On boiling with concentrated HBr, ethyl phenyl ether will give:

Options

A. Phenol and ethyl bromide
B. Bromobenzene and ethanol
C. Phenol and ethane
D. Bromobezene and ethane.

Explanation

To solve the question regarding the reaction of ethyl phenyl ether with concentrated HBr, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Ethyl Phenyl Ether**:
- Ethyl phenyl ether consists of a benzene ring attached to an ethyl group through an oxygen atom. The structure can be represented as:
\[ \text{C}_6\text{H}_5\text{OCH}_2\text{CH}_3 \]

2. **Reaction with Concentrated HBr**:
- When ethyl phenyl ether is treated with concentrated HBr, the ether undergoes protonation. The oxygen atom in the ether has lone pairs that can attract the hydrogen ion (H⁺) from HBr, forming an oxonium ion:
\[ \text{C}_6\text{H}_5\text{OCH}_2\text{CH}_3 + \text{HBr} \rightarrow \text{C}_6\text{H}_5\text{OH}^+ \text{CH}_2\text{CH}_3 \]

3. **Formation of the Oxonium Ion**:
- The protonation leads to the formation of an oxonium ion, where the oxygen carries a positive charge:
\[ \text{C}_6\text{H}_5\text{OH}^+ \text{CH}_2\text{CH}_3 \]

4. **Cleavage of the Weaker C-O Bond**:
- In the oxonium ion, the bond between the oxygen and the ethyl group (C-O) is weaker compared to the bond between the oxygen and the benzene ring due to resonance stabilization in the latter. Therefore, the C-O bond to the ethyl group will break:
\[ \text{C}_6\text{H}_5\text{OH}^+ \text{CH}_2\text{CH}_3 \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{CH}_2\text{Br} \]

5. **Final Products**:
- The reaction results in the formation of phenol (C₆H₅OH) and ethyl bromide (CH₃CH₂Br):
\[ \text{C}_6\text{H}_5\text{OCH}_2\text{CH}_3 + \text{HBr} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{CH}_2\text{Br} \]

### Conclusion:
Thus, on boiling ethyl phenyl ether with concentrated HBr, the products obtained are phenol and ethyl bromide.

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