`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 given question, we need to understand the reaction of the compound \( H_3C-C \equiv CH \) (which is propyne) with water in the presence of sulfuric acid (\( H_2SO_4 \)) and mercuric sulfate (\( HgSO_4 \)). This reaction leads to the formation of an intermediate and a final product through a series of steps.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is propyne (\( H_3C-C \equiv CH \)). We will react it with water (\( H_2O \)), sulfuric acid (\( H_2SO_4 \)), and mercuric sulfate (\( HgSO_4 \)).

**Hint**: Identify the structure of propyne and the role of the reagents.

2. **Protonation of the Alkyne**:
The first step involves the protonation of the alkyne by the \( H^+ \) from sulfuric acid. This creates a carbocation at the more stable position. The proton will add to the terminal carbon, leading to the formation of a more stable secondary carbocation.

**Hint**: Remember that carbocations are more stable when they are more substituted.

3. **Formation of the Carbocation**:
The structure after protonation is:
\[ CH_3-C^+(H)-CH_2 \]
Here, the positive charge is on the second carbon, making it a more stable secondary carbocation.

**Hint**: Draw the structure of the carbocation to visualize the charge distribution.

4. **Nucleophilic Attack by Water**:
Water, acting as a nucleophile, will attack the carbocation. The hydroxyl group (\( OH^- \)) from water will bond to the carbocation, forming an intermediate alcohol.

**Hint**: Consider the mechanism of nucleophilic attack and the stability of the resulting structure.

5. **Formation of the Intermediate (A)**:
The intermediate formed is:
\[ CH_3-C(OH)-CH_2 \]
This is the alcohol intermediate (A).

**Hint**: Identify the functional groups in the intermediate structure.

6. **Tautomerization**:
The next step involves tautomerization, where the hydrogen atom from the hydroxyl group shifts to the adjacent carbon, and the double bond shifts to form a carbonyl group.

**Hint**: Understand the concept of tautomerism and how it changes the structure.

7. **Formation of the Final Product (B)**:
After tautomerization, the final product is:
\[ CH_3-C(=O)-CH_3 \]
This is an enol that converts to a ketone, specifically acetone.

**Hint**: Recognize the final product and its significance in organic chemistry.

### Final Structures:
- **Intermediate (A)**: \( CH_3-C(OH)-CH_2 \)
- **Product (B)**: \( CH_3-C(=O)-CH_3 \) (acetone)

### Conclusion:
The correct answer for the intermediate (A) and product (B) can be matched with the options provided in the question.