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 question regarding the reaction of the phenolic ether `C_(6)H_(5)-O-CH_(2)-CH_(3)` with excess HI under heating, we can follow these steps:

### Step 1: Identify the Reactants
We start with the reactant, which is a phenolic ether: `C_(6)H_(5)-O-CH_(2)-CH_(3)`. This compound consists of a phenyl group (C6H5) attached to an ether group (O-CH2-CH3).

### Step 2: Understand the Reaction Conditions
The reaction is carried out with excess HI (hydroiodic acid) and heating. HI is a strong acid that can protonate the ether oxygen, making the ether bond more susceptible to cleavage.

### Step 3: Protonation of the Ether
When HI is added, the ether oxygen gets protonated, leading to the formation of a more reactive intermediate. This step enhances the electrophilicity of the carbon atoms attached to the ether oxygen.

### Step 4: Cleavage of the Ether Bond
The bond between the oxygen and the ethyl group (O-CH2-CH3) is weaker than the bond between the oxygen and the phenyl group (O-C6H5) due to the presence of a partial double bond character in the latter. The iodide ion (I-) can attack the carbon of the ethyl group, leading to the cleavage of the ether bond.

### Step 5: Formation of Products
Upon cleavage, we obtain two products:
1. **Phenol (C6H5OH)**: This is formed from the phenyl group after the ether bond is broken.
2. **Ethyl iodide (CH3CH2I)**: This is formed from the ethyl group that was attached to the ether.

### Step 6: Final Products
Thus, the products of the reaction are:
- [X] = Phenol (C6H5OH)
- [Y] = Ethyl iodide (CH3CH2I)

### Conclusion
The final answer is:
- [X] = C6H5OH (Phenol)
- [Y] = CH3CH2I (Ethyl iodide)