`CH_(3)C-=CH overset(40% H_(2)SO_(4)) underset(1%HgSO_(4)) toA overset("Isomeristion")to underset(O) underset(||)(CH_(3)-C-CH_(3))`
Structure of A and type of isomerism in the above reaction respectively are

Options

A. prop-1-en-2-ol, metamerism
B. prop-1-en-1-ol, metamerism
C. prop-2-en-2-ol, tautomerism
D. prop-1-en-2-ol, tautomerism

Explanation

To solve the problem step by step, we will analyze the reaction of the given alkyne and determine the structure of compound A and the type of isomerism involved.

### Step 1: Identify the Starting Compound
The starting compound is CH₃C≡CH (1-butyne).

### Step 2: Reaction with Reagents
The reaction involves treating the alkyne with 40% H₂SO₄ and 1% HgSO₄. The mercury(II) sulfate (HgSO₄) acts as a catalyst in the hydration of the alkyne.

### Step 3: Formation of the Mercurinium Ion
When 1-butyne reacts with HgSO₄, a mercurinium ion is formed. The triple bond opens up to form a three-membered ring with mercury.

### Step 4: Nucleophilic Attack by Water
In the presence of water, a water molecule attacks the more stable carbon of the mercurinium ion, leading to the formation of an enol. The OH group is added to one carbon, and the double bond is formed between the other two carbons.

### Step 5: Structure of Compound A
The compound formed after the reaction is 2-buten-1-ol (or 2-butenol), which can be represented as:
```
CH3
|
C=C
|
OH
```
This compound is also known as 2-buten-1-ol.

### Step 6: Tautomerization to Form Ketone
The enol (2-buten-1-ol) undergoes tautomerization to form the corresponding ketone. In this case, it converts to acetone (propan-2-one).

### Step 7: Identify the Type of Isomerism
The isomerism occurring in this reaction is known as tautomerism, specifically keto-enol tautomerism, where the enol form (2-buten-1-ol) converts to the keto form (acetone).

### Final Answer
- **Structure of A**: 2-buten-1-ol (or 2-butenol)
- **Type of Isomerism**: Tautomerism (specifically keto-enol tautomerism)

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