`CH-=CH underset("dil. "H_2SO_4)overset(HgSO_4)to A underset(NaOH)overset("dilute")to B` The compound B is

Options

A. `CH_3-underset(OH)underset(|)(CH)-CH_2-CHO`
B. `CH_3-underset(OH)underset(|)(CH)-CH_2-COONa`
C. `CH_3-underset(OH)underset(|)(CH)-CH_2-overset(O)overset(||)C-CH_3`
D. `CH_3-underset(CH_3)underset(|)overset(OH)overset(|)(CH)-CH_2-overset(O)overset(||)C-CH_3`

Explanation

To solve the problem step by step, we need to analyze the given reactions and determine what compound B is after the reactions have taken place.

### Step 1: Identify the starting compound
The starting compound is an alkyne with the formula CH≡CH (ethyne or acetylene).

### Step 2: Reaction with dilute H₂SO₄
When ethyne (CH≡CH) reacts with dilute sulfuric acid (H₂SO₄) in the presence of mercury(II) sulfate (HgSO₄), it undergoes hydration. This reaction converts the alkyne into a carbonyl compound (an aldehyde or a ketone) through the formation of an enol intermediate that tautomerizes to the carbonyl compound.

The product A from this reaction is acetaldehyde (CH₃CHO).

**Reaction:**
\[ \text{CH} \equiv \text{CH} + \text{H}_2\text{SO}_4 \xrightarrow{\text{HgSO}_4} \text{CH}_3\text{CHO} \]

### Step 3: Reaction with NaOH (dilute)
Next, we take compound A (acetaldehyde, CH₃CHO) and react it with dilute sodium hydroxide (NaOH). In this case, the reaction likely involves an aldol condensation if there is enough concentration of the aldehyde. However, since we are using dilute NaOH, it may lead to the formation of a hemiacetal or a simple aldol product.

In the case of acetaldehyde, the reaction with dilute NaOH will lead to the formation of a β-hydroxy aldehyde (aldol product) through self-condensation.

The product B from this reaction is 3-hydroxybutanal (CH₃CH(OH)CH₂CHO).

**Reaction:**
\[ 2 \text{CH}_3\text{CHO} \xrightarrow{\text{NaOH (dilute)}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO} \]

### Conclusion
The compound B is 3-hydroxybutanal.

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