When anisole is heated with concentrated HI, phenol and methyl iodide are formed, and not iodobenzene and methanol. Which of the following is the reason?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The O–CH3 bond is weaker than the O–C6H5 bond, which has partial double bond character.
HI first protonates anisole to give the methylphenyl oxonium ion, C6H5–O+(H)–CH3. The iodide ion then has two bonds it could break.
The carbon of the phenyl group is sp2 hybridised and its bond to oxygen has partial double bond character, so the I− ion breaks the weaker O–CH3 bond:
\( \displaystyle \ce{C6H5-O-CH3 + HI} \xrightarrow{\Delta} \ce{C6H5OH + CH3I} \)
Why the other options are incorrect
Option (2) The methyl group leaves as a stable methyl carbocation, which then combines with the iodide ion. — Incorrect.
A methyl group does not form a stable carbocation. The iodide ion attacks the methyl carbon directly by an SN2 step.
The carbocation (SN1) route is seen only when one group is tertiary.
Option (3) Phenol is a stronger acid than methanol. — Incorrect. The fact is true, but acidity does not decide which C–O bond breaks; the bond strengths do.
Option (4) The sp2 hybridised carbon of the phenyl group readily undergoes nucleophilic substitution. — Incorrect. It is the reverse: the sp2 carbon cannot undergo nucleophilic substitution, which is also why phenol does not go on to form iodobenzene.
NCERT (2026-27 edition):
7.6.3 Chemical Reactions – “In case of anisole, methylphenyl oxonium ion, C6H5–O+(H)–CH3 is formed by protonation of ether. The bond between O–CH3 is weaker than the bond between O–C6H5 because the carbon of phenyl group is sp2 hybridised and there is a partial double bond character. Therefore the attack by I– ion breaks O–CH3 bond to form CH3I. Phenols do not react further to give halides because the sp2 hybridised carbon of phenol cannot undergo nucleophilic substitution reaction needed for conversion to the halide.”