Chemistry
Alkane
Session objectives
1. Methods of preparation
2. Kolbe’s electrolysis
3. Physical Properties and Chemical
properties
4. Halogenation
5. Refining of petroleum
6. Octane number
General characteristics of
alkene
•Paraffins
•General formula CnH2n+2
•sp3 hybridisation
•C–C bond length 1.15 4 A0
•Chemically unreactive
Methods of preparation
Wurtz reaction
Dry
2CH3CH2Br  2Na 
 CH3CH2CH2CH3  2NaBr
ether
• Follow mainly free radical mechanism
• Useful in preparing an alkane containing
even number of carbon atoms
• Stepping up reaction
Frankland reaction
RX+Zn+Rx R –R +ZnX2
Decarboxylation of sodium or
potassium salt of fatty acid
CaO/Heat
RCOONa +NaOH 
 R - H +Na2CO3
Alkane
Sodium salt of
carboxylic acid
For example
CaO/Heat
CH3COONa +NaOH 
 CH4 +Na2CO3
Sodium acetate
Methane
Kolbe’s electrolysis


2RCOONa  2RCOO  2Na 
Sodium salt of
carboxylic acid
At anode
•
-
2RCOO  2R COO+2e•
•
2RCOO  2R +CO2
•
2R  R -R
Alkane
Kolbe’s electrolysis
For example
2CH3COONa  2CH3COO- +2Na+
Sodium acetate
At anode
•
-
2CH3COO  2CH3COO+2e•
•
2CH3COO  2CH3 +CO2
•
2CH3  CH3 - CH3
Ethane
From Grignard reagent
(RMgX)
RMgX+HOH  RH+Mg(OH)X
RMgX+R'OH  RH+Mg(OR')X
RMgX+R'NH2  RH+Mg(NHR')X
From unsaturated hydrocarbons
Sabatier-Senderens reduction
Ni / 
R  CH  CH2  H2  R  CH2  CH3
Ni / 
R  C  CH  H2  R  CH2  CH3
Methods of preparation
From alkyl halides
Zn-Cu or Sn and HCl or Zn and HCl
RX +2H 
R -H+HX
or HI and Red P or Pd and H2
R – I > R – Br > R – Cl > R – F
From aluminium carbide
Al4C3 +12H2O  3CH4 +4Al(OH)3
Corey House reaction
2RX +Cu+LiI  R2CuLi
R2CuLi+R'X
dialkyl copper lithium
dry ether

R -R'+RCu+LiX
Physical Properties
Boiling point:
CH3 CH2 CH2 CH2 CH3
n-pentane
boiling point = 309 K
H3C — CH — CH2CH3
CH3
iso-pentane
boiling point = 301 K
CH3
H3C — C — CH3
CH3
neo-pentane
boiling point = 282.5 K
Physical Properties
Melting point:
C
C
C
C
C
Zigzag arrangement of
carbon atoms in alkanes
C
Physical Properties
(a) Alkane having even carbon atom
C
C
C
C
C
C
Symmetrical –higher melting point
(a) Alkane having odd carbon atom
C unsymmetrical – lower melting point
C
C
C
C
Chemical properties
Combustion
CH4  2O2  CO2  2H2O
 H  217.0 K cal/mole
Oxidation
Cu
CH4  O2 
 2CH3OH
573 K
Mo O
2 3
CH4  O2 
 HCHO  H2O
Methanal
Halogenation
CH4 + Cl2
h
CH3Cl + HCl
Mechanism
Initiation
h
Cl — Cl
2Cl
Propagation

CH4  Cl

 CH3  HCl

CH3  Cl2  CH3Cl  Cl 
Termination


CH3  CH3  CH3  CH3
Cl  Cl  Cl2

CH3  Cl  CH3Cl
Features of Halogenation
F2 > Cl2 > Br2 > I2.
Attack of or on an alkane is selective
Order of reactivity is 3° > 2° > 1°
Features of Halogenation
Cl
2

 CH3 CH2 CH2 CH2 Cl + CH3 CH2 CHCH3
CH3 CH2 CH2 CH3 
hv
n - Butane
CH3
1°
CH3
1°
Cl
CH3
CH3
3°
1°
CH - CH3
Isobutane
Cl2
C
CH3
Cl
36% (3°)
h
CH3
CH3
CH - CH2 Cl
64%
(1°)
Features of Halogenation
Nitration
CH3 CH2 CH3
450°C
Conc. HNO3
CH3 CH2 CH2 NO2 + CH3 CHCH3 + CH3 CH2 NO2 + CH3 NO2
NO2
Features of Halogenation
Sulphonation
CH3
CH3
CH3
CH CH3
oleum
CH3
C
CH3
SO3 H
tert butyl sulphonic acid
isobutane
Isomerization
CH3
H3 C(CH2 )3 CH3
n-Pentane
AlCl3 / HCl
H3 CCHCH2 CH3
2-Methyl butane
Features of Halogenation
Aromatization
H3C(CH2)4CH3
Hexane
Cr2O3
773 K
10-20 atm
Benzene
Refining of petroleum
Sources of hydrocarbons
Petroleum
Aliphatic hydrocarbons
Coal
Aromatic hydrocarbons
Refining of petroleum
Fraction
Boiling range
Gaseous
113 to 303 K
Approximate
composition
uses
C1 - C5 (2%)
For producing carbon
black and in preparation
of ammonia, methyl
alcohol and gasoline.
Petroleum 303 to 363
ether or
Ligroin
C5 - C7 (2%)
solvent for oils, fats,
rubber and in dry
cleaning.
Gasoline
or petrol
C7 - C12 (32%) Mainly as a motor fuel.
343 to 473
Refining of petroleum
Fraction
Boiling
range
Approximate
composition
uses
Kerosene
448 to 548
C12 - C15(18%) Illuminant fuel and for
preparing petrol gas.
Gas oil,
fuel oil and
diesel oil
523 to 673
C15 - C18(20%) In furnace oil, fuel for
diesel engines and in
cracking.
Lubricating
oils and
petroleum
jelly
623 and up
C16 and up
Used mainly as
lubricants.
Cracking
Heat
Higher alkanes 
Lower alkanes + alkene
Synthetic petrol
Fischer-Tropsch process
Co or Ni
 Mixture of hydrocarbons + H2O
CO+H2 +H2 
Water gas
473 K, 1-10 atm
Bergius process
C+H2
Iron oxide


750 K, 200-250 atm
Mixture of hydrocarbons
Octane number
• Percentage by volume of iso-octane
in the mixture of iso-octane and nheptane which has the same antiknocking qualities as the fuel under
examination.
CH3 CH2 CH2 CH2 CH2 CH2 CH3
Octane number = 0
n-heptane
CH3
CH3
H3 C — C — CH2 — CH — CH3
CH3
iso-octane
(2, 2, 4-trimethyl pentane)
Octane number = 100
Octane number
• Straight chain alkanes have low octane numbers.
The greater the length of the chain, lower is the
octane number.
• Straight chain alkenes and alkynes, and also cyclic
alkanes
have higher octane numbers than their corresponding
alkanes.
• Branched chain hydrocarbons have high octane
numbers.
• Aromatic hydrocarbons have very high octane
numbers.
Class exercise
Class exercise 1
Ethyl iodide + n-propyl iodide
Wurtz reaction


mixture of hydrocarbon.
Which of the following hydrocarbons
will not be formed?
(a) Butene
(b) n-hexane
(c) n-pentane
(d) n-butane
Solution
C 2H5I + CH3 CH2 CH2 I
Ethyl iodide n-Propyl iodide
CH3 – CH2 – CH2 – CH2 – CH3 + CH3 — CH2 — CH2 — CH3
n-pentane
+ CH3 – CH2 – CH2 – CH2 – CH2 – CH3 + C2H4 + C3H8
n-hexane
Hence, the answer is (a).
Class exercise 2
CH3 3 CMgCl+D2O Product.
Product is
(a) CH3 2 C CH2D
(C)
CH3 3 COD
(b) CH3 3 CD
(d) CD3 3 .CH
Solution:
CH3
CH3
CH3 — C — MgCl + D2O
CH3 — C — D + Mg
Br
OD
CH3
Hence, the answer is (a).
CH3
Class exercise 3
Which of the following carbides
are used in preparation of
methane by action of water?
(a) CaC2
(b) Al4C3
(c) Si C
(d) All of these
Solution:
CaC2  H2O  Ca  OH2  C2H2
SiC+H2O  No reaction
Al4C3 +6H2O  2Al2O3 +3CH4
Hence, the answer is (b).
Class exercise 4
Which of the following reactions
will give maximum yield?
UV light
(a) C2H6 +Cl2  excess  


UV light
(b) C2H6  excess +Cl2 


UV light
(c) C2H6 +Cl2 


(d) C2H6 +Cl2 
Solution:
Reaction proceeds via free radical mechanism and
excess of C2H6 is needed as it forms various
products.
Hence, the answer is (b).
Class exercise 5
Which of the following
compounds will have the
highest heat of combustion?
(a)C3H6
(b) C5H12
(c) C6H14
(d) C10H22
Solution:
It has highest number of carbon and hydrogen atoms.
Hence, the answer is (d).
Class exercise 6
Which of the following reactions
will give unsymmetrical alkanes in
good yield?
(a) Frankland reaction
(b) Wurtz reaction
(c) Corey House reaction
(d) All of these
Solution:
Frankland and Wurtz reaction gives alkane
having even number of carbon atoms.
Hence, the answer is (c).
Class exercise 7
CH3 — CH — CH3
Br
2
light
CH3
CH3
CH3 — CH — CH2 — Br + CH3 — C —
B
CH3
(X)
(Y)
Which of the following is true for above
reaction?
(a) Compound X and Y are formed in equal
quantities
(b) Compound Y is formed in excess
(c) Compound X is formed in excess
(d) None of these
Solution
3° free radical is most stable.
3° hydrogens are most easily replaced.
Hence, the answer is (b).
Class exercise 8
The monochlorination of an alkane
(molecular formula C8H18) gives only one
product. The IUPAC name of alkane is
(a) neo octane
(b) iso octane
(c) 2, 2, 3-trimethylpentane
(d) 2, 2, 3, 3-tetramethylbutane
Solution:
CH3 CH3
CH3 — C — C — CH3
CH3 CH3
All the hydrogen present in it are of same type
Hence, the answer is (d).
Class exercise 9
Hydrolysis of calcium carbide
gives a solution with pH
(a) 0
(b) < 7
(c) > 7
(d) 3
Solution:
CaC2 +2H2O  C2H2 +CaOH2
pH is more than 7 due to formation of Ca(OH)2.
Hence, the answer is (c).
Class exercise 10
Sodium benzoate + sodia lime

X. X is
(a) ethane
(c) ethene
(b) benzene
(d) propane
Solution:
COONa
+ NaOH
CaO
Hence, the answer is (b).
+ Na2CO3
Thank you