The product obtained from the following sequence of reactions is
`H_(3)-C-=CH underset(H_(2)SO_(4))overset(HgSO_(4))to A overset (NaBH_(4))rarrB`

Options

A. propanal
B. 2-propanol
C. 1-propanol
D. propane

Explanation

To solve the problem step by step, we will analyze the given reactions and determine the final product.

### Step 1: Identify the starting material
The starting material is propyne, which has the structure:
\[ \text{H}_3\text{C} \equiv \text{C} \text{H} \]

### Step 2: Reaction with H₂SO₄ and HgSO₄
When propyne reacts with sulfuric acid (H₂SO₄) in the presence of mercury(II) sulfate (HgSO₄), it undergoes hydration. According to Markovnikov's rule, the addition of water occurs such that the more substituted carbon gets the hydroxyl group (OH) and the less substituted carbon gets the hydrogen (H).

The reaction can be summarized as follows:
\[ \text{H}_3\text{C} \equiv \text{C} \text{H} + H_2O \xrightarrow{H_2SO_4, HgSO_4} \text{A} \]

The product A formed is propanone (also known as acetone), which has the structure:
\[ \text{CH}_3\text{C(=O)}\text{CH}_3 \]

### Step 3: Reduction with NaBH₄
Next, the product A (propanone) is reduced by sodium borohydride (NaBH₄). Sodium borohydride is a reducing agent that can reduce ketones to secondary alcohols.

The reaction can be summarized as follows:
\[ \text{CH}_3\text{C(=O)}\text{CH}_3 + \text{NaBH}_4 \rightarrow \text{B} \]

The product B formed is 2-propanol, which has the structure:
\[ \text{CH}_3\text{CH(OH)}\text{CH}_3 \]

### Final Product
Thus, the final product obtained from the sequence of reactions is 2-propanol.

### Summary of Steps
1. Start with propyne (H₃C≡CH).
2. React with H₂SO₄ and HgSO₄ to form propanone (A).
3. Reduce propanone with NaBH₄ to form 2-propanol (B).