Consider the following alkyl halides. Choose the correct answer from the options given below for the decreasing order of the ease of the nucleophilic substitution reaction.
A. Methyl Bromide
B. Methyl Chloride
C. Methyl Iodide
D. Methyl Fluoride
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → C, A, B, D
In a nucleophilic substitution reaction of an alkyl halide, the carbon–halogen bond breaks and the halide ion leaves:
\( \displaystyle \ce{CH3-X + OH^- -> CH3-OH + X^-} \)
For the same alkyl group, the ease of substitution depends on:
C–X bond strength: the bond becomes weaker down the group as the halogen atom gets larger, so it breaks more easily.
Leaving group ability: \( \ce{I^{-}} \) is the weakest base and the best leaving group; \( \ce{F^{-}} \) is the poorest.
Bond enthalpy (kJ mol−1): \( \ce{CH3-F} \) 452 > \( \ce{CH3-Cl} \) 351 > \( \ce{CH3-Br} \) 293 > \( \ce{CH3-I} \) 234
Ease of substitution: R–I > R–Br > R–Cl > R–F
Matching the given compounds:
C. Methyl Iodide, \( \ce{CH3I} \): weakest C–X bond
A. Methyl Bromide, \( \ce{CH3Br} \)
B. Methyl Chloride, \( \ce{CH3Cl} \)
D. Methyl Fluoride, \( \ce{CH3F} \): strongest C–X bond
Shortcut: Methyl iodide (C) reacts most easily. Only Option (2) starts with C.
NCERT (2026-27 edition):
Reactions of Haloalkanes – “… For a given alkyl group, the reactivity of the halide, R-X, follows the same order in both the mechanisms R–I > R–Br > R–Cl >> R–F.”