Question: 55: Predict the order of reactivity of the following four isomers towards \mathrm{S}_{\mathrm{N}} 2 reaction.
(I) \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}
(II) \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_{3}
(III) \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{Cl}
(IV) \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}
(1) (IV) > (III) > (II) > (I)
(2) (I) > (II) > (III) > (IV)
(3) (I) > (III) > (II) > (IV)
(4) (IV) > (II) > (III) > (I)
Answer: Option (3)
Explanation:
\mathrm{S}_{\mathrm{N}}2 reactions proceed through a single-step backside attack of the nucleophile.
The rate decreases with increase in steric hindrance around the carbon atom bearing the leaving group.
Primary alkyl halides show highest \mathrm{S}_{\mathrm{N}}2 reactivity,
followed by secondary alkyl halides, while tertiary alkyl halides do not undergo
\mathrm{S}_{\mathrm{N}}2 reactions due to severe steric hindrance.
Compound (I) is a straight-chain primary alkyl chloride, having the least steric hindrance,
so it shows maximum reactivity.
Compound (III) is a branched primary alkyl chloride. Due to branching,
steric hindrance is slightly more than (I), so its reactivity is lower than (I)
but higher than secondary alkyl halide.
Compound (II) is a secondary alkyl chloride, which shows lower
\mathrm{S}_{\mathrm{N}}2 reactivity than primary alkyl halides.
Compound (IV) is a tertiary alkyl chloride and does not undergo
\mathrm{S}_{\mathrm{N}}2 reaction.
Therefore, the correct order of reactivity is:
(\mathrm{I})>(\mathrm{III})>(\mathrm{II})>(\mathrm{IV})