Write the mechanism of the reaction of $HI$ with methoxybenzene.

(N/A) The reaction of methoxybenzene (anisole) with $HI$ proceeds via an $S_N2$ mechanism. The steps are as follows:
Step $1$: Protonation of the ether oxygen atom.
The lone pair on the oxygen atom of methoxybenzene attacks the proton $(H^+)$ from $HI$ to form a protonated ether (oxonium ion).
$C_6H_5-O-CH_3 + H^+ \rightleftharpoons C_6H_5-O^+(H)-CH_3$
Step $2$: Nucleophilic attack by the iodide ion $(I^-)$.
The iodide ion $(I^-)$ acts as a nucleophile and attacks the less sterically hindered methyl carbon atom of the protonated ether. This leads to the cleavage of the $C-O$ bond,resulting in the formation of phenol and methyl iodide $(CH_3I)$.
$C_6H_5-O^+(H)-CH_3 + I^- \xrightarrow{S_N2} C_6H_5-OH + CH_3I$