`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 reaction step by step, we will analyze the transformation of the reactants and products involved.

### Step 1: Identify the starting material
The starting material is ethyne (C2H2), which is an alkyne.

### Step 2: Reaction with HgSO4 and H2SO4
When ethyne is treated with mercuric sulfate (HgSO4) and sulfuric acid (H2SO4), it undergoes hydration. This reaction adds water across the triple bond of the alkyne, leading to the formation of an enol intermediate.

The reaction can be represented as:
\[ \text{C}_2\text{H}_2 + \text{HgSO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3\text{CH}=\text{OH} \]
This compound is known as vinyl alcohol (enol form).

### Step 3: Tautomerization
The enol form (vinyl alcohol) is not stable and undergoes tautomerization to form a more stable aldehyde. The enol rearranges to form acetaldehyde (ethanal).
\[ \text{CH}_3\text{CH}=\text{OH} \rightarrow \text{CH}_3\text{C}(=O)\text{H} \]
Thus, the product X is acetaldehyde (ethanal).

### Step 4: Reaction with LiAlH4
Next, acetaldehyde (X) is treated with lithium aluminum hydride (LiAlH4), which is a strong reducing agent. This reduction converts the aldehyde to a primary alcohol.
\[ \text{CH}_3\text{C}(=O)\text{H} + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} \]
The product Y is ethanol (ethyl alcohol).

### Conclusion
Thus, the final products of the reactions are:
- X = Acetaldehyde (ethanal)
- Y = Ethanol (ethyl alcohol)

### Final Answer:
X is acetaldehyde (ethanal) and Y is ethanol (ethyl alcohol).

---