Question:

Which of the following is most reactive towards \(S_\mathrm{N}2\) reaction?

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Remember the \(S_\mathrm{N}2\) reactivity order: \[ \boxed{ \mathrm{CH_3X} > 1^\circ > 2^\circ \gg 3^\circ. } \] Less steric hindrance gives faster backside attack.
Updated On: Jul 18, 2026
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The Correct Option is B

Solution and Explanation

Step 1: Recall the order of reactivity in \(S_\mathrm{N}2\) reactions. The rate of an \(S_\mathrm{N}2\) reaction decreases with increase in steric hindrance. The general order is \[ \boxed{ \text{Methyl} > 1^\circ > 2^\circ \gg 3^\circ. } \]

Step 2:
Compare the given alkyl bromides. Among the given compounds, \[ \mathrm{CH_3CH_2CH_2CH_2Br} \] (1-bromobutane) is a primary alkyl halide with the least steric hindrance. Therefore, it undergoes the fastest \(S_\mathrm{N}2\) reaction.

Step 3:
Choose the correct option. Hence, \[ \boxed{\text{1-Bromobutane}} \] is the most reactive towards \(S_\mathrm{N}2\) reaction. Therefore, the correct option is \(\boxed{(B)}\).
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