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

Options

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

Explanation

To solve the problem, we need to analyze the reaction of the compound `CH3-oversetoverset(CH3)|C=CH-CH3` (which can be represented as 2-methyl-2-pentene) with `NaIO4` in the presence of `KMnO4`.

### Step-by-Step Solution:

1. **Identify the Compound**:
The compound `CH3-oversetoverset(CH3)|C=CH-CH3` can be rewritten as 2-methyl-2-pentene. It has a double bond between the second and third carbon atoms.

2. **Understand the Reaction**:
The reagents `NaIO4` (sodium periodate) and `KMnO4` (potassium permanganate) are both strong oxidizing agents. They are known to oxidize alkenes to form diols or carbonyl compounds.

3. **Oxidation Mechanism**:
When 2-methyl-2-pentene reacts with `NaIO4` in the presence of `KMnO4`, the double bond will undergo oxidative cleavage. This means that the double bond will break, and the carbons involved in the double bond will be oxidized.

4. **Products Formation**:
The oxidative cleavage of the double bond will result in the formation of carbonyl compounds. Specifically, the double bond between the second and third carbon will be cleaved, resulting in the formation of two products:
- One product will be a ketone from the carbon on the left side of the double bond.
- The other product will be an aldehyde from the carbon on the right side of the double bond.

5. **Final Products**:
The expected products from the reaction of 2-methyl-2-pentene with `NaIO4` and `KMnO4` are:
- 3-pentanone (from the left side of the double bond)
- Acetaldehyde (from the right side of the double bond)

### Summary of the Reaction:
The compound `CH3-oversetoverset(CH3)|C=CH-CH3` reacts with `NaIO4` in the presence of `KMnO4` to yield 3-pentanone and acetaldehyde.