`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 identify the products X and Y from the reaction of C₆H₅-O-CH₂CH₃ with HI, we can follow these steps:

### Step 1: Understand the Reactants
The reactant is C₆H₅-O-CH₂CH₃, which is an ether (specifically ethyl phenyl ether). When it reacts with hydroiodic acid (HI), the ether bond (C-O) can be cleaved.

**Hint:** Recognize that ethers can be cleaved by strong acids like HI.

### Step 2: Protonation of the Ether
When HI is added, the ether oxygen (O) gets protonated, making it a better leaving group. This results in the formation of a protonated ether.

**Hint:** Protonation increases the electrophilicity of the ether, facilitating the cleavage.

### Step 3: Nucleophilic Attack
The iodide ion (I⁻) from HI acts as a nucleophile. According to the SN2 mechanism, the nucleophile will attack the less hindered carbon atom. In this case, the less hindered carbon is the CH₂ group attached to the ethyl group (CH₂CH₃).

**Hint:** Identify which carbon is less hindered for the nucleophilic attack.

### Step 4: Formation of Products
The nucleophilic attack leads to the formation of two products:
1. X: Phenol (C₆H₅OH), which is formed from the cleavage of the ether bond.
2. Y: Ethyl iodide (CH₃CH₂I), which is formed from the other part of the ether.

**Hint:** After the nucleophilic attack, consider what remains from the original ether structure.

### Conclusion
Thus, the products of the reaction are:
- X = Phenol (C₆H₅OH)
- Y = Ethyl iodide (CH₃CH₂I)

### Final Answer
X = C₆H₅OH (Phenol)
Y = CH₃CH₂I (Ethyl iodide)