Equal amount of aqueous solution of `CuSO_4` and alkaline sodium potassium tartarate is mixed and treated with aliphatic aldehyde, then

Options

A. Red brown ppt of `Cu_2O` is formed
B. Red brown ppt of Cuo is formed
C. Blue ppt of CuO is formed
D. There will be no reaction

Answer

To solve the problem, we need to analyze the reaction that occurs when equal amounts of aqueous solutions of copper sulfate (CuSO4) and alkaline sodium potassium tartrate are mixed and treated with an aliphatic aldehyde. This is a classic test for aldehydes known as the Fehling's test.

### Step-by-Step Solution:

1. **Identify the Solutions**:
- We have two solutions:
- **Fehling's Solution A**: A blue solution of copper(II) sulfate (CuSO4).
- **Fehling's Solution B**: A colorless solution of sodium potassium tartrate (C4H4O6) in an alkaline medium.

2. **Mix the Solutions**:
- When equal amounts of these two solutions are mixed, a complex is formed between the copper ions (Cu²⁺) from CuSO4 and the tartrate ions from the sodium potassium tartrate.

3. **Add Aliphatic Aldehyde**:
- When an aliphatic aldehyde (RCHO) is added to this mixture and heated, the aldehyde reduces the copper(II) ions (Cu²⁺) to copper(I) ions (Cu⁺).

4. **Formation of Carboxylate Ion**:
- The aldehyde is oxidized to a carboxylate ion (RCOO⁻) in the process. The reaction can be summarized as:
\[ RCHO + 2Cu^{2+} + 2OH^- \rightarrow RCOO^- + Cu_2O(s) + H_2O \]

5. **Precipitate Formation**:
- The copper(I) oxide (Cu2O) that forms as a result of this reduction is a red-brown precipitate.

6. **Conclusion**:
- Therefore, the correct answer to the question is that a red-brown precipitate of Cu2O is formed.

### Final Answer:
- **Red-brown PPT of Cu2O is formed.**