Glucose on reduction with `NaBH_(4)` gives

Options

A. sorbitol
B. Fructose
C. Glycerol
D. None of these

Explanation

To solve the question "Glucose on reduction with NaBH4 gives," we need to follow these steps:

### Step 1: Understand the structure of glucose
Glucose is a simple sugar (monosaccharide) with the molecular formula C6H12O6. It has an aldehyde group (-CHO) at one end and multiple hydroxyl groups (-OH) attached to its carbon skeleton.

### Step 2: Identify the reducing agent
Sodium borohydride (NaBH4) is a common reducing agent used in organic chemistry. It is known for its ability to reduce aldehydes and ketones to their corresponding alcohols.

### Step 3: Perform the reduction reaction
When glucose is treated with NaBH4, the aldehyde group at the first carbon (C1) is reduced to a primary alcohol. This means that the -CHO group is converted to -CH2OH.

### Step 4: Write the product structure
After the reduction, the structure of glucose changes. The original aldehyde group (C1) becomes a primary alcohol, resulting in the formation of sorbitol, which is a sugar alcohol. The structure of sorbitol can be represented as:
- CH2OH (at C1)
- C2: OH
- C3: OH
- C4: OH
- C5: OH
- C6: CH2OH

### Step 5: Conclusion
Thus, the reduction of glucose with NaBH4 yields sorbitol.

### Final Answer
Glucose on reduction with NaBH4 gives sorbitol.

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