In Fehling's or Benedict's solution, a reducing sugar causes conversion of

Options

A. Ferric to ferrous state
B. Ferrous to ferric state
C. Cuprous to cupric state
D. Cupric to cuprous state.

Answer

To answer the question regarding the conversion caused by a reducing sugar in Fehling's or Benedict's solution, we can break down the solution into clear steps:

### Step-by-Step Solution:

1. **Understanding Reducing Sugars**:
- Reducing sugars are carbohydrates that have free aldehyde or ketone groups. These groups allow them to act as reducing agents. Common examples include glucose and fructose.

2. **Identifying the Solutions**:
- Fehling's solution and Benedict's solution are both used to test for the presence of reducing sugars. They contain copper(II) ions (Cu²⁺) in an alkaline solution.

3. **Chemical Reaction**:
- When a reducing sugar is present, it reduces the cupric ions (Cu²⁺) in the solution to cuprous ions (Cu⁺). This is a key reaction that indicates the presence of reducing sugars.

4. **Color Change**:
- The initial blue color of the copper(II) sulfate solution changes to a reddish precipitate of cuprous oxide (Cu₂O) as the reaction occurs. This color change is a visual indicator of the presence of reducing sugars.

5. **Analyzing the Options**:
- The options provided in the question are:
1. Ferric to ferrous state
2. Ferrous to ferric state
3. Cuprous to cupric state
4. Cupric to cuprous state
- From our understanding, the correct conversion is from cupric (Cu²⁺) to cuprous (Cu⁺).

6. **Conclusion**:
- Therefore, the correct answer is that a reducing sugar causes the conversion of cupric to cuprous state.

### Final Answer:
The correct conversion caused by a reducing sugar in Fehling's or Benedict's solution is **cupric to cuprous state**.

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