Arrange the following alkyl halides in increasing order of reactivity towards $SN^2$ reaction?

(A) $CH_3-Cl$
(B) $CH_3-Br$
(C) $CH_3-F$
(D) $CH_3-I$

Choose the correct answer from the options given below.

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

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

In SN2 reactions, reactivity of alkyl halides (R–X) depends primarily on the leaving group ability of the halide ion (X⁻). Better leaving groups result in faster reactions because the C–X bond breaks more easily in the transition state.

The order of leaving group ability among halides is I⁻ > Br⁻ > Cl⁻ > F⁻, due to:

• Decreasing C–X bond strength down the group (C–F is the strongest, ~485 kJ/mol; C–I is the weakest, ~240 kJ/mol).
• Increasing stability of the halide anion (larger, more polarizable ions like I⁻ are more stable).

Since all compounds here are methyl halides (identical steric hindrance, ideal for SN2), the reactivity order is directly tied to the leaving group:

CH₃F (C) < CH₃Cl (A) < CH₃Br (B) < CH₃I (D)

Thus, increasing order: (C), (A), (B), (D)