Total number of moles of `HlO_(4)` required by one mole of D-glucose when it is subjected to periodic acid oxidation is

Options

A. 3
B. 5
C. 2
D. 4

Answer

To determine the total number of moles of periodic acid (HIO₄) required by one mole of D-glucose during periodic acid oxidation, we can follow these steps:

### Step 1: Understand the Structure of D-Glucose
D-glucose is a six-carbon sugar with the following structure:
- It has an aldehyde group (CHO) at one end.
- There are five hydroxyl (OH) groups attached to the carbon atoms.

### Step 2: Identify the Reaction with Periodic Acid
Periodic acid (HIO₄) oxidizes D-glucose by cleaving carbon-carbon bonds and adding hydroxyl groups (OH) across the resulting carbonyl groups. The reaction can be summarized as follows:
- HIO₄ cleaves the C-C bonds in D-glucose.
- Each cleavage results in the formation of two moles of formic acid (HCOOH) or formaldehyde (CHO).

### Step 3: Determine the Number of Cleavages
In the case of D-glucose, there are five carbon-carbon bonds that can be cleaved:
1. Between C1 and C2
2. Between C2 and C3
3. Between C3 and C4
4. Between C4 and C5
5. Between C5 and C6

### Step 4: Calculate the Number of Moles of HIO₄ Required
Each cleavage of a carbon-carbon bond requires one mole of HIO₄. Since there are five carbon-carbon bonds in D-glucose, the total number of moles of HIO₄ required is:
- Total moles of HIO₄ = Number of cleavages = 5

### Conclusion
Therefore, the total number of moles of HIO₄ required by one mole of D-glucose when subjected to periodic acid oxidation is **5 moles**.