Arrange the following alkyl halides in increasing order of reactivity towards \( S_N2 \) reaction:
(A) 1-Bromobutane
(B) 1-Bromo-2,2-dimethylpropane
(C) 1-Bromo-2-methylbutane
(D) 1-Bromo-3-methylbutane
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (C), (D), (A)
In an \(S_N2\) reaction the nucleophile attacks the carbon carrying the halogen from the back side. Any bulky group near this carbon gets in the way, so steric hindrance decides the rate. All four compounds here are primary bromides, so the difference comes from branching on the next carbons: a branch on the β-carbon (C2) hinders more than one on the γ-carbon (C3), and two branches hinder more than one.
(A) 1-Bromobutane, \( \ce{CH3CH2CH2CH2Br} \): an unbranched chain, the least hindered.
(B) 1-Bromo-2,2-dimethylpropane, \( \ce{(CH3)3CCH2Br} \): two methyl groups on the β-carbon, the most hindered.
(C) 1-Bromo-2-methylbutane, \( \ce{CH3CH2CH(CH3)CH2Br} \): one methyl group on the β-carbon.
(D) 1-Bromo-3-methylbutane, \( \ce{(CH3)2CHCH2CH2Br} \): the branch is on the γ-carbon, farther from the reaction centre.
In increasing order of reactivity towards \(S_N2\):
\( \displaystyle \ce{(CH3)3CCH2Br} \lt \ce{CH3CH2CH(CH3)CH2Br} \lt \ce{(CH3)2CHCH2CH2Br} \lt \ce{CH3CH2CH2CH2Br} \)
So the order is (B), (C), (D), (A).
Shortcut: The least reactive is the neopentyl-type bromide (B), with two methyl groups next to the reacting carbon. Only Option (3) starts with (B).