Arrange the following compounds towards increasing S_N2 reaction? (A)…

The correct answer is Option (4) → (C), (D), (B), (A)

In an \(S_N2\) reaction the nucleophile attacks the carbon carrying the halogen from the back side, in a single step, while the halide leaves. The more bulky groups there are on and near that carbon, the harder it is for the nucleophile to reach it. So the rate is decided by steric hindrance: primary > secondary > tertiary, and among primary halides, branching on the next carbon also slows the attack.

• (C) tert-Butyl bromide, \( \ce{(CH3)3C-Br} \): a tertiary carbon with three methyl groups blocking the back side; the slowest.

• (D) sec-Butyl bromide, \( \ce{CH3CH2CH(Br)CH3} \): a secondary carbon with two alkyl groups; faster than the tertiary halide.

• (B) Isobutyl bromide, \( \ce{(CH3)2CH-CH2Br} \): a primary carbon, but the branched carbon next to it still crowds the line of attack; faster than the secondary halide.

• (A) n-Butyl bromide, \( \ce{CH3CH2CH2CH2Br} \): a primary carbon on an unbranched chain; the least hindered and the fastest.

In increasing order of \(S_N2\) reactivity:

    \( \displaystyle \text{tert-butyl} \lt \text{sec-butyl} \lt \text{isobutyl} \lt \text{n-butyl bromide} \)

Shortcut: The tertiary halide must be the slowest. Only Option (4) starts with (C).

NCERT (2026-27 edition):
Substitution Nucleophilic Bimolecular (S_N2) – “… Since this reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky substituents on or near the carbon atom have a dramatic inhibiting effect. Of the simple alkyl halides, methyl halides react most rapidly in S_N2 reactions because there are only three small hydrogen atoms. Tertiary halides are the least reactive because bulky groups hinder the approaching nucleophiles. Thus the order of reactivity followed is: Primary halide > Secondary halide > Tertiary halide.”