`Ph-C-=C-CH_(2)-CH_(3)-((Hg^(2+))/(D^(o+)))underset((D_(2)SO_(4))/(D_(2)O))toA`, `A` is

Options

A. `Ph-underset(O)underset(||)C-CH_(2)-underset(D)underset(|)CH-CH_(3)`
B. `Ph-underset(O)underset(||)C-underset(D)underset(|)CH-CH_(2)-CH_(3)`
C. `Ph-underset(OH)underset(|)CD-CH_(2)-CH_(2)-CH_(3)`
D. `Ph-underset(O)underset(||)C-CD_(2)-CH_(2)-CH_(3)`

Explanation

To solve the given problem, we will follow the steps involved in the oxymercuration-demercuration reaction using the provided reactants and conditions.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is phenylacetylene (Ph-C≡C-CH₂-CH₃) treated with mercuric ion (Hg²⁺) in the presence of deuterated sulfuric acid (D₂SO₄) and deuterated water (D₂O).

2. **Oxymercuration Reaction**:
The first step involves the oxymercuration of the alkyne. The mercuric ion (Hg²⁺) will add across the triple bond of phenylacetylene. This will lead to the formation of a cyclic mercurinium ion intermediate.

3. **Formation of the Intermediate**:
The cyclic mercurinium ion will form where the mercury atom is bonded to one of the carbons from the triple bond, and the other carbon will have a positive charge. The structure will look like this:
- Ph-C⁺(Hg)-C-CH₂-CH₃

4. **Nucleophilic Attack**:
The deuterated water (D₂O) will act as a nucleophile. The oxygen from D₂O will attack the more substituted carbon (the one bonded to Hg), leading to the opening of the cyclic structure.

5. **Formation of Product**:
After the nucleophilic attack, we will have:
- Ph-CH=CH(OD)-CH₂-CH₃
Here, the hydroxyl group (OD) is attached to the carbon that was initially bonded to mercury.

6. **Demercuration**:
The next step is the demercuration, where the mercury group (Hg) is replaced by a hydrogen atom (or in this case, deuterium). This will lead to the final product:
- Ph-CH=CH(OD)-CH₂-CH₃

7. **Final Product**:
The final product can be represented as:
- Ph-CH=CH(OD)-CH₂-CH₃, which is an enol form with deuterium.

### Conclusion:
The final product A is Ph-CH=CH(OD)-CH₂-CH₃.