Glucose as well as fructose are oxidized by periodic acid. The number of moles of `HCOOH` formed from each mole of glucose and fructose are

Options

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

Answer

To solve the problem of how many moles of formic acid (HCOOH) are produced when glucose and fructose are oxidized by periodic acid (HIO4), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reactants**:
- We have two sugars: glucose and fructose.
- Both are treated with periodic acid (HIO4), which is known to cause oxidative cleavage of carbon-carbon bonds in sugars.

2. **Structure of Glucose**:
- Glucose has the molecular formula C6H12O6 and can be represented structurally as:
```
CHO
|
CH2OH
|
OH
|
OH
|
OH
|
CH2OH
```

3. **Oxidative Cleavage of Glucose**:
- When glucose is treated with periodic acid, it undergoes oxidative cleavage.
- This reaction results in the formation of 5 moles of formic acid (HCOOH) and 1 mole of formaldehyde (HCHO).
- Therefore, for 1 mole of glucose, we produce 5 moles of formic acid.

4. **Structure of Fructose**:
- Fructose has the molecular formula C6H12O6 as well, but its structure is different:
```
CH2OH
|
C=O
|
CHOH
|
CHOH
|
CH2OH
```

5. **Oxidative Cleavage of Fructose**:
- When fructose is treated with periodic acid, it also undergoes oxidative cleavage.
- This reaction results in the formation of 3 moles of formic acid (HCOOH) and 2 moles of formaldehyde (HCHO).
- Therefore, for 1 mole of fructose, we produce 3 moles of formic acid.

6. **Final Results**:
- From the above reactions, we conclude:
- Glucose yields 5 moles of formic acid.
- Fructose yields 3 moles of formic acid.

### Conclusion:
The number of moles of formic acid formed from each mole of glucose and fructose are:
- Glucose: 5 moles of HCOOH
- Fructose: 3 moles of HCOOH

Thus, the correct answer is option C: 5 and 3.