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
4
), ethene (C
2
H
4
), ethyne (C
2
H
2
) are
explained in terms of the use of sp
3
, sp
2
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
3
hybrid orbital. The sp
2
hybrid orbital
is intermediate in s character between sp and
sp
3
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
2
or sp
3
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
ππ
π
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
2
C=CH
2
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
2
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) HCCCH=CHCH
3
(b) CH
2
=C=CHCH
3
Solution
(a) σ
C – C
: 4; σ
C–H
: 6; π
C=C
:1; π CC: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
3
Cl, (b) (CH
3
)
2
CO, (c) CH
3
CN,
(d) HCONH
2
, (e) CH
3
CH=CHCN
Solution
(a) sp
3
, (b) sp
3
, sp
2
, (c) sp
3
, sp, (d) sp
2
, (e)
sp
3
, sp
2
, sp
2
, sp
Problem 12.3
Write the state of hybridisation of carbon
in the following compounds and shapes
of each of the molecules.
(a) H
2
C=O, (b) CH
3
F, (c) HCN.
Solution
(a) sp
2
hybridised carbon, trigonal planar;
(b) sp
3
hybridised carbon, tetrahedral; (c)
sp hybridised carbon, linear.
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336 CHEMISTRY
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
2
H
6
),
ethene (C
2
H
4
), ethyne (C
2
H
2
) and methanol
(CH
3
OH) can be represented by the following
structural formulas. Such structural
representations are called complete structural
formulas.
Similarly, CH
3
CH
2
CH
2
CH
2
CH
2
CH
2
CH
2
CH
3
can be further condensed to CH
3
(CH
2
)
6
CH
3
.
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
3
) 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
3
CH
2
CHCH
2
CH
2
CH
2
CH
2
CH
3
|
CH
3
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:
CH
3
CH
3
H
2
C=CH
2
HC
≡≡
CH CH
3
OH
Ethane Ethene Ethyne Methanol
Ethane Ethene
Ethyne Methanol
(ii) Various ways of representing 2-bromo
butane are:
(a) CH
3
CHBrCH
2
CH
3
(b)
(c)
(b)
(c)
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337ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES
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
3
CH
2
COCH
2
CH
3
(b) CH
3
CH=CH(CH
2
)
3
CH
3
Solution
(b)
Solution
Condensed formula:
(a) HO(CH
2
)
3
CH(CH
3
)CH(CH
3
)
2
(b) HOCH(CN)
2
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
2
CH
2
CH
2
CH(CH
3
)CH(CH
3
)CH
3
(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
4
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
CH
3
CH
3
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
2
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
60
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 ‘-aneand carry
a prefix indicating the number of carbon
atoms present in the chain (except from CH
4
to
C
4
H
10
, 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
2
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
4
becomes -CH
3
and is called methyl group. An
alkyl group is named by substituting ylfor
anein 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
3
–CH–CH
2
–CH
3
CH
3
–CH–CH
2
–CH–CH
3
CH
3
CH
2
CH
3
CH
3
(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
3
-CH- CH
3
-CH
2
-CH- CH
3
-CH-CH
2
-
CH
3
CH
3
CH
3
Isopropyl- sec-Butyl- Isobutyl-
CH
3
CH
3
CH
3
-C- CH
3
-C-CH
2
-
CH
3
CH
3
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
2
C(CH
3
)
3
, 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
3
CH
3
CH
3
CH
3
CH
3
-CH-CH
2
-CH-CH
3
CH
3
CCH
2
CHCH
3
CH
3
2,4-Dimethylpentane 2,2,4-Trimethylpentane
H
3
C H
2
C
CH
3
 
CH
3
CH
2
CHCCH
2
CH
2
CH
3
CH
3
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
3
CH
2
—CH—CH
2
—CH
2
—CH—CH
2
—CH
3
CH
2
CH
3
CH
3
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
3
–CH–CH
2
–CH–
CH
3
CH
3
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
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 CC
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 cycloto 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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344 CHEMISTRY
chemical reactivity in an organic molecule.
Compounds having the same functional group
undergo similar reactions. For example,
CH
3
OH, CH
3
CH
2
OH, and (CH
3
)
2
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
3
H, -COOR (R=alkyl group), COCl,
-CONH
2
, -CN,-HC=O, >C=O, -OH, -NH
2
, >C=C<,
-C
≡≡
C- .
The –R, C
6
H
5
-, halogens (F, Cl, Br, I), –NO
2
,
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
2
(CH
2
)
3
CH
2
COCH
3
will be
named as 7-hydroxyheptan-2-one and not as
2-oxoheptan -7-ol. Similarly, BrCH
2
CH=CH
2
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
2
(OH)CH
2
(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
2
=CH-CH=CH
2
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
th
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 CC 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
3
CH
2
CH
2
CH
2
CH(Cl)CH
3
.
(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
CH
2
=CHCH
2
CH (OH)CH
3
.
(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
2
is prefixed
functional group therefore double bond
gets preference over –NO
2
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
3
CH(OH)CH
2
CH
2
CH
2
CH
2
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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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
(C
6
H
5
-, 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
5
H
12
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)
(c) (d)
Problem 12.10
Write the structural formula of:
(a) o-Ethylanisole, (b) p-Nitroaniline,
(c) 2,3 - Dibromo -1 - phenylpentane,
(d) 4-Ethyl-1-fluoro-2-nitrobenzene.
Solution
CH
3
CH
3
CH
2
CH
2
CH
2
CH
3
CH
3
CHCH
2
CH
3
Pentane Isopentane
(2-Methylbutane)
CH
3
CH
3
C CH
3
CH
3
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
3
H
8
O represents two alcohols:
OH
CH
3
CH
2
CH
2
OH CH
3
CH-CH
3
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
3
H
6
O represents an aldehyde and a
ketone:
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O H
CH
3
C-CH
3
CH
3
CH
2
—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
4
H
10
O
represents methoxypropane (CH
3
OC
3
H
7
) and
ethoxyethane (C
2
H
5
OC
2
H
5
).
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
3
CH
+
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
C
+
H
3
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
3
C
+
H
2
(ethyl cation, a
primary carbocation), (CH
3
)
2
C
+
H (isopropyl
cation, a secondary carbocation), and (CH
3
)
3
C
+
(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
+
H
3
< CH
3
C
+
H
2
< (CH
3
)
2
C
+
H < (CH
3
)
3
C
+
. These
carbocations have trigonal planar shape with
positively charged carbon being sp
2
hybridised. Thus, the shape of C
+
H
3
may be
considered as being derived from the overlap
of three equivalent C(sp
2
) 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
2
)–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
sp
3
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
2
= CH
2
+ Br
2
CH
2
Br CH
2
Br
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.,
π
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
3
C:
). Neutral molecules such
as
etc., can also act as
nucleophiles due to the presence of lone pair
of electrons. Examples of electrophiles
include carbocations (
CH
+
) and neutral
molecules having functional groups like
carbonyl group (>C=O) or alkyl halides
(R
3
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
3
–SCH
3
, (b) CH
3
–CN, (c) CH
3
–Cu
Solution
Problem 12.12
Giving justification, categorise the
following molecules/ions as nucleophile
or electrophile:
Solution
Nucleophiles:
(
)
25 3 2
3
::
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
3
CH=O, CH
3
CN, CH
3
I.
Solution
Among CH
3
HC
*
=O, H
3
C C
*
N, and
H
3
C
*
–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
3
CH
2
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
+
to
δ
end of the polar bond.
δδ
+
δ
+
δ
CH
3
→CH
2
→→
→→
→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
2
),
cyano (- CN), carboxy (- COOH), ester (COOR),
aryloxy (-OAr, e.g. OC
6
H
5
), etc. are electron-
withdrawing groups. On the other hand, the
alkyl groups like methyl (–CH
3
) and ethyl
(–CH
2
–CH
3
) 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
3
C-H, H
3
C-Br
(b) H
3
C-NH
2
, H
3
C-OH (c) H
3
C-OH,
H
3
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
3
CH
2
CH
2
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
3
NO
2
) 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
3
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
2
=CHCHO. Indicate relative stability of
the contributing structures.
Solution
Solution
The two structures are less important
contributors as they involve charge
separation. Additionally, structure I
contains a carbon atom with an
incomplete octe
t.
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
3
COOCH
3
.
(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
2
,
–NHR, –NR
2
, –NHCOR,
R effect: COOH, –CHO, >C=O, CN, –NO
2
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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
32
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
3
)
3
C
+
is more stable than
+
+
32 3
CH CH and C H
is the least stable
cation.
Solution
Hyperconjugation interaction in (CH
3
)
3
C
+
is
greater than in
+
32
CH C H
as the (CH
3
)
3
C
+
has nine C-H bonds. In
+
3
CH
, vacant p
orbital is perpendicular to the plane
in which C-H bonds lie; hence cannot
overlap with it. Thus,
+
3
CH
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
1
) and that due to water (p
2
)
becomes equal to the atmospheric pressure
(p), i.e. p =p
1
+ p
2
. Since p
1
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
f
value (Fig.12.12 b).
R
f
=
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
2
2H + CuO
→
Cu + H
2
O
CO
2
+ Ca(OH)
2
CaCO
3
+ H
2
O
5H
2
O + CuSO
4
→ CuSO
4
.5H
2
O
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
2
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
2+
[Fe(CN)
6
]
4–
3[Fe(CN)
6
]
4–
+ 4Fe
3+
Fe
4
[Fe(CN)
6
]
3
.xH
2
O
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.
S
2–
+ Pb
2+
→ PbS
Black
(b) On treating sodium fusion extract with
sodium nitroprusside, appearance of a
violet colour further indicates the presence
of sulphur.
S
2–
+ [Fe(CN)
5
NO]
2–
→ [Fe(CN)
5
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
Fe
3+
+SCN
[Fe(SCN)]
2+
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
2
S
(C) Test for Halogens
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.
X
+ 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.
Na
3
PO
4
+ 3HNO
3
→ H
3
PO
4
+3NaNO
3
H
3
PO
4
+ 12(NH
4
)
2
MoO
4
+ 21HNO
3
→
Ammonium
molybdate
(NH
4
)
3
PO
4
.12MoO
3
+ 21NH
4
NO
3
+ 12H
2
O
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.
C
x
H
y
+ (x + y/4) O
2
→ x CO
2
+ (y/2) H
2
O
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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
1
and m
2
g respectively;
Percentage of carbon=
2
12 100
44
××
×
m
m
Percentage of hydrogen =
1
2 100
18
××
×
m
m
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.
C
x
H
y
N
z
+ (2x + y/2) CuO →
x CO
2
+ y/2 H
2
O + z/2 N
2
+ (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
1
mL
Room temperature = T
1
K
1
1
1
273
Volume of nitrogen at STP=
760
×
×
p
V
T
(Let it be V mL)
Where p
1
and V
1
are the pressure and volume
of nitrogen, p
1
is different from the atmospheric
pressure at which nitrogen gas is collected. The
value of p
1
is obtained by the relation;
p
1
= Atmospheric pressure Aqueous tension
22400 mL N
2
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
2
28
mL N at STP weighs = g
22400
×
V
V
Percentage of nitrogen =
28 100
22400
××
×
V
m
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
2
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
2
gas. The mixture of gases is collected
over potassium hydroxide solution in which CO
2
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
2
SO
4
→ (NH
4
)
2
SO
4
2
2
4
3
2
22
NaOH
Na SO NH H O
→ ++
2NH
3
+ H
2
SO
4
→ (NH
4
)
2
SO
4
Let the mass of organic compound taken = m g
Volume of H
2
SO
4
of molarity, M,
taken = V mL
Volume of NaOH of molarity, M, used for
titration of excess of H
2
SO
4
= V
1
mL
V
1
mL of NaOH of molarity M
= V
1
/2 mL of H
2
SO
4
of molarity M
Volume of H
2
SO
4
of molarity M unused
= (V - V
1
/2) mL
(V- V
1
/2) mL of H
2
SO
4
of molarity M
= 2(V-V
1
/2) mL of NH
3
solution of
molarity M.
1000 mL of 1 M NH
3
solution contains
17g NH
3
or 14 g of N
2(V-V
1
/2) mL of NH
3
solution of molarity M
contains:
(
)
1
14 M 2 V V / 2
g N
1000
××
(
)
1
14 M 2 V-V /2
100
Percentage of N=
1000
××
×
m
(
)
1
1.4 M 2 - /2
=
××VV
m
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
H
2
SO
4
. Find out the percentage of
nitrogen in the compound.
Solution
1 M of 10 mL H
2
SO
4
=1M of 20 mL NH
3
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
2
SO
4
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
14
20
1000
×
g nitrogen
Percentage of nitrogen =
14 20 100
×
×
×
=
1000
5
56
0%
.
.
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
1
m
m
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
g
1 mol of BaSO
4
= 233 g BaSO
4
= 32 g sulphur
m
1
g BaSO
4
contains
1
32
233
×
m
g sulphur
1
32 100
Percentage of sulphur=
233
××
×
m
m
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
g
1 mol of AgX contains 1 mol of X
Mass of halogen in m
1
g of AgX
1
atomic mass of X g
molecular mass of AgX
×
=
m
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
2
O
5
) when carbon monoxide is
oxidised to carbon dioxide producing iodine.
Compound
heat
→
O
2
+ other gaseous
products
2C + O
2
1373 K
→
2CO]× 5 (A)
I
2
O
5
+ 5CO → I
2
+ 5CO
2
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
1
g
m
1
g carbon dioxide is obtained from
×
1
32
88
m
g O
2
Percentage of oxygen =
××
×
1
32 100
88
m
m
%
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
4
= 137+32+64
= 233 g
233 g BaSO
4
contains 32 g sulphur
0.4813 g BaSO
4
contains
32 0.4813
233
×
g
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
4
)
3
PO
4
.12MoO
3
, by adding ammonia and
ammonium molybdate. Alternatively,
phosphoric acid may be precipitated as
MgNH
4
PO
4
by adding magnesia mixture which
on ignition yields Mg
2
P
2
O
7
.
Let the mass of organic compound taken
= m g and mass of ammonium phospho
molydate = m
1
g
Molar mass of (NH
4
)
3
PO
4
.12MoO
3
= 1877g
Percentage of phosphorus =
1
31 100
1877
××
×
m
m
%
If phosphorus is estimated as Mg
2
P
2
O
7
,
Percentage of phosphorus =
1
62 100
%
222
××
×
m
m
where, 222 u is the molar mass of Mg
2
P
2
O
7
,
m, the mass of organic compound taken, m
1
,
the mass of Mg
2
P
2
O
7
formed and 62, the mass
of two phosphorus atoms present in the
compound Mg
2
P
2
O
7
.
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
3
, sp
2
and sp hybridised orbitals. The sp
3
, sp
2
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
3
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
2
orbital of one carbon with sp
2
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 ?
CH
2
=C=O, CH
3
CH=CH
2
, (CH
3
)
2
CO, CH
2
=CHCN, C
6
H
6
12.2 Indicate the σ and π bonds in the following molecules :
C
6
H
6
, C
6
H
12
, CH
2
Cl
2
, CH
2
=C=CH
2
, CH
3
NO
2
, HCONHCH
3
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
2
CHCH
2
OH
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
3
COCH
3
(c) H–CH=CH
2
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
(c) Hexanedial
12.8 Identify the functional groups in the following compounds
(a)
(b) (c)
12.9 Which of the two: O
2
NCH
2
CH
2
O
or CH
3
CH
2
O
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
6
H
5
OH (b) C
6
H
5
NO
2
(c) CH
3
CH=CHCHO (d) C
6
H
5
–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
+→ +
HO
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371ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES
(b)
()
( )( )
3 3 3
2
CH COCH CH C CN OH
+→
CN
(c)
+→
66 65 3
CH C H COCH
+
3
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
3
3
2 3
2
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
3
CCOOH > Cl
2
CHCOOH > ClCH
2
COOH
(b) CH
3
CH
2
COOH > (CH
3
)
2
CHCOOH > (CH
3
)
3
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
4
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
2
SO
4
. 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
2
= CH – CH
2
CH
2
C CH, the pair of hydridised
orbitals involved in the formation of: C
2
– C
3
bond is:
(a) sp – sp
2
(b) sp – sp
3
(c) sp
2
– sp
3
(d) sp
3
sp
3
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
4
[Fe(CN)
6
] (b) Fe
4
[Fe(CN)
6
]
3
(c) Fe
2
[Fe(CN)
6
] (d) Fe
3
[Fe(CN)
6
]
4
12.38 Which of the following carbocation is most stable ?
(a) (CH
3
)
3
C.
+
C
H
2
(b) (CH
3
)
3
+
C
(c) CH
3
CH
2
+
C
H
2
(d) CH
3
+
C
H CH
2
CH
3
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:
CH
3
CH
2
I + KOH(aq) CH
3
CH
2
OH + KI
is classified as :
(a) electrophilic substitution (b) nucleophilic substitution
(c) elimination (d) addition
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