Methoxybenzene on treatment with HI produces

Options

A. iodobenzene and methanol
B. phenol and methyl iodide
C. jodobenzene and methyl iodide
D. phenol and methanol

Explanation

To solve the question regarding the reaction of methoxybenzene (also known as anisole) with hydroiodic acid (HI), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
Methoxybenzene has a benzene ring with a methoxy group (-OCH₃) attached to it. The chemical formula is C₇H₈O.

2. **Reaction with HI**:
When methoxybenzene reacts with hydroiodic acid (HI), it dissociates into H⁺ and I⁻ ions. The H⁺ ion will interact with the oxygen atom of the methoxy group.

3. **Protonation of the Methoxy Group**:
The lone pair of electrons on the oxygen atom of the methoxy group will attack the H⁺ ion, resulting in the formation of a protonated methoxy group (O⁺H).

4. **Bond Cleavage**:
The protonated methoxy group is now positively charged and can undergo cleavage. The bond between the oxygen and the methyl group (C-O bond) is weaker than the bond between the oxygen and the benzene ring due to the positive charge on the oxygen. Therefore, this bond will break.

5. **Formation of Products**:
When the C-O bond breaks, it releases methanol (CH₃OH) and forms a phenyl cation (C₆H₅⁺). The iodide ion (I⁻) from HI can then react with the phenyl cation to form iodobenzene (C₆H₅I).

6. **Final Products**:
The final products of the reaction between methoxybenzene and HI are iodobenzene (C₆H₅I) and methanol (CH₃OH).

### Summary of Reaction:
Methoxybenzene + HI → Iodobenzene + Methanol