Phenyl acetate `underset((ii)H^(o+))overset((i)CH_3MgBr("excess"))to'P'` 'P' can be

Options

A.
B.
C.
D.

Explanation

To solve the problem, we will analyze the reaction of phenyl acetate with excess methylmagnesium bromide (CH3MgBr) and determine the product 'P'.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is phenyl acetate, which has the structure:
\[ C_6H_5OCOCH_3 \]
and the reagent is methylmagnesium bromide (CH3MgBr).

2. **Understand the Grignard Reaction**:
In a Grignard reaction, the nucleophile (in this case, CH3-) attacks the electrophilic carbon of the carbonyl group in phenyl acetate. The carbonyl carbon is electrophilic due to the electronegativity of the oxygen atom.

3. **Nucleophilic Attack**:
The methyl group from CH3MgBr attacks the carbonyl carbon of phenyl acetate:
\[ C_6H_5OCOCH_3 + CH_3MgBr \rightarrow C_6H_5O^-(C(OH)(CH_3)CH_3)MgBr^+ \]
This results in the formation of a tetrahedral intermediate.

4. **Protonation (Hydrolysis)**:
The tetrahedral intermediate is then hydrolyzed by adding water (or an acid like H+). The oxygen atom that was negatively charged becomes protonated, leading to the formation of a tertiary alcohol:
\[ C_6H_5C(OH)(CH_3)CH_3 \]

5. **Further Reaction with Excess Grignard Reagent**:
Since we have excess CH3MgBr, the alcohol can react again with another equivalent of CH3MgBr:
\[ C_6H_5C(OH)(CH_3)CH_3 + CH_3MgBr \rightarrow C_6H_5C(CH_3)(C(OH)(CH_3)CH_3)MgBr^+ \]
This forms another tetrahedral intermediate.

6. **Final Hydrolysis**:
Again, hydrolysis occurs, leading to the formation of a ketone:
\[ C_6H_5C(CH_3)(C(OH)(CH_3)CH_3) + H_2O \rightarrow C_6H_5C(CH_3)_2 + H_2O \]
The product 'P' is thus a ketone with the structure:
\[ C_6H_5C(CH_3)_2 \]

### Conclusion:
The product 'P' formed from the reaction of phenyl acetate with excess CH3MgBr is **acetophenone**.