`H_(3)C-C-=CH overset(H_(2)O,H_(4)SO_(4))underset(HgSO_(4))to underset(A)("intermediate") to underset(B)("product")`

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to understand the reaction of the alkyne with water in the presence of sulfuric acid and mercuric sulfate. The alkyne given is 1-butyne (H₃C-C≡CH). Let's break down the steps involved in the reaction to identify the intermediate (A) and the final product (B).

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is 1-butyne (H₃C-C≡CH) and the reagents are H₂O, H₂SO₄, and HgSO₄.

2. **Formation of the Mercuric Ion Complex**:
When 1-butyne reacts with HgSO₄, the mercuric ion (Hg²⁺) interacts with the triple bond. This leads to the formation of a non-classical carbocation intermediate. The reaction can be represented as:
\[ H₃C-C≡CH + HgSO₄ \rightarrow \text{Non-Classical Carbocation} \]

3. **Nucleophilic Attack by Water**:
The non-classical carbocation is then attacked by water (H₂O). The oxygen in water acts as a nucleophile and attacks the more substituted carbon of the carbocation. This results in the formation of an oxonium ion (protonated alcohol):
\[ \text{Non-Classical Carbocation} + H₂O \rightarrow \text{Oxonium Ion} \]

4. **Deprotonation**:
The oxonium ion is unstable and loses a proton (H⁺), leading to the formation of an enol:
\[ \text{Oxonium Ion} \rightarrow \text{Enol (C=C-OH)} \]

5. **Tautomerization to Ketone**:
The enol undergoes tautomerization to form a more stable keto form. In this case, the enol will convert to propan-2-one (acetone):
\[ \text{Enol} \rightarrow \text{Keto (Propan-2-one)} \]

6. **Final Product**:
The final product (B) is propan-2-one (acetone, CH₃COCH₃).

### Summary of A and B:
- Intermediate (A): Non-Classical Carbocation
- Product (B): Propan-2-one (Acetone)