Reduction of hexose A (molecular formula `C_(6)H_(12)O_(6)`) with sodium borohydride gives compound B and C. Compound B is optically inactive, whereas compound C is optically active. Which of the following is compound A?

Options

A. D - fructose
B. D - glucose
C. D - mannose
D. D - galactose

Answer

To solve the problem, we need to identify the hexose (compound A) that, when reduced with sodium borohydride (NaBH4), gives one optically inactive compound (B) and one optically active compound (C).

### Step-by-Step Solution:

1. **Understanding the Structure of Hexoses**:
- Hexoses are monosaccharides with the formula C6H12O6. Common hexoses include glucose, fructose, galactose, and mannose.

2. **Identifying the Type of Reaction**:
- Sodium borohydride is a reducing agent that reduces carbonyl groups (aldehydes and ketones) to alcohols. In the case of hexoses, the carbonyl group can be either an aldehyde (as in glucose) or a ketone (as in fructose).

3. **Analyzing the Options**:
- We need to analyze the structures of the common hexoses to see which one can produce two different products upon reduction:
- **D-Fructose**: A ketose (C=O in the second carbon).
- **D-Glucose**: An aldose (C=O at the first carbon).
- **D-Galactose**: An aldose (similar to glucose but with different hydroxyl group arrangements).
- **D-Mannose**: Another aldose.

4. **Reduction of D-Fructose**:
- D-Fructose, when reduced by NaBH4, will yield two products:
- One product will be a secondary alcohol (B) and will be optically inactive due to symmetry.
- The other product will be a different stereoisomer (C) that is optically active due to the presence of an asymmetric carbon.

5. **Conclusion**:
- Since D-Fructose can produce one optically inactive compound (B) and one optically active compound (C) upon reduction, it fits the criteria given in the question. Therefore, compound A is **D-Fructose**.

### Final Answer:
Compound A is **D-Fructose**.