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Chemistry
Alkane
Session objectives
1. Methods of preparation2. Kolbe’s electrolysis3. Physical Properties and Chemical
properties4. Halogenation5. Refining of petroleum6. 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
Dry3 2 3 2 2 3
ether2CH CH Br 2Na CH CH CH CH 2NaBr
• Follow mainly free radical mechanism • Useful in preparing an alkane containing even number of carbon atoms• Stepping up reaction
Frankland reaction
2RX+Zn+Rx R – R+ZnX
Decarboxylation of sodium or potassium salt of fatty acid
CaO/Heat2 3
AlkaneSodium salt ofcarboxylic acid
RCOONa +NaOH R - H +Na CO
For example
CaO/Heat3 4 2 3
MethaneSodium acetateCH COONa +NaOH CH +Na CO
Kolbe’s electrolysis
At anode
Sodium salt ofcarboxylic acid
2RCOONa 2RCOO 2Na
•- -2RCOO 2RCOO+2e
• •
22RCOO 2R +CO•
Alkane2R R - R
Kolbe’s electrolysis
For example- +
3 3Sodium acetate2CH COONa 2CH COO +2Na
At anode•
- -3 32CH COO 2CH COO+2e
• •
3 3 22CH COO 2CH +CO
•
3 3 3Ethane
2CH CH - CH
From Grignard reagent (RMgX)
RMgX+HOH RH+Mg(OH)X
RMgX+R'OH RH+Mg(OR')X
2RMgX+R'NH RH+Mg(NHR')X
From unsaturated hydrocarbonsSabatier-Senderens reduction
Ni /2 2 2 3R CH CH H R CH CH
Ni /2 2 3R C CH H R CH CH
Methods of preparation
From alkyl halides
2
Zn-Cu or Sn and HCl or Zn and HCl
or HI and Red P or Pd and HRX+2H R - H+HX
R – I > R – Br > R – Cl > R – F
From aluminium carbide
4 3 2 4 3Al C +12H O 3CH +4Al(OH)
Corey House reaction
22RX+Cu+LiI R CuLi
dry ether2
dialkyl copper lithiumR CuLi+R'X R - R'+RCu+LiX
Physical PropertiesBoiling point:
H3C — C — CH3
CH3
CH3
neo-pentaneboiling point = 282.5 K
CH3CH2CH2CH2CH3
n-pentaneboiling point = 309 K
H3C — CH — CH2CH3
CH3
iso-pentaneboiling point = 301 K
Physical Properties
Melting point:
C C C
C C CZigzag arrangement of carbon atoms in alkanes
Physical Properties
(a) Alkane having even carbon atom
C C C
C C C
C C
C C C
(a) Alkane having odd carbon atom
Symmetrical –higher melting point
unsymmetrical – lower melting point
Chemical properties
Combustion
4 2 2 2CH 2O CO 2H O
H 217.0 K cal/mole
Oxidation
Cu4 2 3
573 KCH O 2CH OH
2 3Mo O4 2 2
MethanalCH O HCHO H O
Halogenation
CH4 + Cl2 CH3Cl + HClh
Mechanism
Initiation
Cl — Clh
2Cl
Propagation
4 3CH Cl CH HCl
3 2 3CH Cl CH Cl Cl
Termination
3 3 3 3CH CH CH CH
2Cl Cl Cl
3 3CH Cl CH Cl
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
CH3CH2CH2CH3CH3CH2CH2CH2Cl + CH3CH2CHCH3
hv
Cl2
Cln - Butane
1° CH - CH3
Cl2
C
CH - CH2 Cl
CH3
CH3
CH3 CH3
Cl
CH3
CH3
1°
3° 1°CH3
36%
64%
Isobutane
(3°)
(1°)
h
Features of Halogenation
Nitration
CH3CH2CH3
450°C
Conc. HNO3
CH3CH2CH2NO2 + CH3CHCH3 + CH3CH2NO2+ CH3NO2
NO2
Features of Halogenation
Sulphonation
CH3
CH3
CH3CH CH3
SO3H
CH3
CH3Coleum
isobutane tert butyl sulphonic acid
Isomerization
H3C(CH2)3CH3
AlCl3 / HClH3CCHCH2CH3
CH3
n-Pentane 2-Methyl butane
Features of Halogenation
Aromatization
H3C(CH2)4CH3
Cr2O3
Hexane773 K
10-20 atm Benzene
Refining of petroleum
Sources of hydrocarbons
Petroleum Aliphatic hydrocarbons
CoalAromatic hydrocarbons
Refining of petroleum
Fraction Boiling range Approximate composition
uses
Gaseous 113 to 303 K C1 - C5 (2%) For producing carbon black and in preparation of ammonia, methyl alcohol and gasoline.
Petroleum ether or Ligroin
303 to 363 C5 - C7 (2%) solvent for oils, fats, rubber and in dry cleaning.
Gasoline or petrol
343 to 473 C7 - C12 (32%) Mainly as a motor fuel.
Refining of petroleum
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.
Fraction Boiling range
Approximate composition
uses
Cracking
HeatHigher alkanes Lower alkanes + alkene
Synthetic petrol
Fischer-Tropsch process
Co or Ni2 2 2
473 K, 1-10 atmWater gas
CO+H +H Mixture of hydrocarbons + H O
Bergius process
I ron oxide2
750 K, 200-250 atmC+H Mixture of hydrocarb ons
Octane number• Percentage by volume of iso-octane
in the mixture of iso-octane and n-heptane which has the same anti-knocking qualities as the fuel under examination.
CH3CH2CH2CH2CH2CH2CH3
H3C — C — CH2 — CH — CH3
CH3
CH3
CH3
n-heptane
Octane number = 0
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
mixture of hydrocarbon. Which of the following hydrocarbons will not be formed?
(a) Butene (b) n-hexane
(c) n-pentane (d) n-butane
Wurtz reaction
Solution
C2H5I + CH3 CH2 CH2 I
Ethyl iodide n-Propyl iodide
Hence, the answer is (a).
CH3 – CH2 – CH2 – CH2 – CH3 + CH3 — CH2 — CH2 — CH3
n-pentane
+ CH3 – CH2 – CH2 – CH2 – CH2 – CH3 + C2H4 + C3H8
n-hexane
Class exercise 2
3 23CH CMgCl+D O Product.
Product is
3 22(a) CH C CH D 3 3
(b) CH CD
3 3(C) CH COD 3 3
(d) CD .CH
CH3 — C — MgCl + D2O CH3 — C — D + Mg
CH3
CH3
CH3
CH3
OD
Br
Hence, the answer is (a).
Solution:
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
2 2 2 22CaC H O Ca OH C H
2SiC+H O No reaction
4 3 2 2 3 4Al C +6H O 2Al O +3CH
Solution:
Hence, the answer is (b).
Class exercise 4
Which of the following reactions will give maximum yield?
UV light2 6 2(a) C H +Cl excess
UV light2 6 2(b) C H excess +Cl
UV light2 6 2(c) C H +Cl
2 6 2(d) C H +Cl
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:
Hence, the answer is (d).
It has highest number of carbon and hydrogen atoms.
Class exercise 6Which 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
Frankland and Wurtz reaction gives alkane having even number of carbon atoms.
Solution:
Hence, the answer is (c).
Class exercise 7
CH3
CH3 — CH — CH3 Br2
lightCH3
CH3
CH3 — CH — CH2 — Br + CH3 — C —
B
(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 8The 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
CH3 — C — C — CH3
CH3CH3
CH3 CH3
Solution:
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:
Hence, the answer is (c).
2 2 2 2 2CaC +2H O C H +Ca OH
pH is more than 7 due to formation of Ca(OH)2.
Class exercise 10Sodium benzoate + sodia lime
D X. X is
(a) ethane (b) benzene(c) ethene (d) propane
COONa
+ NaOHCaO
+ Na2CO3
Solution:
Hence, the answer is (b).
Thank you
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