Which of the following alkyl halides will undergo S_N2 reaction most…

The correct answer is Option (3) → 1-Bromobutane

In an S_N2 reaction the nucleophile attacks the carbon bearing the halogen from the side opposite to it, in a single step. Bulky groups on or near that carbon hinder the approach of the nucleophile, so the reactivity order is methyl > primary > secondary > tertiary, and among primary halides the one with the least branching near the reaction centre reacts fastest.

All four compounds are primary bromides, so the branching decides:

1-Bromobutane, \( \ce{CH3CH2CH2CH2Br} \), has an unbranched chain; the carbon attacked is the least crowded, so it reacts most readily.

Why the other options are incorrect

• Option (1) 1-Bromo-2,2-dimethylpropane — Incorrect. \( \ce{(CH3)3C-CH2Br} \) (neopentyl bromide); although primary, the carbon next to C–Br carries three methyl groups that block the back-side attack, so it reacts extremely slowly.

• Option (2) 1-Bromo-2-methylbutane — Incorrect. \( \ce{CH3CH2CH(CH3)CH2Br} \); the branch on the carbon next to C–Br hinders the nucleophile.

• Option (4) 1-Bromo-3-methylbutane — Incorrect. \( \ce{(CH3)2CHCH2CH2Br} \); the branch is farther away and hinders only a little, but the compound is still slightly slower than unbranched 1-bromobutane.

NCERT (2026-27 edition):
Substitution Nucleophilic Bimolecular (S_N2) – “The reaction between CH₃Cl and hydroxide ion to yield methanol and chloride ion follows a second order kinetics, i.e., the rate depends upon the concentration of both the reactants. … 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.”