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:

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

The correct answer is Option (3) → (B), (C), (D), (A)

The $S_N2$ reaction rate is governed primarily by steric hindrance. The less crowded the carbon atom being attacked, the faster the reaction proceeds.

All given compounds are primary alkyl bromides, but they differ in the branching of their carbon chains:

1. (A) 1-Bromobutane: A straight-chain primary halide with the least steric hindrance. (Fastest)

2. (D) 1-Bromo-3-methylbutane: Branching is at the third carbon (gamma position), far from the reaction center.

3. (C) 1-Bromo-2-methylbutane: Branching is at the second carbon (beta position), closer to the reaction center, increasing hindrance.

4. (B) 1-Bromo-2,2-dimethylpropane (Neopentyl bromide): Significant branching (two methyl groups) at the beta carbon. This creates massive steric hindrance, making it nearly inert to $S_N2$. (Slowest)

Increasing Order of Reactivity: (B) < (C) < (D) < (A).