`{:(CHO,,,,),(|,,,,),((CH-OH)_(4),,,,),(|,,,,),(CH_(2)-OH,,,,):}overset(HIO_(4))(to)` Product

Options

A. 5 HCHO+HCOOH
B. 5 HCOOH+HCHO
C. `CH_(3)COOH+5HCHO`
D. `5CH_(3)CHO+HCHO`

Answer

To solve the question regarding the reaction of glucose with periodic acid (HIO4), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Glucose**:
- Glucose has the molecular formula C6H12O6. Its open-chain structure consists of an aldehyde group (CHO) at one end and five hydroxyl (OH) groups attached to the carbon atoms.

2. **Understand the Reaction with HIO4**:
- HIO4 is known to cleave carbon-carbon (C-C) and carbon-oxygen (C-O) bonds in sugars. In the case of glucose, it will cleave the C-O bonds, leading to the formation of smaller aldehydes and acids.

3. **Cleavage of C-O Bonds**:
- When glucose reacts with HIO4, the C-O bonds will be cleaved, and oxygen will be added to the resulting fragments. This will lead to the formation of smaller molecules.

4. **Identify the Products**:
- The cleavage of glucose will yield:
- **Formaldehyde (HCHO)** from the cleavage of the terminal aldehyde group.
- **Formic Acid (HCOOH)** from the cleavage of the hydroxyl groups.
- As the reaction proceeds, multiple molecules of formic acid will be produced.

5. **Count the Molecules Produced**:
- From the reaction, we can determine that:
- 1 molecule of formaldehyde (HCHO)
- 5 molecules of formic acid (HCOOH) will be formed due to the cleavage of the C-O bonds.

6. **Final Product Summary**:
- The final products of the reaction of glucose with HIO4 are:
- 1 molecule of formaldehyde (HCHO)
- 5 molecules of formic acid (HCOOH)

### Final Answer:
The final answer is:
- 1 molecule of formaldehyde (HCHO) and 5 molecules of formic acid (HCOOH).

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