Hydrocarbons

Basic concepts of chemistry, alkanes, alkenes, alkynes, benzene, their preparation methods, properties and uses are explained. Isomerism in alkanes and alkynes also discussed.

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HYDROCARBONS
INTRODUCTION
Compounds containing only two elements carbons & hydrogen are known as hydrocarbons. On the
basis of structure & properties, hydrocarbons are divided into two main classes viz. aliphatic & aromatic.
Aliphatic hydrocarbons are further divided into different families namely alkanes, alkenes, alkynes and
their cyclic analogs (cycloalkanes, cycloakenes and cycloalkynes etc.)
ALKANES:
Alkanes are open chain (acyclic) hydrocarbons comprising the homologous series with the general
formula
n 2n 2
C H where ‘n’ is an integer. They have only single bonds & therefore are said to be saturated.
Because of their low chemical activity, alkanes are also known as Paraffins.
NOMENCLATURE:
According to IUPAC system of nomenclature open chain aliphatic compounds are named as al-
kanes.
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
ethane
propane
2,2-Dimethyl propane
H
5
C
2
CH
3
CH
3
CH
3
CH
3
CH
3
2,2,4,4-tetramethylhexane
CH
3
CH
3
CH
3
C
2
H
5
3-ethyl-2-methylpentane
CH
3
CH
3
CH
3
CH
3
CH
3
3,4,7-trimethylnonane
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
2, 3-dimethylpentane 2,5, 6-trimethyloctane
Names of several common alkyl groups

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CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
isopropyl isobutyl sec-butyl
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
tert butyl neopentyl
CONFORMATIONS OF ALKANES
The different arrangement of atoms in space that results from the free rotation of the groups about C –
C bond axis are called conformers or conformational isomers and the phenomenon is known as
conformational isomerism.
Conformations of Ethane
Of the infinite number of possible arrangements of ethane two conformations represent the extremes.
These are called the eclipsed conformation (I) and the staggered conformation (II). Any other arrangement
which will be between these two extreme positions are known as Gauche or Skew form.
Sawhorse Projection
H
H H
H
H
H
H
H
H
H
H
H
Eclipsed form (I) Staggered form (II)
Newman Projection
H
H H
H
HH
H
H H
H
H
H
Eclipsed form (I) Staggered form (II)
dihedral angle = 0 dihedral angle= 180°
Relative Stabilities of the conformations of Ethane
The potential energy of staggered form is minimum and that of eclipsed form is maximum for ethane.
The difference in energy content between the eclipsed and staggered form is 12 kJ mol
–1
. This small
barrier to rotation is called Torsional Barrier. This energy is not large enough to prevent the rotation.
Hence the conformers keep on changing from one form to another. The eclipsed conformation is least
stable where as the staggered conformation is most stable.

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H
H H
H
HH
H
H H
H
H
H
Eclipsed
12 kJ mol
-1
Staggered
H
H H
H
HH
Staggered
Rotation
PE
Conformations of n-Butane
n-Butane molecule can be considered as a dimethyl derivative of ethane in which one H-atom of each
carbon atom is replaced by a methyl group as shown below:
CH
3
C
H
H
C
H
H
CH
3
1 2 3 4
If one of these central carbon atom (C
2
or C
3
) is fixed and the other is rotated round the C
2
– C
3
bond, the
following important conformations are obtained.
CH
3
H H
H
HCH
3
CH
3
H H
H
CH
3
H
Staggered (Anti) (I)
CH
3
H H
CH
3
HH
Eclipsed (II) Gauche (III)
CH
3
H H
H
H
CH
3
Fully eclipsed (IV)
CH
3
H H
H
CH
3
H
Gauche (V) Eclipsed (VI)
CH
3
H
H
CH
3
H
H

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Relative stability of the conformation of n-Butane
Anti  Gauche  Eclipsed  Fully eclipsed
CH
3
H H
H
HCH
3
16 kJ mol
-1
CH
3
H H
CH
3
HH
Rotation
PE
CH
3
H H
H
CH
3
H
CH
3
H H
CH
3
HH
CH
3
H H
H
CH
3
H
CH
3
H H
H
H
CH
3
CH
3
H
H
CH
3
H
H
3.8 kJ mol
-1
19 kJ mol
-1
3.8 kJ mol
-1
16 kJ mol
-1
0 60° 120° 180° 240° 300° 360°
PREPARATION OF ALKANES:
Reduction of Alkenes and Alkynes
Reduction of alkenes: alkenes are reduced into alkanes by hydrogen in the presence of catalyst [Pt,
Pd, Ni/S.
2
PtO]
   
catalyst
2 2 2
R — CH CH — R H R — CH — CH — R
When catalyst is Ni the reaction is known as Sabatier - Sanderen’s reaction
Reduction of alkynes: Alkynes also reduced by hydrogen in presence of catalyst.
   
catalyst
2 2 2
R — C C — R 2H R — CH — CH — R
First alkene forms thereafter alkane.
Reduction of alkyl halides:
Alkyl halides undergo reduction with nascent hydrogen to form alkanes
R X 2H R H HX    
Alkyl halide Alkane
The yields are generally good and the hydrocarbons obtained are pure. The nascent hydrogen for reduc-
tion may be obtained by using any one of the following :
(a) Zinc and dilute hydrochloric acid
(b) Zinc and acetic acid; Zn and NaOH
(c) Zinc-copper couple in ethanol
(d) Red phosphorus and hydrogen iodide

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(e) Al-Hg in ethanol
2 5
Zn HCL
3 4 orZn Cu / C H OH
CH I 2H CH HI


  
2 5
Zn HCL
2 5 2 6 orZn Cu / C H OH
C H I 2H C H HI


  
0
Re dP
3 2 3 3 2 150 C
CH CH Br 2HI CH CH HBr I   
The purpose of use of red phosphorus is to remove iodine. alkyl halides may also be reduced catalyti-
cally by hydrogen using catalyst like nickel, palladium, platinum etc.
Pd
2
R X H R H H X    
Primary and secondary alkyl halides may be conveniently reduced with lithium aluminium hadride in a
dry organic solvent
Organic
4 4 3Sovent
4RX LiAlH 4RH LiAlX (LiX AlX )   
4
LiAlH is not useful for
0
3 alkyl halide which is converted into alkenes. In such case
4
NaBH or TPH
is used.
3
CH
3
CH
3
CH
C
Cl
4
LiAlH

2
CH
3
CH
3
CH
C
Cl
4
NaBH


0
3 2-methylpropane
Alkyl halides
 
0 0 0
1 ,2 ,3 can also be reduced to alkanes with TPH (Triphenyl tin hydride,
3
Ph SnH).
Reduction of alcohols: Alcohols can be reduced to corresponding alkane as


2
P/I /
R — OH R — H

2
P / I /
3 2 3 2
CH CH OH CH CH I

   
Wurtz Reaction: When alkyl halide is treated with sodium in presence of dry ether, higher alkane is
formed.
Na
dry ether
R — X X — R R — R 2NaX  
      R — X R — X R — R R — R R — R
4
CH Cannot be prepared by this method.

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3° R—X do not give this reaction
When a mixture of two different alkyl halides is used a mixture of three alkanes is obtained
   
Ether
3 2 5 3 2 5
CH Br 2Na BrC H CH C H +2NaBr
Ether
3 3 3 3
CH Br 2Na BrCH CH CH 2NaBr    
Ether
2 5 2 5 2 5 2 5
C H Br 2Na BrC H C H C H 2NaBr    
The separation of the mixture into individual members is not always easy. Thus Wurtz reation is not
suitable for the synthesis of alkanes containing odd nubmer of carbon atoms but the method is useful for
the preparation of symmetrical alkanes
Mechanism : .
Mechanism:
R — X Na R NaX  
R R R — R 
Frankland’s reaction
This method is similar to Wurtz reaction. When alkyl halides is heated with zinc in inert solvent higher
alkanes are formed.

  
2 5
Zn/C H OH
2
R — X R — X R — R ZnX

  
2 5
Zn/C H OH
3 3 3 3 2
CH — Br CH — Br CH — CH ZnBr
Decarboxylation of Salts of Carboxylic Acids
Sodium salt of carboxylic acids undergo decarboxylation when heated with sodalime.
R– COONa + NaOH

  
CaO
2 3
R H Na CO
CH
3
—COONa + NaOH

 
CaO
4 2 3
CH Na CO
Mechanism
OH R C
O
O
R
O
O
OH R  HO C
O
O
2
3
CO R H

  
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. An alkyl halide (R–X) which may be primary, secondary
or tertiary 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 halide
(R—X) which must be preferably primary. The reaction follows S
N
2 mechanism. With secondary alkyl
halides the reaction leads to partly substitution forming alkane and partly elimination forming alkene,
with tert alkyl halide only elimination takes place.
  
dry ether
R — X 2Li R — Li LiX
  
2
2R — Li CuI R CuLi LiI

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    
2
R CuLi R — X R — R R — Cu LiX
Li CuI
3 3
Methyl bromide Methyllithium
CH Br CH Li    CH
3
CuLi
CH
3
Lithium dimethylcopper
2 5
C H Br
3 2 3
Pr opane
CH CH CH  
CH
3
Cl
CH
3
sec-Butyl chloride
Li CuI
 
(CH
3
CH
2
CH-)
2
CuLi
CH
3
CH
3
Br
n-Pentyl bromide
CH
3
CH
3
CH
3
3-methyloctane
Illustration - 1 : Prepare 2-methylbutane from chloroethane and 2-chloropropane using Corey-House
synthesis.
Solution:
3 2
CH CH Cl1. Li
3 3 2 3 2 32. cuI
CH CHCl (CH CH) CuLi CH CHCH CH 
3
CH
3
CH
3
CH
2-Chloropropane Lithiumdiisopropyl cuprate
From organo metallic compounds: Grignard reagents and alkyl lithium reacts with water and other
compounds having acidic hydrogen to give hydrocarbon corresponding to alkyl group of the organo
metallic compounds.
 R — Mg — X H — OH R — H
 R — Li H — OH R — H
Kolbe Electrolysis
Alkanes can be prepared by the electrolysis of a concentrated solution of sodium or potassium salt of
carboxylic acid.
2
H O electrolysis
2 2
R —C — O— K R — C— O — K R — R 2CO H 2KOH    
O O

4
CHcan not be prepared by this method.

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–
–e
Anode (Oxidation)
R C
O
O R C
O
O
2
R CO 
R R R — R 
PHYSICAL PROPERTIES
Physical State
i) All are colourless and possess no characteristic odour.
ii) Lower alkanes (C
1
to C
4
) are gases, middle one (C
5
to C
17
) are liquids and highers are solids.
Boiling Points
i) The boiling point of alkanes increases with increase in molecular weight due to increase in van der
Waals forces with increase in molecular weight i.e.Boiling point: pentane  hexane  heptane
ii) Also the branching in alkanes gives a decrease in surface area (as the shape approaches to spherical)
which results in decrease in van der Waals forces. The boiling point of isomeric alkanes show the
order: pentane  isopentane  neopentane.
Melting Points
The melting points of alkanes do not show a regular trend. Alkanes with even number of carbon atoms
have higher melting point than their adjacent alkanes to odd number of carbon atoms.
Melting order:
   
 
 
( 172.0 C) ( 182.5 C)
( 187.7 C)
propane ethane methane
The abnormal trend in melting point is probably due to the fact that alkanes with odd carbon atoms
have their end carbon atom on the same side of the molecule and with even carbon atom alkane, the
end carbon atom on opposite side. Thus alkanes with even carbon atoms are packed closely in crystal
lattice to permit greater intermolecular attractions.
Density: The density of alkanes increases with increase in molecular weight and becomes constant at
0.76g/ml. Thus all alkanes are lighter than water.
Solubility:
i) Alkanes being non polar and thus insoluble in water but soluble in non polar solvents
e.g. C
6
H
6
, CCl
4
, ether etc.
ii) The solubility of alkanes decreases with increase in molecular weight.
iii) Liquid alkanes are themselves good, non polar solvents.
CHEMICAL PROPERTIES OF ALKANES:
Halogenations: One of the important chemical reactions of alkanes is halogenation which is defined
as the replacement of hydrogen atom of the molecule by halogen atom. A mixture of alkanes and
chlorine remains unaffected in the dark but a vigorous reaction occurs when it is exposed to light or at
a higher temperature.

  
hv or
2
R — H Cl R — Cl HCl
As pointed above the reactions takes place by free radical chain mechanism which proceeds in the
following three distinct steps.
a)Chain initiations
hv or
Cl—Cl 2Cl [ H 58 kcal/mole]

   
b)Chain propagation
3 3
Cl H — CH H —Cl CH [ H –1kcal/mole]    

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3 3
CH Cl—Cl CH — Cl Cl [ H –23kcal/mole]    
c)Chain termination
Cl— Cl Cl—Cl
3 3
CH Cl CH — Cl 
3 3 3 3
CH CH CH —CH 
The ease of substitution at various carbon atoms is of the order tertiary > secondary > primary which
is the same as the stability of various alkyl radicals.
Bromination of alkanes has close similarities to chlorination except the rate of reaction is slow.
Fluorination of alkane is so violent that it results in the cleavage of C—C as well as C—H bonds.
 Reactivity of   
2 2 2 2 2
X :F Cl Br I
The potential energy curve for the halogenation (chlorination) of alkane is shown as
Potential energy
Progress of reaction
R H Cl

 
2R Cl


R Cl Cl

 
act
E
REACTIVITY SELECTIVITY PRINCIPLE:
In more complex alkanes, the abstraction of each different kind of H atom gives a different isomeric
product. Three factors determine the relatives yields of isomeric products are
1.Probability factor: This factor is based on the number of each kind of H atom in the molecule. For
example, in CH
3
CH
2
CH
2
CH
3
there are six equivalent 1° H’s and four equivalent 2° H’s. The ratio of
abstracting a 1°H are thus 6 to 4, or 3 to 2.
2.Reactivity of H
*
: The order of reactivity of H 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 closer to those expected from the probability factor.

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2
o
Cl /hv
25 C
 
CH
3
CH
3 CH
3
Cl CH
3
CH
3
Cln-butane
28%
72%
CH
3
CH
3 CH
3
CH
3
CH
3
Cl CH
3
CH
3
CH
3
Cl
36%64%
2
o
Cl /hv
25 C

CH
3
CH
3
CH
3
Br
CH
3
CH
3
Br


2
o
Br , light
127 C

2%
98%
CH
3
Br
CH
3
CH
3
CH
3
CH
3
Br
CH
3
CH
3
CH
3
2
o
Br , light
127 C

trace over 99%
The rate of abstraction of hydrogen atoms is always found to follow the sequence 3°  2°  1°. At room
temperature, for example, the relative rate per hydrogen atom are 5.0 : 3.8: 1.0. Using these values we
can predict quite well the ratio of isomeric chlorination products from a given alkane. For example:
CH
3
Cl
CH
3
CH
3
Cl
2
o
Cl /hv
25 C

CH
3
CH
3

 

o o
o o
n butylchloride no.of 1 H reactivity of 1 H
sec butylchlorideno.of 2 H reactivity of 2 H
  
6 1.0 6 28%
equivalent to
4 3.8 15.2 72%
Inspite of these differences in reactivity, chorination rarely yields a great excess of any single isomer.
The same sequence of reactivity, 3°  2°  1°, is found in bromination, but with enormously larger
reactivity ratios. At 127°C, for example, the relative rates per hydrogen atom are 1600: 82:1. Here,
differences in reactivity are so marked as vastly to outweight probability factors. Hence bromination
gives selective product.
In bromination of isobutene at 127°C,

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 

o o
o o
Isobutylbromide no.of 1 H reactivity of 1 H
tert butylbromideno. of 3 H reactivity of 3 H
 
9 1
1 1600

9
1600
Hence, tert-butyl bromide happens to be the exclusive product (over 99%).
How many monochlorinated products are obtained by the chlorination of
isopentane and what is the percentage of each assuming the reactivity ratio of 3°H : 2°H : 1°H
= 5 : 3.8 : 1.
Solution:Isopentane on chlorination in presence of sun light gives four different monochlorinated products.
hv
3 2 3 2 2 2 3
CH –CH –CH –CH Cl CH Cl –CH –CH –CH  
3
CH
3
CH(A)
3 2 3
CH —CCl— CH —CH 
3
CH
3
CH
(B)
*
3 3
CH —CH —CHCl—CH
(C)
3
CH
3 2 2
CH—CH—CH—CH Cl
(D)
A and C are optical isomers.
Product Reactivity
Factor
 Probability
factor
= Number
of parts
Percentage
(A) 1  6 = 6 27.8%
(B) 5  1 = 5 23.2%
(C) 3.8  2 = 7.6 35.2%
(D) 1  3 = 3 13.8%
Total number of
parts
= 21.6
Nitration
The displacement of an atom of hydrogen by a nitro group (
2
–NO) is called nitration. Alkanes can be
nitrated with nitric acid in the gas phase generally at 400–500°C.

  
400–500 C
2 2 2
R — H HO — NO R — NO H O

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2
HNO
3 2 3 400–500 C
CH — CH — CH


3 2 2 2 3 3
CH — CH — CH — NO CH — CH — CH
2
NO
3 2 2 3 2
CH — CH — NO CH — NO
Sulphonation
Sulphonation of a hydrogen atom of the alkane by sulphuric acid group
3
(–SO H) is called sulphonation.
The ease of substitution is tertiary  secondary > primary..
2
H OCH
3
CH
3
CH
3
H CH
3
CH
3
CH
3
SO
3
H
2 4
H SO 
Isomerisation
Heating of straight chain alkanes with anhydrous aluminum chloride (lewis acid) affords branched
chain hydrocarbons. This process is termed isomerisation.
3
AlCl
3 2 2 3
CH —CH — CH —CH

 CH
3
H
CH
3
CH
3
The process of isomerisation has been of immense utility in petroleum industry for raising the octane
number of a particular petroleum fraction.
Example
3
AlCl (RCl)
25 C


Rearrangement of alkanes are thermodynamically controlled
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 Cr
2
O
3
—Al
2
O
3
as a catalyst.

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2 3 2 3
o
Cr O Al O
600 C

 2
3H

2 3 2 3
o
Cr O Al O
600 C

 2
3H

n-hexane cyclohexane benzene
n-octane ethyl cyclohexane ethyl benzene
Oxidation
All alkanes readily burn in excess of air or oxygen to form carbon dioxide and water.
C
n
H
2n+2
+
 
 
2 2 2
(3n 1) (2n 2)
O (g) nCO (g) H O(l)
2 2
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 C
1
to C
n
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 agents such as potassium permanganate
readily oxidizes a tertiary hydrogen atom to a hydroxyl group. For example, isobutene is oxidized to
tert butyl alcohol.
(CH
3
)
3
CH + [O] 
4
KMnO
3 3
(CH ) COH
ALKENES:
Alkenes are unsaturated hydrocarbons having carbon-carbon double bonds. The general molecular
formula for alkene is
n 2n
C H.
Alkenes and cycloalkanes are isomeric compounds because both have same general molecular formula.
NOMENCLATURE:
The IUPAC names of alkenes are given by replacing the –ane ending of the corresponding alkane with
–ene.


2 2
IUPAC Ethene
Commen Name: ethylene
H C CH


3 2
IUPAC : propene
Commen Name: Pr opylene
CH — CH CH



1 2 3 4
2 2 3
1 Butene
(not 1 2 butene)
H C CH— CH — CH


6 5 4 3 2 1
3 2 2 3
2 Hexene
(not 4 hexene)
CH — CH — CH — CH CH — CH

14. HYDROCARBONS CHEMISTRY
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4 3 2 1
3 2
CH —CH —CH CH
3
CH
3-methyl-1-butene
6 5 4 3
2 2 2 2 2 3
CH —CH —CH — CH — CH — CH —CH
2 1
2
CH CH
6-Bromo-3-propyl-1-hexene
Br
STRUCTURE AND BONDING IN ALKENES:
117.2°
121.4°
134 pm
110 pm
H
H H
H
Ethylene is planar, each carbon is sp
2
hybridized and the double bond is considered to have a 
component and  component. The component arises from overlap of
2
sp hybrid orbital along a line
connecting the two carbon atoms, the component via a side by side overlap of two p orbital. Regions
of high electron density attributed to  electrons, appear above and below the plane of the molecule
and are clearly evident in the electrostatic potential map.
RELATIVE STABILITIES OF ALKENE:
The stabilities of alkene are governed by —
a)Degree of substitution: Double bonds can be classified as monosubstituted, disubstituted, trisubstituted
and tetrasubstituted. According to the number of carbon atoms directly attached to the C=C structural
unit.
Alkenes with more highly substituted double bonds are more stable than isomers with less substituted
double bond.
b)Vander Waal’s strain: Alkenes are more stable when large substituents are trans to each other than
when they are cis.
When two or more atoms are close enough in space that a repulsion occurs between them is one type
of steric effect and therefore trans isomers are more stable than cis isomer.
c)Heat of hydrogenation: The addition of hydrogen is called hydrogenation. Hydrogenation reactions
are exothermic (negative H
0
).
CH
3
CH
3
CH
3
2-methylbut-2-ene
Pt / C
2
H  CH
3
CH
3
CH
3
0
H 26.9 kcal/mol  
CH
2
CH
3
CH
3
Pt / C
2
H  CH
3
CH
3
CH
3
0
H 28.5 kcal/mol  
2-methylbut-1-ene

15. HYDROCARBONS CHEMISTRY
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CH
3
CH
2
CH
3
Pt / C
2
H  CH
3
CH
3
CH
3
0
H 30.3 kcal/mol  
3-methylbut-1-ene
The preceding three catalytic hydrogenation reaction all from the same alkane. So the energy of the
product is the same for the each of three reactions. However, three reactants must have different
energies. The reaction associated with the least negative H
°
has the most stable reactant and the
reaction with the most negative has the least stable reactant.
CH
3
CH
2
CH
3
0
H 30.3 kcal/mol  
CH
2
CH
3
CH
3
CH
3
CH
3
CH
3
Potential Energy
0
H 28.5 kcal/mol  
0
H 26.9 kcal/mol  
The greater the number of alkyl groups attached to the doubly bonded carbon atoms the more
stable the alkene.
Stability of alkenes
R
2
C = CR
2
 R
2
C = CHR  R
2
C = CH
2
 RCH = CHR  RCH = CH
2
 CH
2
= CH
2
PREPARATION OF ALKENES:
Dehydration of Alcohols: Alcohols when heated in presence of H
2
SO
4
, H
3
PO
4
, P
2
O
5
or Al
2
O
3
undergo
loss of water molecule with the formation of alkene.
H
2
H — C —C — OH C C H O

  
Alcohol Alkene

  
2 4
H SO
3 2 2 2 2 160 C
EthyleneEthyl alcohol
CH — CH — OH H C CH H O
2 4
H SO
140 C

OH
2
H O

16. HYDROCARBONS CHEMISTRY
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2 4
H SO

CH
3
C
CH
3
OH
CH
3
CH
2
CH
3
CH
3
2
H O
2 4
H SO and
3 4
H PO are the acids most frequently used in alcohol dehydration.
Regioselectivity in Dehydration of Alcohol
2 4
H SO
80 C
CH
3
CH
3
OH
CH
2
CH
3 CH
2
CH
3
CH
3
CH
3
CH
3
CH
3

2-Methyl-1-butene
(10%)
2-Methyl-2-butene
(90%)
Dehydration of this alcohol is selective in respect to its direction
3 4
H PO


CH
3
OH
CH
3
CH
3

84% 16%
Regioselectivity of  elimination is governed by Zaitsev’s Rule.
Stereoselectivity in Alcohol Dehydration: In addition to being regioselective, alcohol dehydration
are stereoselective. A stereoselective reaction is one which a single starting material can yield two or
more sterioisomeric products, but give one of them in greater amount than other.
2 4
H SO


OH
CH
3
CH
3
CH
3
H H
CH
3

CH
3
H
H
CH
3
cis-2-pentene (25%) trans-2-pentene (75%)
Illustration - 2 : Predict the major product of dehydration of each of:
(A) (CH
3
)
2
C(OH)CH
2
CH
3
(B) (CH
3
)
2
CHCH(OH)CH
3
(C) (CH
3
)C(OH)CH(CH
3
)
2
Solution:
CH
3
C
CH
3
CHCH
3 C C
CH
3
CH
3
CH
3
CH
3
(A) and (B) (C)
Dehydrohalogenations of Alkyl Halide: Dehydrohalogenation is the loss of hydrogen and a halogen
to form an alkene.

17. HYDROCARBONS CHEMISTRY
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H —C —C— X C C HX  
The reaction is carried out in presence of strong base such as sodium ethoxide
2 5
(NaOC H ) in ethyl
alcohol as solvent.
2 5 2 5
H — C— C— X NaOC H C C C H OH NaX    
2 3
NaOCH CH

H
H
Cl
H
H
H
Similarly
3 3 3
NaOCH ,KOC(CH ) are the preferred bases used in this process.

 
3 3
KOC(CH )
3 2 15 2 2 3 2 15 2 DMSO 25 C
CH – (CH ) – CH – CH – Cl CH – (CH ) – CH CH
The regeoselectivity of dehydrohalogenation of alkyl halide follow zaitsev rule of elimination predominates
the direction.
CH
3
CH
3
Br
CH
3
CH
2
CH
3
CH
3

CH
3
CH
3
CH
3
2-methyl-1-butene
(29%)
2-methyl-2-butene
(71%)
Illustration - 3 : Predict the various product formed by dehydrohalogenation of
CH
3
CH
3
Br
Solution:
CH
3
CH
3 CH
3
CH
3
CH
2
CH
3
(major) (minor) (minor)
Dehalogenation of Vicinal Dihalides
There are two types of dihalides namely gem (or geminal) dihalides in which the two halogen atoms are

18. HYDROCARBONS CHEMISTRY
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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.
    
3 2 5
Zn dust
3 2 3 2 CH COOH or C H OH
CH CHBr CH Br CH CH CH
      
NaI
3 3 3 3 Acetone
CH CHBr CHBr CH CH CH CH CH
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:
X


C
X
C
X
C CC C
X
I
IX
I
ii) With Zn-dust and acetic acid
Zn  Zn
2+
+ 2e
–
X


C
X
C
X
C C XC C
X
2e
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 BaSO
4
or CaCO
3
along with quinoline (Lindlar’s
catalyst).
iii) Hydrogen in presence of Ni
2
B (nickel boride).
R—CH
2
—C  CH 
(i), (ii) or (iii)
R—CH
2
CH = CH
2
Alkali metal dissolved in liquid ammonia produces nearly 100% trans alkene by the following mechanism:
Na + liq. NH
3
 Na
+
+ e
s
(solvated electron)

19. HYDROCARBONS CHEMISTRY
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R C C R    C C
R
R
2
H NH
 C C
R
R
H
s
e
s
e
C C
R
R
H
2
H N H
C
R
R
H
H
2
NH
( - 100%)
(trans-alkene)
The dissolution of alkali metal in liquid NH
3
produces solvated electrons. The reaction is initiated by
the attack of sp hybridized carbon atom of alkyne molecule with a solvated electron (e
s
) when the -
electrons move to the other sp hybridized 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
or far away as possible forcing the two alkyl groups to acquire the farthest position. The carbanion then
picks up a proton from NH
3
to produce a vinylic radical. Attack by another solvated electon gives vinylic
anion which produces trans-alkenes by picking up a proton from NH
3
.
Hydrogenation of alkynes by Lindlar’s catalyst or nickel boride produces nearly 100% cis-alkene. The
catalyst provides a heterogeneous surface on which alkyne molecules get absorbed. Hydrogen molecules
collide with adsorbed alkyne to produce cis-alkene in which both the hydrogen atom come from the
same side.
2
Lindlar 's cat.
2 or Ni B
R C R H    
H
R R
H
Hofmann Degradation Method
Alkenes can be prepared by heating quaternary ammonium hydroxide under reduced pressure at a
temperature between 100°C and 200°C
3 3 2 2 2
OH (CH ) N CH CH H O

   CH
3
N
CH
3
CH
3
CH
2
CH
2
H
 
This is the final step of the overall three step reaction called Hofmann degradation. In the first step,
primary, secondary or tertiary amine is treated with enough CH
3
I to convert it to the quaternary ammonium
iodide. In the second step, the iodide in converted into hydroxide by treatment with Ag
2
O.
3
2CH I

2
Ag O



N
H
CH
3
N CH
3
CH
3
CH
3
N CH
2
CH
3
CH
3
HOH N
CH
3
CH
3
I I

20. HYDROCARBONS CHEMISTRY
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The OH ion invariably removes proton from -carbon atom. If two or more alkyl groups have -carbon
atoms, OH removes proton from that -carbon atom which gives more stable carbonion. 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
OH

 2 2
CH CH CH
3
CH
CH
3
CH
2
N
CH
3
H
2
C
CH
2
CH
3
CH
2
CH
3
CH
3
CH
CH
3
CH
2
N
CH
3
H
2
C CH
2
CH
3
PHYSICAL PROPERTIES OF ALKENE:
a) All are colouress and have no characteristic odour. Ethene has pleasant smell.
b) Lower members (C
2
to C
4
) are gases, middle one (C
5
to C
17
) are liquids, highers are solids.
c) The boiling points, melting points, and specific gravities show a regular increase with increase in
molecular weight, however less volatile than corresponding alkanes.
d) A cis isomer has high boiling and melting point than trans because of more polar nature.
e) Like alkanes, these too are soluble in non polar solvents.
f) Alkenes are weak polar. The polaritiy of cis isomer is more than trans which are either non polar or less
polar (e.g. trans butene-2- is non polar; trans pentene-2-is weakly polar).
CHEMICAL PROPERTIES OF ALKENES:
Addition of Hydrogen: Alkenes reacts with hydrogen in the presence of platinum, palladium, rhodium
or nickel catalyst to form the corresponding alkane.
Pt,Pd,Rh or Ni
2 2 2 2 2
R C CR H R CH CHR   
2
H
Pt

Addition of Halogens: Bromine and chlorine adds to alkenes to form vicinal dihalide. A cyclic halonium
ion is an intermediate stereospecific anti addition is observed.
2 2 2
R C CR X   R
R
X
X
R
R

CH
2
CH
3 2
Br
Br
Br
CH
3

21. HYDROCARBONS CHEMISTRY
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Mechanism
Step 1:
CH
3
CH
3
CH
3
CH
3
Br
+
CH
3
CH
3
CH
3
CH
3

–
Br
Br
Br
Bromonium ion Bromide ion
A bromine molecule becomes polarized as it approaches the alkene. The polarized bromine molecule
transfers a positive bromine atom (with six electrons in its valance shell) to the alkene resulting in the
formation of bromonium ion.
Step 2:
Br
Bromonium ion vic-dibromide
C C
Br
+
CH
3
CH
3
CH
3
CH
3
C C
CH
3
CH
3
CH
3
CH
3
Br
Br
Bromide ion
A bromide ion attacks at the back side of one carbon (or the other) of the brominium ion in an S
N
2
reaction causing the ring to open and resulting in the formation of vic-dibromide.
Addition of Hydrogen Halide
Hydrogen halides (HCl, HBr and HI) add to the alkenes to form alkyl halides.
The addition of HX to alkenes is an electrophilic addition reaction. The order of reactivity of hydrogen
halides is HI  HBr  HCl.
HX 
R
X
R
H
Mechanism
Step – 1
slow
RDS
H X  C C
R R
X
Step – 2

22. HYDROCARBONS CHEMISTRY
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Fast
X

  C
R
C
HX
C
R
C
The addition takes place according to Markownikov’s rule.
In the ionic addition of an unsymmetrical reagent to a double bond, the positive part of the reagent
attaches itself to the carbon atom of the double bond so as to yield 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.
The reaction in which rearrangement of carbocation occurs, the overall addition of HX to alkenes does
not follow Markownikov’s rule.
For example the addition of HBr to 3-methylbut-1-ene gives a mixture of 2-Bromo-2-methylbutane and
2-Bromo-3-methylbutane.
Br
HBr


 
CH
2
CH
3
CH
3
CH
3
CH CH
3
CH
3
H shift


Br

 CH
3
CH
3
CH
3
Br
CH
3
C
CH
3
CH
3
Br

 CH
3
CH
3
Br
CH
3
2-Bromo-3-methyl butane
2-bromo-2-methylbutane
Addition of HBr in presence of peroxide (Peroxide effect)
Addition of HBr to unsymmetrical alkenes in presence of organic peroxide is an anti Markownikov’s
addition
ROOR
HBr 
CH
3
H H
H
CH
3
CH
2CH
2
Br
Mechanism
In presence of peroxide addition of HBr to alkenes follows free radical mechanism
    
 
R O O R RO OR
    
 
R O HBr R OH Br

23. HYDROCARBONS CHEMISTRY
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3 2
CH CH CH Br  



CH
3
CH CH
2
Br
CH
3
CH CH
2
Br
(more stable)
(less stable)
H Br  CH
3
CH CH
2
Br
CH
3
CH CH
2
Br
Br

(2°)
(1°)
Note:– Peroxide effect is not observed by HCl and HI
Addition of sulphuric Acid: Alkene reacts with sulphuric acid to form alkyl hydrogen sulphates. A
proton and a hydrogen sulphate ion adds to double bond in accordance with Markovnikov’s rule.
2 2 2 2 2
R C CR HOSO OH R —CH —CR  
2
OSO OH
Halohydrin formation: When treated with bromine or chlorine in aqueous solution alkenes are converted
to vicinal halohydrin. A halonium ion is an intermediate. The halogen adds to the carbon that has the
greater number of hydrogens. Addition is anti.
2 2 2
R — CH CR X H O    OH
R R
R
X HX
2
2
Br
H O
 CH
2
OH
Br
Mechanism
 
  
2 2
H O X HO X HX
CH
3
CH
3
CH
3
CH
3
X
CH
3
CH
3
CH
3
CH
3

 
–
X
X
X
Halide ion
C C
X
CH
3
CH
3
CH
3
CH
3
C C
CH
3
CH
3
CH
3
CH
3
O
X
H
H
O H
H

O H
H

24. HYDROCARBONS CHEMISTRY
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Here however a water molecule is a nucleophile and attacks a carbon of the ring causing the formation
of a protonated halohydrin.
C C
CH
3
CH
3
CH
3
CH
3
O
X
H
O
+
H
H
H
The protonated halohydrin loses a proton (it is transferred to molecule water). This step produces the
halohydrin and hydronium ion.
Epoxidation: Peroxy acids transfer oxygen to the double bond of alkenes to yield epoxides the reaction
is a sterospecific syn addition.
2 2 2 2
R C CR R —C —OOH R C— CR R —COH   
O
O
O
3
CH COOH
CH
3
CH
3
O
3
CH — COOOH 
Mechanism

C
C
CH
3CH
3
CH
3
CH
3
O
O
H O
R
C
C
O
CH
3
CH
3
CH
3
CH
3

C
O
H
R
O
Alkene Peroxy acid Epoxide Carboxylic Acid
The peroxy acid transfers an oxygen atom to the alkene in a cyclic single step mechanism. The result
is the syn addition of the oxygen to the alkene, with formation of an epoxide and a carboxylic acid.
Acid-Catalyzed Hydration: Addition of water to the double bond of an alkene takes place in aquous
acid. Addition occurs according to Markovnikov’s rule. A carbocation is an intermediate and is captured
by a molecule of water acting as a nucleophile.
H
2 2 2 2
RCH CR H O R —CH —CR

  
OH

  
H
2 3 2 2 3 3
H C C(CH ) H O (CH ) COH
Mechanism

25. HYDROCARBONS CHEMISTRY
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Step 1:
CH
2
CH
3
CH
3
HO
+
H
H CH
3
C
CH
3
H
OH
H
slow
The alkene accepts a proton to form more stable 3° carbocation.
Step 2:
C
CH
3
CH
3
CH
3
OH
H
CH
3
C
CH
3CH
3
O
+
H
H
fast
The carbocation reacts with a molecule of water to form a protonated alcohol.
Step 3:
fast
OH
H
CH
3
C
CH
3
CH
3
OH
CH
3
C
CH
3
CH
3
O
+
H
H
HO
+
H
H
A transfer of proton to a molecule of water leads to a product.
Hydroboration-Oxidation: The two step sequence achieves hydration of alkene in a stereospecific
syn manner with a regeoselectivity opposite to Markovnikov’s rule. An organoborane is formed by
electrophilic addition of diborane to an alkene. Oxidation of organoborane intermediate with hydrogen
peroxide completes the process. Rearrangement do not occur.
2 6
–
2 2
B H
2 2H O , OH
R —CH CR R —CH —CH —CR 
OH
 
3
–
2 2
H B. THF
3 2 2 3 2 2 2
H O . OH
(CH ) — CH — CH CH (CH ) — CH — CH — CH — OH

26. HYDROCARBONS CHEMISTRY
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Mechanism
CH
3CH
3
H
H
pi-complex four centered transition state
CH
3CH
3
H
H
B
HH
H

B
HH
H
CH
3CH
3
H
H
H
-
BH
H

–

CH
3
CH
3
H
H
HB
H H
22
–
22
BH
HO,OH

CH
3
CH
3
H
H
HOH




Addition takes place through the initial formation of a  complex which changes into a cyclic four
center transition state with the boron atom adding to less hindered carbon atom. The dashed bonds in
the transition state represent bonds that are partially formed or partially broken.
The transition state posses over to become an alkylborane. The other B—H bonds of alkylborane can
undergo similar additions, leading finally to a tri-alkylborane.
-
3 2 2 2
(BH ) H O , OH
 
CH
3
methylcyclopentane
CH
3
H
H
OH
trans-2-Methyl-1-cyclopentanol
syn-Addition
Oxymercuration-demercuration
Alkenes react with mercuric acetate in the presence of water to give hydroxymercurial compounds
which on reduction yield alcohols.
2 2
Hg(OAc) H O  
OH HgOAc
4NaBH

OHH
Markovnikov addition
Mercuric acetate
The first stage, oxymercuration, involves addition of –OH and HgOAc to the C–C double bond. 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 hydration.
Oxymercuration-demercuration gives alcohols corresponding to Markovnikov addition of water to the

27. HYDROCARBONS CHEMISTRY
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carbon-carbon double bond. (anti addition). For example:
2 2 4
Hg( OAc) , H O NaBH
 
Norbornene
OH
exo-Norborneol
Syn-Hydroxylation
Hydroxylation with permanganate is carried out by reaction at room temperature of the alkene and
aqueous permanganta solution; either neutral or slightly alkaline. Hydroxylation is a very good method
for the synthesis of 1, 2-diols.
2 2 42
3CHCH2KMnO4HO3   
2
2MnO2KOHCH
2
CH
2
OHOH
(ethylene glycol)
4
32
(1)OsO
(2)NaHSO/HO
 CH
2
CH
2
OHOH
CH
3
(propylene glycol)
CH – CH = CH
3 2
The mechanism for the formation of glycols by permanganate ion and osmium tetroxide involves the
formation of cyclic intermediates. Then in several steps cleavage at the oxygen-metal bond takes
place ultimately producing the glycol and MnO
2
or Os metal. The course of these reactions is syn-
hydroxylation.
Hydroxylation can also be carried out with peroxy acids such as peroxy formic acid (HCO
2
OH) mixture
of hydrogen peroxide and formic acid allowed to stand for few hours. Sequently the product is heated
with water give glycol.

28. HYDROCARBONS CHEMISTRY
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2
cold
4 H O/OH
MnO


 
O
Mn
O
O O
2
OH
H O
Several steps


2
MnO
OHOH
2
cold
4 H O/OH
MnO


 
OH
H
OH
H
Oxidative cleavage of Alkenes
Alkenes are oxidatively cleaved by hot alkaline permanganate solution or higher conc.of KMnO
4
or acid
medium (bayer’s reagent)
The terminal CH
2
group of 1-alkene is completely oxidized to CO
2
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 salts of
carboxylic acids. = CH
2
part get oxidized to CO
2
+ H
2
O = CHR part get oxidized to RCOOH
= CR R part get oxidized to RCOR
Example
4
KMnO , OH
3 3 heat
(cis or trans)
CH CH CHCH 2 
H
3 2
2CH CO H

CH
3
O
O
4
KMnO , OH
H

2 2
CO H O 
CH
3
CH
2
CH
3
CH
3
O
CH
3
Ozonolysis of Alkenes
A more widely used method for locating the double bond of an alkene involves the use of ozone (O
3
).
Ozone reacts vigorously with alkenes to form unstable compounds called initial ozonides, which rearranges
spontaneously to form compounds known as ozonides. Ozonides, themselves are unstable and reduced
directly with Zn and water. The reduction produces carbonyl compounds that can be isolated and
identified.
3
O   2
Zn/H O
2O
O
O
O
O O
O
Ozonolysis can be of either of reductive type or of oxidative type. The difference lies in the fact that
products of reductive ozonolysis are aldehydes and/ore ketones while in oxidative ozonolysis, the

29. HYDROCARBONS CHEMISTRY
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products are carboxylic acids and/ore ketones (this is because H
2
O
2
formed would oxidise aldehydes
to carboxylic acids but ketones are not oxidized). In reductive ozonolysis, we add zinc which reduces
H
2
O
2
to H
2
O and thus H
2
O
2
is not present to oxidise any aldehyde formed.
Zn + H
2
O
2
 ZnO + H
2
O
For example,
(CH
3
)
2
CH – CH = CH
2

3 2 2
2
(i) O , CH Cl
(ii) Zn/H O
(CH
3
)
2
CHCHO + HCHO
(CH
3
)
2
C = CH –CH
3

3 2 2
2
(i) O , CH Cl
(ii) Zn/H O
(CH
3
)
2
C = O + CH
3
CHO
Illustration - 4 : Identify the products (x, y) of following reaction:-
3 2
O /H O,Zn
(X) (Y) CH
3
CH
3
CH
3
CH
3
CH
3
Solution:
CH
3
C
CH
3
O HC
O
C
CH
3
CH
3
CH
3
(X) : (Y) :
Substitution Reactions at Allylic Position
Cl
2
+ H
2
C = CHCH
3

High
temperature
H
2
C = CHCH
2
Cl + HCl
Br
2
+ H
2
C = CH–CH
3

2
Low concentration
of Br
CH
2
= CH – CH
2
Br + 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 Cl
2
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 CCl
4
in the presence of peroxides
or light.
4
LightorROOR
2 2CCl
NBr CHCHCHBr
CH = CH – CH
2 3

O
O
O
O
NH
N-Bromosuccinimide (NBS) 3-Bromopropene
(allyl bromide)
Succinimide
Dimerization

30. HYDROCARBONS CHEMISTRY
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acid
CH
3
CH
2
CH
3
 CH
3
CH
2
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
2,4,4-trimethylpent-2-ene

CH
3
CH
3
CH
3
CH
2
CH
3
2,4,4-trimethylpent-1-ene
ALKYNES:
Introduction:
Unsaturated hydrocarbons having a carbon-carbon triple bond are called alkynes. They correspond to
general formula
n 2n–2
C H i.e. they have smaller proportion of hydrogen than alkene. Alkynes are isomeric
with cycloalkenes and alkadienes.
NOMENCLATURE:.
HCC– CH– CH = CH
2 2
pent1-ene4-yne
CH
3
5
4
3
2
CH
1
CH
3
4-methylpent-1-yne
CH
3
1
2
3
4
5
CH
3
6
Cl
2-chlorohex-3-yne
CH
3
CH
CH
3
CH
Cyclopentylethyne
5-methylhex-1-yne
PREPATATIONS OF ALKYNES:
Dehydrohalogenation of vic dihalides: Vicinal dihalides having hydrogens on carbon gives alkynes
with strong base.
2/
NaNH
R—CH—CH—R R—C C—R

 
X X
Kolbe hydrocarbon synthesis: Potassium salt of maleic acid and its alkyl derivatives gives alkynes
on electrolysis.
R—C—COOK
R—C—COOK
electrolysis
2 2
R—C C—R 2CO 2KOH H    

31. HYDROCARBONS CHEMISTRY
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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.
CaC
2
+ 2H
2
O  HC  CH + Ca(OH)
2
Mg
2
C
3
+ 4H
2
O  CH
3
–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

 
3
C C C
. It is pertient to note that aluminium
carbide (Al
4
C
3
) and beryllium carbide (Be
2
C) do not form any alkyne on hydrolysis, instead they form
methane on hydrolysis. This is due to the fact that their anions exist as C
4–
.
PHYSICAL PROPERTIES:
i) Lower members (C
2
to C
4
) are gases; middle one (C
5
to C
12
) are liquids; highers are solids.
ii) The boiling point, melting point and specific gravity of alkynes show a regular increase with increase in
molecular weight; however less volatile than alkene. The order of boiling point in hydrocarbons has
been explained in terms of polarity. Alkynes possess more polarity and thus have higher boiling point.
alkyne  Alkene  Alkane
iii) All are colourless and possess no characteristic odour; however C
2
H
2
has garlic odour due to the
impurities of PH
3
, H
2
S etc. Pure C
2
H
2
has ethereal odour.
iv) Soluble in organic solvents like acetone, alcohol and sparingly soluble in water.
v) The boiling point of acetylene is –84°C. Liquid acetylene is dangerously explosive and therefore storage
and transportation of liquid acetylene is prohibited by law. That is why acetylene is stored and transported
by dissolving it in acetone soaked on porous material like asbestos packed in steel cylinders, under
high pressure.
CHEMICAL PROPERTIES OF ALKYNES:
Acidic Character of alkynes
The hydrogen bonded to the carbon of terminal alkyne is considerably more acidic than those bonded
to carbons of alkenes and alkanes. The pKa values for ethyne, ethene and ethane illustrate this point
H
H H
H
H
H
H
H
H
H
H C C H  
pKa 25 pKa 44 pKa 50
The order of basicity of their anions is opposite that of their relative acidity
Relative Basicity
CH
3
CH
2
–
 CH
2
= CH
–
 HC C
–
From the pKa values it can be concluded that there is a vast difference in the stability of the carbanions.
This difference can be readily explained interms of the character of the orbital occupied by the lone pair
electrons in three anions. In ethyl anion the lone pair electrons are in sp
3
hybrid orbitals, in vinyl anion
the lone pair is in the sp
2
hybrid orbital and in acetylide ion the lone pair is in sp hybrid orbital. The
percentage s-character of the orbital increases in the order sp
3
 sp
2
 sp. Greater is s-character of the
hybrid orbital containing a lone pair of electrons, the less basic is that anion and the more acidic the
corresponding conjugate acid.

32. HYDROCARBONS CHEMISTRY
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Relative Acidity

              
2 2 2 3
16 17 25
15.7 38 44 50
pKa:H OH H OR H C C H H NH H CH CH H CH CH
Relative Basicity

       
2 2 2 3
OH OR C C H NH CH CH CH CH
Reactions due to Acidic Hydrogen
i)Reaction with Sodium : Terminal alkynes react with sodium in liquid ammonia or sodamide to form
sodium alkynide
R – C C – H + Na 
3
NH (l)
R – C C Na + 2
1
H
2
R – C C – H + NaNH
2

3
NH (l)
R – C CNa + NH
3
This reaction is utilized for the preparation of higher alkynes.
R – C C – H 
2
3
NaNH
NH (l)
R – C C – Na
R – C CNa


R X
R – C C - R + NaX
ii)Reaction with ammonical cuprous chloride solution :- When acetylene is passed through ammonical
cuprous chloride solution, a red precipitate of copper acetylide is formed.
H – C C – H + Cu
2
Cl
2
+ 2NH
4
OH 
  
copper acetylide
(Re d ppt)
Cu C C Cu
+ 2NH
4
Cl + 2H
2
O
R – C C – H + Cu
2
Cl
2
+ 2NH
4
OH 
  
Copper alkynide
R C C Cu
+ 2NH
4
Cl + 2H
2
O
iii)Reaction with ammonical silver nitrate solution :- When acetylene is passed through ammonical
silver nitrate solution (Tollen’s reagent) a white precipitate of silver acetylide is formed.
H – C C – H + 2AgNO
3
+ 2NH
4
OH 
  
Silver acetylide ( white ppt)
Ag C C Ag
+ 2NH
4
NO
3
+ 2H
2
O
The alkynes can be regenerated from the insoluble salts and the overall process serves as a method for
purifying terminal alkynes.
iv) Reaction with Grignard reagent: These two reagents reacts with terminal alkyne to form
hydrocarbon and new organometallic compound respectively.
R — C C — H
3
4CH MgBr
R —C C — MgBr CH  
3
CH Li
4
R —C C— Li CH  
Addition of Hydrogen
Alkynes can be reduced directly to alkanes by the addition of
2
H 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 , nickel boride or palladised
charcoal, alkynes can be partially hydrogenated to alkenes
Ni Pt or Pd
3 2 3 2 3
CH C CH 2H CH CH CH   

33. HYDROCARBONS CHEMISTRY
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Lindlar ' s catalyst
3 2 3 2
CH C CH H CH CH CH     
2-butyne when reduced with Lindlar’s catalyst gives nearly 100% cis-isomer while Na in liquid ammonia
gives nearly 100% trans-isomer.
3
CH
3
CH
3
CH
3
CH
3
CH
3
CH
C
C
C
C
C
C
H
H
H
H
Nain
3
liq.NH
+
2
H
(Cis)
(trans)
lindlar’s catalyst
Addition of Halogen Acids
Addition of halogen acids to alkynes occur in accordance with Markovnikoff’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, additon of HI to propyne first gives 2-iodopropane and then
2,2-iodopropane
3
CH C CH HI   
3 2
CHC CH-=
I
HI

3 3
CHC CH--
I
I
2-iodopropene 2,2-diiodopropane
The order of reactivity of halogen acids towards addition reaction is H I H Br H Cl    
Addition of Halogens
One or two molecules of halogens can be added to alkynes giving dihalides and tetra halides respec-
tively. Chloride and bromide add readily to the triple bond while iodine reacts rather slowly
2
Cl
2 2 2
CH CH Cl CHCl CHCl CHCl CHCl     
2
Br
3 2 3 3 2 2
CH C CH Br CH CBr CHBr CH CBr CHBr       
Addition of water (Hydration)
Alkynes cannot be hydrated more easily than alkenes because of their low reactivity towards electro-
philic addition reactions. Further when acetylene is passed through 40%
2 4
H SO containing 1%
4
HgSO at
0
80 C, a water molecule adds upto give acetaldehyde.

34. HYDROCARBONS CHEMISTRY
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CH
CH
+
HOH

2 4
4
40%H SO
1%HgSO
2
CH
CHOH
Vinyl alcohol

Tautomerises
3
CH
CHO
Acetaldehyde
  R C CH HOH


2
4
H SO4 dil
HgSO
CR
2
CH
OH
(Enolic)

Tautomerises
CR
O
3
CH
ketone
Addition of Boron hydrides
Diborane, the simplest hydride of boron reacts with alkyne to form trialkenylborane. Diborane splits into
two
3
BH units and the addition of
3
BH takes place following Markovnikoff’s rule. The addition contin-
ues as long as hydrogen is attached to boron atom.
2 6 2
2R C CH B H 2R CH CH BH      
 
2 2
R C CH R CH CH BH R CH CH BH        
   
2 3
R C CH R CH CH BH R CH CH B       
Trialkenylborane on hydrolysis gives alkene
  
3
CH COOH
2Hydrolysis3 3
R CH CH B 3R CH CH B OH     
Internal alkynes give rise to alkene where geometrical isomerism is possible. Hydroboration followed by
hydrolysis of alkynes gives cis alkene as the major product.
 
2 6 3 2
B H CH CO H
3
R C C R RCH CR B     
CC
H H
(cis)
R R
Oxidation of trialkenylborane with alkaline
2 2
H O results in the formation of carbonyl compounds.
Terminal alkynes give rise to aldehydes whereas internal alkynes gives rise to ketones.

35. HYDROCARBONS CHEMISTRY
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2 2
H O / NaOH
3
R CH CH B  
CHCH
OH
R
2
R CH CHO 
 
  
2 2
H O / NaOH
3
R CH CHR B
CHCH
OH
R

O
CR R
2
CH
Oxidation
Alkynes are oxidised by allkaline
4
KMnO. 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
2
CO and
2
H O.
4
(i)KMnO / OH
3 3 2 2 (ii)H
CH C CH CH COOH CO H O

    
4
(i)KMnO / OH
3 2 3 3 3 2 (ii)H
CH C C CH CH CH COOH CH CH COOH

    
Ozonolysis
Reaction of alkynes with
3
O gives rise to the formation of ozonide. Hydrolysis of ozonide with
2
H O
gives a mixture of two carboxylic acids. This is called oxidative ozonolysis.
3 3
CH C CH O   
O
OO
3
CHCCH
2
H O
 3
CH COOH HCOOH
However, if ozonide is hydrolysed with Zn and
2
H O, a diketone is formed. This is called reductive
ozonolysis
3 2 3 3
CH C C CH CH O    
O
OO
3
CHCC
2 3
CHCH
2
Zn / H O

O
O
3
CH CC
2 3
CHCH
Ethane behaves differently. Ozonolysis followed by oxidative hydrolysis of ethyne gives a mixture of
glyoxal and formic acid.

36. HYDROCARBONS CHEMISTRY
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3
2
1.O
2.H O
HC CH CHO CHO HCOOH   
polymerisation : Acetylene undergoes polymerisation yielding different types of polymeric compounds
under different conditions:
(i) Cyclic polymerisation : When acetylene is passed through a red-hot metallic tube at
0
600 C,
cyclic polymerisation takes place with the formation of benzene
3HC CH 0
Heat
600 C

Benzene
Propyne on heating trimerises under similar conditions and forms mesitylene (1,3,5-trimethyl benzene).
Cu Tube
3 Heat
3CH C CH  
3
CH
3
CH
3
CH
Mesitylene
(ii) Acetylene dissolved in tetrahydrofuran polymerises into cyclo-octa-1,3.5,7-tetrene in presence of

2
Ni CN and under high pressure.

2
Anhydrous Ni CN
Tetrahydrofuran
(Solvent)
4HC CH 
CH CH
CH CH
CH
CH
CH
CH
Cyclo-octatetrene
(iii) Linear polymerisation : When acetylene is passed into cuprous chloride solution containing
4
NH Cl linear polymerisation occurs forming monovinyl acetylene ad divinyl acetylene
2 2
4
Cu Cl
2NH Cl
HC CH HC CH CH CH C CH      
Monovinyl acetylene
HC CH 3rd molecule

2 2
CH CH C C CH CH    
Divinyl acetylene
Vinyl acetylene on reduction with
2
H /Ptin presence of
4
BaSO forms buta-1,3-diene.

37. HYDROCARBONS CHEMISTRY
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
2
4
H / Pt
2 2 2 BaSO
H C CH C CH H C CH CH CH      
Vinyl acetylene 1,3-Butadiene
KEY CONCEPTS
Alkanes
Alkanes are saturated hydrocarbons. They are called paraffins. General formula of alkanes is C
n
H
2n+2
Hydrogenation of alkenes or alkynes in presence of Nickel catalyst at 250°C gives alkanes.
Alkyl halides on reduction with Zn/HCl, Zn/CH
3
COOH or red P/HI or of subjected to Wurtz reaction or
Frankland reaction give alkanes.
Alkanes are formed by the reduction of alcohols, aldehydes, ketones & carboxylic acids with HI & Red
P.
Decarboxylation of sodium or potassium salt of carboxylic acid gives alkanes.
Grignard’s reagent react with compound containing active hydrogen like H
2
O; C
2
H
5
OH, CH  CH, NH
3
etc give alkanes.
Alkanes containing odd number of carbon atoms can be prepared by corey-house synthesis.
Melting points, boiling point and density of alkanes increase with the increase in molecular weights.
Straight chain alkanes have higher boiling point than branched chain alkanes.
Due to strong and non polar C—C and C—H bond alkanes are not very reactive.
Halogenation, nitration and sulphonation of alkanes follow free radical mechanism.
In the halogenation of alkanes the reactivity with respect to halogens follows the order F
2
 Cl
2
 Br
2

I
2
. Reactivity of with respect to C—H follows the order teretiary hydrogen  secondary hydrogen  pri
mary hydrogen.
Alkenes
Alkenes are unsaturated hydrocarbons having one carbon-carbon double bond.
Alkenes are called olefins. General formula of alkenes is C
n
H
2n
.
Dehydration of alcohols leads to the formation of alkenes. Dehydrating agents are H
2
SO
4
, P
2
O
5
, Al
2
O
3
,
H
3
PO
4
and ZnCl
2
. The ease of dehydration follows the order 3°  2°  1°.
Dehydrohalogenation of alkyl halides gives alkenes. The ease of dehydro-halogenation is 3°  2° 1°.
Alkenes can be prepared by the controlled partial hydrogenation of alkynes with Lindlar’s catalyst.
Dehalogenation of vic-dihalides with Zinc dust gives alkenes.
The formation of more substituted alkene as major product is called Saytzeff elimination.
The formation of less substituted alkene as major product is called Hofmann elimation.
The stability of alkenes follow the order R
2
C = CR
2
 R
2
C = CHR  R
2
C = CH
2
 RCH = CHR  R—CH
= CH
2
 CH
2
= CH
2
.
Trans-but-2-ene is more stable than cis-but-2-ene.
Olefines show electrophilic addition reaction.
Alkenes give addition product with halogens to form vic-dihalides.
Alkenes decolourised by
i) Baeyer’s reagent ii) Br
2
water.

38. HYDROCARBONS CHEMISTRY
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These reactions are the test of unsaturation.
Alkenes react with HX to form alkyl halides. The order of reactivity of HX is HI  HBr  HCl  HF.
Addition of HX to unsymmetrical alkenes takes place according to Markownikov’s rule.
Addition of HBr to unsymmetrical alkenes in presence of peroxide follows anti-Markownikov’s addition.
HCl and HI do not give anti Markownikov’s addition.
Alkenes on reductive ozonolysis give carbonyl compounds.
Hydroxylation of alkenes by cold aqueous alkaline KMnO
4
solution (Baeyer’s reagent is a syn-addition.)
Hydroboration-oxidation of alkene is an anti markownikov’s addition of water to a double bond.
Oxymercuration-demercuration is a markownikov’s addition of water to a double bond.
In Oxymercuration-demercuration reaction and hydroboration-oxidation reaction no rearrangement takes
place.
Oxidation of alkenes by acidic KMnO
4
or acidic K
2
Cr
2
O
7
give ketone or carboxylic acid or both.
Periodic acid (HIO
4
) or leadtetra acetate [(CH
3
COO)
4
Pb] oxidizes alkenes into glycols and finally gives
aldehydes or ketones depending upon the nature of alkene.
Oxidation of alkenes by SeO
2
affect the allylic position.
NBS (N-Bromosceccinimide) is used for the bromination of alkenes at the allylic position.
Alkynes
Alkynes are unsaturated hydrocarbons with one carbon-carbon triple bond. General formula C
n
H
2n-2
.
Dehydrohalogenation of vic-dihalides give alkynes.
Dehalogenation of 1, 1, 2, 2-tetrahalides with Zn dust and alcohol give alkynes.
Acetylene can be prepared by the action of water on calcium carbide.
Magnesium carbide on hydrolysis give propyne.
The acidic character of Ethyne, Ethene and Ethane follows the order H – C  C – H  CH
2
= CH
2
 CH
3
– CH
3
.
The basic character of their conjugate base follow the order CH  C
–
 CH
2
= CH
–
 CH
3
—CH
2
–
Alkynes are less reactive towards electrophilic addition as compared to alkenes.
Acetylene and terminal alkynes react with ammonical silver nitrate solution (Tollen’s reagent) to give
corresponding alkynides (white ppt.).
Acetylene and terminal alkynes react with ammonical cuprous chloride solution to give a red precipitate
of corresponding alkylnides.
Hydrogenation of alkynes in presence of Lindlar’s catalyst give cis-alkene where as hydrogenation by
Na/Liq. NH
3
give trans alkenes.
O
Ozonolysis [O]
3 2 2 3 3 2
(F)
(D) CH CH — C — CCH CH CH CH COOH 
O
(E)