By a proper choice of reagent, both symmetrical and unsymmetrical ethers can by prepared by williamson synthesis which involves the reaction between an alkyl halide and an alkoxide ion. The reverse process involving the cleavage of etehrs to give back the original alkyl halide and the alcohol can be carried out by heating the ether with HI at 373K.
Q. Benzyl ethyl ether reacts with HI to form

Options

A. p-iodotoluene annd ethyl alcohol
B. benzyl alcohol and ethyl iodide
C. benzyl iodide ad ethyl alcohol
D. iodobenzene and ethyl alcohol

Explanation

To solve the question regarding the reaction of benzyl ethyl ether with HI, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Benzyl Ethyl Ether**:
- Benzyl ethyl ether can be represented as C6H5CH2-O-CH2CH3. It consists of a benzyl group (C6H5CH2) and an ethyl group (CH2CH3) connected by an ether bond (–O–).

2. **Understand the Reaction Conditions**:
- The ether will be cleaved by heating with hydroiodic acid (HI) at 373 K. This reaction involves the breaking of the ether bond to regenerate the alkyl halide and alcohol.

3. **Cleavage Mechanism**:
- When benzyl ethyl ether reacts with HI, the ether bond (C–O) is broken. The iodine (I) from HI will attach to one of the alkyl groups, while the hydroxyl (OH) group will be formed from the other.

4. **Identify the Products**:
- The benzyl group (C6H5CH2) will react with HI to form benzyl iodide (C6H5CH2I).
- The ethyl group (CH2CH3) will react with HI to form ethanol (CH3CH2OH).

5. **Write the Overall Reaction**:
- The overall reaction can be summarized as:
\[ \text{C6H5CH2-O-CH2CH3} + \text{HI} \rightarrow \text{C6H5CH2I} + \text{CH3CH2OH} \]

### Final Answer:
Benzyl ethyl ether reacts with HI to form benzyl iodide and ethanol.

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