When anisole is heated with HI , the product is :

Options

A. Phenyl iodide and methyl iodide
B. Phenol and methanol
C. Phenyl iodide and methanol
D. Methyl iodide and phenol

Answer

To determine the product when anisole is heated with HI, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Anisole**:
- Anisole is an aromatic compound with the formula C6H5OCH3. It consists of a benzene ring (C6H5) attached to a methoxy group (OCH3).

2. **Reaction with HI**:
- When anisole is heated with hydroiodic acid (HI), a strong acid, the HI dissociates into H⁺ and I⁻ ions. The H⁺ ion acts as a proton (electrophile) that can interact with the oxygen atom in the methoxy group.

3. **Protonation of the Methoxy Group**:
- The lone pair of electrons on the oxygen atom of the methoxy group attacks the H⁺ ion, leading to the formation of a protonated methoxy group (OCH3H⁺). This creates a positive charge on the oxygen.

4. **Cleavage of the O-CH3 Bond**:
- To stabilize the positive charge on the oxygen, the bond between the oxygen and the methyl group (O-CH3) breaks. This results in the formation of a carbocation (C6H5+) and methanol (CH3OH).

5. **Formation of Phenol**:
- The carbocation (C6H5+) can then stabilize itself by losing a proton, resulting in the formation of phenol (C6H5OH).

6. **Iodide Ion Attack**:
- The iodide ion (I⁻), which was generated from the dissociation of HI, can now attack the carbocation. This results in the formation of methyl iodide (CH3I).

7. **Final Products**:
- The final products of the reaction are methyl iodide (CH3I) and phenol (C6H5OH).

### Conclusion:
The products formed when anisole is heated with HI are methyl iodide (CH3I) and phenol (C6H5OH).