`Et-O-overset(O)overset(||)(C)-O-Et underset((2) H_(3)O^(o+))overset((1) HC_(3)MgBr ("excess"))rarr (A)`, Product (A) is :

Options

A. `CH_(3)-overset(O)overset(||)(C)-O-Et`
B. `CH_(3)-overset(O)overset(||)(C)-CH_(3)`
C. `CH_(3)-underset(CH_(3))underset(|)overset(OH)overset(|)(C)-CH_(3)`
D. `CH_(3)-CH_(2)-CH_(3)`

Explanation

To solve the given question, we need to analyze the reaction of an ether with Grignard reagent followed by acidic hydrolysis. The structure provided is an ether with the formula Et-O-C(=O)-O-Et, which can be interpreted as diethyl carbonate.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are diethyl carbonate (Et-O-C(=O)-O-Et) and excess methyl magnesium bromide (HC₃MgBr).

2. **First Reaction with Grignard Reagent**:
- The Grignard reagent (HC₃MgBr) acts as a nucleophile. The carbon atom in the carbonyl group (C=O) of diethyl carbonate is electrophilic and will be attacked by the nucleophile.
- The nucleophilic attack leads to the formation of a tetrahedral intermediate where the carbonyl oxygen becomes negatively charged.

3. **Formation of Alkoxide Intermediate**:
- The tetrahedral intermediate will collapse, leading to the cleavage of the bond between the carbonyl carbon and the oxygen of the ethyl group (Et-O).
- This results in the formation of a new product: Et-O-C(OH)-C(=O)-O-Et and the release of an ethoxide ion (EtO⁻).

4. **Second Reaction with Excess Grignard Reagent**:
- Since excess methyl magnesium bromide is present, it will react again with the carbonyl carbon of the newly formed intermediate.
- This results in another nucleophilic attack, leading to the formation of a second alkoxide intermediate.

5. **Final Product Formation**:
- The alkoxide intermediate will again undergo hydrolysis when treated with H₃O⁺, leading to the formation of a tertiary alcohol.
- The final product will be a tertiary alcohol with the structure: (CH₃)₃C-OH (tert-butanol).

### Final Product:
The final product (A) is a tertiary alcohol, specifically tert-butanol.