In the given sequence of reaction, identify Y
`C_6H_5-overset(O)overset(||)(C )-CH_3 overset(SeO_2)to X underset((ii)H_2O // H^(o+))overset((i)Conc. NaOH//Delta)(rarr)Y`

Options

A. `C_6H_5-CH_2OH`
B. `C_6H_5-COOH`
C. `C_6H_5-CHOH-COOH`
D. `C_6H_5-CO-COOH`

Answer

To solve the problem step by step, we will analyze the given reactions and identify the compound Y.

### Step-by-Step Solution:

1. **Identify the Starting Compound**:
The starting compound is `C6H5-CO-CH3`, which is acetophenone (a phenyl group attached to a carbonyl group and a methyl group).

2. **Reaction with Selenium Dioxide (SeO2)**:
The first reaction involves the oxidation of acetophenone with selenium dioxide (SeO2). SeO2 is an oxidizing agent that will oxidize the methyl group (–CH3) to a carbonyl group (–C=O).
- **Resulting Compound (X)**: After oxidation, the compound X will be `C6H5-CO-COH` (which is benzil).

3. **Reaction with Concentrated NaOH and Heat**:
The next step involves treating compound X with concentrated NaOH and heating it. This reaction typically leads to a rearrangement or condensation reaction.
- The presence of two carbonyl groups allows for nucleophilic attack by the hydroxide ion (OH-) on one of the carbonyl groups, leading to a reaction known as aldol condensation.

4. **Formation of Compound Y**:
During the aldol condensation, the hydroxide ion attacks the carbonyl carbon of one of the carbonyl groups, forming a β-hydroxy ketone. This intermediate can then dehydrate (lose water) to form an α,β-unsaturated carbonyl compound.
- **Final Compound (Y)**: The final product Y will be `C6H5-CO-CH=CH2` (which is styryl ketone).

### Conclusion:
The compound Y formed from the given sequence of reactions is `C6H5-CO-CH=CH2`.