Which of the following `3^@` alcohols does propyl ester give during reaction with `EtMgBr` ?

Options

A.
B.
C.
D.

Explanation

To solve the problem of which `3^@` alcohol is formed when propyl ester reacts with ethyl magnesium bromide (EtMgBr), we can follow these steps:

### Step 1: Understand the Reactants
- **Propyl Ester**: The structure of a propyl ester can be represented as R-C(=O)-O-C3H7, where R is any alkyl group.
- **Ethyl Magnesium Bromide (EtMgBr)**: This is a Grignard reagent, represented as C2H5MgBr.

### Step 2: Reaction Mechanism
1. **Nucleophilic Attack**: The Grignard reagent (EtMgBr) acts as a nucleophile. The ethyl group (C2H5-) from the Grignard reagent will attack the carbonyl carbon of the propyl ester.

- The carbonyl carbon (C=O) becomes positively charged after the nucleophilic attack, and the oxygen becomes negatively charged (alkoxide ion).

2. **Formation of Intermediate**: This results in the formation of a tetrahedral intermediate:
\[ R-C(OH)(C2H5)-O-C3H7 \]

3. **Second Nucleophilic Attack**: The alkoxide intermediate can react again with another equivalent of ethyl magnesium bromide. The ethyl group will again attack the carbonyl carbon.

4. **Formation of Alcohol**: After the second attack, the intermediate will look like:
\[ R-C(OH)(C2H5)(C2H5)-O-C3H7 \]
After hydrolysis, the final product will be:
\[ R-C(OH)(C2H5)(C2H5) \]

### Step 3: Identify the Product
- If we assume R as a methyl group (CH3), the final product would be:
\[ CH3-C(OH)(C2H5)(C2H5) \]
This structure represents a tertiary alcohol with two ethyl groups and one methyl group.

### Step 4: Conclusion
- The product formed when propyl ester reacts with ethyl magnesium bromide is a tertiary alcohol with the structure:
\[ CH3-C(OH)(C2H5)(C2H5) \]
- Therefore, the answer is option A, which corresponds to a tertiary alcohol with two ethyl groups and one methyl group.