The cleavage of an aryl-alkyl ether with cold HI gives

Options

A. alkyl iodide and water
B. aryl iodide and water
C. alkyl iodide, aryl iodide and water
D. phenol and alkyl iodide

Explanation

To solve the question regarding the cleavage of an aryl-alkyl ether with cold HI, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Structure of Aryl-Alkyl Ether**:
- An aryl-alkyl ether has the general structure R-O-Ar, where R is the alkyl group and Ar is the aryl group (like a phenyl group, Ph).
- Example: If R is a methyl group (CH₃) and Ar is a phenyl group (C₆H₅), the ether would be C₆H₅-O-CH₃.

2. **Reaction with Cold Hydroiodic Acid (HI)**:
- When the aryl-alkyl ether is treated with cold HI, the first step involves the protonation of the ether oxygen by the hydrogen ion (H⁺) from HI.
- This protonation increases the electrophilicity of the ether oxygen.

3. **Formation of a Positive Charge**:
- The oxygen atom, now positively charged (due to protonation), can facilitate the cleavage of the ether bond.
- The bond between the oxygen and the aryl group (Ar) is stabilized by resonance, making it less likely to break compared to the bond with the alkyl group (R).

4. **Cleavage of the Ether Bond**:
- The bond between the oxygen and the alkyl group (R) is more likely to break than the bond with the aryl group due to the stability of the resulting products.
- The cleavage leads to the formation of phenol (Ar-OH) and an alkyl iodide (R-I).

5. **Final Products**:
- The products of the reaction are phenol (C₆H₅OH) and the corresponding alkyl iodide (R-I).

### Conclusion:
The cleavage of an aryl-alkyl ether with cold HI gives phenol and an alkyl iodide.