The compound `CH_3-oversetoverset(CH_3)|C=CH-CH_3` on reaction with `NaIO_4` in the presence of `KMnO_4` given :

Options

A. `CH_3COCH_3`
B. `CH_3COCH_3+CH_3COOH`
C. `CH_3COCH_3+CH_3CHO`
D. `CH_3CHO+CO_2`

Answer

To solve the problem, we need to analyze the reaction of the compound `CH3-CH(CH3)=CH-CH3` (isopentylene) with `NaIO4` in the presence of `KMnO4`.

### Step-by-Step Solution:

1. **Identify the Compound**:
The compound given is isopentylene, which has the structure:
\[ CH_3-CH(CH_3)=CH-CH_3 \]
This is a branched alkene.

2. **Understand the Reaction**:
The reaction involves `NaIO4`, which is a strong oxidizing agent that can cleave alkenes. In the presence of `KMnO4`, it further facilitates the oxidation process.

3. **Mechanism of Reaction**:
- The alkene undergoes oxidative cleavage. The double bond between the carbon atoms will be broken.
- The reaction will lead to the formation of carbonyl compounds (aldehydes or ketones) from the alkene.

4. **Products Formation**:
- Upon oxidative cleavage of isopentylene, we can predict the formation of two products:
- The double bond will break between the two carbons, leading to the formation of:
- Acetone (from the central carbon with two methyl groups)
- Acetic acid (from the terminal carbon with a methyl group and a hydrogen).

5. **Final Products**:
The reaction of isopentylene with `NaIO4` in the presence of `KMnO4` will yield:
\[ \text{Products: } CH_3COCH_3 \text{ (acetone) and } CH_3COOH \text{ (acetic acid)} \]

### Conclusion:
The correct products formed from the reaction of isopentylene with `NaIO4` in the presence of `KMnO4` are acetone and acetic acid.