In the given reaction
`C_(6)H_(5)-O-CH_(2)-CH_(3)overset(Excess HI//Delta)(to)[X]+[Y]`
[X] and [Y] will respectively be:

Options

A. `C_(6)H_(5)I` and `CH_(3)CH_(2)I`
B. `C_(6)H_(5)OH` and `CH_(3)CH_(2)-I`
C. `C_(6)H_(5)I` and `CH_(3)CH_(2)OH`
D. `C_(6)H_(5)OH` and `CH_(2)=CH_(2)`

Answer

To solve the problem, we need to analyze the reaction of the ether compound \( C_6H_5-O-CH_2-CH_3 \) with excess HI under heating conditions.

### Step-by-Step Solution:

1. **Identify the Ether Structure**:
The given compound is an ether where \( C_6H_5 \) (phenyl group) is bonded to an oxygen atom, which is further bonded to \( CH_2-CH_3 \) (ethyl group). The structure can be represented as:
\[ C_6H_5-O-CH_2-CH_3 \]

2. **Understanding the Reaction with HI**:
When ethers react with hydrogen iodide (HI), they can undergo cleavage. The presence of an aromatic group (like the phenyl group) can influence the cleavage due to resonance stabilization.

3. **Resonance Effect**:
The lone pair of electrons on the oxygen atom can participate in resonance with the phenyl group. This means that the oxygen atom can stabilize itself by forming a bond with the phenyl group when HI is added.

4. **Cleavage of the Ether**:
Due to the resonance stabilization, the ether bond will break in such a way that the oxygen will bond with the phenyl group, leading to the formation of phenol. The other part of the ether will release an ethyl group with iodide.

5. **Products Formation**:
- The first product, [X], is phenol \( C_6H_5OH \) because the oxygen atom bonds with the phenyl group.
- The second product, [Y], is ethyl iodide \( CH_3CH_2I \) because the ethyl group is left with the iodide from HI.

Thus, the final products of the reaction are:
- [X] = \( C_6H_5OH \) (Phenol)
- [Y] = \( CH_3CH_2I \) (Ethyl iodide)

### Final Answer:
[X] and [Y] will respectively be: \( C_6H_5OH \) and \( CH_3CH_2I \).