Question

Phenyl methyl ether (anisole) reacts with HI to give phenol and methyl iodide and not iodobenzene and methyl alcohol because _____.
A- $I^{-}$ ion prefers to combine with the smaller group in order to minimise steric hindrance
B- $I^{-}$ ion is not reactive towards benzene
C- Phenol is formed as a result of hydrolysis of iodobenzene
D- Methyl alcohol formed during reaction reacts with $I^{-}$ to form methyl iodide

Answer 1

Hint: Try to identify the mechanism of the reaction. Try to figure out how HI starts the reaction. Why ethers take part in the reaction. There will be a nucleophilic substitution. Figure out the mechanism of nucleophilic addition.
Complete step by step answer: We know that HI reacts with ethers. HI is a strong acid so hydrogen ion is readily generated. This generated hydrogen ion attacks the oxygen atom of phenyl methyl ether as oxygen has lone pairs.
When oxygen forms 3 bonds and gains a positive charge it becomes somewhat unstable. So the compound undergoes nucleophilic addition. It undergoes the $S_{N^2}$ mechanism as the charge is unstable on the methyl group and also there is a resonance between oxygen and benzene ring. When there is a nucleophilic attack there will be a transition state where there are 5 bonds for the carbon atom of the methyl group.
We know that iodine ions have a large size and the benzene ring is also large compared to the methyl group. If both the iodine and benzene group approach together there will be a lot of steric repulsions. As a result, phenol and methyl iodide are formed.

Therefore, option A is correct.

Note: When there is a positive charge on the oxygen atom then it will resonate with the benzene ring and stabilize the positive charge. That’s why the oxygen group does not leave the benzene ring. This reaction cannot follow the $S_{N^1}$ mechanism because methyl cation is highly unstable.