`CH_3 -C-= CH underset(1% Hg SO_4 ) overset(40% H_2SO_4 )to A overset("isomerisation ") to CH_3 - underset(O) underset(||)C-CH_3`
Stucture of A and type of isomerism in the above reaction are respectively .

Options

A. Prop-1-en -2 -ol , metamerism
B. Prop-1-en -1-ol , trautomerism
C. prop -2- en -2- ol , geometrical isomerism
D. Prop-1-en- 2-ol tautomerism

Explanation

To solve the given question step by step, we will analyze the reaction and identify the structure of compound A as well as the type of isomerism involved.

### Step 1: Identify the starting compound
The starting compound is CH₃-C≡C-CH₃ (1-butyne). This is an alkyne with a triple bond between the second and third carbon atoms.

### Step 2: Reaction with 1% HgSO₄
When 1% HgSO₄ is added to the alkyne in the presence of sulfuric acid (H₂SO₄), it undergoes hydration. The mercury ion (Hg²⁺) facilitates the formation of a non-classical carbocation.

### Step 3: Formation of the non-classical carbocation
The triple bond can be protonated to form a non-classical carbocation. The structure of this carbocation can be represented as follows:
- The positive charge is delocalized between the two carbons adjacent to the original triple bond.

### Step 4: Nucleophilic attack by water
Water (H₂O) acts as a nucleophile and attacks the carbocation. The more substituted carbon will preferentially bond with the hydroxyl group (OH) from water, leading to the formation of an alcohol.

### Step 5: Formation of compound A
After the nucleophilic attack, the resulting compound is an enol (specifically, prop-1-en-2-ol). The structure of compound A is:
- CH₃-C(OH)=C-CH₃ (Prop-1-en-2-ol)

### Step 6: Tautomerization to form the ketone
The enol can undergo tautomerization to form a ketone. In this case, the enol (A) converts to the ketone:
- CH₃-C(=O)-CH₃ (Acetone or Propan-2-one)

### Step 7: Identify the type of isomerism
The type of isomerism observed in this reaction is tautomerism. Specifically, it is keto-enol tautomerism, where the enol form (A) converts to the keto form (acetone).

### Final Answer
- Structure of A: CH₃-C(OH)=C-CH₃ (Prop-1-en-2-ol)
- Type of isomerism: Tautomerism (Keto-enol tautomerism)

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