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 problem of what products are formed when open-chain glucose is oxidized with periodic acid (HIO₄), we can follow these steps:

### Step 1: Identify the structure of glucose
Glucose is a six-carbon aldose sugar with the molecular formula C₆H₁₂O₆. The open-chain form of glucose has an aldehyde group at one end.

### Step 2: Understand the reaction with HIO₄
Periodic acid (HIO₄) is known to cleave vicinal diols and can oxidize aldehydes. In the case of glucose, it will oxidize the aldehyde group and also cleave the carbon chain.

### Step 3: Determine the products of the reaction
When glucose (C₆H₁₂O₆) is oxidized by HIO₄, it breaks down into smaller carboxylic acids. The reaction yields:
- 5 molecules of formic acid (HCOOH)
- 1 molecule of formaldehyde (HCHO)

### Step 4: Write the balanced equation
The overall reaction can be summarized as:
\[ \text{C}_6\text{H}_{12}\text{O}_6 + \text{HIO}_4 \rightarrow 5 \text{HCOOH} + \text{HCHO} \]

### Step 5: Conclude the answer
Thus, the products of the oxidation of open-chain glucose with HIO₄ are 5 molecules of formic acid and 1 molecule of formaldehyde.

### Final Answer:
The open-chain glucose on oxidation with HIO₄ gives 5 HCOOH and 1 HCHO.

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