The end product `( C)` of the following sequence of reaction is :
`CH -= CH + MeMgBr (1 mol) rarr (A) overset ((i) CO_2) underset((ii)H_3 O^(oplus))rarr (B) overset (HgSO_4//H_2 SO_4)rarr (C)`.

Options

A.
B.
C.
D.

Explanation

To determine the end product (C) of the given sequence of reactions, we will analyze each step carefully.

### Step 1: Reaction of Acetylene with Grignard Reagent
The first step involves the reaction of acetylene (CH≡CH) with methyl magnesium bromide (MeMgBr).

- **Reaction**:
\[ \text{CH} \equiv \text{CH} + \text{MeMgBr} \rightarrow \text{C} \equiv \text{C} \text{MgBr} + \text{CH}_4 \]
- **Product (A)**: The product formed is a magnesium alkynide, which can be represented as C≡C-MgBr.

### Step 2: Reaction of Product A with Carbon Dioxide
Next, product A reacts with carbon dioxide (CO₂) in an acidic medium.

- **Reaction**:
\[ \text{C} \equiv \text{C} \text{MgBr} + \text{CO}_2 \xrightarrow{\text{H}^+} \text{RCOOMgBr} \rightarrow \text{RCOOH} \]
- **Product (B)**: This reaction leads to the formation of a carboxylic acid (RCOOH), where R is the alkyl group from the Grignard reagent. In this case, it will be acetic acid (CH₃COOH).

### Step 3: Reaction of Product B with Acid
The next step involves the reaction of product B (the carboxylic acid) with sulfuric acid (H₂SO₄).

- **Reaction**:
\[ \text{CH}_3\text{COOH} \xrightarrow{\text{H}_2\text{SO}_4} \text{C} = \text{C} \text{(OH)} \text{(unstable alkenol)} \]
- **Product**: This reaction leads to the formation of an alkenol, which is an unstable compound.

### Step 4: Tautomerization of Alkenol
The unstable alkenol undergoes tautomerization to form a more stable ketone.

- **Reaction**:
\[ \text{C} = \text{C} \text{(OH)} \rightarrow \text{C} = \text{O} \text{(C)} \text{COOH} \]
- **Final Product (C)**: The final product is a ketone, specifically 3-oxobutanoic acid (or acetoacetic acid).

### Conclusion
The end product (C) of the sequence of reactions is **3-oxobutanoic acid** or **acetoacetic acid**.