Bromination of ethyl benzene in presence of light gives

Options

A.
B.
C.
D. Both (a) and (b)

Explanation

To solve the question regarding the bromination of ethyl benzene in the presence of light, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure of Ethyl Benzene**:
- Ethyl benzene consists of a benzene ring (C6H5) with an ethyl group (C2H5) attached to it. The molecular formula is C8H10.

2. **Bromination Reaction**:
- Bromination of ethyl benzene involves the addition of bromine (Br2) to the compound. In the presence of light, this reaction occurs via a radical mechanism.

3. **Initiation Step**:
- In the presence of light, Br2 undergoes homolytic cleavage to form two bromine radicals (Br•):
\[ \text{Br}_2 \xrightarrow{\text{light}} 2 \text{Br}• \]

4. **Propagation Steps**:
- The bromine radical can abstract a hydrogen atom from the ethyl group of ethyl benzene, forming an ethyl radical and HBr:
\[ \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 + \text{Br}• \rightarrow \text{C}_6\text{H}_5\text{CH}•\text{CH}_3 + \text{HBr} \]
- The ethyl radical can then react with another bromine molecule to form brominated ethyl benzene:
\[ \text{C}_6\text{H}_5\text{CH}•\text{CH}_3 + \text{Br}_2 \rightarrow \text{C}_6\text{H}_5\text{CHBr}\text{CH}_3 \]

5. **Final Product**:
- The major product of this reaction is 1-bromoethylbenzene, where the bromine atom is attached to the ethyl group.

### Conclusion:
The bromination of ethyl benzene in the presence of light primarily yields 1-bromoethylbenzene.