`CH_(3)-C-= CH underset("dil." H_(2)SO_(4))overset(HgSO_(4))(to) (A)`
`CH_(3) - C-= CHunderset((2) H_(2)O_(2)//HO^(-))overset((1) BH_(3).THF)(to) (B)`
Product (A) and (B) is differentiated by :

Options

A. 2-4 DNP
B. NaOI
C. Na-metal
D. `NaHSO_(3)`

Explanation

To differentiate between the two products A and B formed from the reactions of CH₃-C≡CH, we can follow these steps:

### Step 1: Identify Product A
1. **Reaction with Dilute H₂SO₄ and HgSO₄**:
- The compound CH₃-C≡CH (propyne) reacts with dilute sulfuric acid (H₂SO₄) in the presence of mercuric sulfate (HgSO₄).
- According to Markovnikov's rule, the addition of water (H₂O) occurs such that the hydroxyl (OH) group attaches to the more substituted carbon (the one with fewer hydrogen atoms).
- This results in the formation of an enol, which then tautomerizes to a ketone.

**Product A**: CH₃-C(=O)-CH₃ (acetone)

### Step 2: Identify Product B
2. **Reaction with BH₃/THF followed by H₂O₂/OH⁻**:
- The same starting compound CH₃-C≡CH reacts with borane (BH₃) in tetrahydrofuran (THF) followed by hydrogen peroxide (H₂O₂) in the presence of hydroxide ions (OH⁻).
- This reaction follows the anti-Markovnikov rule, where the hydroxyl group attaches to the less substituted carbon.
- The product formed is an enol, which then tautomerizes to an aldehyde.

**Product B**: CH₃-CH₂-CHO (propionaldehyde)

### Step 3: Differentiation of Products A and B
3. **Distinguishing Tests**:
- **Iodoform Test**: The iodoform test can be used to differentiate between the two products.
- Product A (acetone) contains a methyl ketone group (C(=O)CH₃), which will give a positive iodoform test, producing a yellow precipitate.
- Product B (propionaldehyde) does not have a methyl ketone group and will not give a positive iodoform test.

### Conclusion
- **Product A (acetone)** will give a positive iodoform test, while **Product B (propionaldehyde)** will not. Therefore, the iodoform test can be used to differentiate between the two products.