Write the product formed when glucose is treated with HIO_4

Answer

To determine the product formed when glucose is treated with periodic acid (HIO₄), we can follow these steps:

### Step 1: Understand the Structure of Glucose
Glucose is a six-carbon aldose sugar. Its structure can be represented as follows:
- The open-chain structure of glucose has an aldehyde group (-CHO) at one end and five hydroxyl groups (-OH) attached to the carbon atoms.

### Step 2: Identify the Functional Groups
In glucose, the functional groups present are:
- One aldehyde group (at carbon 1)
- Five hydroxyl groups (at carbons 2, 3, 4, 5, and 6)

### Step 3: Recognize the Reactivity of HIO₄
Periodic acid (HIO₄) is a strong oxidizing agent that can oxidize:
- Aldehyde groups
- Primary alcohols (1° alcohols)
- Secondary alcohols (2° alcohols)

### Step 4: Oxidation Reaction
When glucose is treated with HIO₄, the following oxidation reactions occur:
- The aldehyde group at carbon 1 will be oxidized to formic acid (HCOOH).
- The primary alcohol at carbon 6 will also be oxidized to formaldehyde (HCHO).
- The secondary alcohols at carbons 2, 3, 4, and 5 will be oxidized to formic acid (HCOOH).

### Step 5: Write the Products
The products formed from the oxidation of glucose by HIO₄ are:
- Formic acid (HCOOH) from carbons 2, 3, 4, and 5.
- Formaldehyde (HCHO) from carbon 6.

### Final Product Summary
The overall products formed when glucose is treated with HIO₄ are:
- Formaldehyde (HCHO)
- Formic acid (HCOOH)