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1. HOMOLOGOUS SERIES

Organic compounds can be divided into different groups on the basis of similarity in their structure and

properties. The property by which a number of organic compounds form a homologous series is termed as homology.

The general characteristics of a homologous series are

All compounds in the series have the same functional group.

All compounds in the series can be represented by the same general formula and can be prepared

by the general methods of preparation.

All the homologues show a gradual gradation in their physical and chemical properties.

Successive members of a homologous series differ by CH2 group and by a mass of 14 units.

Table 1a: Homologous series of aliphatic organic compounds

Name of the homologous series

General formula Functional group or substituent

IUPAC name

Alkanes CnH2n + 2 (single bond) Alkanes

Alkenes CnH2n = (double bond) Alkenes

Alkynes CnH2n 2 (triple bond) Alkynes

Monohydric alcohols CnH2n + 1 OH OH Alkanols

Aldehydes CnH2nO CHO Alkanals

Ketones CnH2nO C = O

Alkanones

Monocarboxylic acid CnH2nO2 COOH Alkanoic acids

Ethers CnH2n + 2 O COC Alkoxy alkanes

Primary amines CnH2n + 1 NH2 NH2

Alkanamines

Amides CnH2n + 1 CONH2 CONH2

Alkanamides

Esters CnH2nO2 R COO R Alkyl alkanoate

Cyanides CnH2n + 1CN CN Alkane nitriles

Nitro compounds CnH2n + 1

NO2 N

O

O

Nitroalkanes

Acid chlorides CnH2n + 1 COCl COCl Alkanoyl chlorides

2. NOMENCLATURE OF ORGANIC COMPOUNDS

Nomenclature implies assigning proper name to a particular organic compound on the basis of certain

standard rules so that the study of these compounds may become systematic. In the IUPAC system, the name

of an organic compound consists of three parts. (i) Word root (ii) Suffix (iii) Prefix

WORD ROOT

The word root denotes the number of carbon atoms present in the chain. For example,

Chain length Word root Chain length Word root

C1 Meth C6 Hex

C2 Eth C7 Hept

C3 Prop C8 Oct

C4 But C9 Non

C5 Pent C10 Dec

SUFFIX

The word root is linked to the suffix, which may be primary, secondary or both. Primary suffix

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It indicates the nature of linkages between the carbon atoms. For example,

ane for single bonded compounds, CC ;

ene for double bonded compounds, C=C ;

and yne for triple bonded compounds, CC. Secondary suffix

It indicates the presence of functional group in the organic compound. For example,

Class of organic compound Functional group Secondary suffix

Alcohols OH ol

Aldehydes CHO al

Ketones C = O one

Carboxylic acids COOH oic acid

Esters COOR alkyl…..oate

Acid chlorides COCl oyl chloride

Acid amides CONH2 amide

Nitriles CN nitrile

Amines NH2 amine

PREFIXES

There are many groups, which are not regarded as functional groups in the IUPAC naming of the

compounds. These are regarded as substituents or side chains. These are represented as prefixes and are placed before the word root while naming a particular compound. For example,

if a compound contains more than one functional group, then one of the functional group is regarded as principal functional group and is treated as secondary suffix. The other functional groups are regarded as

substituents and are indicated by prefixes.

Substituent Prefix Substituent Prefix

CnH2n+1 Alkyl NH2 Amino

F Fluoro NO Nitroso

Cl Chloro N=N Diazo

Br Bromo OCH3 Methoxy

I Iodo OC2H5 Ethoxy

NO2 Nitro OH Hydroxy

Thus, a complete IUPAC name of an organic compound may be represented as Prefix + Word root + Primary suffix + Secondary suffix.

HOW TO NAME ORGANIC COMPOUNDS USING THE IUPAC RULES

In order to name organic compounds you must first memorize a few basic names. These names are listed within the discussion of naming alkanes. In general, the base part of the name reflects the number of

carbons in what you have assigned to be the parent chain. The suffix of the name reflects the type(s) of

functional group(s) present on (or within) the parent chain. Other groups which are attached to the parent chain are called substituents.

2.1 RULES FOR NAMING ALKANES – SATURATED HYDROCARBONS

The name of straight chain saturated hydrocarbons for up to 12 carbon chain are shown below. The

names of the substituents formed by the removal of one hydrogen from the end of the chain is obtained by changing the suffix –ane to –yl.

Number of Carbons Name

1 methane

2 ethane

3 propane

4 butane

5 pentane

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6 hexane

7 heptane

8 octane

9 nonane

10 decane

11 undecane

12 dodecane

There are a few common branched substituents which you should memorize. These are shown below.

CH3CH—

CH3

isopropyl

CH3CH2CH—

CH3

sec-butyl

CH3–C—

CH3

tert-butyl

CH3

CH3CHCH2—

CH3

isobutyl

Here is a simple list of rules to follow. Some examples are given at the end of the list.

1. Identify the longest carbon chain. This chain is called the parent chain.

2. Identify all of the substituents (groups attached to the parent chain).

3. Number the carbons of the parent chain from the end that gives the substituents the lowest numbers.

When comparing a series of numbers, the series that is the ―lowest‖ is the one which contains the lowest number at the occasion of the first difference. If two or more side chains are in equivalent

positions, assign the lowest number to the one which will come first in the name.

4. If the same substituent occurs more than once, the location of each point on which the substituent

occurs is given. In addition, the number of times the substituent group occurs is indicated by a prefix

(di, tri, tetra, etc.).

5. If there are two or more different substituents they are listed in alphabetical order using the base name

(ignore the prefixes). The only prefix which is used when putting the substituents in alphabetical order is iso as in isopropyl or isobutyl. The prefixes sec– and tert– are not used in determining alphabetical

order except when compared with each other.

6. If chains of equal length are competing for selection as the parent chain, then the choice goes in series

to : (a) the chain which has the greatest number of side chains.

(b) the chain whose substituents have the lowest– numbers. (c) the chain having the greatest number of carbon atoms in the smaller side chain.

(d) the chain having the least branched side chains. 7. A cyclic (ring) hydrocarbon is designated by the prefix cyclo– which appears directly in front of the

base name. In summary, the name of the compound is written out with the substituents in alphabetical

order followed by the base name (derived from the number of carbons in the parent chain). Commas

are used between numbers and dashes are used between letters and numbers. There are no spaces in the name. Here are some examples:

CH3

4–Ethyl–2–methylhexane

CH3–CH–CH2–CH–CH2–CH3

CH2–CH3 CH3

4–Ethyl–3,3–dimethylheptane

CH3–CH2–CH2–CH––C–CH2–CH3

CH3–CH2

CH3

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1

CH3––CH2––CH––CH––CH––CH––CH3

2,3,5–Trimethyl–4–propylheptane

CH2 CH3 CH3

CH2

CH3

2 3 4

5 6 7

5

6

7

(NOT: 2,3–dimethyl–4–sec–butylheptane)

CH

CH2–CH2

CH3

Methylcyclopropane

CH3

CH3–CH–CH2–CH2–CH–CH2–CH–CH2–CH3

CH3 CH3

CH–CH3

CH2–CH3

5–Sec–butyl–2, 7–dimethylnonane

CH3–CH2–CH–CH–CH2–CH3

CH3 CH2

CH3

3–Ethyl–4–methylhexane

1 2 3 4 5 6 7 8 9

5 4 3 2 1 6

2.2 RULES FOR NAMING ALKENES AND ALKYNES – UNSATURATED HYDROCARBONS

Double bonds in hydrocarbons are indicated by replacing the suffix –ane with –ene. If there is more

than one double bond, the suffix is expanded to include a prefix that indicates the number of double bonds present (–adiene, –atriene, etc.). Triple bonds are named in a similar way using the suffix –yne.

The position of the multiple bond(s) within the parent chain is(are) indicated by placing the number(s) of the first carbon of the multiple bond(s) directly in front of the base name.

Here is an important list of rules to follow: 1. The parent chain is numbered so that the multiple bonds have the lowest numbers (double and

triple bonds have priority over alkyl and halo substituents).

2. When both double and triple bonds are present, numbers as low as possible are given to double and triple bonds even though this may at times give ―–yne‖ a lower number than ―–ene‖.

When there is a choice in numbering, the double bonds are given the lowest number. 3. When both double and triple bonds are present, the –en suffix follows the parent chain directly

and the –yne suffix follows the –en suffix (notice that the ‗e‘ is left off, –en instead of –ene). The location of the double bond(s) is(are) indicated before the parent name as before, and the

location of the triple bond(s) is(are) indicated between the –en and –yne suffixes.

4. For a branched unsaturated acyclic hydrocarbon, the parent chain is the longest carbon chain that contains the maximum number of double and triple bonds. If there are two or more

chains competing for selection as the parent chain (chain with the most multiple bonds), the choice goes to (1) the chain with the greatest number of carbon atoms, (2) the number of

carbon atoms being equal, the chain containing the maximum number of double bonds.

5. If there is a choice in numbering not previously covered, the parent chain is numbered to give

the substituents the lowest number at the first point of difference.

Here are some examples:

CH3–CH=CH–CH2–CH=CH2

1,4–Hexadiene

CHC–CH=CH–CH=CH2

1,3–Hexadien–5–yne

1 2 3 4 5 6 1 2 3 4 5 6

CH3–CH=CH–CCH

3–Penten–1–yne

CH3–C–CH2–CH2–CH=CH2

5,5–Dimethyl–1–hexene

CH3

CH3

1 2 3 4 5 1 2 3 4 5 6

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CHC–C=C–CH=CH2 1 2

CH2

CH2–CH3

3 4 5 6

3,4–Dipropyl–1,3–hexadien–5–yne 1,4,4-Trimethylcyclobutene

CH2–CH2–CH3 1

2

3 4

2.3 RULES FOR NAMING ALKYL HALIDES

The halogen is treated as a substituent on an alkane chain. The halo–substituent is considered of

equal rank with an alkyl substituent in the numbering of the parent chain. The halogens are represented as

F fluoro–

Cl chloro–

Br bromo–

I Iodo–

Here are some examples:

CH3–CH–CH–CH2–CH3

CH3

3–Chloro–2–methylpentane

Cl

CH3–CH–CH–CH3

CH3

2–Bromo–3–methylbutane

Br 1 2 3 4 5 4 3 2 1

2.4 RULES FOR NAMING ALCOHOLS

Alcohols are named by replacing the suffix –ane with –anol. If there is more than one hydroxyl group

(–OH), the suffix is expanded to include a prefix that indicates the number of hydroxyl groups present

(–anediol, –anetriol, etc.). The position of the hydroxyl group(s) on the parent chain is(are) indicated

by placing the number(s) corresponding to the location(s) on the parent chain directly in front of the

base name (same as alkenes).

Here is an important list of rules to follow:

1. The hydroxyl group takes precedence over alkyl groups and halogen substituents, as well as

double bonds, in the numbering of the parent chain.

2. When both double bonds and hydroxyl groups are present, the –en suffix follows the parent

chain directly and the –ol suffix follows the –en suffix (notice that the ‗e‘ is left off, –en instead of

–ene). The location of the double bond(s) is (are) indicated before the parent name as before,

and the location of the hydroxyl group(s) is(are) indicated between the –en and –ol suffixes.

Again, the hydroxyl gets priority in the numbering of the parent chain.

3. If there is a choice in numbering not previously covered, the parent chain is numbered to give

the substituents the lowest number at the first point of difference.

Here are some examples:

CH3–CH–CH–CH2–CH3

OH

CH3

3Methyl2pentanol

CH3–CH–CH–CH3

OH

2,3Butanediol

OH OH

2Cyclopenten1ol

1 2 3 4 5 1 2 3 4 1

2

3

2.5 RULES FOR NAMING ETHERS

The generic IUPAC name of this family is ―alkoxy alkane‖ in which alkane part contains longer carbon

chain in case of unsymmetrical ethers.

CH3OCH2CH3 CH3CH2OCH2CH3

Methoxy ethane Ethoxy ethane

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CH3–OCH2CH2CH3

1–Methoxy propane

CH3–O–CH

2–Methoxy propane

CH3 3 2 1

1

2

3 CH3

2.6 RULES FOR NAMING ALDEHYDES Aldehydes are named by replacing the suffix –ane with –anal. If there is more than one

–CHO group, the suffix is expanded to include a prefix that indicates the number of –CHO groups present (–

anedial –there should not be more than 2 of these groups on the parent chain as they must occur at the ends).

It is not necessary to indicate the position of the –CHO group because this group will be at the end of the

parent chain and its carbon is automatically assigned as C-1. Here is an important list of rules to follow:

1. The carbonyl group takes precedence over alkyl groups and halogen substituents, as well as double bonds, in the numbering of the parent chain.

2. When both double bonds and carbonyl groups are present, the –en suffix follows the parent chain directly and the –al suffix follows the –en suffix (notice that the ‗e‘ is left off, –en instead –ene). The

location of the double bond(s) is(are) indicated before the parent name as before, and the –al suffix

follows the –en suffix directly. Remember it is not necessary to specify the location of the carbonyl group because it will automatically be the first carbon. Again, the carbonyl gets priority in the numbering

of the parent chain. 3. There are a couple of common names which are acceptable as IUPAC names. They are shown in the

examples in parenthesis. Here are some examples:

O

Propanal

O

3–Methylbutanal

O

3–Methyl–3–butenal

HCH

O

Methanal (Common name: Formaldehyde)

CH3CH

O

Ethanal (Common name: Acetaldehyde)

CH

O

Benzaldehyde 2.7 RULES FOR NAMING KETONES Ketones are named by replacing the suffix –ane with –anone. If there is more than one carbonyl group

(C=O), the suffix is expanded to include a prefix that indicates the number of carbonyl groups present (–anedione, –anetrione, etc.). The position of the carbonyl group(s) on the parent chain is(are)

indicated by placing the number(s) corresponding to the location (s) on the parent chain directly in front

of the base name (same as alkenes). Here is an important list of rules to follow:

1. The carbonyl group takes precedence over alkyl groups and halogen substituents, as well as double bonds, in the numbering of the parent chain.

2. When both double bonds and carbonyl groups are present, the –en suffix follows the parent chain directly and –one suffix follows the –en suffix (notice that ‗e‘ is left off, –en instead of –ene). The

location of the double bond(s) is(are) indicated before the parent name as before, and the location of

the carbonyl group(s) is (are) indicated between the –en and –one suffixes. Again, the carbonyl gets priority in the numbering of the parent chain.

3. If there is a choice in numbering not previously covered, the parent chain is numbered to give the substituents the lowest number at the first point of difference. Here are some examples:

O

Propanone

O

(Common name: Acetone)

2–Butanone

O

2,4–Pentanedione 3–Methyl–3–buten–2–one

O O

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2.8 RULES FOR NAMING CARBOXYLIC ACIDS

Carboxylic acids are named by counting the number of carbons in the longest continuous chain including the carboxyl group and by replacing the suffix –ane of the corresponding alkane with –anoic

acid. If there are two –COOH groups, the suffix is expanded to include a prefix that indicates the number of –COOH groups present (–anedioic acid – there should not be more than 2 of these groups

on the parent chain as they must occur at the ends). It is not necessary to indicate the position of the –COOH group because this group will be at the end of the parent chain and its carbon is automatically

assigned as C–1.

Here is an important list of rules to follow: 1. The carboxyl group takes precedence over alkyl groups and halogen substituents, as well as

double bonds, in the numbering of the parent chain.

2. If the carboxyl group is attached to a ring, the parent ring is named and the suffix

–carboxylic acid is added.

3. When both double bonds and carboxyl groups are present, the –en suffix follows the parent chain directly and the –oic acid suffix follows the –en suffix (notice that the ‗e‘ is left off, –en

instead of –ene). The location of the double bond(s) is(are) indicated before the parent name as

before, and the –oic acid suffix follows the –en suffix directly. Remember it is not necessary to specify the location of the carboxyl group because it will automatically be the first carbon. Again,

the carboxyl gets priority in the numbering of the parent chain.

4. There are several common names that are acceptable as IUPAC names. They are shown in

the examples in the parenthesis.

5. If there is a choice in numbering not previously covered, the parent chain is numbered to give the substituents the lowest number at the first point of difference.

HCOH

O

(Common name: Formic acid)

Methanoic acid

CH3COH

O

(Common name: Acetic acid)

Ethanoic acid

HO

3–Methylpentanoic acid

COH

O

Benzoic acid

O

OH O

2Hydroxybenzoic acid (Common name: Salicyclic acid)

COOH HOCCOH

O

Ethanedioic acid (Common name : Oxalic acid)

O

3–Butenoic acid

HO

2.9 RULES FOR NAMING ESTERS

Systematic names of esters are based on the name of the corresponding carboxylic acid. Remember esters look like this:

R–C––O––R

O

acyl group alkyl group The alkyl group is named like a substituent using the –yl ending. This is followed by a space. The acyl

portion of the name (what is left over) is named by replacing the –ic acid suffix of the corresponding carboxylic acid with –ate.

Here are some examples:

CH3CH2COCH3

O

Methyl propanoate

COCH2CH3

O

CH3COC(CH3)3

O

Ethyl benzoate Tert–butyl acetate or

Tertbutylethanoate

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2.10 RULES FOR NAMING AMINES

Primary amines (RNH2) are named as alkanamine. If the compound contain double bond along with

amino group, then it is named alkanamine. Alkyl and halo substituents are named as prefix in

alphabetical order.

For example,

CH2=CHCH2NH2

2–Propenamine

CH3–CH–CH2–CHNH2

1Chloro2methylbutanamine

Cl

CH3

1 2 3 1 2 3 4

CH3CH2–NH2 CH3–CH2–CH2NH2

Ethanamine Propanamine

Secondary amines (R2NH) and tertiary amines (R3N) are named as Nalkyl alkanamine and

N,Ndialkyl alkanamine.

CH3CH2NCH3

N,NDimethylethanamine

CH3CH2CH2–NCH2CH3

NEthylNmethylpropanamine

CH3

CH3CH2–NHCH3 CH3CH2CH2NHCH2CH3

NMethylethanamine NEthylpropanamine

CH3

2.11 RULES FOR NAMING ORGANIC COMPOUNDS CONTAINING ONE OR MORE FUNCTIONAL

GROUPS

The rules for naming an organic compound containing functional groups are exactly same as discussed already for compounds containing double and triple bonds. In this case, the preference of lowest

number is given to carbon atom bearing the functional group. The rules are summarized below:

(i) Select the longest continuous chain containing the carbon atom having functional group(s). (ii) The numbering of atoms in the parent chain is done in such a way that carbon atom bearing the

functional group gets the lowest number.

(iii) It two or more same functional groups are present, these are indicated by using di, tri, tetra as prefixed

to the name of the functional group.

(iv) If the organic compound contains a functional group, multiple bonds, side chain or substituents, the

following order of preference must be followed,

Functional group > Double bond > Triple bond > Side chain. (v) When an organic compound contains two or more functional groups, one group is regarded as the

principal functional group and the other group is treated as the secondary functional group, which may be treated as substituent(s). The following order of preference is used for selecting the principal

functional group,

Carboxylic acids > sulphonic acids > acid anhydrides > esters > acid chlorides > amides > nitriles > aldehydes > ketones > alcohols > amines > imines > ethers >

alkenes > alkynes. Different classes of functional groups including multiple bonded compounds and the suffix or prefix

required to name these compounds are given in the preferential decreasing order in the following table.

Class of compounds Functional group or

substituent Suffix Prefix

1. Carboxylic acids C = O HO

Carboxylic acid/oic acid carboxy

2. Sulphonic acids

O

SOH

O

Sulphonic acid sulfo

3. Acid anhydrides

COC

O O

oic anhydride

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4. Esters C = O RO

(R) … carboxylate /(R)…oate

alkoxy carbonyl

5. Acid halides C = O X

Carbonyl halide/oyl halide

halo carbonyl

6. Amides C = O NH2

Carboxamide/amide carbamoyl

7. Cyanides CN Carbonitrile/nitrile cyano

8. Aldehydes C=O H

Carbaldehyde/al formyl/oxo

9. Ketones C=O

one keto/oxo

10. Alcohols OH ol hydroxy

11. Amines NH2 amine amino

12. Imines =NH imine imino

13. Ethers COC

alkoxy

14. Alkenes = (double bond) ene

15. Alkynes (triple bond) yne

For those functionalities, which have two prefixes and/or suffixes, the first one is used when carbon

atom of the functional group is not a part of the longest continuous chain and the second one is used, when carbon atom is counted in the longest chain. 2.12 RULES FOR NAMING AROMATIC COMPOUNDS

(i) When the benzene ring is present as one of the substituent in the carbon chain, it is named as phenyl

group. (ii) Disubstituted, trisubstituted or tetrasubstituted benzenes are named by using the numbers.

(iii) If different groups are attached to the benzene ring, then the principal functional group is fixed as

number 1. The numbering of the chain is done in any direction (clockwise or anticlockwise), which

gives lowest number to the substituents. The substituents are written in the alphabetical order. Let us see the application of these IUPAC rules to some organic compounds.

(i) (HOCH2CH2O)2CHCO2H Bis(2hydroxyethoxy)ethanoic acid

(ii)

C2H5

CH3

1 2

3

4

1Ethyl4methylcyclohaxane

(iii)

CH3(CH2)4CH2CHCH2CH(CH2)3CH3

CH3CHCH

F F

CH2CH3

7(1,2Difluorobutyl)5ethyltridecane

(iv)

CO2H

O

1 2

3

4

4Oxo1,2,3,4tetrahydronaphthalene

1carboxylic acid

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(v)

CH3CH2CH2CCH2Cl

O

1 2 3 4 5

1chloropentan2one

(vi)

1,2Didehydrobenzene

(vii) CH3CH2OCH2CH2Cl

1 2

1Chloro2ethoxyethane

(viii) OHCCH2CH2CHCH2CHO

CH2CHO

3(Formylmethyl)hexanedial

(ix) OHCCH2CH2CHCH2CHO

CHO

1 2 3 4

Butane1,2,4tricarbaldehyde

(x) CH3CH2CHOCH2CH3

OCH2CH3

1,1Diethoxypropane

(xi) HO2CCH2CH2CHCH2CH2CO2H

CO2H

Pentane1,3,5tricarboxylic acid.

(xii) NCCH2CH2CH2CHCH2CH2CN

CN

Hexane1,3,6tricarbonitrile

(xiii)

OHC 1

2 3

4 CO2H

O

4Formyl2oxocyclohexane1carboxylic acid

(xiv) CH3CSOCOCH2CH3 Propionic thioacetic anhydride

(xv)

CONH2 Cyclohexanecarboxamide

(xvi)

CH3CCH2CH2CH2CO2H

O

1 2 3 4 5 6

5Oxohexanoic acid

(xvii)

CH3CH2CCH2CCH3

O

1 2 3 4 5 6

O

Hexane2,4dione

(xviii) CH2=CHCH2CH(OH)CH3 Pent4en2ol

(xix)

NO2

1

2 3

3Nitrocyclohexene

(xx)

OH

3

2 1

Cyclohex2en1ol

3. ISOMERISM

The existence of different compounds having same molecular formula but different physical and chemical properties is called isomerism. If the difference in properties is due to difference in their structural

formulae, it is called structural isomerism. And if the difference in properties is due to the arrangement of

atoms or group of atoms in space, it is called stereo isomerism. The different compounds are named as isomers.

Isomerism is broadly divided into two types,

(i) Structural isomerism and (ii) Stereoisomerism

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3.1 STRUCTURAL ISOMERISM: Two or more compounds with the same molecular formula,

which differ in the arrangement of atoms within the molecule are called structural isomers and the

phenomenon is called structural isomerism. There are different types of structural isomerism. 3.1.1 CHAIN ISOMERISM

The chain isomers have same molecular formula but they differ in the length of carbon chain (straight or branched). Compounds with four or more carbon atoms can show this kind of isomerism. For example,

(a) C4H10 has two chain isomers.

CH3CH2CH2CH3

(nButane)

CH3CHCH3

CH3 2Methyl propane

(isobutane)

(b) C5H12 has three chain isomers.

CH3CH2CH2CH2CH3

npentane

CH3CHCH2CH3

CH3 2Methyl butane

(isopentane)

CH3CCH3

CH3 2,2Dimethyl propane

(neopentane)

CH3

(c) C4H9OH has two chain isomers.

CH3CH2CH2CH2OH

Butan1ol

(nButyl alcohol)

CH3CHCH2OH

CH3 2Methyl propan1ol

(Isobutyl alcohol)

(d) C4H8 has two chain isomers.

CH3CH2CH=CH2

But1ene (Butylene)

CH3C=CH2

CH3 2Methyl propene

(Isobutylene)

3.1.2 POSITION ISOMERISM The position isomers have same molecular formula but differs in the position of either substituent or functional group on the same carbon skeleton. For example,

(a) The molecular formula C4H8 has two position isomers.

CH3CH2CH=CH2

But1ene (Butylene)

CH3CH=CHCH3

But2ene (Dimethyl ethylene)

(b) The molecular formula C4H6 has two position isomers.

CH3CH2CCH

But1yne (Crotonylene)

CH3CCHCH3

But2yne (Dimethyl acetylene)

(c) The molecular formula C4H9Cl has two position isomers.

CH3CH2CH2CH2Cl

1Chlorobutane

CH3CH2CHCH3

Cl 2Chlorobutane

(d) The molecular formula C6H6O3N has three position isomers. OH

NO2

NO2

OH

NO2 orthonitrophenol

(1, 2substitution)

metanitrophenol

(1, 3substitution) paranitrophenol

(1, 4substitution)

OH

3.1.3 FUNCTIONAL GROUP ISOMERISM The isomers having same molecular formula but different functional groups in the molecule are called

functional isomers. The following classes of organic compounds show functional isomerism among themselves.

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(a) Alcohols and ethers are functional group isomers.

CH3CH2OH and CH3OCH3 Ethanol Methoxy methane

(Ethyl alcohol) (Dimethyl ether)

(Functional group OH) (Functional groupO)

(b) Aldehydes and ketones are functional isomers. They are also functionally isomeric to unsaturated

ethers, unsaturated alcohols, cyclic ethers and cyclic alcohols.

CH3CH2CH2C=O

Butanal (butyraldehyde)

(Functional groupCHO)

H

CH3CH2CCH3

O

Butanone (Functional group C=O)

CH2=CHOCH2CH3, CH2=CHCH2CH2OH

O OH

(c) Carboxylic acids and esters also share functional group isomerism.

CH3COH

Ethanoic acid

O

HCOCH3

Methyl methanoate

O

(d) Dienes, allenes and alkynes are functional isomers. CH2=CHCH=CH2 CH2=C=CHCH3 CH3CH2CCH

Buta1, 3diene Buta1, 2diene (An allene) But1yne

(e) Nitro alkanes and alkyl nitrites are functional group isomers.

CH3CH2NO2 CH3CH2ONO

Nitroethane Ethyl nitrite (f) 1°, 2° and 3° amines share functional group isomerism.

CH3CH2CH2NH2

Propan1amine

(1° amine)

CH3CH2NHCH3

Nmethyl ethanamine

(2° amine)

CH3

N, NDimethyl methanamine (3° amine)

CH3NCH3

(g) Cyanides and isocyanides are functional group isomers.

CH3CN CH3N C

Ethane nitrile Methyl isocyanide

(h) Aromatic alcohols, phenols and ethers are also functional isomers.

Benzyl alcohol

CH2OH OCH3 OH

CH3

Anisole oCresol DEGREE OF UNSATURATION

Deficiency of two hydrogen atoms in a molecule is a result of either a pibond or a ring in the structure

of that molecule. The sum of pibonds and rings in the structure of a compound collectively is called

degree of unsaturation or double bond equivalents in that compound. The most general type of

formula for any organic species is (CaHbNcOd). If the compound contains other atoms also, the

tetravalent atoms are replaced by carbon, monovalent atoms are replaced by hydrogen, divalent atoms

are replaced by oxygen and trivalent atoms are replaced by nitrogen. Then all oxygen and all nitrogen

atoms are removed from the formula. However, for the removal of each N atom, one H atom is also

removed from the molecular formula. If the molecular species is charged, H+ are added or removed to

neutralise the charge. As a result of all these operations, we will get a hydrocarbon. Now this

concluded hydrocarbon is compared with saturated alkane containing same number of carbon atoms to

determine the degree of unsaturation or double bond equivalents.

Example 1: C12H16N2OCl2

Page 14

C12H16N2OCl2 will give C12H18N2O after replacing Cl atoms by H atoms and C12H16 after removing O

and N. Corresponding saturated alkane should be C12H26.

Double bond equivalents = 2

1626 = 5

Example 2: C13H10BrNS

C13H10BrNS will give C13H11NO after replacing Br by H and S by O atoms. Removal of

O and N gives us C13H10. The corresponding saturated alkane will be C13H28.

Double bond equivalents = 2

1028 = 9.

Example 3: C3H8N+

C3H8N+ will give C3H7N after losing one H+ ion. Removal of N gives us C3H6.

The corresponding saturated alkane will be C3H8.

Double bond equivalents = 2

68 = 1.

Example 4: 55HC

55HC will give C5H6 after adding one H+ ion. The corresponding saturated alkane will be C5H12. Double

bond equivalents = 2

612 = 3.

Catalytic hydrogenation (hydrogenation using H2 in presence of Ni or Pd at room temperature) data is used to distinguish between the unsaturation due to a bond or unsaturation due to a ring. Compounds

having unsaturation due to bonds absorb 1 mole of H2 for each mole of a bond while compounds having

unsaturation due to rings do not absorb any hydrogen at room temperature.But compounds like cyclopropane and cyclobutane do absorb 1 mole of H2 at 120°C and 200°C respectively because these rings are highly

strained and in order to get relieve from strain, they do get cleaved by H2 at elevated temperatures. Cyclopentane and higher rings to not absorb any H2 at the experimental temperatures (200300°C) as such

rings are quite stable.

+ H2 CH3CH2CH3 Ni at

120°C

+ H2 CH3CH2CH2CH3 Ni at

200°C

3.1.4 METAMERISM Compounds having same molecular formula but different number of carbon atoms (or alkyl groups) on either side of the multivalent functional group (i.e. –O,S,NH, –CO etc.) are called

metamers and the phenomena is called metamerism. Metamerism occurs among the members of the same family. For example,

CH3COCH

CH3

3methyl butanone

CH3

(a) CH3CH2COCH2CH3 is a metamer of CH3COCH2CH2CH3 and

Pentan3one Pentan2one

The first two compounds are also related as position isomers as the position of C=O in the two

isomers is different.

CH3OCH

CH3

2methoxy propane

CH3

(b) CH3OCH2CH2CH3 is a metamer of CH3CH2OCH2CH3 and

1Methoxy propane

(Methyl npropyl ether)

Ethoxy ethane (Diethyl ether)

2

1

3

Page 15

CH3NHCH

CH3

Methyl isopropyl amine

CH3

(c) CH3CH2NHCH2CH3 is a metamer of CH3NHCH2CH2CH3 and

Diethyl amine (Methyl npropyl amine)

Draw all possible metamers of CH3CH2NCH2CH2CH3.

CH2CH3

3.1.5 TAUTOMERISM

It is a special type of functional isomerism in which an hydrogen atom is shifted from one position

(atom1) to another (atom3). This shift is referred as 1,3shift. Such shifts are common between a

carbonyl compound containing an hydrogen atom and its enol form.

RCCR

R

H O Keto form

RC=CR

R

HO Enol form

1 2

3

1 2

3

H = (CH + CC + C=O) (C=C + CO + OH) = 359 347 = 12 kcal/mol.

Thus, generally keto form is more stable than enol form by 12 kcal. So, in most cases, the equilibrium

lies towards the left.

The term tautomerism is used for isomers that are fairly readily interconvertible and that differ from each other only (a) in electron distribution and (b) in the position of a relatively mobile atom or group,

The mobile atom is generally hydrogen and the phenomenon is then called as prototropy. Both acids and bases catalyse such interconversions. Possible limiting mechanisms are those (a) in

which proton removal and proton acceptance (from the solvent) are separate operations and a carbanion intermediate is involved. i.e. an intermolecular pathway and

(b) in which one and the same proton is transferred intramolecularly.

(a)

R2CCH=Y

B: H

Carbanion intermediate

R2C=CHY (Intermolecular)

R2C=CHY

R2CCH=Y

BH

ROH

H

RO

(b)

R2CCH=Y

H

Transition state

R2C=CHY (Intramolecular) R2C

H H

Y

CH

Mostly the keto form is more stable than enol form but in certain cases, enol form can become the

predominant form. The enol form is predominant in following cases:

1. Molecules in which the enolic double bond is in conjugation with another double bond/phenyl ring. In

such cases, sometimes intramolecular hydrogen bonding also stabilizes the enol.

CCR

R

H O

R

C=CR

HO Enol form

(Crossconjugation) Keto form

CH3CCH2COEt

O

CH3C

O

CH

COEt

O H

O

Enol form

(Crossconjugation and intramolecular hydrogen bonding)

Keto form

Page 16

CH3CCH2CCH3

O

CH3C

O

CH

CCH3

O H

O

Enol form

(Crossconjugation and intramolecular hydrogen bonding)

Keto form

2. Molecules, which contain two bulky aryl groups.

CCH Ar

H O

Keto form (5%)

C=CH

HO Ar

Ar

Ar

Enol form (95%)

where Ar =

Me

Me

Me

In the keto form of 2,2dimesitylethanal, the ArCAr bond angle is 109°, whereby the bulky aryl

groups experience greater steric repulsion. This steric repulsion eases off when the keto form

transforms to enol form, where the ArCAr bond angle widens to 120°.

3. When the enol is aromatic stabilized.

OH O

H

H Keto form Enol form

The extent of enolization is also affected by the solvent, concentration and temperature. Thus,

acetoacetic ester has an enol content of 0.4% in water and 19.8% in toluene. This is because water reduces the enol content by hydrogen bonding with the carbonyl group, making this group less

available for intramolecular hydrogen bonding. The effectiveness of intramolecular hydrogenbonding

in stabilizing the enol, with respect to the keto form is seen on varying the solvent and particularly on transfer to a hydroxylic solvent with MeCOCH2COMe.

Solvent % Enol Thus, the proportion of enol in the nonpolar solvent

(hexane) is the same as in the gas phase and higher than in

the liquid itself, the latter acting as a polar autosolvent. The

proportion drops again in the more polar MeCN and more

dramatically in water. What is happening is the increasing

relative stabilization of the keto form by solvation, this being

particularly marked in water where intermolecular hydrogen

bonding of the keto form‘s C=O groups can now take place

as an alternative to its enolization.

Gas phase 92

Hexane 92

Liquid 76

MeCN 58

H2O 15

Also, the enol content of pentan2,4dione (CH3COCH2COCH3) is found to be 95% and 45% at 27.5° and

275°C respectively. When a strong base is added to a solution of a ketone with hydrogen atom, both the

enol and keto form can lose a proton. The resulting anion is same in both the cases as they differ only in the

placement of electrons. They are not tautomers but canonical forms.

CCR

R

H O

C=CR

HO R

R

R

+H+ H

+ +H+ H

+

CCR R

O

C=CR R

R

R

..

O (Carbanion) (Enolate ion)

Other types of tautomerism

(i) PhenolKeto tautomerism:

Page 17

OH O

H

H Cyclohexadienone Phenol

In this case, enol form is more stable than keto form because of the aromatic stabilization. (ii)

NitrosoOxime tautomerism:

CN

R

H O

C=N

OH

R

R

R

Nitroso form Oxime form

This equilibrium lies far to the right and as a rule nitroso compounds are stable only when there is no

hydrogen atom.(iii) NitroAci tautomerism: Aliphatic nitro compounds are in equilibrium with the aci forms.

CN R

H

CN R

R

R O

O O

O

Nitro form

H

C=N R

R

O

OH

Aci form

The nitro form is much more stable than the aci form because nitro group has resonance. Aci form of nitro

compounds is also called nitronic acids.

(iv) ImineEnamine tautomerism / cyanideenimine tautomerism:

CCR

R

H NR

C=CR

NHR

R

R

R

Imine form Enamine form

Imine form predominates generally. Enamines are stable only when there is no hydrogen atom attached to

nitrogen.

CC

R

H N

C=C

NH

R

R

R

Cyano form Enimine form

Which of the following compound shows tautomerism and also draw their tautomer?

(a) CH3CHO (b) CH3COCH3 (c) CH3COCH2CH3 (d) C6H5CHO

(e) C6H5COCH3 (f) C6H5COC6H5 (g) O O (h)

O

O

3.1.6 RINGCHAIN ISOMERISM

Compounds having same molecular formula but possessing open chain and cyclic structures are called

ringchain isomers and the phenomena is called ringchain isomerism.

For example, (a) C3H6 represents

CH3CH=CH2 and

Propene Cyclopropane

(b) C4H8 represents

CH3CH2CH=CH2 and

But1ene Cyclobutane

or

CH3 H

Methyl cyclopropane

3.2 STEREO ISOMERISM

Page 18

Isomers having the same molecular formula and same functional group but differ in their spatial

arrangement or group of atoms in space are called stereoisomers. They are said to have different configurations or different arrangement of groups in 3D space. Stereoisomerism is of three kinds.

(a) Conformational isomerism (b) Geometrical isomerism and (c) Optical isomerism. 3.2.1 CONFORMATIONAL ISOMERISM

A bond between two carbon atoms is formed by the overlap of sp3 hybrid orbitals of carbon atom along their

internuclear axis as a result of which the electron distribution within the molecular orbital thus formed is cylindrically symmetrical along the internuclear axis. Due to this symmetry, rotation about carboncarbon

single bond is almost free,. As a result number of different arrangements of atoms about CC bond results

which are called conformations or rotational isomers or rotamers.

(a) CONFORMATIONS OF ETHANE

In ethane, the two carbon atoms are connected by a bond. If one of the methyl group in ethane molecule is

kept fixed and the other is rotated about the CC bond, infinite number of possible conformations of ethane

result (while counting the total number of isomers for a given molecular formula, the conforms are ignored).

The two extreme conformers are termed as staggered & eclipsed and the conformations lying between them are termed as gauche or skew conformations. The conformations can be represented by Sawhorse or

Newman projections. (i) Sawhorse projection

In this projection, the molecule is viewed along the axis of the model from above and right. The central CC

bond is drawn as a straight line, slightly tilted to right for the sake of clarity. The line is drawn some what

longer. The front carbon is shown as the lower left hand carbon whereas the rear carbon is shown as the upper right hand carbon. Each carbon has three lines corresponding to three atoms/groups (H atoms in the

case of ethane).

Eclipsed Staggered

H

H

H

Sawhorse projections of ethane

H

H

H

H

H

H

H

H

H

(ii) Newman projection

These projection formulae are obtained by viewing the molecule along the bond joining the two carbon atoms.

The atom near the eye is represented by a point and the three atoms or groups attached to it by three equally spaced radii. The carbon atom farther from the eye is designated by a circle and the three atoms or groups

attached to it by three equally spaced radical extensions.

Staggered Eclipsed

H

H H

H

H H

H

H

H

H

H

H

Newman projections are more widely used to represent conformers. The various conformers of ethane are shown as:

Skew or gauche (II) Eclipsed (I)

60° rotation

60° rotation

Eclipsed (III)

60° rotation

Staggered (IV)

60° rotation H

H

H

H

Hb

H H

Ha

H H

Ha

Hb

Hb H H

Ha

H

H

H

H H

H

Ha

Hb

Skew or gauche (VI) Eclipsed (V)

60° rotation

H

H

H

H Hb

H H

Ha H

H

Ha

Hb

Page 19

There are two types of strain among conformers.

(i) Torsional strainStrain due to repulsion between the bonded electron pair of CH or

CC bonds or other types of bonds. (ii) Vander Waal‘s strain Strain due to repulsion or

crowding between the groups attached to carbon.

Eclipsed form of ethane has both these strains more than in the skew form. Skew form and the staggered form have similar strains, so they are equally stable.

Thus, stability order of conformers of ethane is

[(I) = (III) = (V)] < [(II) = (IV) = (VI)]Dihedral angle is the angle between HaCC and CCHb

planes in ethane.The energy profile of the conformers of ethane is shown as:

Dihedral angle

PE

60° 0° 120° 180° 240° 300° 360°

(I)

(II)

(III)

(IV)

(V)

(VI)

(I)

2.9 kcal

Eclipsed Staggered

(2%) (98%)

Kc = Keq = ]Eclipsed[

]Staggered[=

2

98= 49

(b) CONFORMATIONS OF PROPANE

HC1C2C3H

H

H

H

H

H

H

Rotation about either C1C2 bond or C2C3 bond results in the same conformers. If rotation about C2C3 bond

is considered, the conformers of propane can be drawn by replacing Ha of ethane by CH3 group. Everything

else remains same but the energy difference between eclipsed and staggered form will be 3.2 kcal/mol. (c) CONFORMATIONS OF nBUTANE

In order to examine the conformations of nbutane, it is considered as a derivative of ethane in which one

hydrogen atom of each carbon is replaced by a methyl group. Thus, butane is considered as dimethyl ethane

as shown below,

CH3CCCH3

H H

H H

1 2 3 4

Rotation about C1C2 and C3C4 bond results in the same conformers. If rotation about C1C2 or C3C4 bond

is considered, the conformers of nbutane can be drawn by replacing Hb | Ha of ethane by C2H5 group. The

difference in energy between eclipsed and staggered from will be 3.4 kcal/mol.Rotation about C2C3 bonds

gives six conformers, which are shown as:

rotation by 60°

rotation by 60°

Partially Eclipsed (III) Skew or gauche (II) Fully Eclipsed (I)

H

H

H

H H H H

H

H3C CH3 CH3

H3C

H H H

H H3C

CH3

rotation by 60°

Page 20

H3C

rotation by 60°

rotation

by 60°

Staggered (IV) Partially Eclipsed (V) Skew or gauche (VI)

H

H

H

H

H3C

H

H

H H H H

H H3C

H3C

H

H3C

H3C

Relative stabilities of conformations

Out of the four conformations listed above, anticonformation (IV) is the most stable since in this

conformations the two nonbonded methyl groups (dihedral angle 180°) and the four nonbonded hydrogen atoms are as far apart as possible. The next in order of higher energy come the two gauche conformations ( II

and VI) in which the two nonbonded methyl groups are only 60° apart and hence causing less crowding or

steric strain than eclipsed forms. As a result of this steric strain, the two gauche conformations (II and VI) are

slightly less stable than the anticonformation (IV). However, the two gauche conformations are themselves of

equal energy. Experimentally, it has been found that the gauche conformations are about 0.9 kcal mol1 less

stable than the anti conformation. Next in order of higher energy fall the two partially eclipsed conformations (III and V). In these conformations, there are two methylhydrogen eclipsing interactions and one

hydrogenhydrogen eclipsing interaction. Because of methylhydrogen eclipsing interactions partially eclipsed

conformations of nbutane are less stable than anti and gauche conformations. However, the two partially

eclipsed conformations are themselves of equal energy.Experimentally, it was found that partially eclipsed conformation (III and V) is less stable than gauche conformation (II or VI) by about 2.9 kcal mol1 and than

anti conformation (IV) by about 3.8 kcal mol1. The fully eclipsed conformation (I) is however, the least stable.

This is due to the reason that in this conformation, there is one methylmethyl eclipsing interaction and two

weak hydrogenhydrogen eclipsing interactions. Experimentally, it has been found that fully eclipsed

conformation is about 4.5 kcal mol1 less stable than the most stable anti conformation. Thus, the relative

stabilities of the four distinct conformations of nbutane follow the order,

Anti > Gauche or skew > Partially eclipsed > Fully eclipsed. (IV) (II & VI) (III & V) (I)

The energy profile for the conformers of nbutane by rotation about C2C3 bond is

Dihedral angle

PE

60° 0° 120° 180° 240° 300° 360°

(I)

(II)

(III)

(IV)

(V)

(VI)

(I)

3.8 kcal 3.6 kcal

0.9 kcal

2.9 kcal

4.5 kcal

It is not always that anti or staggered conformation is more stable than skew or gauche. Sometimes,

the skew or gauche conformer is more stable than anti conformer because of stabilization of skew form by

intramolecular hydrogen bonding. For example, ethylene glycol,2chloro ethanol etc.

H

OH H

>

H

Skew form of ethylene glycol

O H

H

H

OH

H

H

Staggered form of ethylene glycol

O H

H

H

Cl H

>

H

Skew form of

2chloroethanol

O H

H

H

H

H

Staggered form of

2chloroethanol

O H

H

Cl

3.2.2 GEOMETRICAL ISOMERISM Hindered rotation about carboncarbon bond

Page 21

A double bond consists of a bond and a bond. The bond is formed by the sideways overlapping

of unhybridized porbitals of two carbon atoms above and below the plane of carbon atoms. If one of the

carbon atoms of the double bond is rotated with respect to the other, the

porbitals will no longer overlap and the bond should break, but the breakage of this bond requires 251 kJ

mol1 of energy which is not provided by the collisions of the molecules at room temperature. Consequently,

the rotation about a carboncarbon double bond is not free but is strongly hindered or restricted. In other

words, a bond prevents free rotation of the carbon atoms of the double bond with respect to each other. Due

to this hindered rotation, the relative positions of atoms or groups attached to the carbon atoms of the double bond get fixed. For example, Ha and Hb cannot exchange their positions by rotations of C1 with respect to C2

without breaking the bond.

Ha

Hb

C C

Ha

Hb

C

C

b a

(cis isomer) (trans isomer)

b a

h or C

C

b a

b a

.

.

rotation

by 180°

C

C

b a

a b

.

.

reformation

of bond

C

C

b a

b a

Energy profile for the conversion of cis isomer to trans isomer can be depicted as

PE

Ea

cis

Progress of reaction

trans

Necessary and sufficient conditions for geometrical isomerism

The necessary conditions for a molecule to exhibit geometrical isomerism are (i) The molecule must have restricted rotation due to the presence of a C=C, C=N, N=N and cyclic

structure. (ii) Each of the two atoms having restricted rotation must be attached to different substituents.

Thus, alkenes of the type abC=Cab and abC=Cde show geometrical isomerism.

a

b

C=C a

b

cisisomer (I)

a

b

C=C b

a

transisomer (II) Both these isomers have the same structural formulae but differ in the relative spatial arrangement of

groups around the double bond. The isomer I, in which the similar atoms or groups lie on the same

side of the double bond is called the cisisomer whereas the isomer II, in which the similar atoms or

groups lie on the opposite sides of the double bond is called the transisomer. It is because of this

reason that geometrical isomerism is also called cistrans isomerism.

In the case of cistrans nomenclature, the atoms or groups attached to the Catoms should be similar

but if all the four groups are different, then E, Z nomenclature comes into picture. According to this

nomenclature, if the atoms or groups of higher priority are on the same side of the double bond, the isomer is designated as Z (zusammen, in German means together) and if the two atoms or groups of

higher priority are on the opposite sides, the isomer is designated as E (Entegegan, in German means opposite). The priority of a group or atom is based on the atomic number but when atomic numbers are

same (for example, isotopes) then priority is decided by atomic mass.

Page 22

a

b

C=C d

e

Zisomer

(If groups ‗a‘ and ‗d‘ have higher priority)

a

b C=C

e

d

Eisomer

(If groups ‗a‘ and ‗d‘ have higher priority)

For example, in 1bromo2chloro2fluro1iodo ethene, C1 has two atoms viz. Br and I. Since I (Z = 53) has higher atomic number than Br (Z = 35), therefore I is assigned priority 1 while Br is

assigned priority 2. Similarly, Cl is assigned priority 1 while F is assigned priority 2 on C2.

Br(2)

(1) C C F(2)

Cl(1)

I

However, geometrical isomerism is not possible, if one or both the doubly bonded carbon atoms carry two

similar substituents. This is because in such cases, the two possible configurations are, identical as shown below.

a

a

C = C d

e

a

a

C = C e

d

It is because of this reason that terminal alkenes such as propene, 1butene,

2methyl propene and alkenes carrying identical substituents on one of the doubly bonded carbon atoms such

as 2methyl 2butene and 2, 3dimethyl 2butene etc. do not show geometrical isomerism. Some other

examples of geometrical isomers are:

C

C

CO2H H

Maleic acid (cis isomer)

CO2H H

C

C

CO2H H

Fumaric acid (trans isomer)

HO2C H

C

C

H3C

cis2pentene

C2H5 H

H C

C

H3C

trans2pentene

C2H5 H

H

Let us see the isomers of 1

CH3CH=CHCH=CHCH3 2

3

4

5

6

(2,4hexadiene). There are

3 geometrical isomers of it, namely ciscis, cistrans or transcis and transtrans. It is a symmetrical diene.

C

C

(ciscis)

C

H3C H

H H

C CH3 H

C

C

(cistrans or transcis)

C

H3C H

H H

C H H3C

C

C

(transtrans)

H

H3C H

C

C CH3 H

H

But 1

CH3CH=CHCH=CHCH2CH3 2

3

4

5

6

7

(2,4heptadiene) shows 4 geometrical isomers, namely ciscis,

cistrans, transcis and transtrans. It is an unsymmetrical diene.

C

C

(ciscis)

C

H3C H

H H

C CH2CH3 H

C

C

(cistrans)

C

H3C H

H H

C H CH3CH2

C

C

(transcis)

C

H

H

C CH2CH3 H

CH3 C

C

(transtrans)

C

H

H

C H

CH3

H

CH3CH2

H

Page 23

Draw and give the stereochemical designation for the geometric isomers of

3,5octadiene.

Number of geometrical isomers

(i) The number of geometrical isomers of an unsymmetrical polyene = 2n (where ‗n‘ is the number of double bonds).(ii) The number of geometrical isomers of symmetrical polyene containing even

number of double bonds = 2n1 + 1

2

n

2

.(iii) The number of geometrical isomers of symmetrical

polyene containing odd number of double bonds = 2n1 + 2

1n

2

.

Geometrical isomerism is also shown by compounds which contains C=NOH , N=N structural units.

Cyclic compounds also exhibit geometrical isomerism.

Compounds containing C=NOH units are commonly called oximes. Oxime of

formaldehyde is incapable of showing geometrical isomerism,

C=NOH H H

while oxime of any other aldehyde

(other than formaldehyde) will exhibit geometrical isomerism.

C

N

H R

OH

(syn isomer)

C

N

H R

HO

(anti isomer)

The nomenclature for aldoximes is syn (when H and OH are present on same side of the double bond) and

anti (when H and OH are present on opposite sides of the double bond). Oximes of symmetrical ketones

C=NOH R R

do not show geometrical isomerism but oximes of unsymmetrical ketones

C=NOH R R

are

capable of showing geometrical isomerism.

C

N

R R

OH

Syn (R) or anti (R)

C

N

R R

HO

Syn (R)

or anti (R)

In this case, syn and anti nomenclature is used but the group with respect to which OH is syn or anti is to be

specified.Diazobenzene also exhibit geometrical isomerism as it fulfills the necessary condition.

N

N

Ph

(cis isomer) (trans isomer)

Ph

N

N

Ph

Ph

Cyclic compounds too have restricted rotation because of the impossibility of rotation around CC single bond

as the conformation of cyclic compound would twist on rotation. Appropriately placed substituents on cycloalkanes would be capable of showing geometrical isomerism.

X

1,2cis

X X

1,2trans X

X

1,3cis

X

X

1,3trans

X

1,4cis

X X

1,4trans

X

X

3.2.3 OPTICAL ISOMERISM

Ordinary light has vibrations in all possible planes. In plane polarized light, the vibrations are only in

one plane. The optical isomers differ in the action on plane polarized light. One of the isomer turn the plane of

Page 24

the polarized light to the right (dextro rotatory), the other turn it to the left

(laevo rotatory). Optical activity is due to molecular asymmetry. Molecular asymmetry implies that the compound should not contain any symmetry element (i.e. plane of symmetry and centre of symmetry). A

compound containing a single chiral atom (attached to four different atom or group of atoms) would always be optically active but compounds with more than one chiral carbon may or may not be optically active.

Conditions for a compound to exhibit optical activity

(a) The compound should be chiral (asymmetrical). The compound should be devoid of any element of

symmetry like plane of symmetry, centre (point) of symmetry etc. (b) The mirror image of the compound should be nonsuperimposable on it.

A compound or an isomer if fulfills the first condition, the second condition will be automatically fulfilled.

If second condition is seen first and is found to be fulfilled, this means that the first condition would

have been obeyed. So, for a compound/isomer, if it is to be checked that it is optically active or not, any one of the above condition can be checked.

(i) Plane of symmetry: A plane of symmetry (also called a mirror plane) is a plane passing through the

molecule such that the molecule is divided into 2 equals parts, one part being the mirror image of other

(the plane acting as a mirror). Plane of symmetry can be at the centre of a CC bond or through the

centre of atom. For example,

a plane of symmetry

a

b

b

d

d

Such an isomer is called meso isomer, which is optically inactive due to the presence of plane of

symmetry. The optical rotation of one half of the molecule is exactly cancelled by the other half. This is called internal compensation. Meso isomers do not have a mirror image as the mirror image is identical

to the isomer itself. (ii) Centre of symmetry: A center of symmetry is a imaginary point within the molecule such that a

straight line drawn from any part or element of the molecule to the center and extended an equal

distance on the other side encounters an equal part or element but this should happen in all directions

to be able to qualify as centre of symmetry. If it happens in one or two direction but not in other

directions then such a point is not a centre of symmetry. For example, 2, 4dimethyl cyclobutane1,

3dicarboxylic acid possess a centre of symmetry, which is the centre of the ring. Centre of symmetry

can be at the centre of molecule or over an atom or midway between a bond.

H

CH3

CO2H

H

CO2H

H H

CH3

Another important example of the compound having a centre of symmetry is the trans form of dimethyl

keto piperazine. The compound exists in two formscis and trans. The cis form of the compound exists

in two enantiomeric forms but the trans form has a centre of symmetry and therefore, it is optically

inactive.

CO

H3C NH

NH CO H

CH3

H

cisdimethyl ketopiperazine Optically inactive

CO H3C

NH

NH CO H CH3

H

transdimethyl ketopiperazine (centre of symmetry)

Optically Active

Similarly, truxillic acid is optically inactive because of the presence of a centre of symmetry.

Page 25

Ph

H

truxillic acid

H

COOH

H

Ph

H

COOH

It must be noted that only evenmembered rings possess a centre of symmetry. It is not found in

oddmembered rings. Centre of symmetry will never exist in a molecule having only one chiral centre.Organic

compounds having atleast one chiral carbon atom or asymmetric carbon atom

(a carbon atom attached to 4 different monovalent atom or groups) are always optically active. For example, lactic acid (2hydroxy propanoic acid) CH3CH(OH)COOH contains a chiral carbon atom and exhibits optical

activity. Lactic acid exists in two forms, one is the mirror image of the other and the two forms are nonsuperimposable.

C

Mirror

CH3

CO2H

H OH

C

CH3

CO2H H HO

These 3dimensional projections of lactic acid can be

represented on 2dimension by Fischer projection

formula‘s. In drawing Fischer projection formula from

a 3dimensional projection, the group at the back of

the plane (COOH) is pulled to bring on the same line

as that of the CH3 and the groups projecting in the front (H and OH) will be pushed to bring them on the

same horizontal line. Thus, Fischer projection of lactic acid would be drawn as

Mirror

CH3

CO2H

OH H

CH3

CO2H

HO H

(Note: Fischer projections are drawn only for those

molecules, which have chiral carbon atom).These two

isomers are called enantiomers. The stereoisomers, which are related as mirror imageobject are called

enantiomers. One isomer and its enantiomer are

mirror images of each other and they are not superimposable. Which of the two forms drawn is

dextro or laevo, cannot be known by looking at their structures,

it can only be determined experimentally using polarimeter.

An equimolar mixture of two enantiomers of lactic acid shows no rotation of plane polarized light, thus it is optically inactive. It is called racemic form or racemic mixture and is designated as (). The optical

rotation of one enantiomer exactly cancels the optical rotation of other so that the net rotation is zero.

So, racemic mixture is optically inactive due to external compensation. The enantiomers can be separated from a racemic mixture and the process of separation is called resolution.

Let us see the optical activity of 2butanol.

Mirror

CH3

CH2CH3

OH H

CH3

HO H

CH2CH3

1

2

3 4

1

2

3 4

Structures I and II are mirror images of each other and are not superimposable. They are enantiomers

of 2butanol. A pair of enantiomers is always possible for molecules that contain one tetrahedral carbon atom

with four different groups attached to it. The carbon atom C2 is called a stereocentre.Let us see the optical

activity of tartaric acid, HO2CCH(OH)CH(OH)CO2H.

Page 26

CO2H

OH H

(1) Meso isomer

(does not have mirror image)

H

H

(2) d or l

Optically active

H

H

H OH

CO2H

CO2H

OH

HO

CO2H

CO2H

CO2H

OH

HO

(3) l or d

Optically active

Stereoisomers (2) and (3) are enantiomers (mirror image isomers) but those stereoisomers which are

not related as mirror imageobject but have same molecular formula are called diastereomers. Stereoisomers

(1) & (2) and (1) & (3) are related as diastereomers. Let us now see the optical activity in 2,3dibromopentane.2,3dibromopentane has two dissimilar asymmetric carbon atoms. The number of

stereoisomers are shown as

CH3

Br H

C2H5

CH3

(1) (2)

Br H

C2H5

Br H

Br H

CH3

Br H

C2H5

(3)

Br H

CH3

(4)

C2H5

Br H

Br H

Structures (1) and (2) are mirror images of each other and so are enantiomers. Structures (3) and (4)

are also mirror images of each other and they form another set of enantiomers, all stereoisomers 1 to 4 are optically active. Structures (1) and (3) are called diastereomers. Structures (1) & (4) and (2) & (4) are also

related as diastereomers.

Number of optical isomers

As it has been discussed above, a compound containing two dissimilar carbon atoms can exist in four optically active forms. Reasoning in the same fashion, we will find that a compound containing three such

asymmetric carbon atoms can exist in eight different configurations, which represent optical isomers. Thus in general, the number of stereoisomers for a compound with n distinct (different) asymmetric carbon atoms in 2n.

When an organic compound contains two similar asymmetric carbon atoms in its molecule, abdCCabd, the

number of optically active isomers would be less than 2n. Thus, tartaric acid [HO2CCH(OH)CH(OH)CO2H] has two similar asymmetric carbon atoms and exists in only three forms, of which two are optically active and one

is optically inactive (meso form). Thus, the general formulas for predicting the number of optical isomers is given as under. 1. When the molecule is unsymmetrical:

Number of d and l isomers (a) = 2n.

Number of meso forms (m) = 0.

Total number of optical isomers (a + m) = 2n.

where n is the number of chiral carbon atom(s).

Common example is CH3CH(Br)CH(Br)COOH. 2. When the molecule is symmetrical and has even number of chiral carbon atoms:

Number of d and l isomers (a) = 2n1.

Number of meso forms (m) = 1

2

n

2

.

Total number of optical isomers = (a + m).

Common example is tartaric acid, HOOCCH(OH)CH(OH)COOH. 3. When the molecule is symmetrical and has an odd number of chiral carbon atoms:

Number of d and l forms (a) = 2n1 2

1n

2

.

Number of meso forms (m) = 2

1n

2

.

Total number of optical isomers = (a + m) = 2n1.

Page 27

Common example is CH3CH(OH)CH(OH)CH(OH)CH3.

Draw all the stereoisomers of 3chloro2pentanol, CH3CH(OH)CH(Cl)CH2CH3 and give the stereochemical relationships of the stereoisomers.

Enantiomers:

Physical and chemical properties of enantiomers are as follows:

1. Enantiomers have identical physical properties like boiling point, melting point, solubility etc.

2. They rotate the plane of polarized light in opposite directions, though in equal amounts. The isomer that

rotates the plane to the left (counterclockwise) is called the laevo isomer and is designated as (), while

the one that rotates the plane to the right (clockwise) is called the dextro isomer and is designated as

(+). Because they differ in this property they are often called optical antipodes. 3. Enantiomers, react at the same rate towards achiral reagents, solvents and catalysts. Towards chiral

reagents, solvents and catalysts, enantiomers react at different rates. The transition states produced

from the chiral reactant and the individual enantiomers are not mirror images. They are diastereomeric

and hence have different enthalpies. The H values are different for the two and hence the rates of

reaction and the amounts of product formed. Their rates may be so far apart that one enantiomer undergoes the reaction at a convenient rate while the other does not react at all. This is the reason that

many compounds are biologically active while their enantiomers are not. Although pure compounds are always optically active, if they are composed of chiral molecules,

mixtures of equimolar amounts of enantiomers are optically inactive since the equal and opposite rotations cancel. Such mixtures are called racemic mixtures or racemates. Their properties are not

always the same as those of the individual enantiomers. The properties in the gaseous or liquid state or

in solution usually are the same, since such a mixture is nearly ideal, but properties involving the solid state, such as melting points, solubilities and heats of fusion, are often different. Thus, racemic tartaric

acid has a melting point of 204206°C and a solubility in water at 20°C of 206 g/litre, while for the (+) or

the () enantiomers, the corresponding data are 170°C and 1390 g/litre. The separation of a racemic

mixture into its two optically active components is called resolution. The most widely used methods for

the resolution of a racemic mixture is chemical separation. Chemical Separation: Pasteur was the first investigator to resolve a racemate chemically and his

method is used even today. For example, an optically pure compound, a (+) base, is reacted with a racemic acid, resulting in two salts: a (+) (+) salt and a () (+) salt. Since these are diastereomers, they

have different solubilities & boiling points and are separable either by fractional crystallization or by

fractional distillation, after which the enantiomers are recovered.

Let us have an enantiomeric pair of CH3CHCO2H.

C2H5

To separate the two isomers of this pair, we treat

them with an optically active (chiral) alcohol,

CH3CHOH.

C2H5

The reaction can be outlined as:

(dform)

H3C COH

H

C2H5

O

HO CH3

H

C2H5

(dform)

(d, d ester)

H3C CO

H

C2H5

O

CH3

H

C2H5

(lform)

HOC CH3

H

C2H5

O

(l, d ester)

H3C CO

H

C2H5 O

CH3

H

C2H5 HO CH3

H

C2H5

(dform)

The esters produced can be seen that they are not enantiomers (mirror image isomers), infact they are

diastereomers. This means that they have different boiling points and thus can be separated by fractional distillation. d, d and l, d esters are collected as different fractions. Each fraction is separately hydrolysed. Let

Page 28

us say fraction I which have d, d ester, on hydrolysis gives d acid and dalcohol, which are different

compounds having different boiling points, so can be further separated by fractional distillation. On the other hand, fraction II has l, d ester, which on hydrolysis yield lacid and dalcohol, having quite different boiling

points, so can be separated by fractional distillation. Thus, the enantiomeric acids (d and lform) have been

separated (resolved). The energy profile for the ester formation reaction of enantiomeric acids with optically active alcohol (dform) can be shown as:

PE

POR

The two enantiomeric acids have same enthalpies but

the diastereomeric esters have different enthalpies, the transition states for the formation of esters also

have different enthalpies. Thus, the activation energy for the formation of d, d ester and l, d ester are

different and so, are their rates of reaction. When we say that a compound is resolvable, it is implied that the compound is optically active and is comprised of d

and lforms.

But when the compound is nonresolvable, it means that the compound is optically inactive (achiral), which

could be either due to absence of a chiral centre and having a plane of symmetry or due to presence of chiral

centre and having a plane or centre of symmetry.Those stereoisomers, which are not mirror images are called diastereomers. Diastereomers have different physical properties. e.g. melting and boiling points, refractive

indices, solubilities in different solvents, crystalline structures and specific rotations. Because of their differences in solubility, they often can be separated from each other by fractional crystallization.

Diastereomers have different chemical properties towards both chiral and achiral reagents. Neither any two

diastereomers nor their transition states are mirror images of each other and so will not necessarily have the

same energies. The H values will be somewhat different and thus the rates of reaction will differ. However,

since the diastereomers have the same functional groups, their chemical properties are not too dissimilar.You

will be astonished to see that the compounds even if they do not have the optically active carbon can still show

optical activity. For example, allenes, biphenyls, spiran etc.By preliminary examination, it seems that the compound possess plane of symmetry and thus, it should be optically inactive. But on closer observation, it is

revealed that it is optically active due to the following reasons.

C=C=C

R1

R2

1 2 3 R3

R4

If we see the structure, C1, C3 are sp2 hybridised but C2 is sp hybridised. The participation

of orbitals for C1 clearly are (s + px + py), for C2 (s + px) and for C3 (s + px + pz) considering

CCCbond is along Xaxis. Therefore, the orbital picture can be shown as

R1

R2

sp2 sp

sp2

R3

R4

2pz

2py

2pz

XY plane

XZ plane

Therefore, within the molecule two distinct planes arise and any one or all of them may be plane of

symmetry or may be none. Very clearly none is the

plane of symmetry as R1, R2, R3, R4 are all different.

If R1 = R2, R3 R4 then XZ plane within the

compound will bisect it into two equal halves. So,

XZ plane will be the plane of symmetry. For the

compounds with cumulative odd number of double bonds and different terminal groups will never show

optical activity because clearly the terminal carbon lie in the same plane.

Therefore, the compounds become optically inactive. Draw the orbital picture and see on your own. Such

compounds will show geometrical isomerism.For the biphenyl compounds having all sp2 hybridised carbon atoms if the o, osubstituents are very bulky then to release steric strain, the rotation around CC bond axis

takes place causing loss of planarity of the compound. For example,

Page 29

C2H5O

NH2 (CH3)3C

(CH3)2CH

Here as the o, o groups are very bulky, biphenyl

rings change their planarity as shown. Now the bulky

groups are situated at 90° angle apart. Hence, none

of the planes XY or XZ are the planes of symmetry.

So, compound is optically active.

OEt

NH2 (CH3)3C

(CH3)2CH

XY plane XZ plane

Similar things happen with spiranes having even number of rings and terminal carbons having different groups. Such a compound will have no plane or centre of symmetry, so it is optically active

R1

R2

R3

R4 (Optically active)

R1

R2

R3

R4 (Geometrically active)

But in the above compound, there is a plane of

symmetry, so it is optically inactive. Such compounds are capable of showing geometrical isomerism.

Optical rotation for a mixture of optically active

isomers is given by obs = ii Xo

Draw all the stereoisomers of (a) 1, 2dimethyl cyclopropane (b) 1, 2dimethyl

cyclobutane (c) 1, 3dimethyl cyclobutane (d) 1, 3dimethyl cyclopentane.

PROFICIENCY TEST

The following 10 questions deal with the basic concepts of this section. Answer the following briefly. Go

to the next section only if your score is greater than or equal to 8.

Do not consult the study material while attempting the questions. 1. True/False. CH3CH2CN can show structural isomerism.

2. True/False. CH3COCH3 can not show tautomerism.

3. True/False. CHDTMe can show optical isomerism. 4. True/False. A mixture of 50% d + 50% l isomers is called racemic mixture.

5. True/False. CH3COCH2COCH3 exists in more enolic form in gaseous phase or nhexane.

6. True/False. Tartaric acid can have 3 optical variety. 7. True/False. Dextro compound rotates the plane polarized light towards left.

8. True/False.Cummulenes with even number of double bonds and each terminal carbon attached to two

different group of atom, is geometrically active. 9. True/False. Geometrical isomers can be referred as diastereomers. 10. True/False. CH3COCH3 will have more enol content than CH3COCH2COCH3 in aqueous solvent.

ANSWERS TO PROFICIENCY TEST

1. True

2. False

3. True 4. True

5. True 6. True

7. False 8. False

9. True

10. False

SOLVED OBJECTIVE EXAMPLES

Page 30

Example 1:

Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauch e) of

2fluoroethanol is (a) Gauche < Eclipse < Anti (b) Eclipse < Anti < Gauche (c) Anti < Gauche < Eclipse (d) Eclipse < Gauche < Anti

Solution:

Eclipsed form of 2fluoroethanol is least stable due to repulsion between the F and OH groups as well as d ue to torsional strain. Gauche form is most stable as the torsional as well as vander Waal‘s strain both are considerably

reduced and the intramolecular Hbond further increases the stability. Therefore, the increasing order of stability is Eclipse < Anti < Gauche.

(b)

Example 2:

The number of structural isomers for C6H14 is (a) 3 (b) 4 (c) 5 (d) 6

Solution: The various isomers possible are:

nhexane, 2methylpentane, 3methylpentane, 2,2dimethylbutane, 2,3dimethylbutane.

(c)

Example 3: STATEMENT1: Molecules that are not superimposable on their mirror images are chiral. because

STATEMENT2: All chiral molecules have chiral centres.

(a) Statement1 is True, Statement2 is True; Statement2 is a correct explanation for Statement1.

(b) Statement1 is True, Statement2 is True; Statement2 is NOT a correct explanation for

Statement1.

(c) Statement1 is True. Statement2 is False.

(d) Statement1 is False. Statement2 is True.

Solution:

Statement-1 is correct. Statement-2 is incorrect because compound can be chiral even in the absence of chiral atoms.

(c)

Example 4

CHCH2CH3 + Cl2 ‘N’ isomers of C5H11Cl CH3

Fractional

distillation

‘n’ distilled products h

CH3

The values of ‘N’ and ‘n’ are (a) 6,6 (b) 6, 4 (c) 4, 4 (d) 3, 3

Solution:

CH3CHCH2CH3

CH3

CH3CHCH2CH2Cl + CH3CHCHCH3 + CH3CCH2CH3

CH3 CH3 +

Cl CH3

Cl

CH2CHCH2CH3

Cl

CH3

Cl2/ h

(d, l) (d, l)

(b)

SOLVED SUBJECTIVE EXAMPLES

Example 1: Find out whether the given compounds are optically active or not.

(i)

SO3H H

H I

(ii)

SO3H C2H5

CH3 I

Page 31

(iii)

C=C=C=C CH3

C2H5

C3H7

CH3

Solution: In compounds (i) and (iii), there is plane of symmetry passing through the compounds

(molecular plane). Therefore, they are optically inactive. Compound (ii) does not have any plane of symmetry as the two phenyl rings are not in the same plane. One of the ring rotates about

CC bond axis because of bulky substituents at o, o positions of two adjacent phenyl rings and the two rings are perpendicular to each other. So, (ii) is optically active.

Example 2: Find whether the following compounds are optically active or not.

(i)

R1

R2

R3

R4 (ii)

R1

R2

R3

R4

C

D T H

(iii)

CH3

H H H

CH3 Cl

H

Cl

Solution: Compound (i) is optically active because there is no plane of symmetry, which can cut the molecule into two

equal halves. Compound (ii) also does not have plane of symmetry so, it is optically active. Compound (iii) is optically inactive because of the presence of centre of symmetry.

Example 3: Given that obs = xii , where xi = Mole fraction of the stable conformer and i= Dipole moment of the

stable conformer. Find the dipole moment of gauche conformer of ZCH2CH2Z if obs = 1.0 D and xanti =

0.82. Also draw the stable Newman projections for meso YCHDCHDY if (i) Y = CH3 and (ii) Y = OH.

Solution:

There are two stable conformers of the compound

ZCH2CH2Z viz, gauche and anti.

x(anti) = 0.82 x(gauche) = 0.18 ; (anti) = 0

obs = (gauche) x(gauche) + (anti) x(anti)

1 = (gauche) 0.18 + 0 0.82 ; (gauche) = 5.56 D.

H

H

H

Z

H

Z

(gauche)

H

H H

Z

H

Z

(anti)

The stable conformer of meso form of YCHDCHDY when Y = CH3 is anti because of least repulsion between methyl groups. When Y = OH, then the Gauche form is more stable because of intramolecular hydrogen bonding.

H

D H

D

CH3

CH3

H

D O

D

O

H

H

(anti) (gauche)

H

EXERCISE – I

AIEEE-SINGLE CHOICE CORRECT

1. How many pairs of diastereoisomers are possible in the following molecule PhCH(Cl)CH(Cl)Ph?

(a) none (b) two (c) four (d) eight

2. Which of the following compound will show optical isomerism?

(a) Butanal (b) 2Chlorobutanol

(c) 2Propanol (d) 1Butene

Page 32

3. The total number of optically active and meso forms possible for the compound [HOOC(CH(OH))3

COOH] respectively are

(a) 4, 4 (b) 4, 2

(c) 3, 2 (d) 2, 3 4. The most stable conformation of chlorohydrin at room temperature is

(a) fully eclipsed (b) partially eclipsed

(c) gauche (d) staggered 5. The dihedral angles between the two CBr bonds in gauche, partially eclipsed, staggered and fully

eclipsed conformations of 1,2dibromoethane respectively are

(a) 60°, 120°, 180°, 0° (b) 0°, 60°, 120°, 180° (c) 120°, 60°, 180°, 0° (d) 60°, 0°, 180°, 120°

6. Which of the following statement is false about cyclopropane1,2dicarboxylic acid?

(a) It has two geometric isomers. (b) It has three stereoisomers.

(c) All the stereoisomers are optically active. (d) Only transisomer shows enantiomorphism.

7. The IUPAC name of Cl3C.CHO is

(a) Trichloroacetaldehyde (b) 1,1,1Trichloroethanal

(c) 2,2,2Trichloroethanal (d) Chloral

8. The name of CH CH

CHO NH2

is

(a) 1Aminoprop2enal (b) 3Aminoprop2propenal

(c) 1Amino2formylethene (d) 3Amino1oxoprop2ene

9. IUPAC name of the compound

BrCH2CHClCHCl2 is

(a) 1Bromo2,3,3trichloropropane (b) 1,1,2Trichloro3bromopropane

(c) 3Bromo1,1,2trichloropropane (d) None of these

10. Indicate the wrongly named compound.

(a) CH3CHCH2CH2CHO

CH3 (4Methylpentanal)

(b) CH3CHCCCOOH

CH3 (4Methyl2pentynoic acid)

(c) CH3CH2CH2CHCOOH

CH3 (4Methylpentanoic acid)

(d)

CH3CH2CH=CHCCH3

O

(3Hexen2one)

11. Which of the following compound is not chiral?

(a) DCH2CH2CH2Cl (b) CH3CH(D)CH2Cl

(c) CH3CH(Cl)CH2D (d) CH3CH2CH(D)Cl 12. Which of the following compound exhibit geometrical isomerism?

(a) CH2=CClBr (b) CH3CH=CClBr

(c) CH3CH=CCl2 (d) (CH3)2C=CClBr 13. Which of the following compound will show metamerism?

(a) CH3OCOC3H7 (b) CH3SC2H5

(c) CH3OCH3 (d) CH3OC2H5

14. Which one of the following compound will not exhibit optical isomerism?

(a) CH3CH(OH)Br (b) CH3CH(OH)CH3

(c)

CH3CHCHCHCH3

CH3 Br

(d) CH3CHCHCH2OH

OH Br

15. What type of isomerism is shown by the following compounds?

Page 33

O OH

(a) Metamerism (b) Positional

(c) Functional (d) Tautomerism 16. Which of the following statements regarding the molecule CH2=C=CH2 is not correct?

(a) Both the two bonds are present in the same plane.

(b) The central carbon atom is sp hybridized while the terminal atoms are sp2 hybridized.

(c) The molecule is not planar. (d) The molecule contains six bonds and two bonds.

17. If two compounds have the same empirical formula but different molecular formulae they must have

(a) different percentage composition (b) different molar mass

(c) same viscosity (d) same vapour density

18. The IUPAC name of the compound is

HO CH3

CH3

(a) 1,1dimethyl3hydroxycyclohexane

(b) 3,3dimethyl1hydroxycyclohexane

(c) 3,3dimethyl1cyclohexanol

(d) 1,1dimethyl3cyclohexanol

19. The compound which is not isomeric with diethyl ether is

(a) npropyl methyl ether (b) 1butanol

(c) 2methyl2propanol (d) butanone

20. The maximum number of isomers for an alkene with the molecular formula C4H8 is

(a) 2 (b) 3 (c) 4 (d) 5

21. The number of isomers of dibromoderivative (molar mass 186 g mol1) of an alkene is

(a) two (b) three

(c) four (d) six 22. The number of optically active compounds in the isomers of C3H5Br3 is

(a) 1 (b) 2

(c) 3 (d) 4 23. The enolic form of acetone contains

(a) 9 sigma bonds, 1 pi bond and 2 lone pairs (b) 8 sigma bonds, 2 pi bond and 2 lone pairs

(c) 10 sigma bonds, 1 pi bond and 1 lone pairs

(d) 9sigma bonds, 2 pi bond and 1lone pairs 24. A molecule is said to be chiral if it

(a) contains a centre of symmetry (b) contains a plane of symmetry

(c) cannot be superimposed on its mirror image (d) exists as cis and transforms

25. Which of the following statements is not correct?

(a) A meso compound has chiral centres but exhibits no optical activity. (b) A meso compound has no chiral centres and thus are optically inactive.

(c) A meso compound has molecules which are superimposable on their mirror images even though they contain chiral centres.

(d) A meso compound is optically inactive because the rotation caused by any molecule is cancelled

by an equal and opposite rotation caused by another molecule that is the mirror image of the first.

EXERCISE – II

IIT-JEE- SINGLE CHOICE CORRECT 1. 1, 2dimethyl cyclohexane has

(a) 2 geometrical isomers and no optical isomers (b) 2 optical isomers and no geometrical isomers

(c) 3 stereoisomers

Page 34

(d) none of these 2. The most stable configuration of ethylene glycol is

(a) Anti (b) Gauche

(c) Partially eclipsed (d) fully eclipsed

3. The compound, C2FClBrI has

(a) 4 geometrical isomers (b) 6 geometrical isomers

(c) 4 optical isomers (d) (b) and (c) both

4. CHOCH(OH)CH(OH)CH2OH has total

(a) 4 optical isomers (b) 2 optical isomers

(c) 3 optical isomers (d) none of these 5. Total number of optical isomers of COOH(CHOH)2COOH is

(a) 2 (b) 3

(c) 4 (d) 6

6. For the compound,

total number of optical isomers will be

(a) 2 (b) 4

(c) 8 (d) 16 7. CH3COCH2COCH3 shows tautomerism, which one of the following will not show?

(a)

O

(b)

O

O

(c)

O

(d)

CH2CN

O

8. CH3CH2CN and CH3CH=C=NH are

(a) functional group isomers (b) geometrical isomers (c) tautomers (d) metamers

9. In which compound, the enol form will be maximum in nhexane?

(a) CH3COCH2CO2C2H5 (b) CH3COCH2Cl

(c) CH3COCH2COCH3 (d) PhCOCH2COCH3 10. Total number of isomeric alcohols with the molecular formula C4H10O is

(a) 2 (b) 3

(c) 4 (d) 5 11. (+) Mandelic acid has a specific rotation of +158°. What would be the observed specific rotation of the

mixture of 25% () mandelic acid and 75% (+) mandelic acid?

(a) +118.5° (b) 118.5°

(c) 79° (d) +79°

12. Which of the following compound can exist in two geometrically isomeric forms?

(a) CH3CCCH3 (b) CH3CHCH=CH2

OH

(c) CH3CH2CH=CHCH3 (d) CH3CH2CH2CH=CH2 13. The total number of isomers possible for trisubstituted C6H3Br2Cl is

(a) Six (b) Four

(c) Three (d) Two

14. In the following sequence of standard stereochemical Fischer formulae, indicate the one that may be

identical with the given compound.

Page 35

NH2

H

CO2H

CH3

(a)

NH2

H

CO2H

CH3 (b)

NH2

H

COOH

H3C

(c)

CH3

H

COOH

H2N (d)

NH2

H

COOH

H3C

15. Which of the follower stereoisomer exhibits a plane of symmetry?

(a)

COOH CH3

CH3

(b)

AsMe3

Br

NO2

(c)

COOH

OH

HOOC OH

(d)

Ph

16. Which of the Newmann projections shown below represents the most stable conformation about the

C1–C2 bond of 1-iodo-2-methylpropane?

(a)

I H3C

CH3 H H

H

(b)

CH3

I

CH3

H

H

H

(c)

CH3

I

CH3

H

H

H

(d)

I

H3C

CH3 H

H

H

17. Vinyl alcohol and acetaldehyde are

(a) geometrical isomers. (b) ketoenol tautomers.

(c) chain isomers. (d) position isomers. 18. Meso tartaric acid and (+)tartaric acid are

(a) position isomers (b) diastereomers

(c) enantiomers (d) racemic mixture 19. Which of the following compound will form geometrical isomers?

(a)

Cl

Cl

(b) CH3CH=NOH

(c)

(d) all of them

20. The number of stereoisomers of the following compound is

C=C

CH3

Cl

H

(CH2)4CH(OH)CH3

Page 36

(a) 8 (b) 4

(c) 6 (d) 2 21. How many structural isomers of a hydroxy acid of formula C4H8O3 show optical activity?

(a) 2 (b) 3 (c) 4 (d) 5 22. Which one of the following compound having molecular formula C7H16 will show optical isomerism?

(a) 2, 3-dimethyl pentane (b) 2, 2-dimethyl pentane (c) 2-methyl hexane (d) none of these 23. At room temperature the eclipsed and the staggered forms of ethane cannot be isolated because

(a) both the conformers are equally stable.

(b) they interconvert rapidly.

(c) there is a large energy barrier of rotation about the -bond.

(d) the energy difference between the conformers is large. 24. The following compound shows

C=C

COOH

H H3C

H3C C H

H3C

(a) optical isomerism only (b) geometrical isomerism only

(c) optical and geometrical isomerism (d) tautomerism only 25. nHeptane on monochlorination gives a number of isomers. How many of them are optically active?

(a) 1 (b) 2

(c) 3 (d) 4

EXERCISE – III

MORE THAN ONE CHOICE CORRECT 1. Which of the following compounds on dichlorination gives 3 different products (may not occur by same

mechanism in all compounds and even low yield products are to be counted).

(a)

(b)

(c)

CH3

(d)

2. Which of the following pairs represents enantiomers?

(a)

NH2

CO2H

H3C H , CH3

NH2

HO2C

H

(b)

CH3

H

CH3

OH

Cl H , H

H H3C

OH

CH3

Cl

(c)

CH3

H

CH3

OH

Cl H ,

H

H3C OH

Cl H3C

H

(d)

NH2

H

CH3

CO2H

OH H , HO

H CO2H

CH3

NH2

H

3. Which of the following compounds show position as well as metamerism?

(a) CH3CH2CCH2CH3

O

(b) CH3OCH2CH2CH3

(c) CH3CH2CH2CHO (d) CH3CH2NH2 4. Which of the following compounds have gauche conformer less stable than staggered conformer?

(a) CH3CH3 (b) FCH2CH2OH

Page 37

(c) CH3CH2CH2CH3 (d) CH3COOCH2CH2OCOOH3

5. Identify among the following, set of functional isomers.

(a) CH3CH2CH2CO2H & CH3CHCH3

CO2H

(b) CH3CH2CO2H & CH3CO2CH3

(c) CH3CH2CH2NH2 & CH3NCH3

CH3

(d) CH3CH2COCH2CH3 & CH3COCH

CH3

CH3

6. Which of these are isomers for an aromatic compound having molecular formula C7H8O?

(a)

CH3 OH

(b)

CH2OH

(c)

CH2OCH3

(d)

OCH3

7. Which of the following compounds are optically active?

(a)

H Cl H Cl

CH3

CH3

(b)

HO H

CH2OH

CH3

(c)

H3C CH2Cl

COOH

COOH

(d)

H OH HO H

CCl3

CCl3

8. Keto-enol tautomerism is observed in

(a) C6H5CHO (b) C6H5COCH3 (c) C6H5COC6H5 (d) C6H5COCH2COCH3 9. Only two structural isomeric monochloro derivatives are possible for

(a) n-Butane (b) 2,4-Dimethyl pentane

(c) Benzene (d) 2-Methyl propane 10. Which of the following compounds are the pair of structural isomers?

(a) (CH3)2CHOC2H5 ; CH3(CH2)2OC2H5

(b) CH3CH2NO2 ; CH2(NH2)COOH (c) (CH3)2CH(CH2)2CH3 ; CH3(CH2)2CH(CH3)2

(d) CH3CH2CO2H ; CH3CO2CH3

11.

CH3CHCH2CH2CH3

CH3

on monochlorination with Cl2 in the presence of light at room temperature

(reaction proceeds by free radical mechanism) gives ‗p‘ isomers. The product mixture on fractional distillation gives ‗q‘ fractions. Which of the following option is correct?

(a) ‗p‘ is 8 (b) ‗p‘ is 7

(c) ‗q‘ is 4 (d) ‗q‘ is 5 12. Which of the following compounds have anti conformer as the most stable conformer?

(a) CH3CH2CH2CH3 (Rotation about C2C3 bond)

(b) ClCH2CH2Br

(c) ClCH2CH2OH

(d) HOCH2CH2CH3

13. Which of the following compounds/isomers have centre of symmetry?

Page 38

(a)

NO2 Me3C

NO2 CMe3

(b)

CO2Et

O

O EtO2C

(trans isomer)

(c)

H3C

CH3

CH3

CH3

(d)

H OH

CH3

CH3

H HO

14. A mixture of racemic C2H5CHCO2H

CH3

can be resolved by reacting with

(a) C2H5CHOH

CH3

(b) C2H5CHNH2

CH3

(c) CH3CHNH2

CH3

(d) CH3CHOH

CH3

15. Which of the following pair represents structural isomers?

(a)

C=C and H

H3C C2H5

H C=C

H H3C

C2H5 H

(b) CH3COC2H5 and CH3CH2CH2COH

O O

(c) CH3CH2OCH2CH2CH3 and CH3OCHCH2CH3

CH3

(d) CH3CH2N=O and CH3CH=NOH

EXERCISE – IV

MATCH THE FOLLOWING

1.

Column I Column II

I. CH3(CH2)3CN and (CH3)3CCN (A) Position isomers

II. CH3CH2CH2OCH3 and CH3CH2OCH2CH3 (B) Chain isomers

III. CH3CHOCH3

CH3

and CH3CH2CH2OCH3 (C) Functional isomers

IV. CH2=CHCH=CH2 and CH3CCCH3 (D) Tautomers

(E) Metamers

2.

Column I Column II

I.

CH3 Br H Br H

C2H5

CH3 H Br H Br

C2H5

and (A) Meso compound

Page 39

II.

CH3 Br H Br H

CH3

CH3 H Br Br H

CH3

and (B) Enantiomers

III.

COOH OH H OH H

COOH

H HO H HO

COOH

COOH

and (C) Tautomers

IV. CH3–CH2–CHO and CH3–CH=CH–OH (D) Diastereomers

(E) Functional isomers

3.

Column I Column II

I. 2,3,5trihydroxy pentanoic acid (A) 2 optically active isomers

II. Lactic acid [CH3CH(OH)CO2H] (B) 4 optical isomers

III. 1,2dimethylcyclohexane (C) No optical isomer

IV. Glycine (NH2CH2CO2H) (D) 3 stereoisomers

(E) Meso isomer

ASSERTION AND REASON Direction: Read the following questions and choose

(A) If both Assertion and Reason are true and Reason is the correct explanation of the assertion.

(B) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion.

(C) If Assertion is true but Reason is false.

(D) If Assertion is false but Reason is true.

1. Assertion: Keto form of acetoacetic ester is more stable than its enolic form.

Reason: Enolic form of acetoacetic ester is stabilized by intramolecular hydrogen bonding.

(a) (A) (b) (B) (c) (C) (d) (D) 2. Assertion: Extent of enolization depends upon the solvent used.

Reason: If the solvent used makes hydrogen bonding with carbonyl group, it decreases the enol

content.

(a) (A) (b) (B) (c) (C) (d) (D) 3. Assertion: Oxime of acetone does not show geometrical isomerism.

Reason: Both the isomers have different configurations around double bond.

(a) (A) (b) (B) (c) (C) (d) (D) 4. Assertion: Disubstituted biphenyls exhibit optical isomerism.

Reason: If the o,o–substituents are very bulky even then there is no loss of planarity of the biphenyl

rings.

(a) (A) (b) (B) (c) (C) (d) (D)

5. Assertion: Racemic mixture and meso forms are optically inactive.

Reason: These are optically inactive due to external and internal compensation respectively.

(a) (A) (b) (B) (c) (C) (d) (D) PASSAGE BASED PROBLEMS

A bond between two carbon atoms is formed by the overlaping of sp3 hybrid orbitals of carbon atom

along their internuclear axis. It gives symmetrical electronic distribution along the internuclear axis and thus

rotation about carbon-carbon single bond is almost free. It results in number of spatial arrangements of atoms

called conformations. In any conformation, the angle between C–C and C–H bonds on adjacent carbon atom

Page 40

is known as dihedral angle. Stability and energy of the conformations depend on the torsional strain and Van

der Waal‘s strain. The order of stability of conformations of nbutane is fully eclipsed < partially eclipsed <

skew or gauche < staggered or anti. But in some cases, skew or gauche is stabilized by intramolecular

hydrogen bonding.Cyclic compounds also follow the same pattern. According to Baeyer strain theory, greater the deviation from the normal tetrahedral angle, greater is the angular strain and lesser is the stability of

cycloalkane. The most commonly found cyclic compounds usually contain six membered ring. They also show

conformations. The most stable conformation of cyclohexane is chair form which is staggered and free from any strain.

1. Dihedral angle in staggered and fully eclipsed conformations of nbutane are

(a) 120°, 0° (b) 180°, 0°

(c) 60°, 120° (d) 180°, 120° 2. Which of the following molecule has the highest deviation from tetrahedral bond angle?

(a) Cyclopropane (b) Cyclobutane

(c) Cyclopentane (d) Cyclohexane 3. The value of equilibrium constant for the equilibrium

Eclipsed form Staggered form

of ethane is

(a) 49 (b) 1

(c) 0.02 (d) 4.9

4. For CH3CHCHCH3

Cl OH

, which conformer is the most stable?

(a) Anti or staggered (b) Fully eclipsed

(c) Gauche or skew (d) Partially eclipsed 5. Which of the following statement is correct?

(a) Three membered ring is more stable than a four membered ring.

(b) Six membered ring is more stable than a five membered ring. (c) Six membered ring has less eclipsing strain than a five membered ring.

(d) All five carbons in cyclopentane are present in the same plane.

EXERCISE – V

SUBJECTIVE PROBLEMS

1. The specific rotation of a pure enantiomer is +12°. What will be its observed rotation

if it is isolated from a reaction with (a) 20% racemization and 80% retention and (b) 80% racemization and 20% inversion?

2. Find out the total stereoisomers for the following compound,

.

3. How many optical isomers are possible for the compound,

C F Cl

Br

?

4. Is the given isomer,

OH

H (I)

H

OH

H

HO

OH

H

, optically active or not?

Page 41

5. The compound,

NH2

O

Cl

cyclises to form one amide linkage. How many optical isomers will be

there for the cyclic product? 6. One or more of the following compound are in their most stable tautomeric forms. Which are they?

(a)

H3C C

H

C C

OCH2CH3

OH O

(b)

N O

H

(c)

H

O

H (d)

CH3C

CH

OH O

CCH3

7. Draw all Newmann projections of FCH2CH2OCH3 and arrange the conformers in their decreasing

order of stability. 8. Draw sawhorse and Newman projection diagrams of the following structures, which are shown as

Fischer projections. Which compound is optically inactive?

(a)

CH3

H

CH3

OH

HO H (b)

OH

H3C

CH3

H

H OH

(c)

H

H3C

CH3

H

OH

OH (d)

CHO

H

CH2OH

HO H

OH

(e)

CH3

H

CH3

Br

H Br (f)

CO2H

H OH

H OH

CO2H

9. Write the structure of a proton tautomer of each of the following compounds.

(a) N

OH (b)

Ph

O

O

CH3 (c)

O

OH

(d)

N=O HO (e)

O O

O

(f)

O

(g)

N

H

(h) CH3NO2 (i)

NO2 H3C

(j)

CN (k)

Ph

O CH3

N

H

10. In what stereoisomeric forms would you expect the following compounds to exist?

Page 42

(a) EtCH(CO2H)Me (b)

O

(c)

(d) Et(Me)C=C=C(Me)Et

(e)

Ph

Ph

CO2H

CO2H

(f)

Br

I

ANSWERS

EXERCISE –I

AIEEE-SINGLE CHOICE CORRECT

1. (b) 2. (b) 3. (b) 4. (c) 5. (a)

6. (c) 7. (c) 8. (b) 9. (c) 10. (c)

11. (a) 12. (b) 13. (a) 14. (b) 15. (c)

16. (a) 17. (b) 18. (c) 19. (d) 20. (c)

21. (b) 22. (a) 23. (a) 24. (c) 25. (b)

EXERCISE – II

IIT-JEE-SINGLE CHOICE CORRECT

1. (c) 2. (b) 3. (b) 4. (a) 5. (b)

6. (b) 7. (b) 8. (c) 9. (d) 10. (d)

11. (d) 12. (c) 13. (a) 14. (d) 15. (d)

16. (c) 17. (b) 18. (b) 19. (d) 20. (b)

21. (b) 22. (a) 23. (b) 24. (a) 25. (b)

EXERCISE – III

MORE THAN ONE CHOICE CORRECT

1. (a, b) 2. (a, c) 3. (a, b) 4. (c, d) 5. (b, c)

6. (a, b, d) 7. (b, d) 8. (b, d) 9. (a, d) 10. (a, b, d)

11. (a,d) 12. (a,b,d) 13. (b,c) 14. (a,b) 15.(b,c,d)

EXERCISE – IV

MATCH THE FOLLOWING

1. I (B) ; II (E) ; III (A) ; IV (C)

2. I (B) ; II (D) ; III (A) ; IV (C)

Page 43

3. I (B) ; II (A) ; III (E) ; IV (C)

ASSERTION AND REASON

1. (d) 2. (a) 3. (c) 4. (d) 5. (a)

PASSAGE BASED PROBLEMS

1. (b) 2. (a) 3. (a) 4. (c) 5. (b)

EXERCISE – V

SUBJECTIVE PROBLEMS 1. (a) obs = 0.80 (+12°) = +9.6°

(b) obs = 0.20 (12°) = 2.4°

2. 4 geometrical isomers which are (trans, trans), (trans, cis), (cis, cis), (cis, trans) and each geometrical

isomers will have 2 optical isomers. So, a total of 8 stereoisomers will be obtained.

3. 2

4. Optically active

5.

NH * ,

*

O 4 optical isomers

6.

CH3C

CH

COEt

O H

O

(Crossconjugation and intramolecular hydrogen bonding)

CH3C

CH

CCH3

O H

O

7.

rotation by 60°

rotation by 60°

Partially Eclipsed (III) Skew or gauche (II) Fully Eclipsed (I)

H

H

H

H H H H H

OCH3 F

H H H

H rotation by 60°

OCH3

F

OCH3

F

rotation by 60°

Staggered (IV)

H

H

H

H

OCH3

F

Partially Eclipsed (V) Skew or gauche (VI)

H

H

H H H H

H OCH3

H

F

rotation by 60°

F

OCH3

rotation by 60°

(I)

Decreasing order of stability of conformers:

(IV) > [(II) = (VI)] > [(III) = (V)] > (I).

Page 44

8.

Sawhorse representation

Newman representation

Optical activity

(a)

CH3

H

CH3

OH

HO H

H OH

H OH

CH3 CH3

H OH

CH3

H OH

H3C

Optically

active

(b)

OH

Me

Me

H

H OH

Me

H OH

Me OH

H

H Me

OH

H OH

Me

Optically inactive

(c)

OH Me

Me

H

H OH

Me

H OH

Me

OH

H

H Me OH

H

OH

Me

Optically active

(d)

CHO

HO

CH2OH

H

H OH

H OH

CH2OH CHO

H OH

H HO

CHO

H OH

HOH2C

Optically

active

(e)

Me

Me

H

H Br

Br

Br

H

Me Me

H Br

H

Me

Br H Br

Me

Optically

inactive

(f)

COOH

H OH H OH

COOH

H OH

COOH COOH

H OH

H

COOH

OH H OH

HOOC

Optically

inactive

9. (a) N

OH N O

(b)

Ph

O

O

CH3 PhC C=CH2

O O

H

(c)

O

OH

O

O

(d)

N=O HO NOH O

(e)

O O

O

OH OH

OH

Page 45

(f)

O OH

(g)

N

H

N

(h)

CH3NO2 CH2=N

OH

O

(i)

NO2 H3C CH2 N

OH

O

(j)

CN C NH

(k)

Ph

O CH3

N

H

PhC=NCH3

OH

10. (a)

Et

H

COOH

Me

Et

H

COOH

Me (b)

CH3

O H

CH3

H and

CH3

O H CH3

H

(c)

Me

H

Me

H

Me

H

Me

H

(d)

Et

C=C=C Me

Me

Et

C=C=C Me

Et Me

Et

(e)

COOH

Ph COOH

H Ph

H

H H

COOH

Ph

COOH H

Ph

H

H H

(f)

Br

I I

Br