`CH=CHoverset(HgSO_(4))underset(H_(2)SO_(4))toXoverset(LiAlH_(4))toY`.
In the above reaction, X and Y are

Options

A. Ethanal, Ethanol
B. Ethanol, Ethanal
C. Ethanal, Ethanoic acid
D. Ethanoic acid, Acetone

Answer

To solve the given question, we will analyze the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is an alkyne represented as CH=CH. This is ethyne (acetylene).

### Step 2: Reaction with HgSO4 and H2SO4
When ethyne (CH≡CH) reacts with mercuric sulfate (HgSO4) in the presence of sulfuric acid (H2SO4), it undergoes hydration to form an enol, which can tautomerize to a ketone.

1. The reaction proceeds through the formation of a non-classical carbocation.
2. Water (H2O) acts as a nucleophile and attacks the carbocation, leading to the formation of an enol.
3. The enol then undergoes keto-enol tautomerism to form a ketone.

The product after this reaction (X) is **ethanol (CH3CHO)**.

### Step 3: Reaction with LiAlH4
Next, the product X (ethanol) is treated with lithium aluminum hydride (LiAlH4), which is a strong reducing agent.

1. LiAlH4 reduces aldehydes to primary alcohols.
2. Therefore, the aldehyde (ethanol) will be reduced to its corresponding alcohol.

The product after this reaction (Y) is **ethanol (CH3CH2OH)**.

### Final Products
- **X**: Ethanal (CH3CHO)
- **Y**: Ethanol (CH3CH2OH)

### Summary of the Reaction
- The reaction starts with ethyne (CH≡CH), which converts to ethanal (X) upon reaction with HgSO4 and H2SO4.
- The ethanal (X) is then reduced to ethanol (Y) by LiAlH4.

### Final Answer
- X = Ethanal (CH3CHO)
- Y = Ethanol (CH3CH2OH)