Write the product formed in the following reaction `:`
`Ph-underset(underset(OH)|)overset(overset(CH_(3))|)(C)-underset(underset(OH)|)(C)H-underset(underset(OH)|)(C)H-overset(overset(CH_(3))|)(C)H-CH_(2)OHoverset(HIO_(4)("excess"))rarr`

Explanation

To solve the given reaction involving the oxidation of a compound with HIO4 (periodic acid), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of the Reactant**:
The reactant contains multiple hydroxyl (OH) groups and is likely a polyol (a compound with multiple alcohol groups). The structure can be represented as:
- Ph-CH(OH)-CH(OH)-CH(2)OH
- This indicates that there are multiple alcohol functional groups present.

2. **Understand the Reactivity with HIO4**:
HIO4 is known to oxidize 2° (secondary) and 3° (tertiary) alcohols to carbonyl compounds (aldehydes or ketones). It does not oxidize 1° (primary) alcohols. In this case, we will focus on the secondary and tertiary alcohols for oxidation.

3. **Determine Which Alcohols Get Oxidized**:
In the given structure:
- The secondary alcohols (the ones attached to two carbon atoms) will be oxidized to ketones or aldehydes.
- The primary alcohol (the one attached to one carbon atom) will remain unchanged.

4. **Perform the Oxidation**:
- The secondary alcohols will be oxidized to form carbonyl groups (C=O).
- The primary alcohol will remain as -CH2OH.

5. **Write the Products**:
After oxidation, we can summarize the products formed:
- From the oxidation of the secondary alcohols, we will get:
- Ph-CH(=O)-CH(=O)-CH2OH
- The primary alcohol remains unchanged.

6. **Final Product Structure**:
The final products can be summarized as:
- Ph-CH(=O)-CH(=O)-CH2OH
- This indicates that we have formed two carbonyl groups from the secondary alcohols and the primary alcohol remains as is.

### Summary of Products:
The products formed from the reaction with excess HIO4 are:
1. Ph-CH(=O) (benzaldehyde)
2. CH(=O) (another carbonyl)
3. CH2OH (remains unchanged)