For SN2 reaction, the increasing order of the reactivity of the following alkyl halides is :
(A) CH3CH2CH2CH2Br
(B) CH3CH2CH(Br)CH3
(C) (CH3)3CBr
(D) (CH3)2CHCH2Br
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) < (B) < (D) < (A)
An SN2 reaction takes place in one step: the nucleophile attacks the carbon atom carrying the halogen from the side opposite to the halogen, while the halide ion leaves. The more crowded this carbon is, the harder it is for the nucleophile to reach it, so steric hindrance decides the rate:
primary halide > secondary halide > tertiary halide
Branching on the carbon next to the C–X carbon also slows the attack, so a branched primary halide reacts more slowly than an unbranched one.
With the given labels:
(C) \( \ce{(CH3)3CBr} \): 2-bromo-2-methylpropane, tertiary, most hindered: first
(B) \( \ce{CH3CH2CH(Br)CH3} \): 2-bromobutane, secondary: second
(D) \( \ce{(CH3)2CHCH2Br} \): 1-bromo-2-methylpropane, primary but branched next to the C–Br carbon: third
(A) \( \ce{CH3CH2CH2CH2Br} \): 1-bromobutane, primary and unbranched, least hindered: last
Shortcut: The tertiary bromide (C) is the slowest and the unbranched primary bromide (A) is the fastest. Only Option (4) starts with (C) and ends with (A).
NCERT (2026-27 edition):
Reactions of Haloalkanes – …
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | 29-05 S1 15-05 S1 | – | – |