The correct increasing order of reactivity of alkyl halides with \( CH_3NH_2 \) is
(A) \( C_2H_5-Br \)
(B) \( C_2H_5-I \)
(C) \( C_2H_5-F \)
(D) \( C_2H_5-Cl \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (C), (D), (A), (B)
Methylamine is a nucleophile; its lone pair attacks the carbon of the alkyl halide and the halide ion leaves (an \( S_N2 \) substitution, the same reaction as the ammonolysis of alkyl halides):
\( \displaystyle \ce{C2H5-X + CH3NH2 -> C2H5-NH-CH3 + HX} \)
The rate depends on how easily the C–X bond breaks. Down the group the halogen atom becomes larger, the C–X bond becomes longer and weaker, and the halide ion becomes a better leaving group. The C–X bond enthalpies (kJ mol−1) are C–F 452, C–Cl 351, C–Br 293 and C–I 234, so the reactivity is R–I > R–Br > R–Cl > R–F.
In increasing order:
\( \displaystyle \ce{C2H5-F} \lt \ce{C2H5-Cl} \lt \ce{C2H5-Br} \lt \ce{C2H5-I} \)
Shortcut: The fluoride (C) is the least and the iodide (B) the most reactive, so the order starts with (C) and ends with (B): only Option (2).
NCERT (2026-27 edition):
Ammonolysis of Alkyl Halides – …
