Reaction of benzyl benzoate with an excess of methyl magnesium bromide generates a mixture of .

Options

A. benzyl alcohol and benzoic acid
B. benzyl methyl ether and 2 - phenylpropan -2- ol
C. benzyl alcohol and 2 - phenylprone - 2- ol
D. ethyl benzene and benzoic acid

Answer

To solve the question regarding the reaction of benzyl benzoate with an excess of methyl magnesium bromide, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Benzyl benzoate is an ester with the structure: C6H5-CO-O-C6H5.
- Methyl magnesium bromide (CH3MgBr) is a Grignard reagent.

2. **Understand the Grignard Reaction**:
- Grignard reagents act as nucleophiles and can attack electrophilic centers, such as the carbonyl carbon in esters.

3. **First Reaction**:
- The carbonyl carbon in benzyl benzoate is attacked by the methyl group (CH3-) from the Grignard reagent.
- This leads to the formation of an intermediate: a tetrahedral alkoxide.

4. **Formation of Acetophenone**:
- The tetrahedral intermediate collapses, resulting in the formation of acetophenone (C6H5-CO-CH3) and benzyl alcohol (C6H5-CH2OH).

5. **Second Reaction with Excess Grignard Reagent**:
- Since we have excess methyl magnesium bromide, the acetophenone can react again.
- The methyl group from CH3MgBr attacks the carbonyl carbon of acetophenone.

6. **Formation of 2-Phenylpropan-2-ol**:
- This results in another tetrahedral intermediate, which upon protonation (addition of H+) leads to the formation of 2-phenylpropan-2-ol.

7. **Final Products**:
- The final mixture of products from the reaction of benzyl benzoate with excess methyl magnesium bromide is:
- Benzyl alcohol (C6H5-CH2OH)
- 2-phenylpropan-2-ol (C6H5-C(OH)(CH3)2)

### Conclusion:
The correct answer to the question is that the reaction generates a mixture of benzyl alcohol and 2-phenylpropan-2-ol.

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