Periodic acid splits glucose and fructose into formaldehyde and formic acid, Ratio of moles of formic acid in glucose and fructose is

Options

A. `1:2`
B. `5:3`
C. `1:1`
D. `2:3`

Answer

To solve the problem of determining the ratio of moles of formic acid produced from glucose and fructose when treated with periodic acid (HIO4), we can follow these steps:

### Step 1: Understand the Reaction
Periodic acid (HIO4) cleaves carbon-carbon bonds in sugars, leading to the formation of smaller molecules, including formaldehyde and formic acid.

### Step 2: Analyze the Structure of Glucose
The structure of glucose can be represented as:
- **Glucose**: C6H12O6
- It has an aldehyde group at the first carbon and hydroxyl groups on the other carbons.

### Step 3: Cleavage of Glucose
When glucose undergoes oxidative cleavage with periodic acid:
- The periodic acid cleaves the carbon-carbon bonds.
- For glucose, it results in the formation of 5 moles of formic acid (one from each of the 5 carbons that can be oxidized) and 1 mole of formaldehyde.

### Step 4: Analyze the Structure of Fructose
The structure of fructose is:
- **Fructose**: C6H12O6
- It has a ketone group at the second carbon and hydroxyl groups on the other carbons.

### Step 5: Cleavage of Fructose
When fructose is treated with periodic acid:
- The periodic acid cleaves the carbon-carbon bonds.
- It results in the formation of 3 moles of formic acid, 2 moles of formaldehyde, and 1 mole of CO2 gas.

### Step 6: Calculate the Ratio of Formic Acid
From the cleavage reactions:
- **From Glucose**: 5 moles of formic acid
- **From Fructose**: 3 moles of formic acid

Thus, the ratio of moles of formic acid produced from glucose to fructose is:
\[ \text{Ratio} = \frac{5 \text{ moles (glucose)}}{3 \text{ moles (fructose)}} = 5:3 \]

### Conclusion
The final answer for the ratio of moles of formic acid produced from glucose and fructose is **5:3**.

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