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Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

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Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

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Case I : When number of double bonds (=n) is even then the number of geometrical isomers = 2n−1 + 2n / 2−1
Cl − CH = CH − CH = CH − CH = CH − CH = CH − Cl

n=4, even

Number of geometrical isomers = 2n−1 + 2(n/ 2)−1 = 23 + 21 = 8 + 2 = 10 .
Case II : When number of double bonds (=n) is odd.

Number of geometrical isomers 2(n−1) 2 n+1  / −1
2 
= +

C6H 6 − CH = CH − CH = CH − CH = CH − C6H5

Number of geometrical isomers = 22 + 22−1 = 22 + 21 = 4 + 2 = 6 .

(5) Geometrical Isomerism in nitrogen compounds C = N − bond are oximes,
(i) Geometrical isomerism due to C = N − bond.

The important class of compounds exhibiting geometrical isomerism due to
nitrones, hydrazones and semicarbazones. But the most common compound is oxime.

Oximes : In aldoxime, when hydrogen and hydroxyl groups are on the same side, the isomer is known as

syn. (analogous to cis) and when these groups are on the opposite side, the isomer is known as anti (analogous to

trans) C6 H5 − C − H C6 H5 − C − H

|| ||

N − OH HO − N

Syn- benzaldoxime Anti − benzaldoxime

In ketoximes the prefixes syn and anti indicate which group of ketoxime is syn or anti to hydroxyl group. For

CH 3 − C − C2 H5 this compound will be named as;


(a) Syn-ethyl methyl ketoxime ⇒ HO and C2H5 are syn or

(b) Anti-methyl ethyl ketoxime ⇒ HO and C2H3 are anti.

Similarly consider the following structure C2H5 −C−CH3



Syn−methyl ethyl ketoxime
or Anti−ethyl methyl ketoxime

(ii) Geometrical isomerism due to N = N bond.

C6 H5 − N C6 H5 − N

|| ||
N −C6 H5
CH − N Anti−azobenzene

(6) Geometrical isomerism show by cumulatrienes : Cumulatrienes (Trienes with three adjacent double
bonds) show only geometric isomerism. This is because their molecule is planar, as such the terminal − CH3 groups

and H- atoms lie in the same plane. Therefore, in this case their planar structure can exist in two diastereoisomeric
forms, cis- and trans- but no enantiomeric forms are possible.

H3C C=C=C=C CH 3 H3C C=C=C=C H
H H H CH 3
cis −Hexa − 2, 3, 4 − triene
trans −Hexa − 2, 3, 4 − triene

(7) Geometrical isomerism in cycloalkanes : Disubstituted cycloalkanes show geometrical isomerism.


H Cis-1,2-cyclohexanediol Trnas-1,2-cyclopentanediol


Note : Certain compounds show geometrical as well optical isomerism. Such type of isomerism is know as

geometrical enantiomerism.

Optical isomerism.

(1) Compounds having similar physical and chemical properties but they have the ability to rotate the plane
of polarised light either to the right (Clockwise) or to the left (Anticlockwise) are termed as optically active or optical
isomer and the properly is called optical activity or optical isomerism.

The optical activity was first observed in organic substances like quartz, rock-crystals and crystals of potassium
chlorate (KClO3), potassium bromate (KBrO3) and sodium periodate (NaIO4 ) .

Biot (In 1815) suggested that optical activity of an organic compound was a molecular phenomenon, i.e., it
was due to constitution of an organic compound rather than its crystalline nature.

The light whose vibrations occur only in one plane is termed plane polarised or simply polarised
light. The device that brings polarisation in light is called a polariser (Nicol prism, made of calcite,
CaCO3) . The apparatus which measures the extent of rotation of polarised light is called polarimeter.

Nicol Organic or
prism compound

Light Ordinary Polarised Rotated Rotated Eye
source light right to right to left

(2) Measurement of optical activity : The measurement of optical activity is done in terms of specific
rotation which is defined as the rotation produced by a solution of length of 10 centimetres (One
decimetre) and unit concentration (1 g/mL) for the given wavelength of the light at the given

[ ]Specific rotation,αt°C = α obs
wavelength l×C

Where αobc is the rotaion observed, l is the length of the solution in decimeters and C is the number of grams
in 1mL of solution. The specific rotation of the sucrose at 20°C using sodium light (D-line, λ=5893Å) is +66.5°C

and is denoted as: [ ]α20°C = +66.5°C(C = 0.02 g / mL water)

+ sign indicates the rotation in clockwise direction.

(3) On the basis of the study of optical activity, the various organic compounds were divided into four types :

(i) The optical isomer which rotates the plane of the polarised light to the rigth (Clockwise) is known as

dextrorotatory isomer (Latin: dextero = right) or d-form or indicated by +ve sign.

(ii) The optical isomer which rotates the plane of the polarised light to the left (Anticlockwise) is known as
laevorotatory isomer (Latin; laevo = left) or l-form or indicated by –ve sign.

(iii) The optical powers of the above two isomers are equal in magnitude but opposite in sign. An equimolar
mixture of the two forms, therefore, will be optically inactive due to external compensation. This mixture is termed
racemic mixture or dl-form or (±) mixture.

(iv) Optical isomer with a plane of symmetry is called meso form. It is optically inactive due to internal
compensation, i.e., the rotation caused by upper half part of molecule is neutralised by lower half part of molecule.

(4) Chirality
(i) Definition : A molecule (or an object) is said to be chiral or dissymmetric, if it is does not possess any
element of symmetry and not superimposable on its mirror image and this property of the molecule to show non-
superimposability is called chirality.

On the other hand, a molecule (or an object) which is superimposable on its mirror image is called achiral
(non-dissymmetric or symmetric).

To understand the term chiral and achiral let us consider the alphabet letters ‘P’ and ‘A’ whereas ‘P’ is chiral,

‘A’ is achiral as shown in fig. Mirror

Non-superimposable Superimposable
(Chiral or dissymmetric) (Achiral or non-dissymmetric)

(ii) Elements of symmetry : There are three elements of symmetry,

(a) Plane of symmetry : It may be defined as a plane which divides a molecule in two equal parts that are

related to each other as an object and mirror image. e.g.,

COOH Plane of symmetry
H − C− OH
H − C − OH

(b) Centre of symmetry : It may be defined as a point in the molecule through which if a line is drawn in one
direction and extended to equal distance in opposite direction, it meets another similar group or atom, eg.

Centre of symmetry

CH 3 CH 3 CH 3 H
| NH − CO | | NH − CO |

C C and C• C

| CO − NH | | CO − NH |

trans−Dimethyl diketo piperazine
cis −Dimethyl diketo piperazine

Since trans form contains a centre of symmetry, it is optically inactive.

(c) Alternating axis of symmetry : A molecule is said to possess an alternating axis of symmetry if an oriention
indistinguishable from the original is obtained when molecule is rotated Q degree around an axis passing through
the molecule and the rotated molecule is reflected in a mirror that is perpendicular to the axis of rotation in step (I).

Cl H Reflection perpendicular Cl H
H F 180o rotation
H F to axis rotation HF
H Cl around axis Cl H
H Cl H Cl


(iii) Symmetric, Asymmetric and Dissymmetric molecules

(a) Symmetric molecules : If any symmetry is present in the molecule then molecule will be symmetric

(b) Dissymmetric molecules : Molecule will be a dissymmetric molecule if it has no plane of symmetry, no
centre of symmetry and no alternating axis of symmetry.

(c) Asymmetric molecules : Dissymmetric molecule having at least one asymmetric carbon is known as
assymmetric molecule. All asymmetric molecules are also dissymmetric molecules but the reverse is not necessarily

| |
H — C∗ — OH H — C∗ — OH
| |
H — C∗ — OH No planeCoHf s3ymmetry
No planeCo6fHsy5mmetry Dissymmet⇓ric molecule
Asymmetr⇓ic molecule
Dissymmet⇓ric molecule
Asymmetr⇓ic molecule

(iv) Chiral or asymmetric carbon atom : A carbon bonded to four different groups is called a chiral
carbon or a chirality centre. The chirality centre is indicated by asterisk. e.g.,

a CH 3

| |

d — C* — b HO — C* — H
| |
Lactic acid

Note : Carbons that can be chirality centres are sp3 -hybridised carbons; sp 2 and sp -hybridised carbons

cannot be chiral carbons because they cannot have four group attached to them.

Isotopes of an atom behave as different group in stereoisomerism.

D H1

| |

H — C* — T Cl 35 — C* — Cl 37
| |

Lactic acid

Carbon of the following groups will not be a chiral carbon

− CH3, − CH2OH, − CHX2, − CHO, − C− Z

Maleic acid (HOOC − CH = CH − COOH) show geometrical isomerism while malic acid
(HOOC − CH 2 − CHOH − COOH) show optical isomerism.

(5) Calculation of number of optical isomers

(i) If molecule is not divisible into two identical halves and molecule has n asymmetric carbon atoms then

Number of optically active forms = 2n = a
Number of enantiomeric pair = a / 2

Number of racemic mixture = a / 2

Number of meso form =0

Examples : C6 H5 − C∗ HOH − C∗ HOH − C∗ HOH − CH 3

This molecule cannot be divided into two identical halves and it has three asymmetric carbons. Hence
number of optical active isomers = a = 2n = 23 = 8 .

CH 2OH − C∗ HOH − C∗ Hn=O4 H− − C∗ HOH − C∗ HOH − CHO
Number of optically active forms = a = 24 = 16

CH 3 − C∗ Hn=O2 H− − C∗ HCl − CH 3

Number of optically active forms = 22 = 4
(ii) If molecule is divisible into two identical halves, then the number of configurational isomers depends on
the number of asymmetric carbon atoms.
Case I : When compound has even number of carbon atoms, i.e., n = 2, 4, 8,10,12,..... :

(i) Number of optically by active forms = a = 2n−1
(ii) Number of enantiomeric pairs = a / 2

(iii) Number of racemic mixture = a / 2

(iv) Number of meso forms = m = 2(n / 2)−1
(v) Total number of configurational isomers = a + m
Example :

(I) (II)

Two idenitcal halves (I) and (II) having n = 2 .
Thus number of optical isomers = a = 22−1 = 2

Number of meso form = m = 2(n / 2)−1 = 2(2 / 2)−1 = 20 = 1
Total number of configurational isomers = 2 + 1 = 3

C6 H5 − C∗ HCl − C∗ HCl− − C∗ HCl − C∗ HCl − C6 H5
n=4, even

a = 24−1 = 23 = 8
m = 2(n / 2)−1 = 21 = 2
Total number of configurational isomers = 8 + 2 = 10
Case II : When compound has odd number of carbon atoms, i.e., n = 3, 5,7, 9,11,...... :

(i) Number of optically active forms = a = 2n−1 − 2(n−1)/ 2
(ii) Number of enantiomeric pairs = a / 2

(iii) Number of racemic mixutre = a / 2

(iv) Number of meso forms = m = 2(n−1)/ 2
(v) Total number of configurational isomers = a + m
Example :

CH 2OH − C∗ HOH− C∗ HOH − C∗ HOH − CH 2OH
(I) (II)

Compound has two identical halves and has three asymmetric carbons.

Thus, a = 2n−1 − 2(n−1)/ 2 = 22 − 21 = 4 − 2 = 2

m = 2(n−1) / 2 = 21 = 2

Hence total number of configurational isomers = 2 + 2 = 4

COOH − C∗ HCl − C∗ HOH − C∗ HBr − C∗ HOH − C∗ HCl − COOH

(I) (II)

(6) Optical activity of compounds containing one asymmetric carbon


Examples : CH3 − C HOH− COOH ; CH3 − C HOH− CHO


CH2OH − C HOH− CHO ; C6 H5 − C HCl− CH3

Any molecule having one asymmetric carbon atom exists in two configurational isomers which are
nonsuperimposible mirror images.

| |
H — C — OH HO — C — H
| |
CH 3 CH 3

(I) (II)

(I) and (II) have the same molecular formula, the same structure but different configurations, hence (I) and (II)
are known as configurational isomers. (I) and (II) are nonsuperimposable mirror images, hence (I) and (II) are
optical isomers. Configurational isomers which are nonsuperimposable mirror images are known as enantiomers.
Thus (I) and (II) are enantiomers. Pair of (I) and (II) is known as enantiomeric pair.

(i) Properties of Enantiomers : All chemical and physical properties of enantiomers are same except two
physical properties.

(a) Mode of rotation : One enantiomer rotates light to the right and the other by an equal magnitude to the
left direction. For example

Enantiomer [α] bp d

(+) 2-methyl-1- + 5.78 128.9 1.41
butanol – 5.78 128.9 1.41

(–2) -methyl-1-

(b) Rate of chemical reaction with an optically active compound : Both the enantiomers of 2-methyl-

1-butanol are converted to 2-methyl butene when treated with conc. H 2SO4 . The rate of the reactions is the same.

CH 3 — C— H conc. H2SO4 → CH 3 − CH 2 C CH 2 K1
∆ − | =
C2 H 5
(+) CH 3

H — C — CH conc. H2SO4 → CH 3 CH 2 C= CH 2 K2
3 ∆ − − |
| CH 3
C2 H 5

K1 = K2

When both these compounds are treated with lactic acid, the rate of the reaction is different.

(+) –2–methyl-1-butanol
K3 (–) lactic acid

(–) –2-methyl-1-butanol
K4 (–) lactic acid

K3 ≠ K4

Thus rate of reactions of enantiomers with optically active compound is different.
(ii) Racemic Mixture : An equimolar mixture of two enantiomers is called a racemic mixture (or racemate,
± form, (dl) form or racemic modification). Such a mixture is optically inactive because the two enantiomers rotate
the plane polarised light equally in opposite directions and cancel each other’s rotation. This phenomenon is called
external compensation.

⇒ Racemic mixture can be separated into (+) and (–) forms. The separation is known as resolution.

⇒ The conversion of (+) or (–) form of the compound into a racemic mixture is called racemisation. It can
becaused by heat, light or by chemical reagents.

⇒ Racemic mixture is designated as being ( ± ) or (dl).
(iii) Enantiomeric Excess : A sample of an optically active substance that consists of a single enantiomer is
said to be enantiomerically pure or to have an enantiomeric excess of 100%. An enantiomeric pure sample of
(+)-2-butanol shows a specific rotation of 13.52o . On the other hand, a sample of (+)-2-butanol that contains less
than an equimolar amount of (–)-2-butanol will show a specific rotation that is less than + 13.52o but greater than
0o .
Such a sample is said to have an enantiomeric excess less than 100%. The enantiomeric excess (ee) is
defined as follows :

% Enantiomeric excess = (moles of one enantiomer – moles of other enantiomer) ×100
Total number of moles of both enantiomers

The enantiomeric excess can be calcualted from optical rotation :

% Enantiomeric excess = Observed specific rotation × 100
Specific rotation of pure enantiomer

Enantiomeric excess is also known as optical purity.

(7) Optical activity of compounds containing two asymmetric carbon
Case I : When molecule is not divisible into two identical halves.

The number of optical isomers possible in this case is four (a = 22 = 4). Further there will be two pairs of
enantiomers and two racemic modifications. In practice also it is found to be so. For example dibromocinnamic acid
exists in the following four optically active forms.


|| ||
H — C— Br Br — C— H H — C— Br Br — C— H
|| ||
H — C— Br Br — C— H Br — C— H H — C— Br
|| ||
C6 H5 C6 H5 C6 H5 C6 H5
(I) (II) (III) (IV)
First pair of enantiomers Second pair of enantiomers

Thus there are two pairs (I), (II) and (III), (IV) of enantiomers. Further, more equimolar mixutre of (I) and (II)
will give one racemic mixture. Similarly, equimolar mixture of (III) and (IV) will give other racemic mixutre.

Now let us examine the relationship between the structures (I) and (III), (I) and (IV), (II) and (III) and (II) and
(IV). These are configurational isomers but these are not mirror images. Configurational isomers which are not
mirror images are known as diastereomers.

Thus (I) and (III) are diastereomers

(I) and (IV) are diastereomers

(II) and (III) are diastereomers

(II) and (IV) are diastereomers

Properties of Diastereomes : Diastereomers have different physical properties, e.g., melting and boiling
points, refractive indices, solubilities in different solvents, crystalline structures and specific rotations. Because of
differences in solubility they often can be separated from each other by fraction crystallisation; because of slight
differences in molecular shape and polarity, they often can be separated by chromatography.

Diastereomers have different chemical properties toward both chiral and achiral reagents. Neither any two
diastereomers nor their transition states are mirror images of each other and so will not neccessarily have the same
energies. However, since the diastereomers have the same functional groups, their chemical properties are not too

Case II : When molecule is divisible into two identical halves.

Number of optical isomers = a = 22−1 = 2

Number of meso forms = m = 20 = 1
Total number of configurational isomers = 3
For example, tartaric acid exists in the following three forms :

| | |
H — C — OH HO — C — H H — C — OH
| | |
HO — C — H H — C — OH H — C — OH
| | |

(I) (II) (III) No non-super imposible mirror
image because it has a plane of

(I) and (II) are enantiomers

(I) and (III) diastereomers

(II) and (III) are diastereomers

(III) is optically inactive due to symmetry of the molecule. It is known as meso form.

(8) Optical activity in compounds containing no assymmetric carbon : Although the largest number
of known optically active compounds are optically active due to the presence of chiral carbon atom, some
compounds are also known which do not possess any chiral carbon atom, but on the whole their molecules are
chiral (such molecules were earlierly called dissymmetric); hence they are optically active. Various types of
compounds belonging to this group are allenes, alkylidene cycloalkanes, spiro compounds (spirans) and properly
substituted biphenyls.

(i) Allenes : Allenes are the organic compounds of the following general formulae.


Allenes exhibit optical isomerism provided the two groups attached to each terminal carbon atom are
different, i.e.,

a aa x

C = C = C or C=C=C
bb by

For example, 2 3-pentadiene shows enantiomerism (optical isomerism)


H3C CH 3 H3C CH 3

Non-superimposable mirror images of 2, 3-pentadiene

(ii) Alkylidene cycloalkanes and spiro compounds : When one or both of the double bonds in allenes

are replaced by one and two rings, the resulting systems are respectively known as alkylidene cycloalkanes are


H3C H a CH2 CH2 a

H COOH b CH2 CH2 b

1-Methylcyclohexylidene-4- acetic Spirans
acid (Alkylidene cycloalkane)

(iii) Biphenyls : Suitably substituted diphenyl compounds are also devoid of individual chiral carbon atom,

but the molecules are chiral due to restricted rotation around the single bond between the two benzene nuclei and

hence they must exist in two non-superimposable mirror images of each other. Such types of stereoisomerism which

is due to restricted rotation about single bond, is known as atropisomerism and the stereoisomers are known and

atropisomers. Examples COOH O2N F HOOC


The above discussion leads to the conclusion that the essential condition for optical isomerism is the molecular

disymmetry or molecular chirality and not the mere presence of a chiral centre. However, it may be noted that the
molecules having only one chiral centre are always chiral and exhibit optical isomerism.

(9) Fischer projection formulae : The arrangement of the atoms or groups in space that characterizes a
stereoisomer is called its configuration.

Emil Fischer (1891) provided an easy method to represent the three dimensional formulae of various
organic molecules on paper. Fischer projection is, thus, a planar representation of the three dimesional

By convention, the following points are followed in writing the Fischer formula.

(i) The carbon chain of the compound is arranged vertically, with the most oxidised carbon at the top.

(ii) The asymmetric carbon atom is in the paper plane and is represented at the interaction of crossed lines.

Carbon atom


(iii) Vertical lines are used to represent bonds going away from the observer, i.e., groups attached to the
vertical lines are understood to be present behind the plane of the paper.

(iv) Horizontal lines represent bonds coming towards the observer, i.e., groups attached to the horizontal lines
are understood to be present above the plane of the paper.

Some Fischer projections are given below :

CH 3
Fischer projection of one of
the enantiomers of Lactic acid COOH Fischer projection of one of the
enantiomers of 2,3-dibromobutane
Fischer projection of one of the
enantiomers of Tartaric acid

(10) Name of optical isomers : Following three nomenclatures are used for optically active compounds,
(i) D,L. System of nomenclature : This nomenclature is mainly used in sugar chemistry or optically active
polyhydroxy carbonyl compounds. This nomenclature was given by Emil Fischer to designate the configurations of
various sugars relative to the enantiomeric (+) and (–) glucose as reference.

All sugars whose Fischer projection formula shows the OH group on the chiral carbon atom adjacent to the
terminal CH2OH group on the right hand side belong to the D -series. Similarly if OH is on the left hand side,
then the sugars belong to the L -series.

| |
H — C — OH HO — C — H

| |

D−series L−series

Examples :


| |

H — C— OH HO — C— H
| |
D(d)glyceraldehyde L(l ) glyceraldehyde
or L(−)glyceraldehyde

⇒ It must be noted that there is no relation between the sign of rotation (+, – or d, l) and the configuration (D
and L) of an enantiomer.

⇒ Any compound that can be prepared from, or converted into D(+) glyceraldehyde will belong to D-series
and similarly any compound that can be prepared from, or converted into L(–) glyceraldehyde will belong to the L-

⇒ This nomenclature is also used in α -amino acids.

(ii) Erythro and Threo System of Nomenclature : This nomenclature is used only in those compounds
which have

(a) Only two chiral carbons and
(b) The following structure, R′ − Cab − Cbc − R′′
i.e., out of six substituents on two asymmetric carbons, at least two should be same.

When two like groups (in the given example, group is b ) in Fischer projection formula are drawn on the same
side of the vertical line, the isomer is called erythro form; if these are placed on the opposite sides, the isomer is said

to be threo form.

R′ CH 3 CH 3

|| |

a—C—b H — C — Cl H — C — Cl

|| |

c—C—b H — C — Br Br — C — H

|| |

R′′ C6 H5 C6 H5

Erythro form Erythro form Threo form

(c) R,S Nomenclature (Absolute configuration)

The order of arrangement of four groups around a chiral carbon (stereocentre) atom is called the absolute
configuration around that atom. System which indicates the absolute configuration was given by three chemists
R.S. Cahn, C.K. Ingold and V. Prelog. This system is known as (R) and (S) system or the Cahn-Ingold Prelog
system. The letter (R) comes from the latin rectus (means right) while (S) comes from the latin sinister (means
left). Any chiral carbon atom has either an (R) configuration or an (S) configuration. Therefore, one enantiomer is
(R) and other is (S). A racemic mixture may be designated (R) (S), meaning a mixture of the two. (R) (S)
nomenclature is assigned as follows :

Step I : By a set of sequence rules given below the atoms or groups connected to the chiral carbon are
assigned a priority sequence.

Sequence Rules for Order of Priority

Rule 1 : If all four atoms directly attached to the chiral carbon are different, priority depends on their atomic
number. The atom having highest atomic number gets the highest priority, i.e., (1). The atom with the lowest
atomic number is given the lowest priority, i.e., (4), the group with next higher atomic number is given the next
higher priority (3) and so on. Thus,

Cl 3


F —C— I ≡ 4—C—1


Br 2

F Cl Br I

Increasing atomicnumbe→r

43 2 1

Increasing priority



H2N — C — Br ≡ 3 — C — 1

OH 2

C N O Br


General organic chemistry

Rule 2 : If two or more than two isotopes of the same element is present, the isotope of higher mass receives the

higher priority. 1 H11H 21H 3

Increasing priority


Rule 3 : If two or more of the atoms directly bonded to the chiral carbon are identical, the atomic number of
the next atoms are used for priority assignment. If these atoms also have identical atoms attached to them, priority is
determined at the first point of difference along the chain. The atom that has attached to it an atom of higher
priority has the higher priority.

I 2H and Br

I − H2C − H2 C − | C↑ H2 − Br

↓ C−
2H and C
CH2 −CH2 −CH3

2H and C

In this example the atoms connected directly to the chiral carbon are iodine and three carbons. Iodine has the
highest priority. Connected, to the three carbons are 2H and Br, 2H and C and 2H and C. Bromine has the highest
atomic number amongst C,H and Br and thus CH2Br has highest priority among these three groups (i.e., priority
no. 2).The remaining two carbons are still identical (C and 2H) connected to the second carbons of these groups
are 2H and I and 2H and aC. Iodine has highest priority amongst these atoms, so that − CH2 − CH2 − I is next in

the priority list and CH2 − CH2 − CH3 has the last priority.

I 1
I CH2 CH2 C− CH2 Br ≡3 2
− − − −
CH2 −CH2 −CH3 4

Rule 4 : If a double or a triple bond is linked to chiral centre the involved atoms are duplicated or triplicated

| | || |
− C = O ≡ − C− O ; − C ≡ N ≡ − C− N ; − C− OH ≡ − C− OH
|| | |

By this rule, we obtained the following priority sequence :


Increasing priority

Step 2 : The molecule is then visualised so that the group of lowest priority (4) is directed away from the
observes (At this position the lowest priority is at the bottom of the plane). The remaining three groups are in a
plane facing the observer. If the eye travels clockwise as we look from the group of highest priority to the groups of
second and and third priority (i.e., 1 → 2 → 3 with respect to 4) the configuration is designated R. If arrangement of
groups is in anticlocwise direction, the configuration designated S.

For example: 2 2
| |

1 −C−3 3 −C−1
| |
4 4

Clockwise arrangement of Anticlockwise arrangement of

1 , 2 and 3 ⇒ R 1 , 2 and 3 ⇒ S

Let us apply the whole sequence to bromochlorofluoro methane.

F 3
1 −C−2
Br − C − H ≡ |
| 4

This Fischer projection the least priority number is not at the bottom of the plane.

In such cases the Fischer projection formula of the compound is converted into another equivalent projection
formula in such a manner that atom or group having the lowest priority is placed vertically downward. This may be
down by two interchanges between four priority numbers. The first interchange involves the two priority numbers,
one is the least priority number and other is the priority number which is present at the bottom of the plane. In the
above case first interchanges will takes place between 2 and 4.

3 First interchange 3
| Between 2 and 3 |

1 −C−4 1 −C−2
| |
2 4

(A) (B)

First interchange of two groups at the chiral centre inverts the configuration and this gives enantiomer of the
original compound. Thus (A) and (B) are enantiomer. The second interchange involves the remaining two groups.

3 Second interchange 1
| between remaining |
groups, i.e., 1 and 4
1 −C−2 3 −C−2
| |
4 4

(B) (A)

Example : Arrangement of 2 , 2 and 3 are
clockwise, hence configuration is R

C 2 2
| O | First interchange | Second interchange 1
| Between 3 and 4 Between 1 and 2
H − C− OH ≡ H− C− OH ≡ 4 −C−1 3 −C−1 32
| | | |
CH2OH CH2OH 3 4 4

CHO 2 (i) First interchange between 3 and 4 1 ⇓
| (ii) Sec. interchange between 1 and 2 Clockwise
| 23 R-configuration
1 −C−4
HO − C− H 4
| 3|
CH2OH Anticlockwise
≡ ⇓


Glyceraldehyde (For example) has one asymmetric carbon, hence it has two configurational isomers (I) and (II).

| |

H − C− OH HO − C− H
| |


(R)glyceral dehyde (S)glyceral dehyde
(I) (II)

One can draw a number other configurations for glyceraldehyde but each of them will be a repetition of either

(I) or (II). In this connection it is important to note that if two projection formulae of differ by an odd number of
interchagnes (1, 3, 5, 7, …..) of positions of groups on the chiral carbon, they are different, but if the two differ by
an even number of interchanges (2, 4, 6, …..) they are identical.
For example :

| |
| H,CH2OH → CH 2OH − Secondinterchange → CH 2 OH C− CHO
First interchange C− H between OH,CHO −
H − C− OH |
| |
(I) H H

(II ) (III )

| |
Third intrerchange → HO − CHO Fourth interchange → HO − H
C− C−
CH 2OH ,OH | H ,CHO |


(IV ) (V )

Thus (I), (III) and (V) are identical. Similarly (II) and (IV) are identical.

(11) Resolution of racemic modifications : The separation of racemic mixture into its enantiomers is
known as resolution. The following methods are used for resolution.
(i) Resolution by mechanical separation : Since (+) and (–) forms have opposite crystalline shapes, this
property can be used in the separation of the two forms from a racemic mixture.

(ii) Biochemical separation : When a living organism is added to or developed to a racemic mixture, they eat
up one of the enantiomers preferentially leaving the other behind. Thus, resolution of forms can be done using this
specific property of living organisms. For example, Penicillium glaucum can be used to remove (d)-ammonium

tartrate from the racemic mixture to leave back only the (l)-ammonium tartrate after a suitable time interval.
(iii) Separation by conversions into diasteremoers : Enantiomers have identical physical properties
while diastereomers have different physical properties. This fact is utilised in separation of enantiomers from

racemic mixture. The racemic mixtrue to be resolved is allowed to react with an optically active compound. This
leads to the formation of two diastereomers. These have different physical properties, so both diastereomers can be
separated by physical methods. Each of the enantiomers may then be regenerated and isolated from the
diastereomers. Resolution of enantiomers through the use of diastereomers is given below:

Pair of diastereomers. Top of the molecules are B
mirror images but bottom s are not.
R, R Chemiocal → A − | D
| | C−
1:1 enantiomers reconversion |
A − C− D D − C− A
BB | | X
|| Physical →
W+ W separation
A − C− D + D − C− A Chemical →
| | reaction
|| B
X X |
U − C− W U − C− W Chemical →
(R) (S) || R, S D C− A
reconversion − |
| (R,R) (R,S)

U − C− W (S)
Separated enantiomers


Single optically active enantiomer

Note :  Group X reacts with group O to give new group W.

(12) Asymmetric synthesis and Walden inversion

(i) Asymmetric synthesis : The synthesis of an optically active compound (asymmetric) from a symmetrical
molecule (having no asymmetric carbon) without resolution to form (+) or (-) isomer directly is termed



asymmetric synthesis. For example the reduction of pyruvic acid (CH3 − C− COOH) in presence of nickel

catalyst gives (±) lactic acid (racemic mixture). On the other hand, pyruvic acid is reduced to (–) lactic acid only by

| |
2H →H − + H − H
C− OH C−
'Ni' | |
C| H3
C=O Reduction → (+) Lacticacid (−) Lacticacid

(Pyruvic acid) |
(2-Keto propanoic acid) 2H → HO − H
Yeast |

(−) Lactic acid

(ii) Walden inversion : The conversion of (+) form into (-) form and vice-versa is called Walden inversion.
When an atom or group directly linked to an asymmetric carbon atom is replaced (inverse); the configuration of the
new compound may be opposite to that of the original, i.e.,


| PCl5 → Cl − | H AgOH → HO − | H PCl5 → H − | Cl AgOH → H − | OH

H − C− OH C− C− C− C−
|| | ||
(+) Malic acid (−) Chloro succinic acid (−)Malic acid (+) Chloro succinic acid (+) Malic acid

Conformational isomerism .

(1) Definition : The different arrangement of atoms in a molecule which can be obtained due to rotation
about carbon-carbon single bond are called conformational isomers (conformers) or rotational isomers (rotamers).
This type of isomerism is found in alkanes and cycloalkanes and their substituted derivatives.

It may be noted that rotation around a C − C sigma bond is not completely free. It is in fact hindered by an
energy barrier of 1 to 20 kJ mol-1 in different bonds. There is a possibility of weak repulsive interactions between the
bonds or electron pairs of the bonds on adjacent carbon atoms. Such type of repulsive interaction is known as
torsional strain.

(2) Difference between conformation and configuration : The term conformation should not be
confused with the configuration which relates to those spatial arrangements of the atoms of a molecule that can be
changed only by the breaking and making of bonds whereas the spatial arrangements in conformation are changed
simply by rotation about a single bond.

(3) Representalion of conformations : Represent conformations we can draw three dimensional pictures.
However, chemists represent conformations in two simple ways. These are : (i) Saw horse representation and (ii)
Newman projection


Back carbon

C H Back carbon

C Front carbon H
Front carbon



Saw horse representation Newman projection

(i) Saw horse representation : In this projection, the molecule is viewed along the axis of the model from
an oblique angle. The central carbon-carbon bond ( C − C ) is draw as a straight line slightly tilted to right for the
sake of clarity. The front carbon is shown as the lower left hand carbon and the rear carbon is shown as the upper
right hand carbon. The three bonds around each carbon atom (C − H in ethane or C − C in higher alkanes) are

shown by three lines.

(ii) Newman projection : This is a simple method to represent the conformations. In this method, he
molecule is viewed from the front along the carbon-carbon bond axis. The two carbon atoms forming the σ-bond
are represented by two circles; one behind the other so that only the front carbon is seen. The front carbon atom is
shown by a point whereas the carbon further from the eye is represented by the circle. Therefore, the C − H bonds
of the front carbon are depicted from the centre of the circle while C − H bonds of the back carbon are drawn from
the circumference of the circle at an angle of 120° to each other.

(4) Conformation in alkanes HH HH

(i) Conformations of ethane : When one of the carbon C C

atom is kept fixed and other is rotated about C − C bond an HH 60° H H
infinite numbers of isomers possible. Out of all the C C
conformations for ethane, only two extreme conformations

are important and these are: (a) Staggered conformation HH HH
(b) Eclipsed conformation Staggered Eclipsed
conformation conformation

(a) Staggered conformation : In this arrangement, the

hydrogens of he two carbon atoms are staggered with respect to one another. As a result, they are at maximum

distance apart and have minimum repulsion between them.

(b) Eclipsed conformation : In this conformation, the hydrogens of one carbon atom are directly behind
those of the other. Consequently, the repulsion in these atoms is maximum. The saw horse projections of these
conformations are represented in fig.

Saw horse representation of staggered and eclipsed conformations of ethane

60° H 60° HH
H H 60°
60° H

Staggered conformation


Newman projections for staggered and eclipsed conformations of ethane

The Newman projections for staggered and eclipsed conformations of ethane are shown in figure. It is clear
that when the staggered conformation is rotated through angle of 60°, it changes to eclipsed conformation and
similarly, when eclipsed conformation is rotated through the same angle, it gives back the staggered conformation.

Relative stabilities of the conformations of ethane : The

two conformations of CH3 HH
ethane differ in their H
relative stabilities. The H H H
staggered conformation H HH H Eclipsed
has minimum repulsions H
(less stable)

between the H-atoms

attached tetrahedrally to

the two carbon atoms. On 14.0 kJ mol 12.5 kJ mol–1

the other hand, the Energy
eclipsed conformation has Energy
maximum force of CH3 CH3 HH
repulsion between H- HH Staggered HH H H Staggered H H
atoms. Therefore, the (stable) H H (stable) H H
staggered conformation is HH HH
more stable than the H H HH

The variation of energy versus rotation The variation of energy versus
about C–C bond in propane rotation about C–C bond

eclipsed conformation. The difference in the energy contents of the staggered and eclipsed conformations is 12.5 kJ

mol-1. This small barrier to rotation is also called torsional barrier of the single bond. However, this energy

difference is not large enough to prevent rotation. Even at ordinary temperatures, the molecules have

thermal of kinetic energy to overcome this energy barrier. Therefore, the two conformations of ethane go on

changing from one form to another and consequently, it is not possible to isolate the different conformations of


The variation of energy versus rotation about the C − C bond has been shown in figure.

(ii) Conformations of propane : The next higher member in alkane series, propane (CH3 − CH2 − CH3)

has also two extreme conformations, staggered conformation and eclipsed conformation. In figure only Newman

projections are shown for simplicity. However, the energy barrier in propane is 14 kJ mol-1, which is slightly higher

than that in ethane. CH3 CHH3
Staggered propane

Newman projection of propane

In the eclipsed conformation of the propane, there are two ethane type H − H interactions and an additional
interaction between C − H bond and C − C bond (of methyl group). The variation of energy versus rotation about
C − C bond as shown in fig.

(iii) Conformations of butane : As the alkane CH3 CH3 CH3 CH3
molecule becomes larger, the conformation situation HH H
becomes more complex. In butane H
(CH3 − CH2 − CH2 − CH3) , for example, the rotation about HH H3C
the single bond between two inner atoms ( C2 and C3 ) is CH3 H HH HH
considered. In this case, all the staggered as well as eclipsed H
conformations will not have same stability and energy I
because of different types of interaction between C − C (Of Anti II III
methyl) and C − H bonds. Eclipsed Skew or Gauche

The lowest energy conformation will be the one, in CHC3 H3 CH3 CH3
which the two methyl groups are as far apart as possible i.e., H CH3 H
180° away form each other.
Fully eclipsed Skew or Gauche Eclipsed
(same as III) (same as II)

This conformation will be maximum staggered and is Different conformations of butane

called anti conformation (marked I). Other conformations can be obtained by rotating one of the C2 or C3

carbon atoms through an angle of 60° as shown ahead. Eclipsed Fully eclipsed
II IV Eclipsed
As is clear from the above Newman projection the VI
Gauche or Skew conformations (III and V) are also staggered.

However, in these conformations, the methyl groups are so

close that they repel each other. This repulsion causes gauche

conformations, to have about 3.8 kJ mol-1 more energy than Energy 16 19

anti conformation. Thi conformations II and VI are eclipsed kJ/mol kJ/mol

conformations. These are unstable because of repulsions. Skew
These are 16 kJ mol-1 less stable than anti 3.8 kJ/mol Skew V

conformation. Conformation IV is also eclipsed and it is least Anti
stable having energy 19 kJ mol-1 more than anti conformation. I
This is because of repulsion between methyl-methyl groups
which are very closed together. It is called fully eclipsed Energy changes during rotation about C2 – C3

bond of n-butane 1 H3 − 2 H2 − 3 H2 − 4 H3



The order of stability of these conformations is, Anti > Skew 24.44°
or Gauche > Eclipsed > Fully eclipsed. deviation

The energy differences between various conformations is 109.5° 60°
shown in figure. 24.44°
(5) Conformations in cycloalkanes deviation
(i) Stability of cycloalkanes : Compounds with three and
four membered rings are not as stable as compounds with five or six
membered rings.

The German chemist Baeyer was the first to suggest that the instability of these small rings compounds was
due to angle strain. This theory is known as Baeyer-strain theory.

Baeyer strain theory was based upon the assumption that when an open chain organic compound having the
normal bond angle 109.5° is convert into a cyclic compound, a definite distortion of this normal angle takes place
leading to the development of a strain in the molecule.

Baeyer assumed that cyclic rings are planar. Assuming that the rings are planar, the amount of strain in
various cycloalkanes can be expressed in terms of angle of deviation (d).

d = 1 109.5 − 2(n − 2) × 90 or d = 1 [109.5 − α]
2 n 2

where n = number of carbon-carbon bonds in cycloalkane ring; α = inner bond angle in the cycloalkane ring.

Angle strain ∝ d ∝ 1 ; Stability ∝ 1 ∝ inner angle(α )
inner angle d

Now let us take the case of three to eight membered cyclic compounds.

αα α α α
Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane α = 135°
α = 108°
α = 60° α = 90° d = 0.44° α = 120° α = 128.6° d = –12.46°

d = 22.44° d = 9.4° d = –5.16° d = –9.33°

The positive and negative values of (d) indicate whether the inner angle is less than or more than the normal
tetrahedral value.

Beayer thus predicted that a five membered ring compound would be the most stable. He predicted that six
membered ring compounds would be less stable and as cyclic compound became larger than five membered ring
they would become less and less stable.

Contrary to what Baeyer predicted, however cyclohexane is more stable than cyclopentane. Furthermore,
cyclic compounds do not become less and less stable as the number of sides increase. Thus Baeyer strain theory is
applicable only to cyclopropane, cyclobutane and cyclopantane.

The mistake that Baeyer made was to assume that all cyclic compounds are planar. In real only cyclopropane
is planar and other cycloalkanes are not planar. Cyclic compounds twist and bend in order to achieve structure
that minimises the three different kinds of strain that can destabilise a cyclic compound.

(a) Angle strain is the strain that results when the bond angle is different from desired tetrahedral bond angle
of 109.5°.

(b) Torsional strain is caused by repulsion of the bonding electrons of one substituent with bonding
electrons of a nearby substituent.

(c) Steric strain is caused by atoms or groups of atoms approaching each other too closely.

(ii) Conformation of cyclohexane : Despite Baeyer’s prediction that five-membered cyclic compounds
would be the most stable, the six membered cyclic compound is the most stable. Six membered cyclic compound
are most stable because they can exist in a conformation that is almost completely free of strain. This conformation
is called the chair conformation. In a chair conformation of cyclohexane all bond angles are 111° which is very
close to the 109.5° and all the adjacent carbon-hydrogen bonds are staggered.

a a HH
cc H CH2 H
c H CH2 H
a a
cc HH
a Newmann projection of the chair conformation

Chair conformation of cyclohexane

⇒ Each carbon in chair conformation has an axial bond and an equatorial bond.

⇒ Axial bonds are perpendicular to the plane of the ring equatorial bonds are in the plane of the ring.

⇒ If axial bond on carbon-1 is above the plane of the ring then axial bond on carbon-2 will be below the plane

of the ring. Thus a a
C − 1,C − 3 and C − 5 axial bonds are above 16 5
C − 2,C − 4 and C − 6 axial bonds are below
a a
a a

⇒ Thus C − 1 axial and C − 2 axial are trans to each other. Similarly C − 1 and C − 5 axials are cis to each

⇒ If axial bond on carbon-1 will be above the plane then equatorial bond on this carbon will be below the



(a) Thus C − 1 equatorial and C − 2 equatorial bonds are trans.

(b) C − 1 axial and C − 2 equatorial will be cis.

⇒ As a result of rotation abut carbon-carbon single bonds cyclohexane rapidly interconverts between two stable
chair conformations. This interconversion is know as ring –flip. When the two chair forms interconvert, axial
bonds become equatorial and equatorial bonds become axial.


⇒ Cyclohexane can also exist in a boat conformation. Like the chair conformation, the boat conformation is free

of angle strain. However, the boat conformation is less stable than the chair conformation by 11 kcal/mole.

Boat conformation is less stable because some of the carbon-hydrogen bonds in boat conformation are

eclipsed. Flagpole


Boat conformation Newmann projection of boat
of cyclohexane conformation

The boat conformations is further destabilised by the close proximity of the flagpole hydrogens. These
hydrogens are 1.8 Å apart but the vander Waal’s radii is 2.4 Å. The flagpole hydrogens are also known as
trans nuclear hydrogens.

Note :  The selective stabilities of the four conformations of cyclohexane decrease in the order:

Chair > twist boat > boat > half chair.

Aliphatic Hydrocarbons

“Organic compounds composed of only carbon and hydrogen are called hydrocarbons. Hydrocarbons are
obtained mainly from petroleum, natural gas or coal. Petroleum is a major source of aliphatic hydrocarbon. The
important fuels like petrol, kerosene, coal gas, oil gas, compressed natural gas [CNG], LPG etc, all are hydrocarbon

and their mixtures”.

Aliphatic or open chain Alicyclic or closed chain or cyclic Aromatic or arenes

Cyclo alkanes Cyclo alkenes

Saturated Unsaturated

Alkanes or paraffines Alkenes or olefins Alkynes or acetylenes

Sources of hydrocarbon.

Mineral oil or crude oil, petroleum [Petra → rock; oleum → oil] is the dark colour oily liquid [do with offensive
odour found at various depths in many regions below the surface of the earth. It is generally found under the rocks
of earth’s crust and often floats over salted water.

(1) Composition
(i) Alkanes : found 30 to 70% contain upto 40 carbon atom. Alkanes are mostly straight chain but some are
branched chain isomers.
(ii) Cycloalkanes : Found 16 to 64% cycloalkanes present in petroleum are; cyclohexane, methyl
cyclopentane etc. cycloalkanes rich oil is called asphattic oil.
(iii) Aromatic hydrocarbon : found 8 to 15% compound present in petroleum are; Benzene, Toluene,
Xylene, Naphthalene etc.
(iv) Sulphur, nitrogen and oxygen compound : Sulphur compound present in the extent of 6% and
include mercaptans [R-SH] and sulphides [R-S-R]. the unpleasant smell of petroleum is due to sulphur compounds.
Nitrogen compounds are alkyl pyridines, quinolines and pyrroles. Oxygen compounds present in petroleum are.
Alcohols, Phenols and resins. Compounds like chlorophyll, haemin are also present in it.

(v) Natural gas : It is a mixture of Methane (80%), Ethane (13%), Propane (3%), Butane (1%), Vapours of
low boiling pentanes and hexanes (0.5%) and Nitrogen (1.3%). L.P.G. Contain butanes and pentanes and used as
cooking gas. It is highly inflammable. This contain, methane, nitrogen and ethane.

(vi) C.N.G. : The natural gas compressed at very high pressure is called compressed natural gas (CNG).
Natural gas has octane rating of 130 it consists, mainly of methane and may contain, small amount of ethane and

(2) Theories of origin of petroleum : Theories must explain the following characteristics associated with

Its association with brine (sodium chloride solution). The presence of nitrogen and sulphur compounds in it.
The presence of chlorophyll and haemin in it. Its optically active nature. Three important theories are as follows.

(i) Mendeleeff’s carbide theory or inorganic theory

(a) Molten metals in hot earth’s crust combine with coal deposits and formed carbides.

(b) Carbides reacted with steam or water under high temperature and pressure to form mixture of saturated
and unsaturated hydrocarbons.

(c) The unsaturated hydrocarbon in presence of metal catalyst, high pressure and high temperature, undergoes
reactions such as hydrogenation, isomerisation and polymerisation to form number of hydrocarbons.

Reactions : Ca + 2C → CaC2 (Calcium carbide); Mg + 2C → MgC2 (Magnesium carbide)

4 Al + 3C → Al4 C3 (Aluminium carbide); CaC2 + 2H 2O → Ca(OH)2 + C2 H 2 (Acetylene)

Al4 C3 + 12H 2O → 4 Al(OH)3 + 3CH 4 (Methane) ; C2 H 2 H2 → C2 H 4 H2 → C2 H6

3[CH ≡ CH] Polymerisation → C6 H 6 ; CH 3 − CH = CH − CH 3 Isomerisation → CH 3 − CH 2 − CH = CH 2
(Benzene) (1-Butene)

Theory fails to account for, The presence of nitrogen and sulphur compounds. The presence of chlorophyll
and haemin derivatives. The presence of optically active compounds.

(ii) Engler’s theory or organic theory : Theory is supported by the following facts,

(a) The presence of brine with petroleum,

(b) The presence of optically active compounds,

(c) The presence of nitrogen and sulphur compounds

(d) The presence of fossils in the petroleum area.

The theory was further supported by the fact that when destructive distillation of fish oil and other animals fats
under high temperature and pressure was carried out, a petroleum like liquid was obtained.

Theory fails to account for, The presence of chlorophyll in the petroleum. The presence of coal deposits found
near the oil fields. The presence of resins in the oil.

(iii) Modern theory : This theory explain nearly all the facts about petroleum.

(a) The presence of chlorophyll and haemin in petroleum.
(b) The presence of coal deposits near oil fields suggesting its vegetable origin.

(c) The presence of nitrogen and sulphur compounds along with optically active compounds in petroleum.

(d) The presence of resins also suggests that oil must have been formed from vegetable substances.

(e) The presence of helium gas in natural gas suggests that radioactive substances must have helped in the
decomposition of organic matter.

(3) Mining of petroleum : Petroleum deposits occurs at varying depth at different places ranging from 500
to 15000 feet. This is brought to the surface by artificial drilling.

(4) Petroleum refining : Separation of useful fractions by fractional distillation is called petroleum refining.

Fraction Boiling range (oC) Approximate Uses
Uncondensed gases Upto room Fuel gases: refrigerants; production of
temperature C1 – C4 carbon black, hydrogen; synthesis of
organic chemicals.
C5 – C10
Crude naphtha on 30 – 150o

refractionation yields, 30 – 70o
70 – 120o
(i) Petroleum ether 120 – 150o C5 – C6 Solvent
150 – 250o C6 – C8 Motor fuel; drycleaning; petrol gas.
(ii) Petrol or gasoline 250 – 400o C8 – C10 Solvent; drycleaning
C11 – C16 Fuel; illuminant; oil gas
(iii) Benzene derivatives C15 – C18 As fuel for diesel engines; converted to
gasoline by cracking.
Kerosene oil

Heavy oil

Refractionation gives, Above 400o C17 – C40
(i) Gas oil
(ii) Fuel oil C17 – C20 Lubrication
(iii) Diesel oil C20 – C30 Candles; boot polish; wax paper; etc
Residual oil on C20 – C30 Toilets; ointments; lubrication.
fractionation by vacuum C30 – C40 Paints, road surfacing
distillation gives, As fuel.
(i) Lubricating oil
(ii) Paraffin wax
(iii) Vaseline
(iv) Pitch
Petroleum coke
(on redistilling tar)

(5) Purification

(i) Treatment with concentrated sulphuric acid : The gasoline or kerosene oil fraction is shaken with
sulphuric acid to remove aromatic compounds like thiophene and other sulphur compound with impart offensive
odour to gasoline and kerosene and also make them corrosive.

(ii) Doctor sweetening process : 2RSH + Na2 PbO2 + S → RSSR + PbS + 2NaOH

(iii) Treatment with adsorbents : Various fractions are passed over adsorbents like alumina, silica or clay
etc, when the undesirable compounds get adsorbed.

(6) Artificial method for manufacture of Petrol or gasoline
(i) Cracking, (ii) Synthesis

(i) Cracking : It is a process in which high boiling fractions consisting of higher hydrocarbons are heated
strongly to decompose them into lower hydrocarbons with low boiling pts. Cracking is carried out in two different

(a) Liquid phase cracking : In this process, the heavy oil or residual oil is cracked at a high temperature (475
– 530oC) under high pressure (7 to 70 atmospheric pressure). The high pressure keeps the reaction product in liquid
state. The conversion is approximately 70% and the resulting petrol has the octane number in the range 65 to 70.

The cracking can be done in presence of some catalysts like silica, zinc oxide, titanium oxide, ferric oxide and
alumina. The yields of petrol are generally higher when catalyst is used.

(b) Vapour phase cracking : In this process, kerosene oil or gas oil is cracked in vapour phase. the temperature
is kept 600 – 800oC and the pressure is about 3.5 to 10.5 atmospheres. The cracking is facilitated by use of a
suitable catalyst. The yields are about 70%.

(ii) Synthesis : Two methods are applicable for synthesis.

(a) Bergius process : This method was invented by Bergius in Germany during first world war.

Coal + H2 Fe2O3 → Mix. Of hydrocarbons or crude oil

45205−05a0t0mo C

(b) Fischer- tropsch process : The overall yield of this method is slightly higher than Bergius process.

H2O + C 1200oC → CO + H2
Water gas

xCO + yH 2 → Mix. Of hydrocarbon + H2O .
The best catalyst for this process is a mixture of cobalt (100 parts), thoria, (5 parts), magnesia (8 parts) and
kieselguhr (200 parts).

Characteristics of hydrocarbons.

(1) Knocking : The metallic sound due to irregular burning of the fuel is termed as knocking.

“The greater the compression greater will be efficiency of engine.” The fuel which has minimum knocking
property is always preferred.

The tendency to knock falls off in the following order : Straight chain alkanes > branched chain
alkanes > olefins > cyclo alkanes > aromatic hydrocarbons.

(2) Octane number : It is used for measuring the knocking character of fuel used in petrol engine. The
octane number of a given sample may be defined as the percentage by volume of iso-octane present in a mixture of

iso-octane and heptane which has the same knocking performance as the fuel itself.

CH3 − CH2 − CH2 − CH2 − CH2 − CH2 − CH3 n-heptane; octane no. = 0

CH3 CH 3
| |

CH3 — C — CH2 — CH — CH3 ; Octane no. = 100



2, 2, 4-Trimethyl pentane or Iso-octane.

For example : a given sample has the knocking performance equivalent to a mixture containing 60% iso-
octane and 40% heptane. The octane number of the gasoline is, therefore, 60.

Presence of following types of compounds increases the octane number of gasoline.

(i) in case of straight chain hydrocarbons octane number decreases with increase in the length of the chain.

(ii) Branching of chain increases the value of octane number

(iii) Introduction of double bond or triple bond increases the value of octane number.

(iv) Cyclic alkanes have relatively higher value of octane number.

(v) The octane number of aromatic hydrocarbons are exceptionally high

(vi) By adding gasoline additives (eg TEL)

(3) Antiknock compounds : To reduce the knocking property or to improve the octane number of a fuel
certain chemicals are added to it. These are called antiknock compounds. One such compound, which is

extensively used, is tetraethyl lead (TEL). TEL is used in the form of following mixture,

TEL = 63%, Ethylene bromide = 26%, Ethylene chloride = 9% and a dye = 2%.

However, there is a disadvantage that the lead is deposited in the engine. To remove the free lead, the
ethylene halides are added which combine with lead to form volatile lead halides.

Pb + Br − CH2 − CH2 − Br → PbBr2 + CH2 = CH2
Ethylene bromide Volatile Ethylene

However, use of TEL in petrol is facing a serious problem of Lead pollution, to avoid this a new compound
cyclopenta dienyl manganese carbonyl (called as AK-33-X) is used in developed countries as antiknocking

(4) Other methods of improving octane number of gasoline

(i) Isomerisation [Reforming] : By passing gasoline over AlCl3 at 200o C .

AlCl3 → C| H3
Pentane 3
(Octane number = 62) (Octane number = 90)

(ii) Alkylation : C| H3 CH 3 H2SO4 → C|H3 C| H3
| CHCH 3
CH CH + CH 2 = C CH 3 CH CCH
3 | Isobutylene 3 | 2

Isobutane Iso-octane
(Octane number = 100)


(iii) Aromatisation : CH3(CH2)5 CH3 Pt /Al2O3 → + 4H2
Heptane 500o C


The octane no. of petrol can thus be improved.
• By increasing the proportion of branched chain or cyclic alkanes.
• By addition of aromatic hydrocarbons (BTX)
• By addition of methanol or ethanol.
• By additon of tetraethyl lead (C2H5)4 Pb

(5) Cetane number : It is used for grading the diesel oils.
CH3 − (CH2)14 − CH3 Cetane → cetane no. = 100


Cetane no. = 0

α-Methyl naphthalene

The cetane number of a diesel oil is the percentage of cetane (hexadecane) by volume in a mixture of cetane
and α -methyl naphthalene which has the same ignition property as the fuel oil

(6) Flash point : The lowest temperature at which an oil gives sufficient vapours to form an explosive
mixture with air is referred to as flash point of the oil.

The flash point in India is fixed at 44o C , in France it is fixed at 35oC, and in England at 22.8oC. The flash
point of an oil is usually determined by means of “Abel’s apparatus”.

Chemists have prepared some hydrocarbons with octane number even less than zero (e.g., n-nonane has
octane number – 45) as well as hydrocarbon with octane number greater than 100 (e.g., 2, 2, 3 trimethyl-butane.
has octane number of 124).

(7) Petrochemicals : All such chemicals which are derived from petroleum or natural gas called
petrochemicals. Some chemicals which are obtained from petroleum are:

Hydrocarbons Compounds derived

Methane Methyl chloride, chloroform, methanol, formaldehyde, formic acid, freon, hydrogen for
synthesis of ammonia.

Ethane Ethyl chloride, ethyl bromide, acetic acid, acetaldehyde, ethylene, ethyl acetate,
nitroethane, acetic anhydride.

Ethylene Ethanol, ethylene oxide, glycol, vinyl chloride, glyoxal, polyethene, styrene, butadiene,
acetic acid.

Propane Propanol, propionic acid, isopropyl ether, acetone, nitromethane, nitroethane, nitropropane.

Propylene Glycerol, allyl alcohol, isopropyl alcohol, acrolein, nitroglycerine, dodecylbenzene, cumene,

Hexane Benzene, DDT, gammexane.

Heptane Toluene

Cycloalkanes Benzene, toluene, xylenes, adipic acid.

Benzene Ethyl benzene, styrene, phenol, BHC (insecticide), adipic acid, nylon, cyclohexane, ABS

Toluene Benzoic acid, TNT benzaldehyde, saccharin, chloramine-T, benzyl chloride, benzal chloride.

Alkanes [Paraffines].

“Alkanes are saturated hydrocarbon containing only carbon-carbon single bond in their molecules.”
Alkanes are less reactive so called paraffins; because under normal conditions alkanes do not react with acids,
bases, oxidising agents and reducing agent.
General formula : CnH2n+2

Examples are : CH 4 , C2 H6 , C3 H8
Methane Ethane Propane

(1) Structure : (i) Every carbon atom is H HH
sp 3 hybridized.

(ii) The bond length between carbon-

carbon and carbon-hydrogen are 1.54 Å and 109o28’ 1.54Å 109o28’
1.112Å respectively. CH
1.112Å 1.112Å C C
(iii) Bond angle in alkanes are tetrahedral
angles having a value of 109.5o (109o.28′) . H

(iv) Alkanes have 3 − D , rather than H H H
planer structure. Methane HH

(v) C − C bond dissociation energy is Ethane
83 k cal / mol .

(vi) C − H bond dissociation energy is 99k cal / mol .

(2) Isomerism : Only chain and structural Isomerism found.

No. of carbon atom in molecule ∝ no. of chain Isomers

Alkanes : C4 H10 C5 H12 C6 H14 C7 H16 C8 H18 C10 H 22
9 18 75
No. of possible Isomer : 2 3 5

(3) General Methods of preparation
(i) By catalytic hydrogenation of alkenes and alkynes (Sabateir and sanderen’s reaction)

Cn H 2n + H2 Ni → Cn H 2n+ 2 ; Cn H 2n−2 + 2H 2 Ni → Cn H 2n+ 2
Alkene heat Alkane Alkyne heat Alkane

Note : Methane is not prepared by this method

(ii) Birch reduction : R − CH = CH 2 1. Na / NH3 → R − CH 2 − CH 3
2. CH3OH

(iii) From alkyl halide

(a) By reduction : RX + H 2 Zn /HCl → RH + HX

(b) With hydrogen in presence of pt/pd : RX + H 2 Pd orPt. → RH + HX

(c) With HI in presence of Red phosphorus : RBr + 2HI → RH + HBr + I 2
Purpose of Red P is to remove I

(iv) By Zn-Cu couple : 2CH 3CH 2OH + Zn Cu →(CH 3CH 2O)2 Zn+ 2H
Zinc ethoxide
Zn-Cu couple

RX + 2H → RH + HX

(v) Wurtz reaction : R X + 2Na + X R′ Dryether → R − R+ 2NaX
Alkyl halide Alkyl halide

Note :  R − Br or RI preferred in this reaction. The net result in this reaction is the formation of even no.

of carbon atoms in molecules.

(vi) Frankland’s reaction : 2RX + Zn → R − R + ZnX 2
(vii) Corey-house synthesis

CH 3 − CH 2 − Cl 1.Li →(CH 3 − CH 2 )2 LiCu CH3 −CH2−Cl → CH 3 − CH 2 − CH 2 − CH 3
2. CuI

Note :  Reaction is suitable for odd number of Alkanes.

(viii) From Grignard reagent

(a) By action of acidic ‘H’ : RMgX + HOH → RH + Mg(OH)X
Alkyl magnesium Water Alkane


(b) By reaction with alkyl halide : R − X + R′MgX → R − R′ + MgX 2

(ix) From carboxylic acids
(a) Laboratory method [Decarboxylation reaction or Duma reaction]

R COONa + NaOH heat → R − H + Na 2 CO3
CaO Alkane

Note :  NaOH and CaO is in the ratio of 3 : 1. O
(b) Kolbe’s synthesis : R − C− O− Na+ Electrolysis R − C− O− + Na+
|| Ionization

At anode [Oxidation] : 2R − C− O− + 2e − → 2R − C− O• → 2 R• + 2CO2
|| ||

2 R• → R − R (alkane)

At cathode [Reduction] : 2Na+ + 2e− → 2Na 2H2O → 2NaOH + H2 ( ↑ )

Note :  Both ionic and free radical mechanism are involved in this reaction.

(c) Reduction of carboxylic acid : CH 3COOH + 6HI Re duction → CH3CH3 + 2H2O + 3I2
Acetic acid p Ethane

(x) By reduction of alcohols, aldehyde, ketones or acid derivatives

CH3OH + 2HI RedP → CH4 + H2O + I2 ; CH3CHO+ 4 HI RedP → C2 H 6 + H2O + 2I2
Methanol 150o C 150o C
Methane Acetaldehyde Ethane
(Methyl alcohol) (Ethanal)

CH3COCH3 + 4 HI RedP → CH 3CH2CH 3 + H2O + 2I2 ;
Acetone 150o C

CH 3 − C− Cl+ 6HI RedP → CH 3 − CH H 2O HCl 3I
Acetyl chloride 200o C Ethane 3 + + + 2

(Ethanoyl chloride)

CH 3 − C− NH 2 + 6HI RedP → CH 3 − CH H2O NH 3 3I 2
Acetamide 200o C 3 + + +

Note :  Aldehyde and ketones when reduced with amalgamated zinc and conc. HCl also yield alkanes.

Clemmensen reduction : CH 3CHO+ 2H 2 Zn−Hg → CH 3 − CH 3 + H2O
Acetaldehyde Ethane

CH3COCH3 + 2H2 Zn−Hg → CH 3CH2CH 3 + H2O
Acetone HCl

Note :  Aldehydes and ketones (> C = O) can be reduced to hydrocarbon in presence of excess of
hydrazine and sodium alkoxide on heating.

Wolff-kishner reduction :
C = O H2 NNH2 → C = NNH2 C2H5ONa → CH2
R′ −H2O 180o C
R′ R′

(xi) Hydroboration of alkenes

(a) On treatment with acetic acid

R − CH = CH2 B2H6 → (R − CH2 − CH2)3 B CH3COOH → R − CH2 − CH3
Alkene Trialkyl borane Alkane

(b) Coupling of alkyl boranes by means of silver nitrate

6[R − CH = CH2] 2B2H6 →[2R − CH 2 − CH2 −]3 B AgNO325oC → 3[ RCH 2CH 2 − CH 2CH 2 R]

(4) Physical Properties
(i) Physical state : Alkanes are colourless, odourless and tasteless.

Alkanes State

C1 − C4 Gaseous state

C5 − C17 Liquid state [Except neo pentane] [gas]

C18 & above Solid like waxes

(ii) Density : Alkanes are lighter than water.

(iii) Solubility : Insoluble in water, soluble in organic solvents, solubility ∝ 1 mass

(iv) Boiling pts and Melting pts : Melting pt. and boiling pts. ∝ Molecular mass ∝ No. of 1

Alkane : C3 H 8 C4 H10 C5 H12 C6 H14 C7 H16 C8 H18
M.P. (K) : 85.9 138 143.3 179 182.5 216.2

Even no. of carbons
Odd no. of Carbons

600 – 300

400 –Temperature (K) 200
200 – Melting point oC
Boiling Pt. 100 ••••• • ••••••••
increases 0 34567 8 9 10 11 12 13 14 15 16

4 8 12 16 20 – 100 •
– 2001• 2

50 | | | | | Number of carbon atoms

No. of C atoms per molecule

Note :  Melting points of even > Odd no. of carbon atoms, this is because, the alkanes with even number

of carbon atoms have more symmetrical structure and result in closer packing in the crystal structure
as compared to alkanes with odd number of carbon atoms.

(5) Chemical properties
(i) Substitution reactions of Alkanes

(a) Halogenation : R − H + X − X → R − X + HX

The reactivity of halogen is : F2 > Cl2 > Br2 > I2

Note :  Fluorine can react in dark Cl2, Br2 require light energy. I2 doesnot show any reaction at room

temperature, but on heating it shows iodination.

 Iodination of methane is done in presence of oxidising agent such as HNO3 / HIO3 / HgO which
neutralises HI .

 Chlorination of methane :

CH 4 + Cl − Cl u.v.light → CH 2 − Cl2 u.v.light → CHCl 3 −HCl → CCl4
− HCl − HCl

(ii) Reaction based on free radical mechanism

(a) Nitration : R−H + HONO2 High → R − NO2 + H2O
Alkane Nitroalkane

Nitrating mixture : (i) (Con. HNO3 + Con. H 2SO4 ) at 250o C

(ii) (HNO3 vapour at 400o − 500o C) .

(b) Sulphonation : Free radical mechanism R − H + HOSO3 H SO3→ R − SO3 H + H2O

Prolonged heating

Note :  Lower alkanes particularly methane, ethane, do not give this reaction.

(iii) Oxidation

(a) Complete Oxidation or combustion : Cn H2n+ 2 +  3n + 1 O2 → nCO2 + (n + 1)H2O + Q
 2

Note :  This is exothermic reaction.

(b) Incomplete combustion or oxidation

2CH 4 + 3O2 Burn → 2CO + 4 H 2O

CH4 + O2 → C +2H2O

(c) Catalytic Oxidation : CH4 + [O] Cu−tube → CH 3 OH

100 atm / 200o C

This is the industrial method for the manufacture of methyl alcohol.

Note :  Higher alkanes are oxidised to fatty acids in presence of manganese stearate.

CH 3 (CH 2 )n CH 3 O2 → CH 3 (CH 2 )n COOH

100 −160 o C

(d) Chemical oxidation : (CH3)3 CH KMnO4 →(CH3)3.C.OH
Isobutane Tertiary butyl alcohol

(iv) Thermal decomposition or cracking or pyrolysis or fragmentation

CH4 1000oC → C + 2H2 ; C2 H 6 500oC → CH2 = CH2 + H2
Methane Ethane Cr2O3 + Al2O3

C3H8 → C2H4 + CH4 or C3H6 + H2

Note :  This reaction is of great importance to petroleum industry.

CH 3
(v) Isomerisation : CH3CH2CH2CH3 AlCl3 +HCl → CH3 CHCH ; AlCl3 +HCl →
n- Butane 200o C, 35atm Isobutane 3 2-Methyl pentane heat
+4H2 2,3 Dimethyl
(vi) Aromatisation : butane


H2C CH3 Cr2O3 /Al2O3 →
600o C / 15 atm

CH2 Benzene


Cr2O3 /Al2O3 → −H2 →
600o C

n-Heptane Methyl cyclo Toluene

(vii) Step up reaction

(a) Reaction with CH2 N2 : R − CH 2 − H + CH 2 N 2 hv → R − CH 2 − CH 2 − H

(b) Reaction with CHCl3 / NaOH : R − CH 2 − H CHCl3 / OH− → R − CH 2 − CHCl2
: CCl2

CH2 =C/∆ →
(c) Reaction with CH 2 =C : R − CH 2 − H :CH2 / −CO R − CH 2 − CH 3


(viii) HCN formation : 2CH 4 N2 / electricarc → 2HCN + 3H 2 or CH4 + NH3 Al2O3 → HCN + 3H2
700o C

(ix) Chloro sulphonation/Reaction with SO2+Cl2

CH3 − CH2 − CH3 + SO2 + Cl2 u.v light → CH3 − CH2 − CH2SO2Cl + HCl

This reaction is known as reed’s reaction.

Note :  This is used in the commercial formation of detergent.

(x) Action of steam : CH4 + H2O Ni /Al2O3 → CO + 3H2
800o C

Individual members of alkanes.

(1) Methane : Known as marsh gas.
(i) Industrial method of preparation : Mathane gas is obtained on a large scale from natural gas by

liquefaction. It can also be obtained by the application of following methods,

(a) From carbon monoxide : A mixture of carbonmonoxide and hydrogen is passed over a catalyst containing

nickel and carbon at 250o C when methane is formed.

CO + 3H2 Ni+C → CH 4 + H2O
250o C

(b) Bacterial decomposition of cellulose material present in sewage water : This method is being used in

England for production of methane.

(C6 H10O5 )n + nH2O → 3nCH4 + 3nCO2

(c) Synthesis :  By striking an electric arc between carbon electrodes in an atmosphere of hydrogen at
1200oC, methane is formed.

C + 2H2 1200oC → CH4

By passing a mixture of hydrogen sulphide and carbon disulphide vapour through red hot copper, methane is

CS2 + 2H2S + 8Cu High temperature → CH4 + 4Cu2S
(ii) Physical properties
(a) It is a colourless, odourless, tasteless and non-poisonous gas.
(b) It is lighter than air. Its density at NTP is 0.71 g/L.
(c) It is slightly soluble in water but is fairly soluble in ether, alcohol and acetone.

(d) Its melting point is − 182.5o C and boiling point is − 161.5o C .
(iii) Uses
(a) In the manufacture of compounds like methyl alcohol, formaldehyde, methyl chloride, chloroform, carbon
tetrachloride, etc.
(b) In the manufacture of hydrogen, used for making ammonia.
(c) In the preparation of carbon black which is used for making printing ink, black paints and as a filler in
rubber vulcanisation.
(d) As a fuel and illuminant.

(2) Ethane
(i) Methods of preparation

(a) Laboratory method of preparation : C2H5I + 2H Zn−Cucouple → C2 H 6 + HI
Ethyl iodide Ethane

(b) Industrial method of preparation : CH2 = CH2 + H2 Ni → CH 3 − CH 3
Ethylene 300o C

(iii) Physical properties
(a) It is a colourless, odourless, tasteless and non-poisonous gas.
(b) It is very slightly soluble in water but fairly soluble in alcohol, acetone, ether, etc.
(c) Its density at NTP is 1.34 g/L
(d) It boils at – 89oC. Its melting point is –172oC.

(ii) Uses
(a) As a fuel. (b) For making hexachloroethane which is an artificial camphor.

(3) Interconversion of Alkanes
Ascent of alkane series,

(i) Methane to ethane : CH 4 Cl2 → CH 3 Cl Wurtz reaction → CH 3 − CH 3
Methane Heat with Na in ether Ethane

(ii) Butane from ethane : C2 H 6 Cl2 → C2 H 5 Cl Wurtz reaction → C2H5 − C2H5
UV Butane
Ethane Ethyl chloride Heat with Na in ether


Descent of alkane series : Use of decarboxylation reaction is made. It is a multistep conversion.

Ethane to methane

C2 H 6 Cl2 → C2 H 5 Cl Aq.KOH → C2H5OH [O] → CH3CHO [O] → CH3COOH NaOH → CH3COONa NaOH /CaO → CH 4
UV Ethyl alcohol Acetaldehyde Acetic acid Sodium acetate heat
Ethane Ethyl chloride Methane

Higher Cl2 → Alkyl Aq. → Alcohol [O] → Aldehyde [O] → Acid NaOH → Sodium salt of NaOH /CaO → Lower alkane
alkane UV halide KOH the acid heat


These are the acyclic hydrocarbon in which carbon-carbon contain double bond. These are also known as

olefins, because lower alkene react with halogens to form oily substances. General formula is CnH2n .

Ex : Ethene C2H4 , Propene C3H6 , Butene C4 H8

(1) Structure π bond Unhybridised orbital
(i) Hybridisation of unsaturated ‘C’ atom is sp2
H sp2 sp2 sp2 H
(ii) Geometry of unsaturated ‘c’ atom is trigonal planer
(iii) C − H bond length is 1.34 Å sp2 sp2
(iv) C = C bond energy is 143.1 K cal/mol sp2
(v) C − H bond length is 1.10 Å H
(vi) C = C bond energy is 108 Kcal/mol H

(2) Isomerism

(i) Chain Isomerism : CH3 − CH2 − CH = CH2 and (CH 3 )2 − C = CH 2

(ii) Position Isomerism : CH2 = CH − CH2 − CH3 and CH3 − CH = CH − CH3

(iii) Functional Isomerism : [Ring chain] CH3 − CH2 − CH = CH2 and CH2 − CH2
1 − butene ||

CH2 − CH2
Cyclo butene

(iv) Geometrical Isomerism : CH3 − C − H and CH3 − C − H
|| ||

CH 3 − C − H H − C − CH3
cis − 2−butene
Trans − 2−butene

CH = CH 2
(v) Optical Isomerism : H − C − CH3

CH 2CH 3

Note :  Cumulated polythene having even no. of double bonds. Which has = C a
system at the both


end can exhibit optical isomerism but cannot exhibit geometrical isomerism.

 Cumulated polythene having odd no. of double bonds which have = C a
system at both end can


exhibit geometrical isomerism but cannot exhibit optical isomerism.

(3) Preparation methods

(i) From Alkynes : R − C ≡ C − H + H2 Lindlar's Catalyst → R − C C− H
Pd. BaSO4 = |


Note :  Poison’s catalyst such as BaSO4 , CaCO3 are used to stop the reaction after the formation of


|| |
(ii) From mono halides : R− C− C− H Alc. KOH −HX → R − C = C− H
| | + |



Note :  If we use alc. NaOH in place of KOH then trans product is formed in majority because of its

stability. According to saytzeff rule.

(iii) From dihalides

(a) From Gem dihalides X Zn X

R – CH + + CH – R ∆ → R – CH = CH – R
− 2 Znx2
Note :  If we take two differentXtypes oZfngemdihaXlides then we different types of alkenes .
get three

Above reaction is used in the formation of symmetrical alkenes only.

(b) From vicinal dihalides : HH HH
|| ||
R− C− C− H Zn dust ∆ → R − C C− H ZnX 2
| | + 300o C = +


Note :  Alkene is not formed from 1, 3 dihalides. Cycloalkanes are formed by dehalogenation of it.

C H2 − CH 2 − C H2 Zn dust → CH2
| |


(iv) By action of NaI on vic dihalide : Br Br NaI → II C=C
|| acetone ||
C− C C− C →

(v) From alcohols [Laboratory method] : CH3CH2OH H2SO4or H3 PO4 → CH2 = CH2 + H2O
Ethyl alcohol 443 K


(vi) Kolbe’s reaction : | + 2H2O Electrolysis →|| + 2CO2 + H2 + 2KOH


Potassium succinate Ethene

(vii) From esters [Pyrolysis of ester] : CH3 − CO − O H Glass wool 450o → CH3 − COOH
| | +
liq. N 2 CH2 = CH2
CH2 − CH2

(viii) Pyrolysis of quaternary ammonium compounds : (C2H5 )4 N+ O−H heat →(C2H5 )3 N + C2H4 + H2O
Tetraethyl ammonium Triethylamine Ethene
hydroxide (Tert. amine)

(ix) Action of copper alkyl on vinyl chloride : H2C = CHCl CuR2 → H2C = CHR
Vinyl chloride


(x) By Grignard reagents : Mg + X − CH = CH2 → MgX2 + R − CH = CH2

(xi) The wittig reaction : (Ph)3 P = CH 2 + CH− R →(Ph)3 P = O+ R − CH
|| ||

O CH 2

(Ph)3 P = CH − R + CH− R →(Ph)3 P = O + R − CH = CH − R

(xii) From β bromo ether [Boord synthesis]

Br O − C2 H5 Br
| | O − C2H5
R − CH− CH Zn → R − CH = CH − R′ + Zn
| C4HgOH


(4) Physical Properties
(i) Alkenes are colourless and odourless.

(ii) These are insoluble in water and soluble in organic solvents.

(iii) Physical state

C1 − C4 → gas

C5 − C16 → liquid

> C16 → solid wax

(iv) B.P. and M.P. decreases with increasing branches in alkene.
(v) The melting points of cis isomers are lower than trans isomers because cis isomer is less symmetrical than
trans. Thus trans packs more tightly in the crystal lattice and hence has a higher melting point.

(vi) The boiling points of cis isomers are higher than trans isomers because cis-alkenes has greater polarity
(Dipole moment) than trans one.

(vii) These are lighter than water.

(viii) Dipole moment : Alkenes are weakly polar. The, π-electron’s of the double bond. Canbe easily
polarized. Therefore, their dipole moments are higher than those of alkanes.

The symmetrical trans alkenes are non polar and hence have zero dipole moments in these alkene the dipole

moment of individual bonds are equal in opposite direction. Therefore these get cancelled resulting zero dipole

moment for the molecule. CH3 H CH3 CH3


HTrans-2-ButenCe H3 HH
Cis –2-Butene

µ=0 µ = 0.25 D

Thus symmetrical and unsymmetricals cis alkene are polar and hence have finite dipole moments

CH3 H CH3 – CH2 H
HH Bute 1-ene
Propene µ = 0.37 D

µ = 0.35 D

(5) Chemical properties

(i) Francis experiment : According to Francis electrophile first attacks on olefinic bond.

CH2 = CH2 + Br – Br CCl4 → CH2 – CH2
Br Br

NaCl → CH2 – CH2 + |CH2 – C| H2
| |
Br Br Br Cl

|| ||
(ii) Reaction with hydrogen : R−C = C− R + H2 Ni → R − C − C− R


(iii) Reduction of alkene via hydroboration : Alkene can be converted into alkane by protolysis

RCH = CH 2 H−BH2 →(R − CH 2 − CH 2 )3 B H+ / H2O → R − CH 2 − CH3

Hydroboration : Alkene give addition reaction with diborane which called hydroboration. In this reaction
formed trialkylborane, Which is very important and used for synthesis of different organic compound

3R − CH = CH2 + BH3 →(R − CH2 − CH2 )3 B

CH3COOH/Zn Trialkyl borane
R – CH2 –CH3
R – CH2 –CH3

R – CH2 –CH2OH

The overall result of the above reaction appears to be antimarkownikoff’s addition of water to a double bond.

(iv) By treatment with AgNO3 + NaOH : This reaction gives coupling

CH 3 CH 3
| |
6CH3 − CH 2 − CH 2 − C = CH 2 B2H6 → 2[CH3 − (CH 2 )2 − C − CH 2 ]3 B Ag /NO3NaOH →


CH 3 CH 3
| |
CH3 − CH 2 − CH2 − C − CH2 − CH2 − C − CH2 − CH2 − CH3


(v) Birch reduction : This reaction is believed to proceed via anionic free radical mechanism.

R − CH = CH 2 Na → R − C H− − C H− 2 Et−O−H → R − CH − CH 3 Na → R − C− H − CH3 Et.−O−H → R − CH2 − CH 3
+e− +e−

(vi) Halogenation

CH3CH = CH2 + Cl2 500oC → ClCH2 − CH = CH2 + HCl
Propene Allyl chloride

or 3-Chloro-1- propene

Note :  If NBS [N-bromo succinimide] is a reagent used for the specific purpose of brominating alkenes at

the allylic position. CH2 – CO CH2 – CO
CH3 CH=CH2 + |
N – Br CH2 – CH = CH2+ | N–H
CH2 – CO | CH2 – CO
Propene NBS Allyl bromide Succinimide

 In presence of polar medium alkene form vicinal dihalide with halogen.

|| ||
R − C = C− H + X − X CCl4 → R − C − C− H


Vicinal dihalide

Reactivity of halogen is F2 > Cl2 > Br2 > I 2

(vii) Reaction with HX [Hydrohalogenation]

C=C + HX → H
alkene C− C


Alkyl halide

According to markownikoff’s rule and kharasch effect.

CH 3 − CH = CH 2 + HBr → CH 3 − C − C− H Markownikoff rule

Br H

|| ||
(Based on F.R.M.) CH 3 − CH CH 2 HBr Peroxide → CH 3 C − C− H + CH 3 C − C− H
= + − || − ||

Br H H Br

(minor) (major)

(viii) Reaction with hypohalous acids : CH 2 = CH 2 + H O− C+l → CH 2OH.CH 2Cl
Ethylene Ethylene chlorohydrin

Note :  In case of unsymmetrical alkenes markownikoff rule is followed.

(ix) Reaction with sulphuric acid : CH 2 = CH 2 + H + HSO4− → CH 3CH 2 HSO4
Ethylene Ethyl hydrogen sulphate

CH 3CH 2 HSO4 → CH 2 = CH 2 + H 2SO4

Note :  This reaction is used in the seperation of alkene from a gaseous mixture of alkanes and alkenes.

(x) Reaction with nitrosyl chloride

C=C + NOCl → NO ( NOCl is called Tillden reagent)

Note :  If hydrogen is attached to the carbon atom of product, the product changes to more stable oxime.

|H + NOCl → (Blue colour)
C− C C − C = NOH ; C=C C −C
| ⇌ || C || C
Cl NO Cl

(xi) Oxidation : With alkaline KMnO4 [Bayer’s reagent] : This reaction is used as a test of unsaturation.

|| ||
R − C = C − H + [O] + H − OH AlkKMnO4 → R − C − C− H
−OH | |



|| ||
With acidic KMnO4 : R−C C H [O] acidic → R C− O H CO2 + H2O
= − + KMnO4 − − +

(xii) Hydroxylation C|H3 H2O2, HCOOH → C|H3
(a) Using per oxy acid : H −|C| or HCO3H H − C − OH

H − C| |
CH3 OH − C| − H

2-Butene CH3

Trans (racemic)


(b) Hydroxylation by OsO4 : C R
||+ OsO4 + I →
Trans R

Note :  If per benzoic acid or peroxy acetic acid is used then oxirane are formed.

R − CH = CH − R C6H5CO3H → R − CH − CH − R −H2O → R − CH − CH − R
or CH 3CO3 H ||



(xiii) Combustion : Cn H 2n + 3n O2 → nCO2 + nH 2O

They burn with luminous flame and form explosive mixture with air or oxygen.

(xiv) Ozonolysis

C=C O3 → C C H2O/ H+ / Zn → ZnO +



 Application of ozonolysis : This process is quite useful to locate the position of double bond in an alkene

molecule. The double bond is obtained by Joining the carbon atoms. of the two carbonyl compounds.


CH3 − C = O + O = C − CH3 → CH3 − CH = CH − CH3


H CH 3 H
| | |
CH3 − C = O + O = C − CH3 → CH3 − C = C − CH3


2-methyl, but- 2-ene

(xv) Oxy – mercuration demercuration : With mercuric acetate (in THF), followed by reduction with
NaBH 4 / NaOH is also an example of hydration of alkene according to markownikoff’s rule.

(CH 3 )3 C − CH = CH 2 + (CH 3COO)2 Hg →(CH 3 )3 C − CH − CH 2 − Hg NaBH4/ NaOH → (CH 3 ) 3 C − CH − CH 3
| THF |


3, 3−Dimethyl−2−butanol

Ex. (CH2)2OH

CH = CH2 BH3 (CH2 – CH2)3 B HOOH/OH –

Antimarkowni -
koff rule

Mercuration markowni - HgOCOCH3
koff rule

H+/H2O H⊕ ⊕ CH2 CH3 H2O/H + CH2 –CH3

Less stable More stable alcohol
Carbocation carbocation

(xvi) Epoxidation O
1 O C–O–O–H
2 |
(a) By O2 / Ag : CH 2 = CH 2 + O2 Ag → CH2 − CH 2


(b) Epoxidation by performic acid or perbenzoic acid : CH2 = CH2 → CH2 − CH2



CH3 − CH = CH 2 H−C−O−O−H → CH3 − CH − CH 2
(xvii) Hydroboration

3R − CH = CH 2 + BH3 →(R − CH 2 − CH 2 )3 B H2O2 / OH− → R − CH 2 − CH 2 − OH + B(OH)3
Tri alkyl borane

(Anti markownikoff’s rule)

(xviii) Hydroformylation : R − CH = CH 2 + CO + H 2 CoH(CO)4 → R − C| − C| − H


Note :  If CO + H2O is taken then respective acid is formed.

R − CH = CH 2 + CO + H 2O CoH(CO)4 → R − CH 2 − CH 2

(xix) Addition of formaldehyde

H2C = O + H⊕ →[H2C = O⊕ H ←→ H2 C⊕ − OH] R−CH =CH2 → R − C⊕ H − CH2 − CH2 − OH HOH
CH2 − H+

R – CH CH2 ⊕
O O ←HCHO/ H 
R − CH − CH 2 − CH 2
| |

Cyclic acetal 1, 3−diol

(xx) Polymerisation

HH  H H H H

||  | | | | 
 
C = C Trace O2 +Catalyst →− C − C − C − C 
1500o / high pressure  
| |  | || | 

HH  H H H H n

Note :  If in polymerisation zeigler- natta catalyst [(R)3 Al + TiCl4 ] is used then polymerisation is known as

zeigler-natta polymerisation. AlCl3 CH3 − CH2 − CH = CH − CH3
(xxi) Isomerisation : CH3 − CH2 − CH2 − CH = CH2

The mechanism proceeds via carbocation.

(xxii) Addition of HNO3 : CH2 = CH2 + HO − NO2 → CH2OH.CH2NO2
Ethene 2- Nitroethanol

(xxiii) Addition of Acetyl chloride : CH2 = CH2 + CH3COCl → CH2ClCH2COCH3
Ethene 4-Chlorobutanone- 2

(6) Uses
(i) For the manufacture of polythene – a plastic material; (ii) For artificial ripening of fruits; (iii) As a general
anaesthetic; (iv) As a starting material for a large number of compounds such as glycol, ethyl halides, ethyl alcohol,
ethylene oxide, etc; (v) For making poisonous mustard gas (War gas); (vi) For making ethylene-oxygen flame.


These are the acyclic hydrocarbons which contain carbon-carbon triple bond are called alkynes. General
formula is Cn H 2n−2 . Ex. Ethyne CH ≡ CH ; Propyne CH3 − C ≡ CH

(1) Structure UNHYBRIDISED
(i) Hybridisation in alkynes is sp . ORBITALS

(ii) Bond angle in alkynes is 180o . H sp sp sp sp H
(iii) Geometry of carbon is linear.
(iv) C − C triple bond length is 120 Å Acetylene

(v) C − H bond length is 108 Å
(vi) C − C triple bond energy is 190 Kcal / mol .

(vii) C − H bond energy is 102.38 Kcal / mol .

Aliphatic hydrocarbons

(2) Directive effect in disubstituted benzene
(i) If the directive effects of two substituents reinforce, then a single product is formed.


Example : Nitration NO2

; +NO2

NO2 (m) NO2 NO2

Thus, both (CH3, NO2) direct further substitution to the same position (Orth).

(ii) If the directing effect of two groups oppose each other strongly activating groups win over deactivating or

weakly activating group. The sequence of directing power is

− NH 2 > −OH > −OCH3 − > NHCOCH3 > −C6 H5 > CH 3 > meta directors

OH OH OH Directs OH OH
(Powerful Br
Example : Br2
; FeBr3


Directs Directs

(iii) There is normally little substitution when the two groups CH3 Too hindered position
are meta to each other. Aromatic rings with three adjacent substituents
are generally prepared by same other routes.


Toluene, methyl benzene or phenyl methane.

Toluene is the simplest homolouge of benzene. It was first obtained by dry distillation of tolubalsam and

hence named toluene. It is commercially known as tolual.

(1) Methods of preparation + CH3Cl AlCl3 CH3 HCl
(i) From benzene [Friedel-craft's reaction] : +

Note :  Alkyl halide employed may undergoBeanzneniesomeric change Toluene

C6 H6 + ClCH2CH2CH3 AlCl3 → C6 H5CH CH 3 + HCl CH3 + 2NaBr
n− Propyl chloride CH 3
Isopropyl benzene (65 −70%)

 Catalysts can be used in place of anhydrous AlCl3 are,

AlCl3 > SbCl3 > SnCl4 > BF3 > ZnCl2 > HgCl2

(ii) Wurtz fitting reaction : Br + 2Na + BrCH3 Ether

Bromobenzene Methyl bromide Toluene

(iii) Decarboxylation : C6 H4 CCOHO3 Na+ NaOH Soda lime → C6 H5CH 3 + Na 2 CO3
(o-,m- or p-)

Sodium toluate

(iv) From cresol : CH3 heat CH3 ZnO
OH +
+ Zn

o-Cresol Toluene

+ H2SO4
(v) From toluene sulphonic acid : + HOH

p-Toluene sulphonic acid

+ N2 + CH3CHO + HCl
(vi) From toluidine : NaNO2 C2H5OH

NH2 N2Cl Toluene

p-Toluidine p-Toluene diazonium chloride CH3
+ MgBr2
(vii) From grignard reagent : MgBr

+ CH3Br

Phenyl magnesium bromide Toluene

(viii) Commercial preparation

From coal tar : The main source of commercial production of toluene is the light oil fraction of coal-tar. The

light oil fraction is washed with conc. H 2SO4 to remove the bases, then with NaOH to remove acidic substances

and finally with water. It is subjected to fractional distillation. The vapours collected between 80 − 110°C is 90%
benzol which contains 70 − 80% benzene and 14 − 24% toluene. 90% benzol is again distilled and the portion

distilling between 108 − 110°C is collected. It is toluene. CH3 CH3
(ix) From n- heptane and methyl cyclohexane |
CH2 Cr2O3 / Al2O3

H2C CH2 500-550°C Toluene
150 atms

(2) Physical properties n-HCepHta2ne

(i) It is a colourless mobile liquid having characteristic aromatic odour.

(ii) It is lighter than water (sp. gr. 0.867 at 20°C).

(iii) It is insoluble in water but miscible with alcohol and ether in all proportions.

(iv) Its vapours are inflammable. It boils at 110°C and freezes at –96°C.

(v) It is a good solvent for many organic compounds.

(vi) It is a weak polar compound having dipole moment 0.4D.

(3) Chemical properties : Toluene shows the behavior of both

CH3 Side chain (Aliphatic)

Benzene ring (Aromatic)

(i) Electrophilic substitution reactions : Aromatic character (More reactive than benzene) due to electron

releasing nature of methyl group. CH3 CH3 CH3
+ E⊕ +


o-Derivative E

Note :  E+ may be chlorine, HNO3, H2SO4,CH3Cl . p-Derivative

(ii) Reactions of side chain CH3 CH2Cl CHCl2 CCl3

(a) Side chain halogenation : Cl2 Cl2 Cl2

Toluene Benzyl chloride Benzal chloride Benzo trichloride

Note :  Benzyl chloride on hydrolysis with aqueous caustic soda forms benzyl alcohol.

C6H5CH2Cl + NaOH → C6H5CH2OH + NaCl

(Phenyl methyl chloride)

 Benzal chloride on hydrolysis forms benzaldehyde.

C6 H5CHCl2 + 2NaOH → C6 H5CH(OH)2 + 2NaCl
(Benzylide chloride)
C6 H5CH↓O+ H 2O

 Benzo trichloride on hydrolysis forms benzoic acid.

C6H5CCl3 + 3NaOH → C6H5C(OH)3 + 3NaCl
(Benzylidyne chloride)

(b) Oxidation : CH3 COOH
• With hot acidic KMnO4 : + H2O
KMnO4 / H+

Toluene Benzoic acid CH3 CHO
+ H2O
• With acidic manganese or chromyl chloride (Etards reaction) : + 2[O] CrO2C2

Toluene Benzaldehyde

Note :  All alkyl benzenes on oxidation with hot acidic KMnO4 or Na2Cr2O7 form benzoic acid. The

length of the side chain does not matter.


(c) Hydrogenation : + 3H2 Na / liquid NH3 – C2H5OH
Birch reduction

Alkyl benzene Alkyl cyclohexane


| |

+ 3H2 Ni H2C CH2
HC CH 200°C

Methyl benzene CH2


(d) Combustion : C6H5CH3 + 9O2 →7CO2 + 4H2O

(e) Ozonolysis : H OO

HC C +3O3 → C CH O +2 + 3H2O2
C Methyl glyoxal Glyoxal


(4) Uses Toluene Triozonide

(i) In the manufacture of benzyl chloride, benzal chloride, benzyl alcohol, benzaldehyde, benzoic acid,
saccharin, etc.

(ii) In the manufacture of trinitrotoluene (TNT), a highly explosive substance.

(iii) As an industrial solvent and in drycleaning.

(iv) As a petrol substitute.

(v) In the manufacture of certain dyes and drugs.

T.N.T. (Tri-nitro toluene)

Preparation : H2SO4 O2N
+ 3FHumNinOg 3 3H2O

Toluene NO2

Properties : It is pale yellow crystalline solid (M.P. = 81°C).

Uses : • It is used as an explosive in shells, bombs and torpedoes under the name trotyl.

• When mixed with 80% ammonium nitrate it forms the explosive amatol.

• TNT is also used as a mixture of aluminium nitrate, alumina and charcoal under the name ammonal.

CH3 T.N.B. (Tri-nitro benzene)
Preparation : CH3 COOH
O2N NO2 O NK2Cr2O7 2 NO2 Soda O2N
H2SO4 lime

Toluene NO2 NO2 NO2


Properties and uses: It is colourless solid (M.P. = 122°C). It is more explosive than T.N.T. and used for

making explosive.

Xylenes (Dimethyl benzene) C6H4(CH3)2.

The molecular formula, C8H10 represents four isomers.

CH3 CH3 CH3 C2H5
Ethyl benzene
o-Xylene CH3 CH3

m-Xylene p-Xylene

These are produced along with benzene, toluene and ethylbenzene when aromatisation of C6 − C8 fraction of
petroleum naphtha is done. The xylenes are isolated from the resulting mixtrue (BTX) by fractional distillation.

These can be prepared by Wurtz – Fittig reaction. A mixture of bromotoluene and methylbromide is treated
with sodium in dry ethereal solution to form the desired xylene.

Br CH3
+ 2Na + BrCH3 + 2Na + BrCH3 + 2NaBr
+ 2NaBr ; Br CH3
o-Xylene m-Bromotoluene m-Xylene


+ 2Na + BrCH3 + 2NaBr

Br p-XCylHen3e


• These can also be obtained by Friedel – craft's synthesis,

+ CH3Cl CH3 +

o-Xylene CH3

• m-Xylene can be obtained from mesitylene. p-Xylene

CH3 CH3 Soda lime CH3
H3C CH3 [O] CH3 + CO2

Mesitylene Mesitylenic acid m-Xylene

Xylenes are colourless liquids having characteristic odour. The boiling points of three isomers are,

o-Xylene = 144°C; m-Xylene = 139°C; p-Xylene = 138°C.

Xylenes undergo electrophilic substitution reactions in the same manner as toluene. Upon oxidation with

KMnO4 or K2Cr2O7 , Xylenes form corresponding dicarboxylic acids. COOH


Phthalic acid Isophthalic acid COOH

Terephthalic acid

Xylenes are used in the manufacture of lacquers and as solvent for rubber. o-Xylene is used for the

manufacture of phthalic anhydride.

Ethyl benzene (C6H5C2H5).

It can be prepared by the following reactions,
(1) By Wurtz-Fittig reaction : C6H5Br + 2Na + BrC2H5 → C6H5C2H5 + 2NaBr
(2) By Friedel-craft's reaction : C6 H5 H + BrC2 H5 AlCl3 → C6 H5C2 H5 + HBr

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