`CH_(3)-C-=CHunderset(Hg^(+2))overset(dil.H_(2)SO_(4))(to)Xunderset(Delta)overset(Ba(OH)_(2))(to)Y,Y` is

Options

A. `CH_(3)-CH_(2)-CH=CH-underset(O)underset(||)C-CH_(3)`
B. `CH_(3)-overset(CH_(3))overset(|)(C)=CH-underset(O)underset(||)C-CH_(3)`
C. `CH_(3)-overset(CH_(3))overset(|)(C)=CH-CH_(2)-CH=O`
D. `CH_(3)-CH_(2)-CH=CH-CH_(2)-CH=O`

Answer

To solve the question, we will go through the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is 1-Butyne (CH₃-C≡CH).

### Step 2: Reaction with Hg²⁺ and dilute H₂SO₄
When 1-Butyne reacts with mercuric ion (Hg²⁺) in the presence of dilute sulfuric acid (H₂SO₄), it undergoes hydration to form a ketone. The reaction proceeds as follows:

1. The triple bond (C≡C) is converted into a double bond (C=C) by the addition of water (H₂O) in the presence of Hg²⁺.
2. The product formed is 2-Butanone (CH₃-CO-CH₂-CH₃).

**Intermediate Product (X):**
\[ \text{X} = \text{2-Butanone (CH}_3\text{-CO-CH}_2\text{-CH}_3) \]

### Step 3: Reaction with Barium Hydroxide (Ba(OH)₂)
Next, we take the intermediate product (2-Butanone) and react it with barium hydroxide (Ba(OH)₂) and heat it. This will lead to an aldol condensation reaction.

1. The barium hydroxide acts as a strong base, which abstracts a proton from the alpha carbon of the ketone (2-Butanone).
2. This generates an enolate ion.
3. The enolate ion then attacks the carbonyl carbon of another molecule of 2-Butanone, forming a β-hydroxy ketone.

**Intermediate Product (after aldol condensation):**
\[ \text{CH}_3\text{-C(OH)(CH}_3)\text{-CH}_2\text{-C(=O)-CH}_3 \]

### Step 4: Dehydration
Upon heating, the β-hydroxy ketone undergoes dehydration (loss of water) to form an α,β-unsaturated ketone.

1. The water is eliminated, resulting in the formation of a double bond between the alpha and beta carbons.

**Final Product (Y):**
\[ \text{Y} = \text{4-Methylpent-3-en-2-one (CH}_3\text{-C(=CH}_2)\text{-C(=O)-CH}_3) \]

### Conclusion
The final product Y is 4-Methylpent-3-en-2-one.