What is the correct order of reactivity of the following alkyl halides towards ammonolysis reaction?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → CH3I > CH3Br > CH3Cl
In ammonolysis, ammonia acts as a nucleophile and replaces the halogen of the alkyl halide (in ethanolic solution, in a sealed tube at 373 K):
\( \displaystyle \ce{CH3-X + NH3 -> CH3-NH3^+X^-} \)
The alkyl group is the same in all three, so the rate depends on how easily the C–X bond breaks. The C–X bond gets longer, and so weaker, from C–Cl to C–I, as the halogen atom gets larger.
First – \( \ce{CH3I} \): iodine is the largest halogen; the C–I bond is the longest and weakest, so it is replaced fastest.
Second – \( \ce{CH3Br} \): the C–Br bond is shorter and stronger than C–I, but longer and weaker than C–Cl.
Third – \( \ce{CH3Cl} \): the C–Cl bond is the shortest and strongest of the three, so it is replaced most slowly.
NCERT (2026-27 edition):
9.4 Preparation of Amines – “The order of reactivity of halides with amines is RI > RBr > RCl.”
6.3 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.”