`underset("(excess)")(CH_(3)MgBr) + Et - O-overset(O)overset(||)(C)-O -Et underset((2) H^(o+))rarr (A)`, Product A is :

Options

A.
B.
C.
D.

Explanation

To solve the problem, we will go through the reaction step by step, starting from the Grignard reagent (CH₃MgBr) and the given ester (Et-O-C(=O)-Et).

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are Methyl Magnesium Bromide (CH₃MgBr) and the ester Ethyl Acetate (Et-O-C(=O)-Et).

2. **Nucleophilic Attack**:
The Grignard reagent (CH₃MgBr) acts as a nucleophile. The carbon atom in the carbonyl group (C=O) of the ester is electrophilic. The nucleophile (CH₃⁻) will attack this electrophilic carbon atom.
- Reaction: CH₃MgBr + Et-O-C(=O)-Et → CH₃-C(=O)-OEt + MgBrOEt

3. **Formation of Tetrahedral Intermediate**:
The attack of the methyl group leads to the formation of a tetrahedral intermediate. The oxygen atom of the ester becomes negatively charged (alkoxide ion).
- Intermediate: CH₃-C(OH)-OEt

4. **Elimination of Ethoxy Group**:
The tetrahedral intermediate can collapse, leading to the elimination of the ethoxy group (OEt) and forming a ketone.
- Reaction: CH₃-C(OH)-OEt → CH₃-C(=O)-Et + OEt⁻

5. **Second Nucleophilic Attack**:
The formed ketone (CH₃-C(=O)-Et) can again react with excess CH₃MgBr. The nucleophile (CH₃⁻) attacks the carbonyl carbon of the ketone.
- Reaction: CH₃-C(=O)-Et + CH₃MgBr → CH₃-C(OH)(CH₃)-OEt

6. **Formation of Tertiary Alcohol**:
The tetrahedral intermediate formed from the second nucleophilic attack can again collapse, leading to the formation of a tertiary alcohol after protonation.
- Final Product: CH₃-C(OH)(CH₃)-OEt

7. **Protonation**:
Finally, the alkoxide ion formed during the reaction will be protonated during the acidic workup to yield the final product, which is a tertiary alcohol.
- Final Product: CH₃-C(OH)(CH₃)-OEt (Tertiary alcohol)

### Conclusion:
The product A is a tertiary alcohol with the structure CH₃-C(OH)(CH₃)-OEt.