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 an alkaline solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compounds**:
- We are dealing with three sugars: D-glucose, D-mannose, and D-fructose. These are all monosaccharides and can interconvert in alkaline conditions.

2. **Understand the Structures**:
- **D-Glucose**: It has an aldehyde group (CHO) at one end and multiple hydroxyl groups (OH) attached to its carbon chain.
- **D-Fructose**: This is a ketose sugar, meaning it has a ketone group (C=O) and also contains hydroxyl groups.
- **D-Mannose**: This is an epimer of D-glucose, differing in the configuration around one specific carbon atom.

3. **Equilibrium in Alkaline Solution**:
- In an alkaline solution, D-glucose can isomerize to D-fructose and D-mannose through a series of reactions. This is due to the presence of hydroxide ions which facilitate the rearrangement of these sugars.

4. **Name the Reaction**:
- The process by which D-glucose, D-mannose, and D-fructose interconvert in an alkaline solution is known as the **Lobry de Bruyn–Alberda van Ekenstein rearrangement**. This rearrangement involves the conversion of aldoses to ketoses and vice versa.

5. **Conclusion**:
- Therefore, the reaction in which D-glucose, D-mannose, and D-fructose are in equilibrium in an alkaline solution is referred to as the **Lobry de Bruyn–Alberda van Ekenstein rearrangement**.