The correct increasing reactivity order for \( S_N2 \) reaction for the following compounds
(A) Benzyl Chloride
(B) Methyl Chloride,
(C) Isopropyl Chloride
(D) Tertiary butyl Chloride
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (D), (C), (B), (A)
Note: None of the given options was correct in the NTA question paper. To provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.
An \( \mathrm{S_N2} \) reaction takes place in one step: the nucleophile attacks the carbon carrying the halogen from the side opposite to the halogen while the halide ion leaves. Two factors decide the rate:
Steric hindrance: bulky groups on that carbon block the approach of the nucleophile, so the reactivity falls as the number of alkyl groups on it rises: methyl > 1° > 2° > 3°.
A neighbouring benzene ring: in a benzylic halide the π electrons of the ring overlap with the carbon being attacked and stabilise the \( \mathrm{S_N2} \) transition state, so a primary benzylic halide reacts very fast.
(D) Tertiary butyl Chloride, \( \ce{(CH3)3CCl} \) — first (least reactive): tertiary, three methyl groups, the most hindered.
(C) Isopropyl Chloride, \( \ce{(CH3)2CHCl} \) — second: secondary, two methyl groups.
(B) Methyl Chloride, \( \ce{CH3Cl} \) — third: no alkyl group on the carbon, the least hindered of the alkyl chlorides.
(A) Benzyl Chloride, \( \ce{C6H5CH2Cl} \) — fourth (most reactive): primary and benzylic; the ring stabilises the transition state, so it reacts even faster than methyl chloride.