`CH_3-undersetunderset(O)(||)C-CH_3+PhMgBrtoAoverset(H_3O^+)toBunderset(Delta)overset(H_3PO_4)toC`
C is :

Options

A. `Ph-undersetunderset(O)(||)C-CH_3`
B. `Ph-undersetunderset(OH)(|)oversetoverset(CH_3)(|)C-CH_3`
C. `Ph-undersetunderset(CH_3)(|)C=CH_2`
D. `Ph-undersetunderset(CH_3)(|)CH-CH_3`

Answer

To solve the problem step by step, we will analyze the reaction of acetone (CH₃COCH₃) with phenylmagnesium bromide (PhMgBr) and the subsequent steps leading to the final product C.

### Step 1: Reaction of Acetone with PhMgBr
- **Starting Material**: Acetone (CH₃COCH₃)
- **Reagent**: PhMgBr (Phenylmagnesium bromide)

In this step, the nucleophile (Ph⁻ from PhMgBr) attacks the electrophilic carbon of the carbonyl group in acetone. The oxygen in the carbonyl group becomes negatively charged.

**Intermediate A**:
\[ \text{CH}_3C(-O^-)CH_3 + \text{Ph}^- \rightarrow \text{CH}_3C(-O^-)CH_3\text{Ph} \]

### Step 2: Protonation of the Intermediate
- **Reagent**: H₃O⁺ (Hydronium ion)

The negatively charged oxygen in the intermediate A is protonated by H₃O⁺, forming a tertiary alcohol.

**Intermediate B**:
\[ \text{CH}_3C(OH)CH_3\text{Ph} \]

### Step 3: Dehydration of the Alcohol
- **Reagent**: H₃PO₄ (Phosphoric acid) and heating

In this step, the hydroxyl group (OH) of the alcohol is protonated, making it a better leaving group (water). The dehydration occurs, resulting in the formation of a double bond.

**Final Product C**:
\[ \text{CH}_3C=CH_2\text{Ph} \]

### Conclusion
The final product C is an alkene with the structure:
\[ \text{Ph}-\text{C}(\text{CH}_3)=\text{CH}_2 \]

### Answer
C is: **Ph-CH=CH₂**