Observe the following sequence of reaction and answer the questions based on it
Phenylacetylene`overset(CH_3MgBr)underset(-CH_4)toxunderset((ii)H^(o+))overset((i)CO_2)toyunderset(HgSO_4)overset(H_2O // H_2SO_4)tozoverset(Delta)tow`
Compound z is

Options

A. `Ph-CH_2-oversetoverset(O)(||)C-COOH`
B.
C. `Ph-oversetoverset(O)(||)C-CH_2-COOH`
D. `Ph-CH_2-COOH`

Explanation

To determine the compound Z from the given sequence of reactions starting with phenylacetylene, we will follow the steps of the reaction sequence carefully.

### Step-by-Step Solution:

1. **Identify the Starting Material**:
The starting compound is phenylacetylene, which has the structure:
\[ \text{C}_6\text{H}_5\text{C} \equiv \text{C}\text{H} \]

2. **Reaction with Grignard Reagent**:
When phenylacetylene reacts with the Grignard reagent (CH₃MgBr), the acidic hydrogen from the terminal alkyne is abstracted, forming a phenylacetylide anion:
\[ \text{C}_6\text{H}_5\text{C} \equiv \text{C}^- + \text{CH}_4 \]

3. **Reaction with Carbon Dioxide (CO₂)**:
The phenylacetylide anion then reacts with CO₂, where the nucleophilic carbon (C⁻) attacks the electrophilic carbon of CO₂:
\[ \text{C}_6\text{H}_5\text{C} \equiv \text{C}^- + \text{CO}_2 \rightarrow \text{C}_6\text{H}_5\text{C} \equiv \text{C}\text{C}(=O)O^- \]

4. **Protonation**:
The negatively charged oxygen (O⁻) from the carboxylate ion is protonated (by H⁺ from the reaction medium), leading to the formation of a carboxylic acid:
\[ \text{C}_6\text{H}_5\text{C} \equiv \text{C}\text{C}(=O)\text{OH} \]
This compound is referred to as Y.

5. **Hydration with H₂SO₄**:
The next step involves treating the compound Y with H₂SO₄, which leads to the formation of a carbocation. The double bond in the alkyne is protonated, forming a more stable carbocation:
\[ \text{C}_6\text{H}_5\text{C}^+\text{C}(=O)\text{OH} \]

6. **Nucleophilic Attack by Water**:
Water attacks the carbocation, leading to the formation of an alcohol:
\[ \text{C}_6\text{H}_5\text{C}(\text{OH})\text{C}(=O)\text{OH} \]

7. **Tautomerization**:
The compound undergoes tautomerization, shifting the hydrogen to form a carbonyl group:
\[ \text{C}_6\text{H}_5\text{C}(=O)\text{C}(=O)\text{OH} \]
This compound is identified as Z.

### Conclusion:
The compound Z is:
\[ \text{C}_6\text{H}_5\text{C}(=O)\text{C}(=O)\text{OH} \]
This corresponds to the structure of a β-hydroxy ketone or a carboxylic acid derivative.