`C_6H_5-O-CH_2CH_3+HI overset(Delta)(rarr)X + Y`
Identify X and Y in the above reaction?

Options

A. `X : C_6H_5OH, Y : CH_3CH_3`
B. `X : C_2H_5I, Y : C_6H_5CHO`
C. `X : C_6H_5I, Y : CH_3CH_2OH`
D. `X:C_6H_5OH, Y : CH_3CH_2I`

Answer

To solve the problem, we need to identify the products (X and Y) formed when the ether \( C_6H_5-O-CH_2CH_3 \) reacts with hydroiodic acid (HI) upon heating.

### Step-by-step Solution:

1. **Identify the Reactants**:
The reactant is an ether, specifically \( C_6H_5-O-CH_2CH_3 \), which consists of a phenyl group \( C_6H_5 \) and an ethyl group \( CH_2CH_3 \) connected by an oxygen atom.

2. **Understand the Reaction Mechanism**:
Ethers react with strong acids like HI through a mechanism that involves protonation of the ether oxygen, leading to the formation of a more reactive species. The protonation makes the ether more susceptible to cleavage.

3. **Protonation of Ether**:
When HI is added to the ether, the oxygen atom of the ether gets protonated, forming a positively charged oxonium ion:
\[ C_6H_5-O^+H-CH_2CH_3 \]

4. **Cleavage of the Ether**:
The oxonium ion can undergo cleavage. The bond between the oxygen and the ethyl group (CH2CH3) is broken, leading to the formation of a carbocation. The bond between the oxygen and the phenyl group remains intact.

5. **Formation of Products**:
- The ethyl group (CH2CH3) will leave as an ethyl cation (which is unstable) and will react with the iodide ion (I-) from HI to form ethyl iodide (CH2CH3I).
- The phenyl group (C6H5) will remain attached to the oxygen, resulting in the formation of phenol (C6H5OH).

6. **Identify X and Y**:
Therefore, the products of the reaction are:
- **X**: Phenol \( (C_6H_5OH) \)
- **Y**: Ethyl iodide \( (C_2H_5I) \)

### Final Answer:
- **X = C_6H_5OH (Phenol)**
- **Y = C_2H_5I (Ethyl iodide)**