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