Explain the reaction of anisole with $HI$.

(N/A) The reaction is as follows:
$C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I$
Explanation:
Anisole is an ether containing $C_6H_5-O$ and $O-CH_3$ bonds.
The $C_6H_5-O$ bond is stronger because the oxygen is attached to an $sp^2$ hybridized carbon of the benzene ring,and resonance gives the $C-O$ bond partial double bond character.
Therefore,the $O-CH_3$ bond is weaker and breaks during the reaction.
Mechanism:
$(i)$ Protonation: The oxygen atom of anisole gets protonated by $H^+$ to form a protonated ether.
$C_6H_5-O(CH_3)-H^+$
(ii) Nucleophilic attack: The iodide ion $(I^-)$ attacks the less sterically hindered methyl group $(CH_3)$,breaking the $O-CH_3$ bond to form phenol $(C_6H_5OH)$ and methyl iodide $(CH_3I)$.
The phenol formed does not react further with $HI$ to form $C_6H_5I$ because the $sp^2$ hybridized carbon of the benzene ring does not undergo nucleophilic substitution with $I^-$.