Read the passage carefully and answer the Questions.
Haloalkanes undergo nucelophilic substitution reactions owing to the polarity of C-X bond. The nucleophile reacts with the haloalkane on the carbon possessing a partial positive charge holding the halogen atom. The halogen atom X is replaced by a nucleophile. Depending on the kinetics and mode of bond breaking, the mechanism can be either \( S_N1 \) or \( S_N2 \) reaction. The rate of \( S_N1 \) reaction is governed by the stability of carbocation and in \( S_N2 \) reaction, the rate of reaction is governed by steric factor. Chirality is the main factor in both \( S_N1 \) and \( S_N2 \). In \( S_N1 \) reaction, the chirality of alkyl halide is accompanied by racemization of the product, while in \( S_N2 \) reaction, the product is characterized by inversion of configuration. The structure of alkyl halide and the nature of the solvent also governs the mechanism of the substitution.
Which of the following bromo isomer of butane will react at a faster rate towards \( S_N2 \) reaction?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → n-butyl bromide
In an \(S_N2\) reaction the nucleophile attacks the carbon carrying the halogen from the back side, opposite to the halogen, in a single step. In the transition state this carbon is bonded to five groups at once, so bulky groups on it, or close to it, hinder the approach of the nucleophile. The rate is therefore decided by steric hindrance: the less crowded the carbon, the faster the reaction (methyl > 1° > 2° > 3°).
The four bromo isomers of butane:
n-Butyl bromide, \( \ce{CH3CH2CH2CH2Br} \): primary, with a straight chain; nothing crowds the back side of C-1.
Isobutyl bromide, \( \ce{(CH3)2CHCH2Br} \): also primary, but the branch on the next carbon (C-2) partly blocks the path of the nucleophile.
sec-Butyl bromide, \( \ce{CH3CH2CH(Br)CH3} \): secondary; two alkyl groups sit on the carbon under attack.
tert-Butyl bromide, \( \ce{(CH3)3CBr} \): tertiary; three methyl groups shield the carbon almost completely.
So n-butyl bromide reacts fastest, for example with hydroxide ion:
\( \displaystyle \ce{CH3CH2CH2CH2Br + OH^- -> CH3CH2CH2CH2OH + Br^-} \)
Why the other options are incorrect
Option (1) t-butyl bromide — Incorrect. As a tertiary halide it is the most hindered and the slowest in \(S_N2\) (though the fastest in \(S_N1\)).
Option (2) Sec-butyl bromide — Incorrect. As a secondary halide it has two alkyl groups on the reacting carbon, so it is slower than both primary isomers.
Option (3) Isobutyl bromide — Incorrect. It is primary, but the branch on the neighbouring carbon hinders the back-side attack, so it is slower than the unbranched n-butyl bromide.
NCERT (2026-27 edition):
Reactions of Haloalkanes – …
