Predict the correct intermediate and product in the following reaction:
`H_(3)C-C-=Choverset(H_(2)O,H_(2)SO_(4))underset(HgSO_(4))to"Intermediate" to "Product"`

Options

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B.
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D.

Answer

To solve the problem, we need to predict the intermediate and the product of the reaction involving the alkyne `H3C-C≡CH` when treated with `H2O`, `H2SO4`, and `HgSO4`. Let's break down the steps:

### Step 1: Identify the Reactants
We start with the alkyne `H3C-C≡CH`, which is 1-butyne. The reagents used are water (`H2O`), sulfuric acid (`H2SO4`), and mercuric sulfate (`HgSO4`).

### Step 2: Protonation of the Alkyne
In the presence of sulfuric acid, the first step is the protonation of the alkyne. The `H+` ion from `H2SO4` will attack the triple bond of the alkyne, leading to the formation of a carbocation.

The reaction can be represented as:
\[ H3C-C≡CH + H^+ \rightarrow H3C-C^+H-CH \]

### Step 3: Formation of the Intermediate (Enol)
Next, water (`H2O`) will react with the carbocation to form an oxonium ion (an intermediate). The reaction can be depicted as:
\[ H3C-C^+H-CH + H2O \rightarrow H3C-C(OH^+)-CH2 \]

This oxonium ion is unstable and will rearrange to form an enol. The enol can be represented as:
\[ H3C-C(OH)-CH2 \]
This enol form is our intermediate.

### Step 4: Tautomerization to Form the Product (Keto Form)
The enol form is less stable than the corresponding keto form. The enol will undergo tautomerization, where the hydroxyl group will shift to form a carbonyl group. This can be represented as:
\[ H3C-C(OH)-CH2 \rightarrow H3C-C(=O)-CH3 \]

The final product of this reaction is the ketone, which is 2-butanone.

### Summary of the Intermediate and Product
- **Intermediate**: Enol form \( H3C-C(OH)-CH2 \)
- **Product**: Keto form \( H3C-C(=O)-CH3 \) (2-butanone)

### Final Answer
- Intermediate: Enol form \( H3C-C(OH)-CH2 \)
- Product: Keto form \( H3C-C(=O)-CH3 \)

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