An organic compound `X` has the molecular formula `C_(2)H_(6)O`. Upon reaction with alkaline `KMnO_(4)` it gets oxidised to compound `Y`. Which of the following reagents can be used to distinguish between compounds `X` and `Y` ?

Options

A. Sodium hydroxide
B. Sodium carbonate
C. Sodium metal
D. All of these

Explanation

To solve the question, we need to identify the compounds X and Y based on the given molecular formula and the reaction with alkaline KMnO4.

### Step-by-Step Solution:

1. **Identify Compound X**:
- The molecular formula of compound X is C2H6O.
- This formula corresponds to ethanol (C2H5OH), which is an alcohol.

2. **Reaction with Alkaline KMnO4**:
- When ethanol (X) is oxidized by alkaline KMnO4, it gets converted into acetic acid (Y).
- The reaction can be summarized as:
\[ \text{C2H5OH} \xrightarrow{\text{KMnO4}} \text{CH3COOH} \]
- Here, CH3COOH is acetic acid.

3. **Distinguishing Between X and Y**:
- To distinguish between ethanol (X) and acetic acid (Y), we need a reagent that reacts differently with each compound.
- Sodium carbonate (Na2CO3) can be used as a distinguishing reagent.

4. **Testing with Sodium Carbonate**:
- When acetic acid (Y) reacts with sodium carbonate, it produces carbon dioxide gas:
\[ \text{CH3COOH} + \text{Na2CO3} \rightarrow \text{CH3COONa} + \text{H2O} + \text{CO2} \uparrow \]
- In contrast, ethanol (X) does not react with sodium carbonate, and no gas is produced.

5. **Conclusion**:
- Therefore, sodium carbonate can be used to distinguish between compound X (ethanol) and compound Y (acetic acid).

### Final Answer:
The reagent that can be used to distinguish between compounds X (C2H6O) and Y (the oxidized product) is sodium carbonate (Na2CO3).