Iupac & Goc.doc

NOMENCLATURE OF ORGANIC COMPOUNDS IUPAC System [International union of pure and applied chemistry] The most important feature of this system is that any given molecular structure has only one IUPAC name which denotes only one molecular structure. Salient features of IUPAC system 1. 2. 3. 4. 5.

A given compound can be assigned only one name. A given name can clearly direct in writing of one and only one molecular structure. The system can be applied in naming complex organic compounds. The system can be applied in naming multifunctional organic compounds. This is simple, systematic and scientific method of nomenclature of organic compounds.

Rule for Naming Prefix (alphabetically) root word (alk) primary suffix (ene, yne) secondary suffix (main functional group) So IUPAC name of any organic compounds essentially consists of two or three parts. (i) Root word (ii) Suffix (iii) Prefix (i) Root Words The basic unit is a series of root words which indicate linear or continuous chains of carbon atoms. Chains containing one to four carbon atoms are known by special root words while chains from C5 onwards are known by Greek number roots.

Chain Length C1 C2 C2 C4 C5 C6 C7 C8 C9 C10

Root word MethEthPropButPentHexHeptOctNonDec-

Chain Length C11 C12 C13 C14 C15 C16 C20 C30 C40 C50

Root word UndecDodecTridecTetradecPentadecHexadecEicosTriacontTetracontPentacont-

In general, the root word for any carbon chain in alk-. (ii) Primary Suffix Primary suffix are added to the root words to show saturation or unsaturation in a carbon chain.

Nature of carbon chain

Primary suffix

Saturated (C – C) Unsaturated (C = C) with one double bond Unsaturated (C  C) with one triple bond Unsaturated with two C = C bonds Unsaturated with two C  C bonds

-ane -ene

Generic name Alkane Alkene

-yne

Alkyne

-diene -diyne

Akladiene Alkadiyne

Unsaturated with three C = C bonds

-triene

Alkatriene

(iii) Secondary Suffixes Suffixes added after the primary suffix to indicate the presence of a particular functional group in the carbon chain are known as secondary suffixes.

Functional Group

Secondary suffix

Alcohol (-OH)

-ol

Aldehyde (-CHO)

-al

Ketone (>CO)

-one-

Carboxylic acid (-COOH)

-oic acid

Sulphonic (-SO3H)

-sulphonic acid

Amine (-NH2)

-amine

Thioalcohol (-SH)

-thiol

Cyanide (-CN)

-nitrile

Ester (-COOR)

-oate

Amide (-CONH2)

-amide

Acid halide (-COX)

-oyl halide

Note: The terminal ‘e’ of the primary suffix is removed when initial letter of secondary suffix is vowel. To illustrate the application of above basic rule, the generic names of few classes of organic compounds are given below:

Homologous series

Root word

Primary suffix

Secondary suffix

Generic name

Alcohols (saturated)

Alk

-ane

-ol

Alkanol

Alcohols (unsaturated) one double bond

Alk

-ene

-ol

Alkenol

Alcohols (Unsaturated) one triple bond

Alk

-yne

-ol

Alkynol

Aldehydes (saturated)

Alk

-ane

-al

Alkanal

Ketones (saturated)

Alk

-ane

-one

Alkanone

Carboxylic acids (Saturated)

Alk

-ane

-oic acid

Alkanoic acid

Acid chlorindes (saturated)

Alk

-ane

-oyl chloride

Alkanoyl chloride

Prefix:

It should always be kept in mind that alkyl groups forming branches of the parent chain are considered as side – chains. Atoms of groups of atoms such as fluoro (-F), chloro (-Cl), bromo (-Br), iodo (-I), nitro (-NO2), nitroso (-NO) and alkoxy (-OR) are referred to as substituents. Roots words are prefixed with the name of the substituent or side chain. Arrangement of Prefixes, Root word and Suffixes These are arranged as follows while writing the name. Prefix (es) + Root word + Primary suffix + Secondary suffix For Example: 4 2 1 3 H3C CH CH2COOH

Br Prefix = Bromo (at position 3), Root word = But, Primary suffix = -ane Secondary suffix = -oic acid Hence, the name of the compound is, 3 – Bromo butanoic acid 4 2 1 5 3 H3C CH CH CHCH2OH

CH3 Prefix = Methyl (at position 4) Root word = Pent, Primary suffix = -ene (at position 2), Secondary suffix = -ol Hence, the name of the compound is, 4 – Methyl pent – 2 – en – 1 – ol The names of simple aliphatic organic compounds containing only straight chains atoms of various homologous series are described in table as to explain the basic rules of IUPAC system. In case of compounds. Other than hydrocarbons, only the saturated compounds have been considered. NAMING OF HYDROCARBON Compound of Hydrogen and Carbon are called hydrocarbon Classifications

Hydrocarbons

Open Chain hydrocarbon(Aliphatic)

Saturated

Unsaturated

Hydrocarbon Alkane

Hydrocarbon

[C - C]

Closed chain (cyclic)

Alicyclic

Alkene

Alkyne

[C = C]

[C = C]

Aromatic

Open Chain Hydrocarbon

Alkane

Alkene

Alkyne

General Formula

CnH2n+2

CnH2n

CnH2n-2

Value of n

C–C

C=C

CC

n=1

CH4 (Methane)

-

-

n=2

C2H6 (Ethane)

C2H4 (Ethene)

C2H2 (Ethyne)

n =3

C3H8 (Propane)

C3H6 (Propene)

C3H4 (Propyne)

n=4

C4H10 (Butane)

C4H8 (Butene)

C4H6 (Butyne)

n=5

C5H12 (Pentane)

C5H10 (Pentene)

C5H8 (Pentyne)

n=6

C6H14 (Haxane)

C6H12 (Hexene)

C6H10 (Hexyne)

n=7

C7H16 (Heptane)

C7H14 (Heptene)

C7H12 (Heptyne)

n=8

C8H18 (Octane)

C8H16 (Octene)

C8H14 (Octyne)

n=9

C9H20 (Nonane)

C9H18 (Nonene)

C9H16 (Nonyne)

n = 10

C10H22 (Decane)

C10H20 (Decene)

C10H18 (Decyne)

Types of carbon and hydrogen atoms in alkanes They may be classified in to four types (i) Primary or 10 carbon: A carbon atom attached to one other carbon atom. (ii) Secondary or 20 carbon: A carbon atom attached to two other carbon atom. (iii) tertiary or 30 carbon: A carbon atom attached to three other carbon atoms. (iv) Ouaternary or 40 carbon: A carbon atom attached to four other carbons atoms. The hydrogen atom attached to 10, 20 and 30 carbon atoms are called primary (1 0), secondary (20) and tertiary (30) hydrogen atoms. Illustration 1. How many 10, 20, 30 and 40 carbon atoms are present in the following hydrocarbons?

CH3 H3C

C

CH2

CH3

Solution:

CH

CH3

CH3

10  five,

20 one, 30  one, 40  four

10 CH3 10 H3C

C 0 CH2 4 20 10 CH3

30 CH

10 CH 3 0

CH3 1

Exercise 1. How many 10, 20, 30 and 40 carbon atoms are present in the following compounds? (CH3)3C – CH(CH3).CH(CH3)2 Alkyl group The removal of one hydrogen atom from the molecule of an alkane gives an alkyl group. General formula CnH2n+1 n = 1, 2, 3---------------------Example: CH3 Methyl C2H5 Ethyl C2H5CH2

Propyl (n – propyl)

CH3 – CHCH3

20 Propyl (Isopropyl or secondary propyl)

C3H7 C3H7CH2 - 10 butyl (n- butyl or primary butyl)

H 5C 2 C

CH3 20 Butyl(sec. butyl)

H

C 4H 9

CH3 H 3C

CH CH2

H 3C

C

2 Methyl -1 - Propyl (Iso butyl)

CH3 2 Methyl -2 - Propyl (ter butyl)

CH3

CH3 H3C

C

CH2

CH3 Nomenclature of Hydrocarbons Alkane

Neo pentyl

The IUPAC system of alkane nomenclature is based on the simple fundamental principle of considering all compounds to be derivatives of the longest single carbon chain present in the compound. The chain is then numbered from one end to the other, the end chosen as number 1 is that which gives the smaller number at the first point of difference. CH3

CH3

H3C H3C

CH3

CH3

2-methylpentane

3-methylhexane

When there are two or more identical appendages - the modifying prefixes di-, tri-, tetra-, penta-, hexa-, and so on are used, but every appendage group still gets its own number. CH3

CH3

H3C

CH3

H3C

2,2-dimethylpentane

CH3

H3C

CH3

CH3

CH3

2,2,4,4-tetramethylpentane

When two or more appendage locants are employed, the longest chain is numbered from the end which produces the lowest series of locants. When comparing one series of locants with another, that series is lower which contains the lower number at the first point of difference. CH3 H3C

CH3 CH3

H3C

H3C

CH3

CH3 CH3

2,3-dimethylpentane

CH3

2,5,6-trimethyloctane (not 3,4,7-trimethyloctane)

H3C

CH3 CH3

CH3

2,2,4-trimethylpentane (not 2,4,4-trimethjylpentane)

Several common groups have special names that must be memorized by the student. CH3 CH3 CH3 | | | CH3 CH– CH3CH CH2– CH3–CH2–CH– isopropyl

CH3 | CH3 C– | CH3 tert-butyl or t-butyl

isobutyl

sec-butyl

CH3 | CH3–C–CH2– | CH3 neo-pentyl

A more complex appendage group is named as a derivative of the longest carbon chain in the group starting from the carbon that is attached to the principal chain. The description of the appendage is distinguished from that of the principal chain by enclosing it in parentheses.

CH3 | H–C–CH3 | H–C–CH3 | CH3CH2CH2CH2CHCH2CH2CH2CH3



5-(1,2-dimethylpropyl)nonane

When two or more appendages of different nature are present, they are cited as prefixes in alphabetical order. Prefixes specifying the number of identical appendages (di, tri, tetra and so on) and hyphenated prefixes (tert-or t, sec-) are ignored in alphabetizing except when part of a complex substituent. The prefixes cyclo-, iso-, and neo-count as a part of the group name for the purposes of alphabetizing. When chains of equal length compete for selection as the main chain for purposes of numbering, that chain is selected which has the greatest number of appendage attached to it. CH3

CH3

CH3

H3C

CH3

H3C

5-(2-ethylbutyl) – 3, 3 – dimethyl decane When two or more appendages are in equivalent positions, the lower number is assigned to the one that is cited first in the name (that is one that comes first in the alphabetic listing). H3C

H3C

CH3

CH3 CH3

H3C

3-ethyl-5-methylheptane

H3C

CH3

CH3

H3C

CH3

4-ethyl-5-isopropyloctane

CH3 H3C

CH3

CH3 CH3

CH3

6-ethyl-3,3-dimethyloctane 3,3,6 is lower than 3,6,6, at first point of difference

H3C

6-ethyl-2-methyloctane 2,6 is lower than 3,7

[The complete IUPAC rules actually allow a choice regarding the order in which appendage groups may be cited. One may cite the appendages alphabetically, as above, or in order of increasing complexity]. Exercise 2. Write the IUPAC name for the following

CH3

CH3

(i)

H3C

CH3 CH3

H3C

CH3

H3C

CH3

(ii)

CH3

CH3 CH3

H3C

(iii) CH3

CH3

Alkenes Nomenclature and Structure: Alkenes (olefins) contains the structural unit

C=C and have the general formula C nH2n. These unsaturated hydrocarbons are isomeric with the saturated cycloalkanes. The IUPAC rules for naming alkenes are similar in many respects to those for naming alkanes.

  

Determine the root word by selecting the longest chain that contains the double bond and change the ending of the name of the alkane of identical length from ane to ene. Number the chain so to include both carbon atoms of the double bond, and begin numbering at the end of the chain nearer the double bond. Designate the location of the double bond by using the number of the first atom of the double bond as prefix : CH3 CH3 H C 3 H C 2

hex-2-ene

but-1-ene



Indicate the locations of this substituent groups by the numbers of the carbon atoms to which they are attached. CH3 CH3

H3C CH3 2,5-dimethylhex-2-ene



CH3 H3C H3C

CH3

H2C

Cl

4-chlorobut-1-ene

(2E)-5,5-dimethylhex-2-ene

Two frequently encountered alkenyl groups are the vinyl group and the allyl group. CH2 = CH — CH2 = CH CH2 — The vinyl group The allyl group (are not included in IUPAC system) The following examples illustrate how these names are employed CH2 = CH — Br CH2 = CH — CH2 Cl vinyl bromide allyl chloride (are not IUPAC name)



The geometry of the double bond of a disubstituted alkene is designated with the prefixes, cis and trans. If two identical group are on the same side of the double bond, it is cis, if they are on opposite sides; it is trans.

Cl

Cl

Cl

H C=C

C=C H H cis - 1,2 - Dichloroethene

H

trans - 1,2 - Dichloroethene

Illustration 2. The IUPAC name for CH2

H3C

Cl

H3C

H3C

CH3

CH3 H3C

CH3 4-(1-methylbutyl)hexa-1,4-diene

H

CH3

CH2 Cyclopropyl ethene

7-ethyl-4,8-dimethyldec-4-ene

H3C

H

H

H3C H

CH3 H H (2Z,4Z)-hexa-2,4-diene

H2C (3E)-penta-1,3-diene

Exercise 3. Give the IUPAC name of the following H3C CH3 (ii (i) H2C ) CH H3C

CH2 CH3

3

(iii ) CH3

H3C

CH3

Alkyne Nomenclature of Alkynes: Alkynes contains the structural unit C C and have the general formula CnH2n-2. The simple alkynes are readily named in the common system as derivatives of acetylene itself. CH3C  CH CH3—CC—CH2—CH3 F3CCCH methylacetylene ethylmethylacetylene trifluoromethylacetylene In the IUPAC system the compounds are named as alkynes in which the final – ane of the parent alkane is replaced by the suffix – yne. The position of the triple bond is indicated by a number when necessary. CH3C  CH (CH3)2CHCCH CH3CH2CH2CHC  CCH3 Propyne 3-methyl –1-butyne CH2CH2CH3 4-propyl-2-heptyne

When both a double and triple bond are present, the hydrocarbon is named an alkenyne with numbers as low as possible given to the multiple bonds. In case of a choice, the double bond gets the lower number. In complex structures the alkynyl group is used as a modifying prefix. C

CH

Cyclopentyl ethyne

Aromatic Hydrocarbons These compounds consists of at least one benzene ring, i.e., a six – membered carbocylic ring having alternate single and double bonds. CH

CH

CH HC

HC

HC

CH

HC

CH CH

CH

CH C

C CH

CH CH

Napthalene (Bicyclic)

C

HC

CH

C CH

Benzene (Monocyclic)

HC

CH C

CH

C CH

CH CH

Anthracine (Tricyclic)

-H � R -H �� � �R - G� ( G is functional group, R is Alkyl group) FUNCTIONAL GROUP � +G CnH2n+1 CnH2n+1 � � �

1. Alkyl Halide The alkyl halides have the general formula C nH2n+1X or RX, where X denotes fluorine, chlorine bromine or iodine. CH3 CH3 H3C

H3C

Br bromoethane

Br 2-bromopropane

H3C

Cl 1-chloro-2-methylpropane H3C

H3C

CH3

CH3 Br

H3C

CH3 Cl 2-chloro-2-methylpropane

H3C CH3 Br 1-bromo-2,2-dimethylpropane 2,4-dibromopentane

Br

H2C

H2C Br 3-bromoprop-1-ene

H2C

CH3

6-bromohepta-1,5-dien-3-yne H

Br

Br Br 1,2-dibromoethane

Br

bromocyclohexane

Cl Phenyl chloro methane

Cl chloroethene

2. Alcohols General formula [CnH2n+1.OH], IUPAC name is alkanol. For the simpler alcohols the common names, are most often used. These consist simply of the name of the alkyl group followed by the word alcohol. For example: OH

CH3

H3C

OH

OH

H3C

CH3 propan-1-ol

H3C

propan-2-ol

2-methylpropan-1-ol

b

CH3 H3C

a

OH

CH3 OH 2-methylbutan-2-ol

O 2N (3-nitrophenyl)methanol

CH3

OH 1-phenylethanol

We should notice that similar names do not always mean the same classification; for example, isopropyl alcohol is a secondary alcohol, whereas isobutyl alcohol is a primary alcohol. Finally, there is the most versatile system, the IUPAC. The rules are: 1. Select as the parent structure the longest continuous carbon chain that contains the -OH group; then consider the compound to have been derived from this structure by replacement of hydrogen by various groups. The parent structure is known as ethanol, propanol, butanol, etc., depending upon the number of carbon atoms; each name is derived by replacing the terminal -e of the corresponding alkane name by -ol. 2. Indicate by a number the position of the -OH group in the parent chain, generally using the lowest possible number for this purpose. 3. Indicate by numbers the positions of other groups attached to the parent chain. H3C H3C

OH

OH

CH3

ethanol

OH 2-phenylethanol

2-methylbutan-1-ol OH

H3C

CH3

CH3

H3C

H3C

CH3

OH 2-methylbutan-2-ol

3-methylbutan-2-ol

OH CH2 OH H3C Cl 2-chloroethanol but-3-en-2-ol

Alcohols containing two hydroxyl groups are called glycols. They have both common names and IUPAC names. CH3

OH

H3C CH3

OH HO Ethane-1, 2 - diol

H3C OH Propane-1, 2 - diol

H HO

OH H 3,4,5-trimethylcyclopentane-1,2-diol

Note: 2 or 3 OH group can not present on same carbon atom, decomposes to give aldehyde/ketone or carboxylic acid respectively. OH R

C

-H2O

OH

R

H

C

H Aldehyde

OH R

O

C

-H2O

OH

R

R

C

O

R Ketone OH

OH R

O

C

-H2O

OH

R

C

O

Carboxylic acid

OH

3. Ether (R  O  R] Ethers are compounds in which two C atoms are connected to a single O atom. In IUPAC nomenclature, name one of the alkyl group plus the O atom (RO) as an alkoxy and comes as a prefix to the parent hydrocarbon. (Oxygen is to be counted with least number of carbon atom) IUPAC name of ether is alkoxy alkane CH3OCH(CH3)2

2 – Methoxy propane (isopropyl methlyl ether)

OCH3

1 – isopropoxy – 2 – methaoxy – cyclohexane OCH(CH3)2

O

H3C

CH3 Methoxy ethane

4. Aldehydes IUPAC names the longest continuos chain including the C of – CH = O and replaces – e of the alkane name by the suffix – al i.e. alkanal. The C of CHO is number 1. For compounds with two – CHO groups, the suffix – dial is added to the alkane name. When other functional groups have naming priority, – CHO is called formyl. Common names replace the suffix –ic (–oic or – oxylic) and the word acid of the corresponding carboxylic acids by – aldehyde. Locations of substituents on chains are designated by Greek letters e.g. e

d

g

b

a

 C C C C C C = O

| H

The terminal C of a long chain is designated w (omega) The IUPAC names of aldehydes follow the usual pattern. The longest chain containing the –CHO group is considered the parent structure and named by replacing –e of the corresponding alkane by –al. The position of the substituent is indicated by a number, the carbonyl carbon always being

considered C-1. Here, as with the carbonyl acids, the C-2 of the IUPAC name corresponds to alpha of the common name. H

H

H

H O formaldehyde

H3C

H3C O acetaldehyde

H

O

H3C

propionaldehyde

O butyraldehyde

H

H O 2N

CHO

H3C

O

O

benzaldehyde

4-nitrobenzaldehyde

4-methylbenzaldehyde

H O H OH salicylaldehyde

O phenylacetaldehyde

CH3

CH3 O

O

O

H3C

H3C

H3C CH3

H 2-methylpentanal

H

H 4-methylpentanal

3-methylpentanal

5. Ketones Common names use the names of R or Ar as separate words, along with the word ketone. The IUPAC system replaces the –e of the name of the longest chain by the suffix –one. In molecules with functional groups such as – COOH, that have a higher naming priority, the carbonyl group is indicated by the prefix keto or oxo. Thus, CH3COCH2CH2COOH is 4-ketopentanoic acid. The simplest aliphatic ketone has the common name acetone. For most other aliphatic ketones we name the two groups that are attached to carbonyl carbon and follow these names by the word ketone. A ketone in which the carbonyl group is attached to a benzene ring is named as phenone, all illustrated below. The positions of various groups are indicated by numbers.

O H3C H3C

CH3

O butan-2-one

acetone

O

CH3 H3C

CH3 pentan-2-one

CH3 H3C

CH3 O pentan-3-one

H3C

CH3 CH3

O 3-methylbutan-2-one

O 1-phenylacetone

CH3 CH3 O 1-phenylethanone

6. Carboxylic acid

O 1-phenylbutan-1-one

O benzophenone

(i) Common or Trivial names: The names of lower members are derived from the Latin or Greek word that indicate the source of the particular acid.

Formula

Source

HCOOH CH3 – COOH CH3 – CH2 – COOH

Red ant (Latin ant – formica) Vinegral (Latin vinegar acetum) Proton-pion (Greek = Proton = first, pion = fat)

Common or trivial names Formic acid Acetic acid Propionic acid

(ii) Derived System: Monocarboxylc acid may be named as alkyl derivative of acetic acid. CH3 H3C

H3C

COOH Methyl acetic acid

COOH Dimethyl acetic acid

(iii) IUPAC System: Acids are named as alkanoic acids. The name is derived by replacing ‘e’ of the corresponding alkane by oic acid. HCOOH Methanoic acid CH3COOH Ethanoic acid CH3CH2COOH Propanoic acid In case of substituted acids, the longest chain including carboxyl group is selected and numbering is done from the carbon of carboxylic group.

Example :

CH3

H3C

COOH

CH3 3,4-dimethylpentanoic acid

Illustration 3.

Give the IUPAC name of the following compounds

O

CH3

(a)

CH2 CH

CH

COOH

(b )

H O

(c) OHC

Solution:

(a) (b) (c)

O

H2C COOH

OH

(d )

COOH

C(OH) - COOH H2C

Pent-2-enoic acid 2-formyl ethanoic acid 4- formyl -2- oxocylo hexananoic acid

COOH

(d) 3-hydroxy propane-1, 2, 3-tricarboxylic acid Exercise 4. Write IUPAC name of the following: (i) X COOH

H3C

(iii)

(ii) OH

(iv )

CHO H3C

H3C

Cl

OH

7. Derivative of Carboxylic acid O

R

C

O

--OH Z

H

Alkanoic acid

O R

C

Z

Acid derative

Z = derivative of carboxylic acid Acyl Group (i) Naming Acyl Groups- Acid halide and Anhydrides The group obtained from a carboxylic acid by the removal of the hydroxyl portion is known as an acyl group. The name of an acyl group is created by changing the - ic acid at the end of the name of the carboxylic acid to –yl, examples: O || H –C –

O || H –C – O – H Formic acid

Formyl group

O ||  C – OH Benzoic acid

O || C– Benzoyl group

(a) Acid Halide IUPAC name of acid halide is alkanoyl halide.

where X may be Cl, Br, I Acid chlorides are named systematically as acyl chlorides.

(b) Acid anhydride

An acid anhydride is named by substituting anhydride for acid in the name of the acid from which it is derived.

IUPAC name of acid anhydride is alkanoic anhydride. O O

R C O Example: O H3C

C

C

R O

O

C

C3 H7

Butanoic, ethanoic anhydride (ii) Naming Salts and Esters General formula of ester RCOOR O

R

C

O

R'

IUPAC name: Alkyl alkanoate Example: CH3COOC2H5 (Ethyl ethanoate) The name of the cation (in the case of a salt) or the name of the organic group attached to the oxygen of the carboxyl group (in the case of an ester) precedes the name of the acid.

(iii) Name of Amides and Imides The names of amides are formed by replacing –oic acid (or –ic acid for common names) by amide or –carboxylic acid by carboxamide. IUPAC name of acid amide is alkanamide O

R

C

NH2

Example:

If the nitrogen atom of the amide has any alkyl groups as substitutents, the name of the amide is prefixed by the capital letter N; to indicate substitution on nitrogen, followed by the name(s) of alkyl group(s).

If the substituent on the nitrogen atom of an amide is a phenyl group, the ending for the name of the carboxylic acid is changed to anilide

Exercise 5. Write IUPAC name of the following (i) CH2(OH), CH(CH3)COCl (ii) CH3COOCOC6H5 (iii) CH3CHOHCHClCHOCH3CONH2 (iv) CH3CHCH3CHOHCH2COCl Some dicarboxylic acids form cyclic amides in which two acyl groups are bonded to the nitrogen atom. The suffix imide is given to such compounds. O

O NH

NH O

O Succinimide

Phthalimide

8. Amine Nomenclature of amines Nomenclature of amines is quite simple. Aliphatic amines are named by naming the alkyl group (or) groups attached to nitrogen , and following that by the word amine.

More complicated amines are often named as prefixing amino - (or-N-methylamino -, N-N, diethyl amino -, etc) to the name of the parent chain.

Aromatic amines - those in which nitrogen is attached to an aromatic ring - are generally named as derivatives of the simplest aromatic amine, aniline. Br N

NH2 Br

Br

CH3 C2H5

(N-ethyl-N-methyl aniline)

(2,4,6-Tribromo aniline)

Salts of amines are generally named by replacing - amine by - ammonium (or - aniline by anilinium), and adding the name of the anion. NH3Cl– (CH3)3NHNO3– (Trimethylammonium nitrate) (Anilinium chloride)

9. Nitro Alkane General formual

CnH2n+1.NO2

CH2

R

N

CH2

or R O

N O

Nitro alkane Example:

NO 2

NO2 H3C

CH

CH3

H3C

CH3

2-nitro propane

H3C

CH3

2 nitro 3, 3 dimethyl pentane NO 2

Alkyl nitrites Common name is alkyl nitrite there is no IUPAC name of nitrite. R–O–N=O Alkyl nitrite CH3CH2 – O NO Ethyl nitrite 10. Alkyl Cyande R–CN

Alkane nitrile

Example: CH3 – C  N

Ethane nitrile H3C CN

2 methyl propane nitrile

H3C

Alkyl iso cyanide RNC

R N C Alkyl iso nitrile or alkyl iso cyanide. There is no specific IUPAC name for alkyl iso cyanide or isonitrile. CH3NC Ethyl isonitrile When a compound contains more than one functional group, the numbering and the suffix in the name of multifunctional compound are determined by nomenclature priority. Preference Order

Class

Carboxylic acid Acid ahydrides

Formula

Suffix (if present as a functional group) oic acid

O C O

OH

C

O

O

Eaters

O C O

O

Acyl halides

X

Amides

C O C

NH2

Nitriles Aldehydes

CN O

Carboxy

oic anhydride

C

Alkyl alkanoate

Alkoxycarbonyl

oyl halide

Haloalkanoyl

-amide

Carbamoyl

Nitrile al

Cyano Formyl

one

Oxo

ol Amine

Hydroxy Amino

H

C

O

Ketones

Prefix (if present as a substituent)

C

Alcohols Amines

OH H/R N H/R

Ethers Alkenes

O

Alkoxy alkane ene

oxy Alkenyl

Alkynes Halides

CC X

-yne 

Alkynyl Halo

NO2

Nitro Alkanes

C

C

Benzene

 ane

Nitro Alkyl

Benzene

Phenyl

BENZENE COMPOUND For naming aromatic compounds, no special rules are required, but are named substituted benzene. The benzene ring is considered to be a parent and alkyl groups, halogens and the nitro group are named as prefix to benzene. Br

CH3

NO2

Bromobenzene

Methyl benzene

Br

Nitrobenzene

Cl

O 2N

Br

NH2

2 – chloro – 4 – nitro aniline

Br

1, 3, 5 – tribromobenzene OH

O2N

COOH

2 – hydroxyl – 4 – nitro – benzoic acid When a benzene ring is attached to an alkane chain with a functional group or to an aklane chain of two or more carbon atoms, then benzene is considered as a substituent (phenyl) instead of a parent. CH2CH2OH OH

HC 2- phenyl ethanol

CH2CH2Br

2- bromo - 1 - phenyl propanol

When more than one group is present on benzene ring then following prefix are giving to certain organic compound. G 2 & 6 = Ortho or ‘o’ 1 3 & 5 = Meta or ‘m’ 2 6 4 = Para or ‘p’ With respect to G. 5

3

4

CH3

CH3

CH3 CH3 CH3

O - Xylene

CH3

m - Xylene

p - Xylene

The following names are given to certain aromatic hydrocarbon residues formed by the loss of one or more hydrogen atoms from the parent hydrocarbon. CH3

C2H5--(Phenyl)

C6H4 (O-Phenylene)

CH2

CH3C6H4--(O-Tolyl)

C

CH

C6C5CH2--(Benzyl or Phenyl methyl))

C6C5CH-

C6C5C

(Benzal or Benzylidene)

(Bezo or Benzylidene)

ALICYCLIC (CYCLIC COMPOUNDS) IUPAC name-cycloalkane (saturated), cyclo alkene, cyclo alkyne (unsaturated)………………. Examples: CH2 or H 2C

CH 2

(cyclo pentene)

(cyclo propane)

(cyclo butyne)

Note: Naming of cyclic compounds containing functional group – same as open chain compound. Exercise 6. Write the IUPAC name of compounds OH

(i)

(ii )

H3C

CH

CH2COOH

BOND CLEAVAGE Organic reactions take place through the formation of reactive intermediates. These intermediates are formed due to cleavage of covalent bonds. These intermediates can be �

(i) free radicals like CH3 �

(ii) carbocation like CH3



(iii) carbanion like CH3 Homolytic (symmetrical) cleavage In which the two electrons shared in a  bond become unpaired as the bond is broken.

CH3

Light/Heat

CH3

CH3

Light/Heat

NH2

CH3

CH3

NH2

CH3

The species formed are called free radicals (i) They are electrically neutral. (ii) They are extremely reactive. Their stability is in the order of

R R R

C

CH > RCH2 > CH3

> R

R 30

20

10

Benzylic and allylic free radicals are resonance stabilized hence are more stable than alkyl free radicals. 





H2C

H2C

H2C

H2C



H2 C

CH

 CH2

Benzyl radical

 H2C

CH

CH2

Allyl radicals �

Thus greater the stability easier will be formation of the species. CH3 (Methyl radical) is

sp2 – hydridized (bearing three  CH bonds and singly occupied p – orbital) with HCH angle 1200 and three C  H bonds coplanar. Thus when a methyl radical is formed in the homolytic cleavage of CH3  X bond, the carbon undergoes a geometric change from tetrahedral to planar and rehybridisation from sp3 to sp2. Heterolytic (unsymmetrical) cleavage When a covalent bond joining two atoms A and B breaks in such a way that both the electrons of the covalent bond (i.e. shared pair) are taken away by one of the bonded atoms, the mode of bond cleavage is called heterolytic cleavage. Heterolytic cleavage is usually indicated by a curved arrow which denotes a two electron displacement. For example

A : B A : B

Heterolytic cleavage

A

B

Heterolytic cleavage

A

B

(When B is more electronegative than A)

(When A is more electronegative than B) As shown above heterolytic fission results in the formation of charged species, i.e. cations and anions. It usually occurs in polar covalent bonds and is favoured by polar solvents.

In the formation of carbocation, we also find that sp 3 hybridised carbon (in CH3 X) changes to sp2 hybridised carbon. An organic ion with a pair of available electrons and a negative charge on the central carbon atom is called carbanion and stability is in order CH3

CH3 > CH3CH2 > H3C

CH > H3C

C

CH3

CH3

Electron attracting group (CN, > C = O) increases stability and electron – releasing group ( CH3 etc) decreases stability of carbanion. Benzyl carbanion is again stabilized by resonance.

H2 C

H2 C

H2 C

H2 C

Exercise 7. The greater the s-character in an orbital the ------------ is its energy (A) Greater (B) Lower (C) Both (D) None REACTION INTERMEDIATES Most of organic reactions occurs through the involvement of certain chemical species. These are generally short lived (10-6 seconds to a few seconds) and highly reactive and hence can not be isolated. These short lived highly reactive chemical species. Through which the majority of the organic reactions occur are called reactive intermediates. These intermediates are detected by spectroscopic methods or trapped chemically or their presence is confirmed by indirect evidence. On the other hand, synthetic intermediate are stable products which are prepared isolated and purified and subsequently used as starting materials in a synthetic sequence. Carbocations (Earlier Called As Carbonium Ions) Carbocations are the key intermediates in several reactions and particularly in nucleophilic substitution reactions and electrophilic addition reaction. (a) Structure: Generally in the carbocations the positively charged carbon atom is bonded to three others atoms and has no nonbonding electrons. It is sp 2 hybridized with a planer structure and bond angles are +

of about 1200. There is a vacant unhybridised p orbital which (e.g in the case of CH3 ) lies perpendicular to the plane of C  H bonds. E m p ty

R 0 1 2 0



C



R

(b) Stability:

R 2 s p -h y b rid iz e dc a rb o no rb ita l & s tru c tu reo fc a rb o c a tio n s

There is an increase in carbocation stability with additional alkyl substitution. Thus one finds that addition of HX to three typical olefins decreases in the order (CH3)2C = CH2 > CH3  CH = CH2 > CH2 = CH2 This is due to the relative stabilities of the carbocations formed in the rate determining step which in turn follows from the fact that the stability is increased by the electron releasing methyl group (+I), three such groups being more effective than two, and two more effective than one. Stabilized by three electron releasing groups

CH3

CH3

C

> H C 3

CH

H2 C

CH3 >

CH3

CH3 +

Stability of carbocations 30 > 20 > 10 > CH3 Electron release: Disperses charge, stabilizasion. Further, any structural feature which tends to reduce the electron deficiency at the tricoordinate carbon stabilizes the carbocation. Thus when the positive carbon is in conjugation with a double bond. The stability is more. This is so, due to resonance the positive charge is spread over two atoms instead of being concentrated only on one. This explains the stability associated with the allylic cations. The benzylic cations are stable, since one can draw canonical forms as for allyl ic. CH2

CH2

CH2

CH2

The benzyl cation stability is affected by the presence of substituents on the ring. Electron donating p – methoxy and p – amino group stabilize. The carbocation by 14 and 26 kcal/mole, respectively. The electron withdrawing groups like e.g, p – nitro destabilize by 20 kcal/mol. A heteroatom with an unshared pair of electrons when present adjacent to the cationic centre strongly stabilizes the carbocation. The methoxy methyl cation has been obtained as a stable +

-

solid CH3 OCH2SbF6 cylopropylmethyl cations are even more stable than the benzyl cations. This special stability is a result of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p orbital of the cationic carbon. The carbocations are planar is shown by the fact these are difficult or impossible to form at bridgeheads, where they can not be planar. +

The stability order of carbocation is explained by hyperconjugation. In vinyl cations (CH2 = CH) , resonance stability lacks completely and therefore are very much less stable. Stability  hyperconjugated structures  number of a hydrogen. Exercise 8. Which of the following carbocations is most stable?

Exercise 9. Which of the following C – Cl bond is weaker.

Cl

(a) Ph – | – CH3 CH

(b) Ph – CH2 – CH2 – Cl

Exercise 10. Arrange the following in the increasing order of C – Br bond energy.

Br | (a) CH3 – C CH2 – CH3 | CH3 Br (b) CH3 – | – CH2 – CH2 – CH3

CH (c) CH3 – CH2 – CH2 – CH2 – CH2 – Br (d) CH2 = CH – Br (e) Ph – CH2 – Br Exercise 11. Which of the following C – I bond is weak and why? (i) CH3 – O – CH2 – I

H

(ii) CH3 – | – CH2 – I N Carbanions Chemical species bearing a negative charge on carbon and possessing eight electrons in its valence shell are called carbonions. These are produced by heterolylic cleavage of covalent bonds in which the shared pair of electrons remain with the carbon atom. (a) Structure: A carbanion posses an unshared pair of electron and thus represents a base. The best likely description is that the central carbon atom is sp 3 hybridized with the unshared pair occupying one apex of the tetrahedron. Carbonions would thus have pyramidal structures similar to those of amines. It is believed that carbanions undergo a rapid interconversion between two pyramidal forms. There is evidence for the sp3 nature of the central carbon and for its tetrahedral structure. At bridgeheads carbon does not undergo reaction in which it is converted to a carbocation. However, the reactions which involve carbanions at such centre take place with ease, and stable bridgehead carbanion are known. In case this structure is correct and if all three R groups on a carbanion are different, the carbanion should give retention of configuration. However, this never happens and has been explained due to an umbrella effect as in amines. Thus the unshared pair and the central carbon rapidly oscillate from one side of the plane to the other. (b) Stability and Generation: The Grignard regent is the best known member of a broad class of substances, called organometallic compounds where carbon is bonded to a metal lithium, potassium sodium, zinc,

mercury, lead, thallium almost any metal known. Whatever the metal it is less electronegative than carbon and the carbon metal bond like the one in the Grignard reagent is highly polar. Although the organic group is not a full fledged carbanion an anion in which carbon carries negative charge, it however, has carbanion character organometallic compounds can serve as a source form which carbon is readily transferred with its electrons. On treatment with a metal, in RX the direction of the original dipole moment is reversed (reverse polarization) d+

d

CH3 CH2  Br

d

Mg

CH3CH2

d

Mg

Br

Considerable carbanion character

R

C

C

Also acetylide ion

(c) Properties: carbanions are nucleophilic and basic and in this behaviour these are similar to amines, since the carbanion has a negative charge on its carbon, to make it a powerful base and a stronger nucleophile than an amine. Consequently is enough basic to remove a proton from ammonia. Illustration 4. Identify the stable carbanion in each pair 

(a) CH3 – CH2 – CH2– and CH3 – CH – CH3 

O

O



(b) CH2 – || – CH3 and CH2 – CH2 – || – H C C (c) Ph – CH2– and Ph – CH2 – CH2– Solution:

(a) CH3 – CH2 – CH2– (because of inductive effect) 

O

(b) CH2 – || – CH3 (because of mesomeric effect) C (c) Ph – CH2– (because of resonance) Free Radicals A free radical is a species which has one or more unpaired electrons. In the species where all electrons are paired the total magnetic moment is zero. In radicals, however, since there are one or more unpaired electrons. There is a net magnetic moment and the radicals as a result are paramagnetic. Free radicals are usually defected by electron spin resonance, which is also termed electron paramagnetic resonance. Simple alkyl radicals have a planar (trigonal) structure i.e., these have sp 2 bonding with the odd electron in a p orbital. The pyramidal structure is another possibility when the bonding may be sp 3 and the odd electron is in an sp 3 orbital. The planar structure is in keeping with loss of activity when a free radical is generated at a chiral centre. Thus, a planar radical will be attacked at either face after its formation with equal probability to give enantiomers unlike carbocations, the free radicals can be generated at bridge. This shows that pyramidal geometry for radicals is also possible and that free radicals need to be planar

H

H

p - orbital

H

H sp3 - hybridized orbital

C

C H

H

Pyramidal structure

Stability As in the case of carbocation, the stability of free radicals is tertiary > secondary > primary and is explained on the basis of hyperconjugation. The stabilizing effects in allylic radicals and benzyl radicals is due to vinyl and phenyl groups in terms of resonance structures. Bond dissociation energies shows that 19 kcal/mol less energy is needed to form the benzyl radicals from toluene than the formation of methyl radical from methane. The triphenyl methyl type radicals are no doubt stalbilized by resonance, however the major cause of their stability is the steric hindrance to dimerization.. C6H5CH3 Toulene

 C6H5CH2

 H

H = +85 kcal

Benzyl radical

+

Ease of formation of alkyl free radicals, benzyl > 30 > 20 > 10 > CH3 > Vinyl Illustration 5: Alkenes undergo electrophilic addition reaction and benzene undergoes electrophilic substitution whereas both proceeds through carbocation intermediate. Explain. Solution:

 electrons are available in case of alkanes whereas they are delocalized in case of benzene. After attack of electrophile a stable delocalized carboncation is formed on benzene ring. Whereas a carbocation which can rearrange is formed by addition of electrophile on alkane.

Exercise 12. Which of the following statement is correct? +

(A) Allyl carbonium ion (CH2=CH– CH2 ) is more stable than propyl carbonium ion (B) Propyl carbonium ion is more stable than allyl carbonium ion (C) Both are equally stable (D) None ELECTRONIC DISPLACEMENT IN COVALENT BONDS The following four types of electronic effects operates in covalent bonds (i) Inductive effect (ii) Electromeric effect (iii) Resonance and mesomeric effect (iv) Hyperconjugation Inductive Effect (Polar Nature Of Covalent Bonds) The displacement of an electron (shared) pair along the carbon chain due to the presence of an electron withdrawing or electron releasing groups in the carbon chain is known as inductive effect (I – effect).  It is a permanent effect which is transmitted along the chain. C……> ……C…>….C……>G (G – Functional group)  This permanent polarity is due to electron displacement due to difference in electronegativities.  This effect weakens steadily with increasing distance from the substitution (electron – withdrawing or electron – donating group) and actually diminishes down after three carbon atoms. There are two types of inductive effect i.e. – I effect and +I effect.

Negative Inductive effect ( I Effect) If the substituent attached to the end of the carbon chain is electron withdrawing (X). The effect is called – I effect.

d

C

d

X I effect decreases as one goes away from groups (electron attacking)

d

d

d

d

C C C X 2 3 1 C1(d+) > C2(dd+) > C3 (ddd+) and other third carbon charge is negligible. I effect is in order. NO2 > F > COOH > Cl > Br > I > OH > C6H5 Due to I effect (electron – with drawing nature) electron density decreases, hence basic nature is decreased and acidic nature is increased. Chloroacetic acid is stronger than acetic acid since Cl shows (-I) effect, electron – density is decreased and O – H bond is weakens causing ionisation of (-COOH) to a greater extent than CH3COOH. O

O Cl

C

CH2

O

CH3

H

C

OH

(-I)

NH3 is a base due to lone pair on nitrogen. Phenyl group is electron – withdrawing. What happens to electron – density of nitrogen in aniline, Naturally electron – density is decreased. Hence aniline is weaker base than NH3 . H

H

N

N

H

H

H

Similarly acidic nature of phenol is greater than H2 O due to electron – withdrawing nature of phenyl group. O

H

O

H

H

Positive inductive effect (+I effect): If the substituent attached to the end of the carbon chain is electron – donating, the effect is called +I effect. This is due to electron – releasing (Y). It develops a negative charge on the chain. d

C

d+

Y

+I effect also decreases as we go away from group Y (electron – releasing)

C 4

ddd

dd

3

2

C

C

d

C 1

d+

Y

C1(d ) > C2 (dd ) > C3 ( ddd) So +I effect is in the order of (CH3 )3 C > (CH3 )2 CH > CH3 CH2  > CH3 

Due to electron – releasing group electron density is increased, hence basic nature is also increased and naturally acidic nature is decreased, thus -I effect +I effect � � Acidic Nature � � Basic Nature Note: Inductive effect is a permanent effect operating in the ground state of the organic molecules and hence is responsible for high melting point, boiling point and dipole moment of polar compounds. Illustration 7. Take the following isomerica,b, g chlorobutyric acid γ

α CH3CH2 C COOH ¯ | Cl (i)

β

CH3 C HCH2COOH ¯ | Cl (II)

C H2CH2CH2COOH ¯ | Cl (III)

Arrange the acids increasing order of acid strength. Solution:

As we have stated, as we go away from the source, electron - withdrawing tendency decreases, hence acidic nature also decrease. Thus (III) < (II) < (I)

Illustration 8. Which of the following carbonyl compound is more acidic? (a) Solution.

O

O

CH3 – || – CH3 (b) CH3 – || – H C C

(b) The acidity of a-hydrogen depends on +ve charge on the carbon atom to which it is attached. In an aldehyde the carbonyl carbon has more positive charge and hence more –I effect.

Illustration 9. CH3CO2H is a stronger acid than CH3CH2 – OH. Explain

O

CH3 – CO2H  CH3 – || – O– + H+ C CH3 – CH2 – OH CH3 – CH2 – O– + H+ In EtO– the negative charge on O atom is increased by +I effect, whereas in CH3CO2–, the negative charge on O atom is decreased by delocalisation. Lesser the charge density on negatively charged atom, weaker is the base. Electromeric Effect Solution:

In presence of an attacking reagent, there is complete transfer of   electrons from one atom to other to produce temporary polarity on atoms joined by multiple bonds, it is called Electromeric effect.

C

added ��electrophile � � �� �� �� �� �� �� �� �� � � electrophileremoved

C

C

C

This effect is temporary and takes place only in the presence of a reagent. As soon as the reagent is removed, the molecule reverts back to its original position. Electromeric effect is of two types, i.e. +E effect and  E effect. Positive Electromeric effect (+E effect): When  electrons transfer takes place C to C (as in alkenes, alkynes etc.), it is called positive electromeric effect (denoted by + E). +



added ��electrophile � � CH2 = CH2 �� �� �� �� �� �� �� �� � �CH2  CH2 electrophile removed +



added ��electrophile � � � CH3  CH = CH2 �� �� �� �� � �� �� �� � �CH3  CH  CH2 electrophile removed

For example, addition of acids to alkenes C

C

+ H+ 

C

C

H (+ E - e ffe c t)

In this, there is also (+I) effect of CH3 group which causes  electron transfer from C2 to C1. What do you think in the following case:

CH = CH

CH3

CH2CH3

 (+I) effect of CH3 CH2  is larger than that of CH3 , hence  electron transfer is from C3 to C2 .

CH3 1

CH 2

CH 3

CH2CH3 4 5

Negative Electromeric effect: When   electrons transfer takes places to more electronegative atom (O, N, S) joined by multiple bonds, it is called Negative Electromeric effect (denoted by E). electrophile added

+  ��� O �� �� �� �� �� �� �� �� �� �C  O electrophile removed

C

electrophile added ��� N �� �� �� �� �� �� �� �� �� � C electrophile removed

C

N

for example, the addition of cyanide ion to the carbonyl group. C

O

CN

C

O

CN (-E effect)

Resonance & Mesomeric Effect There are many organic molecules which can not be represented by a single lewis structure. In turn, they are assigned more than one structure called canonical forms or contributing of

resonating structures. The phenomenon exhibited by such compounds is called resonance. For example, 1, 3 – butadiene has following resonance structure.

H2C

CH

CH

CH2

H2C

CH

CH

H2C

CH2

CH

CH

CH2

and canonical forms of vinyl chloride are

H 2C

CH

Cl

H 2C

CH

Cl

While drawing these canonical forms, the prime thing that has to be kept in mind is that the relative position of any of the atom should not change while we are allowed to change the relative positions of  - bonded electron pair or distribution of charge to other atoms. Also remember that it is not the case that some molecules have one canonical form and some have another form. All the molecules of the substance have the same structure. That structure is always the same all the time and is a weighted average of all the canonical forms. In real sense, these canonical forms have no expect in our imaginations. Now we are in a position to discuss about the conditions necessary for a compound to show resonance. The two essential conditions are (a) There must be conjugation in the molecule. Conjugation is defined as the presence of alternate double and single bonds in the compound like C

C

C

C

(b) The part of the molecules having conjugation must be essentially planar or nearly planar. The first condition of conjugation is not only confined to the one mentioned above but some other systems are also categorized under conjugation. These are C C C (i) (ii) (iii) C C C (iv)

C

C

C

C

C

C

(v)

O C

C

N O

So, any molecules satisfying both the conditions will show resonance. For example, we consider phenol. The structure of phenol is

O

H

By looking at the structure, it must be clear to you that the compound possesses conjugation of the type C C C C As well as the category (iv) because the lone pairs on oxygen are in conjugation with unsaturated (sp 2 hybridised) carbon of the ring. Since, oxygen atom is sp3 hybridized in phenol. The lone pairs on oxygen are nearly planar with respect to the P Z orbital of carbon linked to oxygen. Thus, both the conditions are fulfilled by phenol, therefore it does show resonance and its resonance structures are represented as

O

H

O

H

O

H

O

H

O

H

This has to be borne in mind that resonance always results in different distribution of electron density than would be the case if there were no resonance. It is a permanent effect, also referred as mesomeric effect. Note: The acidity of phenol can be explained by resonance O

H

O

H Phenol

Penoxide ion

O

O

O

O

O

O

d

d

The above structure shows that the phenoxide ion formed is more resonance stabilised than phenol. Hence, the acidity of phenol is explained. Similarly basicity of aniline can be explained. NH2

NH2

NH2

NH2

NH2

The above structure shows that the lone pair present on N – atom undergoes into resonance and is not available for donation. Hence, the basicity of aniline decreases and is less than aliphatic ��

amine R  NH . 2

Resonance (mesomeric) effect is of two types. (i) If the atom or group of atoms is giving electrons through resonance, it is called +R or +M effect. For example, C

C

NH2

(+M effect of NH2 group)

Other groups that shows +M effect are NHR, NR2, OH, OR, NHCOR, Cl, Br, I etc. (ii) If the atom or group of atoms is withdrawing electrons through resonance, it is called R or M effect. For example,

O C

C

N

(-M effect of NO2 group)

O Other groups showing M effect are CN, CHO, COR, CO2H, CO2R, CONH2, SO3H, COCl etc. Now, let us consider resonance in nitrobenzene and its various canonical structures are O

O

O

O

N

N

O

O

O

O

N

N

O

O N

The NO2 group in nitrobenzene has M effect. In general, if any atom (of the group) attached to the carbon of benzene ring bears atleast one lone pair, then the group shows +M effect while if the atom (of the group) linked to the benzene carbon bears either a partial or full positive charge, then the group exhibits M effect. In drawing the canonical forms and deciding about their relative stabilities, following rules are give for your guidance. (i) All the canonical forms must be bonafide lewis structures for example, none of them may have a carbon with five bonds. (ii) All atoms taking part in the resonance must lie in a plane or nearly so. The reason for planarity is to have maximum overlap of the p – orbitals. (iii) All canonical forms must have the same number of unpaired electrons. Thus CH2  CH = CH  CH2 is not a valid canonical form for 1, 3  butadien. (iv) The energy of the hybrid (actual) molecule is lower than that of any canonical form. Obviously then, delocalization is a stabilizing phenomenon. The difference in energy between the hybrid and the most stable canonical structure is called resonance energy. (v) All canonical forms do not contribute equally to the actual molecule. Each form contributes in proportion to its stability, the most stable form contributing the most. (vi) Structures with more covalent bonds are generally more stable than those with fewer covalent bonds. (vii) Structure with formal charges is less stable than uncharged structures. For charged structure, the stability is decreased by an increase in charge separation and the structure with two like charges on adjacent atoms are highly unfavourable. (viii) Structures that carry a negative charge on a more electronegative atom are more stable than those in which the charge is on a less electronegative atom. For example,

CH2

C O (I)

CH3

CH2

C O (II)]

CH3

Structure (II) is more stable than (I). Similarly positive charges are best occupied on atoms of low electronegativity (ix) Those structures in which octet of every atom (expect for hydrogen which have douplet) is complete are more stable than the others with non complete octets. For example,

R

C

O

R

(III)

C

O

(IV)

Structure (IV) is more stable than (III). Resonance Energy: The difference in energy between the hybrid and the most stable canonical structure is called as resonance energy

Illustration10.

Why guanidine is basic in nature. Explain the site of protonation and provide resonating structures.

Solution:

NH

H+

+ NH2

 C NH2 H2N

C H2N

NH2

NH2 NH2

+ H2N

C

NH2

C H2N

+ NH2

Guanidine behaves as strong base because it can provide electron pair easily resulting in three identical resonating structure. Site of protonation is sp 2 hybridised N atom rather than sp3 hybridise N atom resulting in three symmetrical structures. Illustration 11.

Arrange the various resonating structures of formic acid in order of decreasing stability

Solution:

O H

C

O

O O

H > H

C

O

+

H > H

C

O O

H> H

C

O

H

Illustration 12. Explain why phenol is acid while aliphatic alcohols are not. Solution:

After loss of H+ ion from OH group of phenol the remaining part (Phenoxide ion) stabilizes by resonance, hence it will favour the loss of H + ion and hence acidic in nature.

 

 

 

OH

+

OH 

3 more R-structure –H  

  -

O

 

+

(Charge separation is present. Not much stable)  

O

 

 

O

 

 

 

O

 

 

O

 







(No charge separation, more stabilization by Resonance) Illustration 13. Why benzyl carbonium ion is more stable than ethyl carbonium ion Solution:

Due to resonance Benzyl carbonium ion is more stable than Ethyl carbonium ion.

Illustration 14. Among orthochlorophenol and orthofluorophenol, which will be a stronger acid and why? Solution:

The one having a weaker conjugate base will be a stronger acid. If the conjugate base has to be weak, the negative charge has to be delocalised to a larger extent. In o-chlorophenol

Due to the availability of vacant orbitals in chlorine, the negative charge is delocalised to a larger extent. The same cannot take place in case of F as F does not have vacant orbitals. So, o-chlorophenol, having a weaker conjugate base, becomes a stronger acid. Exercise 13. Benzylamine is a stronger base than aniline. Explain. Exercise 14. Why is always the resonance effect dominating over the inductive effect? Exercise 15. Unlike other aromatic amines, why is the following amine strongly basic?

Exercise 16. Arrange the following alcohols in increasing order of acidity. (a) CH2 = CH – OH (b) CH3 – CH2 – OH

Exercise 17. H3C (a)

CH2 OH

Write resonance structure of the given compound. (b) Whether this pairs of compound are tautomers

Hyperconjugation It is delocalisation of sigma electrons. Also known as sigma-pi – conjugation or no bond resonance Hyperconjugation is a permanent effect

  

Occurrence Alkene, alkynes Free radicals (saturated type) carbonium ions (saturated type) Condition Presence of a–H with respect to double bond, triple bond carbon containing positive charge (in carbonium ion) or unpaired electron (in free radicals) Example H

H

C

H

H CH

CH2

H

H

C

CH

CH2

H

H

C H

H

CH

CH 2

H

C

CH

CH2

H

Note: Number of hyperconjugative structures = number of a-Hydrogen. Hence, in above examples structures i,ii,iii,iv are hyperconjugate structures (4-structures). Effects of hyperconjugation

Bond Length: Like resonance, hyperconjugation also affects bond lengths because during the process the single bond in a compound acquires some double bond character and vice-versa. E.g. C —C bond length in propene is 1.488 Å as compared to 1.334Å in ethylene. Dipole moment: Since hyperconjugation causes these development of charges, it also affects the dipole moment of the molecule. Illustration 15. Arrange the following species in increasing order of dipole moment. Cis  2,3  Dichloro  2  butene Cis  1,2  dichloroethene , , (ii) (i) Cis  1,2  dibromo trans  1,2  dichlorethene -1, 2-dichloroethene, (iii) (iv)

Solution:

H3C

H

Cl

Cl

C C H3C

I

Br

Cl

C C Cl

H

Resultant vector

Cl

II

Br

Resultant vector

Cl

H

C

C

C

C

III Cl

Resultant vector

H

IV Cl

Nonet resultant vector

In I, there is addition of vector In II, there will neither be addition nor subtraction of vector In III, there is subtraction of vector In IV, the vectors almost cancel each other. So the increasing order of dipole moment is IV < III < II < I. Stability of carbonium Ions: The order of stability of carbonium ions is as follows Tertiary > Secondary > Primary Above order of stability can be explained by hyperconjugation. In general greater the number of hydrogen atoms attached to a-carbon atoms, the more hyperconjugative forms can be written and thus greater will be the stability of carbonium ions. CH3 Tertiary carbonium ion

CH2H+

H3C—C+

H3C—C CH3

CH3

(9 equivalent forms) CH3 Secondary carbonium Ion

CH2H+

H3C—C+

H3C—C

H

H

(6 equivalent form) CH3 Primary carbonium Ion

CH2H+

H—C+

H—C H

H

(3 equivalent form)

Stability of Free radicals: Stability of Free radicals can also be explained as that of carbonium ion �

�

�

�

(CH3 )3 C > (CH3 )2 CH > CH3 CH2 > CH3 Directive influence of methyl group: The o,p-directing influence of the methyl group in methyl benzene is attributed partly to inductive and partly to hyperconjugation effect. CH3

(orientation influence of the methyl group due to +I effect )

(Orientation influence of methyl group due to hyperconjugation) The role of hyperconjugation in o,p,-directing influence of methyl group is evidenced by the part that nitration of p-iso propyl toluene and p-tert-butyl toluene from the product in which —NO 2 group is introduced in the ortho position with respect to methyl group and not to isopropyl or tbutyl group although the latter groups are more electron donating than Methyl groups H

H

H—C—H

H—C—H

H—C—CH3 CH3

H3C — C—CH3 CH3

i.e., The substitution takes place contrary to inductive effect. Actually this constitutes an example where hyperconjugation overpowers inductive effect. Illustration 16. Which among the following is most acidic? Cl (a) H3C CH2 CH2 CH COOH Cl (b) H3C CH2 CH CH2COOH Cl (c) H3C CH CH2 CH2 COOH Cl CH2 CH2 CH2 CH2 COOH (d) Solution:

As the inductive effect of chlorine decreases with distance (a) will be most acidic because the carboxylate ion which results after the loss of H + can be stabilized by electron withdrawing nature of chlorine (I effect) which is strongest in (a) where chlorine atom is at a - carbon.

Illustration 17. Which among the following is most basic? (a) CH3CH2NH2 (b) CH3CH = NH (c) CH3  C  N Solution:

(a) is most basic because here nitrogen is sp 3 hybridized i.e. p – character is greater than s – character (75% p and 25% s – character), orbital holding lone

pair is more elongated than spherical, the hold of nitrogen nucleus over these electrons is less resulting in more basicity. Illustration 18. Compare the acidic strength of the following CH3COOH ClCH2COOH Cl2HCCOOH CCl3COOH Solution:

CH3COOH <

Cl

H

O

C

C

O

H

< Cl

H

H

O

C

C

O

H

O

H

Cl

< Cl

Cl

O

C

C

Cl Increasing acidic strength due to increasing number of (I) effect group. Illustration 19. Explain why aniline is less basic than ammonia. Solution:

The lone pair present at the nitrogen in aniline is delocalized in the ring (by resonance) and hence, it is not free for protonation, while in ammonia. I t is present at nitrogen all the time, hence it is readily available for ptrotonation.

Illustration 20.

Br

N(CH3)2 —CH3

Br2 AlBr3

Br

CH3

Br –N(CH3)2 and –CH3 both groups are o, p-directing but products are formed only under direction of –(CH3)2. Explain. Solution:

–N(CH3)2 group as well as –CH3 group are o–, p– directing and if we want to place both groups, none of the position will be available for S E.

N(CH3)2 —CH3 o–w.r.t

CH3

m-w.r.t. N(CH3)2 and so on. But o, p-directive effect of the stronger donor (–N(CH 3)2) dominates over that of the weaker donor (–CH 3), hence SE takes place at o-and p-position w.r.t. –N (CH3)2.

N(CH3)2

N(CH3)2

—CH3 O

Br2

—CH3

Br

Br P Illustration 21. 4- nitrophenol is more acidic than 3,5 - dimethyl –4-nitro-phenol. Explain. Solution:

It is explained in terms of inductive effect and hyperconjugation.

Illustration 22: The order of acidic strength of the following compound is OH

OH

C2H5OH (i)

Solution:

O

H3C

(ii)

NO2 (iii)

OH (iv)

(iv) > (iii) > (i) > (ii)

Illustration 23. Write the following Alkenes in increasing order of their stability with explanation R2C=CR2, R2C=CHR, R2C=CH2, RCH=CH2, CH2=CH2 Solution

Let R= CH3

(no. a-H, no. Hyperconjugation)

(a) CH2 = CH2 H3C

H C=C

(b) H

(3a-H, 3-Hyperconjugative structures) H

H3C

H C=C

(c) CH3

(6 a-H, 6H–structures) H

CH3

H3C C=C

(d) CH3

H3C

H

CH3 C=C

(e)

(9a–H, 9H–structures)

(12a–H, 12 H structures) CH3 From (a) to (e) number of hyperconjugative structures increases, hence stability of alkene also increases, hence the correct order is a< b
Exercise 18. (i) Which one is more basic?

O

O H2NH2C

C

CH3

or H2NH2C

C

OCH3

(a) (b) (ii) Why phenol is more acidic than alcohol? (iii) Why aniline is less basic than aliphatic amines? (iv) Which among the following is most acidic? (a) CH3 CH3 (b) CH2 = CH2 (c) HC  CH Illustration 24. Identify the effects operating in each of the molecule. (a) CH3Cl (b) CH3 – CH = CH2

O

(c) CH3 – CH = CH – || – H C (d) CH2 = CH – CH = CH2 Solution:

(a) (b) (c) (d)

Inductive effect Inductive and hyperconjugative effects Inductive, hyperconjugative and mesomeric effects Mesomeric effect and inductive effect

MECHANISM OF ORGANIC REACTION A chemical equation is only a symbolic representation of chemical reaction which indicates the initial reactants and final products involved in a chemical change. Reactants generally consist of two species. (a) Substrate: One which is being attacked in a chemical reaction (b) Reagents: The species which attack the substrate molecule Substrate + Reagent  Products

Br CH3—CH—CH3 + OH–  CH3—CH=CH2 + H2O + Br– substrate

Reagent

Products

It is important to know not only what happens in a chemical reaction but also how it happens. Most of the reactions are complex and take place via reactive intermediates which may be or may not be isolated. The reaction intermediates are generally very reactive which readily react with other species present in the environment to form the products. The detailed step by step description of chemical reaction is called its mechanism. Mechanism is only a hypothesis to explain various facts regarding a chemical reaction. Substrate  Reactive intermediates –— Products O H

B r — C H — C H 3 – B r– S u b s tra te

  OH — C H — C H 3     In te rm e d ia te

— C H — C H 3 P ro d u c t

By knowing the mechanism we can predict the product of a chemical reaction, adjust the experimental conditions to improve the yield of the products or even alter the course of reaction to get the different products.

Most of the attacking reagents carry either positive charge (an electron deficient species) or a negative charge (electron rich species). The positively charged reagents attack the substrate at points of high electron density while (-vely) charged reagents attack the point of low electron density. The organic reactions essentially involve changes in the existing covalent bonds present in the molecules. These changes may involve electronic displacements in covalent bonds breaking of some of the existing bonds (bond fission), formation of new bonds as well as energy change accompanying the bond fission and bond cleavage. We can understand the mechanism of various organic reactions in terms of following well established basic concepts. (i) Electronic displacement in covalent bond (ii) Fission (cleavage) of covalent bonds (iii) Nature of attacking reagents Types of Organic Reactions: All organic reactions can be broadly classified into four catagories. (a) Substitution reactions (b) Addition reactions (c) Elimination reactions and (d) Rearrangement reactions (a) Substitution Reactions In these an atom or a group of atoms in an organic molecule is replaced by another atom or group of atoms without any change in the remaining part of the molecule. These reactions may be initiated by free radical, electrophile or nucleophile. (i) Free radical substitution reaction: This substitution reaction is brought about by free radicals. For example chlorination of methane in presence of diffused sunlight. The mechanism of the reaction is as follows. Cl : Cl  2Cl Chain initiation � CH4 + Cl� � CH� 3 + HCl � � CH� � 3 + Cl2 � CH3 Cl + Cl � � � CH3 Cl + Cl� � CH2Cl + HCl �Chain propagation                � �                � � � Cl�+ Cl� � Cl2

� � �Chain termination � � CH3 + CH3 � CH3 CH3 � � (ii) Nucleophilic substitution reoactions: These reactions are brought about by nucleophile. The reaction can proceed either via S N1 or SN2 mechanism. SN1 mechanism: Rate determining step involves only the species. For example the reaction.

CH3

CH3

CH3—C—Br + OH–  CH3—C—OH + Br– CH3

CH3

takes place as follows Rate (CH3 )C + + Br  ���� � 1st step: (CH3)3CBr determining step Carbocation 2nd step: Attack by nucleophile CH3

(CH3)3C+ + OH–  CH3—C—OH CH3 The stability of carbocation is the controlling factor for this mechanism the formation of 3° carbocation as an intermediate proceeds via this mechanism. In an optically active compound substitution at chiral centre through S N1 mechanism produces recemic mixture (No 100% recemization is observed?). SN2 mechanism: Rate determining step involves two species and reaction proceeds through transition state. CH3 Cl + OH � CH3 OH + Cl H

H

H

H

Rate determinin g step

C———Cl  HOd---------C-------Cld– HO—C----H+ Cl–

–

HO

H

H

H

H

Transition state

Since 1° carbocation is less stable than the transition state formed above, the reaction involving 1° alkyl halides proceed via S N2 mechanism. During reaction configuration of carbon is inverted which is referred to as Walden inversion.

Points to Remember  The higher the polarity of solvent greater the tendency for S N1 reaction.  High concentration of the nucleophile favours SN2 reaction while low concentration favours SN1 reaction.  Rearrangement of the carbocation (formed in S N1 reaction) leading to more stable carbocation is observed in SN1 reaction (discussed latter).  In general SN2 mechanism is strongly inhibited by increasing steric bulk of the reagents. In such case SN1 mechanism is favoured. (iii) Electrophilic substitution reactions: The reaction initiated by an electrophile is known as electrophilic substitution reaction. Aromatic substitution reactions are the examples of this type of reaction. FeCl3 � C6H6 + Cl2 ���� C6H5Cl + HCl or AlCl3

The mechanism of this reaction as follows. + Formation of electrophile: Cl : Cl + AlCl3  Cl+ + AlCl4 Electrophile attack:

Elimination of proton: +

H C l

C l +H C l +A lC l3

 AlCl

4    

(b) Addition Reactions Reactions which involve combination between two molecules to give a single molecule of the product are called addition reactions. Such reactions are typical of compounds containing multiple (double or tripe) bonds. Depending upon the nature of the attacking species (electrophiles, nucleophiles or free radicals) addition reactions are of the following types. (i) Electrophilic addition reaction: These reactions are brought about by electrophiles and are typical reactions of alkenes and alkynes. + + slow C H — C H = C H H H — C H — C H    C 3 2+ 3 3

2 °c a rb o c a tio n Br Fast CH3—+CH—CH3 + Br–   CH3—CH—CH3 2-Bromopropane

(ii) Nucleophilic addition reactions: These reactions are brought about by nucleophiles. The characteristics reaction of aldehyde and ketone are nucleophilic addition reaction i.e., base catalysed addition of HCN to aldehydes or ketones. HO– + HCN  H2O + CN– R

R

R

– – – HCN    — C =O +C N R C — O — C — O H    R

 R K e to n e

 CN 

N u c le o p h ile

C N

C N K e to n ec y n o h y d rin e

(iii) Free radical addition reactions: Addition reactions brought about by free radicals are called free radical addition reactions for example addition of HBr to alkenes in presence of peroxides. CH3  CH2  CH2Br Peroxide� CH3—CH=CH2+HBr ����� � npropyl hydride The reaction proceeds through following mechanism.  2RO° R  O  O  R ��� RO + HBr  ROH + Br slow � CH3—CH —CH2—Br CH3—CH=CH2 + Br ���� HBr CH3—CH—CH2—Br ���� CH3—CH2CH2—Br + Br fast

Br + Br  Br2 Illustration 25. Identify the 1, 2- and 1-4 addition products of Free Radical addition of CBrCl 3

to

1, 3-butadiene.

Solution:

 CCl C H 2= C H – C H = C H 2   3  C l 3 C – C H 2 – C H – C H = C H 2

(I)

Br C l3 C – C H 2 – C H – C H = C H 2 Br 1,2 ad d itio n p ro d uc t

C l3 C – C H 2 – C H = C H – C H 2 Br C l3 C – C H 2 – C H = C H – C H 2 B r 1 ,4 a d ditio n p ro d uc t

Illustration 26. Addition of HCl on 1, 3 butadiene gives two products. Explain. Solution:

Mechanism of addition of HCl on 1, 3-butadiene is as follows

Step:1

+ H+ + CH2 = CH—CH=CH2CH3—+CH—CH=CH2CH3—CH=CH–CH2 An allylic cation equivalent to CH3—CH— CH—CH2 d+ d+ (Resonance hybrid)

(a)

Step:2 CH3—CH— CH—CH2 d+ d+

CH3CH—CH=CH2 1,2 addition

.. + :Cl: .. —

Cl (78%) (b)

CH3CH=CHCH2Cl 1,4 addition (22%)

In step 1 proton adds to one of the terminal cabons of 1, 3 butadiene to form, as usual, the more stable carbonium ion, in this case a resonance stabilized allylic cation. Addition of one of the inner carbon atoms would have produced a much less stable primary cation that could not be stabilized by resonance; +

CH2=CH—CH=CH2  CH2—CH2—CH=CH2 + H In step (2), a chloride ion forms a bond to one of the carbon atoms of the allylic cation that bears a partial positive charge. Reaction at one carbon atom results in the 1, 2 – addition product, reaction at the other gives that 1,4 addition product. Exercise 19. Arrange the following in increasing order of reactivity towards H + addition. (a)

(c )

(b)

(d )

(c) Elimination Reactions An elimination reaction is one which involves the loss of two atoms or groups of atoms from the same or adjacent atoms of a substrate molecule leading to formation of multiple (double or triple) bond. These are of two types.

(i) b - elimination reactions. In these reactions, loss of two atoms or groups occurs from the adjacent atoms of the substrate molecule e.g., acid catalysed dehydration of alcohol and base catalysed dehydrogenation of alkyl halides. H

conc.H SO

H3C

2 4 

H3C

OH

H

CH2 +H O 2

E1 mechanism

H

H

slow — C— C—     — C — C +— + X –

X B:

H fast  — C — C +—  

+ H :B

C = C

E2 Mechanism

: B

H

B H

slow — C — — C — — — C — — C — —     

X

– B C=C +:X +H

X

Transitionstate E1 — CB mechanism

(ii) a - Elimination – In these reactions loss of two atoms or groups occurs from the same atom of the substrate molecule. E.g., base catalysed dehydrohalogenation of chloroform to form dichlorocarbene.

Cl Cl

C

H

Cl Cl

+ OH–-  Cl –H2O Cl

Cl C–  C: + Cl– Cl

Dichlorocarbene is the reactive intermediate involved in carbylamine reaction and Reimer – Tieman reaction. (d) Rearrangement Reaction These reactions involve the migration of an atom or a group of atoms from one atom to another within the same molecule.

Some reactions involving rearrangement. Br AlCl3 � | CH3  CH2  CH2  CH2  Br ���� 575K CH3  CH2  C H  CH3

CH3

CH3

C H OH CH —C—CH —CH CH3—C—CH2—Br    3 2 3 2

5

OCH2CH3

CH3 CH3Br

CH3

Alc .KOH CH3—C—CH2   CH3—C=C—CH3

CH3 CH3

CH3 CH3 HCl

CH3—C—CH=CH2  CH3—C—CH2—CH3 H

Cl

O ||

Br / KOH R  C NH2   R—NH2 + K2CO3 1° amine 1° amide 2

1, 2 Hydride shift C H 3

C H 3

C H 3

C H 3

   Br Br   C H — C — C H — C H C H — C — C H — C H C H — C — C H — C H C H — C — C H — C H          3

3

3

r HB

 H

3

3



2

3

3

3

B r

3 °c a r b o c a t io n

M a jo rp r o d u c t

1, 2 Methyl shift

Illustration 27. Can you think why the below given is order of bases reaction with conc. HCl (in presence of anhydrous ZnCl2, mixture is called Lucas’s reagent). CH3

CH3

O

H < CH3CH2

OH < CH3

CH

OH < CH3

CH3 Solution:

Naturally the one which is the strongest base, will react faster ZnCl2 ROH+ HCl ��� � RCl + H2O base

Conc. acid

white turbidity

C CH3

OH

It is simply a reaction between acid and a base and the strongest base will be reacting fastest. This is called Luca’s test of making distiction between 1 0, 20 and 30 alcohol. Illustration 28. Give the mechanism (i)

Me

Me

OH

Me H Me

CH3

(ii) CH3

Solution:

(i)

Me

Br CH2

H

Me

CH3

HBr

Me

Me

Me

O

Me H

-CH 3

H

shift

Me

Me

CH3

(ii)

CH3

CH3

CH3

H CH3 CH2

Br H3C

Br CH3

Rearrangement of carbocation through ring expansion. Exercise 20. (i) Which of the following intermediate is unstable?

(a)

(b)

(ii) Arrange the following carbocation in the order of decreasing stability

(a)

OH

CH3

CH3

H3C

CH3 (I)

CH3

CH3 H3C OH

H3C

(II)

(b)

+

(III)

OH

+

Why CH3 - O - CH2 is more stable than CH3CH2 CH2 while both are primary carbocations?

Exercise 21. Explain the following observations. – OH (i) CH3 – I ��� � CH3OH + I 

– OH (ii) CF3 – I ��� � CF3H + IO ISOMERISM

In the study of organic chemistry we come across many cases when two or more compounds are made of equal number of like atoms. A molecular formula does not tell the nature of organic compound; sometimes several organic compounds may have same molecular formula. These compounds possess the same molecular formula but differ from each other in physical or chemical properties, are called isomers and the phenomenon is termed isomerism (Greek, isos = equal; meros = parts). Since isomers have the same molecular formula, the difference in their properties must be due to different modes of the combination or arrangement of atoms within the molecule. Broadly speaking, isomerism is of two types. (i) Structural Isomerism (ii) Stereoisomerism (i) Structural isomerism: When the isomerism is simply due to difference in the arrangement of atoms within the molecule without any reference to space, the phenomenon is termed structural isomerism. In other words, while they have same molecular formulas they possess different structural formulas. This type of isomerism which arises from difference in the structure of molecules, includes: (a) Chain or Nuclear Isomerism; (b) Positional Isomerism (c) Functional Isomerism (d) Metamerism and (e) Tautomerism (ii) Stereoisomerism: When isomerism is caused by the different arrangements of atoms or groups in space, the phenomenon is called Stereoisomerism (Greek, Stereos = occupying space). The stereoisomers have the same structural formulas but differ in the spatial arrangement of atoms or groups in the molecule. In other words, stereoisomerism is exhibited by such compounds which have identical molecular structure but different configurations. Stereoisomerism is of two types: (a) Geometrical or cis-trans isomerism; and (b) Optical Isomerism. Thus various types of isomerism could be summarized as follows.

STRUCTURAL ISOMERISM Chain or Nuclear Isomerism This type of isomerism arises from the difference in the structure of carbon chain which forms the nucleus of the molecule. It is, therefore, named as chain, nuclear isomerism or Skeletal isomerism. For example, there are known two butanes which have the same molecular formula (C4H10) but differ in the structure of the carbon chains in their molecules. CH3 CH3

H3C

H3C

n - butane

CH3 isobutane

While n-butane has a continuous chain of four carbon atoms, isobutane has a branched chain. These chain isomers have somewhat different physical and chemical properties, n-butane boiling at -0.5o and isobutane at -10.2o. This kind of isomerism is also shown by other classes of compounds. Thus n-butyl alcohol and isobutyl alcohol having the same molecular formula C4H9OH are chain isomers. CH3 H3C

OH

H3C

n - butyl alcohol

OH isobutyl alcohol

It may be understood clearly that the molecules of chain isomers differ only in respect of the linking of the carbon atoms in the alkanes or in the alkyl radicals present in other compounds. Illustration 29.

How many chain isomers does butane have?

CH3

Solution: H3C

CH2

H3C

n - Butylene (1-Butene)

CH2

Isobutylene (2-Methylpropene)

Illustration 30. How many chain isomers does propyl benzene have? Solution:

CH2CH2CH3

n - Propylbenzene

Illustration 31.

H3C

CH3

Isopropylbenzene

Give the possible chain isomers for ethyl benzene.

Solution:

CH2CH3

CH3 CH3

Ethylbenzene

Xylene (o, m, p)

Positional Isomerism It is the type of isomerism in which the compounds possessing same molecular formula differ in their properties due to the difference in their properties due to difference in the position of either the functional group or the multiple bond or the branched chain attached to the main carbon chain. For example, n-propyl alcohol and isopropyl alcohol are the positional isomers. OH | CH3–CH2–CH2–OH CH3–CH–CH3 n-propyl alcohol isopropyl alcohol Butene also has two positional isomers: CH2=CH–CH2–CH3 CH3–CH=CH–CH3 1-butene 2-butene 1-Chlorobutane and 3-Chlorobutane are also the positional isomers: CH3  CH2  CH2  CH2Cl CH3 CH2  CHCl  CH3 1Chlorobutane 2Chlorobutane Methylpentane also has two positional isomers:

CH3—CH2—CH—CH2—CH3

CH3—CH2—CH2—CH—CH3

CH3

CH3

3-Methylpentane

2-Methylpentane

In the aromatic series, the disubstitution products of benzene also exhibit positional isomerism due to different relative positions occupied by the two substituents on the benzene ring. Thus xylene, C6H4(CH3)2, exists in the following three forms which are positional isomers. CH3

CH3

CH3 CH3

o - Xylene

CH3 m - Xylene

CH3 p - Xylene

Illustration 32. C2H4Cl2 can have two position isomers 1, 2- and 1, 1-dichloro ethane. Each is subjected to following sequence of reactions: H3O+ KCN� ���� C2 H4Cl 2 ���� � ��� What is the end product in each case?

Solution:

CH2Cl

Cl CH3—CH

CH2Cl KCN

CN

CH2CN

CH3CH CN

CH2CN

H3O+

H3O+

COOH CH3—CH

CH2COOH CH2COOH





CH2CO

Cl

KCN

COOH

CH3CH2COOH propanoic acid

O

CH2CO succinic anhydride (if two –COOH groups are at adjacent C-atoms heating would eliminate H2O)

(if two—COOH groups are at same C-atom heating would eliminate CO2)

Exercise 22. Nitrophenol, C6H4(OH) (NO2) can have: (A) No isomer (only a single compound is possible) (B) Two isomers (C) Three isomers (D) Four isomers Functional Isomerism When any two compounds have the same molecular formula but possess different functional groups, they are called functional isomers and the phenomenon is termed functional isomerism. In other words substances with the same molecular formula but belonging to different classes of compounds exhibit functional isomerism. Thus, (1) Diethyl ether and butyl alcohol both have the molecular formula C 4H6O, but contain different functional groups. C2H5–O–C2H5 C4H9–OH diethyl ether butyl alcohol The functional group in diethyl ether is (–O–), while is butyl alcohol it is (–OH). (2) Acetone and propionaldehyde both with the molecular formula C 3H6O are functional isomers. CH3–CO–CH3 CH3–CH2–CHO acetone Propionaldehyde In acetone the functional group is (–CO–), while in propionaldehyde it is (–CHO). (3) Cyanides are isomeric with isocyanides: RCN RNC Alkyl cyanide Alkyl isocyanide (4) Carboxylic acids are isomeric with esters. CH3CH2COOH CH3 COOCH3 Propanoic acid Methyl ethanoate (5) Nitroalkanes are isomeric with alkyl nitrites:

O

R—N

R—O—N=O Alkyl nitrite

O

Nitroalkane

(6) Sometimes a double bond containing compound may be isomeric with a triple bond containing compound. This also is called as functional isomerism. Thus, butyne is isomeric with butadiene (molecular formula C4H6). CH3  CH2C �CH CH2 = CH  CH = CH2 1,3Butadiene 1Butyne (7) Unsaturated alcohols are isomeric with aldehydes. Thus, CH2 = CH  OH CH3 CHO Vinyl alcohol Acetaldehyde (8) Unsaturated alcohols containing three or more carbon atoms are isomeric to aldehydes as well as ketones: CH2 = CH  CH2 OH CH3 CH2 CHO CH3 COCH3 Allyl alcohol Propionaldehyde Acetone (9) Aromatic alcohols may be isomeric with phenols: CH2 OH

CH 3 OH

Benzyl alcohol

o - Cresol

(10)Primary, secondary and tertiary amines of same molecular formula are also the functional isomers. CH3 CH2CH2NH2 CH3  NH  C2H5 n  propylamine (1o )

Ethylmethylamine (2o )

CH3—N—CH3 CH3 Trimethylamine(3°)

(11) Alkenes are isomeric with cycloalkanes:

CH3CH2CH=CH2 CH3

Butene Cyclobutane

Methylcyclopropane

Such isomers in which one is cyclic and other is open chain, are called ring-chain isomers. Alkynes and alkadienes are isomeric with cycloalkenes.

CH3CH2CCH 1-Butyne

CH2=CH—CH=CH2 1,3-Butadiene

Cyclobutene

Exercise 23. The type of isomerism observed in urea molecule is (A) Chain (B) Position (C) Geometrical (D) Functional Metamerism

This type of isomerism is due to the unequal distribution of carbon atoms on either side of the functional group in the molecule of compounds belonging to the same class. For example, methyl propyl ether and diethyl ether both have the same molecular formula. CH3–O–C3H7 C2H5–O–C2H5 methyl propyl ether diethyl ether In methyl propyl ether the chain is 1 and 3, while in diethyl ether it is 2 and 2. This isomerism known as Metamerism is shown by members of classes such as ethers, and amines where the central functional group is flanked by two chains. The individual isomers are known as Metamers. Examples: C4H9—N—CH3 C3H7—N—CH3 C2H5—N—C2H5

C2H5

CH3

Ethylmethylpropylamine

Butyldimethylamine

C2H5 Triethylamine

Tautomerism It is the type of isomerism in which two functional isomers exist together in equilibrium. The two forms existing in equilibrium are called as tautomers. For example, the compound acetoacetic ester has two tautomers – one has a keto group and other has an enol group: CH3—C—CH2—COOC2H5 CH3—C=CH—COOC2H5

O

OH

Keto-form

enol-form

Out of the two tautomeric forms, one is more stable and exists in larger proportion. In above, normally 93% of the keto form (more stable) and only 7% of the enol form (less stable i.e. labile) exist. The equilibrium between the two forms is dynamic, i.e., if one form is somehow removed by making a reaction, some of the amount of the other form changes into the first form so that similar equilibrium exists again. Thus, whole of the acetoacetic ester shows the properties of both ketonic group as well as the enolic group. Thus, it adds on HCN, NaHSO 3 etc. due to the presence of >C=O group and it decolourises bromine water and gives dark colouration with FeCl 3 due the presence of >C—OH group. Due to the presence of keto and enol form this type of tautomerism is known as keto-enol tautomerism. It is the most commonly observed type of tautomerism. Keto-enol tautomerism is generally observed in those compounds in which either a methyl  CH  (CH3—), methylene (—CH2—), or a methyne ( | ) group is present adjacent to a carbonyl (—CO—) group as in acetoacetic ester above. In other words, it can be said that keto-enol tautomerism is possible in only those carbonyl compounds in which atleast one a-hydrogen atom is present so that it may convert the carbonyl group to enol group. Another example of keto-enol tautomerism is:

O

O

CH3—C—CH2—C—CH3

Keto-form 2,4 - Pentanedione

OH

O

CH3—C=CH—C—CH3

It is found that if the a-hydrogen atoms are present on both the carbons attached to carbonyl group, more stable is the enol form and hence more its content. Thus, larger the number of a-

hydrogens in a ketone, more is enol content. Also, if number of a-hydrogen containing carbonyl groups is more, again more is the enol content. Thus the order of enol content is: CH3CHO < CH3COCH3 < CH3COCH2CHO < CH3COCH2COCH3 CH3COCH2COH3 has about 75% enol content. Moreover, the enol form shows acidic nature due to the tendency to liberate proton (H +) from the enol ( C—OH) group. This H is the a-hydrogen in keto – form. Therefore, a-H in carbonyl compound is acidic in nature. More the enol content, more is the acidic nature of a-hydrogens in a carbonyl compound. Thus above is also the increasing order of acidity of a-hydrogens. An interesting observation about the enol content in acetoacetic ester is the fact that above mentioned percentage ratio (93:7) is in its aqueous solution. In liquid state this ratio is nearly 25 : 75. It is because in liquid state the enol form is much stabilised by intramolecular H-bonding. CH2 CH H3C—C

O—OC2H5

O

��� �� � �H3C—C

O

O—OC2H5

O

H

O

enol

In aqueous solution the intermolecular H-bonding with water takes place and it dominates the intramolecular H-bonding, resulting in lower enol content. Keto-enol tautomerism exists in cyclic carbonyl compounds also if they fulfil the condition of presence of a-H. Thus, we have —OH O O

—OH

O

O

OH

OH

O O

OH O

But the compound O

O

OH

O cannot exist as tautomers since a-H is already at unsaturated carbon.

Tautomerism is also termed as desmotropism (desomo= bond, tropis = turn) because in tautomers the bonding changes. O OH || | Every compound having skeleton has its tautomer skeleton e.g., in the  C  CH2   C  CH  O OH || | preparation of a very small amount of enol isomers also forms CH3  C  CH3 CH3  C = CH2 which can be isolated. The two exist in equilibrium with each other and can be separated by suitable methods. A hydroxy group attached to a carbon which is itself attached to another carbon atom by a double bond is known as enolic (en for double bond, ol for alcohol). Its nature becomes acidic as in phenol and unlike OH group in alcohol which is neutral or only very slightly acidic (C 2H5OH + Na 1  C2H5ONa + H2). In the above two examples migration of a proton from one carbon atom to 2 another takes place with simultaneous shifting of bonds.

Hydrocyanic acid, H – C  N and Isohydrocyanic acid H — N  C are also tautomeric isomers or tautomers. Difference between Tautomerism & Resonance (a) In tautomerism, an atom changes place but resonance involves a change of position of pielectrons or unshared electrons. (b) Tautomers are different compounds and they can be separated by suitable methods but resonating structures cannot be separated as they are imaginary structures of the same compound. (c) Two tautomers have different functional groups but there is same functional group in all canonical structures of a resonance hybrid. (d) Two tautomers are in dynamic equilibrium but in resonance only one compound exists. (e) Resonance in a molecule lowers the energy and thus stabilises a compound and decreases its reactivity. But no such effects occur in tautomerism. (f) In resonance, bond length of single bond decreases and that of double bond increases e.g. all six C—C bonds in benzene are equal and length is in between the length of a single and a double bond. (g) Resonance occurs in planar molecule but atoms of tautomers may remain in different planes as well. (h) Tautomers are indicated by double arrow in between the two isomers but double headed single arrow  is put between the canonical (resonating) structures of a resonating molecule. Illustration 33. C4H8 can have so many isomers. Write their structures. What are different tests to make distinction between them? Solution:

C4H8 can have following structures: (a) CH3—CH2—CH=CH2 1-butene (b) CH3—CH=CH—CH3 2-butene, (cis and trans) (c) CH3—C=CH2 2-methyl-1-propene | CH3 (d) Cyclobutane (e) methyl cyclopropane Distinctions: (i) Ozonolysis of (a) forms CH3CH2CHO and HCHO or (b) forms CH 3CHO of (c) forms CH3 CO and HCHO CH3

H

(ii)

CH3—C—H + alkaline KMnO4  CH3—C—H Cis-2-butene

CH3—C—OH CH3—C—OH H Meso-isomer

H CH3—C—H + alkaline KMnO4  H—C—CH3

CH3—C—OH HO—C—CH3

trans-2-butene

H d-and l-

Illustration 34. C4H4O4 can have isomers A, B, C and D (a) is cyclic ester (b) is dicarboxylic acid giving racemic tartaric and with alkaline KMnO 4 (c) is dicarboxylic acid giving meso tartaric acid with alkaline KMnO 4 (d) is also dicarboxylic acid giving another monobasic acid on heating. Identify A, B, C and D. Solution:

(a) Cyclic ester is obtained from dicarboxylic acid and diol. Given formula shows A is cyclic ester of oxalic acid and glycol.

O=C O=C

O CH2 CH2

O

(b), (c) and (d) contain two —COOH groups. Possible structures are COOH COOH

H—C

H—C

CH2=C

C—H

H—C

COOH COOH

COOH

COOH

(fumaric acid)

(maleic acid)

gem (ethene dicarboxylic acid)

On heating d forms another monobasic acid hence d is

C H C 2=

C O O H C O O H

    C H C H — C O O H 2= ge m (ethen edicarb oxylica cid )

when alkaline KMnO4 reacts with maleic acid, there is formation of mesotartaric acid (by syn addition) COOH H—C

COOH H2O + O alkaline KMnO4

H—C COOH (C)

H—C—OH H—C—OH COOH meso

Hence this represents isomer C Alkaline KMnO4 converts fumaric acid into a mixture of d- and l-isomers hence mixture is optically inactive (racemic) (e)

CH3—CH2—C————–––C—CH2CH3 CH2CH3 CH2CH3

O /H O/ Zn � � 3� �2 � ��� 2

O || CH3  CH2  C  CH2 CH3

STEREOISOMERISM The isomers which differ only in the orientation of atoms in space are known as stereoisomerism. It’s of two types. (a) Geometrical isomerism: Isomers which posses the same molecular and structural formula but differ in arrangement of atoms or groups in space around the double bonds, are known as geometrical isomers and the phenomenon is known as geometrical isomerism. Geometrical isomerism are show by the compounds having the structure.

X

C=C

Y

X Y

(i) Cis – trans isomerism: When similar groups are on the same side it is cis and if same groups are on the opposite side it is trans isomerism.

HOOC

C=C

H

(Maleic acid) (cis fom)

COOH

HOOC

H

H

C=C

H COOH

Fumeric acid (trans fom)

(ii) Syn - anti

CH3

CH3

H

C

C

N

N OH

(Syn)

H

OH

(aldoxime)

(anti)

C6H5—N

C6H5—N

N— C6H5

C6H5—N

(anti)

(Syn)

Syn anti isomerism is not possible in ketoxime since only one form is possible two — CH3 groups are at one C.

CH3

CH3

C N OH (iii) In cyclic compounds:

OH

OH

Cis

OH

trans

OH

Differentiating properties of cis-trans isomerism (i) Dipolemoment: Usually dipole moment of cis is larger than the trans-isomer.

H

Cl

H

Cl

H C

C

C

C Cl

Cl

=0

 = 1.84D

H

(ii) Melting point: The steric repulsion of the group (same) makes the cis isomer less stable than the trans isomers hence trans form has higher melting point than cis. (iii) Different chemical properties: Syn-addition makes cis forms into meso and trans into d and l, anti addition makes cis into d – l and trans into meso. COOH COOH COOH H H

OH

C C

H

C

alc. KMnO 4   ( syn, addn)

OH

Br water anti addn.

2   

C H

COOH

H

C

Br

Br

C

H

COOH

COOH d&l

meso

COOH H

C

OH

HO

C

H

alc .KMnO (syn ,addn)

4      

Br water anti addn.

C

2       

C

HOOC

COOH

COOH

COOH

H

H

H

C

H

C

Br Br

COOH m eso

d&l

H — C — C O O H H C C O     O O+H 2 H — C — C O O H H C C O M a le ica c id M a le ica n h y d rid e

H — C — C O O H     n o r e a c t io n H O O C — C — H E and Z nomenclature of Geometrical Isomerism If all the four groups/ atoms attached to C = C double bond are different, then Cis and trans nomenclature fails in such cases and a new nomenclature called E and Z system of nomenclature replace it. H

I

higher priority

H

C

I C

C

C

Br

Cl

Cl

Br

higher priority

E I

Z

higher ranked

II

The group / atom attached to carbon - carbon double bond is given to higher rank, whose atomic weight is higher. If the two higher ranked group are across, it is called E form (E stands for the German word entgegen meaning thereby opposite) and the two higher ranked groups are on the

same side, they are called Z-form (Z stands for German word Zusammen meaning thereby on the same side). In general trans - isomer is more stable then cis isomer because in cis from, there will be more interaction in between groups. Besides suitably substituted alkene and cycloalkane, suitably substituted oximes and azo compounds also exhibit geometrical isomerism. R C=N

H

R OH

H

syn-oxime (Aldoxime)

R

C=N

C=N

OH

OH

R

R-syn-oxime or R-anti-oxime

N=N

OH

Anti-oxime (Aldoxime)

R R

C=N

R-Syn-oxime or R-anti-oxime

R

R

N=N R

Cis-azo

anti-azo

Illustration 35. How will you distinguish between maleic acid and fumaric acid? Solution.

Maleic acid forms an anhydride where as fumaric acid does not.

Exercise 24: Assign E / Z configuration to each of the following compounds. CH3 (a) H3C (b) Br H

H

H3C

H3C

(c) Br

HO

CH3

(d)

CH3

Cl

H

Br

D

CH3

CH3

Exercise 25. Write the isomeric structures of an alkene having m.f. C 4H8. Indicate which is more stable. Exercise 26.

Why does 2-butene exhibit cis-trans isomerism but but-2-yne does not? Exercise 27. Geometrical isomerism is possible in: (A) Butene-2 (C) Propane

(B) Ethene (D) Propene

(b) Optical isomerism: Any substance which rotates the plane polarised light (PPL) is said to be optically active. If a substance is optically active, it is non - superimposable on its mirror image. If a molecule of substance is superimposable on its mirror image, it cannot rotate PPL and hence optically inactive. The property of non-superimposibility on mirror image is called chirality. The ultimate criterion for optical activity is chirality, i.e., non-superimposibility on its mirror image. If a molecule of organic compound contains 'one' chiral carbon, it must be chiral and hence optically active. Chiral carbon: If all the four bonds of carbon are satisfied by four different atoms / groups, it is chiral. Here it should be noted that isotopes are regarded as different atoms / groups) chiral carbon is designated by an asterisk (*). Optical isomerism in bromochloroiodomethane: The structural formula of bromochloroiodomethane is Br * H—C—Cl I

The molecule has one chiral carbon as designated by star. So molecule is chiral. It is non superimposable on its mirror image. According to Van't Hoff rule, Total number of optical isomers should be = 2n; when n is number of chiral centre The Fischer projections of the two isomers are

Br

Br H——— I

I——— H

Cl (i)

Cl (ii)

Mirror

Stereoisomers which are mirror - image of each other are called enantiomers or enantiomorphs. (i) and (ii) are enantiomers. All the physical and chemical properties of enantiomers are same except two: (i) They rotate PPL to the same extent but in opposite direction. One which rotates PPL in clockwise direction is called dextro-rotatory (dextro is Latin word meaning thereby right) and is designated by d or (+). One which rotates PPL in anti-clockwise direction is called laevo rotatory (means towards left) and designated by  or (–). (ii) They react with optically active compounds with different rates. Illustration 36. State whether the following compound is optically active or inactive.

H H3C—C—CH2—CH3 2

H

H

H

Solution:

H3C—C—CH2—CH3 or 2

H

H3C—C—CH 2—CH3 * D

The molecule contains one chiral carbon and so must be optically active. Isotopes are regarded as different atom. Optical isomerism in compounds having more than one chiral carbons If an organic molecule contains more than one chiral carbons then the molecule may be chiral or achiral depending whether it has element of symmetry or not. Elements of symmetry: If a molecule have either (a) a plane of symmetry, and / or (b) centre of symmetry, and / or (c) n-fold alternating axis of symmetry If an object is superimposable on its mirror image; it cannot rotate PPL and hence optically inactive. If an object can be cut exactly into two equal halves so that half of its become mirror image of other half, it has plane of symmetry.

superimposable

Plane of symmetry

Non-superimposable

Centre of symmetry: It is a point inside a molecule from which on travelling equal distance in opposite directions one takes equal time.

A

C

B A

B

B B

C

Thus, if an organic molecule contains more than one chiral carbons but also have any elements of symmetry, it is superimposable on its mirror - image, cannot rotate PPL and optically inactive. If the molecule have more than one chiral centres but not have any element of symmetry, it must be chiral. Stereoisomerism in 2, 3-dibromopentane The structural formula of 2, 3-dibromopentane is

H

H

* * H3C—C—C—CH 2—CH3 Br Br The molecule contains two chiral carbons and hence according to Van't Hoff rule the total number of optical isomers should be 2n = 22 = 4 and it is. The four optical isomers are.

H3C

CH3

H3C

CH3

H———Br

Br———H

H———Br

Br———H

H———Br

Br———H

Br———H

H———Br

CH2CH3

H3CH2C

(I)

(II)

CH2CH3 (III)

 Nonsuperimposable   Rotate PPL   Optically active  I, II, III and IV are four stereoisomers of 2, 3-dibromopentane. I and II are enantiomers. III and IV are also enantiomers

H3CH2C (IV)

Non -superimposable Rotate PPL Optically active

What is relation between I and III; or I and IV; or II and III; or II and IV? All these pairs are diastereomers, stereoisomers which are not mirror - image of each other are called diastereomers. Therefore, I and III are diastereomers I and IV are diastereomers II and III are diastereomers II and IV are diastereomers Stereoisomerism in Tartaric Acid The IUPAC name of tartaric acid is 2, 3 - dihydroxy butandioic acid. The structural formula is *

*

HOOC  CHOH  CHOH  COOH The molecule contains two chiral carbon and the number of optical isomers should be 2 n=22=4; but number of optical isomers reduces to 3 because one molecule has plane of symmetry. Stereoisomers of Tartaric Acid

H———OH

HO———H

HO———H

H———OH

(ii)

(i)

   

H———OH ---------------- Plane of symmetry H———OH

Non - superimposable Superimposable Can rotate PPL Optically active

HO———H ---------------HO———H HOOC

HOOC

HOOC

COOH

HOOC

COOH

HOOC

COOH

(iv)

(iii)

III and IV are same Rotation of IV by 180° yield III. Superimposability means same in identity

I and II are enantiomers III is meso-form of tartaric acid A meso compound is one which is optically inactive although have more than one chiral carbons. Number of Optical Isomers Number of possible optical isomers in compounds containing different no. of asymmetric atoms. (1) The molecule has no symmetry The no. of d and l – forms a = 2n n = no. of asymmetric atoms The no. of meso l- forms m = 0 Total no. of optical isomers = a + m = 2n (2) The molecule has symmetry and n is even The no. of d and l forms a = 2n–1 n

Meso forms m = 2 2 1 Total = a + m n

= 2n–1 + 2 2 1 Tartaric acid COOH H—C—OH

Plane of syn n =2

H—C—OH COOH

a = 2n–1 = 22–1 n

m = 2 2 1 = 21–1 = 2° = 1 Total = 2 +1 = 3

COOH H—C—OH HO—C—H COOH I

COOH HO—C—H

COOH H—C—OH

H—C—OH

H—C—OH

COOH II

COOH III

(meso forms)

One of asymmetric centre is rotating plane polarised light towards right other towards left so total optical rotation is zero. I & III or II & III which are not mirror image to each other are called diastereomers. (3) The molecule has symmetry & n is odd The no. of d and l forms a = 2n–1 - 2 Meso forms m = 2

( n 1) 2

( n 1) 2

Total = a + m = 2( ) e.g. Lactic acid n 1

CH3 H—C—OH COOH n = 1 no symmetry d & l forms = 2n = 2 meso form = 0 total = 2

CH3

CH3

OH—C—H

H—C—OH COOH d-form

Mirror images are also called as enantiomers

COOH l-form

Difference between racemic mixture and meso compound A racemic mixture contains equimolar amounts of enantiomers. It is optically inactive due to external compensation. It can be resolved into optically active forms. A meso compound is optically inactive due to internal compensation. Optically active compounds having no chiral carbon The presence of chiral carbon is neither a necessary nor a sufficient condition for optical activity, since optical activity may be present in molecules with no chiral atom and since molecules with two or more chiral carbon atoms are superimposable on their mirror images and hence inactive. (i) Any molecule containing an atom that has four bonds pointing to the corners of a tetrahedron will be optically active if the four groups are different. 16

O

—CH2—S—

—CH3

18O

(ii) Atoms with pyramidal bonding might be expected to give rise to optical activity if the atom is connected to three different groups, since the unshared pair of electron is analogous to a fourth group.

Z N X

Y

Many attempts have been made to resolve such compounds, but until recently all failed because of umbrella effect, also called pyramidal inversion. The umbrella effect is rapid oscillation of the unshared pair from one side of XYZ plane to the other. (iii) Biphenyls, containing four large groups in the ortho position so that there is restricted rotation, are optically active if the rings are asymmetrical. If either or both rings are symmetrical, the molecule has plane of symmetry and optically inactive. Cl

'A'

NO2 HOOC

NO 2 HOOC

COOH O 2N

Cl

Cl

'B'

COOH O 2N

Cl

'B'

Mirror Ring B is symmetrical (having plane of symmetry) superimposable optically inactive. Cl

NO2 HOOC

COOH O2N

COOH O 2N

NO2 HOOC

Cl

Non - superimposable optically active

Allenes, with even number of cumulative double bonds are optically active if both sides are dissymmetric. H3C

H3C

C=C=C Inactive

H H

H H3C

C=C=C Active

CH3 H

Specific Rotation The specific rotation [ a ] l is an inherent physical property of an enantiomer, which varies with the T

solvent used, temp (in °C) and wavelength of the light used. It is calculated from the observed rotation a as follows. a T [ a] l = l �c Where  = length of tube in decimeters (dm) C = Concentration in gram cm–3, for a solution density in g cm–3, for a pure liquid. Illustration 37. The specific rotation of R–2–bromooctane is –36°. What is percentage composition of a mixture of enantiomers of 2-bromooctane where rotation is +18°. Solution:

Let X = mole fraction of R,  (1–X) = mole fraction S.

X (–36°) + (1–X) 36° = 18° 1 or, X = 4 The mixture has 25% R and 75% S, it is 50% racemic and 50% S. Nomenclature of Optical Isomers (1) Absolute and Relative Configuration While discussing optical isomerism, we must distinguish between relative and absolute configuration (arrangement of atoms or groups) about the asymmetric carbon atom. Let us consider a pair of enantiomers, say (+)- and (–)- lactic acid. COOH

H––– C–––OH

COOH

HO–––C–––H

CH3 CH3 (+)-lactic acid (–)-lactic acid We know that they differ from one another in the direction in which they rotate the plane of polarized light. In other words, we know their relative configuration in the sense that one is of opposite configuration to the other. But we have no knowledge of the absolute configuration of the either isomer. That is, we cannot tell as to which of the two possible configuration corresponds to (+) - acid and which to the (–) - acid. (2) D and L system The sign of rotation of plane-polarized light by an enantiomer cannot be easily related to either its absolute or relative configuration. Compounds with similar configuration at the asymmetric carbon atom may have opposite sign of rotations and compounds with different configuration may have same sign of rotation. Thus d-lactic acid with a specific rotation + 3.82o gives l-methyl lactate with a specific rotation -8.25°, although the configuration (or arrangement) about the asymmetric carbon atom remains the same during the change. CO2H CO2CH3 | | H––C––OH H––C––OH | | CH3 CH3 + 3.82 - 8.25o Obviously there appears to be no relation between configuration and sign of rotation. Thus DL-system has been used to specify the configuration at the asymmetric carbon atom. In this system, the configuration of an enantiomer is related to a standard, glyceraldehyde. The two forms of glyceraldehyde were arbitrarily assigned the absolute configurations as shown below. CHO CHO | | H––C––OH HO––C––H | |

CH2OH CH2OH (+)-glyceraldehyde (–)-glyceraldehyde D configuration L configuration If the configuration at the asymmetric carbon atom of a compound can be related to D (+)glyceraldehyde, it belongs to D-series; and if it can be related to L(–)-glyceraldehyde, the compound belongs to L-series. Thus many of the naturally occurring a-amino acids have been correlated with glyceraldehyde by chemical transformations. For example, natural alanine (2-aminopropanoic acid) has been related to L(+)-lactic acid which is related to L(–)glyceraldehyde. Alanine, therefore, belongs to the L-series. In general, the absolute configuration of a substituent (X) at the asymmetric centre is specified by writing the projection formula with the carbon chain vertical and the lowest number carbon at the top. The D configuration is then the one that has the substituent 'X' on the bond extending to the 'right' of the asymmetric carbon, whereas the L configuration has the substituent 'X' on the 'left'. Thus, R1 R1 | | R2 ––C––X X––C––R2 | | R3 R3 D configuration

L configuration

When there are several asymmetric carbon atoms in a molecule, the configuration at one centre is usually related directly or indirectly to glyceraldehyde, and the configurations at the natural (+)-glucose there are four asymmetric centres (marked by asterisk). By convention for sugars, the configuration of the highest numbered asymmetric carbon is referred to glyceraldehyde to determine the overall configuration of the molecule. For glucose, this atom is C–5 and, therefore, OH on it is to the right. Hence the naturally occurring glucose belongs to the D-series and is named as Dglucose. However, the above system of nomenclature based on Fischer projection formulae, has certain disadvantages. Firstly before a name can be assigned to a compound, we must specify how its projection formula is oriented. Secondly, sometimes the two asymmetric carbon atoms having the same kind of arrangements of substituents are assigned opposite configurational symbols. Thus for (–)-2, 3-butanediol we have

FISCHER PROJECTION German chemist Emil Fischer, proposed the “Fischer projection” concept to simplify the displaying of stereo – chemical relationship of carbohydrates. Fischer projection is the representation of tetrahedral carbon atoms as a two – dimensional structure on paper. In this projection, molecule arranged with the horizontal bonds along with chiral centre projecting above the plane of the page. The vertical bonds projecting behind the page. Main chain of molecule arranged by vertical lines and most important functional group is placed at the top.

CHO

CHO



CHOH

H

C

CH2OH

CHO OH



C H

CH2OH

CH2OH OH

CHO H

OH CH2OH

Projected form (Fischer projection)

Illustration 38. A racemic mixture of (±) 2phenyl propanoic acid on esterification with (+) 2butanol gives two esters. Mention the stereochemistry of the two esters produced. Solution.

H

H3C

H3C

H3C

H

Ph

( )

+ H

COOH

(+ )

CH3

O Ph

OH

O

(+)

C 2H 5

H

(+ )

H

H3C

(Racemic mixture)

H 5C 2

CH3

O Ph O

H 5C 2

()

H

(+ )

The bonds attached to the chiral carbon in both the molecules are not broken during the esterification reaction. (+) Acid reacts with (+) alcohol to give an (++) ester while (–) acid reacts with (+) alcohol to give (+ –) ester. These two esters are diastereoisomers. Illustration 39. Identify each of the following as R or S CH3 (a) (b) H

C

H3C

CH2Cl

H

C

a – S, b – S, c – R, d – S

C

C

CH2OH COOH

(d) OH

CH2OH

Solution:

CH3

CH3

CHO

(c)

CN

NH2

C CH3

H

CH

Exercise 28. Which of the following will be a stronger acid and why? Cl 3CH and F3CH Exercise 29. HC  C– is a weaker base than CH2 = CH–. Why? Exercise 30. Which of the following compounds will exhibit optical isomerism? (A) 2-butene (B) 2-butyne (B 2-butanol (D) Butanol Exercise 31. Dichloro ethylene shows. (A) Geometrical isomerism (C) Both

(B) Position isomerism (D) None

Exercise 32. The compound having molecular formula C4H10O can show (A) Metamerism (B) Functional isomerism (C) Positional isomerism (D) All Exercise 33. A compound contains 2 dissimilar asymmetric carbon atoms. The number of optical isomers is: (A) 2 (B) 3 (C) 4 (D) 5 Exercise 34. Number of possible isomers of glucose are (A) 10 (B) 16

(B) 14 (D) 20

ANSWERS TO EXERCISES Exercise 1: 1° - six 3° - two 4° - one Exercise 2: (i) (ii) (iii)

2, 3, 5 – tri methyl heptane 3, 4, 5, 6 tetra methyl octane 3 ethyl 3 propyl heptane

Exercise 3: (i) (ii) (iii)

3, 3 – dimethyl but – 1 – ene 3- (1, 2 – dimethyl propyl) – 1, 3 – hexadiene 4 – cyclopentyl – 2 – butane

Exercise 4: (i) (ii) (iii)) (iv)

2, 2 dimethyl propane 3, 3 – diethyl pentane 6 hydroxy 4, 5 – dimethyl hexanoic acid 3 chloro 3 hydroxy butanal

Exercise 5:

(i) (ii) (iii)

3 hydroxy, 3 methyl propanyl chloride Benzoic, ethanoic anhydride H H H H O

H

C

5

H

C 4

C

OH

Cl

C 2

3

O

C 1

NH2

CH3 (iv)

3 chloro, 4 - hydroxy, 2 methoxy pentanamide H H H O H3C 5

C

C 3

C

CH3 OH

H

4

2

C 1

Cl

3 hydroxy 4 methyl pentanoyl chloride Exercise 6: (i) (ii)

Cyclo propanol 3 cyclo propyl butanoic acid

Exercise 7: Bond energy order sp–sp > sp2-sp2 > sp3-sp3 Hence, (A) is correct. Exercise 8: (a) Exercise 9:

is most stable as it is a 3° carbocation as well as allylic.

(a) In this structure if we break C – Cl bond heterolytically we end up with a C + ion; a benzyl carbocation. Exercise 10: e
Exercise 12: Allyl carbonium ion undergoes resonance stabilization. Hence, (A) is correct. Exercise 13: In aniline the lone pair of electrons on N atom are delocalised in the benzene ring and hence the basicity decreases. Exercise 14: Resonance effect operates through the -electrons and inductive effect operates through the -electrons. -electrons can be easily perturbed and hence that effect is dominateng. Exercise 15: Due to the presence of bulky –NO 2 groups on its ortho positions, the –NMe 2 group goes outside the plane of resonance to avoid steric repulsion. The C – N bond rotates and hence the lone pair of N goes perpendicular to the plane of benzene ring. As a result the resonance is stopped and hence the lone pair is readily available as a base. Exercise 16. a rel="nofollow">b The negative charge on ‘O’ atom is delocalized to a larger extent. CH3 – CH2 – O–, the negative charge on ‘O’ atom is increased by +I effect. Exercise 17: (a)

H3C

CH2 OH

(b) Exercise 18: (i) (ii)

H3C

CH2 OH

Yes (a) Phenoxide ion (C6H5O) which results after loss of H+ from phenol can be resonance stabilized whereas alkoxide ion (RO ) which results after the loss of H+ from alcohol can not stabilize it through resonance.

(iii)

Lone pair on nitrogen of aniline are less available for a base because they are involved in the resonance the ring. NH2

(iv)

(c) is most acidic as triple bonded carbon atom are sp hybridized i.e. orbital holding bond pairs of C  H bond has 50% s and 50% p – character, the greater is s – character greater is hold of electrons of carbon nucleus over bon d pair electrons which makes loss of H+ easier.

Exercise 19. d>c>a>b The order depends on the carbocation stability.

Exercise 20. (i) (ii)

(a) because for geometrical reasons this carbocation cannot attain planarity. (a) I > III > II (I is most stable due to resonance effect of OH group) (b) In CH3  O  CH2+, positive charge over carbon is stabilized with the help of lone pair of electrons present in adjacent oxygen atom. +

CH3  O  CH2+ ��� CH3  O = CH2 ��

Exercise 21: In the first case I is more electronegative than C and hence the positive charge is an the C atom. So OH– attacks C and displaces I–. In the second case due to the presence of three F atoms, the electron deficiency of C increases. Hence it gives a positive charge on I. Hence OH– attacks I and displaces CF3.

Exercise 22: o-, m- and p-isomers, i.e., position isomers Hence, (C) is correct. Exercise 23: NH4CNO is functional isomer of urea. Hence, (D) is correct. Exercise 24: (a) (c)

Z Z

(b) (d)

E Z

Exercise 25:

The stability order is I > III > II. Exercise 26: Alkynes are all linear molecules and hence there is no chance of exhibiting geometrical isomerism. Exercise 27:

H

H C=C

CH3

CH3

If either of the two doubly bonded carbon atoms has same group or atoms attached on it, it will not show geometrical isomerism. Hence, (A) is correct. Exercise 28: CHCl3 is a stronger acid. Cl3CH  Cl3C– + H+ In this case the negative charge on C is delocalised in the vacant orbitals of Cl, which is absent in F3C–. Exercise 29: HC  C–, in this the negative charge is on a sp hybridised carbon atom which has less tendency to give electrons. Exercise 30: Due to the presence of asymmetric carbon atom. Hence, (B) is correct. Exercise 31: CH2 = CCl2 and CHCl = CHCl are position isomers; CHCl = CHCl also show geometrical isomerism. Hence, (C) is correct. Exercise 32: Alcohols show position isomerism; Ethers show metamerism; Alcohol and ethers shows functional isomerism. Hence, (D) is correct. Exercise 33: a = 2n; where n is no. of dissimilar asymmetric carbon atoms. Hence, (C) is correct. Exercise 34. Glucose has four dissimilar asymmetric carbon atoms; a = 24. Hence, (C) is correct.

MISCELLANEOUS EXERCISES Exercise 1:

Write the structural formula of 2 – methyl pentan – 1 – ol.

Exercise 2:

Write the structural formula of the compound 4 – methyl pent – 2 – ene.

Exercise 3:

Write the structural formula of the compound 3 – methyl Butanoic acid.

Exercise 4:

Write the structural formula of the compound.

5 - Chloro- 7 - hydroxy - 3 methoxy, oct - 5 - enal Exercise 5:

Write the structural formula of the compound 2 – Bromo butanoyl chloride.

Exercise 6:

Explain what is homolytic fission?

Exercise 7:

What is inductive effect?

Exercise 8:

Why inductive effect is called transmission effect?

Exercise 9:

What is difference between a free radical and an ion?

Exercise 10:

What is hyper conjugation?

Exercise 11:

Why hyperconjugation is called no bond resonance?

Exercise 12:

Name the kind of effect that operates to explain the stability of carbocations?

Exercise 13:

What is electromeric effect?

Exercise 14:

Define and explain the terms chirality. Which of the following molecule are chiral and which are achiral? (i) ClCH ( Br ) F (ii) CH3 CH2  CH ( Cl) CH2CH3 (iii) CH3CHOHCH2CH3 (iv) ClCH2CH2CH2CH3 (v) CH3CH2CHOHCH2CH2CH3

Exercise 15:

A compound C4H10O shows optical activity. Identify the compound and write the possible structures.

ANSWER TO MISCELLANEOUS EXERCISES CH3

Exercise 1:

OH

CH3

Exercise 2: CH3 CH3

O

Exercise 3: OH CH3

Exercise 4:

OH

Cl

O

O

H

O

Exercise 5: Cl Br

Exercise 6:

In this type of cleavage each fragment formed as the result of cleavage gets one electron from the shared pair of electrons. �

�

hn Cl � M �Cl ����� � Cl + Cl or 523  725K

Exercise 7:

Permanent displacement of electrons along a carbon chain when some atom or group of atoms with different electronegativity than carbon is attached to carbon chain is called inductive effect.

Exercise 8:

Because the effect is carried on from one carbon atom to the next two consecutive carbon atoms.

Exercise 9:

Free radical is electrically neutral form by homolytic cleavage of co-valent bond but ions are positively & negatively charged species.

Exercise 10:

Hyper conjugation is a special kind of resonance in which decolaziation of electrons takes place through overlap between  bond orbital and  bond orbital or p orbital.

Exercise 11:

Since there is no bond between carbon and hydrogen atoms in hyperconjugation structures.

Exercise 12:

Resonance, inductive effect, hyper conjugation.

Exercise 13:

The complete transfer of the shared pair of  - electrons of a multiple bond to one of the atoms in the presence of the attacking agent is called electromeric effect.

Exercise 14:

(i), (iii) & (V)

SOLVED PROBLEMS Subjective: Prob 1.

Sol.

Write the IUPAC name H3C CH2 CH CH2 CH

1 2 H3C CH2

CH3

H2 C

3 CH

4 5 CH2 CH

CH3

H2 C

CH2 CH2 CH3 CH3 7 8 6 CH2 CH2 CH3 CH3

The longest chain is of 8 carbon atoms. The root word for the chain is oct. (b) It is a saturated chain. Thus, the primary suffix-ane should be affixed with root word. Oct + ane = Octane Root Primary suffix (c) One methyl group and one ethyl group are present as substituents. The numbering of the chain is done as indicated above. The names of substituents along with locants in alphabetical order are written before the root word. Hence, the name of the compound is

Prob 2.

CH3 H3C C CH3

CH2 CH2 CH H3C

C

CH2 CH2 CH2 CH3 CH3

CH H3C

Sol.

CH3 1 2 H3C C CH3

CH3

4 5 3 CH2 CH2 CH H3C 3' H3C

1' C 2' CH

7 8 6 9 CH2 CH2 CH2 CH3 CH3 Branched chain

CH3

The compound is a derivative of nonane. It has two methyl groups and one branched chain as substituents. The branched chain is also numbered. The name of the branched chain is 1’, 2’, 2’ – trimethylpropyl. Hence, the name of the compound is: 2, 2-Dimethyl-5-(1’, 1’, 2’ – trimethylpropyl) nonane.

Prob 3.

H3C

CH

C

C

CH3

3 C

2 C

1 CH3

Cl Sol.

5 H3C

4 CH

Cl It is a derivative of unsaturated hydrocarbon. The parent chain consisting the triple bond is of five carbon atoms. The numbering is done from R.H.S as it gives lowest number to triple bond. Cl is a substituent at carbon – 4. Hence, the name of the compound is: 4 – Chloro pent – 2 – yne Prob 4.

Sol.

H3C CH2

1 2 H3C CH2

CH

CH

Cl

Br

3 CH

4 CH

Cl

Br

CH

CH2 CH2 CH3 NO2

5 CH

7 8 6 CH2 CH2 CH3 NO2

The compound is a derivative of octane. All the three groups are subsituents. The numbering is done from L.H.S as it gives lowest numbers to two subtituents. The names of the substituents are arranged in alphabetical order. Hence, the name of the compound is: 4 – Bromo – 3 – chloro – 6 – nitro – octane

Br

Prob 5.

Sol.

H3C

CH

C

C

CH3

5 H3C

Br 4 CH

3 C

2 C

1 CH3

The compound is a derivative of an unsaturated hydrocarbon consisting a double bond. The numbering is done from R.H.S as it gives lowest number to carbon atom linked by a double bond in the chain. Br is the subtituent at carbon – 4. Hence, the name of the compound is: 4 – Bromo pent – 2 – ene

Prob 6.

H3CH2CHC

CHCOOH

Sol.

5 4 3 H3CH2CHC

2 1 CHCOOH

The compound consists of two functional groups, COOH group is the principal functional group. The numbering is thus, done from R.H.S. Root word = Pent, double bond = 2 – ene, functional group = -oic acid Thus, the name of the compound is: 2 – Penten – 1 – oic acid

Prob 7.

Sol.

CH2

CH

CH2

CN

CN

CN

1 CH2

2 CH

3 CH2

CN CN CN There are three CN groups. All group of one kind which occurs in a single molecule should be given the same treatment as far as possible (Like Things Alike). Propane – 1, 2, 3 – tricarbonitrile C6 H5

Prob 8.

H3C C

O CH

CH3

C

CH3

CHCl CH3

C6 H5

Sol.

H3C C 4 5 CH3

O CH 3

C 2

CH3 1

CHCl 1'

CH3 The compound is a ketone. Numbering is done at R.H.S as to give lowest number to the functional group. The substituents are C  3 and C  4. Root word = Pent, Primary suffix = -ane, functional group = -one, Substituent names = phenyl, methyl, 1’ – chloroethyl. Thus, the name of the compound is: 3 – (1’ chloroethyl) – 4 – methyl – 4 – phenylpentan- 2 – one O

Prob 9.

H3C Sol.

5 H3C

C

O CH2

CH2

O 4 C

3 2 CH2 CH2

C

OH

O 1 C

OH

Two functional groups are present but COOH group ranks higher in the priority table. The compound is named as acid. The numbering is done from R.H.S. as to give minimum number to the functional group carbon. The carbonyl group act as a substituent. Root word = pent, primary suffix = -ane, secondary suffix = -oic acid and substituent = oxo or keto. Thus, the name of compound is: 4 – Oxo – pentanoic acid.

H 2C

Prob 10.

H3C CH Sol.

CH

CH2 CH

CH

CH

CH2 CH3

CH2 CH2 COOH

H2 C CH CH CH2 CH3 9 7 8 5 6 H3C CH CH CH2 CH CH2 CH2 COOH 3 2 1 4 The longest chain in the compound consists 10 carbon atoms and two double bonds but it does not include the principal functional group, COOH. Hence, the longest chain of nine carbon bond is selected. The numbering is started from the end of COOH group as to give minimum number to carbon formatting the COOH group. Thus, the compound is a derivative of nonene. The compound has a side chain which is named 2 – butenyl. Root word = -non, primary suffix = -ene, principal functional group (secondary suffix) = -oic acid, substituent (prefix) = 2’ – butenyl. Hence, the name of the compound is: 4 – (2’ – Butenyl) – non – 6 – enoic acid

Prob 11. Arrange the following compounds in the increasing order of basicity O

N N

H (II)

(I)

Sol.

N

N

H

H

(III)

(IV)

In (II), (III) and (IV) the lone pair present in sp 3 orbital but in II, the electron pair on nitrogen is involved in delocalization making it least basic of all. In (I), the unshared electron pair is present in sp 2 orbital which makes it less basic than (III) and (IV) is less basic (III) due to I effect of oxygen atom. Thus, the increasing order of basic strength would be II < I < IV < III.

Prob 12. Compare the stabilities of phenyl (C6H5) and cyclohexyl (a) cations and (b) anions. Sol.

(a) C6H5+ is a vinyl carbocation and is less stable than C6H11+, a 20 carbocation. (b) In C6H5 :, the electron pair is in an sp 2 hybrid orbital. The carbanion has more S – character and is more stable than C6H11 : whose unshared electron pair is in an sp3 hybrid orbital. H

sp2

C 6H 5

C6H11

C6H5

H sp3

C6H11

Prob 13. Of the ethanol and trifluro ethanol, which is more acidic? Sol.

As the methyl group is weakly +I while the CF3 group is strongly I, Thus, due to I effect of CF3 group. The electron pair in O  H bond will be drawn in towards the oxygen atom, thereby facilitating the release of the hydrogen as a proton.

Prob 14. Compare the acidic strength of propenoic acid, propanoic acid and propynoic acid. Sol.

HC  C  CO2H > CH2 = CH  CO2H > CH3  CH2  CO2H

Prob 15. Out of phenyl acetic acid and acetic acid, which is stronger? Sol.

Phenylacetic acid is stronger than acetic acid due to I effect of phenyl group.

Prob 16. Give the correct order of acidic strength of O , m  and p  methyl phenols. Sol.

m - methyl phenol > p – methyl phenol > O – methylphenol

Prob 17. Arrange 4 – nitrophenol, 2, 4 - dinitrophenol and 2, 4, 6 – trinitrophenol in the decreasing order of acid strength. Sol.

2, 4, 6 – trinitrophenol > 2, 4- dinitrophenol > 4 – nitrophenol CH3

Prob 18.

CH3

is more stable than

X

Why?

H

X

(a)

Sol.

H (b)

Here carbocation (a) is stabilized most as the CH 3 group is attached to the carbon bearing positive charge.

Prob 19. Compare the acidic strength of the following CH 3COOH, ClCH2COOH, Cl2CHCOOH, CCl3COOH. Sol. CH3COOH < Cl

H

O

C

C

O

H < Cl

H

H

O

C

C

O

H

O

H

Cl Cl O < Cl

C

C

Cl

Increasing acidic strength due to increasing number of similar (I) groups. Prob 20. Compare the basic strength of the following NH3, CH3NH2, (CH3)2NH, (CF3)3N. Sol.

The increasing order of basic strength is as follows:

CH3 CF3 F 3C

H

N

< H

H

N

CF3

<

H

H

(I)

< H3 C

N

N

H

CH3 (IV)

(III)

(II)

Prob 21. Explain that a - methyl acetyl acetone undergoes enolization to smaller extent than acetylacetone O

Sol.

H

O O

H3C

C

CH2 C

CH3 CH3

Acetyl acetone

O

C

H Enolization is greater due to less strain

(Keto - form) O

H

O O

H3C

C

CH3

C C

CH

C

CH3 CH3

CH3

O

C

CH3

C C CH3

a- methyl acetyl acetone keto-form

Enolization is smaller due to high strain

Prob 22. Write the possible isomers of the formula C5H10 O2 . Sol.

(a) Carboxylic acids H3C CH2 (i)

CH2

CH2

COOH

CH3

(ii)

H3C

CH

CH2 COOH

CH3

(iii)

H3C

C

COOH

CH3 (b) Esters: (i)

O H3C

CH2

CH2

OCH3

C

O

(iii) H3C

C

O

(ii) H3C

CH2 O

C

OCH2

CH3

H3C

C

O

CH

CH3

(iv) O

CH2

CH2

CH3

CH3

And

So on.

(c) Hydroxy aldehydes (d) Hydroxy ketones etc. Prob 23. Which among the following is most acidic

(a) HCOOH (b) CH3COOH (c) (CH3)2CCOOH (d) (CH3)3CCOOH Sol.

–CH3 group is electron releasing group, it decreases polarity of –OH bond in –COOH group. Hence CH3 group decreases the acidic strength of the carboxylic acid. �

�

Prob 24. Why allylic free radical CH2 = CH – CH2 is more stable than CH3– CH2 - CH2 while both are primary free radicals. Sol.

Allylic free radical is resonance stabilized while propyl free radical is not resonance stabilized. �

�

d�

K

K d�

CH2 = CH – CH2  CH2 – CH = CH2  CH2  CH  CH2 Prob 25. Though enol form is less stable than keto form, phenol exists in enol form, why? Sol.

OH

O H H

Enol form is much more stable than keto form because of great stability associated with aromatic ring which is absent in keto form.

Objective: Prob 1.

The IUPAC name of the compound CH3

CH3CH2CH2CH

CH

C

CH CH3

(A) (B) (C) (D)

CH2CH3 CH2CH3

is

CH2CH3

CH3

3, 3 – diethyl – 4 – methyl – 5 – isopropyl octane 3, 3 – diethyl – 5 – isopropyl – 4 – methyl octane 4 – isopropyl – 5 – methyl – 6, 6 – diethyl octane 6, 6 – diethyl – 4 – isopropyl – 5 – methyl octane

Sol.

(B)

Prob 2.

The IUPAC name of the compound CH3CH =CH - CH =CH - C �CCH 3 is (A) 4, 6 – octadien – 2 – yne (C) 2 – octyn – 4, 6 – diene

(B) 2, 4 – octadien – 6 – yne (D) 6 – octyn – 2, 4 – diene

Sol.

(B)

Prob 3.

The correct IUPAC name of the compound O

CH3

CH

C

CH3

CH2

CH2

CH is : CH3

CH3

(A) (B) (C) (D) Sol. Prob 4.

2, 5 – dimethyl heptan – 4 – one 3, 6 – dimethyl heptan – 4 – one 2 – ethyl – 5 – methyl hexan – 3 – one 1, 1 – dimethyl – 4 – ethyl pentan – 3 – one

(A) The IUPAC name of

CH3

CH

CH2

OH

OH

(A) 1, 1 – dimethyl – 1, 3 – butanendiol (C) 1, 3, 3 – trimethyl – 1, 3 – propane diol Sol.

The IUPAC name of the compound (A) (B) (C) (D)

Sol.

(B) 2 – methyl – 2, 4 – pentane diol (D) 4 – methyl – 2, 4 – pentane diol

(B) HO

Prob 5.

C(CH3)2 is

(C)

1, 1 – dimethyl – 3 – hydroxy cyclohexane 3, 3 – dimethyl – 1 – hydroxy cyclohexane 3, 3 – dimethyl cyclohexanol 1, 1 – dimethyl – 3 – cyclohexanol

CH3 CH3

is

Prob 6.

Which of the following compounds represents 2, 2, 3 – trimethylhexane? (A) CH3C( CH3 )2 CH 2CH 2CH( CH 3 )2 (B) CH3C( CH 3 )2 CH 2CH( CH3 ) CH 2CH 3 (C) CH3C( CH 3 )2 CH( CH 3 ) CH 2CH 2CH 3 (D) CH 3C( CH 3 )2 CH 2C( CH 3 )2 CH 3

Sol.

(C)

Prob 7.

O

The IUPAC name of the compound CH3

(A) 1 – hydroxy butan – 1, 2 – dione (C) 3 – oxo butanoic acid Sol. Prob 8.

OH

C CH2 C O is (B) 4 – hydroxy butan – 2, 4 – dione (D) 1 – hydroxy – 1, 3 – dioxo butane

(C) The IUPAC name of the compound.

CH2COCH3 is

(CH3)2C CN

(A) (B) (C) (D) Sol. Prob 9.

4 – cyano – 4 – methyl – 2 – oxo pentane 2 – cyano – 2 – methyl – 4 – oxo pentane 2, 2 – dimethyl – 4 – oxo pentanenitrile 4 – cyano – 4 – methyl – 2 – pentanone

(C) The IUPAC name of

(C 2H 5)2 N C H 2C H

COOH is

Cl

(A) (B) (C) (D) Sol.

2 – chloro – 4 – (N – ethyl) pentanone acid 2 – chloro – 3 – (N, N – diethyl amino) propanoic acid 2 – chloro – 2 – oxo diethyl amine 2 – chloro – 2 – carboxy – N – ethyl ethane

(B)

Prob 10. Which of the following compound contains isopropyl group? (A) 2 – methyl pentane (B) 2, 2, 3 – tri methyl pentane (C) 2, 2, 3, 3 – tetra methyl pentane (D) 2, 3 – dimethyl pentane Sol.

(A)

Fill in the blanks Prob 11. Isomerism which are ………….. and mirror image are known as ……………….. . Sol.

Non super imposable, enantiomeis

Prob 12. Geometrical isomerism is due to ………………….. rotation round a covalent bond. Sol.

Hundred

Prob 13. A meso compound is made up of ……………. molecules that contain ………… centres.

Sol. Achiral, chiral Prob 14. The (+) and () forms of tartaric acid are …………………….. . Sol.

enentiomersm

Prob 15. Tautomerism is …………………. isomerism. Sol.

dynamic

Prob 16. The possible number of derivatives of propane are …………….. . Sol.

Four

Prob 17. d and l isomers of a compound are mirror image to each other and thus know as ……………. . Sol.

mirror image isomerism

Prob 18. Glucose and fructose are ……………. isomerism. Sol.

Functional

Prob 19. Metamers include ………… class of compounds. Sol.

Same

Prob 20. Dipole moment of trans but -2-ene is ……………………… Sol.

zero

State true and false Prob 21. Homologues can be isomers. Sol.

[F]

Prob 22. Tartaric acid possesses two asymmetric carbon atom in its molecule. Sol.

[T]

Prob 23. A mixture of cis and trans isomers can be separated by fractional distillation. Sol.

[T]

Prob 24. Geometrical isomers must contain a carbon – carbon double bond. Sol.

[F]

Prob 25. A ring can make a molecule rigid and causes geometrical isomers. Sol.

[T]

Prob 26. But-2- ene-1, 4 – dioic acid shows geometrical isomerism. Sol.

[T]

Prob 27. Optical isomers have different physical and chemical properties. Sol.

[F]

Prob 28. cis and trans isomers have different dipole moments. Sol.

[T]

Prob 29. Alkanes show isomerism with alkamines. Sol.

[T]

Prob 30. Only organic compound are optically active. Sol.

[F]

ASSIGNMENT PROBLEMS Subjective: Level – O 1. Explain why the give names in the following are wrong Give a correct name in each case: (a) 1 – methylpentane (b) 2, 3 – dichloropropane (c) 1 – chloro – 1 – methylpropan – 2 – ol (d) 1, 1, 2, 2 – tetramethylethane (e) 2 – keto – pentan – 5 – oic acid 2. Write IUPAC name of tertiary butyl methyl ether. 3. Write structure and IUPAC name of aceto nitrile. 4. Find number of primary, secondary, tertiary and quaternary carbon atom in 2, 2, 4 tri methyl hexane. 5. Write IUPAC name of CH 2CHO CH3

6. What is +E and E effect? Explain with one example in each case. 7. What are free radicals? Discuss the stability order or primary, secondary and tertiary free radicals? 8. Arrange the different carbanions ions in the decreasing order of stability giving reasons? 9. Explain the stability order of carbocations giving reasons. 10. For each of the following bond cleavages, use curved arrows to describe the electron flow (a)  CH O OCH 3

CH O OCH 3 3

3

(b)

E E

CH3

(c) H3C

CH3 Br

H3C

Br CH3

CH3

O

(d)

O

H3C

OH

H

C

H 2O

11. What is hyper conjugation effect? How does it differ from resonance effect. Briefly discuss the significance of hyperconjugation effect. 12. D – (+) – glyceral dehyde is oxidised to () glyceric acid, OHCH2.CH(OH)COOH. Give the D – L designation of the acid. 13. Draw the (a) Newman & (b) Fisher projection for enantiomer of CH 3CHIC2H5. 14. Halogen group attached to benzene nucleus is o – and p – directing but is considered as deactivating group. Explain. 15. What are end products of the following? NO2 (a) Fe Br2

(b)

NH2 Fe Br 2

(C)

OH

CH3Cl

AlCl 3

Level – I Write IUPAC name of the following compounds: 1.

2. OH

3.

O O OH

4. In which C  C bond of CH3CH2CH2Br, the inductive effect is expected to be the least? 5. Write the resonance structures of (a) CH 3COO and (b) C6H5NH2. Show the movement of electrons by curved arrows. 6. Give reasons why the following two structures (I and II) can not be the major contributors to the real structure of CH3COOCH3. O O

H3C

C

O

H3C

CH3

C

O

CH3

(II)

(I) 7. Explain acidic nature of vinyl alcohol.

8. Explain the basic character of following compounds. NH3, CH3NH2, (CH3)2NH, C6H5NH2 9. Explain the acidic nature of a - hydrogen of aldehydes and ketones. 10. Arrange the following compounds in order of their increasing acidity COOH

COOH

(I)

COOH

NO 2

NH2

(II)

(III)

11. Arrange the following compounds in order of their increasing acidic characters H3C

H2C

CH

(I)

COOH

COOH

COOH (II)

(III)

12. Draw example of (a) a meso alkane having the molecular formula C 8H18 and (b) the simplest alkane with a chiral carbon.

13. Predict the preferred regiochemistry for the addition of HCl to each of the following compounds on the basis of carbocation stability. (a) (b) CH3

H3C

H3C

(c)

(d)

Ph

CH2 H3C

H3C

CH3

14. Explain mechanism of 1:4 addition of HBr on 1, 3 – butadiene. 15. Explain which compound is the weaker base: NH2 NH2 (a)

or

NO 2

or

(b)

O (c)

O (d)

O

O

C

C

OH

OH

or HO OH

CH3

CF3 or

O

O

C

C

OH

Level – II Me

1.

Me

NH2

Me

Me

H2C

2.

H3C H3C CH3

3. Which of the following compounds are correctly named? (A) CH3 CH2CH2 CH(Cl)CO2H; 2  chloropen tanoic acid (B) CH3 C �CCH(CH3 )CO2H; 2  methyl haxenoic acid  3 (C) CH3CH2CH = CHCOCH3 ; Hex  3  en  2  one (D) (CH3 )2 CHCH2CH2CHO; 4  methylpen tanal 4. Draw resonance structures for (i) C6H5F and (ii) C6H5NO2 showing which carbon of the ring bear the partial charge induced in the ring by extended  - bonding. 5. NH2 in benzene is orth-para directing but when passed in acidic solution it becomes meta directing why? 6. C  O bond length in carbonate ion is longer than C  O bond length in formate ion. Explain in brief. 7.

Math list II with list I

List – I (Statement)

List - II (reason in terms of effect) (a) The alkene CH3 CH = CH  CH3 is (a) P - d bonding effect more stable its isomer CH3 CH2  CH = CH2 + Hyper conjugation (b) the carbocation CH3  O  H2  (b) C

is less stable than its another canonical form CH3  O+ = CH2  (c) The carbanion: : CCl3 is more stable (c) Fulfillment of octel

than : CF3 . (d) Phenol is more acidic than alcohol

(d) Resonance effect

8. Arrange the following in increasing order of resonance stability.

(II)

(I)

(III)

9. Arrange the following in increasing order of acidic nature. OH

OH

OH

(III)

(I)

OH

NO2

NO2

CH3

(II)

(IV)

10. Arrange the following in increasing order of carbocation stability. CH

(I)

CH

(II)

CH

C

(III)

(IV)

11. Arrange the following in increasing order of carbanion stability. CH2

CH2

CH2

(I)

(II)

(III)

CH 2

NO 2 (IV)

12. Arrange the following in increasing order of basic property. NH2

NMe2

NMe2 Me

(I)

(II)

Me

(III)

13. Write the increasing order of stability of following cations

CF3 , CH2Cl , C6H5CH2 , C6H5 14. Arrange the following in increasing acid strength

O

(a)

O

O

C

OH

C

C

OH

OH

C

CH3

O (I)

(III) C O

CH3 (II)

(b)

OH

OH

CH3OH (V) (IV) CN (VI)

(c)

O H3C

C

HO

OH

O

O

C

C

OH

HO

C

(II)

(I)

(b)

N

H (I)

(II)

NH2

NH2

(II)

NH2

(IV)

NH2 CH3

(c)

NH2

(III)

NH2

(I)

NH2 NO2

(III) NH2

(IV) NH2

CH2NH2

CH3

(I)

(II)

CH2

CH2 C (III)

15. Rank the following amines in increasing basic nature (a)

N

O

O

CH3

(IV) CH3 (III)

OH

Objective: Level – I 1. Which one is the wrong statement? (A) saturated hydrocarbons are called alkenes (B) open – chain compounds are called aliphatic (C) unsaturated hydrocarbons contain double or triple bond (bonds) between carbon atoms (D) aromatic compounds posses a characteristic aroma 2. Which is the correct statement? (A) the prefix are written before the name of the compound (B) the suffix are written after the name of the compound (C) the IUPAC name of a compound is always written as one word (D) all the above are correct 3. Which of the following statement is wrong? (A) the IUPAC name of alkenes ends with suffix –ene (B) the IUPAC name of alkynes ends with suffix –yne (C) the substituents gets lower number in comparison to functional group (D) the IUPAC name of acid amides is alkanamide 4. The correct decreasing order of preference of functional groups during the IUPAC nomenclature of poly functional compounds is (A) COOH, CHO, OH, NH2 (B) NH2, OH, CHO, COOH (C) COOH, OH, NH2, CHO (D) COOH, NH2, CHO, OH 5. In which of the following compounds the carbon atom chain has been correctly numbered 7 3 2 1 6 5 4 (A) H3C CH CH CH CH2 CH2 CH3

1 (B)

H3C

5 (C)

H3C

2 CH2

4 CH2

3 C 4 H 2C 3 CH H 2C

(D)

C 2H 5

CH3

CH 5 CH3 2 C

CH3

CH2

CH3

1 CH

CH3 H3C 5

C CH 4 3 CH3

CH CHO 2 1

6. IUPAC name of CH3 CHO is: (A) Acetaldehyde (C) Methyl aldehyde

(B) Ethanal (D) Formalin

7. IUPAC name of CH3 CH(OH)CH2CH2COOH is: (A) 4 – hydroxy pentanoic acid (C) 1 – carboxy – 4 – butanol

(B) 1 – carboxy – 3 – butanol (D) 4 – carboxy – 2 – butanol

8. The IUPAC name of the following compound is: H2 C CH CH2

CN CN CN (A) 1, 2, 3 – tricyano propane (C) 1, 2, 3 – propane tricarbonitrile

(B) Propane tricarbylamine (D) 3 – cyanopropane – 1, 5 – dinitrile

9. The structure of 4 – methyl – 2 – penten – 1 – ol is: (A) (CH3 )2 C = CHCH2 CH2OH (B) (CH3 )2 CHCH = CHCH2 OH CH CH CH = CHCH OH (C) (D) CH3 CHOHCH = C(CH3 )2 3 2 2 10. The IUPAC name for the following compound is: OH

CH3

OH

C

CH2

CH

CH3

CH3

(A) (B) (C) (D)

1, 1 – dimethyl – 1, 3 – butandiol 2 – methyl – 2, 4 – pentandiol 4 – methyl – 2, 4 – pentandiol 1, 3, 3 – trimethyl – 1, 3 – propandiol

11. What is the decreasing order of stability of the ions? +

(I) CH3  CH CH3 (A) I > II > III (C) III > I > II

+

(II) CH3  CH OCH3

+

(III) CH3  CH COCH3 (B) II > III > I (D) II > I > III

12. The C  H bon distance is longest in (A) C2H2 (C) C2H6

(B) C2H4 (D) C2H2Br2

13. The most stable free radical among the following �

(A) C6H5 CH2 CH2 �

(C) CH3 C H2

�

(B) C6H5 C HCH3 �

(D) CH3 C HCH3

14. Theorder of decreasing stability of the carbanions (CH3)3C (I) ; (CH3)2CH (II); CH3CH2 (III); C6H5CH2 (IV) is (A) I > II > III > IV (B) IV > III > II > I (C) IV > I > II > III (D) I > II > IV > III 15. The most stable carbanion among the following is H 2C CH2 (A)

(B)

CH2

CH2

(C)

(D)

OCH3

CH2

NO 2

16. Among the following alkene: 1 – butene (I), cis 2- butene (II), trans 2 – butene (III), the decreasing order of stability is (A) III > II > I (B) III > I > II (C) I > II > III (D) II > I > III 17. The arrangement of (CH3)3C-, (CH3)2CH-, CH3CH2- when attached to benzene or unsaturated group in increasing order of activating effect is (A) (CH3)3C < (CH3)2CH  < CH3CH2  (B) CH3CH2  < (CH3)2CH  < (CH3)3C  (C) (CH3)2CH < (CH3)3C  < CH3CH2  (D) (CH3)3C < CH3CH2  < (CH3)2CH  18. Decreasing I power of given groups is (1) CN (2) NO2 (3) NH3 (A) 2 > 1 > 4 > 3 (C) 3 > 2 > 4 > 1

(4) F (B) 2 > 3 > 4 > 1 (D) 3 > 2 > 1 > 4

19. A free radical is (A) neutral is character (C) paramagnetic

(B) shortly lived (D) all the above

20. Which amongst the following will be most stable? +

+

(B) ( CH3 ) 2 CH

(A) CH3 CH2

+

+

(C) ( CH3 ) 3 CCH2

(D) ( CH3 ) 3 C

21. The most stable carbocation amongst the following is +

+

(A) ( CH3 ) 2 CH

(B) Ph3 C +

+

(C) CH3 CH2

(D) CH2 = CH  CH2

22. Which of the following exerts +I effect? (A) CH3  (C) NH2

(B) Cl (D) NO2

23. Which of the following species has/have electron releasing effect? (A) CHO (B) NO2 (C) CH3 (D) C6H5 24.

+R power of the given groups

(1) (4) (A) 1 > 2 > 3 > 4 (C) 1 > 3 > 2 > 4

O NHCOCH3

(2)

NH2

(3)

in decreasing order is (B) 4 > 3 > 2 > 1 (D) 1 > 4 > 3 > 2

OH

25. Which statement is correct for inductive effect? (A) it is a permanent effect (B) it is the property of single bond (C) it causes permanent polarization in the molecule (D) all are correct

Level – II 1. The IUPAC name of Cl3 CCH2 CHO is (A) Chloral (C) 1, 1, 1 – trichloropropanal

(B) 3, 3, 3 – trichloropropanal (D) 2, 2, 2 – trichloropropanal

2. The formula of propanenitrile is (A) CH3 CN (C) C3H7 CN

(B) C2H5 CN (D) C2H5NC

3.

CH3

The correct name of

CH2 C

CH2 is : C

CH

(A) 3 – hexyn – 5 – ene (C) 3 – hexyn – 1 – ene

(B) 5 – hexen – 3 – yne (D) 1 – hexen – 3 – yne

4. Which compound is 2, 2, 3 – trimethylhexane CH3 CH3 (A) CH3 C CH CH2 CH3

CH3

(B)

CH3

CH3 CH3

(C) CH3

CH

CH2

CH2

CH3

5.

IUPAC name of

(A) (B) (C) (D)

C

CH3

CH3

H3C

CH

CH2

CH2

CH

CH3

CH3 CH3 CH3

(D) CH3

C

CH3

C

CH

CH2

CH2

CH3

CH3 CH

CH2Cl is :

C 2H 5 OH 1 – chloro – 4 – methyl – 2 – hexanol 1 – chloro – 4 – methyl – hexanal – 2 1 – chloro – 4 – ethyl – 2 – pentanol 1 – chloro – 2 – hydroxy – 4 – methyl hexane

6. In which class of compounds will the following compound be named? ClCH2 CH CH2 C CH2NO2

OH O (A) Nitro compounds (C) Alkyl halides

(B) Alcohols (D) Ketones

7. Which one of the following IUPAC names is incorrect? (A) Ethanoic acid (B) Ethanal (C) pent – 3 – ene (D) 3 – methyl – pentan – 2 – ol 8. The IUPAC name of the compound CH3 CONHBr is: (A) 1 – bromoacetamide (B) ethanoyl bromide (C) N – bromoethanamide (D) None of these

CH2 9.

The IUPAC name of

C2 H5 C (A) 4 – amino – 2 – ethyl – 1 – pentene (C) amino – 4 – pentene

CH3 CH2

CHNH2 is : (B) 2 – ethyl – 4 – pentanamine (D) 4 – ethyl – 4 – penten – 2 – amine

O C2H5

10.

C O is :

The IUPAC name of CH3

C O

(A) ethoxy methanone (C) ethanoic propanoic anhydride

(B) ethyl – 2 – methyl propanoate (D) 2 – methyl ethoxy propanone

11. Which statement is correct for electromeric effect? (A) it is a temporary effect (B) it is the property of  bond (C) it takes place in presence of reagent, i.e. electropile or nucleophile (D) all are correct 12. Which one of the following is most acidic? (A) phenol (C) 3 – chlorophenol

(B) 2 – chlorophenol (D) 4 – chlorophenol

13. Which one of the following is most – acidic? (A) 2 – propanol (C) ethanol

(B) 2 – methyl – 2 – propanol (D) methanol

14. Consider the following carbanions

H3C CH2 CH (1) (2) H2C (3) HC Correct order of stability of these carbanions in decreasing order is (A) 1 > 2 > 3 (B) 2 > 1 > 3 (C) 3 > 2 > 1 (D) 3 > 1 > 2

C

15. Consider the following carbocations (1)

H3C

CH2

(2)

H2C

CH

(3)

H2 C

CH

CH2

C6H5CH2 (4) Stability of these carbocation in decreasing order is (A) 4 > 3 > 1 > 2 (B) 4 > 3 > 2 > 1 (C) 3 > 4 > 2 > 1 (D) 3 > 4 > 1 > 2 16.

Which among the following carboncations is most stable?

(A)

(C)

(B)

C

C6 H5

CH

C6 H5

(D)

C6 H5

CH2

H3 C

C CH3

CH3

17. Arrange basicity of the given compounds in decreasing order (1) CH3  CH2  NH2 (2) CH2 = CH  NH2 (3) CH  C  NH2 (A) 1 > 2 > 3 (C) 3 > 2 > 1

(B) 1 >3 > 2 (D) 2 > 3 > 1

18. Arrange the following groups in order of decreasing deactivating effect (1) NO2 (2) SO3H (3 CF3 (4) CHO (A) 1 > 3 > 2 > 4 (B) 1 > 2 > 3 > 4 (C) 1 > 4 > 3 > 2 (D) 4 > 3 > 2 > 1 19.

Consider the following compounds NH2

NH2

NH2

NH2

(I)

NO 2

CN

CH3

(II)

(III)

(IV)

Arrange these compounds in deceasing order of their basicity (A) 1 > 2 > 3 > 4 (B) 2 > 3 > 1 > 4 (C) 4 > 1 > 3 > 2 (D) 4 > 1 > 2 > 3 20.

Which one of the following is vinyl carbocation

(A) (C)

21.

H5 C6

CH2

H2 C

CH

(B) (D)

CH2

CH

CH2

All of these

Which one of the following is most basic?

NH2

(A)

NH2

(B)

CH3

NH2

(C) H3C

NH2

(D) CH3

H3C

22. Which of these species are electrophiles? (A) FeCl3 (C) AlCl3

(B) BF3 (D) all of these

CH3

23. Arrange given compounds in order of decreasing acicity CH3  NO2 (1) (2) NO2  CH2  NO2 NO 2 (4) O 2N

(A) 4 > 2 > 1 > 3 (C) 3 > 1 > 2 > 4

CH

(3)

CH3  CH2  NO2

NO 2

(B) 4 > 2 > 3 > 1 (D) 3 > 1 > 4 > 2

24. Which one of the following has the highest nucleophilicity? (A) F (B) OH (C) CH3 (D) NH2 25. The stability of given free radical in decreasing order is (1) (2) H3C H3C CH2 CH CH3

(3)

H3C

C CH3

(4)

CH3

(A) 3 > 4 > 1 > 2 (C) 3 > 2 > 4 > 1

(B) 1 > 2 > 3 > 4 (D) 3 > 2 > 1 > 4

CH3

ANSWERS TO ASSIGNMENT PROBLEMS Subjective: Level - O

1. (a)

(b)

(c)

H

H

H

CH3 H

H

H

H

C

C

C

C

C

H H

H H

H H

H

H

C

C

C

H H

Cl H

H CH3

C

C

H

(d) H3C

H Hexane

Cl 1, 2 dichloro propane

C

Cl 3 chloro butan - 2 - ol

OH H CH3 CH3

C

C

H

H

H

CH3 2, 3 dimethyl butane H

H

(e) H

C

C

C

C

H

O

H

H

COOH

4 keto pentanoic acid

CH3

2.

H3C

C

O

CH3

2 methoxy 2 methyl propane

CH3 H

3.

H

C

CN Ethane nitrile

H

4. 1° - five, 2° - two, 3° - one, 4 – one 5. 2 (2- methyl) phenyl ethanal 12. Oxidation of D – (+) – aldehyde does not affect any of the double bonds to the chiral C. The acid has the same configuration even through the sign of rotation is changed. CHO COOH

H

C

OH

CH2OH D-( )-glyceraldehyde

H

C

OH

CH2OH D-(-)-glyceric acid

13.

CH3

CH3

CH3 I

I

H

H

H H

I

I C 2H 5

C 2H 5

C2H5

CH3

C2H5

14. The high electron negativity of a halogen polarizes a C  X bond so that carbon bears partial positive. This leaves the ring electron deficient and makes electrophilic attach difficult. The inductive deactivation is greater at the ortho – than at the para – position because its influence decrease with distance from the electronegative halogen atom

Cl d 

+d E

15. NO2 group is m – directing while NH2 and OH groups are O  and p  directing. NH2 NO 2 (a) (b) Br

m - bromonitrobenzene Br

O - (also p-) bromoaniline

OH

(c)

CH3

O - (also p-) cresol

Level – I 1. Cyclo dodecane 2. 4 – ethyl – 2 – cyclopropyl – 1 – hexene 3. 1, 2 – ethanedioic acid 4. The magnitude of inductive effect decreases with distance and hence the effect is least in C 2 C3 bond 2 1 3 H3C CH2 CH2 Br 5. (a)

O

O H3C

H3C

C

O

O

(b)

NH 2

C

NH 2

NH 2

NH2

NH2

6. Both these structures involve sepetration of charge and hence do not contribute substanitially towards the resonance hybrid. Further, the contribution of I is lower than that of structure II since C atom has only a sextet of electrons. 7. (a)

H2 C

CH

CH

OH

less stable due to charge seperation.

-H

H 2C

H 2C

OH

CH

H 2C

O

CH

O

More stable due to absence of charge seperation After loss of H+ ion vinyl alcohol stabilizes by resonance, hence it will favour loss of H + and hence acidic in nature. 8. Order of basic strength is as follows:

9.

H3C

C

R

CH2

-H

C

R

CH2

O

O

C

R

O

Since after the loss of a - H remaining part of alhehyde, or ketone stabilizes by resonance, hence, it will favaour the loss of H+ ions hence it is acidic in nature. 10. III < I < II The conjugate bases of the 3 compounds are stabilized by equivalent resonance as shown below O

O

O

O

O

O

O

O

-M

O

M NH2

N

N

(I)

O

O

O

O

O

O

O

NH2

(III)

(II)

11. III < II < I As we go from I  II  III S character of the hybridized orbital of a - c decreases (sp  sp2  sp3) hence the acidic character decreases. 12. (a)

CH3 CH3 CH3CH2C

C H

CH2CH3 3, 4 dimethy hexane

(b) The chiral C in this alkane must be attached to the four simplest alkyl group. CH3 H2CH3C

C

H

CH2 CH2 CH3 3 methyl hexane

13. (a) HCl

Cl ( 1, 4 - addition product)

CH3

(b)

Cl

CH3 CH3

H3C

HCl

CH3

H3C

Cl

CH3

H3C H3C

H3C

(c)

H3C HCl

CH2 H3C

(d)

H3C Cl

CH3

H3C H3C

H3C

Ph

Ph

Ph HCl

H3C

CH3

Cl

Cl H3C

Cl H3C

CH3

CH3

14. H2C

CHCH

CH2

H 2C CH CHCH2 HCH2C CHCH2 Due to mesomeric effect, there is polarity on C 1 and C4 resulting in the 1, 4 addition at these sites. This type of behaviours is generally obtained in conjugated systems. H2C

CH

CH

CH2

H

H2C CH CH CH3 Alkyl carbocation (resonance stabilized)

Br H2C

15.

CH

(a)

CH

BrH2C CH CH CH3 1, 4 - addtion product

CH3 NH2

(b)

NO2

(c)

O HOC

O COH

(d)

OH CH3

Level – II 1. 1 amino 4, 5 diethyl 2 methyl non – 3 - ene 2. 3 ethyl 3 methyl pentene 3. A, C & D 4. (a)

d or

d

d F

(b)

F

O

O

F

O

O

N

F

O

O

N

F

O

O

O

O

N

N

N

d

d

or

d

In each case the para and two ortho positions bear some of the charge (ve charge in fluorobenzene and +ve charge in nitrobenzene) 5. Because the nitrogen atom of NH2 contains lone pairs of electrons it donates electron density to the benzene ring, so it is ortho para directing but when it is passed in acidic medium then it converts into. Which is meta – directing?

NH3

(anilinium ion)

6. CO32 ion is a resonance hybrid of the following three resonating structure of equal stability. O

O

O

O

O

O

O

O

O

2

O

O

O

HCOO- is a resonance hybrid of only two resonating structure of (as shown below) of equal stability. O

H

O

O

H

O

O

H

O

Thus is the resonance hybrid of carbonate ions each C  O bond has 2/3rd signal bond character and 1/3rd doubled bond character while HCOO  ion each C  O bond has half single bond and half doubled bond character. Greater the single bond character of any bond larger that bond. 7.

8. II < I < III

List – I (a) (b) (c) (d)

List – II (b) (c) (a) (d)

9. IV < I < II < II 10. IV < I < II < III 11. II < III < I < IV 12. I < II < III 13.

C6H5 < CF 3 < CH2Cl < C6H5CH2

14. (a) I < III < II (Intramolecular H – bonding in III makes it less acidic than p – isomer) (b) II < I < III (c) I < III < II 15. (a) IV < II < III < I (b) III < I < II < IV (c) II < III < I < IV

Objective:

Level - I 1.

A

2.

D

3.

C

4.

A

5.

D

6.

B

7.

A

8.

C

9.

B

10.

B

11.

D

12.

C

13.

B

14.

B

15.

D

16.

A

17.

A

18.

A

19.

D

20.

D

21.

B

22.

A

23.

C

24.

A

25.

D

Level - II 1.

B

2.

B

3.

D

4.

D

5.

A

6.

D

7.

C

8.

C

9.

D

10.

C

11.

D

12.

B

13.

D

14.

C

15.

A

16.

A

17.

A

18.

A

19.

C

20.

C

21.

C

22.

D

23.

A

24.

C

25.

D