Question

When iodoethane is boiled with alc. KOH, an alkene X is formed.

(a) Write the reaction and identify the alkene.
(b) What happens when X is treated with HBr?
(c) Give a chemical test for X.
(d) How will you convert X into polyethene?
A. {CH}_3{CH} = {CH}_2 )
B. {CH}_2 = {CH}_2 )
C. {CH}_3{CH} = {CH}_{2} )
D. {CH}_4 )

Answer 1

When iodoethane is boiled with alcoholic KOH, ethylene (ethene) is formed. Ethylene reacts with HBr to form bromoethane and with bromine water to indicate the presence of an alkene. Ethylene can be polymerized to form polyethylene under high pressure and temperature with a catalyst.

Step-by-step explanation:

When iodoethane is boiled with alcoholic KOH, an elimination reaction occurs to form an alkene. The specific reaction can be represented as:

CH3CH2I + KOH(alcoholic) → CH2=CH2 + KI + H2O

The alkene formed here is ethylene (ethene), which is option B: CH2=CH2.

(b) When ethylene is treated with HBr, a haloalkane is formed, specifically bromoethane:

CH2=CH2 + HBr → CH3CH2Br

(c) A chemical test for ethylene would be its reaction with bromine water, where the brownish-red color of the solution disappears, indicating the presence of an alkene.

(d) To convert ethylene to polyethylene, a polymerization reaction is carried out, typically with a catalyst under high pressure and temperature:

n CH2=CH2 → (-CH2-CH2-)n

This process is known as the polymerization of ethylene to form polyethene (also known as polyethylene).