334 CHEMISTRY

ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES

AND TECHNIQUES

After studying this unit, you will be

able to

••

• understand reasons for

tetravalence of carbon and

shapes of organic molecules;

••

• write structures of organic

molecules in various ways;

••

• classify the organic compounds;

••

• name the compounds according

to IUPAC system of

nomenclature and also derive

their structures from the given

names;

••

• understand the concept of

organic reaction mechanism;

••

• explain the influence of

electronic displacements on

structure and reactivity of

organic compounds;

••

• recognise the types of organic

reactions;

••

• learn the techniques of

purification of organic

compounds;

••

• write the chemical reactions

involved in the qualitative

analysis of organic compounds;

••

• understand the principles

involved in quantitative analysis

of organic compounds.

In the previous unit you have learnt that the element

carbon has the unique property called catenation due to

which it forms covalent bonds with other carbon atoms.

It also forms covalent bonds with atoms of other elements

like hydrogen, oxygen, nitrogen, sulphur, phosphorus and

halogens. The resulting compounds are studied under a

separate branch of chemistry called organic chemistry.

This unit incorporates some basic principles and

techniques of analysis required for understanding the

formation and properties of organic compounds.

12.1 GENERAL INTRODUCTION

Organic compounds are vital for sustaining life on earth

and include complex molecules like genetic information

bearing deoxyribonucleic acid (DNA) and proteins that

constitute essential compounds of our blood, muscles and

skin. Organic compounds appear in materials like clothing,

fuels, polymers, dyes and medicines. These are some of

the important areas of application of these compounds.

Science of organic chemistry is about two hundred

years old. Around the year 1780, chemists began to

distinguish between organic compounds obtained from

plants and animals and inorganic compounds prepared

from mineral sources. Berzilius, a Swedish chemist

proposed that a ‘vital force’ was responsible for the

formation of organic compounds. However, this notion

was rejected in 1828 when F. Wohler synthesised an

organic compound, urea from an inorganic compound,

ammonium cyanate.

4 22

Heat

NH CNO NH CONH

   →

Ammonium cyanate Urea

The pioneering synthesis of acetic acid by Kolbe (1845)

and that of methane by Berthelot (1856) showed

conclusively that organic compounds could be synthesised

from inorganic sources in a laboratory.

UNIT 12

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335ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

The development of electronic theory of

covalent bonding ushered organic chemistry

into its modern shape.

12.2 TETRAVALENCE OF CARBON:

SHAPES OF ORGANIC COMPOUNDS

12.2.1 The Shapes of Carbon Compounds

The knowledge of fundamental concepts of

molecular structure helps in understanding

and predicting the properties of organic

compounds. You have already learnt theories

of valency and molecular structure in Unit 4.

Also, you already know that tetravalence of

carbon and the formation of covalent bonds

by it are explained in terms of its electronic

configuration and the hybridisation of s and

p orbitals. It may be recalled that formation

and the shapes of molecules like methane

(CH

), ethene (C

), ethyne (C

) are

explained in terms of the use of sp

, sp

and

sp hybrid orbitals by carbon atoms in the

respective molecules.

Hybridisation influences the bond length

and bond enthalpy (strength) in compounds.

The sp hybrid orbital contains more s

character and hence it is closer to its nucleus

and forms shorter and stronger bonds than

the sp

hybrid orbital. The sp

hybrid orbital

is intermediate in s character between sp and

and, hence, the length and enthalpy of the

bonds it forms, are also intermediate between

them. The change in hybridisation affects the

electronegativity of carbon. The greater the s

character of the hybrid orbitals, the greater is

the electronegativity. Thus, a carbon atom

having an sp hybrid orbital with 50% s

character is more electronegative than that

possessing sp

or sp

hybridised orbitals. This

relative electronegativity is reflected in several

physical and chemical properties of the

molecules concerned, about which you will

learn in later units.

12.2.2 Some Characteristic Features of

ππ

Bonds

In a π (pi) bond formation, parallel orientation

of the two p orbitals on adjacent atoms is

necessary for a proper sideways overlap.

Thus, in H

C=CH

molecule all the atoms must

be in the same plane. The p orbitals are

mutually parallel and both the p orbitals are

perpendicular to the plane of the molecule.

Rotation of one CH

fragment with respect to

other interferes with maximum overlap of p

orbitals and, therefore, such rotation about

carbon-carbon double bond (C=C) is

restricted. The electron charge cloud of the π

bond is located above and below the plane of

bonding atoms. This results in the electrons

being easily available to the attacking

reagents. In general, π bonds provide the most

reactive centres in the molecules containing

multiple bonds.

Problem 12.1

How many σ and π bonds are present in

each of the following molecules?

(a) HC≡CCH=CHCH

(b) CH

=C=CHCH

Solution

(a) σ

C – C

: 4; σ

C–H

: 6; π

C=C

:1; π C≡C:2

(b) σ

C – C

: 3; σ

C–H

: 6; π

C=C

: 2.

Problem 12.2

What is the type of hybridisation of each

carbon in the following compounds?

(a) CH

Cl, (b) (CH

)

CO, (c) CH

CN,

(d) HCONH

, (e) CH

CH=CHCN

Solution

(a) sp

, (b) sp

, sp

, (c) sp

, sp, (d) sp

, (e)

, sp

Problem 12.3

Write the state of hybridisation of carbon

in the following compounds and shapes

of each of the molecules.

(a) H

C=O, (b) CH

F, (c) HC≡N.

Solution

(a) sp

hybridised carbon, trigonal planar;

(b) sp

hybridised carbon, tetrahedral; (c)

sp hybridised carbon, linear.

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12.3 STRUCTURAL REPRESENTATIONS

OF ORGANIC COMPOUNDS

12.3.1Complete, Condensed and Bond-line

Structural Formulas

Structures of organic compounds are

represented in several ways. The Lewis

structure or dot structure, dash structure,

condensed structure and bond line structural

formulas are some of the specific types. The

Lewis structures, however, can be simplified

by representing the two-electron covalent

bond by a dash (–). Such a structural formula

focuses on the electrons involved in bond

formation. A single dash represents a single

bond, double dash is used for double bond

and a triple dash represents triple bond. Lone-

pairs of electrons on heteroatoms (e.g.,

oxygen, nitrogen, sulphur, halogens etc.) may

or may not be shown. Thus, ethane (C

ethene (C

), ethyne (C

) and methanol

(CH

OH) can be represented by the following

structural formulas. Such structural

representations are called complete structural

formulas.

Similarly, CH

can be further condensed to CH

(CH

)

For further simplification, organic chemists

use another way of representing the

structures, in which only lines are used. In

this bond-line structural representation of

organic compounds, carbon and hydrogen

atoms are not shown and the lines

representing carbon-carbon bonds are drawn

in a zig-zag fashion. The only atoms

specifically written are oxygen, chlorine,

nitrogen etc. The terminals denote methyl

(–CH

) groups (unless indicated otherwise by

a functional group), while the line junctions

denote carbon atoms bonded to appropriate

number of hydrogens required to satisfy the

valency of the carbon atoms. Some of the

examples are represented as follows:

(i) 3-Methyloctane can be represented in

various forms as:

(a) CH

CHCH

These structural formulas can be further

abbreviated by omitting some or all of the

dashes representing covalent bonds and by

indicating the number of identical groups

attached to an atom by a subscript. The

resulting expression of the compound is called

a condensed structural formula. Thus, ethane,

ethene, ethyne and methanol can be written

as:

C=CH

≡≡

≡CH CH

Ethane Ethene Ethyne Methanol

Ethane Ethene

Ethyne Methanol

(ii) Various ways of representing 2-bromo

butane are:

(a) CH

CHBrCH

(b)

(c)

(b)

(c)

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In cyclic compounds, the bond-line formulas

may be given as follows:

Cyclopropane

Cyclopentane

chlorocyclohexane

Problem 12.4

Expand each of the following condensed

formulas into their complete structural

formulas.

(a) CH

COCH

(b) CH

CH=CH(CH

)

Solution

(b)

Solution

Condensed formula:

(a) HO(CH

)

CH(CH

)CH(CH

)

(b) HOCH(CN)

Bond-line formula:

(a)

(b)

Problem 12.5

For each of the following compounds,

write a condensed formula and also their

bond-line formula.

(a) HOCH

CH(CH

)CH(CH

)CH

(b)

(a)

Problem 12.6

Expand each of the following bond-line

formulas to show all the atoms including

carbon and hydrogen

(a)

(b)

(c)

(d)

Solution

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338 CHEMISTRY

Framework model

Ball and stick model

Space filling model

Fig. 12.2

12.3.2 Three-Dimensional

Representation of Organic

Molecules

The three-dimensional (3-D) structure of

organic molecules can be represented on

paper by using certain conventions. For

example, by using solid (

) and dashed

(

) wedge formula, the 3-D image of a

molecule from a two-dimensional picture

can be perceived. In these formulas the

solid-wedge is used to indicate a bond

projecting out of the plane of paper, towards

the observer. The dashed-wedge is used to

depict the bond projecting out of the plane of

the paper and away from the observer. Wedges

are shown in such a way that the broad end

of the wedge is towards the observer. The

bonds lying in plane of the paper are depicted

by using a normal line (—). 3-D representation

of methane molecule on paper has been

shown in Fig. 12.1.

Fig. 12.1 Wedge-and-dash representation of CH

Molecular Models

Molecular models are physical devices that

are used for a better visualisation and

perception of three-dimensional shapes of

organic molecules. These are made of wood,

plastic or metal and are commercially

available. Commonly three types of molecular

models are used: (1) Framework model, (2)

Ball-and-stick model, and (3) Space filling

model. In the framework model only the

bonds connecting the atoms of a molecule

and not the atoms themselves are shown.

This model emphasizes the pattern of bonds

of a molecule while ignoring the size of atoms.

In the ball-and-stick model, both the atoms

and the bonds are shown. Balls represent

atoms and the stick denotes a bond.

Compounds containing C=C (e.g., ethene) can

best be represented by using springs in place

of sticks. These models are referred to as ball-

and-spring model. The space-filling model

emphasises the relative size of each atom

based on its van der Waals radius. Bonds

are not shown in this model. It conveys the

volume occupied by each atom in the

molecule. In addition to these models,

computer graphics can also be used for

molecular modelling.

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339ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

12.4 CLASSIFICATION OF ORGANIC

COMPOUNDS

The existing large number of organic

compounds and their ever-increasing

numbers has made it necessary to classify

them on the basis of their structures. Organic

compounds are broadly classified as follows:

I. Acyclic or open chain compounds

These compounds are also called as aliphatic

compounds and consist of straight or

branched chain compounds, for example:

(homocyclic).

Cyclopropane Cyclohexane Cyclohexene

Sometimes atoms other than carbon

are also present in the ring (heterocylic).

Tetrahydrofuran given below is an example of

this type of compound:

Tetrahydrofuran

These exhibit some of the properties similar to

those of aliphatic compounds.

(b) Aromatic compounds

Aromatic compounds are special types of

compounds. You will learn about these

compounds in detail in Unit 13. These include

benzene and other related ring compounds

(benzenoid). Like alicyclic compounds,

aromatic comounds may also have hetero

atom in the ring. Such compounds are called

hetrocyclic aromatic compounds. Some of the

examples of various types of aromatic

compounds are:

Benzenoid aromatic compounds

Benzene Aniline Naphthalene

Non-benzenoid compound

Tropone

Isobutane

Acetaldehyde

Acetic acid

Ethane

II Cyclic or closed chain or ring

compounds

(a) Alicyclic compounds

Alicyclic (aliphatic cyclic) compounds contain

carbon atoms joined in the form of a ring

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340 CHEMISTRY

Heterocyclic aromatic compounds

Furan Thiophene Pyridine

Organic compounds can also be classified

on the basis of functional groups, into families

or homologous series.

12.4.1 Functional Group

The functional group is an atom or a group of

atoms joined to the carbon chain which is

responsible for the characteristic chemical

properties of the organic compounds. The

examples are hydroxyl group (–OH), aldehyde

group (–CHO) and carboxylic acid group

(–COOH) etc.

12.4.2 Homologous Series

A group or a series of organic compounds each

containing a characteristic functional group

forms a homologous series and the members

of the series are called homologues. The

members of a homologous series can be

represented by general molecular formula and

the successive members differ from each other

in molecular formula by a –CH

unit. There are

a number of homologous series of organic

compounds. Some of these are alkanes,

alkenes, alkynes, haloalkanes, alkanols,

alkanals, alkanones, alkanoic acids, amines etc.

It is also possible that a compound contains

two or more identical or different functional

groups. This gives rise to polyfunctional

compounds.

12.5 NOMENCLATURE OF ORGANIC

COMPOUNDS

Organic chemistry deals with millions of

compounds. In order to clearly identify them, a

systematic method of naming has been

developed and is known as the IUPAC

(International Union of Pure and Applied

Chemistry) system of nomenclature. In this

systematic nomenclature, the names are correlated

with the structure such that the reader or listener

can deduce the structure from the name.

Before the IUPAC system of nomenclature,

however, organic compounds were assigned

names based on their origin or certain

properties. For instance, citric acid is named

so because it is found in citrus fruits and the

acid found in red ant is named formic acid

since the Latin word for ant is formica. These

names are traditional and are considered as

trivial or common names. Some common

names are followed even today. For example,

Buckminsterfullerene is a common name given

to the newly discovered C

cluster (a form of

carbon) noting its structural similarity to the

geodesic domes popularised by the famous

architect R. Buckminster Fuller. Common

names are useful and in many cases

indispensable, particularly when the

alternative systematic names are lengthy and

complicated. Common names of some organic

compounds are given in Table 12.1.

Table 12.1 Common or Trivial Names of Some

Organic Compounds

12.5.1 The IUPAC System of Nomenclature

A systematic name of an organic compound is

generally derived by identifying the parent

hydrocarbon and the functional group(s)

attached to it. See the example given below.

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341ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

By further using prefixes and suffixes, the

parent name can be modified to obtain the

actual name. Compounds containing carbon

and hydrogen only are called hydrocarbons. A

hydrocarbon is termed saturated if it contains

only carbon-carbon single bonds. The IUPAC

name for a homologous series of such

compounds is alkane. Paraffin (Latin: little

affinity) was the earlier name given to these

compounds. Unsaturated hydrocarbons are

those, which contain at least one carbon-

carbon double or triple bond.

12.5.2 IUPAC Nomenclature of Alkanes

Straight chain hydrocarbons: The names

of such compounds are based on their chain

structure, and end with suffix ‘-ane’ and carry

a prefix indicating the number of carbon

atoms present in the chain (except from CH

, where the prefixes are derived from

trivial names). The IUPAC names of some

straight chain saturated hydrocarbons are

given in Table 12.2. The alkanes in Table 12.2

differ from each other by merely the number

of -CH

groups in the chain. They are

homologues of alkane series.

In order to name such compounds, the names

of alkyl groups are prefixed to the name of

parent alkane. An alkyl group is derived from

a saturated hydrocarbon by removing a

hydrogen atom from carbon. Thus, CH

becomes -CH

and is called methyl group. An

alkyl group is named by substituting ‘yl’ for

‘ane’ in the corresponding alkane. Some alkyl

groups are listed in Table 12.3.

Table 12.3 Some Alkyl Groups

Table 12.2 IUPAC Names of Some Unbranched

Saturated Hydrocarbons

Branched chain hydrocarbons: In a

branched chain compound small chains of

carbon atoms are attached at one or more

carbon atoms of the parent chain. The small

carbon chains (branches) are called alkyl

groups. For example:

–CH–CH

–CH

–CH–CH

  

(a) (b)

Abbreviations are used for some alkyl

groups. For example, methyl is abbreviated as

Me, ethyl as Et, propyl as Pr and butyl as Bu.

The alkyl groups can be branched also. Thus,

propyl and butyl groups can have branched

structures as shown below.

-CH- CH

-CH

-CH- CH

-CH-CH

  

Isopropyl- sec-Butyl- Isobutyl-

 

-C- CH

-C-CH

 

tert-Butyl- Neopentyl-

Common branched groups have specific trivial

names. For example, the propyl groups can

either be n-propyl group or isopropyl group.

The branched butyl groups are called sec-

butyl, isobutyl and tert-butyl group. We also

encounter the structural unit,

–CH

C(CH

)

, which is called neopentyl group.

Nomenclature of branched chain alkanes:

We encounter a number of branched chain

alkanes. The rules for naming them are given

below.

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342 CHEMISTRY

separated from the groups by hyphens

and there is no break between methyl

and nonane.]

4. If two or more identical substituent groups

are present then the numbers are

separated by commas. The names of

identical substituents are not repeated,

instead prefixes such as di (for 2), tri

(for 3), tetra (for 4), penta (for 5), hexa (for

6) etc. are used. While writing the name of

the substituents in alphabetical order,

these prefixes, however, are not considered.

Thus, the following compounds are

named as:

   

-CH-CH

CCH

CHCH



2,4-Dimethylpentane 2,2,4-Trimethylpentane

C H

 

CH

CHCCH

CH



3-Ethyl-4,4-dimethylheptane

5. If the two substituents are found in

equivalent positions, the lower number is

given to the one coming first in the

alphabetical listing. Thus, the following

compound is 3-ethyl-6-methyloctane and

not 6-ethyl-3-methyloctane.

1 2 3 4 5 6 7 8

— CH

—CH—CH

—CH

—CH—CH

—CH

 

6. The branched alkyl groups can be named

by following the above mentioned

procedures. However, the carbon atom of

the branch that attaches to the root

alkane is numbered 1 as exemplified

below.

4 3 2 1

–CH–CH

–CH–

 

1,3-Dimethylbutyl-

1. First of all, the longest carbon chain in

the molecule is identified. In the example

(I) given below, the longest chain has nine

carbons and it is considered as the parent

or root chain. Selection of parent chain as

shown in (II) is not correct because it has

only eight carbons.

1 2 3 4 5

1 2 3 4 5 6 7

2. The carbon atoms of the parent chain are

numbered to identify the parent alkane and

to locate the positions of the carbon atoms

at which branching takes place due to the

substitution of alkyl group in place of

hydrogen atoms. The numbering is done

in such a way that the branched carbon

atoms get the lowest possible numbers.

Thus, the numbering in the above example

should be from left to right (branching at

carbon atoms 2 and 6) and not from right

to left (giving numbers 4 and 8 to the

carbon atoms at which branches are

attached).

1 2 3 4 5 6 7 8 9

C  C  C  C  C  C C  C  C

 

C CC

9 8 7 6 5 4 3 2 1

C  C  C  C  C  C  C  C  C

 

C C  C

3. The names of alkyl groups attached

as a branch are then prefixed to the

name of the parent alkane and position

of the substituents is indicated by the

appropriate numbers. If different alkyl

groups are present, they are listed in

alphabetical order. Thus, name for the

compound shown above is: 6-ethyl-2-

methylnonane. [Note: the numbers are

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343ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

The name of such branched chain alkyl group

is placed in parenthesis while naming the

compound. While writing the trivial names of

substituents’ in alphabetical order, the

prefixes iso- and neo- are considered to be

the part of the fundamental name of alkyl

group. The prefixes sec- and tert- are not

considered to be the part of the fundamental

name. The use of iso and related common

prefixes for naming alkyl groups is also allowed

by the IUPAC nomenclature as long as these

are not further substituted. In multi-

substituted compounds, the following rules

may aso be remembered:

• If there happens to be two chains of equal

size, then that chain is to be selected which

contains more number of side chains.

• After selection of the chain, numbering is

to be done from the end closer to the

substituent.

5-(2-Ethylbutyl)-3,3-dimethyldecane

[and not 5-(2,2-Dimethylbutyl)-3-ethyldecane]

5-sec-Butyl-4-isopropyldecane

5-(2,2-Dimethylpropyl)nonane

Problem 12.7

Structures and IUPAC names of some

hydrocarbons are given below. Explain

why the names given in the parentheses

are incorrect.

3-Ethyl-1,1-dimethylcyclohexane

(not 1-ethyl-3,3-dimethylcyclohexane)

2,5,6- Trimethyloctane

[and not 3,4,7-Trimethyloctane]

3-Ethyl-5-methylheptane

[and not 5-Ethyl-3-methylheptane]

Cyclic Compounds: A saturated monocyclic

compound is named by prefixing ‘cyclo’ to the

corresponding straight chain alkane. If side

chains are present, then the rules given above

are applied. Names of some cyclic compounds

are given below.

Solution

(a) Lowest locant number, 2,5,6 is lower

than 3,5,7, (b) substituents are in

equivalent position; lower number is given

to the one that comes first in the name

according to alphabetical order.

12.5.3 Nomenclature of Organic

Compounds having Functional

Group(s)

A functional group, as defined earlier, is an

atom or a group of atoms bonded together in a

unique manner which is usually the site of

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chemical reactivity in an organic molecule.

Compounds having the same functional group

undergo similar reactions. For example,

OH, CH

OH, and (CH

)

CHOH — all

having -OH functional group liberate hydrogen

on reaction with sodium metal. The presence

of functional groups enables systematisation

of organic compounds into different classes.

Examples of some functional groups with their

prefixes and suffixes along with some

examples of organic compounds possessing

these are given in Table 12.4.

First of all, the functional group present

in the molecule is identified which determines

the choice of appropriate suffix. The longest

chain of carbon atoms containing the

functional group is numbered in such a way

that the functional group is attached at the

carbon atom possessing lowest possible

number in the chain. By using the suffix as

given in Table 12.4, the name of the compound

is arrived at.

In the case of polyfunctional compounds,

one of the functional groups is chosen as the

principal functional group and the compound is

then named on that basis. The remaining

functional groups, which are subordinate

functional groups, are named as substituents

using the appropriate prefixes. The choice of

principal functional group is made on the basis

of order of preference. The order of decreasing

priority for some functional groups is:

-COOH, –SO

H, -COOR (R=alkyl group), COCl,

-CONH

, -CN,-HC=O, >C=O, -OH, -NH

, >C=C<,

-C

≡≡

≡C- .

The –R, C

-, halogens (F, Cl, Br, I), –NO

alkoxy (–OR) etc. are always prefix

substituents. Thus, a compound containing

both an alcohol and a keto group is named as

hydroxyalkanone since the keto group is

preferred to the hydroxyl group.

For example, HOCH

(CH

)

COCH

will be

named as 7-hydroxyheptan-2-one and not as

2-oxoheptan -7-ol. Similarly, BrCH

CH=CH

is named as 3-bromoprop-1-ene and not 1-

bromoprop-2-ene.

If more than one functional group of the

same type are present, their number is

indicated by adding di, tri, etc. before the class

suffix. In such cases the full name of the

parent alkane is written before the class suffix.

For example CH

(OH)CH

(OH) is named as

ethane–1,2–diol. However, the ending – ne of

the parent alkane is dropped in the case of

compounds having more than one double

or triple bond; for example, CH

=CH-CH=CH

is named as buta–1,3–diene.

Problem 12.8

Write the IUPAC names of the compounds

i-iv from their given structures.

Solution

• The functional group present is an

alcohol (OH). Hence the suffix is ‘-ol’.

• The longest chain containing -OH has

eight carbon atoms. Hence the

corresponding saturated hydrocarbon

is octane.

• The -OH is on carbon atom 3. In

addition, a methyl group is attached

at 6

carbon.

Hence, the systematic name of this

compound is 6-Methyloctan-3-ol.

Solution

The functional group present is ketone

(>C=O), hence suffix ‘-one’. Presence of

two keto groups is indicated by ‘di’,

hence suffix becomes ‘dione’. The two

keto groups are at carbons 2 and 4. The

longest chain contains 6 carbon atoms,

hence, parent hydrocarbon is hexane.

Thus, the systematic name is Hexane-

2,4-dione.

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Table 12.4 Some Functional Groups and Classes of Organic Compounds

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Solution

Here, two functional groups namely

ketone and carboxylic acid are present.

The principal functional group is the

carboxylic acid group; hence the parent

chain will be suffixed with ‘oic’ acid.

Numbering of the chain starts from

carbon of – COOH functional group. The

keto group in the chain at carbon 5 is

indicated by ‘oxo’. The longest chain

including the principal functional

group has 6 carbon atoms; hence the

parent hydrocarbon is hexane. The

compound is, therefore, named as

5-Oxohexanoic acid.

Solution

The two C=C functional groups are

present at carbon atoms 1 and 3, while

the C≡C functional group is present at

carbon 5. These groups are indicated by

suffixes ‘diene’ and ‘yne’ respectively. The

longest chain containing the functional

groups has 6 carbon atoms; hence the

parent hydrocarbon is hexane. The name

of compound, therefore, is Hexa-1,3-

dien-5-yne.

Problem 12.9

Derive the structure of (i) 2-Chlorohexane,

(ii) Pent-4-en-2-ol, (iii) 3- Nitrocyclohexene,

(iv) Cyclohex-2-en-1-ol, (v) 6-Hydroxy-

heptanal.

Solution

(i) ‘hexane’ indicates the presence of

6 carbon atoms in the chain. The

functional group chloro is present at

carbon 2. Hence, the structure of the

compound is CH

CH(Cl)CH

(ii) ‘pent’ indicates that parent

hydrocarbon contains 5 carbon atoms in

the chain. ‘en’ and ‘ol’ correspond to the

functional groups C=C and -OH at carbon

atoms 4 and 2 respectively. Thus, the

structure is

=CHCH

CH (OH)CH

(iii) Six membered ring containing a

carbon-carbon double bond is implied by

cyclohexene, which is numbered as

shown in (I). The prefix 3-nitro means that

a nitro group is present on C-3. Thus,

complete structural formula of the

compound is (II). Double bond is suffixed

functional group whereas NO

is prefixed

functional group therefore double bond

gets preference over –NO

group:

(iv) ‘1-ol’ means that a -OH group is

present at C-1. OH is suffixed functional

group and gets preference over C=C

bond. Thus the structure is as shown

in (II):

(v) ‘heptanal’ indicates the compound to

be an aldehyde containing 7 carbon

atoms in the parent chain. The

‘6-hydroxy’ indicates that -OH group is

present at carbon 6. Thus, the structural

formula of the compound is:

CH(OH)CH

CHO. Carbon

atom of –CHO group is included while

numbering the carbon chain.

12.5.4 Nomenclature of Substituted

Benzene Compounds

For IUPAC nomenclature of substituted

benzene compounds, the substituent is

placed as prefix to the word benzene as

shown in the following examples. However,

common names (written in bracket below)

of many substituted benzene compounds

are also universally used.

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347ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

2-Chloro-4-methylanisole

4-Ethyl-2-methylaniline

1-Chloro-2,4-dinitrobenzene

(not 4-chloro,1,3-dinitrobenzene)

If benzene ring is disubstituted, the

position of substituents is defined

by numbering the carbon atoms of

the ring such that the substituents

are located at the lowest numbers

possible.For example, the compound(b) is

named as 1,3-dibromobenzene and not as

1,5-dibromobenzene.

Substituent of the base compound is

assigned number1 and then the direction of

numbering is chosen such that the next

substituent gets the lowest number. The

substituents appear in the name in

alphabetical order. Some examples are given

below.

2-Chloro-1-methyl-4-nitrobenzene

(not 4-methyl-5-chloro-nitrobenzene)

3,4-Dimethylphenol

Methylbenzene Methoxybenzene Aminobenzene

(Toluene) (Anisole) (Aniline)

Nitrobenzene Bromobenzene

1,2-Dibromo-

benzene

1,3-Dibromo-

benzene

1,4-Dibromo-

benzene

(a)

(b)

(c)

In the trivial system of nomenclature the

terms ortho ( o), meta (m) and para (p) are used

as prefixes to indicate the relative positions

1,2;1,3 and 1,4 respectively. Thus,

1,3-dibromobenzene (b) is named as

m-dibromobenzene (meta is abbreviated as

m-) and the other isomers of dibromobenzene

1,2-(a) and 1,4-(c), are named as ortho (or just

o-) and para (or just p-)-dibromobenzene,

respectively.

For tri - or higher substituted benzene

derivatives, these prefixes cannot be used and

the compounds are named by identifying

substituent positions on the ring by following

the lowest locant rule. In some cases, common

name of benzene derivatives is taken as the

base compound.

When a benzene ring is attached to an

alkane with a functional group, it is

considered as substituent, instead of a parent.

The name for benzene as substituent is phenyl

-, also abbreviated as Ph).

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348 CHEMISTRY

different carbon skeletons, these are referred to

as chain isomers and the phenomenon is termed

as chain isomerism. For example, C

represents three compounds:

Isomerism

Structural isomerism

Stereoisomerism

Chain

isomerism

Position

isomerism

Functional

group

isomerism

Metamerism Geometrical

isomerism

Optical

isomerism

12.6 ISOMERISM

The phenomenon of existence of two or more

compounds possessing the same molecular

formula but different properties is known as

isomerism. Such compounds are called as

isomers. The following flow chart shows

different types of isomerism.

12.6.1 Structural Isomerism

Compounds having the same molecular

formula but different structures (manners in

which atoms are linked) are classified as

structural isomers. Some typical examples of

different types of structural isomerism are given

below:

(i) Chain isomerism: When two or more

compounds have similar molecular formula but

(a) (b)

Problem 12.10

Write the structural formula of:

(a) o-Ethylanisole, (b) p-Nitroaniline,

(d) 4-Ethyl-1-fluoro-2-nitrobenzene.

Solution



−CHCH

Pentane Isopentane

(2-Methylbutane)



 C CH



Neopentane

(2,2-Dimethylpropane)

(ii) Position isomerism: When two or more

compounds differ in the position of substituent

atom or functional group on the carbon

skeleton, they are called position isomers and

this phenomenon is termed as position

isomerism. For example, the molecular

formula C

O represents two alcohols:



OH CH

−CH-CH

Propan-1-ol Propan-2-ol

(iii) Functional group isomerism: Two or

more compounds having the same molecular

formula but different functional groups are

called functional isomers and this

phenomenon is termed as functional group

isomerism. For example, the molecular

formula C

O represents an aldehyde and a

ketone:

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349ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

O H

 

−C-CH

−CH

—C= O

Propanone Propanal

(iv) Metamerism: It arises due to different alkyl

chains on either side of the functional group

in the molecule. For example, C

represents methoxypropane (CH

) and

ethoxyethane (C

12.6.2 Stereoisomerism

The compounds that have the same

constitution and sequence of covalent bonds

but differ in relative positions of their atoms

or groups in space are called stereoisomers.

This special type of isomerism is called as

stereoisomerism and can be classified as

geometrical and optical isomerism.

12.7 FUNDAMENTAL CONCEPTS IN

ORGANIC REACTION MECHANISM

In an organic reaction, the organic molecule

(also referred as a substrate) reacts with an

appropriate attacking reagent and leads to the

formation of one or more intermediate(s) and

finally product(s)

The general reaction is depicted as follows :

understanding the reactivity of organic

compounds and in planning strategy for their

synthesis.

In the following sections, we shall learn

some of the principles that explain how these

reactions take place.

12.7.1 Fission of a Covalent Bond

A covalent bond can get cleaved either by : (i)

heterolytic cleavage, or by (ii) homolytic

cleavage.

In heterolytic cleavage, the bond breaks

in such a fashion that the shared pair of

electrons remains with one of the fragments.

After heterolysis, one atom has a sextet

electronic structure and a positive charge and

the other, a valence octet with at least one

lone pair and a negative charge. Thus,

heterolytic cleavage of bromomethane will give

and Br

–

as shown below.

A species having a carbon atom possessing

sextext of electrons and a positive charge is

called a carbocation (earlier called carbonium

ion). The

ion is known as a methyl cation

or methyl carbonium ion. Carbocations are

classified as primary, secondary or tertiary

depending on whether one, two or three

carbons are directly attached to the positively

charged carbon. Some other examples of

carbocations are: CH

(ethyl cation, a

primary carbocation), (CH

)

H (isopropyl

cation, a secondary carbocation), and (CH

)

(tert-butyl cation, a tertiary carbocation).

Carbocations are highly unstable and reactive

species. Alkyl groups directly attached to the

positively charged carbon stabilise the

carbocations due to inductive and

hyperconjugation effects, which you will be

studying in the sections 12.7.5 and 12.7.9. The

observed order of carbocation stability is: C

< CH

< (CH

)

H < (CH

)

. These

carbocations have trigonal planar shape with

positively charged carbon being sp

hybridised. Thus, the shape of C

may be

considered as being derived from the overlap

of three equivalent C(sp

) hybridised orbitals

Organic

molecule

(Substrate)

[Intermediate] Product(s)

Attacking

Reagent

Byproducts

Substrate is that reactant which supplies

carbon to the new bond and the other reactant

is called reagent. If both the reactants supply

carbon to the new bond then choice is

arbitrary and in that case the molecule on

which attention is focused is called substrate.

In such a reaction a covalent bond

between two carbon atoms or a carbon and

some other atom is broken and a new bond is

formed. A sequential account of each step,

describing details of electron movement,

energetics during bond cleavage and bond

formation, and the rates of transformation

of reactants into products (kinetics) is

referred to as reaction mechanism. The

knowledge of reaction mechanism helps in

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350 CHEMISTRY

with 1s orbital of each of the three hydrogen

atoms. Each bond may be represented as

C(sp

)–H(1s) sigma bond. The remaining

carbon orbital is perpendicular to the

molecular plane and contains no electrons.

[Fig. 12.3(a)].

Fig. 12.3 (a) Shape of methyl carbocation

The heterolytic cleavage can also give a species

in which carbon gets the shared pair of

electrons. For example, when group Z

attached to the carbon leaves without

electron pair, the methyl anion is

formed. Such a carbon species carrying a

negative charge on carbon atom is called

carbanion. Carbon in carbanion is generally

hybridised and its structure is distorted

tetrahedron as shown in Fig. 12.3(b).

Carbanions are also unstable and reactive

species. The organic reactions which proceed

through heterolytic bond cleavage are called

ionic or heteropolar or just polar reactions.

In homolytic cleavage, one of the electrons

of the shared pair in a covalent bond goes with

each of the bonded atoms. Thus, in homolytic

cleavage, the movement of a single electron

takes place instead of an electron pair. The

single electron movement is shown by ‘half-

headed’ (fish hook:

) curved arrow. Such

cleavage results in the formation of neutral

species (atom or group) which contains an

unpaired electron. These species are called free

radicals. Like carbocations and carbanions,

free radicals are also very reactive. A homolytic

cleavage can be shown as:

Alkyl

free radical

Alkyl radicals are classified as primary,

secondary, or tertiary. Alkyl radical stability

increases as we proceed from primary to

tertiary:

Methyl Ethyl Isopropyl Tert-butyl

free free free free

radical radical radical radical

Organic reactions, which proceed by

homolytic fission are called free radical or

homopolar or nonpolar reactions.

12.7.2 Substrate and Reagent

Ions are generally not formed in the reactions of

organic compounds. Molecules as such

participate in the reaction. It is convenient to

name one reagent as substrate and other as

reagent. In general, a molecule whose carbon is

involved in new bond formation is called

substrate and the other one is called reagent.

When carbon-carbon bond is formed, the choice

of naming the reactants as substrate and

reagent is arbitrary and depends on molecule

under observation. Example:

(i) CH

= CH

+ Br

Br – CH

Substrate Reagent Product

(ii)

Nucleophiles and Electrophiles

Reagents attack the reactive site of the substrate.

The reactive site may be electron deficient

Fig. 12.3 (b) Shape of methyl carbanion

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351ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

portion of the molecule (a positive reactive site)

e.g., an atom with incomplete electron shell or

the positive end of the dipole in the molecule. If

the attacking species is electron rich, it attacks

these sites. If attacking species is electron

deficient, the reactive site for it is that part of the

substrate molecule which can supply electrons,

e.g.,

electrons in a double bond.

A reagent that brings an electron pair to the

reactive site is called a nucleophile (Nu:) i.e.,

nucleus seeking and the reaction is then called

nucleophilic. A reagent that takes away an

electron pair from reactive site is called

electrophile (E

) i.e., electron seeking and the

reaction is called electrophilic.

During a polar organic reaction, a

nucleophile attacks an electrophilic centre of

the substrate which is that specific atom or part

of the substrate which is electron deficient.

Similarly, the electrophiles attack at

nucleophilic centre, which is the electron

rich centre of the substrate. Thus, the

electrophiles receive electron pair from the

substrate when the two undergo bonding

interaction. A curved-arrow notation is used

to show the movement of an electron pair from

the nucleophile to the electrophile. Some

examples of nucleophiles are the negatively

charged ions with lone pair of electrons such

as hydroxide (HO

–

), cyanide (NC

–

) ions and

carbanions (R

–

). Neutral molecules such

etc., can also act as

nucleophiles due to the presence of lone pair

of electrons. Examples of electrophiles

include carbocations (

) and neutral

molecules having functional groups like

carbonyl group (>C=O) or alkyl halides

C-X, where X is a halogen atom). The

carbon atom in carbocations has sextet

configuration; hence, it is electron deficient

and can receive a pair of electrons from the

nucleophiles. In neutral molecules such as

alkyl halides, due to the polarity of the C-X

bond a partial positive charge is generated

on the carbon atom and hence the carbon atom

becomes an electrophilic centre at which a

nucleophile can attack.

Problem 12.11

Using curved-arrow notation, show the

formation of reactive intermediates when

the following covalent bonds undergo

heterolytic cleavage.

(a) CH

–SCH

, (b) CH

–CN, (c) CH

–Cu

Solution

Problem 12.12

Giving justification, categorise the

following molecules/ions as nucleophile

or electrophile:

Solution

Nucleophiles:

(

)

25 3 2

HS,CHO,CH N,HN

− −

−

These species have unshared pair of

electrons, which can be donated and

shared with an electrophile.

Electrophiles:

3 3 2

BF ,Cl,CH C O,NO

−=

Reactive sites have only six valence

electrons; can accept electron pair from

a nucleophile.

Problem 12.13

Identify electrophilic centre in the

following: CH

CH=O, CH

CN, CH

Solution

Among CH

=O, H

C C

≡N, and

–I, the starred carbon atoms are

electrophilic centers as they will have

partial positive charge due to polarity of

the bond.

12.7.3 Electron Movement in Organic

Reactions

The movement of electrons in organic reactions

can be shown by curved-arrow notation. It

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352 CHEMISTRY

shows how changes in bonding occur due to

electronic redistribution during the reaction.

To show the change in position of a pair of

electrons, curved arrow starts from the point

from where an electron pair is shifted and it

ends at a location to which the pair of electron

may move.

Presentation of shifting of electron pair is

given below :

(i)

from π bond to

adjacent bond position

(ii)

from π bond to

adjacent atom

(iii)

from atom to adjacent

bond position

Movement of single electron is indicated by

a single barbed ‘fish hooks’ (i.e. half headed

curved arrow). For example, in transfer of

hydroxide ion giving ethanol and in the

dissociation of chloromethane, the movement

of electron using curved arrows can be

depicted as follows:

12.7.4 Electron Displacement Effects in

Covalent Bonds

The electron displacement in an organic

molecule may take place either in the ground

state under the influence of an atom or a

substituent group or in the presence of an

appropriate attacking reagent. The electron

displacements due to the influence of

an atom or a substituent group present in the

molecule cause permanent polarlisation of the

bond. Inductive effect and resonance effects are

examples of this type of electron displacements.

Temporary electron displacement effects are

seen in a molecule when a reagent approaches

to attack it. This type of electron displacement

is called electromeric effect or polarisability

effect. In the following sections we will learn

about these types of electronic displacements.

12.7.5 Inductive Effect

When a covalent bond is formed between atoms

of different electronegativity, the electron

density is more towards the more

electronegative atom of the bond. Such a shift

of electron density results in a polar covalent

bond. Bond polarity leads to various electronic

effects in organic compounds.

Let us consider cholorethane (CH

Cl)

in which the C–Cl bond is a polar covalent

bond. It is polarised in such a way that the

carbon-1 gains some positive charge (δ

) and

the chlorine some negative charge (δ

–

). The

fractional electronic charges on the two atoms

in a polar covalent bond are denoted by symbol

δ (delta) and the shift of electron density is

shown by an arrow that points from δ

–

end of the polar bond.

δδ

−

→CH

→→

→Cl

2 1

In turn carbon-1, which has developed

partial positive charge (δ

) draws some electron

density towards it from the adjacent C-C bond.

Consequently, some positive charge (δδ

)

develops on carbon-2 also, where δδ

symbolises relatively smaller positive charge

as compared to that on carbon – 1. In other

words, the polar C – Cl bond induces polarity

in the adjacent bonds. Such polarisation of σ-

bond caused by the polarisation of adjacent

σ-bond is referred to as the inductive effect.

This effect is passed on to the subsequent

bonds also but the effect decreases rapidly as

the number of intervening bonds increases and

becomes vanishingly small after three bonds.

The inductive effect is related to the ability of

substituent(s) to either withdraw or donate

electron density to the attached carbon atom.

Based on this ability, the substitutents can be

classified as electron-withdrawing or electron

donating groups relative to hydrogen. Halogens

and many other groups such as nitro (- NO

cyano (- CN), carboxy (- COOH), ester (COOR),

aryloxy (-OAr, e.g. – OC

), etc. are electron-

withdrawing groups. On the other hand, the

alkyl groups like methyl (–CH

) and ethyl

(–CH

–CH

) are usually considered as electron

donating groups.

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353ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

Problem 12.14

Which bond is more polar in the following

pairs of molecules: (a) H

C-H, H

C-Br

(b) H

C-NH

, H

C-OH (c) H

C-OH,

C-SH

Solution

(a) C–Br, since Br is more electronegative

than H, (b) C–O, (c) C–O

Problem 12.15

In which C–C bond of CH

Br, the

inductive effect is expected to be the least?

Solution

Magnitude of inductive effect diminishes

as the number of intervening bonds

increases. Hence, the effect is least in the

bond between carbon-3 and hydrogen.

12.7.6 Resonance Structure

There are many organic molecules whose

behaviour cannot be explained by a single

Lewis structure. An example is that of

benzene. Its cyclic structure

containing alternating C–C single

and C=C double bonds shown is

inadequate for explaining its

characteristic properties.

As per the above representation, benzene

should exhibit two different bond lengths, due

to C–C single and C=C double bonds. However,

as determined experimentally benzene has a

uniform C–C bond distances of 139 pm, a

value intermediate between the C–C single(154

pm) and C=C double (134 pm) bonds. Thus,

the structure of benzene cannot be represented

adequately by the above structure. Further,

benzene can be represented equally well by the

energetically identical structures I and II.

be adequately represented by any of these

structures, rather it is a hybrid of the two

structures (I and II) called resonance

structures. The resonance structures

(canonical structures or contributing

structures) are hypothetical and

individually do not represent any real

molecule. They contribute to the actual

structure in proportion to their stability.

Another example of resonance is provided

by nitromethane (CH

) which can be

represented by two Lewis structures, (I and II).

There are two types of N-O bonds in these

structures.

However, it is known that the two N–O bonds

of nitromethane are of the same length

(intermediate between a N–O single bond

and a N=O double bond). The actual

structure of nitromethane is therefore a

resonance hybrid of the two canonical

forms I and II.

The energy of actual structure of the molecule

(the resonance hybrid) is lower than that of any

of the canonical structures. The difference in

energy between the actual structure and the

lowest energy resonance structure is called the

resonance stabilisation energy or simply

the resonance energy. The more the number

of important contributing structures, the more

is the resonance energy. Resonance is

particularly important when the contributing

structures are equivalent in energy.

The following rules are applied while writing

resonance structures:

The resonance structures have (i) the same

positions of nuclei and (ii) the same number of

unpaired electrons. Among the resonance

structures, the one which has more number of

covalent bonds, all the atoms with octet of

electrons (except hydrogen which has a

duplet), less separation of opposite charges, (a

negative charge if any on more electronegative

atom, a positive charge if any on more

electropositive atom) and more dispersal of

charge, is more stable than others.

Benzene

Therefore, according to the resonance theory

(Unit 4) the actual structure of benzene cannot

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354 CHEMISTRY

Problem 12.16

Write resonance structures of CH

COO

–

and show the movement of electrons by

curved arrows.

Solution

First, write the structure and put

unshared pairs of valence electrons on

appropriate atoms. Then draw the arrows

one at a time moving the electrons to get

the other structures.

Problem 12.17

Write resonance structures of

=CH–CHO. Indicate relative stability of

the contributing structures.

Solution

The two structures are less important

contributors as they involve charge

separation. Additionally, structure I

contains a carbon atom with an

incomplete octe

12.7.7 Resonance Effect

The resonance effect is defined as ‘the polarity

produced in the molecule by the interaction of

two π-bonds or between a π-bond and lone pair

of electrons present on an adjacent atom’. The

effect is transmitted through the chain. There

are two types of resonance or mesomeric effect

designated as R or M effect.

(i) Positive Resonance Effect (+R effect)

In this effect, the transfer of electrons is away

from an atom or substituent group attached

to the conjugated system. This electron

displacement makes certain positions in the

molecule of high electron densities. This effect

in aniline is shown as :

Stability: I > II > III

[I: Most stable, more number of covalent

bonds, each carbon and oxygen atom has

an octet and no separation of opposite

charge II: negative charge on more

electronegative atom and positive charge

on more electropositive atom; III: does not

contribute as oxygen has positive charge

and carbon has negative charge, hence

least stable].

Problem 12.18

Explain why the following two structures,

I and II cannot be the major contributors

to the real structure of CH

COOCH

(ii) Negative Resonance Effect (- R effect)

This effect is observed when the transfer of

electrons is towards the atom or substituent

group attached to the conjugated system. For

example in nitrobenzene this electron

displacement can be depicted as :

The atoms or substituent groups, which

represent +R or –R electron displacement

effects are as follows :

+R effect: – halogen, –OH, –OR, –OCOR, –NH

–NHR, –NR

, –NHCOR,

– R effect: – COOH, –CHO, >C=O, – CN, –NO

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355ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

The presence of alternate single and double

bonds in an open chain or cyclic system is

termed as a conjugated system. These systems

often show abnormal behaviour. The examples

are 1,3- butadiene, aniline and nitrobenzene

etc. In such systems, the π-electrons are

delocalised and the system develops polarity.

12.7.8 Electromeric Effect (E effect)

It is a temporary effect. The organic

compounds having a multiple bond (a double

or triple bond) show this effect in the presence

of an attacking reagent only. It is defined as

the complete transfer of a shared pair of

π-electrons to one of the atoms joined by a

multiple bond on the demand of an attacking

reagent. The effect is annulled as soon as the

attacking reagent is removed from the domain

of the reaction. It is represented by E and the

shifting of the electrons is shown by a curved

arrow (

). There are two distinct types of

electromeric effect.

(i) Positive Eelctromeric Effect (+E effect) In this

effect the π−electrons of the multiple bond are

transferred to that atom to which the reagent

gets attached. For example :

system or to an atom with an unshared

p orbital. The σ electrons of C—H bond of the

alkyl group enter into partial conjugation with

the attached unsaturated system or with the

unshared p orbital. Hyperconjugation is a

permanent effect.

To understand hyperconjugation effect, let

us take an example of

CH CH

(ethyl cation)

in which the positively charged carbon atom

has an empty p orbital. One of the C-H bonds

of the methyl group can align in the plane of

this empty p orbital and the electrons

constituting the C-H bond in plane with this p

orbital can then be delocalised into the empty

p orbital as depicted in Fig. 12.4 (a).

(ii) Negative Electromeric Effect (–E effect) In this

effect the π - electrons of the multiple bond are

transferred to that atom to which the attacking

reagent does not get attached. For example:

When inductive and electromeric effects

operate in opposite directions, the electomeric

effect predominates.

12.7.9 Hyperconjugation

Hyperconjugation is a general stabilising

interaction. It involves delocalisation of

σ electrons of C—H bond of an alkyl group

directly attached to an atom of unsaturated

Fig. 12.4(a) Orbital diagram showing

hyperconjugation in ethyl cation

This type of overlap stabilises the

carbocation because electron density from the

adjacent σ bond helps in dispersing the positive

charge.

In general, greater the number of alkyl

groups attached to a positively charged carbon

atom, the greater is the hyperconjugation

interaction and stabilisation of the cation.

Thus, we have the following relative stability

of carbocations :

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Hyperconjugation is also possible in

alkenes and alkylarenes.

Delocalisation of electrons by

hyperconjugation in the case of alkene can be

depicted as in Fig. 12.4(b).

Problem 12.19

Explain why (CH

)

is more stable than

32 3

CH CH and C H

is the least stable

cation.

Solution

Hyperconjugation interaction in (CH

)

greater than in

CH C H

as the (CH

)

has nine C-H bonds. In

, vacant p

orbital is perpendicular to the plane

in which C-H bonds lie; hence cannot

overlap with it. Thus,

lacks

hyperconjugative stability.

12.7.10 Types of Organic Reactions and

Mechanisms

Organic reactions can be classified into the

following categories:

(i) Substitution reactions

(ii) Addition reactions

(iii) Elimination reactions

(iv) Rearrangement reactions

You will be studying these reactions in

Unit 13 and later in class XII.

12.8 METHODS OF PURIFICATION OF

ORGANIC COMPOUNDS

Once an organic compound is extracted from

a natural source or synthesised in the

laboratory, it is essential to purify it. Various

methods used for the purification of organic

compounds are based on the nature of the

compound and the impurity present in it.

The common techniques used for

purification are as follows :

(i) Sublimation

(ii) Crystallisation

(iii) Distillation

(iv) Differential extraction and

(v) Chromatography

Finally, the purity of a compound is

ascertained by determining its melting or

boiling point. Most of the pure compounds

have sharp melting points and boiling points.

New methods of checking the purity of an

organic compound are based on different types

Fig. 12.4(b) Orbital diagram showing

hyperconjugation in propene

There are various ways of looking at the

hyperconjugative effect. One of the way is to

regard C—H bond as possessing partial ionic

character due to resonance.

The hyperconjugation may also be

regarded as no bond resonance.

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357ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

of chromatographic and spectroscopic

techniques.

12.8.1 Sublimation

You have learnt earlier that on heating, some

solid substances change from solid to vapour

state without passing through liquid state. The

purification technique based on the above

principle is known as sublimation and is used

to separate sublimable compounds from non-

sublimable impurities.

12.8.2 Crystallisation

This is one of the most commonly used

techniques for the purification of solid organic

compounds. It is based on the difference in the

solubilities of the compound and the

impurities in a suitable solvent. The impure

compound is dissolved in a solvent in which it

is sparingly soluble at room temperature but

appreciably soluble at higher temperature.

The solution is concentrated to get a nearly

saturated solution. On cooling the solution,

pure compound crystallises out and is

removed by filtration. The filtrate (mother

liquor) contains impurities and small quantity

of the compound. If the compound is highly

soluble in one solvent and very little soluble

in another solvent, crystallisation can be

satisfactorily carried out in a mixture of these

solvents. Impurities, which impart colour to the

solution are removed by adsorbing over

activated charcoal. Repeated crystallisation

becomes necessary for the purification of

compounds containing impurities of

comparable solubilities.

12.8.3 Distillation

This important method is used to separate (i)

volatile liquids from nonvolatile impurities and

(ii) the liquids having sufficient difference in

their boiling points. Liquids having different

boiling points vaporise at different

temperatures. The vapours are cooled and the

liquids so formed are collected separately.

Chloroform (b.p 334 K) and aniline (b.p. 457

K) are easily separated by the technique of

distillation (Fig 12.5). The liquid mixture is

taken in a round bottom flask and heated

carefully. On boiling, the vapours of lower

boiling component are formed first. The

vapours are condensed by using a condenser

and the liquid is collected in a receiver. The

vapours of higher boiling component form later

and the liquid can be collected separately.

Fractional Distillation: If the difference in

boiling points of two liquids is not much, simple

distillation cannot be used to separate them.

The vapours of such liquids are formed within

the same temperature range and are condensed

simultaneously. The technique of fractional

distillation is used in such cases. In this

technique, vapours of a liquid mixture are

passed through a fractionating column before

condensation. The fractionating column is

fitted over the mouth of the round bottom flask

(Fig.12.6, page 358).

Vapours of the liquid with higher boiling

point condense before the vapours of the liquid

with lower boiling point. The vapours rising

up in the fractionating column become richer

in more volatile component. By the time the

Fig.12.5 Simple distillation. The vapours of a

substance formed are condensed and

the liquid is collected in conical flask.

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358 CHEMISTRY

vapours reach to the top of the fractionating

column, these are rich in the more volatile

component. Fractionating columns are

available in various sizes and designs as shown

in Fig.12.7. A fractionating column provides

many surfaces for heat exchange between the

ascending vapours and the descending

condensed liquid. Some of the condensing

liquid in the fractionating column obtains heat

from the ascending vapours and revaporises.

The vapours thus become richer in low boiling

component. The vapours of low boiling

component ascend to the top of the column.

On reaching the top, the vapours become pure

in low boiling component and pass through

the condenser and the pure liquid is collected

in a receiver. After a series of successive

distillations, the remaining liquid in the

distillation flask gets enriched in high boiling

component. Each successive condensation

and vaporisation unit in the fractionating

Fig.12.7 Different types of fractionating columns.

column is called a theoretical plate.

Commercially, columns with hundreds of

plates are available.

One of the technological applications of

fractional distillation is to separate different

fractions of crude oil in petroleum industry.

Fig.12.6 Fractional distillation. The vapours of lower boiling

fraction reach the top of the column first followed by

vapours of higher boiling fractions.

Distillation under reduced

pressure: This method is used

to purify liquids having very

high boiling points and those,

which decompose at or below

their boiling points. Such liquids

are made to boil at a

temperature lower than their

normal boiling points by

reducing the pressure on their

surface. A liquid boils at a

temperature at which its vapour

pressure is equal to the external

pressure. The pressure is

reduced with the help of a

water pump or vacuum pump

(Fig.12.8). Glycerol can be

separated from spent-lye in

soap industry by using this

technique.

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359ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

Fig.12.8 Distillation under reduced pressure. A liquid boils at a temperature below its

vapour pressure by reducing the pressure.

Steam Distillation: This technique is

applied to separate substances which are

steam volatile and are immiscible with water.

In steam distillation, steam from a steam

generator is passed through a heated flask

containing the liquid to be distilled. The

mixture of steam and the volatile organic

compound is condensed and collected. The

compound is later separated from water

using a separating funnel. In steam

distillation, the liquid boils when the sum

of vapour pressures due to the organic

liquid (p

) and that due to water (p

)

becomes equal to the atmospheric pressure

(p), i.e. p =p

+ p

. Since p

is lower than p,

the organic liquid vaporises at lower

temperature than its boiling point.

Thus, if one of the substances in the

mixture is water and the other, a water

insoluble substance, then the mixture will boil

close to but below, 373K. A mixture of water

and the substance is obtained which can be

separated by using a separating funnel.

Aniline is separated by this technique from

aniline – water mixture (Fig.12.9, Page 360).

12.8.4 Differential Extraction

When an organic compound is present in an

aqueous medium, it is separated by shaking

it with an organic solvent in which it is more

soluble than in water. The organic solvent and

the aqueous solution should be immiscible

with each other so that they form two distinct

layers which can be separated by separatory

funnel. The organic solvent is later removed by

distillation or by evaporation to get back

the compound. Differential extraction is carried

out in a separatory funnel as shown in

Fig. 12.10 (Page 360). If the organic compound

is less soluble in the organic solvent, a very

large quantity of solvent would be required to

extract even a very small quantity of the

compound. The technique of continuous

extraction is employed in such cases. In this

technique same solvent is repeatedly used for

extraction of the compound.

12.8.5 Chromatography

Chromatography is an important technique

extensively used to separate mixtures into their

components, purify compounds and also to

test the purity of compounds. The name

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360 CHEMISTRY

chromatography is based on the Greek word

chroma, for colour since the method was first

used for the separation of coloured substances

found in plants. In this technique, the mixture

of substances is applied onto a stationary

phase, which may be a solid or a liquid. A

pure solvent, a mixture of solvents, or a gas is

allowed to move slowly over the stationary

phase. The components of the mixture get

gradually separated from one another. The

moving phase is called the mobile phase.

Based on the principle involved,

chromatography is classified into different

categories. Two of these are:

(a) Adsorption chromatography, and

(b) Partition chromatography.

a) Adsorption Chromatography: Adsor-

ption chromatography is based on the fact that

different compounds are adsorbed on an

adsorbent to different degrees. Commonly

used adsorbents are silica gel and alumina.

When a mobile phase is allowed to move

over a stationary phase (adsorbent), the

components of the mixture move by varying

distances over the stationary phase. Following

are two main types of chromatographic

techniques based on the principle of differential

adsorption.

(a) Column chromatography, and

(b) Thin layer chromatography.

Fig.12.10 Differential extraction. Extraction of com-

pound takes place based on difference

in solubility

Fig.12.9 Steam distillation. Steam volatile component volatilizes, the vapours con-

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361ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

Column Chromatography: Column

chromatography involves separation of a

mixture over a column of adsorbent

(stationary phase) packed in a glass tube. The

column is fitted with a stopcock at its lower

end (Fig. 12.11). The mixture adsorbed on

adsorbent is placed on the top of the adsorbent

column packed in a glass tube. An appropriate

eluant which is a liquid or a mixture of liquids

is allowed to flow down the column slowly.

Depending upon the degree to which the

compounds are adsorbed, complete separation

takes place. The most readily adsorbed

substances are retained near the top and others

come down to various distances in the column

(Fig.12.11).

plate is then placed in a closed jar containing

the eluant (Fig. 12.12a). As the eluant rises

up the plate, the components of the mixture

move up along with the eluant to different

distances depending on their degree of

adsorption and separation takes place. The

relative adsorption of each component of the

mixture is expressed in terms of its retardation

factor i.e. R

value (Fig.12.12 b).

Distance moved by the substance from base line (x)

Distance moved by the solvent from base line (y)

Fig.12.11 Column chromatography. Different

stages of separation of components

of a mixture.

Thin Layer Chromatography: Thin layer

chromatography (TLC) is another type of

adsorption chromatography, which involves

separation of substances of a mixture over a

thin layer of an adsorbent coated on glass plate.

A thin layer (about 0.2mm thick) of an

adsorbent (silica gel or alumina) is spread over

a glass plate of suitable size. The plate is known

as thin layer chromatography plate or

chromaplate. The solution of the mixture to

be separated is applied as a small spot about

2 cm above one end of the TLC plate. The glass

The spots of coloured compounds are visible

on TLC plate due to their original colour. The

spots of colourless compounds, which are

invisible to the eye but fluoresce in ultraviolet

light, can be detected by putting the plate under

ultraviolet light. Another detection technique is

to place the plate in a covered jar containing a

few crystals of iodine. Spots of compounds,

which adsorb iodine, will show up as brown

spots. Sometimes an appropriate reagent may

also be sprayed on the plate. For example,

amino acids may be detected by spraying the

plate with ninhydrin solution (Fig.12.12b).

Fig.12.12 (b) Developed chromatogram.

Fig.12.12 (a) Thin layer chromatography.

Chromatogram being developed.

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362 CHEMISTRY

The spots of the separated colourless

compounds may be observed either under

ultraviolet light or by the use of an appropriate

spray reagent as discussed under thin layer

chromatography.

12.9 QUALITATIVE ANALYSIS OF

ORGANIC COMPOUNDS

The elements present in organic compounds

are carbon and hydrogen. In addition to these,

they may also contain oxygen, nitrogen,

sulphur, halogens and phosphorus.

12.9.1 Detection of Carbon and Hydrogen

Carbon and hydrogen are detected by heating

the compound with copper(II) oxide. Carbon

present in the compound is oxidised to carbon

dioxide (tested with lime-water, which develops

turbidity) and hydrogen to water (tested with

anhydrous copper sulphate, which turns blue).

C + 2CuO

∆

→

2Cu + CO

2H + CuO

∆

→

Cu + H

+ Ca(OH)

→ CaCO

↓

+ H

O + CuSO

→ CuSO

.5H

White Blue

12.9.2 Detection of Other Elements

Nitrogen, sulphur, halogens and phosphorus

present in an organic compound are detected

by “Lassaigne’s test”. The elements present

in the compound are converted from covalent

form into the ionic form by fusing the

compound with sodium metal. Following

reactions take place:

Na + C + N

∆

→

NaCN

2Na + S

∆

→

Na + X

∆

→

Na X

(X = Cl, Br or I)

C, N, S and X come from organic

compound.

Cyanide, sulphide and halide of sodium so

formed on sodium fusion are extracted from

the fused mass by boiling it with distilled water.

This extract is known as sodium fusion extract.

(A) Test for Nitrogen

The sodium fusion extract is boiled with

iron(II) sulphate and then acidified with

Partition Chromatography: Partition

chromatography is based on continuous

differential partitioning of components of a

mixture between stationary and mobile

phases. Paper chromatography is a type

of partition chromatography. In paper

chromatography, a special quality paper

known as chromatography paper is used.

Chromatography paper contains water trapped

in it, which acts as the stationary phase.

A strip of chromatography paper spotted

at the base with the solution of the mixture is

suspended in a suitable solvent or a mixture

of solvents (Fig. 12.13). This solvent acts as

the mobile phase. The solvent rises up the

paper by capillary action and flows over the

spot. The paper selectively retains different

components according to their differing

partition in the two phases. The paper strip

so developed is known as a chromatogram.

The spots of the separated coloured

compounds are visible at different heights from

the position of initial spot on the chromatogram.

Fig.12.13 Paper chromatography.

Chromatography paper in two different

shapes.

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363ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

concentrated sulphuric acid. The formation

of Prussian blue colour confirms the presence of

nitrogen. Sodium cyanide first reacts

with iron(II) sulphate and forms sodium

hexacyanidoferrate(II). On heating with

concentrated sulphuric acid some iron(II) ions

are oxidised to iron(III) ions which react with

sodium hexacyanidoferrate(II) to produce

iron(III) hexacyanidoferrate(II) (ferriferrocyanide)

which is Prussian blue in colour.

6CN

–

+ Fe

→ [Fe(CN)

]

4–

3[Fe(CN)

]

4–

+ 4Fe

[Fe(CN)

]

.xH

Prussian blue

(B) Test for Sulphur

(a) The sodium fusion extract is acidified with

acetic acid and lead acetate is added to it.

A black precipitate of lead sulphide

indicates the presence of sulphur.

2–

+ Pb

→ PbS

Black

(b) On treating sodium fusion extract with

sodium nitroprusside, appearance of a

violet colour further indicates the presence

of sulphur.

2–

+ [Fe(CN)

NO]

2–

→ [Fe(CN)

NOS]

4–

Violet

In case, nitrogen and sulphur both are

present in an organic compound, sodium

thiocyanate is formed. It gives blood red colour

and no Prussian blue since there are no free

cyanide ions.

Na + C + N + S → NaSCN

+SCN

–

→ [Fe(SCN)]

Blood red

If sodium fusion is carried out with excess

of sodium, the thiocyanate decomposes to yield

cyanide and sulphide. These ions give their

usual tests.

NaSCN + 2Na → NaCN+Na

The sodium fusion extract is acidified with nitric

acid and then treated with silver nitrate. A white

precipitate, soluble in ammonium hydroxide

shows the presence of chlorine, a yellowish

precipitate, sparingly soluble in ammonium

hydroxide shows the presence of bromine and

a yellow precipitate, insoluble in ammonium

hydroxide shows the presence of iodine.

–

+ Ag

→ AgX

X represents a halogen – Cl, Br or I.

If nitrogen or sulphur is also present in the

compound, the sodium fusion extract is first

boiled with concentrated nitric acid to

decompose cyanide or sulphide of sodium

formed during Lassaigne’s test. These ions

would otherwise interfere with silver nitrate test

for halogens.

(D) Test for Phosphorus

The compound is heated with an oxidising

agent (sodium peroxide). The phosphorus

present in the compound is oxidised to

phosphate. The solution is boiled with nitric

acid and then treated with ammonium

molybdate. A yellow colouration or precipitate

indicates the presence of phosphorus.

+ 3HNO

→ H

+3NaNO

+ 12(NH

)

MoO

+ 21HNO

→

Ammonium

molybdate

(NH

)

.12MoO

+ 21NH

+ 12H

Ammonium

phosphomolybdate

12.10 QUANTITATIVE ANALYSIS

Quantitative analysis of compounds is very

important in organic chemistry. It helps

chemists in the determination of mass per cent

of elements present in a compound. You have

learnt in Unit-1 that mass per cent of elements

is required for the determination of emperical

and molecular formula.

The percentage composition of elements

present in an organic compound is determined

by the following methods:

12.10.1 Carbon and Hydrogen

Both carbon and hydrogen are estimated in

one experiment. A known mass of an organic

compound is burnt in the presence of excess

of oxygen and copper(II) oxide. Carbon and

hydrogen in the compound are oxidised to

carbon dioxide and water respectively.

+ (x + y/4) O

→ x CO

+ (y/2) H

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364 CHEMISTRY

The mass of water produced is determined

by passing the mixture through a weighed

U-tube containing anhydrous calcium

chloride. Carbon dioxide is absorbed in

another U-tube containing concentrated

solution of potassium hydroxide. These tubes

are connected in series (Fig.12.14). The increase

in masses of calcium chloride and potassium

hydroxide gives the amounts of water and

carbon dioxide from which the percentages of

carbon and hydrogen are calculated.

Let the mass of organic compound be

m g, mass of water and carbon dioxide

produced be m

and m

g respectively;

Percentage of carbon=

12 100

××

Percentage of hydrogen =

2 100

××

Problem 12.20

On complete combustion, 0.246 g of an

organic compound gave 0.198g of carbon

dioxide and 0.1014g of water. Determine

the percentage composition of carbon and

hydrogen in the compound.

Solution

12 0.198 100

Percentage of carbon =

44 0.246

= 21.95%

××

Percentage of hydrogen =

2 0.1014 100

18 0.246

××

= 4.58%

12.10.2 Nitrogen

There are two methods for estimation of

nitrogen: (i) Dumas method and (ii) Kjeldahl’s

method.

(i) Dumas method: The nitrogen containing

organic compound, when heated with copper

oxide in an atmosphere of carbon dioxide,

yields free nitrogen in addition to carbon

dioxide and water.

+ (2x + y/2) CuO →

x CO

+ y/2 H

O + z/2 N

+ (2x + y/2) Cu

Traces of nitrogen oxides formed, if any,

are reduced to nitrogen by passing the gaseous

mixture over a heated copper gauze. The

mixture of gases so produced is collected over

an aqueous solution of potassium hydroxide

which absorbs carbon dioxide. Nitrogen is

collected in the upper part of the graduated

tube (Fig.12.15).

Let the mass of organic compound = m g

Volume of nitrogen collected = V

Room temperature = T

273

Volume of nitrogen at STP=

760

(Let it be V mL)

Where p

and V

are the pressure and volume

of nitrogen, p

is different from the atmospheric

pressure at which nitrogen gas is collected. The

value of p

is obtained by the relation;

= Atmospheric pressure – Aqueous tension

22400 mL N

at STP weighs 28 g.

Fig.12.14 Estimation of carbon and hydrogen. Water and carbon dioxide formed on oxidation of

substance are absorbed in anhydrous calcium chloride and potassium hydroxide solutions

respectively contained in U tubes.

365ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

mL N at STP weighs = g

22400

Percentage of nitrogen =

28 100

22400

××

Problem 12.21

In Dumas’ method for estimation of

nitrogen, 0.3g of an organic compound

gave 50mL of nitrogen collected at 300K

temperature and 715mm pressure.

Calculate the percentage composition of

nitrogen in the compound. (Aqueous

tension at 300K=15 mm)

Solution

Volume of nitrogen collected at 300K and

715mm pressure is 50 mL

Actual pressure = 715-15 =700 mm

273 700 50

Volume of nitrogen at STP =

300 760

= 41.9 mL

××

22,400 mL of N

at STP weighs = 28 g

28 41.9

41.9 mL of nitrogen weighs= g

22400

28 41.9 100

Percentage of nitrogen =

22400 0.3

=17.46%

××

Fig.12.15 Dumas method. The organic compound yields nitrogen gas on heating it with

copper(II) oxide in the presence of CO

gas. The mixture of gases is collected

over potassium hydroxide solution in which CO

is absorbed and volume of

nitrogen gas is determined.

(ii) Kjeldahl’s method: The compound

containing nitrogen is heated with concentrated

sulphuric acid. Nitrogen in the compound gets

converted to ammonium sulphate (Fig. 12.16).

The resulting acid mixture is then heated with

excess of sodium hydroxide. The liberated

ammonia gas is absorbed in an excess of

standard solution of sulphuric acid. The

amount of ammonia produced is determined

by estimating the amount of sulphuric acid

consumed in the reaction. It is done by

estimating unreacted sulphuric acid left after

the absorption of ammonia by titrating it with

standard alkali solution. The difference between

the initial amount of acid taken and that left

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366 CHEMISTRY

after the reaction gives the amount of acid reacted

with ammonia.

Organic compound + H

→ (NH

)

NaOH

Na SO NH H O

 →    ++

2NH

+ H

→ (NH

)

Let the mass of organic compound taken = m g

Volume of H

of molarity, M,

taken = V mL

Volume of NaOH of molarity, M, used for

titration of excess of H

= V

mL of NaOH of molarity M

= V

/2 mL of H

of molarity M

Volume of H

of molarity M unused

= (V - V

/2) mL

(V- V

/2) mL of H

of molarity M

= 2(V-V

/2) mL of NH

solution of

molarity M.

1000 mL of 1 M NH

solution contains

17g NH

or 14 g of N

2(V-V

/2) mL of NH

solution of molarity M

contains:

(

)

14 M 2 V V / 2

g N

1000

×× −

(

)

14 M 2 V-V /2

100

Percentage of N=

1000

××

(

)

1.4 M 2 - /2

××VV

Kjeldahl method is not applicable to

compounds containing nitrogen in nitro and

azo groups and nitrogen present in the ring

(e.g. pyridine) as nitrogen of these compounds

does not change to ammonium sulphate

under these conditions.

Problem 12.22

During estimation of nitrogen present in

an organic compound by Kjeldahl’s

method, the ammonia evolved from 0.5 g

of the compound in Kjeldahl’s estimation

of nitrogen, neutralized 10 mL of 1 M

. Find out the percentage of

nitrogen in the compound.

Solution

1 M of 10 mL H

=1M of 20 mL NH

1000 mL of 1M ammonia contains 14 g

nitrogen

20 mL of 1M ammonia contains

Fig.12.16 Kjeldahl method. Nitrogen-containing compound is treated with concentrated H

to get

ammonium sulphate which liberates ammonia on treating with NaOH; ammonia is absorbed

in known volume of standard acid.

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367ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

1000

g nitrogen

Percentage of nitrogen =

14 20 100

1000

12.10.3 Halogens

Carius method: A known mass of an organic

compound is heated with fuming nitric acid in

the presence of silver nitrate contained in a hard

glass tube known as Carius tube, (Fig.12.17)

Percentage of halogen

atomic mas

molecular

s of X

mass of AgX

100

Problem 12.23

In Carius method of estimation of

halogen, 0.15 g of an organic compound

gave 0.12 g of AgBr. Find out the

percentage of bromine in the compound.

Solution

Molar mass of AgBr = 108 + 80

= 188 g mol

-1

188 g AgBr contains 80 g bromine

0.12 g AgBr contains

80 012

188

g bromine

80 0.12 100

Percentage of bromine=

188 0.15

= 34.04%

××

12.10.4 Sulphur

A known mass of an organic compound is

heated in a Carius tube with sodium peroxide

or fuming nitric acid. Sulphur present in the

compound is oxidised to sulphuric acid. It is

precipitated as barium sulphate by adding

excess of barium chloride solution in water.

The precipitate is filtered, washed, dried and

weighed. The percentage of sulphur can be

calculated from the mass of barium sulphate.

Let the mass of organic

compound taken = m g

and the mass of barium

sulphate formed = m

1 mol of BaSO

= 233 g BaSO

= 32 g sulphur

g BaSO

contains

233

g sulphur

32 100

Percentage of sulphur=

233

××

Fig. 12.17 Carius method. Halogen containing

organic compound is heated with fuming

nitric acid in the presence of silver

nitrate.

in a furnace. Carbon and hydrogen present in

the compound are oxidised to carbon dioxide

and water. The halogen present forms the

corresponding silver halide (AgX). It is filtered,

washed, dried and weighed.

Let the mass of organic

compound taken = m g

Mass of AgX formed = m

1 mol of AgX contains 1 mol of X

Mass of halogen in m

g of AgX

atomic mass of X g

molecular mass of AgX

Problem 12.24

In sulphur estimation, 0.157 g of an

organic compound gave 0.4813 g of

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368 CHEMISTRY

percentage composition (100) and the sum of the

percentages of all other elements. However, oxygen

can also be estimated directly as follows:

A definite mass of an organic compound is

decomposed by heating in a stream of nitrogen

gas. The mixture of gaseous products containing

oxygen is passed over red-hot coke when all the

oxygen is converted to carbon monoxide. This

mixture is passed through warm iodine

pentoxide (I

) when carbon monoxide is

oxidised to carbon dioxide producing iodine.

Compound

heat

 → 

+ other gaseous

products

2C + O

1373 K

 →  

2CO]× 5 (A)

+ 5CO → I

+ 5CO

]× 2 (B)

On making the amount of CO produced in

equation (A) equal to the amount of CO used in

equation (B) by multiplying the equations (A) and

(B) by 5 and 2 respectively; we find that each

mole of oxygen liberated from the compound will

produce two moles of carbondioxide.

Thus 88 g carbon dioxide is obtained if 32 g

oxygen is liberated.

Let the mass of organic compound taken be m g

Mass of carbon dioxide produced be m

∴ m

g carbon dioxide is obtained from

g O

∴Percentage of oxygen =

××

32 100

The percentage of oxygen can be derived

from the amount of iodine produced also.

Presently, the estimation of elements in an

organic compound is carried out by using

microquantities of substances and automatic

experimental techniques. The elements,

carbon, hydrogen and nitrogen present in a

compound are determined by an apparatus

known as CHN elemental analyser. The

analyser requires only a very small amount

of the substance (1-3 mg) and displays the

values on a screen within a short time. A

detailed discussion of such methods is

beyond the scope of this book.

barium sulphate. What is the percentage

of sulphur in the compound?

Solution

Molecular mass of BaSO

= 137+32+64

= 233 g

233 g BaSO

contains 32 g sulphur

0.4813 g BaSO

contains

32 0.4813

233

sulphur

32 0.4813 100

Percentage of sulphur=

233 0.157

= 42.10%

××

12.10.5 Phosphorus

A known mass of an organic compound is

heated with fuming nitric acid whereupon

phosphorus present in the compound is

oxidised to phosphoric acid. It is precipitated

as ammonium phosphomolybdate, (NH

)

.12MoO

, by adding ammonia and

ammonium molybdate. Alternatively,

phosphoric acid may be precipitated as

MgNH

by adding magnesia mixture which

on ignition yields Mg

Let the mass of organic compound taken

= m g and mass of ammonium phospho

molydate = m

Molar mass of (NH

)

.12MoO

= 1877g

Percentage of phosphorus =

31 100

1877

××

If phosphorus is estimated as Mg

Percentage of phosphorus =

62 100

222

××

where, 222 u is the molar mass of Mg

m, the mass of organic compound taken, m

the mass of Mg

formed and 62, the mass

of two phosphorus atoms present in the

compound Mg

12.10.6 Oxygen

The percentage of oxygen in an organic compound

is usually found by difference between the total

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369ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

SUMMARY

In this unit, we have learnt some basic concepts in structure and reactivity of organic

compounds, which are formed due to covalent bonding. The nature of the covalent bonding

in organic compounds can be described in terms of orbitals hybridisation concept,

according to which carbon can have sp

, sp

and sp hybridised orbitals. The sp

, sp

and

sp hybridised carbons are found in compounds like methane, ethene and ethyne

respectively. The tetrahedral shape of methane, planar shape of ethene and linear shape

of ethyne can be understood on the basis of this concept. A sp

hybrid orbital can overlap

with 1s orbital of hydrogen to give a carbon - hydrogen (C–H) single bond (sigma, σ bond).

Overlap of a sp

orbital of one carbon with sp

orbital of another results in the formation

of a carbon–carbon σ bond. The unhybridised p orbitals on two adjacent carbons can

undergo lateral (side-by-side) overlap to give a pi (π) bond. Organic compounds can be

represented by various structural formulas. The three dimensional representation of

organic compounds on paper can be drawn by wedge and dash formula.

Organic compounds can be classified on the basis of their structure or the functional

groups they contain. A functional group is an atom or group of atoms bonded together

in a unique fashion and which determines the physical and chemical properties of the

compounds. The naming of the organic compounds is carried out by following a set of

rules laid down by the International Union of Pure and Applied Chemistry (IUPAC). In

IUPAC nomenclature, the names are correlated with the structure in such a way that

the reader can deduce the structure from the name.

Organic reaction mechanism concepts are based on the structure of the substrate

molecule, fission of a covalent bond, the attacking reagents, the electron displacement

effects and the conditions of the reaction. These organic reactions involve breaking and

making of covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic

fashion. A heterolytic cleavage yields carbocations or carbanions, while a homolytic

cleavage gives free radicals as reactive intermediate. Reactions proceeding through

heterolytic cleavage involve the complimentary pairs of reactive species. These are electron

pair donor known as nucleophile and an electron pair acceptor known as electrophile.

The inductive, resonance, electromeric and hyperconjugation effects may help in

the polarisation of a bond making certain carbon atom or other atom positions as places

of low or high electron densities.

Organic reactions can be broadly classified into following types; substitution,

addition, elimination and rearrangement reactions.

Purification, qualitative and quantitative analysis of organic compounds are carried

out for determining their structures. The methods of purification namely : sublimation,

distillation and differential extraction are based on the difference in one or more physical

properties. Chromatography is a useful technique of separation, identification and

purification of compounds. It is classified into two categories : adsorption and partition

chromatography. Adsorption chromatography is based on differential adsorption of various

components of a mixture on an adsorbent. Partition chromatography involves continuous

partitioning of the components of a mixture between stationary and mobile phases. After

getting the compound in a pure form, its qualitative analysis is carried out for detection

of elements present in it. Nitrogen, sulphur, halogens and phosphorus are detected by

Lassaigne’s test. Carbon and hydrogen are estimated by determining the amounts of

carbon dioxide and water produced. Nitrogen is estimated by Dumas or Kjeldahl’s method

and halogens by Carius method. Sulphur and phosphorus are estimated by oxidising

them to sulphuric and phosphoric acids respectively. The percentage of oxygen is usually

determined by difference between the total percentage (100) and the sum of percentages

of all other elements present.

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370 CHEMISTRY

EXERCISES

12.1 What are hybridisation states of each carbon atom in the following compounds ?

=C=O, CH

CH=CH

, (CH

)

CO, CH

=CHCN, C

12.2 Indicate the σ and π bonds in the following molecules :

, C

, CH

=C=CH

, CH

, HCONHCH

12.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-

one.

12.4 Give the IUPAC names of the following compounds :

(a)

(b) (c)

(d) (e) (f) Cl

CHCH

12.5 Which of the following represents the correct IUPAC name for the compounds

concerned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-

Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or

4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne.

12.6 Draw formulas for the first five members of each homologous series beginning

with the following compounds. (a) H–COOH (b) CH

COCH

12.7 Give condensed and bond line structural formulas and identify the functional

group(s) present, if any, for :

(a) 2,2,4-Trimethylpentane

(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid

12.8 Identify the functional groups in the following compounds

(a)

(b) (c)

12.9 Which of the two: O

NCH

–

or CH

–

is expected to be more stable and

why ?

12.10 Explain why alkyl groups act as electron donors when attached to a π system.

12.11 Draw the resonance structures for the following compounds. Show the electron

shift using curved-arrow notation.

(a) C

OH (b) C

CH=CHCHO (d) C

–CHO (e)

65 2

C H CH

−

(f)

3 2

CH CH CHC H

12.12 What are electrophiles and nucleophiles ? Explain with examples.

12.13 Identify the reagents shown in bold in the following equations as nucleophiles or

electrophiles:

(a)

3 3 2

CH COOH CH COO H O

−

+→ +

–

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371ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

(b)

()

( )( )

3 3 3

CH COCH CH C CN OH

+→

–

(c)

+→

66 65 3

CH C H COCH

CH C O

12.14 Classify the following reactions in one of the reaction type studied in this unit.

(a)

32 32

CH CH Br HS CH CH SH Br

− −

+→ +

(b)

(

)

(

)

3 2 3 3

2 2

CH C CH HCl CH ClC CH

=+→ −

(c)

32 2 22

CH CH Br HO CH CH H O Br

− −

+→ =++

(d)

CH C CH OH HBr CH CBrCH CH CH H O

2 3

223 2

()

− +→

()

12.15 What is the relationship between the members of following pairs of structures ?

Are they structural or geometrical isomers or resonance contributors ?

(a)

(b)

(c)

12.16 For the following bond cleavages, use curved-arrows to show the electron flow

and classify each as homolysis or heterolysis. Identify reactive intermediate

produced as free radical, carbocation and carbanion.

(a)

(b)

(c)

(d)

12.17 Explain the terms Inductive and Electromeric effects. Which electron displacement

effect explains the following correct orders of acidity of the carboxylic acids?

(a) Cl

CCOOH > Cl

CHCOOH > ClCH

COOH

(b) CH

COOH > (CH

)

CHCOOH > (CH

)

C.COOH

12.18 Give a brief description of the principles of the following techniques taking an

example in each case.

(a) Crystallisation (b) Distillation (c) Chromatography

12.19 Describe the method, which can be used to separate two compounds with

different solubilities in a solvent S.

12.20 What is the difference between distillation, distillation under reduced pressure

and steam distillation ?

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372 CHEMISTRY

12.21 Discuss the chemistry of Lassaigne’s test.

12.22 Differentiate between the principle of estimation of nitrogen in an organic

compound by (i) Dumas method and (ii) Kjeldahl’s method.

12.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present

in an organic compound.

12.24 Explain the principle of paper chromatography.

12.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing

halogens?

12.26 Explain the reason for the fusion of an organic compound with metallic sodium

for testing nitrogen, sulphur and halogens.

12.27 Name a suitable technique of separation of the components from a mixture of

calcium sulphate and camphor.

12.28 Explain, why an organic liquid vaporises at a temperature below its boiling point

in its steam distillation ?

12.29 Will CCl

give white precipitate of AgCl on heating it with silver nitrate? Give

reason for your answer.

12.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved

during the estimation of carbon present in an organic compound?

12.31 Why is it necessary to use acetic acid and not sulphuric acid for acidification of

sodium extract for testing sulphur by lead acetate test?

12.32 An organic compound contains 69% carbon and 4.8% hydrogen, the remainder

being oxygen. Calculate the masses of carbon dioxide and water produced when

0.20 g of this substance is subjected to complete combustion.

12.33 A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s

method. The ammonia evolved was absorbed in 50 ml of 0.5 M H

. The residual

acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the

percentage composition of nitrogen in the compound.

12.34 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius

estimation. Calculate the percentage of chlorine present in the compound.

12.35 In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur

compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur

in the given compound.

12.36 In the organic compound CH

= CH – CH

– CH

– C ≡ CH, the pair of hydridised

orbitals involved in the formation of: C

– C

bond is:

(a) sp – sp

(b) sp – sp

– sp

(d) sp

– sp

12.37 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue

colour is obtained due to the formation of:

(a) Na

[Fe(CN)

] (b) Fe

[Fe(CN)

]

[Fe(CN)

] (d) Fe

[Fe(CN)

]

12.38 Which of the following carbocation is most stable ?

(a) (CH

)

(b) (CH

)

(d) CH

H CH

12.39 The best and latest technique for isolation, purification and separation of organic

compounds is:

(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography

12.40 The reaction:

I + KOH(aq) → CH

OH + KI

is classified as :

(a) electrophilic substitution (b) nucleophilic substitution

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