Arrange the following compounds in increasing order of their reactivity towards \( S_N2 \) reaction:
(A) \( CH_3CH_2CH_2CH_2Br \)
(B) \( (CH_3)_2CHCH_2Br \)
(C) \( CH_3CH_2CH(Br)CH_3 \)
(D) \( (CH_3)_3CBr \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (D), (C), (B), (A)
In an \(S_N2\) reaction the nucleophile attacks the carbon carrying the halogen from the side opposite to the halogen, and the C–Br bond breaks as the new bond forms, in a single step. The nucleophile must be able to reach that carbon, so bulky groups on or near it slow the reaction (steric hindrance). Primary halides react fastest, then secondary, then tertiary; among primary halides, a branch on the next carbon also gets in the way.
(A) \( \ce{CH3CH2CH2CH2Br} \): 1-bromobutane, a primary bromide with an unbranched chain; the least hindered.
(B) \( \ce{(CH3)2CHCH2Br} \): 1-bromo-2-methylpropane, also primary, but the branch on the neighbouring carbon crowds the back of the reacting carbon.
(C) \( \ce{CH3CH2CH(Br)CH3} \): 2-bromobutane, a secondary bromide with two alkyl groups on the reacting carbon.
(D) \( \ce{(CH3)3CBr} \): 2-bromo-2-methylpropane, a tertiary bromide; three methyl groups shield the reacting carbon almost completely.
In increasing order of reactivity towards \(S_N2\):
\( \displaystyle \ce{(CH3)3CBr} \lt \ce{CH3CH2CH(Br)CH3} \lt \ce{(CH3)2CHCH2Br} \lt \ce{CH3CH2CH2CH2Br} \)
Shortcut: The tertiary bromide (D) is the least reactive, and only Option (1) starts with (D).
NCERT (2026-27 edition):
Reactions of Haloalkanes – …
