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 find 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 cleaves the carbon-carbon bonds in sugars, leading to the formation of formaldehyde (HCHO) and formic acid (HCOOH).

### Step 2: Analyze glucose
Glucose has the molecular formula C6H12O6. When glucose is treated with periodic acid, it undergoes oxidative cleavage.

- The breakdown yields:
- 5 moles of formic acid (HCOOH)
- 1 mole of formaldehyde (HCHO)

### Step 3: Analyze fructose
Fructose also has the molecular formula C6H12O6, but its structure is different from glucose. When fructose is treated with periodic acid, it also undergoes oxidative cleavage.

- The breakdown yields:
- 3 moles of formic acid (HCOOH)
- 2 moles of formaldehyde (HCHO)
- Additionally, carbon dioxide (CO2) is produced from the cleavage of the carbon skeleton.

### Step 4: Calculate the ratio of formic acid
Now we can calculate the ratio of moles of formic acid produced from glucose to that from fructose:

- Moles of formic acid from glucose = 5
- Moles of formic acid from fructose = 3

Thus, the ratio of moles of formic acid from glucose to fructose is:

\[ \text{Ratio} = \frac{\text{Formic acid from glucose}}{\text{Formic acid from fructose}} = \frac{5}{3} \]

### Final Answer
The ratio of moles of formic acid in glucose to fructose is **5:3**.

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