Write the mechanism of the reaction of HI with methoxybenzene.

Explanation

To write the mechanism of the reaction of HI with methoxybenzene (C6H5O-CH3), we can break it down into several steps:

### Step 1: Protonation of Methoxybenzene
The first step involves the protonation of the ether oxygen by the hydrogen ion (H⁺) from HI. The lone pair of electrons on the oxygen atom of methoxybenzene attacks the H⁺ ion, leading to the formation of a protonated ether.

**Reaction:**
\[ \text{C}_6\text{H}_5\text{O-CH}_3 + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{OH}^+-\text{CH}_3 \]

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