The open - chain glucose on oxidation with `HIO_(4)` gives

Options

A. `5HCOOH+H_(2)C=O`
B. `4HCOOH+2H_(2)C=O`
C. `3HCOOH+3H_(2)C=O`
D. `2HCOOH+4H_(2)C=O`

Answer

To solve the question regarding the oxidation of open-chain glucose with periodic acid (HIO4), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Glucose**:
- Glucose in its open-chain form is an aldohexose with the structure:
\[ \text{CHO-CHOH-CHOH-CHOH-CHOH-CH}_2\text{OH} \]
- This structure consists of six carbon atoms, with the aldehyde group (CHO) at one end.

2. **Reaction with Periodic Acid (HIO4)**:
- Periodic acid is known to cleave vicinal diols (two hydroxyl groups on adjacent carbons) and oxidizes them.
- In glucose, there are multiple vicinal diols present.

3. **Understanding the Cleavage**:
- When glucose reacts with HIO4, it cleaves the carbon chain at the vicinal diols, leading to the formation of smaller molecules.
- The cleavage will occur between the hydroxyl groups, resulting in the formation of formic acid (HCOOH) and formaldehyde (HCHO).

4. **Counting the Products**:
- For each cleavage, we can expect to produce formic acid from the carbons involved in the cleavage.
- Specifically, the reaction of glucose with HIO4 will yield:
- 5 moles of formic acid (HCOOH)
- 1 mole of formaldehyde (HCHO)

5. **Final Products**:
- Therefore, the final products of the oxidation of open-chain glucose with periodic acid are:
\[ 5 \text{ moles of formic acid} + 1 \text{ mole of formaldehyde} \]

### Conclusion:
The open-chain glucose on oxidation with HIO4 gives 5 moles of formic acid and 1 mole of formaldehyde.

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