The document provides comprehensive study material on alkanes, including their structure, properties, and methods of preparation. It discusses various reactions such as reduction of alkyl halides and hydrogenation of alkenes, as well as techniques like the Wurtz reaction and Corey-House synthesis. The document emphasizes the importance of reaction mechanisms and characteristics of alkanes, alongside their physical properties and boiling points related to molecular structure.

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Key Features
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 NCERT & Advanced Level Solved Examples

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Chapter – 15

Alkanes are saturated hydrocarbons which can be represented by the general formula CnH2n + 2.
They are also known as paraffins after their poor affinity towards common reagents e.g., acids, bases,
oxidising and reducing agents. The molecular formula of alkanes suggests that each individual member
differs from its neighbour by CH2 group. Such a series of compounds is known
as homologous series and the individual members being known as homologues.
1.1 METHODS INVOLVING NO CHANGE IN THE CARBON SKELETON
1.1.1 Reduction of alkyl halides, RX where X = F, Cl, Br or I:
(Substitution of halogen by hydrogen)
This may be done in three different ways
(i) Reduction by dissolving metals: Reduction by dissolving metals e.g. zinc and acetic or
hydrochloric acid, zinc and sodium hydroxide, zinccopper couple and ethanol, etc. In this
reaction, earlier ‘nascent’ hydrogen was considered to be the reducing agent. Now it is believed
that there is an electrontransfer from the metal to the substrate leading to the formation of
carbanion, which is followed by the abstraction of a proton from the solvent. Thus, reduction with
a zincethanol couple may be formulated as
Zn  Zn2+
+ 2e
RX + e
 X
+ R. 


e
R:
R:
+ C2H5OH  RH + 
OC2H5
(ii) Reduction by reducing agents like LiAlH4, NaBH4 etc.: Primary and secondary alkyl halides are
readily reduced to alkanes by lithium aluminium hydride (LiAlH4) while reduction of tertiary
halides with LiAlH4 gives mainly alkenes. On the other hand, sodium borohydride (NaBH4)
reduces secondary and tertiary halides, but not primary, whereas triphenyltin hydride (Ph3SnH)
reduces all three types of alkyl halide. So each reducing agent is specific in its action.
(a) 4RX + LiAlH4  4RH + LiX + AlX3(X  F)
or RX + H:()
 RH + X
(H
comes from LiAlH4)
(b) RX + (nC4H9)3SnH  RH + (nC4H9)3SnX
(iii)Using organometallic compounds like Grignard Reagent: Alkyl halides react with either Mg or
Li in dry ether to give organometallic compounds having a basic carbanionic site.
RX + 2Li 

 

ether
dry
R
Li+
+ LiX then R
Li+
+ H2O  RH + LiOH
RX + Mg 

 

ether
dry
R
Mg2+
X–
then, RMgX + H2O  RH + MgX(OH)
Grignard Reagent strong acid weak acid
The net effect is replacement of MgX by H. This reaction can be afforded with any compound that
is more acidic than alkane e.g., alcohols, NH3, terminal alkynes etc. Thus the net effect of the
METHODS PREPARATION OF ALKANES
1
HYDROCARBON
ALKANES

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reaction is the displacement of a weak acid from its salt by a strong acid. Using this reaction, we
determine the number of active hydrogens present in a given compound (the one which reacts with
Grignard reagent). This quantitative estimation of number of active hydrogens is called
Zerrwittnoff’s method. For example,
3RMgX + HCCCH(OH)CO2H  3RH + XMgCCCHC
OMgX
O
OMgX
Three moles of alkane formed shows that the compound contains three active (acidic) hydrogens.
1.1.2. Hydrogenation of alkenes in the presence of Pd, Pt or Ni:
This addition is an example of heterogeneous catalysis involving synaddition.
CH3C=CH2 + H2
Pt
CH3
CH3CH  CH3
CH3
The stereospecificity of the reaction is that the addition of hydrogen to the double bond occurs in
syn fashion without disturbing the configuration at the chiral carbon. The mechanism of this
reaction will be dealt in the topic “alkenes”. For example,
CH3
CH=CH2
H OH
D2/Ni
CH3
CH2D
H OH
H D
+
CH3
CH2D
H OH
D H
Diastereomers
The addition of both the deuterium atoms occurs from the same side. In some molecules, the attack
is from the bottom side and in other molecules, D2 attacks from the top side leading to the
formation of 2 isomers called diastereomers.
Raney Nickel is more reactive than the Nickel catalyst. It consists of an alloy containing equal
amounts of Ni and Al digested with NaOH, residual part containing mainly nickel is washed, dried
and stored under ethanol.
1.1.3. Reduction of alcohols, carbonyl compounds, acids and acid derivatives:
Alcohols, aldehydes, Ketones, carboxylic acids and their derivatives like acid halides and acid
amides can be reduced by HI and red phosphorous to alkanes.
RCH2OH + 2HI 
 

P
red
RCH3 + I2 + H2O
RCHO + 4HI 
 

P
red
RCH3 + 2I2 + H2O
RCOR + 4HI 
 

P
red
RCH2R + 2I2 + H2O
RCOOH + 6HI 
 

P
red
RCH3 + 3I2 + 2H2O
RCOCl + 6HI 
 

P
red
RCH3 + 3I2 + + HCl + H2O
RCONH2 + 6HI 
 

P
red
RCH3 + 3I2 + NH4OH
Carbonyl compounds can also be reduced to alkanes by (i) ZnHg amalgam and HCl
(Clemmensen Reduction) and (ii) H2NNH2 and KOH (Wolff Kishner Reduction)
RCHO + Zn/Hg + HCl  RCH3 + H2O
RCOR + NH2NH2 + KOH  RCH2R + N2 + H2O
(The mechanism of these reactions will be taken up in the chapter of aldehydes and ketones).

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1.2 METHODS INVOLVING CHANGE IN THE CARBON SKELETON
1.2.1 Methods in which number of carbon atoms increases w.r.t. starting compound:
(a) Wurtz Reaction:
An ethereal solution of an alkyl halide (preferably the bromide or iodide) is treated with sodium,
when alkane is obtained. For example,
R1
X + R2
X + 2Na  R1
R2
+ 2NaX
In this reaction, two R groups are coupled by reacting RBr, RCl or RI with Na or K.
The yields of the product are best for 1° alkyl halides (60%) and least for 3° alkyl halides (10%).
Looking further in the above reaction, it was found that in addition to the desired alkane R1
R2
,
there will also be present the alkanes R1
R1
and R2
R2
. Unsaturated hydrocarbons are also obtained.
Obviously, then, the best yield of an alkane will be obtained when R1
& R2
are same, i.e., when the alkane
contains an even number of carbon atoms and is symmetrical. It has been found that the Wurtz reaction
gives good yields only for ‘even carbon’ alkanes of high molecular weight, and that the reaction generally
fails with tertiary alkyl halides.
[Note: Metals other than sodium, which can be employed in Wurtz reaction are
Ag and Cu in finally divided state]
The reaction probably involves the formation of carbanions as intermediate.
NaBr
Na
H
C
Na
Br
H
C 


 

5
2
5
2 2
NaBr
H
C
H
C
Br
H
C
Na
H
C N
S







5
2
5
2
2
5
2
5
2
The support for such a mechanism involving carbanions is provided by the observation that
optically active halides demonstrate inversion of configuration at the carbon atom undergoing nucleophilic
attack. The carbanion can also act as a base and promote elimination.
NaBr
CH
CH
CH
CH
Br
CH
CH
H
H
C
CH
Na 









2
2
3
3
2
2
2
3
This is often observed as a side reaction to the normal Wurtz reaction proceeding by SN2
mechanism.
(b) CoreyHouse Synthesis:
A superior method for coupling is the CoreyHouse Synthesis which could be employed for
obtaining alkanes containing odd number of carbon atoms (unsymmetrical alkanes). An alkyl halide (RX)
is first converted into alkyl lithium by treating with lithium. The alkyl lithium is then reacted with cuprous
halide to get lithium dialkyl cuprate. The complex is then treated with another alkyl halides (RX), which
must be preferably primary. The reaction follows SN2 mechanism. With secondary alkyl halides the
reaction leads, to partly substitution forming alkane and partly elimination forming alkene, with tertiary
alkyl halide only elimination takes place.
For example,
RX + 2Li  R
Li+
+ LiX
2R
Li+
+ CuX  (R)2CuLi + LiX
R2CuLi + 2RX  2RR + LiX + CuX
Illustration
Question: Prepare 2methylbutane from chloroethane and 2chloropropane using Corey
House synthesis.

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Solution: 3
2
|
3
Cl
CH
CH
2
|
3
Cu
.
2
Li
.
1
|
3 H
C
HCH
C
CH
CuLi
)
H
C
CH
(
HCl
C
CH
3
3
3 H
C
CH
CH
2
3



 


 

I
2Chloropropane Lithiumdiisopropylcuprate
Illustration
Question: Predict the products in the following reactions:
(i) Cyclopropane + H2
C
120
Ni


 ?
(ii) C = C
CH3
D
CH3
D
+ H2 

Pt
?
Solution: (i) Propane (ii) CC
H3C
D
CH3
D
H H
(Meso)
(c) Kolbe’s electrolytic method:
A concentrated solution of the sodium or potassium salt of a carboxylic acid or a mixture of
carboxylic acids is electrolysed. For example,
R1
CO2K + R2
CO2K + 2H2O  R1
R2
+ 2CO2 + H2 + 2KOH
If R1
and R2
are different, then hydrocarbons R1
R1
and R2
R2
are also obtained along with R1
R2
.
Earlier several mechanisms have been proposed for the Kolbe’s reaction. The freeradical theory is the one
now favoured, having strong evidences in support of it. For example, when sodium propionate is
electrolysed, nbutane, ethylene and ethyl propionate are obtained.
C2H5CO2Na  
2
5
2 CO
H
C + Na+
At anode, the propionate ion discharges to form a free radical.

2
5
2 CO
H
C  
2
5
2 CO
H
C + e
This propionate free radical then breaks up into the ethyl free radical and carbon dioxide.

2
5
2 CO
H
C  
5
2H
C + CO2
Then, ethyl radicals undergo chain termination by recombination forming butane. They may also
undergo disproportionation reaction forming ethane and ethene as by products. Another possible by
product is ethyl propionate formed by recombination of ethyl radical and propionate radical.
(i) 
5
2H
C
2  C4H10
(ii) 
5
2H
C + 
5
2H
C  C2H6 + C2H4
(iii) 
5
2H
C + 
2
5
2 CO
H
C  C2H5CO2C2H5
Thus at anode, gases evolved are CO2, ethane, ethene and butane. At cathode,
H+
accepts an electron and is converted to Hatom. Two of the Hatoms combine to form H2 gas.
H+
+ e  H
H + H  H2
1.2.2 Methods in which number of carbon atoms decreases w.r.t. starting compound:
Decarboxylation of carboxylate salts:
By heating a mixture of the sodium salt of a carboxylic acid and sodalime, alkanes can be
obtained.
RCO2Na + NaOH(CaO) 


RH + Na2CO3

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This process of eliminating CO2 from a carboxylic acid is known as decarboxylation.
This reaction can be employed for decreasing the length of carbon chain i.e. to descend a homologous
series. This decarboxylation reaction probably involves following mechanistic steps.
OCR H
R
R
CO
fast
H
slow











2
O
(I) The first four alkanes (methane to butane) are colourless gases, the next thirteen (pentane to
heptadecane) are colourless liquids and those containing 18 carbon atoms or more are solids at
ordinary temperatures.
(II) Their boiling points show gradual rise as the carbon content increases. In general,
the boiling point difference between two successive members of the homologous series (except for
the first few members) is about 2030°C. Among the isomeric alkanes,
the straight chain (i.e. normal) isomer has a higher boiling point than the branched chain isomer.
The greater the branching of the chain, the lower the boiling point. For example,
Alkane CH3CH2CH2CH3 CH3(CH2)3CH3 CH3(CH2)4CH3
(nButane) (nPentane) (nHexane)
(b.p. in °C) 0 36 69
Alkane (CH3)2CHCH3 (CH3)2CHCH2CH3 (CH3)4C
(Isobutane) (Isopentane) (Neopentane)
(b.p. in °C) 11.5 28 9.5
In fact, the lowering of boiling point with the branching of the carbon chain is a feature
characteristic of all the families of organic compounds.
The vander Waal’s forces which hold nonpolar molecules are weak and have a very short range.
Therefore, within a family of compounds the strength of intermolecular forces would be directly
proportional to the size (or the surface area) of the molecule. In other words, larger the molecule,
the stronger would be the intermolecular forces. The process of boiling requires overcoming these
intermolecular forces of a liquid and a solid. As the molecules become larger, the intermolecular
forces increase and the boiling points should rise with increase in the number of carbon atoms. As
the branching increases in a molecule, its shape approaches that of a sphere and there is a reduction
in surface area. This renders the intermolecular forces weaker and they are overcome at relatively
lower temperature. Therefore, a branchedchain isomer should boil at a temperature lower than
that of a straightchain isomer.
(III)Their melting points also show a rise with the increasing number of carbon atoms, but the rise is
not as regular as in the case of boiling points. For example,
nAlkane C4H10 C5H12 C6H14 C7H16 C8H18
m.p. °C 135 130 95 90 57
It is, however, significant that as we move from an alkane having an odd number of carbon atoms
to the next higher alkane, the rise in melting point is much higher than that when we move up from
an alkane with an even number of carbon atoms.
The intermolecular forces in a crystal depend not only on the size of the molecules but also on how
they are packed into a crystal. During melting, these intermolecular forces have to be overcome.
Since breaking of crystal structure is a more complicated process, it is understandable that the rise
GENERAL PHYSICAL PROPERTIES OF ALKANES
2

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in melting point with increasing molecular weight is not as regular as in the case of boiling points.
The structure/geometry of the alkane is of considerable importance.
(IV) Alkanes are made up of carbon and hydrogen atoms only. Since these two elements have almost
similar electronegativities, alkanes are nonpolar. Therefore, nonpolar alkanes are soluble in
nonpolar solvents like carbon tetrachloride, benzene, etc. but insoluble in polar solvents like
water, alcohol, etc.
(V) The densities of alkanes show a definite rise with increasing molecular weight, but they reach a
limiting constant value of about 0.8 g/ml with nhexadecane (C16H34). Thus, alkanes are always
lighter than water.
The alkanes are generally stable towards common reagents at room temperature.
The minimum energy required to cause homolytic cleavage of almost nonpolar CH and CC bonds is
not available at room temperature. However, alkanes undergo substitution reactions at high temperature.
As the temperature is increased, the CH and CC bonds break forming free radicals as intermediates. The
reactivity of alkanes is decided on the basis of stability of the free radicals. As the stability of free radical
increases, energy required to produce them i.e, energy of activation decreases and hence the rate of
reaction increases.
Benzyl radical )
H
C
H
C
( 2
5
6

 and allyl radical (CH2=CHCH2) are stabilised by resonance.
CH2 CH2 CH2 CH2 CH2
CH2=CHCH2 CH2CH=CH2
Benzyl radical has five resonating structures, while allyl radical has two resonating structures.
Therefore, benzyl radical is slightly more stable than allyl radical and hence toluene is more reactive than
propene towards substitution reactions. Alkyl radicals, on the other hand, are stabilised by
hyperconjugation. Their stability may be compared by the number of hyperconjugation structures. In the
reaction given below a primary radical is stabilised by two hydrogen atoms, a secondary radical is
stabilised by six hydrogen atoms and a tertiary radical is stabilised by nine hydrogen atoms.
CH3CH2CH3 CH3CHCH3 + H
secondary or 2° radical
CH3CHCH3 CH3CCH3 + H

 
CH3


CH3CH2CH3 CH3CH2CH2 + H
npropane primary or 1° radical

CH3
tertiary or 3° radical
isobutane
Since resonance superseedes hyperconjugation, the stability order of various free radicals is as
given below:
C6H5CH2 > CH2=CHCH2 > (CH3)3C > (CH3)2CH > CH3CH2 > CH3 > CH2=CH
3.1 HALOGENATION
Chlorination may be brought about by photo irradiation, heat or catalysts and the extent of
chlorination depends largely on the amount of chlorine used. A mixture of all possible isomeric
GENERAL CHEMICAL PROPERTIES OF ALKANES
3

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monochlorides is obtained, but the isomers are formed in unequal amounts, due to difference in reactivity
of primary, secondary; and tertiary hydrogen atoms.
The order of ease of substitution is
Tertiary Hydrogen > Secondary Hydrogen > Primary Hydrogen
Chlorination of isobutane at 25°C gives a mixture of two isomeric monochlorides.
CH3CHCH2Cl
CH3
and CH3CCH3
CH3
Cl
(64%) (36%)
The tertiary hydrogen is replaced about 5 times as fast as primary hydrogen.
Bromination is similar to chlorination, but not so vigorous. Iodination is reversible, but it may be
carried out in the presence of an oxidising agent such as HIO3, HNO3 etc., which destroys the hydrogen
iodide as it is formed and so drives the reaction to the right.
CH4 + I2  CH3I + HI
5HI + HIO3 3I2 + 3H2O
Iodides are more conveniently prepared by treating the chloro or bromo derivative with sodium
iodide in methanol or acetone solution. For example,
RCl + NaI 

 

acetone
RI + NaCl
This reaction is possible because sodium iodide is soluble in methanol or acetone, whereas sodium
chloride and sodium bromide are not. This reaction of halide exchange is known as ConantFinkelstein
reaction. It will be dealt in detail in chapter of Alkyl Halides.
Direct fluorination is usually explosive. So special conditions are necessary for the preparation of
the fluorine derivatives of the alkanes.
RH + X2



 

or
ht
lg
li
uv
RX + HX
Reactivity of X2: F2 > Cl2 > Br2 > I2
The mechanism of chlorination of methane is as follows.
Chain initiation step:
ClCl



 

or
ht
lg
li
uv
2Cl
; H = + 243 kJ mol1
The required enthalpy comes from ultraviolet (uv) light or heat supplied.
Chain propagation step:
(i) H3CH + Cl



 

or
ht
lg
li
uv
H3C
+ HCl ; H = 4 kJ mol1
(ii) H3C
+ ClCl  H3CCl + Cl
; H = 96 kJ mol1
The sum of the two chain propagation steps in the overall reaction is
CH4 + Cl2  CH3Cl + HCl ; H = 100 kJ mol1
In propagation steps, the same free radical intermediates, here Cl
and H3C
, being formed and
consumed.
Chain termination step:
Chains terminate on those rare occasions when any two freeradical intermediates collide to form
a covalent bond.
Cl
+ Cl
 Cl2
H3C
+ Cl
 CH3Cl
H3C
+ 
CH3  H3CCH3
Radical inhibitors stop chain propagation by reacting with free radical intermediates.
For example,
H3C
+ O O  CH3O O
 
 
 
 
 
 
 
 
The potential energy curve for the halogenation (chlorination) of alkane is shown as

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Potential
energy
Eact
RH + Cl
R
+ Cl2
R Cl + Cl
Progress of reaction
In more complex alkanes, the abstraction of each different kind of hydrogen atom gives a different
isomeric product. Three factors determine the relative yields of isomeric product.
1. Probability Factor: This factor is based on the number of each kind of hydrogen atoms in the
alkane molecule. For example, in CH3CH2CH2CH3 there are six equivalent 1° H and four
equivalent 2° H. The probability of abstracting a 1° H to 2° H is 6 to 4, or 3 to 2.
2. Reactivity of H
: The order of reactivity of hydrogen atoms is 3° > 2° > 1°.
3. Reactivity of X
: The more reactive Cl
is less selective and more influenced by the probability
factor. The less reactive Br
is more selective and less influenced by the probability factor, as
summarized by the ReactivitySelectivity Principle. If the attacking species is more reactive, it
will be less selective and the yields will be determined by the probability factor as well as
reactivity of hydrogen atoms while if the species attacking is less reactive and more selective, the
yield of the product is governed exclusively by reactivity of hydrogen atoms.
CH3CH2CH2CH3
C
25
h
/
Cl2



 


CH3CH2CH2CH2Cl + CH3CH2CH(Cl)CH3
nbutane (28%) (72%)
(CH3)2CHCH3
C
25
h
/
Cl2



 


(CH3)2CHCH2Cl + (CH3)3CCl
isobutane (64%) (36%)
CH3CH2CH2CH3
C
127
h
/
Br2



 


CH3CH2CH2CH2Br +CH3CH2CH(Br)CH3
nbutane (2%) (98%)
(CH3)2CHCH3
C
127
h
/
Br2



 


(CH3)2CHCH2Br + (CH3)3CBr
isobutane traces (over 99%)
In the chlorination of isobutane abstraction of one of the nine primary hydrogens leads to the
formation of isobutyl chlorides, whereas abstraction of a single tertiary hydrogen leads to the formation of
tertbutyl chloride. The probability favours formation of isobutyl chloride by the ratio of 9 : 1. But the
experimental results show the ratio roughly to be 2 : 1 or 9 : 4.5. Evidently, about
4.5 times as many collisions with the tertiary hydrogen are successful as collisions with the primary
hydrogens. The Eact is less for abstraction of a tertiary hydrogen than for the abstraction of a primary
hydrogen.
The rate of abstraction of hydrogen atoms is always found to follow the sequence
3° > 2° > 1°. At room temperature (25°C), the relative rates in chlorination are 5.0 : 3.8 : 1.0 respectively
for 3°, 2° and 1° hydrogen atoms. Using these values, we can predict quite well the ratio of isomeric
chlorination products from a given alkane. For example,

10. All right copy reserved. No part of the material can be produced without prior permission
CH3CH2CH2CH3
C
25
,
light
Cl2


 
 CH3CH2CH2CH2Cl + CH3CH2CHClCH3
%
72
%
28
6
.
7
3
8
.
3
0
.
1
4
6
2
1
2
1
sec
to
equivalent
H
of
reactivity
H
of
reactivity
H
of
number
H
of
number
chloride
butyl
chloride
butyl
n











Inspite of these difference in reactivity, chlorination rarely yields a great excess of any single
isomer. In most cases, both the products are formed in considerable amounts.
The same sequence of reactivity, 3° > 2° > 1°, is found in bromination, but with enormously larger
reactivity ratios. At 127°C the relative rates per hydrogen atom in bromination are
1600 : 82 : 1 respectively for 3°, 2° and 1° hydrogen atoms. Here, differences in reactivity are so marked
that it outweighs probability factor. Hence bromination almost exclusively gives selective product.
In bromination of isobutane at 127°C,
H
3
of
reactivity
H
1
of
reactivity
H
3
of
number
H
1
of
number
bromide
butyl
tert
bromide
isobutyl







%
5
.
99
%
5
.
0
to
equivalent
1600
9
1600
1
1
9



Hence, tertbutyl bromide happens to be the exclusive product (over 99%) with traces of isobutyl
bromide.
The reason for the higher selectively in bromination as compared to chlorination is due to the
following explanation.
According to the general principle, for comparable reactions, the more endothermic (or less
exothermic) reaction has a transition state (TS), which more closely resembles the intermediate and may
more closely resemble the ground state (reactants). Since attack by Br
on an alkane is more endothermic
than attack by Cl
, its TS shows more CH bond breaking and more HBr bond formation. Any
stabilization in the intermediate radical also occurs in the corresponding TS. Therefore, a TS leading to a
3° R
, has a lower enthalpy than one leading to a 2°R
, which in turn has a lower enthalpy than one leading
to a 1°R
, the relative rates of H abstraction by Br
are 3° > 2° > 1°. The TS for Habstraction by Cl
has
less CH bond breaking and less HCl bond formation. The nature of the incipient radical has less effect
on the enthalpy of the TS and on the rate of its formation. Hence there is less difference in the rate of
formation of the three kinds of R
s. In the attack by the comparatively unreactive bromine atom, the
transition state is reached late in the reaction process, after the alkyl group has developed considerable
radical character.
In the attack by the highly reactive chlorine atom, the transition state is reached early, when the alkyl group
has gained very little radical character. Thus bromination is more selective than chlorination.
RH + Br
R…….H…….Br


R
+ H  Br
low reactivity,
high selectivity
Transition state
reached late, much
radical character
RH + Cl
R…….H…….Cl


R
+ H  Cl
Transition state
reached early, little
radical character
Illustration
Question: How many monochlorinated products are obtained by the chlorination of
isohexane and what is the percentage of each assuming the reactivity ratio of
3°H : 2°H : 1°H = 5 : 3.8 : 1.

11. All right copy reserved. No part of the material can be produced without prior permission
Solution: Isopentane on chlorination in presence of sun light gives five different monochlorinated
products.
CH3CHCH2CH2CH3 + Cl2
h
CH3
CH2ClCHCH2CH2CH3
CH3 (A)
+
CH3CClCH2CH2CH3
CH3 (B)
+ CH3CHCHClCH2CH3
+
CH3 (C)
+
CH3CHCH2CHClCH3
CH3
(D)
+ CH3CHCH3CH2CH2Cl
CH3
(E)
Product Reactivity factor  Probability factor = number of parts Percentage
(A) 1  6 = 6 20.6%
(B) 5  1 = 5 17.1%
(C) 3.8  2 = 7.6 26.0%
(D) 3.8  2 = 7.6 26.0%
(E) 1  3 = 3 10.3%
Total number of parts = 29.2
3.2 OXIDATION
All alkanes readily burn in excess of air or oxygen to form carbon dioxide and water.
CnH2n+2 + )
g
(
O
2
)
1
n
3
(
2

 nCO2(g) + )
(
O
H
2
)
2
n
2
(
2 l

On the other hand, controlled oxidation under various conditions, leads to different products.
Extensive oxidation gives a mixture of acids consisting of the complete range of C1 to Cn carbon atoms.
Less extensive oxidation gives a mixture of products in which no chain fission has occurred. Under
moderate conditions mixed ketones are the major products and oxidation in the presence of boric acid
produces a mixture of secondary alcohols. The oxidation of alkanes in the vapour state occurs via free
radicals, e.g. alkyl (R
), alkylperoxy (ROO
) and alkoxy (RO
). Oxidising reagents such as potassium
permanganate readily oxidise a tertiary hydrogen atom to a hydroxyl group. For example, isobutane is
oxidised to tbutanol.
(CH3)3CH + [O] 

 
 4
KMnO
(CH3)3COH
3.3 SULPHONATION
It is the process of replacing hydrogen atom by a sulphonic acid group, SO3H. Sulphonation of a
normal alkane from hexane onwards may be carried out by treating the alkane with oleum (fuming
sulphuric acid). The order of ease of replacement of H atoms in tertiary compounds is very much easier
than secondary and in secondary compounds replacement of Hatoms by sulphonic acid group is easier
than primary. Replacement of a primary; hydrogen atom in sulphonation is very slow indeed. Isobutane,
which contains a tertiary hydrogen atom, is readily sulphonated to give tbutyl sulphonic acid.
(CH3)3CH + H2SO4/SO3  (CH3)3CSO3H + H2SO4
3.4 NITRATION
Under certain conditions alkanes react with nitric acid, when a hydrogen atom will be replaced by
a nitrogroup, NO2. This process is known as nitration. Nitration of the alkanes may be carried out in the
vapour phase between 150° and 475°C, when a complex mixture of mononitroalkanes is obtained. The
mixture consists of all the possible mononitroderivatives and the nitrocompounds formed by every
possibility of chain fission of the alkane. For example, propane gives a mixture of 1nitropropane, 2
nitropropane, nitroethane and nitromethane.

12. All right copy reserved. No part of the material can be produced without prior permission
CH3CH2CH3
HNO3
400°C
CH3CH2CH2NO2 + CH3CHCH3 + C2H5NO2 + CH3NO2
NO2
3.5 ISOMERISATION
It is a process by which nalkane is converted into a branched alkane containing a methyl group in
the side chain by heating the nalkane with AlCl3HCl at 300°C.
For example,
CH3  CH  CH3
Isobutane
CH3
CH3CH2CH2CH3
nbutane
AlCl3HCl
300°C
CH3  CH  CHCH3
2, 3dimethyl butane
CH3
CH3CHCH2CH2CH3
Isohexane
CH3
AlCl3HCl
300°C
CH3
The isomerisation is believed to be an ionic chain reaction initiated by a carbonium ion followed
by 1, 2 shift of hydride or methyl group.
3.6 AROMATISATION
Aromatisation of nalkanes containing six or more carbon atoms into benzene and its homologues
takes place at high temperature (600°C) in presence Cr2O3Al2O3 as a catalyst.
Cr2O3Al2O3
600°C
nhexane
3H2
cyclohexane benzene
Cr2O3Al2O3
600°C
noctane
3H2
ethylcyclohexane ethyl benzene
3.7 PYROLYSIS OR CRACKING
The thermal decomposition of organic compounds is known as pyrolysis and the process of
cleavage of complex hydrocarbons into simpler molecules by the application of heat is known as cracking,
i.e. the thermal decomposition is called cracking, but when induced by catalyst it is called catalytic
cracking. Methane is the most stable hydrocarbon because of a higher CH bond energy through red hot
(500600°C) tube in absence of air we get a mixture of lower compounds.
The products depend on the following factors:
(a) structure of alkane, (b) extent of temperature and pressure, (c) absence or presence of catalysts
like SiO2Al2O3, etc.
C2H6 


 
  C
600
500 o
C2H4 + CH4 + H2
Cracking involves either breaking of CC or CH bond or both.
500600°C
CH3CH2CH3 CH3CH=CH2 + H2 (CH fission)
500600°C
CH2=CH2 + CH4 (CC fission)

13. All right copy reserved. No part of the material can be produced without prior permission

CH3CH2CH2CH3 CH3CH=CH2 + CH4 + H2
CH3CH2CH=CH2 + H2
CH3CH3 + CH3CH3
Alkenes are unsaturated hydrocarbons having one double bond. They are represented by the
general formula CnH2n. They are also known as olefins since ethene, the first member of the homologous
series forms oily liquid when treated with chlorine.
1.1 DEHYDROHALOGENATION OF ALKYL HALIDES
Alkyl halides when treated with a strong base like hot alcoholic solution of KOH undergo
elimination of hydrogen halide leading to the formation of alkene. The yield of alkene depends on the
nature of alkyl halide used. It is fair with primary and very good with secondary and tertiary alkyl halides.
For example tertiary butyl bromide when heated with alcoholic KOH results in the formation of isobutene
with the elimination of hydrogen halide.
CH3CBr + alc. KOH heat
CH3
CH3
CH3C=CH2 + KBr + H2O
CH3
In this type of elimination reaction the two leaving groups are lost from the adjacent carbon atoms.
One of the leaving groups is the halogen atom and the Catom from which halogen is lost is usually
designated as 1 or ()carbon atom. Hydrogen, the other leaving group is lost from the neighbouring
carbon which is designated as 2 or () carbon atom and the reactions are referred to
1, 2elimination or elimination ( is commonly omitted). If the alkyl halide contains only one 
carbon atom, only one alkene is formed as in the abovesited example. However, if the alkyl halide has
two or more carbon atoms, two or more alkenes are possible. According to
METHODS PREPARATION OF ALKENES
1
ALKANES

14. All right copy reserved. No part of the material can be produced without prior permission
Saytzeff rule, dehydrohalogenation of alkyl halides leads to the formation of that alkene as the major
product which has maximum number of alkyl groups attached to C = C . It is the stability of the alkene
that decides the major product. For example,
CH3CHCHCH3
alc. KOH
CH3
Br
CH3CH=CCH3 +
  
CH3
(major)
CH2=CHCHCH3
CH3
(minor)
alc. KOH
CH3
Cl
CH3
(major)
+
CH2
(minor)
However, if the size of base is increased, it finds it relatively easier to abstract proton from a less
substituted carbon atom than from more substituted carbon atom of alkyl halide. Therefore, less
stable alkene becomes the major product. This is known as Hoffmann’s rule.
For example,
Et3N
Br CH3
(a bulky base)
CH2
+
CH3
(major) (minor)
KOH dissolved in tertiary butanol gives potassium tertiary butoxide, which is another bulky base
used in Hoffmann elimination reaction.
KOH+ (CH3)3COH (CH3)3COK+
+ H2O
CH3CH2CI
(CH3)3CO
CH3
CH3
CH3CH=C +
CH3
in (CH3)3COH
CH3
CH3CH2C=CH2
CH3
(minor) (major)
In place of haloalkanes, sulphonyl derivatives of alkanes can also be used in the base catalysed
elimination reaction for the preparation of alkenes e.g.,
CH3CHCH2OS
alc. KOH
O
CH3 CH3C=CH2 + K OS CH3
O
O
O
CH3
isobutyltosylate isobutene
CH3
pot. tosylate
+ –
The groups commonly used in this class of compounds include:
OS CH3
O
O
(OTs)
OSCH3
O
O
(OMe)
OS
O
O
(OTf)
CF3 OS
O
O
(OBs)
Br
The mechanism of 1, 2 elimination involves simultaneous removal of a proton from the carbon
atoms by the base and the other leaving group (L) (halogen or sulphonyl group as the case may be) from
the carbon atom. The two leaving groups must align themselves at 180°C to each other and in the same
plane before they are lost.

15. All right copy reserved. No part of the material can be produced without prior permission
C C
B:  H
L
C C
B H
L
C C + BH + L
(+) ()
. .
It is a single step bimolecular elimination reaction which passes through a transition state. The rate
determining step involves cleavage of the CH and CL bonds. The energy required to break these bonds
comes from the energy released due to the formation of B:H bond and bond. The cleavage of CL
bond in the rate determining step implies that the reactivity of alkyl halides should follow the order: RI >
RBr > RCl to match, the bond dissociation energy of carbonhalogen bond. This has indeed been found
to be so. The above mechanism helps us to predict the stereochemistry of the alkene wherever possible.
For example 2bromo3phenyl butane contains two dissimilar chiral carbon atoms and hence exists in
four optically active isomers (2 pair of enantiomers) as shown below.
PhCH
Me
HCBr
Me
HCPh
Me
BrCH
Me
(A) (B)
PhCH
Me
Me
HCPh
Me
Me
(C) (D)
BrCH HCBr
What type of alkene is obtained on dehydrohalogenation of compound (A)? In order to answer this
question we shall rewrite the structure of compound (A) according to Newmann projection and ensure that
the two leaving groups are at 180° to each other and in the same plane.
PhCH
Me
HCBr
Me
(A)

Ph
H
MeMe
H
Br

Ph
Br
H
HBr
Me
Me H
Ph Me
Me H
or
C
C
Me H
Ph Me
(trans)
You will be curious to know the stereochemistry of alkenes obtained from (B), (C) and (D). Try it
out in the same manner to get the answer. (The enantiomers give the same alkene while diastereomers give
different alkenes)
Dehydrohalogenation of alkyl halildes with alcoholic KOH invariably leads to the formation of
both the Saytzeff product (more stable alkene) as well as Hoffmann product (less stable alkene) with
Saytzeff product as the major product for alkyl iodides, alkyl bromides and alkyl chlorides. However with
alkyl fluorides, Hoffmann product is the major product. Infact as we move from iodide  bromide 
chloride  fluoride, the percentage of Saytzeff product gradually decreases and that of Hoffmann product
gradually increases. The change over takes place with fluoride. How do we explain this?
The decrease in the percentage of Saytzeff product is directly linked with carbon halogen bond
dissociation energy. With the increase in carbonhalogen bond dissociation energy it becomes more and
more difficult for the leaving group (L) to leave as L()
. As a result the reaction has a tendency to follow
another mechanism called Elimination from conjugate base. In this mechanism the base first removes
proton from the carbon atom forming a carbanion as the intermediate. The carbanion then attacks the 
carbon atom causing the removal of F.
For example.

16. All right copy reserved. No part of the material can be produced without prior permission
CH3CHCHCH2
H H
F
alc. KOH
CH3CHCHCH3
CH3CH2CHCH2
F
F
CH3CH=CHCH3
(minor)
CH3CH2CH=CH2
(major)
(Less stable)
(more stable)
1.2 DEHYDRATION OF ALCOHOLS
Alcohols when heated in presence of H2SO4, H3PO4, P2O5, Al2O3 or BF3 undergo loss of water
molecule with the formation of alkene.
RCH2CH2OH
Conc. H2SO4
RCH=CH2 + H2O
(180°C)
H3PO4 or P2O5
(200°C)
Al2O3
(350°C)
RCH=CH2 + H2O
RCH=CH2 + H2O
The reaction mechanism involves the following steps:
(i) In the first step OH group of the alcohol is protonated in a fast reversible reaction. Unlike OH
group, protonated OH group is a good leaving group.
(ii) In the second step, water molecule is lost with the formation of a carbonium ion. This is the rate
determining step.
In any reaction if a carbonium ion is formed as an intermediate and there is a possibility of
rearrangement in which an atom or a group of atoms migrates from
an adjacent carbon atom or a cyclic ring expands so that a more stable carbonium ion is formed, that
rearrangement will take place.
(iii)In the final step carbonium ion loses proton from its adjacent carbon atom which results in more
stable alkene. The anions of the acid or another alcohol molecule will function as a base and
facilitate loss of proton.
(a) CH3COH
HA
CH3
CH3
CH3COH2 + A
CH3
CH3
+
(Fast)
(b) CH3COH2
RDS
CH3
CH3
CH3C + H2O
CH3
CH3
+
(Slow)
+
(c) CH3C+ A
CH2H
CH3
CH3C + HA
CH2
CH3
(fast)
The ease of dehydration follows the order
tertiary > secondary > primary alcohols.
This is reflected in the reaction conditions used to carry out dehydration of alcohols.
(i) Primary alcohols require the most stringent conditions to undergo dehydration i.e. use of conc.
H2SO4 and high temperature (180  200°C).
CH3CH2CH2OH
conc. H2SO4
180200°C CH3CH=CH2 + H2O

17. All right copy reserved. No part of the material can be produced without prior permission
(ii) Secondary alcohols can be dehydrated under relatively milder conditions by the use of 85% H3PO4
and a temperature of 160°C
85% H3PO4
160°C
CH3CHCH3
OH
CH3CH=CH2 + H2O
(iii)Tertiary alcohols can be easily dehydrated by using 25% H2SO4 at 85°C.
H3C
25% H2SO4
(minor)
OH
85°C
CH3
+
(major)
CH2
Examples of dehydration of alcohols involving rearrangement.
1. CH3  C  CH2OH
(i) H+
CH3
CH3
CH3  C  CH2
CH3
CH3
+
(ii) H2O
methanide
shift
CH3  C  CH  H
H
+
CH3
CH3  C CHCH3
CH3
(major)
+
CH3
2.
OH
(i) H
+
(ii) H2O
ring
expansion
+
or
H
+
+ H
(major)
+
1.3 DEHALOGENATION OF VICINAL DIHALIDES
There are two types of dihalides namely gem (or geminal) dihalides in which the two halogen
atoms are attached to the same carbon atom and vic. (or vicinal) dihalides in which the two halogen atoms
are attached to the adjacent carbon atoms.
Dehalogenation of vic dihalides can be effected by either NaI in acetone or zinc in presence of
acetic acid or ethanol.
CH3CHBrCH2Br
OH
H
C
or
COOH
CH
dust
Zn
5
2
3

 
 CH3CH=CH2
CH3CHBrCHBrCH3
Acetone
Na


 I
CH3CH=CHCH3
The reaction mechanism involves loss of the two halogen atoms in two steps. The two halogen
atoms align themselves at 180° and in the same plane before they are lost.
(i) With NaI in acetone:
C C
X:
X
X
C C
.
C C + IX
X+
I
I
(ii) With Zn dust and acetic acid:
Zn  Zn2+
+ 2e

18. All right copy reserved. No part of the material can be produced without prior permission
C C
X
X
X
2e
C C C C + X
X
..
1.4 CLEAVAGE OF ETHERS
Olefins can be formed by the treatment of ethers with very strong bases such as alkylsodium,
alkyllilthium or sodamide.
C C

RNa
OR
C C + RONa + RH
H

The reaction is aided by electron withdrawing groups in the position. For example
C2H5OCH2CH(COOC2H5)2 forms alkene just by heating without any base at all.
C2H5OCH2CH(COOC2H5)2 


CH2=C(COOC2H5)2 + C2H5OH
The mechanism probably involves a cyclic intermediate
PhCH O
H
CH2
CH2
CH2=CH2 + PhCH2O
()
()
PhCH2OCH2CH3
B:
BH
()
1.5 PYROLYSIS OF ESTERS
Thermal cleavage of an ester usually acetate involves the formations of a six membered ring as the
transition state leading to the elimination of acid leaving behind alkene.
R2C 500°C
H2C
O
C
O
H
R2C
H2C
H
O
C
O Me
R2C
H2C
+
H
O
C
O Me
Me
As a direct consequence of cyclic transition state, both the leaving groups namely proton and
carboxylate ion are in the cis position. This is an example of cis elimination.
1.6 PARTIAL REDUCTION OF ALKYNES
Reduction of alkyne to alkene is brought about by any one of the following reducing agents.
(i) Alkali metal dissolved in liquid ammonia.
(ii) Hydrogen in presence of palladium poisoned with BaSO4 or CaCO3 along with quinoline
(Lindlar’s catalyst).
(iii)Hydrogen in presence of Ni2B (nickel boride).
RCH2CCH 



 

)
iii
(
or
)
ii
(
),
i
(
RCH2CH=CH2
Alkali metal dissolved in liquid ammonia produces nearly 100% trans alkene by the following
mechanism:
Na + liq NH3  Na+
+ es (solvated electron)

19. All right copy reserved. No part of the material can be produced without prior permission
RCCR
es
C=C
R
()
R
HNH2
C=C
R
R
es
H
NH2 + C=C
R
H
H
R
C=C
R H
R
H2NH ..
()
(~ 100%)
(trans alkene)
The dissolution of alkali metal in liquid NH3 produces solvated electrons. The reaction is initiated
by the attack of sphybridised carbon atom of alkyne molecule with a solvated electron (es) when the 
electrons move to the other sp hybridised carbon atom. In order to acquire greater stability the single
electron on one carbon atom and the pair of electrons on the adjacent carbon atom orient themselves as far
away as possible forcing the two alkyl groups to acquire the farthest position. The carbanion then picks up
a proton from NH3 to produce a vinylic radical. Attack by another solvated electron gives vinylic anion
which produces trans alkene by picking up a proton from NH3.
Hydrogenation of alkynes by Lindlar’s catalyst or nickel boride produces nearly 100% cis alkene.
The catalyst provides a heterogenous surface on which alkyne molecules get adsorbed. Hydrogen
molecules collide with the adsorbed alkyne to produce cis alkene in which both the hydrogen atom come
from the same side.
RCCR + H2 C=C
H H
R
Lindlar’s cat.
R
or Ni2B
1.7 HOFFMANN DEGRADATION METHOD
Alkenes can be prepared by heating quaternary ammonium hydroxide under reduced pressure at a
temperature between 100°C and 200°C.
CH3NCH2CH2H OH
CH3

CH3
(CH3)3N + CH2=CH2 + H2O
+  
This is the final step of the overall three step reaction called Hoffmann degradation. In the first
step, primary, secondary on tertiary amine is treated with enough CH3I to convert it to the quaternary
ammonium iodide. In the second step, the iodide is converted into hydroxide by treatment with Ag2O.
2CH3I
CH3
N
H
CH3
N
+
I
CH3 CH3
Ag2O
CH2H OH
N
+
I
CH3 CH3

CH3 CH3
N
The H
O

ion invariably removes proton from carbon atom. If two or more alkyl groups have 
carbon atoms, H
O

removes proton from that  carbon atom which gives more stable carbanion. That
means if tetra alkyl ammonium halide contains ethyl group as one of the alkyl groups then ethene will be
the major product in this reaction. For example,
CH3CHCH2NCH2CH3
CH3

CH3
+ OH
CH2CH2CH3
CH3CHCH2N + CH2=CH2
CH3
CH3
CH2CH2CH3

20. All right copy reserved. No part of the material can be produced without prior permission
1.8 WITTIG REACTION
This method involves a convenient method of converting aldehydes and ketones into alkenes by
using a special class of compounds called phosphorous yields, also called
Wittig reagent. Primary or secondary alkyl halide is first treated with triphenyl phosphine the phosphonium
halide produced in the above reaction is converted into phosphorane by adding a strong base like C6H5Li or
nC4H9Li. Phosphorane is stabilised by resonance.
RCHX + Ph3P
R
RCHPPh3
R X
+
C6H5Li
RC PPh3
R
+
(ylide)
RC=PPh3
R
RC PPh3
R
+
The Triphenyl group of phosphorane has a strong tendency to pull oxygen atom of the aldehyde or
ketone forming alkene.
C=O
R
R
RC PPh3
R
C O
R
R
RC PPh3
R
C=C
R
R
R
R
+ Ph3P=O
+
–
(R,R, R and R may be hydrogen or any alkyl group)
The first three alkenes are gases, the next fourteen members are liquids and the higher ones are
solids. They are colourless and odourless (except ethylene which has a faint sweet smell), practically
insoluble in water but fairly soluble in nonpolar solvents like benzene, petroleum ether, etc. They show a
regular gradation in physical properties, such as boiling points, with increasing carbon content. The boiling
point of two successive members of the homologous alkene series differ by about 2030°C, except for very
small homologues. The branched chain alkenes have lower boiling points than the corresponding straight
chain alkenes. Like alkanes, alkenes are generally nonpolar, but certain alkenes are weakly polar. For
example, dipole moment of propene and 1butene is about 0.35D due to their unsymmetrical geometry.
cisAlkenes, in contrast to transalkenes, also have a small dipole moment. Therefore cisalkenes,
boil at somewhat higher temperature than the transalkene. cisalkenes have poorer symmetry and as such
do not fit into the crystalline lattice, with respect to the transisomers. Consequently, cisalkenes have
generally lower melting points.
3.1 HYDROGENATION
Alkenes are readily hydrogenated under pressure in presence of a catalyst.
RCH=CH2 + H2 

 

Catalyst
RCH2CH3
The following catalysts have been used satisfactorily in the above reaction.
GENERAL PHYSICAL PROPERTIES OF ALKENES
2
GENERAL CHEMICAL PROPERTIES OF ALKENES
3

21. All right copy reserved. No part of the material can be produced without prior permission
(i) Finely divided platinum and palladium are effective at room temperature. Platinum or palladium
black i.e., metals in a very finely divided state, may be prepared by reducing their soluble salts
with formaldehyde.
(ii) Nickel requires a temperature of 200300°C. Raney nickel is effective at room temperature and
atmospheric pressure. (The method for the preparation of Raney nickel has been described
elsewhere in the module).
One molecule of hydrogen is adsorbed for each double bond present in the unsaturated compound.
The rate of hydrogenation of alkenes at room temperature and atmospheric pressure is
CH=CH2 > CH=CH or a ring double bond
Alkenes of the type R2C=CR2 or R2C=CHR are difficult to hydrogenate under the above reaction
conditions.
The mechanism of catalytic hydrogenation is not known with certainty. It is widely accepted that
hydrogen is adsorbed on the surface of heterogeneous catalyst and is present as atomic hydrogen
(H2  2H). The alkene is also adsorbed on the catalytic surface.
It appears that the adsorption is more chemical than physical. The chemisorption has converted
hydrogen molecule into hydrogen atoms and has broken the weak bond of alkene, this is how
the catalyst lowers the activation energy of hydrogenation reaction.
The adsorption is then followed by addition of both the hydrogen atoms from the same side of the
double bond. The addition of hydrogen to an alkene is predominantly stereo selectively syn.
Various steps involved in the catalytic hydrogenation reaction are given below. The asterisks (*)
indicate metallic sites.
H2 + CH2=CH2 H H + CH2CH2 H + CH2CH3 CH3CH3
* * * * * *
For example,
C=C
CH3
D
D
+ H2
(trans)
CH3
CC
CH3
D
D
()
CH3
H H
Pt
Hydrogenation of alkenes can also be effected by the use of a homogeneous catalyst,
[RhCl(Ph3P)3], also called Wilkinson’s catalyst. The overall reaction proceeds in four steps. In the first
step, H2 adds to the rhodium complex and one Ph3P group is lost, thereby rhodium changing its oxidation
state from +1 to +3. The coordination number of rhodium in the compound (A) has also increased from 4
to 5. In the 2nd
step alkene attacks (A) forming a complex (B) which undergoes rearrangement in the 3rd
step of a Hatom to one of the carbon atoms of the double bond, the other carbon forming a sigma bond
with Rh. In the 4th
step the second hydrogen atom is transferred to the other carbon and the alkane is lost
with the regeneration of the catalyst.
Ph3P H2
Rh
Ph3P
PPh3
Cl
PPh3
Ph3P
Rh
PPh3
Cl
H
H
C=C Ph3P
Rh
PPh3
Cl
H
H
(A)
(B)
C=C

22. All right copy reserved. No part of the material can be produced without prior permission
Ph3P +PPh3
Rh
HCC
PPh3
Cl
Ph3P
Rh
PPh3
Cl
(C)
+ CHCH
H
Ph3P
3.2 ADDITION OF HYDROGEN HALIDES
Hydrogen halides (HCl, HBr and HI) add to the double bond of alkenes.
R
C = C + HX CC
X H
R
Mechanisms for addition of hydrogen halide to an alkene involves the following two steps.
Step 1.
R
C = C + HX CC +X

H
R Slow
(RDS)
Step 2.
R
CC
H
Fast
CC + X

R
H X
The addition of HBr to some alkenes gives a mixture of the expected alkyl bromide and an isomer
formed by rearrangement.
H
H+
CH2=CHCHCH3

CH3
CH3CHCCH3
CH3
CH3CHBrCHCH3
CH3
2Bromo3methylbutane
Br 
~ H
:
CH3CH2CCH3
Br 
CH3CH2CCH3
CH3
Br
2Bromo2methylbutane
3°

CH3
With this understanding of the mechanism for the ionic addition of hydrogen halides to alkenes, a
statement can be made as:
In the ionic addition of an unsymmetrical reagent to a double bond, the positive portion of the
reagent attaches itself to a carbon atom of the double bond so as to yield the more stable carbocation as
intermediate. Because this is the step that occurs first, it is the step that determines the overall orientation
of the reaction. In those cases where rearrangement does not occur, the addition of HX to alkenes follows
Markownikov’s rule, according to which “the negative part of the unsymmetrical reagent goes to that
carbon atom which bears lesser number of hydrogen atoms”.
But in those cases, where rearrangement occurs, the overall addition of HX to alkenes does not
follow Markownikov’s rule.
When HI is added to 1butene the reaction leads to the formation 2iodobutane, that contains a
stereocentre.
*
CH3CH2CH=CH2 + HI
I
CH3CH2CHCH3

23. All right copy reserved. No part of the material can be produced without prior permission
The product, therefore, can exist as a pair of enantiomers. The carbocation that is formed in the
first step of the addition is trigonal planar (sp2
hybridized) and is achiral. When the iodide ion reacts with
this flat carbocation, reaction is equally likely at either face. Thus reaction leads to the formation of two
enantiomers and both the enantiomers are produced in equal amounts. Thus, product of the reaction is a
racemic mixture.
3.3 ADDITION OF HYDROGEN BROMIDE IN PRESENCE OF PEROXIDE
The addition of HBr to propene, MeCH=CH2(1), under polar conditions (in absence of peroxide)
yields 2bromopropane. However, in the presence of peroxide (or under other contitions that promote
radical formation), the addition proceeds via a rapid chain reaction to yield 1bromopropane. This addition
of HBr in presence of peroxide is generally referred to as the peroxide effect leading to antiMarkownikov
addition. This difference in orientation of HBr addition is due to the fact that in the first (polar) case, the
reaction is initiated by H+
and proceeds via the more stable (secondary) carbocation while in the second
(radical) case, it is initiated by 
Br and proceeds via the more stable (secondary radical).
Chain initiation step:
ROOR  
RO
2 (OObond is weak)

RO + HBr  ROH + 
Br
Chain propagation step:
MeCH = CH2 + Br  MeCHCH2Br + MeCHCH2 (1° radical)
Br + MeCH2CH2Br
Br
(4)
(1) (2)
HBr
(3)
[Br generated
continues the chain]
The initiation is by 
Br , as hydrogen abstraction by 
RO from HBr is energetically much more
favourable than the alternative of bromine abstraction to form ROBr + 
H . The alternative addition of 
Br
to (1) to form MeCH(Br) 
2
CH (4) does not occur, as secondary radical )
2
(
B
HCH
C
Me 2
 is more stable
than primary radical.
HBr is the only one of the four hydrogen halides that will add readily to alkenes via a radical
pathway. The reason for this is reflected in the H values (kJ mol1
) for the two steps of the chain reaction
for addition of HX to CH2 = CH2. For example,
(1) 
X + CH2 = CH2 (2) XCH2 
2
CH + HX
HF 188 +155
HCl 109 +21
HBr 21 46
HI+29113
Only for HBr, both the chain steps are exothermic while for HF the second step is highly
endothermic, reflecting the strength of the HF bond and the difficulty of breaking it. For HCl, it is again
the second step that is endothermic (though not to such a great extent) while for HI it is the first step that is
endothermic, reflecting the fact that the energy gained in forming the weak IC bond is not as much as that
lost in breaking the bond. Thus only a few radical additions of HCl are known, but the C=C reactions are
not very rapid and the reaction chains are short at ordinary temperatures.

24. All right copy reserved. No part of the material can be produced without prior permission
Even with HBr addition, the reaction chains tend to be rather short, much shorter than those in
halogen addition and more than a trace of peroxide is thus needed to provide sufficient initiator radicals.
For preparative purposes up to 0.01 mol peroxide per mol of alkene is required. Once initiated, reaction by
this pathway is very much faster than any competing addition via the polar pathway and the anti
Markownikov product like (3) will thus predominate. If the Markownikov product, e.g. MeCH(Br)CH3
from propene, is required it is necessary either to purify the alkene rigorously before use or to add
inhibitors (good radical acceptors such as phenols, quinines, etc) to mop up any radicals. Essentially
complete control of orientation of HBr addition, in either direction, can thus be achieved, under preparative
conditions, by incorporating either peroxides (radical initiators ) or radical inhibitors in the reaction
mixture . This is particularly useful as such control is not confined purely to alkenes themselves. For
example, CH2 = CHCH2Br can be converted into 1,2 or 1,3dibromopropane at will.
3.4 ADDITION OF WATER
3.4.1 ACID CATALYZED HYDRATION
The acid catalyzed addition of water to the double bond of an alkene is a method of preparation of
low molecular weight alcohols. The addition of water to the double bond follows Markownikov’s rule in
those cases where rearrangement is not involved.
CH3C=CH2 + HOH
CH3
H
+
25°C
CH3CCH3
OH
CH3
As the reactions follow Markownikov’s rule acid catalyzed hydration of alkenes do not yield
primary alcohols except in the special case of the hydration of ethene. The occurrence of carbocation
rearrangements limits the utility of alkene hydration as a laboratory method for preparing alcohols.
Acid catalysed hydration of an alkene is the reversal of the similarly acid catalysed dehydration
(by the E1 pathway) of alcohols to alkenes.
MeCH = CH2 MeCHCH2 MeCHCH2 MeCHCH2
H
(1)
H
H2O H
H H
OH

OH2

The formation of the carbocationic intermediate (1), either directly or via an initial
 complex, appears to be rate limiting and the overall orientation of addition is Markownikov (in the
present case).
Acids that have weakly nucleophilic anions (like 
4
HSO from dilute aqueous H2SO4) are chosen
as catalysts, so that their anions will offer little competition to the actual nucleophile H2O. In case, if any
ROSO3H is formed, it will be hydrolysed to ROH under the conditions of the reactions.
3.4.2 OXYMERCURATIONDEMERCURATION
In the overall OxymercurationDemercuration reaction, H2O is added to the double bond. The
reaction is free from rearrangement (as it does not involve carbocation intermediate) and involves syn
addition using Markownikov’s rule.
Alkenes react with mercuric acetate in the presence of water to give hydroxymercurial compounds
which on reduction yield alcohols.
OH
C = C + H2O + Hg(OAc)2 RCCH
H
R Oxymercuration
H
H
H
HgOAc
H
NaBH4
Demercuration
OH
RCCH
H H
H

25. All right copy reserved. No part of the material can be produced without prior permission
The first stage, oxymercuration involves addition to the carboncarbon double bond of OH and
Hg(OAc)2. The electrophile of Hg(OAc)2 is AcOHg+
that adds C=C to form a mercurinium ion similar to a
bromonium ion. The mercurinium ion then reacts with H2O (not OAc
) at the more substituted carbon.
HgOAc
H2O
OH
OAc
[RCHCH2] RCHCH2 + HOAc
+
HgOAc
Then, in demercuration, HgOAc is replaced by H. The reaction sequence amounts to hydration of
the alkene, but is much more widely applicable than direct acid catalysed hydration.
Oxymercurationdemercuration gives alcohols corresponding to Markownikov addition of water
to the carboncarbon double bond. For example,
(i) CH3(CH2)3CH=CH2
OH
Hg(OAc)2, H2O
CH3(CH2)3CHCH2
HgOAc
NaBH4
OH
CH3(CH2)3CHCH3
(ii) CH3CCH=CH2
Hg(OAc)2, H2O
CH3
CH3
NaBH4
CH3CCHCH3
CH3
H3C OH
3.4.3 HYDROBORATIONOXIDATION
In the overall HydroborationOxidation reaction, H2O is added to the double bond. The reaction is
free from rearrangement (as it does not involve carbocation intermediate) and involves syn addition using
antiMarkownikov’s rule overall.
With the reagent diborane, B2H6, or disubstituted borane ,
BH
'
R
( 2 in THF solvent, alkenes undergo
hydroboration to yield trialkylboranes. The addition follows Markownikov’s rule.
2R–CH=CH2+B2H62RCH 2CH2BH2
R CH2CH2BH2+R–CH=CH2(R CH 2CH2)2BH
(R CH2CH2)2BH+R–CH=CH2(RCH 2CH2)3B
Oxidation of trialkylborane is carried out by alkaline H2O2 which results in the formation of
alcohol as if water has been added to alkene according to anti Markownikov’s rule.
O
H
O
O
H
H
O
H
O
O
H 2










B CH2CH2R
CH2CH2R
RCH2CH2
OH
O

B–
CH2CH2R
CH2CH2R
RCH2CH2
OH
O 

 


 H
O
B
CH2CH2R
RCH2CH2
OCH2CH2R





 

H
O
2
HOO
2
B
OCH2CH2R
OCH2CH2R
RCH2CH2O
2
H
O


B–
OCH2CH2R
OCH2CH2R
RCH2CH2O 
O

 


 H
O
CH2CH2R
RCH2CH2O OCH2CH2R
..
H
H
OH
B– +
H
 RCH2CH2OH
B
OCH2CH2R
OH
RCH2CH2O

 
 O
H
2 2
2RCH2CH2OH+H3BO3

26. All right copy reserved. No part of the material can be produced without prior permission
For example,
(i)
(ii) CH3CH=CCH3
B2H6
CH3
H2O2/OH
CH3CHCHCH3
CH3
OH



 


 


H
O
/
O
H
H
B 2
2
6
2
HO H
Me
H
(Syn addition)
Me
3.5 ADDITION OF BROMINE AND CHLORINE
Alkenes react rapidly with bromine at room temperature and in the absence of light.
If bromine in CCl4 is added to an alkene, the redbrown colour of the bromine disappears almost instantly
as long as the alkene is present in excess. This serves as a classical test for detection of unsaturation of
C=C or CC type.
(a) C=C + Br2
CCl4
Br
(b) CH3CH=CHCH3 + Cl2
9°C
CH3CHCHCH3
Cl
CC
Br
VicDibromide
rapid decolorization of
Br2/CCl4 is a test for
alkenes and alkynes
Cl
Mechanism:
The mechanism proposed for halogen addition is an ionic mechanism.
In the first step the exposed electrons of the  bond of the alkene attack the halogen in the
following way.
C = C
R
1 2
Br2 in
CCl4
CC
R
1 2
Br 
+ 
C C
R Br
Br
+

C C
R
Br
Br
Br

Br
+
Cyclic bromonium ion Vicinal dibromide
In this reaction, when reagent (bromine) approaches alkene, the temporary polarization develops
on the alkene with C2 atom gaining a negative charge and C1 atom acquiring positive charge (as it can be
compensated by the +I effect of R group). The alkenes being electron rich compounds (due to the presence
of electron cloud) are attacked by the electrophile (Br+
) to give a cyclic bromonium ion. Here, the
formation of cyclic bromonium ion as intermediate is possible because bromine is of considerably large
size having lone pairs to be bonded to both the carbons simultaneously. The cyclic bromonium ion is then
attacked by Br 
from the top (as lower side is already blocked) whereby the three membered ring is
cleaved by trans opening giving vicinal dibromide as the product. Thus, the overall addition of Br2 to
alkene follows trans stereoselectivity.
When cyclopentene reacts with bromine in CCl4, antiaddition occurs and the products of the
reaction are trans1,2dibromocyclopentane enantiomers (as a racemate).

27. All right copy reserved. No part of the material can be produced without prior permission
Br2
CCl4
Br
H Br
H + H
Br H
Br
When cis2butene adds bromine, the product is a racemic form of 23dibromobutane. When
trans2butene adds bromine, the product is the meso compound.
Thus we find that a particular stereoisomeric form of the starting material react in such a way that
it gives a specific stereoisomeric form of the product. Thus the reaction is stereospecific.
3.6 HALOHYDRIN FORMATION
If the halogenation of an alkene is carried out in aqueous solution (rather than in CCl4), the major
product of the overall reaction is a haloalcohol called halohydrin. In this case, the molecules of the
solvent become reactant.
C = C + X2 + H2O
X
CC +  CC + HX
OH X X
(major) (minor)
X2 = Cl2 or Br2
Halohydrin formation can be explained by the following mechanism
C = C + X  X CC + X
X+
CC + H2O CC
H
+
X+
X
+
OH2
CC
X
OH
The addition of X and OH occurs in the trans manner, as the reaction proceeds by the formation of
halonium ion intermediate.
If the alkene is unsymmetrical, the halogen adds up on the carbon atom with greater number of
hydrogen atoms i.e. the addition follows Markownikov’s rule.
C = CH2
Br2 , H2O
or HOBr
OH
H3C
H3C
CCH2
CH3
CH3
Br
3.7 HYDROXYLATION
There are a number of reagents that can add two OH groups to alkenes. The two OH groups can be
either added from the same side (syn hydroxylation) or from the opposite side (anti hydroxylation).
3.7.1 SYN HYDROXYLATION
Osmium tetroxide (OsO4) adds to alkene to form cyclic osmic ester (2) which can be made to
undergo ready hydrolytic cleavage of their OsO bonds to yield the vicdiol(3).
H
Me
Me
H
OsO4
Me H
O Me H
O
Os
O
O
2H2O
Me H
HO Me H
OH
+ (HO)2OsO2
(1) (2) (3)
cis 2butene (1) thus yields the meso butan-2, 3diol (3), i.e. the overall hydroxylation is
stereoselectively syn, as would be expected from OsO cleavage in a necessarily cis cyclic ester (2). The

28. All right copy reserved. No part of the material can be produced without prior permission
disadvantage of this reaction as a preparative method is expense and toxicity of OsO4. However, this can be
overcome by using it in catalytic quantities in association with H2O2 which reoxidises the osmic acid,
(HO)2OsO2, formed to OsO4.
Alkaline permanganate, 
4
18
O
Mn (a reagent used classically to test for unsaturation), will also
effect stereoselective syn addition and this by analogy with the above, is thought to proceed via cyclic (cis)
permanganic ester. It has not proved possible actually to isolate such species but use of 
4
18
O
Mn , was
found to lead to a vicdiol in which both oxygen atoms were O18
labeled. Thus both were derived from

4
MnO , and neither from the solvent H2O, which provides support for a permanganic analogue of (2) as an
intermediate, provided that 
4
18
O
Mn undergoes no O18
exchange with the solvent H2O under these
conditions. The disadvantage of 
4
MnO for hydroxylation is that the resultant 1,2diol is very much
susceptible to further oxidation by it.
3.7.2 ANTIHYDROXYLATION
Peroxyacids, RCOOOH will also oxidize alkenes, e.g. trans 2butene (4), by adding an oxygen
atom across the double bond to form an epoxide (5).
H
Me Me
H
Me H
O
Me
H
O
(4)
HO+
C
R
O

H
O
C
R
O

Me H
Me
H
O + RCO2H
(5)
Epoxides (though uncharged) have a formal resemblance to cyclic bromonium ion intermediates,
but unlike them are stable and may readily be isolated. However, they undergo nucleophilic attack under
either acid or base catalysed conditions to yield the 1,2diol. In either case attack by the nucleophile on
carbon atom will be from the opposite side of the oxygen bridge in (5). Such attack on the epoxide will
involve inversion of configuration.
OH
Me
H
Me
H
O
O
Me H
(5)
OH
H Me
OH H2O
HO
Me H
H Me
OH
(7)
Me
H
Me
H
O
(6)
H2O H
+
OH2
H
H
H Me
OH

OH2
Me H
H Me
OH
OH
(7)

Me H
Attack has been shown on only one of the two possible carbon atoms in (5) and (6), though on
different ones in the two cases. In each case, attack on the other carbon will lead to the same product, the
meso vicdiol (7). By comparing the configuration of (7) with that of the original alkene (4), it can be seen

29. All right copy reserved. No part of the material can be produced without prior permission
that in overall terms setereoselective anti hydroxylation has been effected. Thus by suitable choice of
reagent, the hydroxylation of alkenes can be made stereoselectively syn or anti at will.
For example,
3CH2=CH2 + 2KMnO4 + 4H2O
(ethylene glycol)
OH
3CH2CH2 + 2MnO2 + 2KOH
OH
CH3CH=CH2
(1) OsO4
OH
CH3CHCH2
OH
(2) NaHSO3/H2O
(propylene glycol)
C=C
CH3
H
CH3
H
cis2butene
RCO3H
H
+
, H2O
CH3CCCH3 + enantiomer
HO H
H OH
(trans1,2diol)
+ enantiomer
3.8 OXIDATIVE CLEAVAGE BY HOT ALKALINE KMnO4
Alkenes are oxidatively cleaved by hot alkaline permanganate solution. The terminal CH2 group of
1alkene is completely oxidized to CO2 and water. A disubstituted atom of a double bond becomes
C=O group of a ketone. A monosubstituted atom of a double bond becomes aldehyde group which is
further oxidized to salt of carboxylic acid.
For example,
CH3CH=CHCH3
KMnO4, 
OH
(cis or trans)
O
Heat 2CH3C
O
Acetate ion
H
+
2CH3CO2H
(a)
CH3CH2C=CH2
KMnO4, 
OH
H
+
(b)
CH3
CH3CH2C=O + CO2 + H2O
CH3
CH3C=CHCH2CH2CH3
(c)
CH3
KMnO4, 
OH
H
+ CH3C=O + HO2CCH2CH2CH3
CH3
CH3CH2CH2HC=CHCH
(d)
CH3
KMnO4, 
OH
H
+ CH3CH2CH2CO2H + CHCO2H
CH3
CH3
CH3
3.9 OZONOLYSIS
A more widely used method for locating the position of double bond in an alkene involves the use
of ozone (O3). Ozone reacts vigorously with alkene to form unstable compound called molozonide, which
rearranges spontaneously to form a compound known as ozonide. Ozonides, themselves are unstable and
reduced directly with Zn and water. The reduction produces carbonyl compounds (aldehydes and ketones)
that can be isolated and identified.

30. All right copy reserved. No part of the material can be produced without prior permission
CC
O O
O
C
O
C
O O
2 C=O + ZnO + H2O
Ozonide Aldehydes and / or Ketones



 3
O
C
C
H2O
Zn
Ozonolysis can be either of reductive type or of oxidative type. The difference lies in the fact that
products of reductive ozonolysis are aldehydes and/or ketones while in oxidative ozonolysis, the products
are carboxylic acids and/or ketones. This is because H2O2 formed would oxidize aldehydes to carboxylic
acids but ketones are not oxidized. In reductive ozonolysis, we add zinc which reduces H2O2 to H2O and
thus H2O2 is not present to oxidize any aldehyde formed.
Zn + H2O2  ZnO + H2O
For example,
(CH3)2CHCH=CH2
(i) O3, CH2Cl2,
(ii) Zn/H2O
(CH3)2CHCHO + HCHO
(CH3)2C=CHCH3
(i) O3, CH2Cl2,
(ii) Zn/H2O
(CH3)2C=O + CH3CHO
3.10 SUBSTITUTION REACTIONS AT ALLYLIC POSITION
Cl2 + H2C = CHCH3
High
Temperature
H2C = CHCH2Cl + HCl
Br2 + H2C=CHCH3
Low concentration
of Br2
CH2=CHCH2Br + HBr
These halogenations are like free radical substitution of alkanes. The order of reactivity of H
abstraction is allyl>3°>2°>1°>vinyl.
Allylic substitution by chlorine is carried out using Cl2 at high temperature and alkene (with 
carbon) in gaseous phase. Allylic bromination can be carried out using NBromosuccinimide. Propene
undergoes allylic bromination when it is treated with Nbromosuccinimide (NBS) in CCl4 in the presence
of peroxide or light.
CH2=CHCH3+
O
Light or ROOR
CCl4
O
NBr CH2=CHCH2Br +
O
O
NH
NBromosuccinimide (NBS) 3Bromopropene
(allyl bromide)
Succinimide
The reaction is initiated by the formation of a small amount of Br (possibly formed by
dissociation of Br2 molecule). The chain propagation steps for this reaction are the same as for
chlorination.
CH2=CHCH2H + Br CH2=CHCH2 + HBr
CH2=CHCH2 + BrBr CH2=CHCH2Br + Br
NBromosuccinimide is nearly insoluble in CCl4 and provides a constant but very low
concentration of bromine in the reaction mixture. It does this by reacting very rapidly with the HBr formed

31. All right copy reserved. No part of the material can be produced without prior permission
by the reaction of NBS with traces of H2O present in it. Each molecule of HBr is replaced by one molecule
of Br2.
O
O
NBr + HBr
O
O
NH + Br2
Under these conditions, that is, in a nonpolar solvent and with a very low concentration of
bromine, very little bromine adds to the double bond, instead it undergoes substitution and replaces an
allylic hydrogen atom.
A question must have arisen in your mind that why does a low concentration of bromine favour
allylic substitution over addition? To understand this we must recall the mechanism for addition and notice
that in the first step only one atom of the bromine molecule becomes attached to the alkene in a reversible
step.
BrBr +
C
C
C
C
Br
+
+ Br C
C
Br
Br
The other atom (now the bromide ion) becomes attached in the second step. Now, if the
concentration of bromine is low, the equilibrium for the first step will lie far to the left. Moreover, even
when the bromonium ion forms, the probability of its finding a bromide ion in its vicinity is also low.
These two factors slow the addition to such an extent that allylic substitution competes successfully.
The use of a nonpolar solvent also slows addition. Since there are no polar molecules to solvate
(and thus stabilize) the bromide ion formed in the first step, the bromide ion uses a bromine molecule as a
substitute.
2Br2 +
C
C
C
C
Br
+
+ Br3

Nonpolar
Solvent
This means that in a nonpolar solvent the rate equation is second order with respect to bromine
rate = k C=C [Br2]2
and that the low bromine concentration has an even more pronounced effect in slowing the rate of addition
whereby increasing the tendency to undergo substitution.
To understand why a high temperature favours allylic substitution over addition requires a
consideration of the effect of entropy changes on equilibria. The addition reaction has a substantial
negative entropy change because it combines two molecules into one. At low temperatures, the TS term
in G = H  TS, is not large enough to offset the favourable H term. But as the temperature is
increased, the TS term becomes more significant, G becomes more positive, and the equilibrium
becomes more unfavourable for addition and subsequently favours allylic substitution.
Illustration
Identify compounds (A) to (F) in the following sequence of reactions.
CH3CH2CH3
Br2/h
(A) aq. KOH (B)
Na
(C)
alc. KOH
(D)
NBS
(E)
+(C)
(F)

32. All right copy reserved. No part of the material can be produced without prior permission
Solution:
CH3CH2CH3
Br2/ h
CH3CHCH3
Br
aq. KOH
CH3CHCH3
OH
Na
CH3CHCH3
ONa
(A) (B) (C)
alc. KOH
CH3CH=CH2
NBS
CH2BrCH=CH2
+(C)
CH3CHOCH2CH=CH2
CH3
(A)
(D) (F)
(E)
Illustration
Identify the unknown alkenes (A) to (E) in the following reactions.
(a) CH3COCH3 + CH3CH(CH3)CHO
(A)
(i) O3
(ii) Zn/H2O
(b) CH3COCH2CH2CH2CH2COOH
(B)
(i) KMnO4/OH
(ii) H3O+
(c) 2CH3CH2CO2H
(C) (i) O3
(ii) H2O
(d) CH3CH2COOH + CH3CH2CH2CH2COOH
(D)
(i) KMnO4/ OH
(ii) H3O+
(e) + CH2O
(E)
(i) O3
(ii) Zn/H2O
O
Solution:
(a) Alkene (A) on ozonolysis followed by reductive hydrolysis gives two different carbonyl
compounds. Thus, it must be an unsymmetrical alkene. The alkene can be identified by
removing the oxygen atom of the carbonyl groups and inserting a double bond between the
two carbon atoms of the carbonyl groups. Thus, structure of alkene (A) is CH3C=CHCHCH3
CH3
CH3
(b) Oxidative cleavage of alkene (B) by hot alkaline KMnO4 followed by acidification gives a keto
acid. The alkene must be a cyclic alkene. The ring opens up on oxidation of double bond.
The alkene (B) is
CH3
(c) Ozonolysis followed by oxidative hydrolysis of alkene (C) of a single carboxylic acid. It must
be a symmetrical diene. Thus, alkene (C) is
CH3CH2CH=CHCH2CH3
(d) Oxidation of alkene (D) gives a mixture of two different carboxylic acids. It must be an
unsymmetrical alkene. Thus, alkene (D) is

33. All right copy reserved. No part of the material can be produced without prior permission
CH3CH2CH2CH2CH=CHCH2CH3
(e) Ozonolysis of alkene (E) followed by reductive hydrolysis gives cyclopentanone and
formaldehyde. Thus, the structure of alkene (E) is
CH2
3.11 ACID CATALYZED DIMERIZATION OF ALKENES
I. In case of monoalkene, two alkenes dimerize to form a larger alkene.
2CH3C=CH2
H3PO4
CH3C=CHCCH3 + CH2=CCH2CCH3
CH3 CH3 CH3
CH3
CH3 CH3
CH3
(major)
Mechanism:
CH3C=CH2
H+
CH3C–CH3
CH3 CH3 CH3
CH3

CH3C=CH2
CH3
CH3C–CH2CCH3
CH3

H+
CH3
CH3
CH3CCH=CCH3 + CH3CCH2C
CH3
.
CH2
CH3
CH3
CH3
(major) (minor)
II. In case of diene, ring formation takes place depending upon the structure of diene.
H3PO4
CH3C=CHCH2CH2CH=CCH3
CH3 CH3
(minor)
Me
Me
+
Me
Me
CH2
CH3 CH3 CH3
(major)
If the ring formed is five or sixmembered, this reaction occurs with great ease.
Mechanism:
H+
+
Me
Me
CH3
(major)
Me
Me
Me
Me
Me
Me
CH3 CH3

Me
Me
H3C CH2
(minor)
H3C

Me
Me
CH3
H2C
H+

34. All right copy reserved. No part of the material can be produced without prior permission
Alkynes are hydrocarbons having four hydrogen atoms less than the corresponding alkane. They
have two degrees of unsaturation in the form of a triple bond between two carbon atoms. They are isomeric
with dienes and cycloalkene and have the general formula CnH2n2. The most important member of the
alkyne series is acetylene and hence alkynes are also known acetylenes.
1.1 HYDROLYSIS OF CARBIDES
Some of the lower members of alkyne series can be synthesized by the hydrolysis of carbides. For
example, calcium carbide on hydrolysis gives acetylene and magnesium carbide on hydrolysis gives
propyne.
CaC2 + 2H2O  HCCH + Ca(OH)2
Mg2C3 + 4H2O  CH3CCH+ 2Mg(OH)2
The difference in the behaviour of calcium carbide and magnesium carbide is due to the
differences in their structures. Both the carbides are ionic in nature. In calcium carbide, the anion exists as
_
_
C
C while in magnesium carbide, the anion exists as
_
_
.
C
C
_
C
3
 It is pertinent to note that aluminium
carbide (Al4C3) and beryllium carbide (Be2C) do not form any alkyne on hydrolysis, instead they form
methane on hydrolysis. This is due to the fact that their anions exist as C4
.
1. 2FROM ACETYLENE
The two hydrogen atoms of acetylene are acidic in nature and can be replaced by a strong base like
sodium or sodamide.
HCCH + Na  
 Na
C
HC
_
+ ½ H2
or HCCH + NaNH2  
 Na
C
HC
_
+ NH3
Sodium acetylide when treated with primary alkyl halide gives 1alkynes following nucleophilic
substitution reaction by SN2 mechanism. Secondary alkyl halide gives poor yield of 1alkyne because
substitution reaction is accompanied by elimination reaction with acetylide ion )
C
HC
(
_
 functioning as a
strong base. Tertiary alkyl halides do not undergo any substitution reaction because of steric hindrance.
Instead they undergo elimination reaction easily forming alkene as the only product. For example,
(i) HCC + CH2CH2CH3 HCCCH2CH2CH3 + Cl
Cl (1°RX) 1pentyne
GENERAL METHODS OF PREPARATION OF ALKYNES
1
ALKYNES

35. All right copy reserved. No part of the material can be produced without prior permission
(ii) HCC + CH3CHCH3 HCCCHCH3 + Cl
Cl
(2°RX)
3methyl1butyne
(poor yield)
CH3
HCC
+ HCH2CHCH3 CHCH + CH3CH=CH2 + Cl
Cl 3methyl1butyne
(poor yield)
(iii) HCC H CH2 HCCH + CH3C + Cl
CH3CCl
(3°RX)
CH3
CH3
CH2
1alkyne still has one more acidic hydrogen and by repeating the same set of reactions,
it is possible to introduce the same alkyl group or different alkyl group at the other sp hybridised carbon
atom.
CH3CH2CH2CCH + NaNH2  CH3CH2CH2CC
Na+
+ NH3
CH3CH2CH2CC CH2CH3
Cl
CH3CH2CH2CCCH2CH3 + Cl
1.3 DEHYDROHALOGENATION OF VICINAL OR GEMINAL DIHALIDES
Both vicinal dihalides and geminal dihalides undergo dehydrohalogenation reaction with a strong
base to give alkyne in fairly good yield. The reaction follows 1, 2elimination mechanism.
In the first step, the base employed is alcoholic KOH while in the subsequent step, we need a strong base
like sodamide as vinyl halide is less reactive towards elimination.
H
CH3C CH
alc. KOH
Br
H
Br
H
CH3C CH
Br
NaNH2
CH3CCH
vic. dibromide
CH3C CH2CH3
alc. KOH
Br
CH3C CHCH3
Br
NaNH2
CH3CCCH3
gem. dibromide
Br
NaNH2 is preferred over alc. KOH in the second step as alc. KOH finds it difficult to eliminate a
hydrogen halide molecule as the two leaving groups are attached to sp2
hybridised carbon atoms.
1.4 DEHALOGENATION OF VICINAL TETRAHALIDES
Vicinal tetrahalides are compounds containing two halogen atoms attached to each of the two
adjacent carbon atoms. They lose all the four halogen atoms forming alkynes when treated with either (i)
NaI in acetone/methanol, or (ii) Zn dust and ethanol. The mechanism is very similar to the one discussed in
the lesson on “Alkenes”.
CH3C C–CH3
(i) NaI in Me2CO/CH3OH
Br Br
CH3CCCH3
vic. tetrabromide
Br Br
(ii) Zn dust in C2H5OH
or
GENERAL PHYSICAL PROPERTIES OF ALKYNES
2

36. All right copy reserved. No part of the material can be produced without prior permission
Being compounds of low polarity, the alkynes have physical properties that are essentially the
same as those of the alkanes and alkenes. They are insoluble in water but quite soluble in the usual organic
solvents of low polarity like ether, benzene, carbon tetrachloride etc. Their densities are lower than that of
water. Their melting points and boiling points show the usual increase with increase in number of carbon
atoms and the usual effects of chain branching.
The CH3C bond in propyne is formed by overlap of an sp3
hybrid orbital from methyl carbon with
a sp hybrid orbital from acetylenic carbon. The bond is between sp3
sp carbon.
Since one orbital has more ‘s’ character than the other and is thereby more electronegative, the electron
density in the resulting bond is not symmetrical. The unsymmetrical electron distribution results in a dipole
moment larger than alkene but still relatively small. Symmetrically disubstituted alkynes have zero dipole
moment.
The triple bond of alkynes consists of one sigma bond and two pie bonds, which are perpendicular
to each other. The two pie bonds get mixed up and all the four pie electrons are cylindrically distributed
around the two sp hybridised carbon atoms. Being unsaturated, alkynes undergo addition reactions to form
alkene derivatives with the addition of one molecule and saturated compounds with the addition of two
molecules. Under suitable conditions, it is possible to isolate the intermediate alkene.
Acetylene is less reactive than ethylene towards most of the electrophilic reagents. This is
unexpected in view of the fact that the electron density in a triple bond is higher than that in a double
bond. Possible reason for decreased reactivity of a triple bond towards electrophiles may be the fact that
the bridged halonium ion from acetylene is more strained than the bridged halonium ion from ethylene.
CC
Br2
C C + Br
Br+
C C Br2
C C + Br
Br+
Towards hydrogenation (which do not involve electrophilic attack), alkynes are more reactive than
alkenes.
3.1 HYDROGENATION
Alkynes can be reduced directly to alkanes by the addition of H2 in the presence of Ni, Pt or Pd as
a catalyst. The addition reaction takes place in two steps. It is not possible to isolate the intermediate alkene
under the above reaction conditions. By using Lindlar’s catalyst [Pd on CaCO3 + (CH3COO)2Pb], nickel
boride or palladised charcoal, alkynes can be partially hydrogenated to alkenes.
CH3CCH + 2H2 



 

Pd
or
Pt
or
Ni
CH3CH2CH3
CH3CCH + H2 



 

.
cat
s
'
Lindlar
CH3CH=CH2
2butyne when reduced with Lindlar’s catalyst gives nearly 100% cis isomer while Na in liquid
ammonia gives nearly 100% trans isomer.
GENERAL CHEMICAL PROPERTIES OF ALKYNES
3

37. All right copy reserved. No part of the material can be produced without prior permission
CH3CCCH3 + H2
Lindlar’s cat.
Na in
C=C
H H
H3C CH3
(cis)
C=C
H
H
H3C
CH3
(trans)
liq. NH3
3.2 ADDITION OF HALOGEN ACIDS
Addition of halogen acids to alkynes occur in accordance with Markownikoff’s rule. Addition of
one molecule of halogen acid gives an unsaturated halide, which then adds another molecule of hydrogen
halide to form gem dihalides. For example, addition of HI to propyne first gives 2iodopropene and then
2,2diiodopropane.
CH3CCH + HI
2iodopropene
CH3C=CH2
I
HI
CH3CI2CH3
2,2diiodopropane
The order of reactivity of halogen acids towards addition reaction is HI > HBr > HCl.
Peroxides have the same effect on addition of HBr to alkyne as that on alkene and the reaction
follows free radical mechanism.
RO + HBr  ROH + Br
CH3CCH + Br  CH3 CHBr
C


 

HBr
CH3CH=CHBr + Br
 CH3 2
CHBr
H
C 



HBr
CH3CH2CHBr2 + Br
3.3 ADDITION OF HALOGENS
One or two molecules of halogens can be added to alkynes giving dihalides and tetra halides
respectively. Chlorine and bromine add readily to the triple bond while iodine reacts rather slowly.
CHCH + Cl2  CHCl=CHCl 
 
 2
Cl
CHCl2CHCl2
CH3CCH + Br2  CH3CBr=CHBr 
 
 2
Br
CH3CBr2CHBr2
(colourless)
This reaction can be used as a test to detect unsaturation (both alkenes and alkynes)
as Br2 in CCl4 is reddish brown in colour while the product obtained is colourless.
3.4 ADDITION OF WATER
Water adds to alkyne when alkyne is treated with 40% H2SO4 containing 1% HgSO4
(as a catalyst) to form a carbonyl compound. The addition of water follows Markownikoff’s rule forming
enol as intermediate, which tautomerizes to give a more stable carbonyl compound.
For example, acetylene gas when passed through dil. H2SO4 containing HgSO4 initially forms vinyl
alcohol, which tautomerizes to acetaldehyde.
HCCH + H2O
)
(
2 ]
[
4
4
2
alcohol
Vinyl
HgSO
SO
H
OH
CH
CH 


 

tautomerizes
)
(
3
de
acetaldehy
CHO
CH 
The reaction is believed to take place via the formation of a three membered ring involving Hg2+
ion.

38. All right copy reserved. No part of the material can be produced without prior permission
CC +Hg2+
C C
Hg2+
H2O
C=C
Hg+
+
OH2
H
+
C=C
Hg+
OH
H3O
+
Hg
2+
CH=C
OH
CH2C
O
Acetylene is the only alkyne forming an aldehyde in this reaction. Higher homologues of acetylene
either form a single ketone or a mixture of ketones. For example, 2pentyne gives a mixture of 2
pentanone and 3pentanone.
CH3CCCH2CH3 + H2O
CH3C=CHCH2CH3
H
+
/Hg
2+
CH3CCH2CH2CH3
OH O
CH3CH=CCH2CH3 CH3CH2CCH2CH3
OH O
tautomerizes
tautomerizes
3.5 ADDITION OF BORON HYDRIDES
Diborane, the simplest hydride of boron reacts with alkyne to form trialkenylborane. Diborane
splits into two BH3 units and the addition of BH3 takes place following Markownikoff’s rule. The addition
continues as long as hydrogen is attached to boron atom.
2RCCH + B2H6  2RCH=CHBH2
RCCH + RCH=CHBH2  (RCH=CH)2BH
RCCH + (RCH=CH)2BH  (RCH=CH)3B
Trialkenylborane on hydrolysis gives alkene.
(RCH=CH)3B
hydrolysis
COOH
CH


 
 3
3RCH=CH2 + B(OH)3
Internal alkynes give rise to alkenes where geometrical isomerism is possible,. Hydroboration
followed by hydrolysis of alkynes gives cis alkene as the major product.
RCCR +B2H6 (RCH=CR)3B
CH3CO2H
C C
R R
H H
(cis)
Oxidation of trialkenylborane with alkaline H2O2 results in the formation of carbonyl compounds.
Terminal alkynes give rise to aldehydes whereas internal alkynes give rise to ketones.
(RCH=CH)3B
H2O2/NaOH
RCH=CH RCH2CHO
OH
tautomerizes
(RCH=CR)3B
H2O2/NaOH
RCH=CR RCH2CR
OH O
tautomerizes
3.6 DIMERISATION
Acetylene dimerises when treated with a mixture of Cu2Cl2 and NH4Cl to give vinyl acetylene.
2CHCH 


 
  Cl
NH
Cl
Cu 4
2
2
CH2=CHCCH
The dimer undergoes addition reactions just like any other unsaturated hydrocarbon.
The addition reaction preferably takes place at the triple bond and not at the double bond inspite of the fact
that alkynes are less reactive than alkenes towards electrophilic addition reactions.
For example, addition of HCl to vinyl acetylene gives chloroprene.

39. All right copy reserved. No part of the material can be produced without prior permission
CH2=CHCCH + HCl CH2=CHC=CH2
Cl
(Chloroprene)
3.7 OXIDATION
Alkynes are oxidised by hot alkaline KMnO4, which causes cleavage of CC resulting in the
formation of salts of carboxylic acids. The salts on acidification are converted into acids. Internal alkynes
give mixture of carboxylic acids while terminal alkynes give a carboxylic acid and the terminal Catom is
oxidised to CO2 and H2O.
CH3CCH 






 
 
H
)
ii
(
/
OH
/
KMnO
)
i
( 4
CH3COOH + CO2 + H2O
CH3CCCH2CH3







 
 
H
)
ii
(
/
OH
/
KMnO
)
i
( 4
CH3COOH + CH3CH2COOH
3.8 OZONOLYSIS
Reaction of alkynes with O3 gives rise to the formation of ozonide. Hydrolysis of ozonide with
H2O gives a mixture of two carboxylic acids. This is called oxidative ozonolysis.
CH3CCH + O3 CH3C CH
O
H2O
O O
CH3COOH + HCOOH
However, if ozonide is hydrolysed with Zn and H2O, a diketone is formed. This is called reductive
ozonolysis.
CH3CCCH2CH3 + O3 CH3C CCH2CH3
O
Zn/H2O
O O
CH3CCCH2CH3
O
O
Ethyne behaves differently. Ozonolysis followed by oxidative hydrolysis of ethyne gives a mixture
of glyoxal and formic acid.
HCCH
O
H
.
2
O
.
1
2
3

 
 CHOCHO + HCOOH.
3.9 POLYMERISATION REACTIONS
(i) Hydrochloric acid adds to acetylene in the presence of Hg2+
ion as catalyst to form vinyl chloride.
Polymerisation of vinyl chloride results in the formation of polyvinyl chloride (PVC).
HCCH + HCl 
 


2
Hg
CH2=CHCl 


 

tion
Polymeriza
CH2CH
Cl n
(PVC)
(ii) Addition of HCN to ethyne is catalysed by Cu2Cl2 in HCl. The product obtained is acrylonitrile,
which on polymerisation gives polyacrylonitrile (PAN).
HCCH + HCN
HCl
Cl
Cu 2
2


 
 CH2=CHCN 


 

tion
Polymeriza
CH2CH
CN n
(PAN)
Acrylonitrile is also used in the manufacture of a synthetic rubber called BuNaN
(a copolymer of butadiene and acrylonitrile) and a thermoplastic called ABS (a terpolymer of
acrylonitrile, butadiene and styrene).
(iii)Acetic acid adds to ethyne in the presence of Hg2+
ion to give vinyl acetate, which is used as
monomer in the preparation of polyvinyl acetate (PVA).

40. All right copy reserved. No part of the material can be produced without prior permission
HCCH + CH3COOH 
 


2
Hg
CH2=CHOOCCH3
(vinyl acetate)
CH2=CH
OOCCH3
Polymerization
CH2CH
OOCCH3 n
Polyvinylacetate (PVA)
(iv) Acetylene when passed through a hot metallic tube polymerizes to give benzene.
3CHCH Red hot
tube
Higher homologues of acetylene also polymerize under similar conditions to give derivatives of
benzene.
3CH3CCH
Red hot
tube
(1,3,5trimethyl benzene)
3CH3CCCH3
Red hot
tube
(1, 2, 3, 4, 5, 6hexamethyl benzene)
3.10 ADDITION OF HYPOHALOUS ACID
Alkynes react with hypohalous acid in the molar ratio of 1 : 2 to give dihalo ketones.
Acetylene forms dihaloaldehyde.
RCCH + HOX  RC(OH)=CHX 
 
HOX
RC(OH)2CHX2 2
||
2
CHX
C
R
O
O
H



 


3.11 ACIDITY OF ALKYNES
The acidic nature of hydrogen in acetylene is characteristic of hydrogen in the group CH and it is
because the H
C
 bond has considerable ionic character due to resonance.
HCCH  HCCH  HCCH  HCCH
 + +  +   +
There is, evidence that the electronegativity of a carbon atom depends on the number of bonds by
which it is joined to its neighbouring carbon atom. Since electrons are more weakly bound than 
electrons, the electron density around a carbon atom with bonds is less than that when only bonds are
present. Thus a carbon atom having one bond has a slight positive charge compared with a carbon atom,
which has only bonds. Hence, the electronegativity of an sp2
hybridised carbon atom is greater than that
of an sp3
hybridised carbon atom. Similarly, a carbon atom, which has two bonds, carries a small
positive charge, which is greater than that carried by a carbon atom with only one bond. Thus the
electronegativity of an sp hybridized carbon atom is greater than that of an sp2
hybridised carbon atom.
Thus, more the ‘s’ character a bond has, the more electronegative is that carbon atom. Therefore the
attraction for electrons by hybridized carbons will be sp > sp2
> sp3
.
Therefore, the hydrogens in terminal alkynes are relatively acidic. Acetylene itself has a pKa of
about 25. It is a far weaker acid than water (pKa =15.7) or the alcohols (pKa =16 to 19) but it is much more
acidic than ammonia (pKa =34). A solution of sodium amide in liquid ammonia readily converts acetylene
and other terminal alkynes into the corresponding carbanions.
RCCH + 
2
NH  RCC
+ NH3
This reaction does not occur with alkenes or alkanes. Ethylene has a pKa of about 44 and methane
has a pKa of about 50, which means that they are weaker acid that NH3.

41. All right copy reserved. No part of the material can be produced without prior permission
From the foregoing pKa’s we see that there is a vast difference in the basic character of the
carbanions RCC
, CH2=CH
and 
3
CH . This difference can be explained in terms of the character of the
orbital occupied by the lonepair electrons in the three anions. Methyl anion has a pyramidal structure with
the lonepair electrons in an orbital that is approximately sp3






p
4
3
and
s
4
1
. In vinyl anion, the lonepair
electrons are in an sp2
orbital 





p
3
2
and
s
3
1
. In acetylide ion, the lone pair is in an sp orbital






p
2
1
and
s
2
1
.
C
H
H
H
methyl anion
C C
H
H
H
vinyl anion
sp
3
sp
2
HCC
acetylide ion
sp
Electrons in sorbitals are held closer to the nucleus than they are in porbitals.
This increased electrostatic attraction means that selectrons have lower energy and greater stability then
pelectrons. In general, the greater the amount of scharacter in a hybrid orbital containing a pair of
electrons, the less basic is that pair of electrons and more acidic is the corresponding conjugate acid.
Alternatively, since greater the electronegativity of an atom, the more readily it can accommodate a
negative charge and hence less basic the species would be. Basicities of the following carbanions follow
the order: CHC
< CH2=CH
< 
3
CH and hence the order of acidic strength would be HCCH > CH2=CH2
> CH4.
Of course, the foregoing argument applies to hydrogen cyanide as well. In this case, the conjugate
base, 
CN, is further stabilized by the presence of the electronegative nitrogen. Consequently, HCN is
sufficiently acidic (pKa 9.2) that it is converted to its salt with hydroxide ion in water.
HCN + OH
CN
+ H2O
Alkynes are also quantitatively deprotonated by alkyl lithium compounds, which may be viewed as
the conjugate base of alkanes.
CH3(CH2)3CCH + 

 Li
H
C
n 9
4  CH3(CH2)3CC
Li+
+ nC4H10
These transformations are simply an acidbase reaction, with 1hexyne being the acid and n
butyllithium being the base. Since the alkyne is a much stronger acid than the alkane (by over 20 pK
units!), equilibrium lies essentially completely to the right.
Terminal alkynes give insoluble salts with a number of heavy metal cations such as Ag+
and Cu+
.
The alkyne can be regenerated from the salt and the overall process serves as a method for purifying
terminal alkynes. However, many of these salts are explosively sensitive when dry and should always be
kept moist.
CCH + M+
 CCM + H+
For example,
HCCH + 2[Ag(NH3)2]+
 AgCCAg + 
4
NH
2 Identification of
silver acetylide terminal alkynes
(white ppt.)

42. All right copy reserved. No part of the material can be produced without prior permission
CH3CCH + [Cu(NH3)2]+
 CH3CCCu + 
4
NH + NH3
Cuprous methylacetylide
(red ppt.)
HCCH + Na 

 
 3
. NH
liq
HCC:
Na+
+ ½ H2
Sodium acetylide
CH3CHCCH + NaNH2
CH3
ether
CH3CHCC:
Na+
+ NH3
CH3
Sodium isopropylacetylide
3.12 DIELS-ALDER REACTION
This is a reaction involving cyclo addition of alkene or alkyne, commonly referred to as the
dienophile with a conjugated diene system. The product formed in the reaction is usually a six membered
ring and the addition takes place in the 1, 4-position.
C
C
C–R
–C
C
–C
C
C
C
C–R
–C
–C
C
C
C–R
C
C
C
C
C–R
–C
–C
–C
–C
Illustration
Question: Identify (A), (B), (C) and (D) in the following reactions
CCCH3
HCO3H
(A)
CrO3
in AcOH
aq. KMnO4
O3/H2O
(B)
(C)
(D)a mixture
Solution: (i) Peroxy formic acid hydroxylates double bond but not triple bond. Therefore, compound
(A) is CCCH3
OH
HO

43. All right copy reserved. No part of the material can be produced without prior permission
(ii) Oxidation of triple bond is very much slower than that of double bond. Therefore, by
using a suitable oxidising agent it is possible to selectively oxidise double bond in
presence of triple bond. CrO3 in presence of AcOH is one such oxidising agent, which
oxidises double bond and not triple bond. Thus, compound (B) is
COOH COOH
CCCH3
(iii)Aqueous KMnO4 hydroxylates both the double bond as well as triple bond. The
compound (C) is
C CCH3
OH
HO
O O
(iv) Both the double bond and triple bond undergo ozonolysis followed by oxidative
hydrolysis reaction. The product (D) is a mixture of acetic acid and carboxylic acids as
given below
COOH
COOH
COOH + CH3COOH
SOLVED OBJECTIVE EXAMPLES
Example 1:
A sample of 4.50 mg of unknown alcohol is added to CH3MgBr when 1.68 mL of CH4 at
1atm. pressure and 273 K is obtained. The unknown alcohol is
(a) methanol (b) ethanol
(c) 1propanol (d) 1butanol
Solution:
All the four options given are monohydric alcohols. When a monohydric alcohol is treated with
CH3MgBr. The number moles of CH4 gas produced is equal to the number of moles of alcohol.
ROH + CH3MgBr  CH4 + Mg
OR
Br
If M is the molecular weight of alcohol

44. All right copy reserved. No part of the material can be produced without prior permission
4
.
22
68
.
1
M
50
.
4

On solving, M = 60. Therefore, the unknown alcohol is 1propanol.
Example 2:
A compound (X) when passed through dil H2SO4 containing HgSO4 gives a compound (Y)
which on reaction HI and red phosphorous gives C2H6. The compound (X) is
(a) ethene (b) ethyne
(c) 2butene (d) 2butyne
Solution:
The compound (X) is likely to be alkyne which reacts with water in presence of H2SO4 and HgSO4
as catalyst to form a carbonyl compound. HI and red phosphorous can reduce a carbonyl
compound to alkane having same number of carbon atoms.
Therefore (Y) is likely to be acetaldehyde which is the hydration product of ethyne.
HCCH + H2O
H2SO4
HgSO4
[CH2=CHOH] CH3CHO
(X) (Y)
CH3CHO + 4HI
red P
CH3CH3 + 2I2 + H2O
 (X) is ethyne.
Example 3:
When 2methylbutane is monochlorinated, the percentage of (CH3)2CHCH2CH2Cl is
………… assuming reactivity ratio of 3°H : 2°H : 1°H = 5 : 3.8 : 1
(a) 28% (b) 35%
(c) 23% (d) 14%
Solution:
Monochlorination of 2methylbutane gives four products
CH3CHCH2CH3 + Cl2
CH3
h
CH2ClCHCH2CH3 + CH3CClCH2CH3
+ CH3CHCHClCH3 + CH3CHCH2CH2Cl
CH3 CH3
CH3 CH3
(A) (B)
(C) (D)
The reactivity ratio of 3°H : 2°H : 1°H towards chlorination is 5 : 3.8 : 1
Compound Reactivity factor  Probability factor = No. of parts
(A) 1  6 = 6
(B) 5  1 = 5
(C) 3.8  2 = 7.6
(D) 1  3 = 3
Total numbers of parts 21.6
Percentage of 14%


 100
6
.
21
3
D
Example 4:
During debromination of mesodibromobutane, the major compound formed is
(a) nbutane (b) 1butene

45. All right copy reserved. No part of the material can be produced without prior permission
(c) cis2butene (d) trans2butene
Solution:
HCBr
CH3
HCBr
CH3
H H Br
Br
CH3
H3C
H
Br
CH3
H3C
Br
H
H
CH3
H3C
H
trans
–2Br–
In the debromination reaction we have to ensure that the two leaving groups i.e., the two Bratoms
are 180° to each other in the same plane before they are lost. From the above conformations of
mesodibromobutane drawn on Newman projection, it is clear that debromination results in the
formation of trans2butene.
Example 5:
The addition of HCl to 3,3,3trichloropropene gives
(a) Cl3CCH2CH2Cl (b) Cl3CCH(Cl)CH3
(c) Cl2CHCH(Cl)CH2Cl (d) Cl2CHCH2CHCl2
Solution:
CCl3 is a strongly electron withdrawing group. The addition of HCl to C=C double bond of
3,3,3trichloropropene does not follow the Markownikoff’s rule because intermediate secondary
carbonium ion is destabilized by the I effect of CCl3 group.
CCl3CH=CH2
H+
CCl3CHCH3
+
Less stable
Instead the addition follows anti Markownikoff’s rule because primary carbonium ion becomes
relatively more stables.
CCl3CH=CH2
H+
CCl3CH2CH2
+
relatively more
stables
Cl
CCl3CH2CH2Cl
Example 6:
The treatment of propene to Cl2 at 500600°C produces
(a) 1,2dichloropropene (b) allyl chloride
(c) 2,3dichloropropene (d) 1,3dichloropropene
Solution:
At high temperature Cl2 does not add to alkenes. Chlorine molecule undergoes homolytic cleavage
at high temperature forming Cl radicals which initiate substitution reaction at the allylic position.
Therefore, when propene reacts with Cl2 at high temperature, allyl chloride is formed.
CH3CH=CH2 + Cl2
500600°C
CH2ClCH=CH2 + HCl
Example 7:
The main product produced in the dehydrohalogenation of 2bromo3, 3dimethylbutane is
(a) 3, 3dimethylbut1ene (b) 2, 3dimethylbut1ene
(c) 2, 3dimethylbut2ene (d) 4methylpent2ene
Solution:

46. All right copy reserved. No part of the material can be produced without prior permission
CH3CHCCH3  CH3CHCCH3 + Br 

+
CH3
CH3
H+
CH3C=CCH3 CH3CHCCH3
CH3
+
CH3
CH3
CH3
CH3
Br CH3
The reaction follows E1 mechanism. In the first step a secondary C+
ion is formed which rearranges
to give more stable tert C+
ion. In the second step tert C+
ion loses H+
from its adjacent Catom to
give more stable alkene i.e., 2,3dimethylbut2ene.
Example 8:
The intermediate during the addition of HCl to propene in the presence of peroxide
would be
(a) Cl
HCH
C
CH 2
3

(b) 3
3 HCH
C
CH

(c) 2
2
3 H
C
CH
CH

(d) 2
2
3 H
C
CH
CH

Solution:
Addition of HCl to propene follows Markownikoff’s rule because peroxide free radicals are unable
to break HCl bond due to its high bond dissociation energy.
CH3CH=CH2
H+
Cl
+
CH3CHCH3
Cl
CH3CHCH3
The reaction proceeds via ionic mechanism with 3
3 HCH
C
CH

as intermediate.
Example 9:
The product of the reaction,
CH3CHO + HC  CD 


 



Na
O
CH3
is
would be
(a) CH3CHOH + NaC  CD
OCH3
(b) CH3CHC  CD
OH
(c) CH3CHC  CH
OD
(d) H2C=CDCH2CHO
Solution:
HC  CD OH
CH
CD
C
Na 3
Na
O
CH3





 




CH3O
removes H+
from HC  CD and not D+
because CH bond energy is lower than CD. The
anion attacks the Catom of carbonyl group.
CH3C + CCD
H
O
CH3OH
CH3CHCCD CH3CHCCD
O OH
Example 10:
2methyl pent2ene on ozonolysis will give
(a) propanal only (b) propanal and ethanal

47. All right copy reserved. No part of the material can be produced without prior permission
(c) propanone and propanal (d) propan2ol and ethanal
Solution:
CH3
C
CH3C=CHCH2CH3 + O3
O
CH
CH2CH3
O O
CH3
CH3
Zn/H2O
C=O + CH3CH2CHO
CH3
CH3
The products of the above reaction are propanone and propanal.

48. All right copy reserved. No part of the material can be produced without prior permission
SOLVED SUBJECTIVE EXAMPLES
Example 1:
Identify (A) and (B) in the following reactions.
CH3CH2CH2CH3 + Br2
h
(A)
(i) Mg/Ether
(ii) D2O
(B)
Solution: Secondary hydrogen is much more reactive than primary hydrogen towards bromination.
Therefore, (A) is mainly 2bromobutane. The subsequent reaction involves formation of
Grignard reagent which picks up D+
from D2O.
CH3CH2CH2CH3+ Br2
h
(A)
Mg/Ether
D2O
CH3CHCH2CH3
Br
CH3CHCH2CH3
MgBr
(B)
CH3CHCH2CH3
D
Example 2:
Tetrachloroethene (CCl2=CCl2) is non reactive towards Cl2 but addition of AlCl3 makes it
reactive. Explain.
Solution:
The four Clatoms attached to C=C considerably reduce the electron density in ethene molecule
due to I effect of the Cl atoms. As a result, Cl2 does not add on to the tetrachloroethane molecule.
Addition of AlCl3, a Lewis acid, produces a more reactive Cl+
(chloronium ion) by reacting with
Cl2.
AlCl3 + Cl2  
4
AlCl + Cl+
Chloronium ion initiates the addition reaction.
CCl2=CCl2 
 


Cl
2
Cl
C

CCl3 
 


4
AlCl
CCl3CCl3 + AlCl3.
Example 3:
Isobutene dimerises in the presence of conc. H2SO4. What are the possible structures of the
dimer?
Solution:
Conc. H2SO4 protonates isobutene molecule to form carbonium ion.
CH3C=CH2 + H+
(from H2SO4)
CH3
CH3CCH3
CH3
+
The carbonium ion thus formed attacks another isobutene molecule.
CH3C=CH2 + CCH3
CH3
CH3CCH2CCH3
CH3
+
CH3 CH3
CH3
H+
CH3

CH3C=CHCCH3 +
CH3
CH2=CCH2CCH3
CH3 CH3
CH3
CH3
CH3
(A) (B)

49. All right copy reserved. No part of the material can be produced without prior permission
(A) and (B) are the two possible structures of the dimer of isobutene, with (A) being more
stable than (B).
Example 4:
Predict the products in the following sequence of reactions with suitable explanation.
CHCH (A)
Cu2Cl2
NH4Cl
+
H2 + Pd / BaSO4
(B) HCCH

(C)
Solution:
Cu2Cl2 reacts with NH4Cl to form 

Cl
)
NH
(
Cu 2
3 , which causes dimerization of acetylene to
vinylacetylene (A). Hydrogen in presence of palladium supported on BaSO4 (Lindlar’s catalyst)
partially reduces triple bond of vinylacetylene to double bond thus forming 1, 3butadiene (B).
The last reaction involves cyclisation of 1, 3butadiene with acetylene to form
1, 4cyclohexadiene (C). All the reactions are as given below
CHCH
Cu2Cl2
NH4Cl
+ CH2=CHCCH H2 + Pd / BaSO4
CH2=CHC=CH2
CH CH

H2C CH2
CHCH
CH CH
CH2
CH CH
CH2
(C)
Example 5:
Outline the synthesis of
(a)
H
C C
CH3
H
CH2=CH
(trans)
from CHCH(2 moles) and CH3Br
(b) () 2, 3 dibromobutane from 2butyne.
Solution:
(a) Acetylene is first dimerised to vinylacetylene by using a mixture of Cu2Cl2 and NH4Cl. Vinyl
acetylene is treated with NaNH2 followed by CH3Br when the acidic hydrogen of vinyl
acetylene is substituted by CH3 group to get pent1en3yne. Finally pen1en3yne is
partially reduced to get the desired trans isomer.
C C
CH3
H
2CHCH
Cu2Cl2 + NH4Cl
CH2=CH
NaNH2
CH2=CHCCH
NH3
CH2=CHCCNa+
CH3Br
CH2=CHCCCH3
H Na / liq. NH3
pent1en3yne
(trans)
(b) 2butyne is first converted into cis2butene by partial hydrogenation using Lindlar’s catalyst.
Bromination of cis2butene undergoes anti addition and results in  2, 3dibromobutane
C C
CH3
H
CH3CCH3 + H2
PdBaSO4
CH3
H
Br2
C C Br
CH3
H
H3C
H
Br
()

50. All right copy reserved. No part of the material can be produced without prior permission
MIND MAP
ALKENES

Me3COK
CH3CHCH=CH2
CH3
in Me3COH
(Hoffmann prod.)
CH3CHCHCH3
Br
CH3
alc. KOH
CH3C=CHCH3
CH3
(Saytzeff prod.)
Elimination generally occurs by E2 mechanism forming more stable alkene as the
major product (Saytzeff rule). The two leaving groups align themselves 180° to each
other in the same plane before they are lost.
Bulky base removes proton from less crowded Catom forming less stable alkene
as the major product (Hoffmann rule)
 CH3CH2CHCH3
F
alc. KOH
CH3CH2CH=CH2
Alkyl fluoride undergoes elimination from conjugate base as F
is a poor leaving
group forming less stable alkene as the major product.
 CH3C CHCH3
OH
CH3
H
+
CH3C=CCH3
CH3
H2O
CH3 CH3
Reaction follows E1 mechanism forming C+ as intermediate. Rearrangement of C+
takes place if that results in more stable C+ ion.
 RCH(X)CH(X)R
CO
Me
in
Na
or
EtOH
or
AcOH
Zn
2
I
/




 
 RCH=CHR
 cis RCH=CHR
 
B
Ni
or
cat
s
Lindlar
H
2
2 '




 


RCCR 

 
 3
/ LiqNH
Na
trans RCH=CHR
 H
O
CH
CH
N
R



 3
2
3 


R3N + CH2=CH2 + H2O
H
O

removes proton from that carbon atom which produces more stable carbanion.
 R2C=O + 3
2 Ph
P
)
R
(
C


  R2C=CR2 + Ph3P=O

RCH=CH2 Ni / Pt / Pd
+H2
RCH2CH3
H+
/ H2O
+HX
RCH(OH)CH3
RCH(X)CH3
The addition of polar molecules follows Markownikoff’s rule with the possibility of
rearrangement.
 RCH=CH2 + HBr 

 
peroxide
RCH2CH2Br (Anti Markownikoff’s rule)
 RCH=CH2 + H2O 

 
 2
)
(OAc
Hg

 
 4
NaBH
RCH(OH)CH3 (Markownikoff’s addn.)
 RCH=CH2 + B2H6  


 


H
O
or
O
H 2
2
RCH2CH2OH (Anti Markownikoff’s addition)
 RCH=CH2 + X2  RCH(X)CH2(X) (anti addition)
 RCH=CH2
Or OsO4 / OH
RCHCH2
(i) RCO3H
RCHCH2 (ii) H3O+
OH
OH
aq. KMnO4
OH OH

51. All right copy reserved. No part of the material can be produced without prior permission
MIND MAP
ALKYNES
 2
|
|
H
C
H
C
R
X
X

 or RCH2CHX2 

 

KOH
.
alc
RCH=CHX 

 
 2
NaNH
RCCH
RCCH
Na
RCCNa
–½ H2
NaNH2
NH3
RCCH
 +
RX
(1° only)
RCCR


 RCX2CHX2
O
C
Me
in
Na
or
EtOH
or
AcOH
Zn





 

2
I
/
RCCH
 RCCH + H2 


 
 Pd
or
Pt
or
Ni
RCH=CHR 



 

Pd
or
Pt
or
Ni
H ,
2
RCH2CH3
 RCCR + Na / liq. NH3  trans RCH=CHR
 RCCR + H2 


 
 .
' cat
s
Lindlar
cis RCH=CHR
 RCCR + HX  RCX=CH2 

HX
RCX2CH3
 RCCH + HBr 

 
Peroxide
RCH=CHBr
Peroxide
HBr


 RCH2CHBr2
 RCCH + H2O
4
4
2
.
HgSO
SO
H
dil


 
 3
||
CH
C
R
O


 RCCH + 2HOX  2
||
X
CH
C
R
O


 RCCR






 

H
ii
OH
KMnO
i
)
(
/
)
( 4
RCOOH + RCOOH
 RCCR
O
H
)
ii
(
O
)
i
(
2
3
RCOOH + RCOOH.

52. All right copy reserved. No part of the material can be produced without prior permission
1. Thermal decomposition of
a) b) c) d)
2. Which of the following is not a petroleum product?
a) Petrol b)Paraffin wax c) Bees wax d)Kerosene
3. A knocking sound is produced more in the engine when the fuel contains mainly:
a) 𝑛-alkanes b)CO2 c) CO d)Lubricating oil
4. Reaction of HBr with propene in presence of peroxides gives:
a) Isopropyl bromide b)3-bromopropane c) Allyl bromide d)𝑛-propyl bromide
5. The next higher homologue of C6H14 is:
a) C7H14 b)C7H16 c) C7H10 d)C7H12
6. The reaction conditions used for converting 1,2-dibromopropane to propylene are
a) KOH, alcohol ∆
⁄ b) KOH, water ∆
⁄ c) Zn, alcohol ∆
⁄ d) Na, alcohol ∆
⁄
7. A gas formed by the action of alcoholic KOH on ethyl iodide, decolourises
alkalineKMnO4. The gas is
a) C2H6 b) CH4 c) C2H2 d) C2H4
8. Alkyne, C7H12, when reacted with alkaline KMnO4 followed by acidification with HCl gives a
mixture of (CH3)2CHCOOH + CH3CH2COOH,The alkyne 𝐶7𝐻12 is
a) 3-hexyne b)2-methyl-2-hexene c) 2-methyl-3-hexene d)3-methyl-2-hexyne
9. The relationship between acetylene and benzene is comparable to the relationship
between propyne and
a) Dimethyl benzene b)Neoprene c) Propyl benzene d)Mesitylene
10. Complete oxidation of one mole of an alkane forms 3 moles ofCO2. The alkane is
a) CH4 b)C2H6 c) C3H8 d)C6H14
11. The ozonolysis of ethylene, acetylene and propylene respectively gives:
a) HCHO, CHO— CHO and CH3CHO + HCHO
b)CHO— CHO, HCHO and CH3CHO
c) HCHO + CH3CHO, CHO— CHO and HCHO
d)CHO— CHO, CH3CHO + HCHO and HCHO
12. The reaction, CH2 = CH2 + CH3COCl
AlCl3
→ gives the product:
a) CH3COCH2CH2Cl
b)CH3. CH2. CH2Cl
c) CH3COCH2. CH2COCH3
d)ClCH2CH2Cl
13. Alkyl halides react with dialkyl copper reagents to give
a) Alkenyl halides b)Alkanes
c) Alkyl copper halides d)Alkenes
14. The gas which is used for the artificial ripening of fruits is:
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 1

53. All right copy reserved. No part of the material can be produced without prior permission
a) C2H6 b)C2H2 c) C2H4 d)Marsh gas
15. CH3— C ≡ CH reacts with HCI to give:
a) 2,2-dichloropropane b)1,1-dichloropropane c) 1,2-dichloropropane d)1-chloropropene
16.
CH3CH3 + HNO3
675 K
→ ?
a) CH3CH2NO2 b) CH3CH2NO2 + CH3NO2
c) 2CH3NO2 d) CH2 = CH2
17. Which of the following is produced when coal is subjected to destructive distillation?
a) Methane b)Ethane c) Acetylene d)Coal gas
18. The product of the following reaction are:
a) CH3COOH + CH3COCH3
b)CH3COOH + CH3CH2COOH
c) CH3CHO + CH3CH2CHO
d)CH3COOH + CO2
19. Methyl bromide heated with zinc in closed tube produces:
a) Methane b)Ethane c) Ethylene d)Methanol
20. Aqueous solution of an organic compound, ′𝐴′ on electrolysis liberates acetylene and
CO2 at a node. ′𝐴′ is
a) Potassium acetate b)Potassium succinate
c) Potassium citrate d)Potassium maleate
21. The reaction of alkanes with halogen is explosive in the case of:
a) F2 b)Cl2 c) I2 d)Br2
22. Which of the following is unsymmetrical alkene?
a) 1-butene b)2-hexene c) 1-pentene d)All of these
23. Which of the statement is wrong for alkanes?
a) Most of the alkanes are soluble in water
b)Their density is always less than water
c) At room temperature some alkanes are liquid, some solid and other are gases
d)All alkanes burn
24. Propane cannot be prepared from which reaction?
a) CH3 − CH = CH2
B2H6
→
OH−
b) CH3CH2CH2I
HI
→
P
c) CH3CH2CH2COONa
NaOH CaO,∆
⁄
→
d)None of the above
25. Nitrating mixture is
a) Fuming nitric acid
b)Mixture of conc. H2SO4 and conc. HNO3
c) Mixture of nitric acid and anhydrous zinc chloride
d)None of the above
26. Cyclohexene on reaction with OsO4 followed by reaction with NaHSO3 gives
a) 𝑐𝑖𝑠 − diol b) 𝑡𝑟𝑎𝑛𝑠 − diol c) Epoxy d)Alcohol
27. Al4C3 on hydrolysis yields

54. All right copy reserved. No part of the material can be produced without prior permission
a) Nitrogen gas b)Methane gas c) Hydrogen gas d)Carbon dioxide
28. The compounds 𝑃, 𝑄and𝑆
where separately subjected to nitration using HNO3 H2SO4
⁄ mixture. The major product
formed in each case respectively, is
a)
b)
c)
d)

55. All right copy reserved. No part of the material can be produced without prior permission
29. Which of the following is not a mixture of hydrocarbons?
a) Candle wax b)Kerosene c) Vegetable oils d)Paraffin oil
30. C8H10(𝐴)
O3/H2O
→ acid(𝐵)
C3H5MgBr(C)
CO2,H3O+
→ acid𝐵
Identify 𝐴, 𝐵 and 𝐶
a) b)
c)
CH3 − CH2 − CH2
− CH3, CH3CH2CH2COOH, CH2
= CH − CH2MgBr
d)
31. Which of the following has the maximum heat of hydrogenation?
a) b) c) d)
32.
CH3CH2CH3
400−600℃
→ 𝑋 + 𝑌, 𝑋and𝑌are
a) Hydrogen and methane b)Hydrogen and ethylene
c) Ethylene and methane d)Any of these
33. Position of double bond in alkenes is identified by
a) Ozonolysis b)Bromine water
c) Ammonical silver nitrate d)None of these
34. Consider the following reaction
I. H2/Ni2B
II. H2/Pd − CaCO3 in quinoline
III. Na/NH3 or LiAIH4
This reaction takes place by
a) I or II b)I or III c) II or III d)I, II or III
35. Which of the following reagent can distinguish between 1-butyne and 2-butyne?
a) Aqueous NaOH
b)Bromine water
c) Fehling’s solution
d)Ammoniacal AgNO3
36. CH4 is formed when:
a) Sodium acetate is heated with soda lime
b)Iodo methane is reduced
c) Aluminium carbide reacts with water
d)All of the above
37. Reaction of HBr with propene in the presence of peroxide gives

56. All right copy reserved. No part of the material can be produced without prior permission
a) 𝑖𝑠𝑜-propyl bromideb)3-bromo propane c) Allyl bromide d)𝑛-propyl bromide
38. Predict structure of 𝑋 in following reaction
a) b) c) d)
39. The middle oil fraction of coal-tar distillation contains:
a) Benzene b)Anthracene c) Naphthalene d)Xylene
40. On halogenation, an alkane (C5H12) gives only one monohalogenated product. The alkane is
a) 𝑛-pentane b)2-methyl butane
c) 2, 2-dimethyl propane d)Cyclopentane
41. Acrylic emulsion in paints is a polymer of:
a) CH2 = CH − COOCH3
b)CH3 − CH = CH − COOCH3
c) CH2 = CH − COOH
d)CH2 = C(CH3) − COOCH3
42. A hydrocarbon X adds on one mole of hydrogen to give another hydrocarbon and decolourised
bromine water. X react with KMnO4 in presence of acid to give two mole of the same carboxylic
acid. The structure of X is:
a) CH3CH = CHCH2CH2CH3
b)CH3CH2CH = CHCH2CH3
c) CH3CH2CH2— CH = CHCH3
d)CH2 = CH— CH2CH2CH3
43. An anaesthetic narcylene is commercial name of:
a) C2H4 b)C2H2 c) CHCI3 d)ether
44. By which one of the following compounds both CH4 and CH3 − CH3 can be prepared in
one step?
a) CH3I b) CH3OH c) CH3CH2I d) C2H5OH
45. What volume of methane (NTP) is formed from 8.2 g of sodium acetate by fusion with sodalime?
a) 10 litre b)11.2 litre c) 5.6 litre d)2.24 litre
46. When methyl iodide is treated with sodium in ethereal solution, it gives
a) Methane b)Ethane
c) Methyl sodium iodide d)Sodium methoxide
47. 2-methylpentene 2 on ozonolysis will give:
a) Only propanal
b)Propanal and ethanal
c) Propanone-2 and ethanal
d)Propanone-2 and propanal
48. The reaction,
a) Eglinton’s reaction

57. All right copy reserved. No part of the material can be produced without prior permission
b)Glaser reaction
c) Gomberg-Beckmann’s reaction
d)Leuckart reaction
49. 2-Hexyne gives 𝑡𝑟𝑎𝑛𝑠-2-hexene on treatment with:
a) Li/NH3 b)Pd/BaSO4 c) LiAlH4 d)Pt/H2
50. Which of the following will give three mono-bromo derivatives?
a) CH3CH2CH2CH(CH3)CH3 b)CH3CH2C(CH3)2CH3
c) CH3CH3(CH3)CH (CH3)CH3 d)All the above can give
1) a 2) c 3) a 4) d
5) b 6) c 7) d 8) c
9) d 10) c 11) a 12) a
13) b 14) b 15) a 16) b
17) d 18) b 19) b 20) d
21) a 22) d 23) a 24) a
25) b 26) a 27) b 28) c
29) c 30) a 31) c 32) c
33) a 34) a 35) d 36) d
37) d 38) d 39) c 40) c
41) a 42) b 43) b 44) a
45) d 46) b 47) d 48) a
49) c 50) b
1 (a)
The formation of the alkene in an elimination reaction is called Hofmann elimination
(Thermal decomposition). Elimination of hydrogen occurs from the β-carbon. So,
2 (c)
Bees wax is myricyl palmitate, 𝑖. 𝑒., C15H31COOC30H61.
3 (a)
The knocking order is:
Straight > branched >olefins>arenes.
chain alkane chain alkane
4 (d)
Follow peroxide effect.
5 (b)
Successive homologous differ by —CH2gp.
6 (c)
1, 2-dihalogen (𝑣𝑖𝑐𝑖𝑛𝑎𝑙) derivatives of the alkanes on reaction with zinc dust and
methanol produces alkenes by loss of two halogen atoms (dehalogenation).
CH3 − CH − CH2 + Zn
Alcohol ∆
⁄
→ CH3CH = CH2
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 1 (ANSWERS)

58. All right copy reserved. No part of the material can be produced without prior permission
| | propylene
Br Br
1,2-dibromopropane
7 (d)
Ethylene is formed by dehydrohalogenation of alkyl halide in presence of alcoholic
KOH. Ethylene decolourise alkaline KMnO4 due to get oxidized by it.
CH3 − CH2I
Alc.KOH
→ CH2 = CH2
Ethylene
8 (c)
(CH3)2CH − C ≡ C − CH2CH3
[O]
→ (CH3)2CH2COOH + CH3CH2COOH
9 (d)
Benzene is obtained by the polymerisation of acetylene,. Similarly, mesitylene is
obtained by the polymerisation of propyne.
10 (c)
C3H8 + 5O2 → 3CO2 + 4H2O
11 (a)
Follow cleavage of two bonds at multiple bonding position during ozonolysis.
12 (a)
CH2 = CH2 + CH3COCl
AlCl
→ CH3COCH2CH2Cl.
13 (b)
It is a Corey House synthesis of alkanes.
14 (b)
C2H2is used for artificial ripening of fruits. C2H4 for natural ripening.
15 (a)
Follow Markownikoff’s rule for addition.
16 (b)
Ethane gives a mixture of nitroethane and nitromethane.
CH3 − CH3 + HNO3
Ethane
673 K
→
−H2O
CH3 − CH2 − NO2 + CH3NO2
nitro ethane (minor)
(major)
During nitration chain fission of alkanes also takes place, so CH3NO2 is also obtained
along with CH3CH2NO2.
17 (d)

59. All right copy reserved. No part of the material can be produced without prior permission
Coal gives coal gas.
19 (b)
Frankland reaction: 2CH3Br
Zn
→ C2H6.
20 (d)
CHCOOKCH
||
Electrolysis
→ ||| + 2CO2 + 2KOH + H2
CHCOOKCH cathode
Potassium maleate acetylene anode
21 (a)
F2reacts violently even in dark.
22 (d)
𝑒. g., CH3CH2CH = CH2 is unsymmetrical. CH3CH = CHCH3 is symmetrical. Note the
positions of carbon atoms on two sides of double bond.
23 (a)
Due to non-polar nature, alkanes are insoluble in water because water is a polar
solvent.
24 (a)
(a)CH3 − CH = CH2
B2H6
→
(CH3 − CH2 − CH2)3B
OH−
→ CH3CH2CH2OH
Hydroboration of alkenes followed by hydrolysis in basic medium yield alcohol.
(B)CH3 − CH2 − CH2I
HI P
⁄
→ CH3 − CH2 − CH3
propane
Reduction of alkyl halides yield alkane.
(c)CH3CH2CH2COONa + NaOH
CaO
→ CH3CH2CH3 + Na2CO3
Propane
Decarboxylation of sodium salt of fatty acid yield alkane having one carbon atom
less than parent acid salt.
25 (b)
Nitrating, mixture is conc. HNO3 + conc. H2SO4.
It produces NO2
+
electrophile which carried out electrophilic substitution reaction.
26 (a)
OsO4 is a valuable oxidising agent. It oxidises alkenes to give 𝑐𝑖𝑠 − diols.

60. All right copy reserved. No part of the material can be produced without prior permission
27 (b)
Al4C3 on hydrolysis gives methane gas.
Al4C3 + 12H2O → 4Al(OH)3 + 3CH4
29 (c)
Vegetable oils are esters of glycerol or glycerides.
31 (c)
As the conjugation increases, heat of hydrogenation decreases. Thus, alkene (c) with
two isolated double bonds has the highest heat of hydrogenation.
32 (c)
CH3CH2CH3
400−600℃
→ CH2 = CH2 + CH4
(𝑋)(𝑌)
33 (a)
The position of the double bond in alkene is identified by ozonolysis. Bromine water is used
to detect the presence of π-bond whereas ammoniacal silver nitrate AgNO3 is used to detect
the presence of terminal alkynes or – CHO group
34 (a)
While with Na/NH3 or LiAlH4, 𝑡𝑟𝑎𝑛𝑠 alkene is obtained, 𝑖𝑒, 𝑎𝑛𝑡𝑖-addition product
35 (d)
H3C − CH2C ≡ CH
AgNO3
→
NH4OH
CH3CH2C ≡ CAg
(1-butyne) (silver-1 butynide)
H3C − C ≡ C − CH3
AgNO3
→
NH4OH
No reaction
2-butyne
36 (d)
CH3COONa
Soda lime
→ CH4

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Al4C3 + 12H2O ⟶ 4Al(OH)3 + 3CH4
CH3I
2H
→ CH4 + HI.
37 (d)
Reaction of HBr with propene in the presence of peroxide gives 𝑛-propyl bromide.
This addition reaction is an example of 𝑎𝑛𝑡𝑖-Markownikoff’s addition reaction.
(𝑖. 𝑒., it is completed in form of tree radical addition.)
CH3 − CH = CH2 + HBr
Peroxide
→ CH3 − CH2 − CH2Br
𝑛-propyl bromide
38 (d)
Friedel-Craft reaction proceeds 𝑣𝑖𝑎 most stable carbocation
39 (c)
Follow text.
41 (a)
The polymer is
42 (b)
Symmetrical alkenes on ozonolysis give same product during ozonolysis.
43 (b)
C2H2 is commercially named narcylene.
44 (a)
CH3I + 2H
Zn+HCl
→
or Zn−Cu/C2H5OH
CH4 + HI
methane
CH3I + 2Na + ICH3
Dry ether
→ CH3 − CH3 + 2NaI
ethane
45 (d)
CH3COONa + NaOH
CaO
→ CH4 + Na2CO3
82 g CH3COONa gives 22.4 litre CH4.
46 (b)
2CH3I + 2Na
Ether
→ C2H6 + 2NaI
48 (a)
It is the name of reaction.
49 (c)
Na/Liq. NH3or LiAlH4 reduce hex-2-yne to trans-hex-2-ene.
50 (b)
The number of di-and poly-halogenation products depends upon (i) and the number of
different types of hydrogens present in an alkane and (ii) the number of halogens
introduced

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