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