properly placed to displace Cl① from — COCl. O C

OCH3

O

∫ Cl Cl II

:OCH3

H

C H

C O

Cl H

(e, a)

:OCH3

H (a, e)

No reaction

3.175

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

1.

Explain the following observation: *

S

2.

Ag2O

Cl

S

H2O

*

OH +

S *

OH (* == 14C)

Predict the product of the following reactions and account for their formations: (a)

OH HO CH 2CH 2Br æææÆ H O *

(b)

2

OH

(c)

(d)

H

OH H 2N(CH 2 )4 Br æææÆ H O *

2

18OH

OH

H3PO4 H2O

OH H Cl

(e)

HS Me

[Hint: (e)

OH H2O

(f)

Br(CH2)3 COOH

OH H2O

3.176

3.

Organic Chemistry—A Modern Approach

Account for the fact that the two isomeric bromoethers (A and B) undergo solvolysis in acetic acid to give the same mixture of products (C and D). Me Me

MeO

AcOH

Me

MeO

Br

AcOH

MeO Br

OAc C(40%) +

A

B

Me MeO D(60%)

4.

Explain the following observation: HO—CEt2 Me

5.

Br

HO—CEt2

C——Cl

OH H2O

Me

C—OH + Cl

H H [Hint: The retention of configuration is a result of two successive inversions (SN2).] Give the mechanism of the following reaction and account for the relative rate of acetolysis of the following two tosylates (I and II): TsO

H

TsO AcOH

I rate = 104

H

; OAc

II rate = 1

[Hint: The C == C bond of the syn-tosylate I does not assist the ionization of the substrate because it is not properly situated for participation as a neighbouring group. The solvolysis occurs through the formation of a homoallylic carbocation which rearranges to an allylic carbocation. The allylic carbocation then reacts with AcOH to give an acetate. The high reactivity (104 times) of I compared to II is due to participation of s electron of the two allylic s bonds (C-1 — C-6 and C-4 — C-5).

3.177

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

6.

Explain why the compound I undergoes solvolysis at a rate much slower than the compound II. O O CH3O—

—S—O

CH3O—

H

—S—O

O

H

O CF3 CF3 I

7.

[Hint: The ability of C ==C to serve as a neighbouring group depends on its electron density.] Arrange the following compounds in order of increasing rate of solvolysis and explain the order: H

OCOC6H4NO2-p

I

8.

II

H

II

OCOC6H4NO2-p

H

OCOC6H4NO2-p

III

The brosylate I undergoes acetolysis ~ 140,000 times faster than its saturated analog II. Explain.

3.178

Organic Chemistry—A Modern Approach

OBs

H

9.

10.

OBs

H

I II [Hint: The double bond in I is geometrically fixed in an especially favourable position for backside attack on the carbon bearing the leaving group and so, there is a very large rate enchancement.] When 5-chloro-2-hexyl tosylate is solvolyzed in acetic acid, there is little participation by the Cl, but when the solvent is changed to trifluoroacetic acid, neighbouringgroup participation by the Cl becomes the major reaction pathway. Explain. [Hint: Cl is a very weak neighbouring group and can be shown to act in this way only when the solvent does not interfere.] When G ==OCH3, about 93% of the products of solvolysis of I can be attributed to the participation of the neighbouring phenyl group. But, when G==NO2, essentially none of the products come from neighbouring phenyl group participation. Explain. HCOOH

G

11.

OTs

I The compound I undergoes solvolysis in 80% aqueous ethanol 105 – 106 times faster than its cis-isomer II. Explain.

[Hint: The high solvolysis rate of I is due to participation of the neighbouring thiophenoxy group leading to a cyclic intermediate.] 12.

The tosylate I solvolyzes 3 ¥ 105 times faster than the tosylate II. Explain. TsO

TsO

I

13.

II

Explain why ‘mustard gas’, S(CH2CH2Cl)2, a high boiling liquid used in World War I as a combat gas in the form of aerosol, hydrolyses much more rapidly to give HCl and a diol than its sulphur free analog, 1,5–dichloropentane.

3.179

Nucleophilic Subs tu on Reac ons at Saturated Carbon Atom

[Hint: The reason for the high rate of hydrolysis of mustard gas is due to participation of sulphur atom as the neighbouring group in an intramolecular SN2 reaction

14.

In the nucleophilic substitution reaction of the following 14C labelled compound with water, what labelling pattern is expected to be observed in the product (a) if the neighbouring group participation does not take place and (b) if neighbouring group participation does take place? S

15.

Cl (* ==14C)

Although there is a substantial difference in the rate at which I and II solvolyze (I undergoes solvolysis 4.4¥104 times faster than II is acetic acid), both compounds give products of completely retained configuration. Br

H

I

16.

*

Br

H

II

Offer a mechanistic interpretation of the above observation. Suggest a mechanism for the following reaction:

3.180

Organic Chemistry—A Modern Approach

[Hint:

17.

Which reaction in each pair would be expected to be faster? Give your reasoning. CH3 CH3 S S (a)

CF3—

–C—O—C—Ph or

CH3—

CH3

H

H

CH2CH2ONoS

(b)

–C—O—CPh (solvolysis in 100% ethanol) CH3

CH2CH2ONoS

or

(solvolysis in acetic acid)

(c)

PhS(CH2 )3 Cl or PhS(CH2 )4 Cl (solvolysis in methanol)

(d)

PhO2C—

—CHBrPh or

—CHBrPh CO2Ph

(solvolysis in acetic acid)

4

ELIMINATION REACTIONS

CHAPTER

Chapter Outline 4.1

4.2

The E2 Reaction 4.1.1 Example of E2 Reaction 4.1.2 Kinetics of E2 Reaction 4.1.3 Mechanism of E2 Reaction 4.1.4 Stereochemistry of E2 Reaction 4.1.5 Evidence in Favour of the E2 Mechanism 4.1.6 Factors Influencing E2 Reaction Rate or E2 Reactivity 4.1.7 Factors that Govern the Proportions of E2 and SN2 Reactions 4.1.8 Regioselectivity in b-elimination Reactions (Orientation of p Bond in the Product Alkene) 4.1.9 Hofmann Exhaustive Methylation or Hofmann Degradation 4.1.10 Fragmentations 4.1.11 Summary of the E2 Reaction The E1 Reaction 4.2.1 Example of E1 Reaction 4.2.2 Kinetics of E1 Reaction 4.2.3 Mechanism of E1 Reaction 4.2.4 Stereochemistry of E1 Reaction 4.2.5 Evidence in Favour of the E1 Mechanism 4.2.6 Factors Influencing E1 Reaction Rate or E1 Reactivity 4.2.7 Regioselectivity of E1 Reactions

4.2.8

Rearrangement of the Carbocation Intermediate Involved in an E1 Reaction 4.2.9 Acid-Catalyzed Dehydration of Alcohols 4.2.10 Dehydration using POCl3 and Pyridine 4.2.11 Factors Influencing the Extent of E1 and E2 Reactions 4.2.12 Factors that Govern the Proportions of E1 and SN1 Reactions 4.3

The E1CB Reaction 4.3.1 Example of E1cB Reaction 4.3.2 Kinetics of E1cB Reaction 4.3.3 Mechanism of E1cB Reaction 4.3.4 The Nature of the Substrate 4.3.5 To Distinguish between E1cB and E2 Mechanisms

4.4

a- or 1,1-Elimination 4.4.1 Example of a- or 1,1-elimination Reaction 4.4.2 Kinetics of a- or 1,1-elimination Reaction 4.4.3 Mechanism of a- or 1,1-elimination Reaction 4.4.4 Structure of the Substrate Involved in a-elimination Reactions

Organic Chemistry—A Modern Approach

4.2

INTRODUCTION Elimination reactions belong to one of the major class of organic reactions involving the loss of two atoms or groups from a substrate molecule. Thus, it is the reverse process of addition reaction. Alkyl halides, in addition to the nucleophilic substitution reactions, also undergo elimination reactions. The loss of one H atom and the Br atom (dehydrobromination) from 2-methyl-2-bromopropane by the action of alcoholic alkali is a typical example of elimination reaction. alcohol (CH 3 )3CBr + KOH ææææ Æ (CH 3 )2 C == CH 2 + KBr + H 2O D 2-Methyl-2-bromopropane 2-Methylpropene

The carbon atom that bears the leaving group (e.g., the halogen atom in the above reaction is called an alpha (a) carbon atom and any carbon atom adjacent to it is called a beta (b) carbon atom. A hydrogen atom attached to the b carbon atom is called a b hydrogen atom. The carbon next to b is called a g carbon atom. On the basis of the relative positions of the eliminating groups or atoms, the elimination reaction may be classified as: (a) 1,1 or a-elimination reaction, (b) 1,2- or b-elimination reactions and (c) 1,3 or g -elimination reactions. (a)

1,1- or a-Elimination reactions: When the two atoms or groups are lost from the same atom of a substrate molecule in an elimination reaction, the reaction is said to be a 1,1- or a-elimination reaction. Such a reaction leads to the formation of very reactive intermediate, carbene, rather than an alkene. For example, when chloroform (CHCl3) is heated with alcoholic KOH solution, it undergoes a 1,1- or a-elimination reaction to yield dichlorocarbene (:CCl2). 1(a )

alc. KOH :CCl2 + H 2O + KCl CHCl3 ææææÆ D

(b)

1,2 - or b-Elimination reaction: When the two atoms or groups are lost from the two adjacent atoms of a substrate molecule in an elimination reaction, the reaction is said to be 1,2-elimination reaction. In the older ab-terminology, a is commonly omitted, and the reactions are referred to as b-eliminations. In a 1,2 - or b-elimination reaction a p bond is formed.

The formation of cyclohexene from cyclohexyl chloride by the action of sodium ethoxide in ethanol is an example of 1,2- or b-elimination reaction.

Elimina on Reac ons

(c)

4.3

1,3- or g-Elimination reaction: When in an elimination reaction the two atoms or groups are lost from two different atoms which remain separated by another atom, the reaction is called a 1,3- or g-elimination reaction. These less common elimination reactions lead to the formation of three-membered rings. For example, when the following quaternary ammonium iodide is treated with sodamide in liquid ammonia it undergoes 1,3-elimination reaction to form a cyclopropane derivative.

Depending upon the mechanism involved 1,2- or b-elimination may be classified as E1, E2 and E1cB. Most common b-eliminations follow E2 path. • Competition between nucleophilic substitution and b-elimination reactions Because the reactive part of both nucleophiles and bases is an unshared electron pair, all nucleophiles are potential bases and all bases are potential nucleophiles. For this reason, nucleophilic substitution reaction and elimination reaction often compete with each other. When the substrate is a primary halide and the base is strong and unhindered, like ethoxide ion (OEt * ), substitution in highly favoured and this is because the base can easily approach the carbon bearing the leaving group. For example: * ≈

C H OH

2 5 CH 3CH 2 O Na + CH 3CH 2Br æææææ Æ CH 3CH 2OCH 2CH 3 + CH 2 == CH 2 55∞ C Primary E2 Product SN 2 Product Minor(10%) substrate Major(90%)

With secondary substrates, however, a strong base favours elimination reaction because substitution becomes more difficult due to steric hindrance in the substrate. For example: Br | * ≈ C2H5OH C2H5 ONa + CH 3 CH CH 3 æææææ Æ CH 2 == CHCH 3 + (CH 3 )2 CH — O — C2H5 55∞ C Secondary E2 Product SN 2 Product substrate Major (79%) Minor (21%) With tertiary halides, an SN2 reaction cannot take place because steric hindrance in the substrate is severe and thus the elimination reaction is highly favoured, especially when the reaction is carried out at higher temperature. Any substitution that occurs must take place through an SN1 reaction. For example:

Organic Chemistry—A Modern Approach

4.4

At room temperature *

≈

C H OH

2 5 C2H5 O Na + (CH 3 )3 C — Br æææææ Æ (CH 3 )2 C == CH 2 + (CH 3 )3 C — O — C2H5 25∞ C SN1 Product Tertiary E2 Product substrate Major (91%) Minor (9%)

At higher temperature *

≈

C H OH

2 5 C2H5 O Na + (CH 3 )3 C — Br æææææ Æ (CH 3 )2 C == CH 2 55∞ C Tertiary OnlyE2 + E1 Product substrate (100%)

4.1

THE E2 REACTION

Example, kinetics, mechanism, stereochemistry, evidence in favour of the mechanism, the factors influencing E2 reactivity, i.e., the E2 reaction rate and other relevant topics.

4.1.1 Example of E2 Reaction Dehydrobromination of isopropyl bromide caused by the strong base sodium ethoxide (NaOEt) to yield propene is a typical example of E2 reaction, i.e., a reaction which proceeds through the E2 mechanism. EtO : * +

4.1.2

EtOH (CH 3 )2 CHBr ææææ Æ CH 3 CH == CH 2 + EtOH + :Br :* Isopropyl bromide Propene (79%)

Kinetics of E2 Reaction

The E2 reaction exhibits second-order kinetics (first order in the substrate and first order in the base) and obeys the following rate law: Rate = k ÎÈ(CH3 )2 CHBr ˚˘ [EtO :* ] The rate of elimination thus depends on the concentration of both the substrate and the base. Therefore, the reaction is bimolecular, i.e., the substrate (Me2CHBr) and the base ( : OEt * ) are involved in the transition state of the rate-determining step.

4.1.3

Mechanism of E2 Reaction

The most straight-forward explanation for the observed second-order kinetics is a concerted reaction, i.e., bond breaking and bond making occur simultaneously. The mechanism of this reaction is designated as E2 (E stands for elimination and 2 stands for bimolecular) because the two reactants (the substrate and the base) are involved in the transition state of the rate-determining step (the only step).

Elimina on Reac ons

4.5

In the E2 reaction, the base (EtO :* ) abstracts a proton from the b-carbon of the isopropyl bromide and simultaneously the leaving group ( :Br :* ) departs from the a-carbon with the formation of a double bond. The electron pair in the b C — H bond forms the new p bond and the leaving group Br① comes off with the electron pair in the C — Br bond. Thus, it is a concerted elimination, i.e., a one-step process which passes through a single transition state. In the transition state, there is a partially formed O — H bond, a partly broken H — C bond, a partly formed C == C bond and a partly broken C — Br bond. The process can be represented as follows:

Entropy favours the products of an E2 reaction because two molecules of starting material are formed from three molecules of product. The energy profile diagram for this E2 reaction is as follows:

4.1.4

Stereochemistry of E2 Reaction

There is a stereoelectronic requirement of an E2 process and the requirement is that the groups to be eliminated should be anti-periplanar (i.e., conformationally trans to one another with a dihedral angle of 180°) and not syn-periplanar (i.e., conformationally cis to one another with a dihedral angle of 0°).

Organic Chemistry—A Modern Approach

4.6

The reasons for such stereoelectronic requirement of anti-periplanarity are as follows: (i) If the sigma orbitals of the departing groups become coplanar, there occurs maximum overlap of the developing p orbitals. This causes the transition state to have minimum energy and maximum stability. (ii)

Unlike syn geometry the anti geometry allows the b C — H bonding electrons to displace the leaving group from the backside. Like SN2 this familiar backside attack is energetically advantageous from the electronic point of view.

(iii)

The attacking base and the leaving group remain far apart from each other in anti geometry but not in syn geometry and, therefore, the steric interactions are minimized.

Elimina on Reac ons

(iv)

4.7

Approach of the base towards the anti-periplanar conformation, but not towards the syn-periplanar conformation, avoids electrostatic repulsion between the negatively charged base and the leaving group.

The E2 is a stereospecific reaction, because different stereoisomers of the starting material react to give different stereoisomers of the product. This sterospecificity results from anti-coplanar transition state that is usually involved in the E2 reaction. The stereospecificity in E2 reaction may be observed in the following examples. (1) (1R, 2R)-1-Bromo-1,2-diphenylpropane undergoes base-induced dehydrobromination to give cis-1,2-diphenylpropene but not trans-1,2-diphenylpropene. The diastereoisomeric bromide (1R, 2S)-1-bromo-1,2-diphenylpropane, on the other hand, undergoes base-induced dehydrobromination to give trans-1,2-diphenylpropene but not cis-1,2-diphenylpropene. Also, the former reaction takes place at a much slower rate than the latter.

4.8

Organic Chemistry—A Modern Approach

Elimina on Reac ons

4.9

In both of these two reactions, the anti-elimination product is obtained exclusively. This stereospecificity is the proof of the existence of a stereoelectronic requirement or stereoelectronic constraint upon the E2 reaction. The molecule of 1-bromo-1,2-diphenylpropane starts to flatten with the extent of formation of the double bond. In the transition state I, the bulky phenyl groups crowded together on the same side of the molecule, whereas in the transition state II, they lie in roomier positions on opposite sides of the molecule. Because of greater van der Waals strain, the T.S. I is less stable than the T.S. II. Thus, the free energy of activation for the formation of the cis-alkene is greater than that for the formation of the trans-alkene and because of this, the (1R,2R)-isomer undergoes dehydrobromination at a much slower rate than the (1R, 2S)-isomer does. (The ground states of these two diastereoisomeric bromides also differ in stability, but the effect is not considerable and may be neglected in assessing relative free energies of activation.)

4.10

(2)

Organic Chemistry—A Modern Approach

Iodide catalyzed debromination of (2R, 3S)-2,3-dibromo-2,3-dideuteriobutane produces trans-2,3-dideuterio-2-butene, but of (1R, 2S)-1,2-dibromo-1,2-dideuterioethane produces cis-1,2-dideuterioethene. In (2R, 3S)-2,3-dibromo-2,3-dideuteriobutane, both of the Br atoms are attached to secondary carbon atoms and so, the compound undergoes E2 reaction (instead of SN2) to give only trans-2,3-dideuterio-2-butene by an anti-elimination process. The reaction proceeds as follows:

On the other hand, (1R, 2S)-1,2-dibromo-1, 2-dideuterioethane does not undergo debromination directly. Since the Br atoms are attached to primary carbon atoms, it undergoes first an SN2 reaction (displacement of Br① by I①) with inversion of configuration and then an E2 reaction to yield cis-1,2-dideuterioethene by an antielimination process.

Elimina on Reac ons

(3)

4.11

The following two diasteoisomeric bromides undergo E2 reaction to give different products when treated with sodium methoxide in methanol and this is because of anti stereospecificity of E2 reaction.

In the trans-isomer of the bromide, the H and Br atoms are anti-coplanar to each other and it is the most suitable geometry for base-promoted E2 elimination. On the other hand, in the cis-isomer, the vinylic hydrogen is cis to Br, but the allylic hydrogens at C-3 can easily adopt the preferred anti conformation. For these reasons, the trans-isomer undergoes base-promoted E2 elimination to yield oct-4yne (an alkyne), while the cis-isomer undergoes base-promoted E2 elimination to yield octa-3,4-diene (an allene, i.e., a compound containing two double bonds that share a single carbon atom).

4.12

Organic Chemistry—A Modern Approach

(4)

When the erythro and the threo isomers of sodium 2,3-dibromo-3-pnitrophenylpropanoate are heated, they undergo decarboxylative debromination (E2) to give cis- and trans-1-bromo-2-p-nitrophenylethene, respectively. It thus follows that these elimination reactions must have taken place in anti fashion.

(5)

When (3R, 4R)-3,4-dibromohexane is heated with zinc in methanol, it undergoes debromination (E2) to yield cis-3-hexene. However, when (3R, 4S)-3,4dibromohexane is similarly treated, debromination takes place to give trans-3hexene.

Elimina on Reac ons

(6)

4.13

In cyclohexane derivatives, the stereoelectronic requirement of E2 reaction, i.e., the anti-periplanar orientation of the departing groups is satisfied only when the groups are in diaxial (a, a) positions but not in diequatorial (e, e) or equatorial-axial (e, a) positions. In fact, when they are equatorial-axial, they are syn to each other and also not coplanar. When they are diequatorial, they are anti to each other but not coplanar. They are anti as well as coplanar only when they are diaxial. Therefore, the diaxial conformer of cyclohexane derivatives is very susceptible to E2 reaction as the stereoelectronic effect is helpful for this concerted process.

(a)

When the cis-isomer of 4-tert-butylcylohexyl tosylate is heated with NaOEt in EtOH at 75°C, it undergoes a facile E2 reaction to yield 4-tert-butylcyclohexene. However, the trans-isomer reacts slowly with NaOEt to give the same product when similarly treated.

4.14

Organic Chemistry—A Modern Approach

The bulky tert-butyl group ( —< ) in both cis- and trans-4-tert-butylcyclohexyl tosylate always assumes an equatorial position, holding the ring in a single conformation. The cis-isomer exists nearly exclusively in the axial-tosylate conformation, while the trans-isomer exists nearly exclusively in the equatorial-tosylate conformation. In the thermodynamically stable axialtosylate conformation of the cis-isomer, the -OTs group and a b-H atom (either at C-2 or at C-6) are anti-periplanar. Because of such arrangement of the departing groups in which the stereoelectronic requirement is satisfied, this isomer undergoes a facile E2 reaction, i.e., the loss of TsOH occurs at a faster rate.

On the other hand, in the stable equatorial tosylate conformation of the transisomer, the -OTs group cannot become coplanar with a b-H atom (either at C-2 or at C-6), i.e., in this conformation the stereoelectronic requirement is not fulfilled. Although the stereoelectronic requirement is satisfied in the energetically unfavourable axial-tosylate conformation, flipping to this conformation, which suffers from strong 1,3-diaxial interactions, does not practically take place. For these reasons, the trans-isomer does not undergo E2 reaction under the same reaction conditions. Slow elimination does take place, but by a unimolecular (E1) mechanism.

Elimina on Reac ons

(b)

4.15

When neomenthyl chloride is treated with sodium ethoxide in ethanol, it undergoes E2 elimination of HCl at a much faster rate (about 200 times faster) than does menthyl chloride under similar reaction conditions. Again, neomenthyl chloride produces a mixture of 75 percent 3-menthene and 25 percent 1-menthene, while menthyl chloride produces 2-menthene as the only product.

4.16

Organic Chemistry—A Modern Approach

In the thermodynamically more stable conformation (the two relatively bulky alkyl groups are in equatorial positions) of neomenthyl chloride, the chlorine atom is axial and there are axial hydrogens at both C-2 and C-4. This substrate, therefore, reacts readily with ethoxide ion (EtO①) through this energetically favourable conformation in which the stereoelectronic requirement for E2 reaction is satisfied and gives a mixture of 2- and 3-menthenes. 3-Menthene predominates in the product mixture because it is more substituted and thermodynamically more stable than 2-menthene. The reaction is governed by the Saytzeff rule.

On the other hand, the thermodynamically more stable conformation of menthyl chloride has all three groups (including the chlorine) equatorial. However, this conformation is not suitable for facile E2 elimination of HCl because chlorine is equatorial. For the chlorine to become axial, menthyl chloride has to assume a conformation in which the large isopropyl group (—CHMe2) and the methyl group are also axial. This conformation is of much higher energy, and the free energy of activation for the reaction is large because it includes the energy required for the conformational change. For this reason, menthyl chloride undergoes E2 dehydrochlorination very slowly. In this conformation, the only axial hydrogen b to the axial chlorine is at C-2. The reaction, therefore, leads to the formation of only 2-menthene (the Hofmann product).

Elimina on Reac ons

(c)

4.17

Both the trans- and the cis-isomer of 1,2-dibromocyclohexane produces cyclohexene when treated with KI in ethanol.

The trans-isomer of 1,2-dibromocyclohexane undergoes debromination by the iodide ion (I①) through the more stable diaxial conformation (in which the stereoelectronic requirement for E2 is satisfied) to yield cyclohexene.

The cis-isomer of 1,2-dibromocyclohexane, on the other hand, does not undergo E2 debromination, since the two Br atoms are not anti-periplanar. In fact, it yields the same product by a merged substitution-elimination (SN2 – E2)

4.18

Organic Chemistry—A Modern Approach

reaction. As a nucleophile I① displaces the axial Br atom by an SN2 reaction. The resultant bromo-iodo compound then undergoes anti-elimination (E2) of IBr to yield cyclohexene.

(7)

The lactone I undergoes dehydrobromination to give the lactone II when treated with NaOMe in MeOH, but the lactone III undergoes dehydrobromination to give the lactone IV when treated similarly.

In the lactone I, the bromine atom and the b-H atom at the ring junction are antiperiplanar. Therefore, it undergoes E2 elimination of HBr smoothly to yield the lactone II (the Saytzeff product) when treated with NaOMe/MeOH. On the other hand, in the lactone III, the Br atom is syn to the junction b-H atom and also they are not coplanar. However, it can be anti-periplanar with a methyl b-hydrogen. Therefore, this lactone undergoes E2 dehydrobromination involving a methyl hydrogen smoothly to give the lactone IV (the Hofmann product).

Elimina on Reac ons

(8)

4.19

That E2 elimination is stereospecifically trans (anti) can be well demonstrated by the following example. The trans-decaline derivative I in which Cl atom is axial undergoes dehydrochlorination to give two isomeric alkenes (Cl is anti-periplanar with two axial b–H atoms) when treated with sodium ethoxide in ethanol.

On the other hand, the trans-decaline derivative II in which Cl atom is equatorial does not undergo E2 dehydrochlorination because Cl is not anti-periplanar with any b-H atom. Furthermore, a trans-decaline system cannot flip and so, chlorine cannot become axial for a smooth E2 elimination. H

H

H H Cl

NaOEt/EtOH

No elimination reaction

H

II

(9)

Due to anti stereochemical disposition of H and Br, trans-2-bromobut-2-ene undergoes E2 dehydrobromination at 50 times faster rate than the cis-isomer

Organic Chemistry—A Modern Approach

4.20

during treatment with NaNH2 in liquid NH3. Because of less favourable synelimination, the latter reaction is slow.

•

Syn-elimination

There is experimental evidence that compounds in which the departing groups are coplanar on the same side of the molecule (i.e., eclipsed) may also undergo concerted elimination. The groups are said to be syn-periplanar, and the process is called syn-elimination. Syneliminations take place only under forcing conditions and only when the departing groups cannot achieve anti-periplanarity. (1)

syn-elimination most often occur in certain rigid cyclic systems. In the following bicyclic compound, for example, the quaternary ammonium group and the b-H atom cannot become coplanar (the dihedral angle is about 120°). The syn deuterium atom is, however, coplanar with the ammonium group (the dihedral angle is about 0°) and so, it is lost in the elimination reaction (syn-elimination) that takes place when this hydroxide is heated.

Elimina on Reac ons

4.21

Similarly, the following deuterated norbornyl bromide undergoes base-induced syn-elimination of HBr to yield the corresponding deuterated alkene.

Stereoelectronic considerations suggest that syn-periplanarity of the departing groups allows a reasonable conversion of the reactant molecular orbitals to the p orbital of the resulting alkene. Coplanarity of the departing groups, whether anti or syn, is the most important factor for reactions that proceed by the E2 mechanism. However, the syn-elimination taking place from the higher energy ‘eclipsed’ conformation is considerably less favourable than anti-elimination taking place from the lower energy ‘staggered’ conformation. That coplanarity of the departing groups, whether syn or anti, is most important for reaction that proceed by the E2 mechanism can be established by the following observations. The trans-isomer of 11,12-dechloro-9,10-dihydro-9,10-ethanoanthracene undergoes base-induced dehydrochlorination at a rate faster than its cis-isomer. In the cisisomer, the dihedral angle between the departing groups is about 120°. Because of regidity of the molecule, the departing groups cannot achieve a dihedral angle of 180°, i.e., the departing groups cannot become anti-periplanar. It thus undergoes anti-elimination with a dihedral angle of about 120°. In the trans-isomer, on the other hand, the dihedral angle between the departing groups is about 0°, i.e., the groups are syn-periplanar. This isomer thus undergoes base-induced syn-elimination of HCl. Since coplanarity is a very important requirement for E2, therefore, synelimination with a dihedral angle of about 0° is energetically more favourable than anti-elimination with a dihedral angle of about 120°. Dehydrochlorination, therefore, takes place more rapidly from the trans-isomer than from the cisisomer.

4.22

(2)

Organic Chemistry—A Modern Approach

Syn-elimination may also take place in acyclic molecules when anti-periplanar transition state is highly destabilized by the steric strain caused due to eclipsing of the two very bulky alkyl groups. For example, both (4S, 5R)- and (4R, 5R)-4-bromo-3,3,5,6,6-pentamethyloctane undergo base-induced E2 dehydrobromination to give the same alkene.

In the (4S, 5R)-diastereoisomer of this bromoalkene, the b-H atom and the Br atom can be anti-periplanar with the very bulky tert-pentyl groups anti to each other. Thus, the transition state for anti-elimination is considerably stable (unencumbered by any steric strain from the tert-pentyl groups) and so, this

Elimina on Reac ons

4.23

diastereoisomer undergoes smooth E2 dehydrobromination through the favoured staggred conformation to yield (E)-3,3,4,6,6-pentamethyl-4-octene (the alkene with trans tert-pentyl groups). On the other hand, the transition state for anti-elimination of HBr from the (4R, 5R)-diastereoisomer is very much destabilized because the bulky tert-pentyl groups are crowded together on the same side of the molecule and become involved in server steric interaction. However, syn-coplanarity of Br and H atom can be attained with the bulky tert-pentyl groups anti to each other. This diasteroisomer, therefore, undergoes syn-elimination of HBr involving a relatively stable transition state to form the same diasterioisomeric alkene.

Organic Chemistry—A Modern Approach

4.24

(3)

There are a number of organic compounds (e.g., esters like acetates and xanthaltes, amine oxides, etc.) that undergo syn-elimination through a cyclic transition state when heated with no other reagent present. The formation of cyclic transition states in these reactions is evidenced by a large decrease of entropy. These reactions, which are run either in inert solvents or in the gas phase, are referred to as Ei eliminations (elimination, intramolecular). The degree of syn stereoselectivity also reflects the exent to which they proceed via cyclic transition states.

(a) Pyrolysis of acetates

The pyrolytic elimination of an acetate containing at least one b-H atom proceeds concertedly through a six-membered cyclic transition state. The cyclic pathway required a syn-coplanar arrangement of the departing groups and so, the reaction is a highly stereoselective syn-elimination. The stereochemical course and the mechanism of this intramolecular E2 reaction may be shown as follows:

Elimina on Reac ons

4.25

When there are two possibilities of such intramolecular elimination, the major product is that which is obtained through a more stable transition state. The following acetate, for example, undergoes pyrolytic syn-elimination to yield the alkene I as the major product and the alkene II as the minor product.

The six-membered cyclic transition state leading to the formation of I is planar and geometrically as well as thermodynamically more favourable, while the six-membered transition state leading to II is not planar (C — O and b C — H bonds are not coplanar) and thermodynamically less favourable.

4.26

Organic Chemistry—A Modern Approach

When there are two diastereotopic b-hydrogen in an acetate, that hydrogen will be eliminated which leads to the formation of a more stable transition state to give the major product. For example (1S, 2R)-2-deuterio-1,2-diphenylethyl acetate undergoes pyrolytic elimination to give trans-stilbene predominantly, while the (1S, 2S)-isomer undergoes pyrolytic elimination to give trans-1-deuteriostilbene predominantly.

Since these pyrolytic syn-eliminations involve expulsion of the acetoxy group with a cis b-deuterium, therefore, the (1S, 2R)-isomer of 2-deuterio-1,2-diphenylethyl acetate is expected to give trans-stilbene as well as cis-1-deuteriostilbene. However, because of steric interaction between two bulky and eclipsing phenyl groups, the transition state leading to the formation of cis -1-deuteriostilbene is less stable than that leading to the formation of trans-stilbene. For this reason, trans-stilbene is formed readily and predominantly. Due to the same basic reason, trans-1-deuteriostilbene instead of cis-stilbene is obtained predominantly form the (1S, 2S)-isomer.

Elimina on Reac ons

Mechanism:

4.27

4.28

Organic Chemistry—A Modern Approach

(b) Pyrolysis of xanthates Pyrolysis of xanthates to alkenes is known as the Chugaev reaction. The following reaction sequence involving the Chugaev reaction is an indirect method of dehydration of alcohols.

Elimina on Reac ons

4.29

In the reaction sequence, the alcohol is first converted to the corresponding alkoxide ion by the action of alkali. The alkoxide then reacts with carbon disulphide to yield O-alkyl sodium xanthate. The O-alkyl sodium xanthate subsequently reacts with methyl iodide by an SN2 mechanism to form a xanthate ester. The xanthate ester then undergoes pyrolytic elimination (Ei elimination), a concerted reaction that proceeds through a six-membered cyclic transition state to yield an alkene. The reaction exhibits higher degree of syn stereoselectivity. Mechanism:

(c) Pyrolysis of tertiary amine oxides When tertiary amine oxides containing at least one b-H atom are heated, they undergo decomposition to yield an alkene and a dialkylhydroxylamine. This is called Cope elimination reaction. For example: ≈

D PhCH 2CH 2 N(CH 3 )2 ææææ Æ Ph CH == CH 2 + (CH 3 )2 NOH ª 150∞ C | Styrene Dimethylhydroxyl - amine :O* A tertiary amine oxide

In this Ei reaction, the negatively charged oxygen of the amine oxide is the base that removes a proton from the b-carbon. The reaction proceeds through a planar five-membered cyclic transition state. The mechanism of the reaction requires the departing groups to be syn-coplanar and in fact, the reaction exhibits the greatest degree of syn stereoselctivity of any of the Ei reactions.

4.30

Organic Chemistry—A Modern Approach

The Cope elimination reaction is very useful for the preparation of many olefins because at relatively low reaction temperature, possible isomerization of the resulting alkene is minimized and also because of mild conditions, the side reactions are few. The five-membered cyclic transition state involved in the Cope elimination reaction must be completely planar can be well demonstrated by the following reactions.

Since it is relatively easier to force an axial and an equatorial bond into a place than two equatorial bonds, therefore, an equatorial and an axial substituent (i.e., cis to each other) become involved in pyrolytic elimination more easily (low activation energy) than two equatorial substituents (i.e., trans to each other). The cis-amine oxide undergoes unfavourable anti-elimination of the equatorial — NMe2 Æ O group and the equatorial b-H at C-2 to yield 1-phenycyclohexene as the minor product. However, it undergoes favourable syn-elimination of the equatorial — NMe2 Æ O group and the axial b-H at C-6 to yield 3-phenylcyclohexene as the major product (almost exclusively).

Elimina on Reac ons

4.31

The pyrolytic elimination of trans-amine oxide leading to the formation of both 1- and 3-phenylcyclohexene involve a favourable syn-elimination of an equatorial — NMe2 Æ O group and an axial b-H (at C-2 or C-6). An equal amount of both the alkenes is expected to be obtained from this isomer. In fact, 1-phenylcyclohexene is obtained predominantly and this is because 1-phenylcyclohexene, being a conjugated alkene, is thermodynamically more stable than the isomeric 3-phenylcyclohexene.

In the case of acyclic systems, the Cope elimination have a carbanion-like transition state. That is, the major product of the Cope elimination, like that of Hofmann elimination, is the one obtained by removing a hydrogen from the b-carbon bonded to the most hydrogens. For example: CH3 |! D CH 3CH 2 NCH 2CH 2CH 3 ææ Æ CH 2 == CH 2 + CH 3N(OH)CH 2CH 2CH 3 | :O @ However, in the case of cyclic compounds, the transition state of the Cope elimination possesses considerable double bond character, i.e., the Saytzeff rule is followed. For this reason, although the formation of both the alkenes I and II from the following tertiary amine oxide involves planar five-membered transition state, the transition state leading to the formation of the more substituted alkene I is relatively more stable than the transition state leading to the formation of the less substituted alkene II. Because of this, the thermodynamically more stable alkene I is obtained nearly exclusively.

Organic Chemistry—A Modern Approach

4.32

4.1.5 (a) (b)

(c)

Evidence in Favour of the E2 Mechanism The 1,2- or b-elimination reactions which are designated as E2 reactions have been found to follow second-order kinetics. If the C — H bond cleavage take place in the rate-determining step of a reaction, then the same reaction of the deuterated compound will take place at a slower rate (C — D bond is stronger than the C — H bond). The variation in the reaction rate is called primary kinetic isotope effect. For second-order E2 eliminations, an isotope effect of 3 to 8 has been observed. This provides a very strong support for the ratedetermining cleavage of C — H (or C — D) bond in E2 reactions. In a second-order elimination, two-step mechanism involving reversible formation of a carbanion may also operate.

If a reaction which proceeds by this mechanistic pathway is carried out in a solvent which could act as a deuterium source (e.g., EtOD) and the reaction is interrupted before completion, the recovered substrate is expected to contain deuterium.

Elimina on Reac ons

(d)

(e)

4.1.6 (a)

(b)

4.33

The recovered substrates in typical second-order elimination reactions are actually found to be free of any deuterium. It is inconsistent with the mechanism in which carbanions are formed reversibly but consistent with the concerted E2 mechanism in which there is no scope of deuterium incorporation. However, the mechanism in which the substrate slowly loses a proton to form a carbanion and the carbanion then rapidly loses the halide ion (X①) cannot be ruled out in the absence of hydrogen exchange. The first step is essentially irreversible. The rate of C — X bond cleavage must have no effect on the overall reaction rate if this mechanism operates. Actually, the rates of typical second-order elimination reactions were found to depend on the strength of the C — X bond, i.e., the rate varies with X. It is inconsistent with the (irreversible) carbanion mechanism but consistent with the E2 mechanism in which the C — X bond breaking is involved in the rate-determining step. Rearranged products are never obtained in typical second-order E2 eliminations. This is consistent with the concerted (one-step) mechanism because it provides no opportunity for rearrangement. E2 reactions are stereospecific. The threo isomers give trans-olefins and the erythroisomers give cis-olefins usually by the E2 pathway. This indicates that not only a trans-elimination occurs but also the process of elimination is a concerted one.

Factors Influencing E2 Reaction Rate or E2 Reactivity The effect of substrate structure on rate: The major structural features that can increase the rate of E2 reaction are those that serve to stabilize the resultant alkene or more particularly, the transition state that precedes it. Therefore, increasing alkyl substituent at both a- and b-carbon atoms of the substrate molecule facilitates the E2 reaction, because the corresponding alkene is thermodynamically more stable and the competing SN2 reactions become progressively slower due to steric reasons. The effect of base on rate: Since the base appears in the rate equation of E2 reaction, therefore, the reaction rate increases as the strength and concentration of the base increases. Stronger the base faster the rate of the reaction. E2 reactions ① are generally run with strong negatively charged bases like OH①, EtO①, NH2 , ① etc. Since their basicity decreases in the order: NH2 > OEt① > OH①, therefore, the E2 reaction rate decreases in the same order. Strong sterically hindered bases like Me3CO①, DBN, DBU, etc. also favour E2 reaction. The bulky t-butyl group of the tert-butoxide ion inhibits the reaction by substitution, allowing elimination reaction to take precedence.

Organic Chemistry—A Modern Approach

4.34

(c)

The effect of leaving group on rate: Because the bond to the leaving group is partially broken in the transition state, the better the leaving group the higher the rate of the E2 reaction. The increasing order of reactivity of R — X is R — F < R — Cl < R — Br < R — I. When PhCH2CH2X (X = halogen), for example, is treated with EtO①/EtOH, the following relative rates have been observed. PhCH 2CH 2F 1

(d)

(e)

Ph CH 2CH 2 Cl 70

PhCH 2CH 2Br 4.2 ¥ 103

PhCH 2CH 2l 2.7 ¥ 104

The effect of solvent on rate: The E2 reactions are favoured by a decrease in the polarity of the solvent because the charge in the E2 transition state is considerably dispersed. In fact, polar aprotic solvents increase the rate of E2 reactions. Because polar aprotic solvents like HCONMe2 (DMF), Me2SO(DMSO) and Me2CO (acetone) do not solvate anions well, a negatively charged base is not trapped by the strong interaction with the solvent molecules and the base is stronger. A stronger base increases the reaction rate. The effect of temperature on rate: The rate of E2 reaction increases with rise in temperature. The number of particles increases in E2 reactions and so, E2 reactions have the more favourable entropy term, and because this (DS=| ) is multiplied by T in the relation of free energy of activation, DG=| (DG=| = DH=| – DS=| ), it increasingly outweighs a less favourable DH=| as the temperature rises.

4.1.7 Factors that Govern the Proportions of E2 and SN2 Reactions Since the reactive part of both nucleophiles and bases is an unshared pair of electrons, therefore, all nucleophiles are potential bases and all bases are potential nucleophiles. Then, it should not be surprising that nucleophilic substitution reactions and elimination reactions often compete with each other.

Elimina on Reac ons

4.35

The following factors govern the proportion of E2 and SN2 reactions: (a) Substrate structure: When the substrate is a primary (1°) halide and the base is strong and unhindered (e.g., EtO①), substitution is highly favoured because the base can easily approach the carbon bearing the halogen atom (the leaving group). For example: C H OH

2 5 CH 3CH 2O* Na≈ + CH 3CH 2Br æææææ Æ CH 3CH 2OCH 2 CH 3 + CH 2 == CH 2 55∞ C Primary E2 - product SN 2 - product ( - NaBr) substrate (major; 90%) (minor; 10%)

When the substrate is a secondary (2°) halide, a strong base favours elimination and this is because steric hindrance in the substrate makes substitution more difficult. Br | C2H5OH CH 3CH 2O* Na≈ + CH 3 CHCH 3 æææææ Æ (CH 3 )2 C — O — CH 2 CH 3 + CH 2 == CH CH 2 55∞ C E2 - product Secondary ( - NaBr) SN 2 - product substrate (minor; 79%) (major; 20%) When the substrate is a tertiary (3°) halide, an SN2 reaction cannot take place due to severe steric hindrance in the substrate. Elimination reaction is highly favoured, especially when the reaction is carried out at higher temperatures. If any substitution occurs, it must take place through an SN1 mechanism. At room temperature: C H OH

2 5 CH 3CH 2O* Na≈ + (CH 3 )3 C — Br æææææ Æ CH 2 == C(CH 3 )2 + (CH 3 )3 C — O — CH 2 CH 3 25∞ C Tertiary SN1 - product E2 + E1- product substrate (major; 91%) (major; 9%)

At higher temperature: C H OH

2 5 CH 3CH 2O* Na≈ + (CH 3 )3 C — Br æææææ Æ CH 2 == C(CH 3 )2 55∞ C E2 + E1- product Tertiary substrate (100%)

(b) Base: One way of favourably influencing an elimination reaction of an alkyl halide is to use a strong sterically hindered base such as the tert-butoxide ion (Me3CO①). The bulky methyl groups of the Me3CO① ion inhibit its reaction by SN2 and as a consequence, elimination reaction takes place. This can be well demonstrated by the following two reactions. The relatively unhindered methoxide ion (MeO①) reacts with 1-bromooctadecane mainly by substitution (SN2), whereas the bulky tert-butoxide ion reacts with the same substrate mainly by elimination (E2).

4.36

Organic Chemistry—A Modern Approach CH OH

3 CH 3O* + CH 3 (CH 2 )15 CH 2CH 2Br ææææ Æ 65∞ C an unhindered 1 - Bromooctadecane base/nucleophile CH 3 (CH 2 )15 CH 2CH 2OCH 3 + CH 3 (CH 2 )15 CH == CH 2 SN 2 - product E2 - product (major; 99%) (minor, 1%)

CH OH

3 (CH 3 )3 CO* + CH 3 (CH 2 )15 CH 2CH 2Br ææææ Æ 65∞ C a hindered 1 - Bromooctadecane base/nucleophile CH 3 (CH 2 )15 CH == CH 2 + CH 3 (CH 2 )15 CH 2CH 2OC(CH 3 )3 E2 - product SN 2 - product (major; 85%) (minor, 15%)

The relative rates of E2 and SN2 reaction also depend on the relative basicity and polarizability of the base/nucleophile. The likelihood of elimination (E2) increases when a strong, weakly polarizable base (the attacking atom is more electronegative and smaller in size) such as OH①, NH2① or RO① (especially a hindered one) is used. On the other hand, the likelyhood of substitution (SN2) increases when weakly basic ions such as Cl①, CHCOO① or weakly basic and highly polarizable ions (the attacking atom is less electronegative and larger in size) such as Br①, I① or RS① are used. For example, the acetate ion reacts with 2-bromopropane nearly exclusively by the SN2 mechanism. Br | CH 3COO* + CH 3 CHCH 3 ææ Æ CH 3COOCH(CH 3 )2 + : Br:* 2-Bromopropane weakly SN 2 - product basic (∼ 100%) On the other hand, the more strongly basic ethoxide ion (EtO①) reacts with 2-bromopropane mainly by the E2 mechanism. (c) Solvent: The bimolecular reactions are favoured by decrease in the polarity of the solvent and since the charge in the E2 transition state is more dispersed than in the SN2 transition state due to involvement of five atoms, decreasing solvent polarity favours elimination (E2) more than substitution (SN2).

Elimina on Reac ons

4.37

The influence of solvent on the E2 – SN2 competition is illustrated by the following reaction. The E2 dehydrobromination of 2-bromopropane (CH3CHBrCH3) increases as the medium is changed from polar aqueous ethanol to less polar ethanol. Br | 60% EtOH + 40% H2O CH 3 CHCH 3 + NaOH æææææææææ Æ CH2 == CHCH 3 + (CH 3 )2 CHOC2H5 + (CH 3 )2 CHOH 55∞ C E2 – product SN 2 - product (54%) (46%)

Br | C2H5OH CH 3 CHCH 3 + NaOEt æææææ Æ CH2 == CHCH 3 + (CH 3 )2 CHOC2H5 55∞ C E2 - product SN 2 - product (71%) (29%) (d) Temperature: Elimination – whether E1 or E2 – is favoured with respect to substitution by rise in temperature. Because more bonding changes occur during elimination, these reactions have greater free energies of activation than substitution reactions. When the reaction is carried out at a higher temperature, the proportion of molecules able to surmount the barrier of activation energy for elimination increases more than the proportion of molecule able to undergo substitution, even though the rate of both reactions will be increased. Furthermore, elimination reactions are favoured entropically over substitution because the products of an elimination reaction are greater in number than the reactants, i.e., elimination leads to an increase in the number of particles, whereas substitution does not. Since DS=| is multiplied by T in the relation for the free energy of activation, DG=| (DG=| = DH=| – TDS=| ), therefore, it will increasingly outweigh a less favourable DH=| term as the temperature rises. (e) Leaving group: In second-order reactions, the elimination/substitution ratio is not greatly dependent on a halide leaving group, although a slight increase in elimination in the order I > Br > Cl has been observed. When OTs① is the leaving group, there is usually much more substitution. For example, when n–C18H37 Br is treated with t-BuOK, 85% elimination occurs. However, n–C18H37OTs gave, under the same reaction conditions, 99% substitution. The positively charged leaving groups, on the other hand, increase the amount of substitution.

4.1.8

Regioselectivity in b-elimination Reactions (Orientation of p Bond in the Product Alkene)

Symmetrical substrate undergoes b-elimination reaction to yield a single olefinic product. But, unsymmetrical substrates, i.e., substrates containing two types of b-hydrogens, undergo E2 elimination reaction to yield more than one alkene. Two empirical rules, the Hofmann and the Saytzeff rules, govern the orientation of elimination of these reactions. The Saytzeff rule: This rule states that neutral substrates (alkyl halides except fluorides and sulphonates) possessing two different types of b-hydrogens yield predominantly the

4.38

Organic Chemistry—A Modern Approach

more substituted alkene. For example, when 2-bromobutane is treated with NaOEt in EtOH, it undergoes E2 dehydrobromination following the Saytzeff rule because it produces 75% of the more substituted alkene but-2-ene and 25% of the less substituted alkene, but1-ene.

The predominant formation of Saytzeff product can be explained as follows: Neutral substrates (except fluorides) undergo ideal E2 elimination reaction in which both the C — H and C — LG bonds are being broken and the carbon–carbon double bond is being formed simultaneously. For this reason, the transition state possesses considerable double bond character. Therefore, any effect that stabilizes the product alkene also stabilizes the transition state. Because of hyperconjugation effect, the stability of a double bond increases progressively with increase in the number of alkyl groups bonded to it, i.e., with increase in the number of a-H atoms. The disbustituted alkene, but-2-ene (with six hyperconjugable a-H-atoms) is, therefore, thermodynamically more stable than the monosubstituted alkene, but-1-ene (with only two hyperconjugable a-H atoms) and so, but-2-ene is formed more rapidly through the transition state of lower energy than but-1ene, i.e., the reaction produces the more substituted alkene but-2-ene predominantly.

Thus, an E2 reaction is regioselective because more of one constitutional isomer is formed than the other. The Hofmann rule: This rule states that E2 elimination produce predominantly the less-substituted alkene if the eliminations are carried out with charged substrates. For ≈

example, when sec-butyltrimethylammonium hydroxide, (CH 3 )3 NC(CH 3 )CH 2CH 3 :OH * ,

Elimina on Reac ons

4.39

is heated, 95% of the less substituted alkene, but-1-ene and 5% of the more substituted alkene, but-2-ene are obtained.

The predominant formation Hofmann product can be explained as follows: ≈

≈

Substrates containing charged leaving group (e.g ., — N(CH3 )3 ), — SMe2 , etc.) undergo E2 elimination reaction in which breaking of the b C — H bond starts well before the breaking of the C— LG bond. For this reason, the transition state of such a reaction possesses little alkene character but considerable carbanionic character. Therefore, any factor that stabilizes a carbanion also stabilizes the transition state. The stabilities of carbanions decrease progressively with increase in the number of electron-releasing (+I) alkyl groups attached with the negative carbon, i.e., the order of carbanion stability is methyl > primary (1°) > secondary (2°) > tertiary (3°). The abstraction of b-H by base thus takes place preferentially form the methyl group because this leads to the formation of a more stable (lower energy) carbanion-like transition state. The reaction thus produces a predominance of the less substituted alkene but-1-ene.

The formation of the less substituted terminal alkene but-1-ene may also be explained in terms of steric effect. The charged groups are generally bulkier than the neutral ones. From the Newman projections for the conformation required for anti-elimination it becomes ≈

clear that the bulky — N(CH 3 )3 group comes into steric interaction with the adjacent methyl group (gauche to each other) in the conformations B and C leading to the formation of the more substituted Saytzeff product. B also suffers from an additional butane–gauche interaction, consequently, a large number of molecule react through the less crowded and more stable conformation A to give the Hofmann product but-1-ene predominantly.

4.40

Organic Chemistry—A Modern Approach

Steric effect due to branching in the alkyl group, in the group which is going to leave *

with its bonding electrons and in the base, i.e., R, LG and B :, plays an important role in increasing the proportion of Hofmann product over Saytzeff. It has been found that an increase in the size of LG, or more particularly branching in it, leads to an increasing proportion of Hofmann elimination with the same alkyl group. For example:

The amount of Hofmann product is also found to increase with increasing branching in the alkyl group of the substrate (with the same LG and base), and with increasing branching in the base used. For example, the proportion of Hofmann elimination increases with increasing branching in the base from a bromide substrate like 2-bromo-2,3-dimethyl butane from which preferential Saytzeff elimination would normally be expected.

These several steric effects are explainable on the basis of the fact that any crowding, irrespective of its origin, will make the T.S. I that involves the removal of proton (2) form the substrate R2CHCMe2LG (Saytzeff elimination) relatively more crowded than the T.S. II that involves removal of proton (I) (Hofmann elimination). The differential increases as the crowding increases in R, LG or B① and Hofmann elimination will be progressively favoured over Saytzeff.

Elimina on Reac ons

4.41

Saytzeff elimination:

Regioselectivity in E2 dehydrohalogenation The distribution of products obtained from the E2 reaction of MeONa/MeOH and 2-halohexanes are given in the following table: More substituted product Less substituted product CH 3 (CH 2 ) CH 2 CHCH 3 CH 3ONa/CH 3OH 2 | ææææææææ æ Æ CH3(CH2)2CH==CHCH3 + CH3CH2CH2CH== CH2 E2 X Hex-2-ene Hex-1-ene (dehydrohalogenation) 2- Halohexane (mixture of cis and trans)

X I Br Cl F

Conjugate acid HI HBr HCl HF

pKa –10 –9 –7 2.2

81% 72% 67% 30%

19% 28% 33% 70%

When 2-halohexanes undergo dehydrohalogenation in the presence of MeO①, the major product of the chloride, bromide and iodide is the more substituted alkene, but that of fluoride is the less substituted alkene. Also, there is a steady increase in the fraction of the less substituted alkene hex-1-ene along the series I, Br, Cl and F. In the case of 2-iodohexane, breaking of both the C—H and C—I bonds occurs simultaneously in the transition state, i.e., the transition state possesses considerable double bond character. So, the formation of the more thermodynamically stable, i.e., more substituted alkene hex-2-ene is preferred and the orientation is Saytzeff. As the series is traversed, the C—X

4.42

Organic Chemistry—A Modern Approach

bond becomes progressively stronger (as evidenced by their pKa values) and its tendency to be broken in the transition state decreases. At the same time, the electron-attacking –I effect of X increases. This progressively favours the development of negative charge on b-carbon. Therefore, the carbanionic transition state is gradually favoured and so, the percentage of the less substituted alkene hex-1-ene is increased. Due to much stronger –I effect of fluorine and much stronger C—F bond, the C—H bond-breaking occurs considerably before C—F bond breaking. Therefore, the transition state possesses little alkene character but considerable carbanionic character. Since a methyl hydrogen is more acidic, it is preferentially abstracted by the base to give the less substituted alkene hex-1ene predominantly. Hence, the orientation is Hofmann.

Violation of Saytzeff Rule In cyclic systems, the usual simple requirements of Saytzeff or Hofmann rules may be overridden by other special requirements of the system. For example, there is a preference for elimination from the trans-diaxial conformation in cyclohexane derivative. Another such limitation is that it is not normally possible to carry out on elimination reaction so as to introduce a double bond on a bridgehead carbon atom in a bridged bicylic compound with small ring (Bredt’s rule). For example, when the bromide I is treated with NaOEt in EtOH it produces the Hofmann product II exclusively instead of III (the Saytzeff product).

This is because the developing p orbitals in this E2 elimination, far from being coplanar, would be virtually at right angles to each other, and so, significant orbital overlap does not take place to allow development of a double bond. The bicyclic ring system is rigid enough to make the distortion required for effective p orbital overlap, i.e., such distortion is energetically unattainable. With bigger rings, for example, the bicyclononene (IV), or more flexible system (V), sufficient distortion is possible to allow the introduction of a double bond at the bridgehead position by an E2 reaction.

In the following reaction, Saytzeff rule does not lead to the more substituted alkene because the rule does not take into account the fact that conjugated double bonds are more stable than isolated double bonds. Because the conjugated alkene is the thermodynamically more

Elimina on Reac ons

4.43

stable alkene, it is the one that is more easily and readily formed and is, therefore, the major product of the reaction. Hence, if there is a double bond (or a benzene ring) in the alkyl halide, Saytzeff rule is not to be used to predict the major product of the elimination reaction. *

OH CH 3CH == CHCH 2CH ClCH(CH 3 )2 æææ Æ CH 3CH == CH — CH == CHC(CH 3 )2 5-Chloro-6-methyl-2-heptene 6-Methyl-2,4-heptandiene (a conjugated diene) major product + CH 3CH == CHCH 2CH == C(CH 3 )2 6-Methyl-2,5-heptadiene (an isolated diene) minor product + H2O + Cl*

4.1.9

Hofmann Exhaustive Methylation or Hofmann Degradation

Hofmann exhaustive methylation or Hofmann degradation is a widely used process for structure elucidation of alkaloids. It is actually a three-step process. In the first step, a 1°, 2° or 3° amine is converted to a quaternary ammonium iodide salt by treatment with an excess of methyl iodide. For example:

In the second step, the iodide salt is treated with moist silver oxide (Ag2O) to yield the corresponding hydroxide. +

2

–

—CH2CH2N(CH3)3I + Ag2O + H2O Æ 2

+

–

—CH2CH2N(CH3)3OH + AgIØ

In the third step, an aqueous or alcoholic solution of the hydroxide is distilled. As a result, the hydroxide undergoes thermal decomposition to form an alkene, a tertiary amine and water. A specific E2 reaction in which a quaternary ammonium hydroxide undergoes thermal decomposition to yield an alkene, a tertiary amine and water is called Hofmann elimination.

4.44

Organic Chemistry—A Modern Approach

Complete Hofmann degradation of some secondary amines containing two types of b-H atoms are shown below:

Elimina on Reac ons

4.1.10

4.45

Fragmentations

When carbon is the positive leaving group (the electrofuge) in an elimination reaction, the reaction is called fragmentation. These processes take place on substrates of the type ≈

≈

W-C-C-X, where X is a normal nucleofuge (e.g., halogen, OH 2 , OTs, NR 3 , etc.) and W is a positive carbon electrofuge. In general, W is HO— C or R 2N— C, so that the positive charge on carbon is stabilized by the unshared pair on the oxygen or nitrogen atom. For example:

4.46

Organic Chemistry—A Modern Approach

The mechanism are mostly E2 or E1. Some examples of fragmentation reaction are given below: (1) When the following 1,4-dibromide is treated with zinc, it undergoes debromination accompanied by cleavage of a C — C bond to give two molecules of 2-methyl-1m-tolylpropene. This takes place through the conformation in which each of the departing Br atoms is anti to the C-2 — C-3 bond which breaks.

(2)

When trans-1,4-dibromo cyclohexane is treated with zinc, it undergoes fragmentation reaction to give hexa-2,5-dideuterio-1,5-diene. Since in the stable diequatorial conformation of this dibromide, each C — Br bond is anti to the C-2 — C-3 bond, therefore, it undergoes debromination by zinc accompained by fission of the C-2 — C-3 bond.

[Debromination does not take place through the trans-diaxial conformation because in that conformation anti relationship between the C-2 — C-3 bond and the axial C — Br bonds is not maintained.] (3)

When the following bromo derivative of quinuclidine (a bicyclic compound) is heated, it undergoes fragmentation reaction to give an iminium salt and this is because the C — C bond is anti to both the sp3 orbital of nitrogen containing an unshared electron pair and the departing Br atom.

Elimina on Reac ons

4.1.11

4.47

Summary of the E2 Reaction

The E2 reaction is a b-elimination reaction. The key points about this reaction are as follows: 1. The rates of E2 reaction are second order overall: first order in base and first order in the alkyl halide. 2. E2 reactions normally occur with anti stereochemistry. 3. The E2 reaction occurs at a faster rate with better leaving groups, i.e., those that give the weakest bases as products. 4. The rate of E2 reactions shows substantial primary kinetic isotope effects (deuterium isotope effects at the b-H atoms). 5. When an alkyl halide has more than one type of b-H atoms, more than one alkene products can be formed. In that case, the major product is the more substituted alkene (the Saytzeff product) unless the base is large, the alkyl halide is an alkyl fluoride and the alkyl halide contains one or more double bonds. 6. E2 reactions compete with SN2 reactions. Elimination is favoured by alkyl branches in the alkyl halide at the a- or -b-carbon atoms, by alkyl branches in the base, by stronger base and by rise in temperature.

1.

When (R)-1,2-di-p-tolyl-1-chloroethane is treated with NaOEt in EtOH, it undergoes base-induced dehydrochlorination to give trans-1,2-di-ptolylethene as the major product and cis-1,2-di-p-tolylethene as the minor product. Explain this observation.

Solution (R)-1,2-Di-p-tolyl-1-chloroethane contains two b-H atoms. So, anti-elimination of HCl may take place from two different conformations. The gauche-conformation (the two bulky p-CH3C6H4-groups are gauche to each other) gives rise to cis-1,2-di-p-tolylethene via the transition state I, while the anti-conformation (the two bulky p-CH3C6H4-groups are anti to each other) gives rise to trans-1,2-di-p-tolylethene via the transition state II. The stereochemical course of the reaction may be shown as follows:

4.48

Organic Chemistry—A Modern Approach

The population of the anti-conformation (B) is greater than the gauche-conformation (A) because the former conformation is relatively more stable. On that basis, it is normally expected that the trans-product would predominate. However, this reasoning is incorrect. The activation energy of the reaction is very large compared to the barrier to internal rotation. In such a case according to Curtin–Hammett principle, the relative amounts of the two disastereoisomeric products do not depend on the relative populations of the two

Elimina on Reac ons

4.49

conformations, but depends only on the relative stabilities of the transition states (i.e., on the values of activation energies) leading to the products. Because of steric strain between the two bulky p-tolyl groups on the same side of the molecule, the transition state I is less stable than the transition state II in which they are placed in opposite sides. The activation energy for the formation of the trans-alkene is, therefore, lower than for the formation of the cis-alkene and consequently, the former alkene is formed readily as the major product. 2. Predict the two elimination products expected to be formed in the following reaction: *

OEt /EtOH CH 3CH 2CHDCH 2Br ææææææ Æ E2

Which is formed as greater yield and why? Solution The substrate CH3CH2CHDCH2Br undergoes base-induced dehydrobromination (loss of HBr) and dedeuteriobromination (loss of DBr) to yield 2-deuteriobut-1-ene and but-1-ene, respectively:

Since the C — D bond is stronger than C — H bond, therefore, the activation energy for the elimination of DBr is greater (less stable T.S.) than that for the elimination of HBr. Because of this, elimination of HBr takes place at a faster rate to produce 2-deuteriobut1-ene in greater yield. 3.

(1R,2R)-1-2-Dibromo-1,2-dicyclohexylethane undergoes dehydrobromination in the presence of pyridine

(

)

N: to yield trans-1-bromo-1,2-dicy-

clohexylethene while (1R, 2S)-1,2-dibromo-1,2-dicyclohexylethene undergoes debromination to yield trans-1,2-dicylohexylethene. Explain these observations.

4.50

Organic Chemistry—A Modern Approach

Solution In the presence of the base pyridine, the (1R, 2R)-isomer of 1,2-dibromo-1,2dicyclohexylethene undergoes dehydrobromination instead of debromination because the reaction proceeds through a more stable transition state in which the two bulky cyclohexyl groups are anti to each other (too far from each other to become involved in steric interaction).

The (1R, 2S)-isomer, on the other hand, undergoes debromination instead of dehydrobromination because this proceeds through a more stable transition state where the cyclohexyl groups are anti.

Elimina on Reac ons

4.

4.51

Dehydrohalogenation of vic-dibromides leads to the formation of alkynes. Suggest a mechanism of the following reaction:

4.52

Organic Chemistry—A Modern Approach

Solution The reaction proceeds through the steps as follows:

Step 1: The strongly basic amide ion brings about an E2 reaction to form a bromalkene.

Step 2: A second E2 reaction produces the alkyne.

In both the steps, the reaction takes place smoothly because the stereoelectronic requirement of E2 process is satisfied. 5. Show how you might synthesize cyclohexylacetylene from cyclohexyl methyl ketone. Solution Cyclohexyl methyl ketone is first treated with phosphorus pentachloride to yield a gem-dichloride. The gem-dichloride is then heated with 3 equivalents of NaNH2 in the presence of mineral oil. The terminal alkyne thus obtained is deprotonated by the third equivalent of base. Aqueous NH4Cl is finally added to the reaction mixture to convert the sodium alkynide to the desired product, cyclohexylacetylene.

Elimina on Reac ons

6.

4.53

Suggest a mechanism for each of the following reactions along with the stereochemical course involved when it is appropriate: (a) HO —

— OH

H

+

(b)

≈

H Ph 2C(OH)CMe2 (OH) Me2 æææ Æ

H

(c)

(d)

Me N

Cl

D

Solution (a) The cyclic diol fragments in the presence of acid (acid-catalyzed dehydration involving fission of a C — C bond) to give a b, g-unsaturated aldehyde.

(b)

This 1,3-diol undergoes fragmentation reaction in the presence of acid to yield the alkene Ph2C == CMe2 and acetone. Due to stabilizing effect to the two phenyl groups, this particular — OH is expelled as water.

(c)

The tosylate undergoes fragmentation reaction in the presence of the base t-BuO① and this is because the C-2—C-3 bond is anti to both the O — H and C — OTs bonds.

Organic Chemistry—A Modern Approach

4.54

(d)

This heterocyclic ring system undergoes concerted fragmentation reaction because the stereoelectronic effect is helpful in this case also. 7.

How would you carry out the following transformations?

(a)

(b)

(c) CH3CH2CH2CH2CH2Br Æ CH3CH2CH2COCH3 (d) CH3CH2CH == CH2 Æ CH2 == CH—CH == CH2 Solution

(a)

(b)

(c)

(d)

Elimina on Reac ons

8.

4.55

Predict the major product in each of the following reactions and account for its formation: (a)

(b) H

OH H Ph

OH

OH

–

Ph

H H

(c)

–

H H

Br

OH

Br

(d) H

OH H Ph

Br H

H

OH

OH

–

Ph

Br H

OH

–

or Ag2O/H2O

H

Solution (a) The –OH group is in proper position for backside attack and therefore, SN2 type displacement leading to the formation of an epoxide is the main reaction.

(b)

Since H and Br are anti-periplanar, therefore, E2 elimination is stereoelctronically favourable. An enol is formed as an intermediate, which readily tautomerizes to the more stable keto form (a cyclohexanone derivative).

(c)

This more stable conformation of the compound has no proper stereoelctronic situation for either epoxidation or E2 elimination. Elimination of HBr is possible only from the other conformation in which the stereoelectronic requirement is satisfied. However, the rate is slow because this conformation is thermodynamically less stable as it suffers from syn-axial interactions. The intermediate enol readily tautomerizes to the more stable ketone.

Organic Chemistry—A Modern Approach

4.56

(d)

In the presence of OH①, the compound undergoes ring inversion followed by an internal SN2 attack to give an epoxide. However, because of syn-diaxial interactions, the conformation is less stable and as a consequence, the reaction is slow.

In the presence of Ag2O (Ag≈), a cyclopentanecarboxaldehyde derivative is obtained by bond migration and this is because Ag≈ induces the departure of equatorial bromine by acting as a Lewis acid.

9.

Draw the structure of an isomer of benzene hexachloride which does not undergo E2 dehydrochlorination and explain why it is unreactive.

Solution The (e,e,e,e,e,e) isomer of benzene hexachloride or its conformational diastereoisomer (a,a,a,a,a,a) does not undergo E2 dehydrochlorination. This is because in both forms, no Cl and H atoms are diaxially placed, i.e., in none of two forms, the stereoelectronic requirement for the E2 elimination is satisfied.

Elimina on Reac ons

10.

4.57

The compound A readily forms an alkene with I①, but the isomeric compound B does not. Explain these observations.

Solution The tert-butyl group ( ) is particularly bulky and due to sever 1,3-diaxial interactions it cannot take up the axial position. A tert-butyl group-substituted cyclohexane, therefore, remains fixed in the chair conformation where the Me3C-group remains in the equatorial position. Hence, the compound A and its isomer B containing equatorial tertbutyl group exist in these particular conformations and they are expected to react through these conformations. Now, in A, the two Br atoms are in axial positions, i.e., they are antiperiplanar. Therefore, the stereoelectronic requirement for E2 elimination is satisfied. Hence, the compound A readily undergoes debromination by I① to produce an alkene. On the other hand, in the compound B, the two Br atoms are in equatorial positions. Therefore, the stereoelectronic requirement of E2 debromination is not satisfied in this case. For this reason, the compound B does not undergo debromination to yield an alkene.

11.

Predict the products A and B and suggest the mechanism for the conversion of A to B:

Solution

Organic Chemistry—A Modern Approach

4.58

The thiocarcbonate A is converted into B (cyclohexene) through the steps as follows:

12.

Predict the product and suggest a mechanism for each of the following reactions: (a)

*

OH /H2O (b) ClCH 2C ∫∫ CCH 2Cl æææææ Æ D

(d) (f)

*

OH HOCMe2CH 2CMe2Br æææÆ

Mg (c) BrCMe2CMe2OCH 3 æææÆ ether

(e) Me2C

CMe2 Mg ether

C Br

Br

Solution (a)

Since the methylene hydrogen adjacent to the electron-withdrawing p-nitrophenyl group is more acidic than the other methylene hydrogen, therefore, the transition state leading to this alkene is relatively more stable than the transition state leading to the formation of the isomeric alkene trans-p-CH3OC6H4CH == CHCH2– C6H4NO2-p and so, this alkene is obtained predominantly by an E2 process.

Elimina on Reac ons

4.59

(b)

This dichloride undergoes repeated E2 dehydrochlorination in the presence of alkali to form butadiyne.

(c)

When 2-bromo-3-methoxy-2,3-dimethylbutane is treated with Mg in ether, a Grignard reagent is obtained. Since the carboanionic centre of this Grignard reagent has a good leaving group (— OMe) adjacent to it, therefore, it is unstable and so, it undergoes ready elimination (E2) of MgBr(OMe) to give 2,3-dimethyl-2-butene.

(d)

The compound undergoes dehydrobromination along with fragmentation in the presence of alkali to yield 2-methylpropene (Me2C ==CH2) and acetone (Me2C ==0). (e)

The initially formed Grignard reagent is unstable. It readily dissociates to yield an allene.

Organic Chemistry—A Modern Approach

4.60

(f)

When treated with acetic anhydride (Ac2O) in the presence of acid, the aldoxime undergoes acetylation. The — OCOCH3 group then undergoes protonation and becomes a very good neutral leaving group. Subsequent E2 elimination of AcOH and H% yields acetonitrile (CH 3C ∫∫ N).

13. (a)

Account for the following observations: 3-Bromobutanal undergoes base-promoted dehydrobromination at a very much faster rate than 2-bromobutanal under similar conditions. NaOEt/EtOH

fast

CH3CHBr CH2 CHO CH3 CH2 CHBr CHO 3-Bromobutanal 2-Bromobutanal CH3CH == CH — CHO 2-Butenal

NaOEt/EtOH

slow

(b) D

D

NaOEt

D ;

EtOH

Cl

Cl D I

(c)

(d)

D

D II

NaOEt EtOH

D

The following compound gives no elimination product when treated with NaOMe in MeOH:

Elimina on Reac ons

4.61

Solution (a) The electron-attracting group is placed a to the bromine atom in 2-bromobutanal, but b to the bromine atom in 3-bromobutanal. The — CHO group could stabilize the incipient double bond by conjugation from either position. However, it can make the b-H atom in 3-bromobutanal somewhat acidic due to its – I effect. As a result, the transition state for E2 dehydrobromination of 3-bromobutanal is relatively more stable than that for dehydrobromination of 2-bromobutanal and so, 3-bromobutanal undergoes base-promoted dehydrobromination much faster than 2-bromobutanal under similar reaction conditions.

(b)

The compound I undergoes base-induced E2 elimination of HCl through that conformation in which Cl atom and an adjacent H(or D) atom are anti-periplanar, i.e., through that conformation in which the stereolectronic requirement is satisfied, to form 1,3-dideuteriocyclohexene.

4.62

Organic Chemistry—A Modern Approach

Similarly, the diasteremeric compound II undergoes base-induced elimination of DCl to yield 3-deuteriocyclohexene.

(c)

There is no axial b-H atom in the conformation B in which Br is axial. For this reason, this compound III does not undergo E2 dehydrobromination when treated with NaOMe in MeOH. (d)

Elimina on Reac ons

14.

4.63

The compound IV undergoes E2 dehydrobromination because in this case the stereoelectronic requirement of E2 reaction is satisfied. On the other hand, because of its rigidity, the isomeric compound V cannot flip and so, Br cannot go to the axial position. Since the stereoelectronic requirement is not satisfied, therefore, this compound does not undergo base-induced dehydrobromination. Explain the following observations: (a) 2-Bromo-3-methyl-1-phenylbutane reacts with NaOEt in EtOH to give 3-methyl-1-phenylbut-1-ene as the major product and 3-methyl-1-phenylbut-2-ene as the minor product, i.e., the less and more substituted products become the major and minor products, respectively.

(b) 2-Bromo-2-methylbutane undergoes dehydrobromination in the presence of t-BuO① to give an excess of the less substituted alkene, 2-methylbut-1-ene, even though the substrate is a bromide.

Solution

(a)

The transition state of this E2 elimination involving an alkyl bromide possesses considerable double bond character. Since a phenyl group stabilizes a double bond by conjugation, the transition state leading to 3-methyl-1-phenylbut-1-ene (the Hofmann product) is much more stable than that leading to 3-methyl-1-phenylbut2-ene (the Saytzeff product). It is for this reason, the less substituted product 3-methyl-1-phenylbut-1-ene is obtained predominantly.

4.64

(b)

Organic Chemistry—A Modern Approach

Because of the steric bulk of the base t-BuO①, the transition state of the expulsion of one of the less exposed methylene hydrogens leading to the formation of the more substituted alkene (the Saytzeff product) becomes highly crowded. Since the steric effect then outweighs the hyperconjugation effect, therefore, the transition state energy become higher than that of the expulsion of the more exposed methyl hydrogen leading to the formation of less substituted alkene (the Hofmann product). As a consequence, the less substituted alkene (the Hofmann product) is obtained as the major product, even though this bromide substrate is expected to give the more substituted alkene predominantly.

Elimina on Reac ons

15.

4.65

Out of the three diasteroisomers of 2-chloro-1,3-dimethycyclohexane only one cannot undergo E2 dehydrochlorination and the only product obtained from the other two is 1,3-dimethylcyclohexene. Identify these diastereoisomers and account for these observations.

Solution The three diastereoisomers of 2-chloro-1,3-dimethylcyclohexane are as follows:

Among the three diastereoisomeric substrates only II does not undergo dehydrochlorination because the vicinal hydrogens are equatorial and anti-periplanarity with Cl cannot be achieved. Also, an axial Cl atom cannot be syn-coplanar with an adjacent-equatorial H atom. In I the axial Cl atom is anti-periplanar with two axial H atoms and in III, it is anti-periplanar with one axial H atom. So, they undergo E2 dehydrochlorination to yield 1,3-dimethylcyclohexene.

Organic Chemistry—A Modern Approach

4.66

16.

Predict the major product and suggest a mechanism for each of the following reactions: (a)

(b) Solution (a) Charged substrates violated Hofmann rule when an electron-withdrawing group remains attached to one of the b-carbons. In this quaternary ammonium hydroxide, the elelctron-attracting C == O group makes methylene b-hydrogens more acidic than methyl b-hydrogens. Furthermore, due to conjugation, the carbanion-like transition state leading to the elimination product, CH2== CHCOCH3 is relatively more stable than the carbanionic transition state leading to CH2== CH2, the other elimination product. So, this quaternary ammonium hydroxide on thermal decomposition produces CH2== CHCOCH3 predominantly.

Elimina on Reac ons

4.67

(b)

Due to — I effect of the — C6H4NO2-p group, the hydrogens on the carbon containing the p-nitrophenyl group are somewhat more acidic than methyl hydrogens of the ethyl group. Also, due to conjugation, the transition state leading to the formation of the elimination product p-O2N—C6H4—CH== CH2 is thermodynamically more stable than the transition state leading to the other product CH2== CH2. Therefore, in this case also, the quaternary ammonium hydroxide on thermal decomposition produces the more substituted alkene p-O2NC6H4CH== CH2 predominantly.

17.

3-Bromopentane undergoes E2 dehydrobromination at a rate much faster in DMSO-tert-BuOK than in EtOH-KOH. Explain.

Solution A less polar solvent favours E2 over SN2. Since ethanol (e = 24) is less polar than DMSO (e = 45), elimination is expected to be favoured more in ethanol. But dehydrobromination of 3-bromopentane (CH3CH2CHBrCH2CH3) occurs much more readily in DMSO-t-BuOK than in EtOH-KOH. This can be explained in terms of the strength of the base which is a more important factor that promotes elimination reaction. A strong base favours E2 more than SN2. t-BuOK is a stronger base than OH① or OEt① and in the aprotic solvent DMSO, its effectiveness as a base increases because solvation through H-bonding does not take place. Elimination is, therefore, favoured very much by using t-BuOK in DMSO. Furthermore, E2 is also favoured because backside attack by the bulky base (nucleophile) Me3CO: * to bring about substitution (SN2) is disfavoured.

Organic Chemistry—A Modern Approach

4.68

18.

The following reaction is not an useful method for the synthesis of t-butyl isopropyl ether (Me3COCHMe2) — Why? *

≈

EtOH (CH 3 )2 CH — O Na + (CH 3 )3 C — Br ææææ Æ (CH 3 )3 C — O — CH(CH 3 )2

Solution When a tertiary halide is treated with a strong base, it give almost exclusively the elimination product mainly by E2 mechanism. This is due to the fact that tertiary substrates are very much reluctant to undergo an SN2 reaction due to steric hindrance. A very small amount of substitution product is obtained by an SN1 reaction. t-Butyl bromide, therefore, reacts with sodium isopropoxide (a strong base) to produce 2-methylpropene as the major product by the E2 mechanism. Hence, the given reaction is not an useful one for the synthesis of t-butyl isopropyl ether.

19.

How would you carry out each of the following transformations: Me

Me

C== C Et Et (a) Me C—C Me NMe2 H

Et

cis

Me

Et

C== C Me

Et trans

(b)

Et

Elimina on Reac ons

4.69

Solution (a)

(b)

The alkene I can be prepared by syn-elimination of –NMe2 and the b-D as follows:

The alkene II can be prepared by anti-elimination of –NMe2 and the b-H as follows:

Organic Chemistry—A Modern Approach

4.70

20.

Explain the following observation:

Solution When dehydrobromination of 1-bromo-1-methylcyclopentane is carried out with the bulky base potassium tert-butoxide (t-BuOK) in tert-butyl alcohol (t-BuOH), the formation of the less substituted alkene methylencyclopentane is favoured. This is due to the steric bulk of the base and to the fact that in tert-butyl alcohol the base is associated with solvent molecules and thus made even larger. The large tert-butoxide ion faces much difficulty in removing one of the ring b-H atoms because of greater crowding at that site in the transition state. It removes one of the more exposed methyl hydrogens instead to give methylenecyclopentane as the major product.

Elimina on Reac ons

21.

4.71

Selenoxides have recently been introduced as excellent alkene precursors. They readily undergo syn-elimination at or near room temperature to give good yields of alkenes. How would you carry out the following transformation through the formation of a selenoxide?

Give the mechanism of syn-elimination process involving the selenoxide. Solution This transformation may be carried out as follows:

Organic Chemistry—A Modern Approach

4.72

22.

Suggest a mechanism for each of the following pyrolytic eliminations: Br

(a)

(b)

Me H

H Me

H An alkyl bromide

(c)

Solution

(a)

(b)

600°C

Elimina on Reac ons

4.73

(c)

   

23.

When heated, the salt I undergoes decarboxylative debromination at a rate faster than its isomeric salt II. Explain.

Solution In the salt I, the Br atom and the — COONa group are anti to each other. Since the stereoelectronic requirement for normal E2 elimination in anti-fashion is satisfied, therefore, it undergoes decarboxylative debromination through a more stable transition state. On the other hand, in the salt II, the Br atom and the — COONa group syn to each other. So, it undergoes decarboxylative debromination through a less stable syn-coplanar transition state. It is for this reason, the salt I undergoes decarboxylative debromination as a rate faster than does the salt II.

Organic Chemistry—A Modern Approach

4.74

1.

2.

3.

Chlorofumaric acid undergoes dehydrochlorination at about 50 times faster than does chloromalic acid to give the common product acetylenedicarboxylic acid. Explain this observation. [Hint: Chlorofumaric acid is (Z)-2-chlorobutenedioic acid in which H and Cl are anti to each other and chloromalic acid is (E)-2-chlorobutenedioic acid in which H and Cl are syn to each other.] Account for the following observations:

Predict the product and suggest a mechanism for each of the following fragmentation reactions: (a)

(c) 4.

(b) D R 2NCR 2CH 2 CR 2¢ Br ææ Æ

@

OH HOCR 2CH 2CR 2¢ Br æææÆ

(d)

Dehydrohologenation of vic-dihalides (with the elimination of 2 moles of HX) normally leads to the formation of an alkyne rather than to the formation of a conjugate diene. However, when 1,2-dibromocyclohexane is dehydrolalogenated, 1,3-cyclohexadiene is obtained in good yield. Account for this observation. [Hint: A six-membered ring cannot accommodate a triple bond because of severe strain that would be introduced.

Elimina on Reac ons

5.

Predict the product and give the mechanism of each of the following reactions: CH3 CH3 | Me Me | Zn Zn (a) (b) Br CH 2C == CCH 2 Br æææ Æ Br

(c) 6.

4.75

Br

Zn Br CH 2C ∫∫ CCH 2Br æææ Æ

Br O | || Zn Me2C — C — Br æææ Æ

(d)

When the compound I (and its enantiomer) is treated with C2H5OH in C2H5OH, cis-2-butene is obtained without loss of deuterium and trans-2-butene is obtained with loss of deuterium. However, its diastereoisomer II (and its enantiomer) gives cis-2-butene with loss of deuterium and trans-2-butene without loss of deuterium. How do you account for these findings. H H

CH3 D Br

D H

CH3 I

7.

CH3 II

When the following deuterium-labelled compound is subjected to dehydrobromination using NaOEt in EtOH, the only alkene product is 3-methylcyclohexene (which contains no deuterium). Provide an explanation for this result. Br H H H3C

8.

CH3 H Br

H D

Predict the product of the following reaction? What is the purpose of sulfuric acid in the second step? Br H HO2C

CO2H H

1. KOH / EtOH, D 2. dil. H2SO4

Br

9.

Compare the substitution/elimination ratio for the reactions of ethyl bromide (CH3CH2Br) and 2,2,2-trideuterioethyl bromide (CD3CH2Br) with potassium tert-butoxide in tert-butyl alcohol.

10.

On E2 elimination of HCl with n-C5H11ONa/n-C5H11OH to yield 2-chloronorbornene (I), II reacts about 100 times as fast as its diastereoisomer III. Account for this observation.

Organic Chemistry—A Modern Approach

4.76

11.

Dehydrochlorination of 1-chloro-1-methylcyclopropane produces two alkenes (I and II). Explain why I is obtained as the major product despite the fact that it contains less substituted double bond. –

+

CH3

Me3COK

Cl

Me3COH

1-Chloro-1-methylcyclopropane

+ I (major)

CH3 II (minor)

[Hint: The alkene II containing two sp2 hybridized carbon in a three-membered ring is thermodynamically very unstable due to severe strain that has been introduced.] O Et 12.

Threo- PhCHDCHPhO—C—

, when treated with NaOEt in EtOH, retained

Et practically all the D. However, the erythro-isomer, when similarly treated, lost practically all its D. Write a mechanism consistent with this observation.

[Hint:

13.

Which isomers of C4H9Br yield only a single alkene on dehydrohalogenation? Give the structures of the alkenes.

Elimina on Reac ons

4.77

14.

Convert: (a) Ethylbenzene Æ Styrene; (b) 2-Butyne Æ 1-Butyne.

15.

Write the product when b-phenylethyl bromide is treated with sodium ethoxide in the presence of C2H5OD. Give the mechanism of the reaction. Describe how you can convert (Z)-2-butene to (E)-2-butene.

16.

[Hint: Epoxidation of (Z)-2-butene by PhCOOOH produces an oxirane which on treatment with triphenyl phosphine (Ph3P) gives a charge-separated intermediate (a betaine). This undergoes cyclization followed by elimination of Ph3PO to give (E)-2-butene.

17.

Explain the following observations: (a)

(b) [Hint: (a)

(b)

Organic Chemistry—A Modern Approach

4.78

18.

E2 elimination of HX involving enantiotropic hydrogens must produce homomers, while that involving diastereotopic hydrogens must produce diastereoisomers. Illustrate with suitable examples.

19.

Write a mechanism of the following transformation:

[Hint:

The thermodynamic stability of the acetylide anion will drag the overall reaction far to the right.] 20.

Which of the following compounds (A or B) eliminates HBr more readily when heated with a base and why?

21.

1-Bromobicyclo[2.2.1]heptane does not undergo dehydrobromination when heated with a base. Explain.

22.

Explain the following reaction and write the product: Br | Et3CO:@ Me2CCH 2CH 3 æææææ Æ (only the major product) Et COH 3

23.

Which alkyl halide and what conditions should be used to prepare the following alkene in good yield by an E2 elimination? Explain your choice. ==

Elimina on Reac ons

4.79

[Hint: 1-Bromo-1-methylcyclohexane is to be treated with a sterically hindered base like Me3CO@ or Et3CO@ in the corresponding alcohol.] 24.

Which alkyl halide would you expect to be more reactive in an E2 reaction? (a)

—CH2CHBrCH2CH3 or

—CH2CH2CHBrCH3

or

(b) Br Br

25.

26.

(c) Me2CCHBrCH2CH3 or Me2CHCH2CHBrCH3 [Hint: Alkyl halides leading to the formation of conjugated double bonds and more substituted double bond are relatively more reactive.] Write down the major elimination product expected to be obtained from an E2 elimination of each of the following alkyl halides with OH@ ion: (a) CH3CHClCH2CH == CH2 (b) (CH 3 )2CHCHCH 2CH 3 (c) | Br Br Br (d) (e) CH3CHFCH2CH3 CH3 Predict the product of each of the following reactions: (a)

(b)

D

(c)

27.

(d)

Me

Me2CH

SO2

N—NHTs

1. RLi (2 equiv.) 2. H2O

Explain the results of the following reactions: H

(a)

Ph H OC—SCH3 erythro S

D

Ph

Ph +

3-Phenylcyclohexene (>96%)

1-Phenylcyclohexene (<4%)

Organic Chemistry—A Modern Approach

4.80

(b)

28.

Give the steps involved in the complete Hofmann degradation of each of the following compounds: (a)

NH

(b) Me H3C

N

CH3

(c) NHMe H

MeHN

(d)

H

29.

Write down the structures of the three alkenes I, II and III when the following quaternary ammonium hydroxide is heated. Arrange the products in the increasing order of their relative amounts and explain this observation.

30.

Explain the distribution of products in the following dehydrobromination using various bases:

31.

Arrange the following compounds in the order of decreasing E2 reaction rate and explain the order:

Elimina on Reac ons

4.81

CH3

CH3

,

Br I

CH3

,

Br

Br CH3 III

II

I, II, III Æ; this is because the conformation in which [Hint: ææææææææææ E2 reaction rate decreases

32.

the stereo-electronic requirement of E2 is satisfied becomes gradually more unstable.] Dehydrohalogenation of isopropyl bromide, which requires several hours of refluxing in alcoholic KOH, is brought about in less than a minute at room temperature by t-BhOK in DMSO. Suggest a possible explanation for this.

33.

Explain why (CH3)2CHCHBrCH2CH3 reacts faster than (CH3)2 CHCH2CHBrCH3 in an E2 reaction, even though both alkyl halides are secondary.

34.

DBN and DBU are two strong sterically hindered nitrogen bases useful for E2 reactions. Write their structures and explain why they are strong bases. Give the products that would be obtained by treating the following tertiary amines with hydrogen peroxide followed by heat:

35.

(a)

(CH3)2NCH2CH2CH3

(b) (d)

(c)

CH3 | CH 3 CH 2NCH 2 CH (CH 3 )2

H3C—

N

—N(CH3)CH2CH2CH3

CH3

CH3

36.

How would you expect the ratio of SN2 to E2 product formed from the reaction of CH3CH2CH2Br with CH3O@ in CH3OH to change when the nucleophile is changed to CH3S@? [Hint: CH3O@ is a strong base, but a weak nucleophile. On the other hand, CH3S@ is a strong nucleophile, but a weak base. Therefore, the SN2/E2 ratio is higher in the presence of CH3S@, but lower in the presence of CH3O@.]

37.

Which member of each of the following pairs of compounds results in higher ratio of elimination (E2) to substitution (SN2) when treated with NaOEt/EtOH? (a) (CH3)2CHBr or (CH3)3 CBr Br Br (b) or Me Me (c) CH CH CHBrCH or CH CH CHCH 3 2 3 3 2 3 | !SMe I@ 2

Organic Chemistry—A Modern Approach

4.82

[Hint: (a) (CH3)3CBr gives higher E2/SN2 ratio because SN2 is inhibited due to steric hindrance. Br (b)

38.

gives higher E2/SN2 ratio because the resulting double bond is Me stabilized by conjugation. (c) Being a charged substrate CH3CH2CH(CH3)S≈Me2I@ gives higher E2/SN2 ratio.] The following two isomeric compounds yield the same products when treated with Me3CO@ in MeCOH. Predict the products and sketch the mechanisms involved. H OTs (b) (a)

OH

H

[Hint:

(a)

(b)

39.

Which one of the following hexachlorocyclohexanes is the least reactive in an E2 reaction and why? Cl Cl Cl

Cl Cl

Cl

Cl

Cl

Cl Cl

Cl

Cl

Cl

Cl Cl

Cl

Cl

Cl

Cl Cl

Cl

Cl

Cl

Cl

I

II

III

IV

Elimina on Reac ons

40.

41.

[Hint: I; because in both of its conformations there is no diaxially placed Cl and H atoms.] Give the dehydrochlorination products obtained when the following isomeric substrates are treated with high concentration of MeO@: (a) (2S, 3R)-2-Chloro-3-methylpentane (b) (2R, 3R)-2-Chloro-3-methylpentane Indicate which of the species in each pair would give a higher substitution/ elimination ratio when it reacts with isopropyl bromide (MeCHBrMe): (a)

42.

4.83

@

@

(b) Cl@ or Br@

OCN or SCN

(c)

C2H5O@ or Me3CO@

Explain why 2-bromopropane gives different amounts of substitution and elimination products when the nucleophile/base is changed from acetate (CH3COO@) to ethoxide (C2H5O@). Me2CHBr + CH 3COO@ ææ Æ Me2CH OCOCH 3 (100%)

43.

44.

Me2CHBr + C2H5O@ ææ Æ Me2CHOC2H5 + MeCH == CH 2 (80%) (20%) Predict the products of the following reaction:

[Hint: One of the products is Me3SiOCH3.] Would you expect the following two structures to form the identical or different elimination product? D OTs Ph H I

Ph H D OTs II

45.

E2 elimination of the compound I is ten times slower than that of the compound II. Explain. CH 3CH 2CH C (CH 3 )3 CH 3 — CHC(CH 3 )3 | | @ ! !N(CH ) OH N(CH3 )OH@ 3 3 I II

46.

The rate of pyrolytic elimination of 1-phenylethyl acetate is about eleven times faster than for 2-phenylethyl acetate. Explain.

Organic Chemistry—A Modern Approach

4.84

47.

Which one of the following bromides will undergo dehydrobromination at a faster rate in sodium hydroxide-isopropanol and why? H H Br H C==C C==C H Br O2N

48.

NO2 A B Suggest a mechanism for the following reaction:

[Hint:

49. 50.

Suggest a mechanism for the preparation of (E)-1,2-diphenypropene form erythro1,2-diphenyl-1-propanol by the Chugaev reaction. Predict the major and minor products obtained in the following reaction and explain such distribution of products: ª 150∞C CH 3CH 2CH CH 3 ææææ Æ | ! @ (CH3 )2 N — O:

51.

When optically active coniine is subjected to Hofmann degradation, the major amine product (C10H21N) is found to be optically active. Predict the structure of the product and give the mechanism of the reaction involved. N CH2CH2CH3 H Coniine (an alkaloid)

52.

Explain the following observations: (a)

Elimina on Reac ons

(b)

4.85

+

N(CH3)3OH

D –

H

H

(E)-Cyclooctene (principal product)

53.

Explain why weakly ionizing solvent promotes syn-elimination when the leaving group is uncharged but anti-elimination when the leaving group is charged. [Hint: Ion pairing occurs extensively in non polar solvents. This can cause synelimination with an uncharged leaving group through the following fashion:

] The effect is reversed when the leaving group is positively charged. Ion pairing favours anti-elimination in this case. A relatively free base can be attracted to the charged-leaving group and as a result, it places itself in a favourable position for attack on the syn b-H atom. Ion pairing reduces this attraction and thereby disfavours the arrangement for syn-elimination as shown below:

54.

The ratio of syn-to-anti elimination when 1,1,4,4-tetramethyl-7-cyclodecyl tosylate is treated with t-BuOK in the nonpolar solvent benzene is much higher. But when the crown ether dicyclohexano-18-crown-6 is used, the syn/anti ratio decreases to a much lower value. Explain these observations. [Hint: Ion pairing causes syn-elimination somewhat greater than anti-elimination because the leaving group is uncharged and the reaction is carried out in nonpolar @ !

solvent benzene. The crown ether selectively removes K≈ from t-BuO K by forming a stable complex with K≈. The free base t-BuO@ then causes predominant antielimination.] 55.

Predict the products in each of the following reaction sequences:

(a)

Organic Chemistry—A Modern Approach

4.86

(b)

Give the mechanism of the last step in each sequence. [Hint: C fi

56.

Ph

D

H

Ph

;

F fi

Ph

H

H

Ph

Explain the following observation:

[Hint: The b-D seems to be sandwiched between the bulky t-butyl groups and the methyl groups of trimethylamine so that an external base (i.e., OH@) cannot approach it but an intramolecular attack by the anion (obtained by abstraction of a proton from the — CH3 group attached to positive nitrogen) is favoured.]

4.2

THE E1 REACTION

Example kinetics, mechanism, stereochemistry, evidence in favour of the mechanism, factors influencing E1 reactivity, i.e., the E1 reaction rate and other relevant topics.

4.2.1 Example of E1 Reaction When tert-butyl chloride is treated with 80 percent aqueous ethanol at 65°C, 36 percent 2-methylpropene is obtained by unimolecular elimination (E1) and 64 percent tert-butyl alcohol and tert-butyl ethyl ether by unimolecular substitution (SN1).

Elimina on Reac ons

4.87

E1 reactions almost always accompany SN1 reactions.

4.2.2

Kinetics of E1 Reaction

The E1 reaction exhibits first-order kinetics and obeys the following rate law: Rate = k[(CH3)3 Cl] Like the SN1 mechanism, the kinetics suggests that the mechanism of this elimination reaction involves more than one step, and that the slow step is unimolecular, involving only the alkyl halide.

4.2.3

Mechanism of E1 Reaction

The most straightforward explanation for the observed first-order kinetics is a two-step reaction in which the bond to the leaving group breaks followed by the p bond is formed. The mechanism of this reaction is designated as E1 (E stands for Elimination and 1 stands for unimoleculare) because only one reactant (the substrate) is involved in the transition state of the rate-determining step. The E1 mechanism of the formation of 2-methylpropene form tert-butyl chloride involves the following two steps: Step 1:

Step 2:

In the first step, tert-butyl chloride slowly dissociates under the influence of solvent forces to form tert-butyl cation (a stable 3° carbocation) and chloride ion. It is the rate-determining step of the reaction and it is the same first step as the SN1 mechanism. In the second step, the solvent H2O (or Cl@) acts as a base and removes a proton from a carbon adjacent to the positive carbon of the carbocation (a b-carbon) to form the alkene. During proton loss

Organic Chemistry—A Modern Approach

4.88

the vacant p orbital of the positive carbon and the C — H bond become coplanar because in such a geometry maximum interaction between the C — H bonding electrons and the vacant p orbital occurs. The energy profile diagram for this E1 reaction is as follows:

4.2.4

Stereochemistry of E1 Reaction

The carbocation formed in the first step of an E1 reaction is planar. This means that the electrons from a departing proton can move towards the positively charged carbon from either side. Therefore, both syn- and anti-elimination can take place. Because both synand anti-elimination can occur, an E1 reaction forms both the E and Z products regardless whether the b-carbon atom from which the proton is removed is bonded to one or two hydrogens. The major product is the one with the bulkiest groups on opposite sides of the double bond because this alkene is thermodynamically more stable. For example:

Elimina on Reac ons

4.89

(i)

(ii)

It is to be noted that an E2 reaction forms the E and Z products only if the b-carbon from which the proton is expelled is bonded to two hydrogens. If it contains only one hydrogen, only one product is obtained because anti-elimination takes place. The configuration of the product then depends on the configuration of the substrate.

Organic Chemistry—A Modern Approach

4.90

4.2.5 (a)

(b) (c)

(d)

(e)

4.2.6 (a)

(b)

(c)

(d)

(e)

4.2.7

Evidence in Favour of the E1 Mechanism The E1 reactions have been found to follow first-order kinetics and obeys the rate law: rate = k[RX]. This is consistent with the E1 mechanism in which only one species, the substrate, is involved in the rate-determining step (the first step). Primary isotope effect is not observed in first-order eliminations. This is consistent with the E1 mechanism where the proton is lost in the second, fast step. The first-order eliminations show the same effect of structure on reactivity as SN1 reactions do. E1 involves exactly the same first step as SN1. Since the first step is rate-determining, it follows that the order of reactivity of alkyl halides (R–X) in E1 reaction must be same as in SN1 reaction (i.e., 3°> 2°> 1°). The substitution/elimination ratio remains the same when two substrates which differ only in the leaving group undergo first-order elimination reaction. This supports the E1 mechanism in which the competition between substitution and elimination cannot be influenced by the nature of the leaving group. Where the structure permits, first-order eliminations often result in formation of rearranged products. A rearrangement product can be obtained if the intermediate is a carbocation which can rearrange to give a more stable carbocation. This is also in agreement with the E1 mechanism which involves rate-determining formation of a carbocation.

Factors Influencing E1 Reaction Rate or E1 Reactivity The effect of substrate structure on rate: The order of reactivity is primary < secondary < tertiary and this is because the stability of the carbocations formed in the rate-determining step increases on going from primary to tertiary due to +I and hyperconjugative effects. The effect of the base on rate: Since the base is not involved in the ratedetermining step of the E1 reaction, therefore, the E1 reaction rate does not depend on base concentrations. Weak bases like H2O and EtOH favour E1 reactions. The effect of the leaving group on rate: Since a better leaving group causes facile ionization in the rate-determining step, the more the leaving ability of a group increases the more the E1 reaction is favoured. The effect of solvent on rate: E1 reactions are favoured by increasing the polarity and ion-solvating ability of the solvent. This is because the transition states (step 1) of E1 reactions involving ionic intermediates are considerably polar. The effect of temperature: Increasing the temperature increases the percentage of elimination product. This is because elimination leads to an increase in the number of particles and thus has the more favourable entropy term.

Regioselectivity of E1 Reactions

In an E1 reaction, the leaving group departs in the first step and a proton is lost from an adjacent carbon in the second step. If the substrate contains more than one type of

Elimina on Reac ons

4.91

b-hydrogen, then which alkene is produced predominantly is clearly determined by which b-proton is lost faster from the carbocation in the second step. The transition step of the second step possesses considerable alkene character. Therefore, the transition state leading to a more substituted alkene is more stable than the transition state leading to a less substituted alkene and because of this, the more substituted alkene is produced predominantly. Thus, for E1 reactions, the Saytzeff rule governs the orientation of elimination. For example, the E1 elimination of HBr from 1-bromo-1-methylcyclopentane yields 1-methylcyclopentene as the major product and methylenecyclopentane as the minor product.

Similarly, 2-bromo-2-methylbutane produces 2-methylbut-2-ene as the major product and 2-methylbut-1-ene as the minor product.

4.92

Organic Chemistry—A Modern Approach

If in an E1 reaction the transition state leading to the more substituted alkene suffers from severe steric strain, it becomes the minor product. For example, 2-chloro-2,4,4trimethylpentane undergoes E1 reaction to form an excess of the less substituted alkene 2,4,4-trimethylpent-1-ene (the Hofmann product).

Elimina on Reac ons

4.93

When a substituted cyclohexane undergoes an E1 reaction, the two groups that are eliminated do not have to both be in axial positions because the elimination reaction is not concerted. The carbocation formed in the first step of the following reaction loses a proton to yield the Saytzeff product predominantly.

4.2.8

Rearrangement of the Carbocation Intermediate Involved in an E1 Reaction

Because the E1 reaction involves the formation of a carbocation intermediate, rearrangement of the carbon skeleton may take place before the loss of a proton. For example, the secondary carbocation obtained from 3-chloro-2-methyl-2-phenylbutane undergoes a 1,2-methyl shift to form a more stable tertiary benzylic cation

4.94

Organic Chemistry—A Modern Approach

In the following example, the secondary carbocation undergoes a 1,2-hydride shift to yield a more stable secondary allylic cation.

Similarly, in the following reaction, the secondary carbocation undergoes a 1,2-hydride shift, forming a more stable tertiary carbocation.

Some examples of carbocations which have the potential to undergo rearrangement to form a more stable carbocation are given below: (a)

Elimina on Reac ons

(b)

4.95

CH3

– +

(CH3)2 C——CHCH3 A 2° carbocation (less stable)

CH3- shift

+

(CH3)2 C——CH(CH3)2 A 3° carbocation (more stable)

(c)

(d)

(e)

4.2.9

Acid-Catalyzed Dehydration of Alcohols

Most alcohols undergo dehydration (loss of a molecule of H2O) to form an alkene when heated with a strong acid.

Organic Chemistry—A Modern Approach

4.96

This is an elimination reaction which is favoured at higher temperature. The most commonly used acids are sulphuric acid (H2SO4) and phosphoric acid (H3PO4). The reaction temperature and the concentration of acid required to dehydrate an alcohol depends on the structure of the alcohol used. (a) Primary alcohols are very much difficult to dehydrate. For example, dehydration of ethyl alcohol requires concentrated H2SO4 and a temperature of 180°C. conc.H2SO4 CH 3 — CH 2 — OH ææææææ Æ CH 2 == CH 2 + H 2O 180∞C Ethylene Ethyl alcohol (a 1° alcohol)

(b)

Secondary alcohols usually require milder conditions for dehydration. Dehydration of cyclohexanol, for example, requires 85 percent H3PO4 and a temperature of 165–170°C.

(c)

Dehydration of tertiary alcohols is usually so easy that extremely mild conditions can be used. tert-Butyl alcohol, for example, undergoes dehydration in only 20% aqueous sulphuric acid at a temperature of 85°C. 20% H2SO4 (CH 3 )3 C — OH æææææ æ Æ CH 2 == C(CH 3 )2 + H 2O 85∞C tert-Butyl alcohol Methylpropene (a 3° alcohol)

Thus, the relative ease with which alcohols undergo dehydration is in the following order: R 3C — OH > R 2CH — OH > RCH 2OH 2∞ alcohol 1∞ alcohol 3∞ alcohol This behaviour is, in fact, related to the relative stabilities of the intermediate carbocations.

Mechanism of dehydration The dehydration of alcohols is generally an E1 reaction of their conjugate acids and involves the steps as follows: Step 1: H

H +

—C——C— + H—OH2 OH

—C——C— +

+ H2O

OH2

In this step, protonation converts the very poor leaving group (OH@) into a good neutral leaving group (H2O).

Elimina on Reac ons

4.97

Step 2:

In this step (the rate-determining step), the protonated alcohol dissociates to form a carbocation and water. Step 3:

In this step, a base in the reaction mixture (either H2O or HSO4@) removes a proton from a carbon adjacent to the positively charged carbon, forming the alkene product and regenerating the acid catalyst. Because the rate-determining step is the formation of a carbocation, the ease of dehydration follow the same order as the ease of fromation of carbocations: 3° > 2° > 1°. Hence, the case of dehydration of alcohols follows the order: 3° > 2° > 1°. In fact, the primary (1°) carbocations are too unstable to be fromed and dehydration of primary alcohols actually takes place by the E2 mechanism (as in the case of CH3CH2OH) or by the E1 mechanism in which the incipient 1° carbocation undergoes rearrangement. Acid-catalyzed dehydration of primary alcohols is rarely a good method for converting them to 1-alkenes because rearrangement and isomerization of the products in the presence of acid are very common with primary alcohols. 1-Butanol, for example, undergoes acidcatalyzed dehydration with rearrangement to give a mixture of 1-butene and 2-butene. The more substituted alkene 2-butene is thermodynamically more stable and so it is obtained as the major product following the Saytzeff rule.

4.98

Organic Chemistry—A Modern Approach

The formation of 2-butene as the major product can also be explained by an E2 elimination to form 1-butene followed by addition of a proton from the acidic solution to the double bond of 1-butene following Markownikoff rule, thereby forming a stable carbocation and finally loss of a proton form the carbocation following the Saytzeff rule.

Rearrangement during dehydration Dehydration of secondary and tertiary alcohols involves the formation of a carbocation intermediate. If that carbocation is less stable it may rearrange to a more stable one. For example:

An example of a ring-expansion rearrangement may be given as follows. Both the initially formed carbocation and the carbocation it rearranges to give are 2° carbocations. However, because of strain in the four-membered ring, the initially formed carbocation is less stable. Rearrangement relieves this strain. The rearranged 2° carbocation can further rearrange by a 1,2-hydride shift to form a relatively more stable 3° carbocation.

Elimina on Reac ons

4.2.10

4.99

Dehydration using POCl3 and Pyridine

The relatively harsh conditions (acid and heat) required for dehydration of alcohols can cause low yield of a desired alkene. The sever conditions required for dehydration of an alcohol can be avoided, however, if the reaction is carried out in the presence of phosphorus oxychloride (POCl3) and pyridine (C H N) . For example: 6

5

OH | POCl3 CH 3CH 2CHCH 3 ææææææ Æ CH 3 CH == CH CH 3 pyridine,0∞C 2-Butene 2-Butanol (major product)

Organic Chemistry—A Modern Approach

4.100

Mechanism: The mechanism of the reaction is as follows:

@

Phosphorus oxychloride (POCl3) converts the — OH group of the alcohol into O POCl2 (a good leaving group) in the presence of pyridine. The resulting compound undergoes E2 reaction under the mildly basic reaction conditions to give 2-butene as the major product and 1-butene (CH3CH2CH == CH2) as the minor product.

4.2.11

Factors Influencing the Extent of E1 and E2 Reactions

The following factors affect the extent of E1 and E2 eliminations: (a) Substrate structure: The increasing order of reactivity by both the mechanisms is primary < secondary < tertiary. (In fact, primary alkyl halides undergo only E2 reactions or E1 reactions with rearrangement. They cannot undergo normal E1 reactions because of the difficulty encountered in forming primary carbocation.) The reasons for the reactivity order are, however, different. The rate of elimination by E2 increases mainly because of the alkene being formed increases as the number of alkyl group attached to the doublebonded carbon increases. The rate of E1 reaction, on the other hand, increases because the stability of the carbocation to be formed in the rate-determining step increases. (b) Base: The base is involved in the rate-determining step of an E2 reaction but not in the rate-determining step of an E1 reaction. Because of this, it is only the rate of E2 reaction which depends on the concentration of the base used. In fact, an E2 reaction is favoured by a high concentration of a strong base. In E1 reaction, it is generally the solvent which acts as the base. Hence, when external bases are added, the mechanism is shifted towards E2. (c) Leaving group: Since a better leaving group causes facile ionization, the more the leaving ability of a group increases, the more the E1 reaction is favoured over the E2 reaction. (d) Solvent: E1 reactions are favoured by increasing the polarity and ion-solvating ability of the solvent, i.e., these are generally favoured by protic polar solvents like H2O, AcOH ROH etc. E2 reactions, on the other hand, are favoured by a decreasse in the polarity of

Elimina on Reac ons

4.101

the solvent. This is because the transition states of E1 reactions (step 1) involving ionic intermediates are more polar than the transition states of E2 reactions. In fact, aprotic polar solvents like DMSO, DMF, etc. have a very pronounced effect to increase the rate of the E2 reaction.

4.2.12

Factors that Govern the Proportions of E1 and SN1 Reactions

The following factors govern the proportion of E1 and SN1 reactions: (a) Substrate structure: The first step of these two unimolecular reactions is the same (formation of a carbocation) and it is the second step which determines the proportion of elimination to substitution. The extent of E1 elimination depends upon the fact that how fast the carbocation intermediate loses a proton, and the extent of SN1 substitution depends upon the fact that how fast the carbocations undergoes attack by a nucleophile. The tendency of a carbocation to lose a proton increases in the order: primary < secondary < tertiary. This is because the stability of the resulting alkanes increases progressively on going from primary to secondary to tertiary. The tendency of the carbocation to be attacked by the nucleophile (usually the solvent), on the other hand, increases in the order: tertiary < secondary < primary. This is because the stability of the carbocation decreases progressively, i.e., the carbocation becomes progressively more reactive on going from tertiary to secondary to primary. The other reason for such order is purely steric. The bond angles of the carbocation remain the same on elimination (120°–120°), but decreases on substitution (120°–109.5°). Therefore, steric crowding increases when a carbocation with a trigonal planar arrangement of three groups is converted to the substitution product, with a tetrahedral arrangement of four groups. The extent of substitution thus decreases as the size of the groups attached to the positive carbon increases, i.e., on going from primary to secondary to tertiary and as a consequence, the extent of elimination increases in this order. (b) Base: At low base concentrations or in the absence of base altogether, in ionizing solvents, unimolecular mechanisms are favoured, and it is SN1 which is favoured with respect to E1. (c) Leaving group: For first-order reactions, the leaving group has nothing to do with the competition between elimination and substitution, since it is gone before the decision is made as to which path to take. However, where ion pairs are involved, this is not true, and it has been observed that the nature of the leaving group does affect the product ratio. (d) Solvent: The unimolecular reactions are favoured by an increase in the polarity and ion-solvating ability of the solvent and in most solvents SN1 reactions are favoured over E1. However, the extent of elimination somewhat increases by a decrease in solvent polarity and this is because the charge is more dispersed in the transition state leading to elimination than in the transition state leading to substitution. (e) Temperature: In most unimolecular reactions, the SN1 reaction is favoured over the E1 reaction, especially at lower temperatures. However, increasing the temperature of

Organic Chemistry—A Modern Approach

4.102

the reaction favours the E1 reaction at the expense of the SN1 reaction. This is because elimination leads to an increase in the number of particles and thus has the more favourable entropy term.

1.

Some primary and secondary alcohols undergo rearrangement of their carbon skeleton during dehydration. Write structures of the alkenes expected to be formed in each of the following dehydration reactions and provide mechanism for their formations. CH3 | H2SO4 H2SO4 (a) CH 3CH 2CHCH 2OH ææææ (b) (CH 3 )2CH CHOH CH 3 ææææ Æ Æ D D (c)

H2SO4 (CH 3 )2CCHOH CH 3 ææææ Æ D

(d)

Solution (a) 2-Methyl-1-butanol undergoes dehydration in the presence of concentrated H2SO4 to give 2-methyl-2-butene as the major product and 2-methyl-1-butene as the minor product. CH3 CH3 CH3 | | | H2SO4 CH 3CH 2CHCH 2OH ææææ CH CH C — CH CH CH C Æ == + 3 3 3 2 == CH 2 D 2-Methyl-1-butanol 2-Methyl-2-butene 2-Methyl-1-butene (major product) (minor product)

Mechanism: H

CH3 CH3CH2CHCH2OH

H—OSO3H D

– +

CH3CH2C—CH2—OH2 CH3

H shift —H2O

+

CH3CH2CCH3 CH3 A 3° carbocation (very stable)

Elimina on Reac ons

4.103

It is to be noted that carbocations rearrange only if they become more stable as a result of rearrangement. In this reaction, the protonated alcohol undergoes ionization assisted by a 1,2-hydride shift to give a stable 3° carbocation. This carbocation undergoes loss of a proton forms a b-carbon to yield either 2-methyl-2-butene or 2-methyl-1-butene. The protonated alcohol may also undergo an E2 reaction to give 2-methyl-1-butene. The more substituted alkene 2-methyl-2butene is thermodynamically more stable and so, it is obtained as the major product. (b)

3-Methyl-2-butanol undergoes acid-catalyzed dehydration to give 2-methyl-2butene as the major product and two 1-butenes as the minor products. H2SO4 (CH 3 )2CH CH OH CH 3 ææææ Æ (CH 3 )2C == CH CH 3 + (CH 3 )2CH CH == CH 2 D 3-Methyl-2-butanol 2-Methyl-2-butene 3-Methyl-1-butene (major product) CH3 | + CH 3 CH 2C == CH 2 2-Methyl-1-butene

Mechanism:

Organic Chemistry—A Modern Approach

4.104

The initially formed 2° carbocation rearranges to give a more stable 3° carbocation by a 1,2-hydride shift. The resulting 3° carbocation loses different b-hydrogens to yield 2-methyl-2-butene and 2-methyl-1-butene. The 2° carbocation also loses different b-hydrogens to yield 2-methyl-2-butene and 3-methyl-1-butene. The more substituted alkene 2-methyl-2-butene is thermodynamically more stable and so, it is obtained as the major product. (c)

3,3-dimethyl-2-butanol undergoes acid-catalyzed dehydration to yield 2,3-dimethyl2-butene as the major product and two 1-butenes as minor products. CH3 | H2SO4 (CH 3 )3CCHOH CH 3 ææææ (CH )C C(CH ) + (CH ) CH C Æ == == CH 2 3 3 2 3 2 D 3,3-Dimethyl-2-butanol 2,3-Dimethyl-2-butene 2,3-Dimethyl-1-butene (major product) + CH 2 == CH — C(CH 3 )3 3,3-Dimethyl-1-butene

Mechanism:

Elimina on Reac ons

4.105

The initially formed 2° carbocation rearranges to give a more stable 3° carbocation by 1,2-methanide shift. This 3° carbocation loses different b-hydrogens to yield 2,3-dimethyl-2-butene and 2,3-dimethyl-1-butene. The 2° carbocation may also lose a proton to produce 3,3-dimethyl-1-butene. The more substituted alkene 2,3dimethyl-2-butene is thermodynamically more stable and so, it is obtained as the major product. (d)

1-Cyclohexylethanol undergoes dehydration in the presence of concentrated H2SO4 to yield 1-ethylcyclohexene as the major product.

   

Mechanism:

Organic Chemistry—A Modern Approach

4.106

The initially formed 2° carbocation rearranges to give a more stable 3° carbocation by a 1,2hydride shift. The resulting 3° carbocation loses a b-hydrogen to yield 1-ethylcyclohexene. The 2° carbocation may also lose different b-hydrogens to yield the two minor products. 2.

Suggest a mechanism for each of the following reactions: C2H5 + C2H5 H (a) CHC2H5 heat C2H5 OH

(b)

(c)

OH H

(d)

+

heat

CH3 Br

HBr

(e)

CH3 OH CH3

CH3 H

(f)

+

heat

O

O

CH2OH

CH3

OH CH3 CH3

(g)

CH3

H

+

heat

H3C

H3C

CH3

(h)

CH3 OH

+

H heat

CH3

CH3 CH3

Elimina on Reac ons

4.107

CH2OH

(i) Solution

(a)

(b)

(c)

+

H heat

4.108

Organic Chemistry—A Modern Approach

(d)

This process is especially favourable because relief in ring strain occurs. (e)

(f)

Elimina on Reac ons

(g)

(h)

4.109

Organic Chemistry—A Modern Approach

4.110

(i)

3.

Give mechanistic explanation for each of the following observations:

(a)

(b)

Solution (a) trans-1-Bromo-2-ethylcyclohexane undergoes E2 dehydrobromination when it is heated with sodium ethoxide in ethanol. For a facile E2 elimination involving a low energy T.S., the ring must flip to the diaxial conformation in which the departing groups are anti-periplanar. Since in this conformation the only axial hydrogen b to the axial bromine atom is situated at C-6, therefore, the reaction leads to the formation of 3-ethylcyclohexene exclusively.

Elimina on Reac ons

4.111

When trans-2-ethylcyclohexanol is heated in the presence of an acid, it undergoes acid-catalyzed dehydration by the E1 mechanism. The alcohol becomes protonated and then loses water to form a carbocation. The carbocation may lose two b-protons (from C-2 or C-6) to form two alkenes. Since the transition state leading to the more substituted alkene (the Saytzeff product) 1-ethylcyclohexene is relatively more stable than the transition state leading to the less substituted alkene 3-ethylcyclohexene, therefore, the former alkene is obtained as the major product.

Organic Chemistry—A Modern Approach

4.112

4.

Explain why alkenes are prepared by heating alkyl halides with concentrated alcoholic, rather than aqueous solution of KOH.

Solution Substitution is more favoured under E1/SN1 conditions than under E2/SN2 conditions. Hence, for a given alkyl halide, a higher yield of alkene is expected to be obtained by the E2 mechanism than by the E1 mechanism. Strong bases at high concentrations and solvents with low ionizing power favour bimolecular mechanism and it is E2 which is favoured over SN2. Thus, an ethanolic KOH solution (solvent of low ionizing power and a very strong base :OEt @ ) gives better yield of alkene than does an aqueous KOH solution (solvent of high ionizing power and a relatively weak base, :OH @ ). The reactions are carried out at higher temperature because higher temperature always favours elimination instead of substitution. This is due to the fact that elimination leads to an increase in the number of particles and thus has the more favourable entropy term.

Elimination: KOH (CH 3 )3C —Br ææææææÆ (CH 3 )2C == CH 2 + NaBr + C2H5OH C2H5OH,55°C tert-Butyl bromide 2-Methylpropene

Substitution: KOH (CH 3 )3 C — Cl ææææææ Æ (CH 3 )3 C — OH + Cl@ H2O/acetone tert-Butyl chloride tert-Butyl alcohol

5.

Predict the products expected to be formed when 2-bromo-3-methylbutane is heated in ethanol.

Solution

2-Bromo-3-methylbutane is a secondary alkyl halide. The conditions involve no nucleophile or base other than the solvent, and a polar protic solvent. Because the solvent ethanol is a poor nucleophile and a weak base, neither SN2 nor E2 reactions can take place. Because polar protic solvents promote the SN1 and E1 reactions, these will be the only reactions taking place to give the products: Br OC2H5 | | C2H5OH CH 3 — CH — CH(CH 3 )2 ææææ CH — CH — CH(CH 3 )2 Æ 3 D 2-Bromo-3-methylbutane SN1 Product + CH 2 == CH — C(CH 3 )2 + CH 3CH == C(CH 3 )2 E1 Products OC2H5 CH3 | | + CH 3CH 2 — C — CH 3 + CH 3CH 2 — C == CH 2 | CH3 rearrangement products

Elimina on Reac ons

Mechanism:

6.

Write a reasonable mechanism for the following transformation:

Solution

4.113

Organic Chemistry—A Modern Approach

4.114

The protonated alcohol undergoes loss of water to form a 3° carbocation. Intramolecular cyclization produces another 3° carbocation. This finally loses a proton to yield the diene.

1.

Propose a mechanism for each of the following reactions:

(a)

2.

3.

(b)

Give the major product obtained when each of the following alcohols is heated in the presence of H2SO4: (a)

CHOHCH3

(b)

(d)

CH3 | CH 3CH 2CH 2CCH (CH 3 )2 | OH

(e)

(CH3)3C CHOHCH3

(c)

(f) O

CH2OH

OH

CH3 CH3 CH3 OH

Suggest a mechanism for the following reaction: H

H

4.

!

H Me3CCH == CH 2 æææ Æ

+

CH2OH

Suggest a mechanism for each of the following reactions: O

(a)

(b)

OH CH==CH2

5.

H

+

CH3 The following alcohol undergoes dehydration to produce three alkenes when heated with sulphuric acid. Give mechanisms to account for the formation of these alkenes.

Elimina on Reac ons

4.115

[Hint: A ring-contraction rearrangement of the initially formed 3° carbocation takes place to form another 3° carbocation. This undergoes proton loss to form II and CH3@ shift followed by proton loss to give III.] 6.

7.

Arrange the following alcohols in the order of decreasing their reactivity towards acid-catalyzed dehydration: (a) 2-Methylbut-2-ol (b) Pent-1-ol (c) 3-Methylbut-2-ol Provide a mechanistic explanation for each of the following reactions: (a)

(b)

H2SO4

(c)

heat

OH CH3OH (d) CH 3CH == CHCH 2CHBrCH 2CH 3 ææææ Æ CH 3CH == CH — CH == CHCH 2CH 3

H

H CH3

(e) Br

H CH3

CH3 CH3OH

CH3

(f) CH3 CH3 | | ! EtOH/H2O CH3CHC — S (CH3 )2 Br @ æææææÆ CH CH == C(CH ) + CH CH C == CH2 + S(CH3 )2 3 3 2 3 2 D 8.

Predict the major product that would be formed when each of the following alkyl halides is allowed to undergo an E1 reaction: Cl CH3

(a)

O

(d) H3C

Cl

CH3 | (b) CH 3CH 2 CH CH CH 2 CH 3 | I CMe3 (e) Br

Br | (c) CH CH CH CH CHCH 3 2 2 2 3 Br

(f)

Organic Chemistry—A Modern Approach

4.116

(g)

Cl

(h)

CH3

CH3 Cl CH3

(i)

CH2CH2CH3 Cl

9.

When the following alkyl bromide is treated with methanol, three substitution products and three elimination products are obtained. Account for the formation of these products.

10.

Give detailed mechanism to account for the formation of four products in the following reaction: CH3 CH3 CH3 H3C Cl H3C OCH3 CH3OH

+

(—HCl)

11.

+

+

OCH3

When the following alkyl halide is allowed to react with methoxide ion in a solvent that favours SN2/E2 reactions, only one product is obtained. However, when this compound is treated with MeO@ ion in a solvent that favours SN1/E1 reactions, eight products are obtained. Identify the products obtained under two different sets of conditions and account for their formations. CH3 Cl CH3

12. 13.

Why are 1° alcohols poor starting materials for the synthesis of 1-alkenes? When the following compound is heated in methanol, four products are obtained. Propose mechanisms to account for these products. CH3 CH2Br OCH3 CH3OH + + OCH3 + D

14.

1-Bromomethylcyclohexene undergoes solvolysis in ethanol to yield three major products. Give mechanisms to account for these products. CH2Br CH2OC2H5 CH3OH + + D OC2H5

Elimina on Reac ons

15.

4.117

Solvolysis of bromomethylcyclopentane in methanol produces a mixture of five products. Give mechanisms to account for these products. CH2Br

CH3 CH3OH

+

D

OCH3

CH2 +

H3C

OCH3

+

+

16.

When neopentyl alcohol, Me3CCH2OH, is heated in the presence of acid, it is converted into a 85:15 mixture of two alkenes. Propose mechanism to account for the formation of these alkenes (C5H10). Which one is expected to be the major product and why?

17.

Which alcohol of each pair would you expect to be more easily dehydrated when heated in the presence of an acid? (a) CH3CH2CH2CH2CH2CH2OH or CH3CH2CH2CH2CH(OH)CH3 (b) (CH3)2C(OH)CH2CH2CH3 or (CH3)2CH CH (OH) CH2CH3 (c) (CH3)2CH C(OH) (CH3)2 or (CH3)2CH CH (CH3) CH2OH How would you carry out the following conversion?

18.

n-C4 H 9OH ææ Æ CH 3C ∫∫ CCH 3 19.

Give the major product obtained in each of the following reactions: (a)

H2SO4 CH 3CH 2CH 2CH 2OH ææææ Æ D

OH

(c)

OH H

+

H

H O!

3 Æ (b) (CH)3 CH — CH — CH 2CHO ææææ D | OH O

H

20.

Branching at the b-carbon atom favours E1 elimination. Explain. [Hint: For example, MeCH2CMe2Cl yields only 34% alkene, while Me2CHCMe2Cl yields 62%. The reason may be party steric: the greater the branching, the more crowding strain is released on going from sp3 hybridized halide to the sp2 hybridized carbocation intermediate. Strain is reintroduced on substitution (SN1), but not on proton loss (E1) to yield an alkene. It is due also to the fact that the more b-substituted alkyl halide will yield a more substituted and hence thermodynamically more stable alkene.]

21.

Explain the following observations: (a)

Organic Chemistry—A Modern Approach

4.118

(b)

4.3

THE E1cB REACTION

Example, kinetics, mechanism and the nature of the substrate, i.e., the substrate structure.

4.3.1 Example of E1cB Reaction Base-promoted dehydrofluorination of 2,2-dichloro-1,1,1-trifluoroethane proceeds by the E1cB mechanism. :OEt@ / EtOH

F3C — CHCl2 ææææææÆ

4.3.2

F2C == CCl2 + EtOH + :F:@ 1,1-Dichloro-2,2difluoroethene

Kinetics of E1cB Reaction

The E1cB reaction exhibits second-order kinetics (first order in the substrate and first order in the base) and obeys the following rate law (same as E2): Rate = k[F3CCHCl2] [OEt@]

4.3.3

Mechanism of E1cB Reaction

Although the E1cB mechanism is the least common of the elimination pathways, it provides a reasonable extension of the elimination mechanism that is not related to substitution. This is, in fact, a base-promoted unimolecular reaction taking place through the formation of the conjugate base (cB) of the substrate. The mechanism is designated as E1cB (Elimination Unimolecular conjugate Base) because it is the conjugate base of the substrate that gives up the leaving group and the conjugate base is the only species involved in the rate-determining step of the reaction (i.e., unimolecular). An E1cB reaction proceeds through the following two steps: Step 1: The base abstracts the most acidic b-hydrogen atom from the substrate to form an intermediate carbanion. This step is fast and reversible.

Elimina on Reac ons

4.119

Step 2: The carbanion loses the leaving group to form the alkene product. This step is slow (ratedetermining). –

–

C====C

—C——C—

+ LG

LG

4.3.4 The Nature of the Substrate Substrates containing acidic b-hydrogens (groups like –NO2, –CHO, –COR, –CN, –COOR, @

@

@

–SO2Ar etc. should present at Cb) and poor leaving groups like F@ , Cl@ , SePh,O Ph,OH, etc. are very prone to react by the E1cB mechanism. In the given example, the halogens enhance the acidity of the b-hydrogen atom and fluoride ion (F@) is a poor leaving group. The reaction occurs as follows: Step 1:

Step 2:

4.3.5 To Distinguish between E1cB and E2 Mechanisms In many cases, it is not possible to distinguish between E1cB and E2 mechanism. This can be established by the following example. Dehydrobromination of 2-phenylethyl bromide using NaOEt as a base in ethanol can be shown to take place by E1cB or E2 mechanism using kinetic studies. E1cB mechanism: Step 1: H fast

–

EtO + Ph—CH—CH2Br

–

EtOH + Ph CH CH2Br

Step 2: –

slow

Ph—CH—CH2 Br

–

Ph CH==CH2 + Br

Organic Chemistry—A Modern Approach

4.120

Since the reaction is carried out in ethanol medium, the concentration of the ethanol formed in the first step of the reaction can be neglected. Hence, the equilibrium constant. @

[Ph C H CH 2Br] K¢ = [Ph CH 2CH 2Br][EtO@ ] The rate-determining step of the reaction is the loss of Br@ from the carbanion (conjugate base). Therefore, the rate of the reaction is given by @

rate = k[Ph CH CH2 Br] = k K ¢ [Ph CH2CH2 Br][EtO@ ] Since k and K ¢ are both constants, their product is also a constant (say K). Therefore,

rate = k[Ph CH2CH2Br] [EtO@]

Thus, kinetically the E1cB reaction can be interpreted as a second-order reaction. E2 mechanism: It is a one-step process and occurs through the participation of both the substrate, PhCH2CH2Br and base, EtO@. H –

EtO + Ph—CH—CH2

PhCH==CH2 + EtOH + Br

–

Br Therefore, E1cB and E2 mechanisms cannot be kinetically distinguished.

Deuterium labelling can be used to distinguish the E1cB from the E2 pathway. The reaction is carried out in a solvent which could act as a deuterium source (e.g., EtOD) and the substrate recovered after half completion of the reaction is analysed for its deuterium content. There is no scope of incorporation of deuterium into the substrate if the reaction proceeds by the concerted E2 mechanism. Thus, if the substrate is found to contain no deuterium, the reaction must take place by the E2 mechanism. On the other hand, deuterium incorporation is expected to occur in an E1cB reaction because it involves a carbanion intermediate. Therefore, if the recovered substrate is found to contain deuterium, the reaction must proceed through the E1cB pathway. E2 reaction:

Elimina on Reac ons

4.121

E1cB reaction:

In actual experiment, it was observed that there is no deuterium incorporation when PhCH2CH2Br is allowed to react with NaOEt/EtOH. That is, the reaction occurs by the E2 mechanism.

1.

Propose a mechanism for each of the following reactions: F NaNH2 – (a) + NH3 + F liq.NH3

(b)

(c)

(d)

Ph MeO

H NO2

MeO

–

MeOH

Ph

NO2 + MeOH + MeO

–

O O || || @ EtO CH 3 — S — CH 2CH 2OPh ææææ Æ CH 3 — S — CH == CH 2 + EtOH + PhO@ EtOH

Solution (a) Since F@ is a poor leaving group, :NH 2@ is a very strong base and ortho-H atom is

considerably acidic (attached to an sp2 hybridized carbon and adjacent to the electro negative fluorine atom), therefore, fluorobenzene undergoes dehydrofluorination by the E1cB mechanism.

4.122

Organic Chemistry—A Modern Approach

(b)

When a cyanohydrin is treated with a base, it undergoes elimination reaction by the E1cB mechanism to give the corresponding carbonyl compound and this is because the hydrogen attached to electronegative oxygen is considerably acidic and the CN@ ion is not a good leaving group.

(c)

Since the hydrogen atom attached to carbon containing the electron-withdrawing –NO2 group is considerably acidic and MeO@ ( a strong base) is not a good leaving group, therefore, this compound undergoes elimination of MeOH by the E1cB mechanism.

(d)

Since the hydrogen atom attached to the carbon a to the electron-withdrawing S ==O group is considerably acidic and PhO@ is not a good leaving group, therefore, this compound undergoes elimination of PhOH by the E1cB mechanism when treated with EtO@/EtOH.

2.

Reaction of Cl2C == CHCl with NaOD in D2O affords ClC ∫∫ CCl. When the reaction is stopped short of completion, the recovered olefin contains deuterium. Suggest a mechanism for the reaction consistent with this observation.

Solution The observation suggests that the reaction follows the E1cB pathway and this is because an E1cB reaction involves reversible formation of a carbanion which can take up D≈ form D2O to form deuterium-labelled alkene.

The substrate is suitable for an E1cB reaction because the b-hydrogen is considerably acidic (it is attached to an sp2 carbon and halogens enhance its acidity) and Cl@ ion is not a good leaving group. The reaction occurs as follows:

Elimina on Reac ons

3.

4.123

Explain the mechanism of the following base-catalyzed dehydration reactions: OH

(a)

–

(—H2O)

HO

(b)

Ph

Ph :OH

@

CH 3CH(OH)CH 2CHO ææææ Æ CH 3 CH == CHCHO (–H O) 2

Solution

(a)

The b-H of the substrate is considerably acidic because the conjugate base is a very stable (aromatic) cyclopentadienyl anion system. Also, the leaving group OH@ (a strong base) is a very poor one. Therefore, this alcohol undergoes base-catalyzed dehydration by the E1cB mechanism.

(b)

Since the b-H of this substrate (a b-hydroxyaldehyde or aldol) is considerably acidic (because the conjugate base is stabilized by resonance involving the —CHO group) and OH@ (a strong base) is a very poor leaving group, therefore, it undergoes basecatalyzed dehydration by the E1cB mechanism.

4.124

4.

Organic Chemistry—A Modern Approach

The threo as well as the erythro-isomer of 1,2-diphenylpropyltrimethyl ammonium salt yields the same alkene, (E)-1,2-diphenylpropene when treated with t-BuOK / t-BuOH. Explain.

Solution Since the two diasteroisomeric salts undergo E1cB reaction when treated with t-BuOK / t-BuOH, therefore, they yield the same stable alkene, (E)-1,2-diphenylpropene.

This is a case of stereoconvertent reaction because substrates of different stereochemistry, i.e., diastereoisomers, yield the same product. 5.

E1, E2 and E1cB make a spectrum of elimination reactions. Discuss this statement.

Solution The loss of HLG in an E2 reaction is a concerted process in which two bonds are formed and two bonds are cleaved at the same time. However, the C — H bond cleavage may precede C — LG bond cleavage an 'E1cB-like' E2 process) or the C — LG bond cleavage may precede C — H bond cleavage (an 'E1-like' E2 process). In fact, the extent of concertedness varies from one extreme, at which the C — H bond is cleaved but the C — LG bond is intact, to another, at which C — LG bond is cleaved but the C — H bond is intact. These extremes are clearly nothing more than a pure E1cB and pure E1 mechanistic course. Hence, all the three mechanistic possibilities are related, differing mainly in the timing of various

Elimina on Reac ons

4.125

bond-breaking and bond-making processes and, therefore, provide a continuum of reaction pathway for elimination.

1.

When dehydrofluorination of 1,1-dichloro-2,2,2-trifluoroethane is carried out by treating with NaOMe in MeOD, 1,1-dichloro-1-deuterio-2,2,2-trifluoroethane is recovered from the reaction mixture. How does this observation fit into the E1cB mechanism proposed for this elimination reaction?

2.

Propose a mechanism for each of the following reactions:

(a)

(b)

(c)

(d)

O O || || t -BuOK p-ClC6 H 4CCHClCH ClC6H 4Cl-p ææææÆ p-ClC6 H 4 C — C == CH — C6 H 4Cl-p t -BuOH | Cl + t - BuOH + Cl@

3.

Explain why the following b-hydroxyketone does not undergo base-catalyzed dehydration to give the corresponding a, b-unsaturated ketone.

Organic Chemistry—A Modern Approach

4.126

Me O OH

4.

Predict the mechanism of the reaction from the following observation regarding primary kinetic isotope effect: @

EtO CH3CH Br CH3 ææææ Æ CH3CH == CH2 EtOH

kH /kD = 6.7

@

EtO CD3CH Br CD3 ææææ Æ CD3CH == CD2 EtOH

5.

Suggest a mechanism for the following reaction: – H3C OEt OH C H3C OH [Hint: The E1cB elimination is the usual mechanism by which a hemiacetal is converted to a carbonyl compound under basic condition.]

4.4 a- or 1,1-ELIMINATION Example, kinetics, mechanism and substrate structure.

4.4.1 Example of a- or 1,1-Elimination Reaction When heated with an aqueous solution of KOH, chloroform undergoes a-elimination via a two-step 1,1-E1cB pathway to yield dichlorocarbene, a highly reactive neutral intermediate. KOH/H2O :CCl2 + H 2O + KCl CHCl3 ææææÆ D

4.4.2

Kinetics of a- or 1,1-Elimination Reaction

The reaction follows second-order kinetics, i.e., rate = k[CHCl3] [OH@]

4.4.3

Mechanism of a- or 1,1-Elimination Reaction

The reaction proceeds through the steps as follows: Step 1: The base OH@ takes up the acidic proton of chloroform to form a carbanion (the conjugate base).

Elimina on Reac ons

Step 2:

4.127

The carbanion is converted to dichlorocarbene (:CCl2) by the loss of Cl@.

Therefore, this is a case of elimination reaction where both the departing groups are lost from the same carbon. E2 and a-elimination reactions are kinetically indistinguishable. These two can be differentiated by isotope labelling. The reaction shows the incorporation of D in the unreacted substrate when D2O was used as a solvent.

4.4.4

Structure of the Substrate Involved in a-Elimination Reactions

Substrates in which a-H is strongly acidic and contains no b-H undergo a- or 1,1-elimination reaction in the presence of strong base.

1.

Explain why 2,2-dichlorobutane does not form a Grignard reagent when treated with Mg in dry ether.

Solution Since the compound is a gem-dihalide, therefore, the initially formed Grignard reagent readily lose a chloride ion to form a carbene, i.e., an a- or 1,1-elimination reaction takes place.

2.

What happens when sodium trichloroacetate (Cl3CCOONa) is heated?

Solution When sodium trichloroacetate is heated, it undergoes an a-elimination reaction to form dichlorocarbene (:CCl2), CO2 and NaCl. The reaction, which is initiated due to @

stability of Cl3 C : ion, occurs as follows:

Organic Chemistry—A Modern Approach

4.128

3.

Predict the products when CHClBr2 and CHF2Br are treated separately with the base t-BuOK.

Solution When CHClBr2 and CHF2Br are treated with the base t-BuOK, bromchlorocarbene (:CClBr) and difluorocarbene (:CF2) are obtained respectively. This occurs because Br@ is a better leaving group than Cl@ or F@.

1.

2.

Propose a mechanism for each of the following reactions: (a)

PhCH 2Br + PhLi ææ Æ PhCH + PhH + LiBr

(b)

Me2C == CH Br + Na NH 2 ææ Æ Me2C == C: + NH 3 + NaBr

Identify the products obtained in each of the following reactions: (a)

D or hn CH 2N2 ææææ Æ A+B

(b)

hn CH 2 == C == O æææ Æ C+D

(c)

hn Ph2CN2 æææ Æ E+F

(d)

D or hn N2CHCO2Et ææææ Æ G+H

!

CH2

@

(e)

D Me2 S— C H 2 ææ Æ I+J

(g)

O Ph CH—CHPh

hn

M + N

(f)

Ph—CH—CH2

hn

K + L

INDEX

A Acetylacetone 1.224 Acid strength and pKa 1.47 Acid-catalyzed dehydration of alcohols 4.95 Acid-catalyzed tautomerism 1.221 Acidic character of alcohols 1.63 character of aromatic acids 1.126 character of imides 1.92 character of phenols (Ar-OH) 1.86 Acidity of carboxylic acids 1.84 Acids and bases 1.46 Acraldehyde 1.16 Acrylonitrile 1.16 AgSCN 3.85 Allyl bromide 3.11 Allylic rearrangement 3.18 Alternating axis of symmetry 2.37 Ambident nucleophiles 3.27 Ammonia 2.39 Anchimeric assistance 3.147 Angle strain 1.126, 3.94 Anthracene 2.41 Antiaromatic carbocation 3.93 Antiaromatic Compounds 1.242 Aromatic carbocation 3.93 Aromaticity 1.240 Asymmetric and dissymmetric molecules 2.38 Atropisomerim 2.87 Axial approach of a nucleophile 3.72 Axial chirality 2.86 Azulene 1.26

B 1-Bromotryptecene 3.135

1,3-Butadiene 1.104 2-Bromohydrins 3.154 (Z)-2-Butene to (E)-2-butene 4.77 p Bond as neighbouring group 3.157 Base 4.35, 4.100, 4.101 Base-catalyzed racemization 2.190 Base-catalyzed tautomerism 1.221 Basic character of arylamines 1.93 character of guanidine 1.97 strength of amines 1.65 Basicity 3.25 Bicyclic systems 3.10 Boiling point 1.150 Bond angle 1.10 dissociation enthalpy or bond dissociation energy 1.9 energy 1.10 length 1.9, 1.120 polarity 1.22 polarizability 1.36 Boron tifluoride 1.12 Branching at the b-carbon 3.8 Bredt’s rule 1.135 Bridgehead carbon 3.93 Brönsted-Lowry theory 1.46 Butane-gauche interaction 2.217

C C—C bond as a neighbouring group 3.161 Captodative ethylene 1.109 Captodative radical 1.217 Carbanionic character 4.42 Carbanions 1.178

Index

I.2

Carbenium ion 1.178 Carbocation rearrangements 3.105 Carbocations 1.172 Carbonium ion 1.178 Carbontetrachloride 2.30 Catalysis 1.283 Centre of symmetry 2.36 Charge-transfer complex 1.26 Chiral and achiral molecules 2.84 Chiral axis 2.86 Chirotopic and achirotopic atom 2.131 Chloroacetone 3.12 Chloroform 1.92 Chlorofumaric acid 4.74 Chugaev reaction 4.28 CIP chirality rules 2.115 Cis-1,2-dibromocyclohexane 4.17 Cis-1,2-dichlorocyclohexane 1.29 Cis-1,2-dideuterioethane 2.30 Cis-trans or geometric isomers 2.2 Competition experiments 3.96 Concerted reaction 1.172 Configuration 2.108 Configurational nomenclature 2.108 Conformation of acyclic organic molecules 2.211 chloroethane 2.218 ethane 2.213 propane 2.214 Constitutional isomers 2.2, 2.58 Contribution of resonance structures towards resonance hybrid 1.78 Coordination with an electrophilic species 3.35 Copper acetylide 2.31 Corey–House method 3.66 Criteria for aromaticity 1.240 Crown ethers 3.31 Curtin–Hammett principle 4.48 Cyclohexyl methyl ketone 4.52 Cyclohexylacetylene 4.52 Cyclopropenyl cation 3.136

D 2,2-Dimesityl ethanol 1.226 1,2-Dichloroethane 1.27, 2.218 1,2-Dideuterioethene 2.30 1,2-Dimethylcyclobutane 2.40 1,3-Dibromoallene 2.40 1,3-Dibromocyclobutane 2.31 1,3-Di-sec-butylyclobutane 2.36

2,6-Di-tert-butylpyridine 3.74 p-Diacetylbenzene 1.40 p-Dinitrobenzene 1.40 D L-system 2.109 DBN 4.33 DBU 4.33 Debromination 4.49 Debye unit 1.23 Decarboxylative debromination 4.73 Dehydration using POCl3 and pyridine 4.99 Dehydrobromination 4.49 Deuterium oxide 2.29 Dextrorotatory 2.182 Diastereoisomers 2.60 Diastereotopic faces 2.138 Diastereotopic ligands 2.135 Diaxial conformation 4.13 Diazomethane 3.61 Dielectric constants 3.21 Difluoromethane 2.30 Dihedral angle 2.212 Dimethyl sulphoxide (DMSO) 3.18 Dimethylacetamide (DMA) 3.18 Dioxane 3.53 Dipole moment 1.23, 1.121 of chlorobenzene 1.25 of disubstitute benzenes 1.32 Dipole–Dipole interactions 1.147 Dodecadeuterocyclohexane 2.41 Double bond character 4.41 Double Bond Equivalent (DBE) 1.44

E a-Effect 3.78 a-Elimination 4.127 a-or 1,1-Elimination 4.126 1,1- or a-Elimination reactions 4.2 1,2- or b-Elimination reactions 4.2 1,2-Ethanediol 1.27 1,3- or g-Elimination reactions 4.3 E1 reaction 4.86 E1cB and E2 mechanisms 4.119 E1CB reaction 4.118 E2 reaction 4.4, 4.5 Eclipsed conformation 2.10 Effect of base 4.33 hyperconjugation 1.119 resonance on the properties of molecules 1.81 solvent 3.18, 4.34

Index

on rate 3.99 substrate structure 3.7, 4.33 temperature 4.34 the leaving group 3.32, 4.34 on rate 3.102 the nucleophile 3.22 on rate 3.101 Ei eliminations 4.24 Electromeric effect 1.67 Electronegativity 1.22 Electrophiles and nucleophiles 1.193 Electrophilic and nucleophilic radicals 1.192 Electrophilic catalysis 3.102 Element effect 1.49 Enantiomeric excess 2.187 Enantiomers 2.60 Enantiotopic faces 2.137 Enantiotopic ligands 2.134 Energy diagram 1.278 Energy profile diagram 3.4 Entropy 4.5 Erythro and threo nomenclature 2.124 Ether cleavage 3.39 Ethyl acetoacetate 1.219 Evidence in favour of E1 mechanism 4.90 E2 mechanism 4.32 SN1 mechanism 3.90 Example of NGP 3.146 E-Z system of designating alkene diastereoisomers 2.125

F Face strain or F-strain 1.133 Factors affecting acidity 1.49 Factors influencing E1 reaction rate 4.90 E2 reaction rate 4.33 SN1 reaction 3.91 SN2 reaction rate 3.7 the extent of E1 and E2 reactions 4.100 Field effect 1.67 Fischer projection 2.4 Fluoroform 1.92 Flying-Wedge projection 2.3 Formal charge 1.33 Formation of ethylene 1.6 methane 1.4 Fragmentations 4.45

I.3

Furan 1.26

G Gabriel synthesis 3.36 Gas phase nucleophilicity 3.25 Gas-phase acidity of alcohols 1.64 Gem-dichloride 4.52 Geometric enantiomerism 2.127 Grignard reagent 4.127

H Halocyclohexanes 3.12 Halocyclopropanes 3.12 Hammond postulate 1.284 Hard and soft nucleophiles and electrophiles 3.26 Hard centres 3.28 Heat of combustion 1.119 Heteroatoms as neighbouring groups 3.148 Heterolytic fission 1.171 Hexadeuteriobenzene 2.32 Hexamethylphosphoramide (HMPA) 3.18 Hofmann exhaustive methylation or Hofmann degradation 4.43 Hofmann rule 4.38 HOMO 3.6 Homoallylic carbocation 1.178 Homoaromatic compounds 1.251 Homolytic fission 1.171 Homomers 2.2 Homotopic faces 2.137 Homotopic ligands 2.134 Hybrid orbital number 1.3 Hybridization effect 1.2, 1.51 Hydrogen bonding 1.148 Hyperconjugation 1.117 Hyperconjugative effect 1.118

I Imidazole 1.104, 3.58 Index of Hydrogen Deficiency (IHD) 1.44 Induced moment 1.31 Inductive and electrometric effects 1.60 Inscribed polygon method 1.245 Intermolecular dehydration 3.117 forces 1.147 hydrogen bonding 1.148 Intramolecular elimination 4.25

Index

I.4

hydrogen bonding 1.149 SN2 reaction 3.15 Inversion of configuration 2.140 Invertomerism 2.220 Ion pairing in determining nucleophilicity 3.25 Isomers 2.2 Isotope effects and salt effects 3.113 Isovalent and heterovalent resonance 1.81

K Ketene 1.16 Keto-Enol tautomerism 1.219 Kinetic control versus thermodynamic control 1.286 Kinetics of a- or 1,1-elimination 4.126 E1 reaction 4.87 E1cB reaction 4.118 E2 reaction 4.4 SN1 reaction 3.86 SN2 reaction 3.3 KSCN 3.85

rotation 2.185 Monochromatic light 2.182

N N-bromosuccinimide (NBS) 1.217 N,N-dimethylformamide 3.18 Neighbouring group participation 3.146 Neomenthyl chloride 4.15 Neopentyl bromide 3.9 Newman projection 2.11 Nitroform 1.92 Nitrotoluenes 1.32 No bond resonance 1.118 Nonaromatic carbocation 3.93 compounds 1.242 Nonclassical carbocation 1.177 Nonpolar covalent bonds 1.22 Nucleophilic substitution 3.2, 4.3 Nucleophilicity 3.25 in protic solvents 3.57

O L Leaving group 4.37, 4.100, 4.101 Levorotatory 2.182 Lewis Acid-Base theory 1.52 Lucas test 3.130 LUMO 3.6

M Mechanism of a- or 1,1-elimination 4.126 dehydration 4.96 E1 reaction 4.87 E1cB reaction 4.118 E2 reaction 4.4 Keto-enol tautomerism 1.221 NGP 3.146 SN1 reaction 3.86 SN2 reaction 3.3 Melting point 1.150 Menthyl chloride 4.15 Meso Compounds 2.89 Meso-Tartaric acid 2.36 Methods of determining mechanisms of reactions 1.298 Modern definition of aromaticity 1.243 Molecular chirality 2.84 orbital energy diagram 1.244

Optical activity 2.182 purity 2.187 Order of basicity of halides 3.25

P Percentage of ionic character 1.36 Phase-transfer catalysts 3.30 Phloroglucinol 3.79 Plane of polarization 2.182 Plane of symmetry 2.33 Plane-polarized light 2.182 Polar and apolar solvents 3.20 Polar covalent bond 1.22 Polarimeter 2.183 Principal axis 2.32 Prochiral and prostereogenic centres 2.135 Pro-R and pro-S designation 2.137 Prostereoisomerism and topicity 2.133 Protic and aprotic solvents 3.20 Proton sponges 1.132 Pyramidal inversion (umbrella effect) 2.186 Pyrolysis of acetates 4.24 tertiary amine oxides 4.29 xanthates 4.28 Pyrrole 1.26

Index

Q Quinclidinone 1.146

R R,S-system 2.111 Ra,Sa-nomenclature of allenes 2.121 Ra,Sa-nomenclature of biphenyls 2.122 Racemic modification 2.187 Racemization 2.188 Radical inhibitors or ‘scavengers’ 1.192 Re and Si designation 2.138 Reactive intermediates 1.170, 1.172 Rearrangement during dehydration 4.98 Rearrangement of the carbocation intermediates 4.93 Regiochemistry of epoxide ring opening 3.118 Regioselectivity in b-elimination reactions 4.37 Regioselectivity of E1 reactions 4.90 Resolution of racemic modification 2.192 Resonance and resonance effect or mesomeric effect 1.75 energy 1.77 structures 1.76 theory 1.76 Retention of configuration 2.139 Ring-chain tautomerism 1.228 Rotational axis of symmetry 2.29 Rules for writing meaningful resonance structures 1.78

S Sacrificial and isovalent hyperconjugation 1.119 Sawhorse projection 2.10 Saytzeff rule 4.37 Scale of acidities and basicities 1.48 Selenoxides 4.71 Sequence rules 2.115 Simple axis of symmetry 2.29 SN1 reaction 3.86 SN1¢ mechanism 3.112 SN2 reaction 3.3 SN2¢ mechanism 3.18 SNi mechanism 3.109 SNi¢ mechanism 3.112 SODAR 1.44 Sodium D line 2.185 Sodium trichloroacetate 4.127 Soft centres 3.28 Source of chirality 2.85 Specific rotation 2.184

I.5

Stabilities of carbanions 1.66 Stabilities of carbocations 1.66 Staggered conformation 2.10 Stepwise reaction 1.172 Stereoaxis 2.86 Stereocentre or stereogenic centre 2.88 Stereochemistry of E1 reaction 4.88 E2 reaction 4.5 SN1 reaction 3.87 SN2 Reaction 3.5 Stereoelectronic requirement 4.6 Stereoisomers 2.59 Stereospecific reaction 4.7 Steric acceleration 1.134 effect 1.125 retardation 1.134 strain 1.125 Sulphonic acids 1.104 Super acid 1.194 Super leaving groups 3.33 Symmetric, asymmetric and dissymmetric 2.38 Symmetry elements 2.29 Syn-anti momenclature for aldols 2.128 Syn-axial interactions 4.55 Syn-elimination 4.20

T a-Truxillic acid 2.44 1,2,3-Triphenylcyclopropane 1.106 1,3,5-Trihydroxybenzene 1.227 Tautomerism 1.218 Tetramethylallene 2.43 Thermal racemization 2.189 Thermodynamically stable conformation 4.16 Thiosulfate 3.79 Torsional strain 1.126 Trans-1,2-dibromocyclohexane 4.17 Trans-2-bromocyclopentanol 3.16 Trans-4-tert-butylcyclohexyl tosylate 4.14 Trichloroacetic acid 1.63 Triphenyl phosphine 4.77 Triphenylmethyl bromide 3.135 Triphenylmethyl chloride 1.205 Triphenylpyrylium salt 3.70 Tri-p-nitrophenylamine 1.104, 1.107 Triptycenyl cation 1.205 Trityl cation 1.205 Trivial axis 2.33

Index

I.6

U

W

Unreactive nature of phenyl halides 3.15

Weakly polarizable base 4.36 Williamson synthesis of ethers 3.17, 3.43

V Valence tautomerism 1.228 van der Waals Forces 1.149 VESPR theory 1.11 Vinylic and aryl halides 3.14, 3.95 Violation of Saytzeff rule 4.42

X Xenon tetrafluoride 2.35