In the given reaction `C_(6)H_(5)-O-CH_(2)-CH_(3)overset("Excess"Hl//Delta)rarr(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 given reaction \( C_6H_5OCH_2CH_3 \overset{\text{Excess } HI}{\rightarrow} (X) + (Y) \), we will analyze the reaction step by step.

### Step 1: Identify the Reactants
The reactant is ethyl phenyl ether, which is represented as \( C_6H_5OCH_2CH_3 \). This compound consists of a phenyl group (\( C_6H_5 \)) attached to an ethyl group (\( CH_2CH_3 \)) via an oxygen atom.

**Hint:** Look for the functional groups in the reactant to understand the reaction mechanism.

### Step 2: Understand the Role of Hydrogen Iodide (HI)
Hydrogen iodide is a strong acid that can protonate the ether oxygen. In the presence of excess HI, the ether bond can be cleaved, leading to the formation of products.

**Hint:** Consider how acids interact with ethers and what products can be formed from such reactions.

### Step 3: Protonation of the Ether
When \( HI \) is added, the oxygen atom in the ether will be protonated, forming an oxonium ion. This makes the ether more reactive.

**Hint:** Remember that protonation increases the electrophilicity of the ether.

### Step 4: Cleavage of the Ether
The protonated ether can undergo cleavage. The bond between the oxygen and the ethyl group (\( CH_2CH_3 \)) is broken. This can lead to the formation of phenol and an alkyl iodide.

**Hint:** Think about which bond is more likely to break based on stability and the nature of the groups involved.

### Step 5: Formation of Products
After the cleavage, we will have two products:
1. Phenol (\( C_6H_5OH \)) from the phenyl group.
2. Ethyl iodide (\( CH_3CH_2I \)) from the ethyl group.

Thus, we can identify:
- \( X = C_6H_5OH \) (Phenol)
- \( Y = CH_3CH_2I \) (Ethyl iodide)

**Hint:** Check the structure of the products to ensure they match the expected outcome from the reaction.

### Conclusion
The final products of the reaction are:
- \( X = C_6H_5OH \)
- \( Y = CH_3CH_2I \)

### Final Answer
The products \( X \) and \( Y \) will respectively be:
- \( X = C_6H_5OH \) (Phenol)
- \( Y = CH_3CH_2I \) (Ethyl iodide)