The correct increasing reactivity order for $S_N2$ reaction for the following compounds

(A) Benzyl Chloride
(B) Methyl Chloride,
(C) Isopropyl Chloride
(D) Tertiary butyl Chloride

Choose the correct answer from the options given below:

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

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

$\text{SN}_2$ reactivity is governed by steric hindrance (bulkiness). The reaction needs a clear path for a ''backside attack.'' Less bulk = faster reaction.

(A) Benzyl (1°): Also has low bulk, but it's even faster than methyl. The adjacent benzene ring helps stabilize the transition state, lowering the reaction's energy.(Most Reactive)

(B) Methyl: Least bulky (only small H atoms). Very fast $\text{SN}_2$.

(C) Isopropyl (2°): Very bulky. Very slow $\text{SN}_2$.

(D) Tertiary butyl (3°): Most bulky. No $\text{SN}_2$ reaction. (Least Reactive)