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.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C), (A), (B), (D)
In an \( \mathrm{S_N2} \) reaction the C–X bond breaks as the nucleophile attacks, so for the same alkyl group the reactivity depends on how easily the halide ion leaves. Down the group from F to I the C–X bond becomes longer and weaker, and the larger halide ion is a better leaving group:
\( \displaystyle \ce{R-I} \gt \ce{R-Br} \gt \ce{R-Cl} \gt\gt \ce{R-F} \)
The given halides all have the methyl group, so only the halogen differs:
(C) \( \ce{CH3-F} \): strongest bond, least reactive.
(A) \( \ce{CH3-Cl} \)
(B) \( \ce{CH3-Br} \)
(D) \( \ce{CH3-I} \): weakest bond, most reactive.
Shortcut: The fluoride (C) is the least and the iodide (D) the most reactive; only Option (4) starts with (C) and ends with (D).
NCERT (2026-27 edition):
Reactions of Haloalkanes – …
Nature of C-X Bond – “… As we go down the group in the periodic table, the size of halogen atom increases. Fluorine atom is the smallest and iodine atom is the largest. Consequently the carbon-halogen bond length also increases from C—F to C—I.”