Write the mechanism of the reaction of HI with methoxybenzene.

Methoxybenzene reacts with hydroiodic acid HI to form phenol and iodomethane.

Mechanism of the above reaction is as follows:

Methoxybenzene is an alkyl aryl ether. There are two bonds linked with an oxygen atom in methoxybenzene. One is an oxygen-methyl group( O-alkyl bond) and the other one is Oxygen-Aryl group( O-Aryl bond). Out of these, Oxygen-Aryl bond is more stable due to resonance having partial double bond character. So in this reaction, the oxygen-methyl bond gets cleaved to form methyl iodide and phenol.

The reaction of HI with methoxybenzene takes place in two steps.

Step1:

The oxygen of ether gets protonated (addition of proton) by capturing H⁺ of HI to form protonated ether molecule, known as oxonium ion.

Step 2:

In this step, SN² reaction occurs since Iodide ion (I^-) is a good nucleophile. I^- ion preferably attacks the least substituted carbon of protonated ether molecule ( in this case methyl group) to form methyl iodide and phenol.