`underset(" "CH_(2)OH)underset(|)underset(" "(CHOH)_(3))underset(|)overset(" "CH_(2)OH)overset(|)(" "C=O) overset(NaBH_(4))(rarr) A+B`
Fructose
The product A and B in the a above reaction are not

Options

A. Diastereomers
B. `C-2` epimers
C. Anomers
D. Optically active hexahydroxy compounds

Explanation

To solve the question regarding the reduction of fructose with NaBH4 and the identification of products A and B, we will follow these steps:

### Step 1: Identify the Structure of Fructose
Fructose is a ketohexose with the following structure:
- It has a ketone group (C=O) at the second carbon.
- The structure can be represented as:
```
CH2OH
|
(CHOH)3
|
C=O
```

### Step 2: Understand the Reaction with NaBH4
Sodium borohydride (NaBH4) is a reducing agent that can reduce ketones and aldehydes to alcohols. In the case of fructose, the ketone group at the second carbon will be reduced to a hydroxyl group (–OH).

### Step 3: Draw the Products
After the reduction, the structure of fructose will change. The ketone at C2 will be converted to an alcohol, resulting in two possible stereoisomers (due to the presence of chirality at C2). The products can be represented as:
- Product A:
```
CH2OH
|
(CHOH)3
|
CH2OH
```
- Product B (the epimer of A at C2):
```
CH2OH
|
(CHOH)3
|
CH2OH
```

### Step 4: Identify the Relationship Between the Products
Products A and B differ only in the configuration around the second carbon atom. This means they are epimers, specifically C2 epimers, because they differ at the second carbon.

### Step 5: Determine the Nature of the Products
Since A and B are epimers, they are also classified as diastereomers. Additionally, both products are optically active due to the presence of multiple chiral centers.

### Conclusion
The products A and B in the above reaction are not:
- Enantiomers (since they are not mirror images)
- Identical (since they differ in configuration at C2)
- Monosaccharides (since they are both polyhydroxy compounds with multiple hydroxyl groups)

Thus, the correct answer to the question is that the products A and B are not enantiomers.

---