`PhMgBr+CH_(3)-CN underset(H_(3)O^(o+))rarr(A)Ph-overset(O)overset(||)(C)-O-Hunderset((2)H_(3)O)overset((1) "excess" CH_(3)-Li)rarr(A)`
Same product (A) will form in both reactions. A is :

Options

A. `Ph-underset(CH_(3))underset(|)overset(OH)overset(|)(C)-CH_(3)`
B. `Ph-CHO`
C. `Ph-overset(O)overset(||)(C)-CH_(3)`
D. `Ph-CH_(2)-CO_(2)H`

Explanation

To solve the problem step by step, we will analyze the two reactions involving the Grignard reagent (PhMgBr) and the product formed (A) in both cases.

### Step 1: Analyze the First Reaction
1. **Reactants**: PhMgBr + CH₃CN + H₃O⁺
2. **Mechanism**:
- The Grignard reagent (PhMgBr) dissociates into Ph⁻ and MgBr⁺.
- The Ph⁻ ion attacks the carbon of the nitrile (CH₃CN), which is electrophilic due to the electronegative nitrogen.
- This results in the formation of a carbocation intermediate.
- The Ph group attaches to the carbon of the nitrile, yielding a new intermediate: CH₃C(Ph)N.
3. **Hydrolysis**:
- Upon adding H₃O⁺, the nitrile group is hydrolyzed.
- The nitrogen is protonated, and the carbon-nitrogen triple bond is converted into a carbonyl group (C=O) and an alcohol (OH).
- The final product after dehydration (removal of water) is:
\[ \text{Ph-C(=O)-CH₃} \]

### Step 2: Analyze the Second Reaction
1. **Reactants**: CH₃C(=O)OH + excess CH₃Li
2. **Mechanism**:
- CH₃Li dissociates into CH₃⁻ and Li⁺.
- The CH₃⁻ ion attacks the carbonyl carbon of the carboxylic acid (CH₃C(=O)OH).
- This forms a new intermediate with a negative charge on the oxygen (alkoxide).
- A second CH₃Li adds to the carbonyl oxygen, forming a more complex intermediate.
3. **Protonation**:
- Upon adding H₃O⁺, the alkoxide is protonated to form the alcohol.
- The final product after dehydration is also:
\[ \text{Ph-C(=O)-CH₃} \]

### Conclusion
Both reactions yield the same product (A):
\[ \text{A} = \text{Ph-C(=O)-CH₃} \]

### Final Answer
The product A formed in both reactions is **Ph-C(=O)-CH₃**.

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