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HYDROCARBONS
 The compounds containing hydrogen and carbon with essentially a covalent bond between them
 Broadly classified as
a. Aliphatic

i. Saturated: which contains only single bonds e.g. Alkane

ii. Unsaturated: which contains at least one double or triple bonds e.g. Alkene and alkyne

b. Aromatic hydrocarbons: e.g. Benzene and derivatives

ALKANES
 Saturated hydrocarbon where, C atom is sp 3 hybridized

General formula – C2H2n+ 2 e.g. CH4, C2H6, C3H8 etc.

Note: Methane is also called marsh gas  found mostly in marshy places. Alkanes exhibit chain isomerism.

Alkanes containing 4 or more atoms exhibit chain isomerism.

Preparation
i. Reduction of unsaturated hydrocarbons [Sabatier – Sanderen’s reduction] / Catalytic Hydrogenation of alkenes and
alkynes

e.g. HC  CH + H2 Ni
or  H2C = CH2 Ni
Pt or Pd
/   CH3 – CH3
Pt / Pd

ethyne (200 – 250)0C ethane H2 ethane

This process is exothermic and energy released is k/a heat of hydrogenation. Greater the stability of unsaturated
hydrocarbon, lower will be the energy associated with them hence, lower will be the heat of hydrogenation.

Note: Methane can’t be obtained by this method

If this reaction is carried out in presence of Lindlar’s catalyst (i.e. Pd/BaSO4 in presence of sulphur); then alkenes can be
produced as the further reduction of alkene is checked by Lindlar’s catalyst.

ii. Reduction of alcohol,Red

aldehyde,
P ketone and carboxylic acid [Red P + HI used]

CH3 – CH2 OH + HI 

 CH3CH3 + I2 + H2O
(Alcohol)
O
CH3 – C – CH3 + HI 

Red P

CH3CH3 + I2 + H2O
(aldehyde)
O
CH3 – C – CH3 + HI 

Red P


CH3CH2CH3 + I2 + H2O
O
Red P
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CH3 – C – OH + HI 


CH3CH3 + PI2 + H2O

iii. Special reduction of Aldehydes and ketones

Clemenson and Wolfkisher reduction is reduction of carbonyl compound to alkanes.
a. Clemenson reduction [ Zn/Hg + HCl] [@ Clemenson HCl]

Zn/Hg
CH3 – CHO HCl

 CH3CH3 + H2O
ZnHg
CH3 – CO –CH3  CH3CH2CH3 + H2O
HCl

b. Wolfkisher reduction [Hydrazine + glyc KOH][@Wolfkisher KOH]

NH2NH2
CH3 – CHO  CH3CH3 + H2O + N2
KOH

NH2NH2
CH3 – CO – CH3 KOH
 CH3CH2CH3+ H2O + N2

iv. Soda lime decarboxylation

* Sodium salt of carboxylic acid when heated in presence of 3:1 solution of NaOH (sodium hydroxide) and CaO(lime); alkane
is produced

i.e. RCOONa + NaOH + CaO R – H + Na2CO3

3 : 1

Soda lime

* The number of carbon in alkane is reduced by 1

The use of CaO is to keep NaOH dry and hence enhance reaction

v. Kolbe's electrolysis

Alkali metal salts of carboxylic acid in their aqueous solution when electrolysed give alkanes.

2CH3 – COONa + 2H2O   

 CH3 – CH3 +2CO2+2NaOH + H2
electrolysis

anode in cathod
e
Note: This reaction shows free radical mechanism. solution

This reaction is not suitable if two different salts are used to obtain unsymmetrical alkane (i.e. with odd number of C
atom) since, the mixtures of alkane if formed.
The solution obtained after electrolysis of sod. Salt of carboxylic acid is

a. Acidic b. Basic c. Neutral d. Depends on medium

Ans. b (due to NaOH).

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Note: Methane can’t be obtained by this process

vi. Wurtz reaction

R – X + 2Na + X – R dry
  R – R + 2NaX
ether

(Alkyl halides) alkanes

Mechanism Free radical mechanism

Note: Methane can’t be prepared by Wurtz reaction.

 Alkanes having odd number of C atoms [unsymmetrical] cant be prepared if 2 different alkyl halides are taken than mixture of
alkanes is obtained.

 This method is suitable for the preparation of symmetrical alkanes.

eg: 2CH3CH2 I + 2Na  CH3–CH2–CH2–CH3+ 2NaI
(symmetric alkane)
 But if CH3 CH2 I and CH3I are used then mixture of propane, butane and ethane are formed.
 CH3CH2I + 2Na + CH3I  CH3CH2CH3 + 2NaI
(propane)
 CH3CH2I + 2Na + CH3CH2I  CH3CH2CH2CH3 + 2NaI
(butane)
 CH3I + 2Na + CH3I  CH3CH3 + 2NaI
(ethane)

Note: If ‘Na’ in this reaction is replaced by ‘Zn’ then, this reaction in known as Frankland’s reaction.

Reactivity order of different alkyl halides: RI > RBr > RCl > RF

Note: If tertiary alkyl halides are used then alkenes are produced instead of alkanes

vii. From Gignard’s reagent

R – MgX + H – OH  R – H + MgXOH
R – MgX + H – C  C – H  R – H + MgX (C  CH)
* R – MgX (Gingnard’s reagent) when treated with compounds having active hydrogen ( H – OH, R – OH, R – COOH, R – NH2,
HC  CH etc) give alkanes

viii. Hydrolysis of carbides (Methane Production)

Al4C3 + 12H2O  3CH4 + 4Al(OH)3 [MOE model]
(Aluminium carbide) methane [MOE 2002, IOM 1996]
Be2C + 4H2O  CH4 + 2Be(OH)2
(Berilium carbide) methane
But,
CaC2 + 2H2O  C2H2 + Ca(OH)2
(Calcium carbide) Ethyne
ix. Corey – House reaction (coupling of organometallic compounds)
R2CuLi + R1 – X ether
 RR1 + RCu + LiX
(alkane)
The advantage of this reaction is that both symmetrical and unsymmetrical alkanes can be prepared
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ix. By the reduction of alkyl halides with various agents (catalytic reduction).
Reducing
R – X + 2[H]  R – H + H – X
agents
 Reducing agents:
i. Zn + HCl or Zn + NaOH
ii. Zn + CH3COOH
iii. Zn – Cu couple and alcohol
iv. Aluminium amalgam and alcohol (AL– Hg)
 Also,
Alkyl iodides are reduced by heating with HI and red phosphorus in a sealed tube.
150°C;press
R – I + HI  R – H + I2
+P
 Also,
Alkyl halides can be reduced with complex metal hydrides like LiAlH4, NaBH4, Triphenyl stannic hydride (Ph3SnH).
(Note: Ph = Phenyl)
for
Note: LiAlH4  1° and 2° alkyl halides
for
NaBH4  2° and 3° alkyl halides
for
Ph3SnH  1°, 2° and 3° all

x. Corey-House synthesis
R
|
R – CuLi + Rx  R – R' + RCu + LiX
(Gilman reagent) Alkylhalide alkane
OR
Lithium dialkylcopper
 involves coupling of Gilman reagent with alkyl halide.
1
 For good synthesis R X should be methylhalide or 1° halide.
 Note: This method is also useful for synthesis of assymetrical hydrocarbon. Thus, it is said to be superior to Wurtz reaction.

PROPERTIES

Physical

a. State C1 to C4gas

C5 to C17 liquid

C18 & greater solid

b. Soluble in organic solvents but not in water because they are non-polar

c. Melting and boiling point

 with  in molecular weight (OR no of carbons due to increase in molecular size and Vander Waals force.
  in bpt is uniform.
  in mpt show oscillation or alternation effect
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e.g. Propane butane pentane

86K 138K 143K

50K 5K (Difference)
This fluctuation in mpt is due to the fact that fitting of molecules in crystal lattice is different for symmetrical and un-
symmetrical hydrocarbons Since, symmetrical are best fitted so difference in mpt is very high when we move from
unsymmetrical to symmetrical.

d. From C5 to C17 (liquid state) the viscosity of alkanes increases with increase in C- no.
i.e. n – Hexane will be more viscous than n-pentane.

Chemical
* Alkanes are extremely inert chemically so k/a paraffins no affinity

* Their chemical inertness is due to the fact that the Electro negativity difference between C and H is very less, there fore C–H
bond is non-polar. Moreover, the bonds i.e. C–C and C–H present in alkanes are stronger and hence reagents can’t break
them easily.
* Alkanes show free radical substitution reactions because of their non polar nature

a. Halogenation:


CH4 + X2 hf/
  CH3X + CH2X2 + CHX3 + CX4 + HX

Mechanism: Free radical

Florination: CH4 + F2 

N
2
CH3F + HF

Note: Fluorination is violent. Thus, N2 is added to dilute the reagents and thus decrease the violentness of the reaction.
Chlorination: CH4 + Cl2  to  C + HCl
direct exposure
sunlight

CH4 + Cl2    CH3Cl +CH2Cl2 + CHCl3 + CCl4 + HCl

diffused light

Bromination: slower

CH4 + Br2 

hf
CH3Br + CH2Br2 + CHBr3 + CBr4 + HBr

Iodination: Reversible so carried out in presence of oxidizing agents like HIO 3

CH4 + I2 CH3I + HI

HI + HIO3 H2O + I2

So, backward reaction will be ceased in absence of HI

Note: In case of unsymmetrical alkanes, isomeric alkyl halides will be formed. The more stable alkyl radical will form major product

Order of halogenation:
3° > 2° > 1°> methane
b. Sulphonation:

 Lower alkanes don’t undergo sulphonation

 Hexane & onwards give sulphonation reaction in presence of H 2SO4 or oleum

400  500) C 0
C6H14 (  
H SO
 C6H13SO3H + H2O
2 4
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c. Nitration:

 Same as sulphonation but HNO3 used instead of H2SO4

C6H14 + HNO3 C6H13NO2 + H2O

NOTE: Lower alkanes also undergo nitration but it should be in vapour phase i.e. vapourphase nitration.

CH3

d. Isomerisation: |

AlCl /HCl
CH3CH2CH2CH3  
3
 CH3 – CH – CH3

e. Armatisation: CH3(CH2)4CH3 2  

Cr O 3 / MnO


Aromatisation increases octane number Benzene

Octane number is less for st chain alkanes than branched ones.

f. Combustion / oxidation

i. Complete oxidation

 3n  1 
 
CnH2n+2 + 
2 
O2 
 nCO2 + (n + 1) H2O + Heat

[IOM 2010]

e.g. C3H8 + 5O2 

 3CO2 + 4H2O + Heat

This is basic principle in biogas production.

ii. Controlled oxidation

Cu tubes
a. CH4 + O2  CH3OH
100atom‚200°c
9:1 (alcohol)
MoO
b. CH4 + O2  HCHO + H2O
275°c
(aldehyde)
(CH3COO)2Mn
c. 2CH4 + 3O2  2HCOOH + H2O
(acids)
Note: instead of Manganese acetate; any transitional metal acetates can be used.
alk.KMnO4
iv. (CH3)3CH  (CH3)3COH

isobutene terbutanol
g. Pyrolysis/ Cracking/ Thermal decomposition

Higher alkanes heated

 lower alkanes + alkenes

Also, may be called as

Conversation of higher boiling hydrocarbons to lower boiling [IOM 03]

CH3CH2CH3 

CH4 + CH2 = CH2 [Higher alkane lower alkane alkene]
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Note:

 Octane number: Octane number of any fuel is defined as % of isooctane present in a mixture of n-heptane and iso-octane
which will have the same antiknocking property as the fuel.

 Cetane number: It is a special term to describe the quality of diesel.

For Extra Knowledge

Conformation
Groups bonded to C – C single bond can rotate freely about C-C bond axis.
 The different arrangement of atoms or groups in space that can be formed by the rotation of groups about carbon-
carbon bond axis is known as conformations.
 Possible conformations of some important alkanes
Methane: only one tetrahedral form
Ethane: eclipsed form, staggered form, skewed form
[Note: Relative stability: staggered > skew> eclipsed.]
 Propane: Eclipsed staggered, skew form
 Butane: 2 Eclipsed, 2 staggered and skew form.
 Cyclohexane: chair form and Boat form
Note: for knowledge
Eclipsed form: In which rear-CH3 group (i.e. the group behind) is completely eclipsed.
If;

Rear carbon
Front carbon

Then,
H H
H
H

H
H
Eclipsed form of ethane
Staggered form: Atoms are perfectly staggered due to rotation of rear –CH3 group.
H
H*
H*

H H
H*
Staggered form of ethane
Skewed form: All conformations between eclipsed and staggered form.
Conformations of cyclohexane
Since, it is not planar molecule and bond angle is not 120°.
120°
is not a true structure

 Therefore, its two conformations, which satisfy the stable bond angle of 109° 28’ are

chair form Boat form