Consider the following reaction sequence,
`CH_(3)C-= CH overset(aq.H_(2)SO_(4))underset(Hg SO_(4))rarr A overset(PCl_(5))underset(Heat)rarrB`
The products `(A)` and `(B)` are, respectively,

Options

A. `CH_(3)COCH_(3)` and `CH_(3)C CI_(2)CH_(3)`
B. `CH_(3)CH_(2)CHO` and `CH_(3)CH_(2)CHCI_(2)`
C. `CH_(3)CHOHCH_(3)` and `CH_(3)CHCICH_(3)`
D. `CH_(3)CH_(2)CH_(2)OH` and `CH_(3)CH_(2)CH_(2)CI`

Answer

To solve the given reaction sequence, we need to analyze each step carefully.

### Step 1: Identify the starting compound
The starting compound is 1-butyne, which can be represented as CH₃C≡CH.

### Step 2: First reaction with aqueous H₂SO₄ and HgSO₄
The reaction of 1-butyne with aqueous sulfuric acid (H₂SO₄) in the presence of mercuric sulfate (HgSO₄) is known as the Kucherov reaction. This reaction leads to the formation of a ketone through the following steps:

1. **Formation of Enol**: The alkyne first reacts with H₂SO₄ to form an enol. The enol form of 1-butyne is CH₃C(OH)=CH₂.
2. **Tautomerization**: The enol undergoes tautomerization to form the more stable ketone. In this case, the enol tautomerizes to yield acetone (CH₃C(=O)CH₃).

Thus, the product A is **acetone** (CH₃C(=O)CH₃).

### Step 3: Second reaction with PCl₅ and heat
The next step involves the reaction of acetone with phosphorus pentachloride (PCl₅) under heating conditions.

1. **Reaction with PCl₅**: PCl₅ reacts with the carbonyl group (C=O) of acetone. The chlorine atoms from PCl₅ replace the oxygen in the carbonyl group, leading to the formation of a dichlorinated product.
2. **Formation of Product B**: The reaction results in the formation of 2,2-dichloropropane (CH₃CCl₂CH₃) by substituting the oxygen with two chlorine atoms.

Thus, the product B is **2,2-dichloropropane** (CH₃CCl₂CH₃).

### Final Products
- Product A: Acetone (CH₃C(=O)CH₃)
- Product B: 2,2-Dichloropropane (CH₃CCl₂CH₃)

### Summary
- The products (A) and (B) are, respectively: **Acetone and 2,2-Dichloropropane**.