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Correct increasing order of reactivity towards S_N2 reactions of butyl bromides is
A = n–butyl bromide
B = Iso butyl bromide
C = 3° butyl bromide
D = 2° butyl bromide

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C, D, B, A

C, D, A , B

C, A, D, C

D, C, B, A

answer is B.

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Detailed Solution

${CH}_{3} - {CH}_{2} - {CH}_{2} - {CH}_{2} - Br (1^{0}) > {({CH}_{3})}_{2} CH - {CH}_{2} - Br (1^{0}) > {CH}_{3} - {CH}_{2} - CH (Br) - {CH}_{3} (2^{0}) > {({CH}_{3})}_{3} C - Br (3^{0})$

Lesser the steric hindrance on carbon bonded to halogen greater will be the reactivity towards S_N2 reactions.

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