In the given reaction `C_6H_5-CH_2-CH_3overset(NBS)rarr(X)overset(Alc.KOH//Delta)rarr(Y)` In the given reaction X and Y respectively are

Options

A. `C_6H_5-CHBr-CH_3 and C_6H_5-C-=CH`
B. `C_6H_5-CBr_2-CH_3 and C_6H_5-C-=CH`
C. `C_6H_5-CBr_2-CH_3 and C_6H_5-CBr=CH_2`
D. `C_6H_5-CHBr-CH_3and C_6H_5-CH=CH_2`

Answer

To solve the given reaction sequence, we will break it down step by step.

### Step 1: Identify the starting compound
The starting compound is ethyl benzene, which can be represented as:
\[ C_6H_5-CH_2-CH_3 \]

### Step 2: Reaction with NBS
When ethyl benzene is treated with N-bromosuccinimide (NBS), it undergoes benzylic bromination. This means that the bromine atom will attach to the benzylic carbon (the carbon adjacent to the phenyl group). The reaction can be represented as:
\[ C_6H_5-CH_2-CH_3 \xrightarrow{NBS} C_6H_5-CHBr-CH_3 \]
This compound is referred to as compound X:
\[ X = C_6H_5-CHBr-CH_3 \]

### Step 3: Reaction with alcoholic KOH
Next, compound X is treated with alcoholic KOH and heated. This reaction leads to dehydrohalogenation, where a hydrogen atom and the bromine atom are eliminated, resulting in the formation of an alkene. The reaction can be represented as:
\[ C_6H_5-CHBr-CH_3 \xrightarrow{Alc.KOH, \Delta} C_6H_5-CH=CH_2 \]
This compound is referred to as compound Y:
\[ Y = C_6H_5-CH=CH_2 \]

### Final Result
Thus, the compounds X and Y formed in the reaction are:
- \( X = C_6H_5-CHBr-CH_3 \)
- \( Y = C_6H_5-CH=CH_2 \)

### Conclusion
Based on the analysis, the correct answer is:
- **X**: \( C_6H_5-CHBr-CH_3 \)
- **Y**: \( C_6H_5-CH=CH_2 \)