Products of the following reaction
`CH_(3)C-=CCH_(2)CH_(3) overset((i)O_(3)//"ether")underset((ii) H_(2)O//Zn)to` are

Options

A. `CH_(3)-overset(O)overset(||)C-overset(O)overset(||)C-CH_(2)-CH_(3)`
B. `CH_(3)COOH + HOOC.CH_(2)CH_(3)`
C. `CH_(3)CHO + CH_(3)CH_(2)CHO`
D. `CH_(3)COOH + CH_(3)COCH_(3)`

Answer

To solve the given problem, we need to determine the products of the reaction involving the compound CH₃C≡CCH₂CH₃ when treated with ozone (O₃) followed by water (H₂O) in the presence of zinc (Zn). This process is known as reductive ozonolysis.

### Step-by-Step Solution:

1. **Identify the Compound**:
The compound given is CH₃C≡CCH₂CH₃, which is 1-pentyne. It contains a triple bond between the second and third carbon atoms.

2. **Ozonolysis Reaction**:
When 1-pentyne undergoes ozonolysis, ozone (O₃) reacts with the triple bond. The triple bond breaks, and ozone adds across the bond, forming a molozonide intermediate.

3. **Formation of Carbonyl Compounds**:
The molozonide rearranges to form carbonyl compounds. In the case of terminal alkynes, this typically results in the formation of two carbonyl groups (ketones).

4. **Reduction with Water and Zinc**:
The molozonide is then treated with water in the presence of zinc, which reduces the intermediate and leads to the formation of the final products. In this case, the reaction will yield two molecules of aldehydes or ketones.

5. **Final Products**:
For 1-pentyne, the ozonolysis will yield two molecules of propan-2-one (acetone) and one molecule of formaldehyde (methanal). The overall reaction can be summarized as:
- CH₃C≡CCH₂CH₃ + O₃ → 2 CH₃COCH₃ (acetone) + CH₂O (formaldehyde)

### Conclusion:
The products of the reaction are acetone and formaldehyde.