In alkalin solution D-glucose, D-mannose and D-fructose are in equilibrium. This reaction is known as

Options

A. Fries rearrangement
B. Lobry de Bruyn-van Enkelstein rearrangement
C. Hofmann rearrangement
D. Kolbe's reaction

Answer

To solve the question regarding the equilibrium of D-glucose, D-mannose, and D-fructose in alkaline solution, we need to follow these steps:

### Step-by-Step Solution:

1. **Understanding the Compounds**:
- D-glucose, D-mannose, and D-fructose are all monosaccharides (simple sugars) that can interconvert in alkaline solutions.
- D-glucose is an aldohexose, while D-mannose is an epimer of glucose, and D-fructose is a ketohexose.

2. **Equilibrium in Alkaline Solution**:
- In an alkaline solution, these sugars can undergo a series of reactions that allow them to convert from one form to another, establishing an equilibrium between them.

3. **Identifying the Reaction**:
- The process by which D-glucose, D-mannose, and D-fructose interconvert in alkaline solution is known as the **Lobry de Bruyn–Van Ekenstein transformation**. This reaction involves the rearrangement of the sugar structures.

4. **Conclusion**:
- Therefore, the answer to the question is that the reaction is known as the **Lobry de Bruyn–Van Ekenstein rearrangement**.

### Final Answer:
The reaction is known as the **Lobry de Bruyn–Van Ekenstein rearrangement**.

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