Arrange the following compounds towards increasing $S_N2$ reaction?

(A) n-Butyl bromide
(B) Isobutyl bromide
(C) Tertiary butyl bromide
(D) Sec-Butyl bromide

Choose the correct answer from the options given below:

(C), (D), (B), (A)

The correct answer is Option (4) → (C), (D), (B), (A)

SN2 reactions proceed via backside attack and are favored by less steric hindrance around the carbon bearing the leaving group.

• Primary alkyl halides → fastest SN2
• Secondary alkyl halides → slower
• Tertiary alkyl halides → very slow (almost no SN2)

The general order of SN2 reactivity is: Primary > Secondary > Tertiary

Given compounds:

1. n-Butyl bromide (A) → primary → fastest
2. Isobutyl bromide (B) → primary but slightly branched → slower than n-butyl
3. Sec-butyl bromide (D) → secondary → slower
4. Tertiary butyl bromide (C) → tertiary → very slow

Increasing order of SN2 reactivity:

$\text{C (slowest)} < \text{D} < \text{B} < \text{A (fastest)}$

Or written from slowest to fastest (increasing reactivity):

$\text{C}, \text{D}, \text{B}, \text{A}$