When pent-2-yne is treated with dilute `H_2SO_4` and `HgSO_4` the product formed is

Options

A. pentan-1-ol
B. pentan-2-ol
C. pentan-2-one
D. pentan-3-one

Explanation

To solve the problem of what product is formed when pent-2-yne is treated with dilute H₂SO₄ and HgSO₄, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Pent-2-yne**:
- Pent-2-yne is an alkyne with the structure:
\[ CH_3 - C \equiv C - CH_2 - CH_3 \]
The triple bond is located between the second and third carbon atoms.

2. **Reaction with Dilute H₂SO₄ and HgSO₄**:
- When pent-2-yne is treated with dilute sulfuric acid (H₂SO₄) and mercuric sulfate (HgSO₄), it undergoes hydration. The reaction adds water across the triple bond.

3. **Markovnikov's Rule**:
- According to Markovnikov's rule, the hydrogen (H) will add to the carbon with more hydrogen atoms (the less substituted carbon), and the hydroxyl group (OH) will add to the more substituted carbon. In this case, the H will add to the terminal carbon (CH₃) and the OH will add to the second carbon.

4. **Formation of Enol**:
- The initial product of this reaction is an enol:
\[ CH_3 - C(OH) = CH - CH_2 - CH_3 \]
This is an unstable intermediate.

5. **Tautomerization**:
- The enol will undergo tautomerization, which is a rearrangement of the structure to form a more stable ketone. In this case, the hydrogen from the hydroxyl group will migrate to the adjacent carbon, resulting in the formation of a carbonyl group (C=O).
- The final product after tautomerization will be:
\[ CH_3 - C(=O) - CH_2 - CH_3 \]
This compound is 2-pentanone.

6. **Final Answer**:
- Therefore, the product formed when pent-2-yne is treated with dilute H₂SO₄ and HgSO₄ is **2-pentanone**.