The compound `CH_(3)-overset(CH_(3))overset(|)C=CH-CH_(3)`
on reaction with `NalO_(4)` in the presence of `KMnO_(4)` gives

Options

A. `CH_(3)COCH_(3)`
B. `CH_(3)COCH_(3)+CH_(3)COOH`
C. `CH_(3)COCH_(3)+CH_(3)CHO`
D. `CH_(3)CHO+CO_(2)`

Answer

To solve the problem, we need to analyze the reaction of the compound \( CH_3C(CH_3)=CH-CH_3 \) with sodium periodate (\( NaIO_4 \)) in the presence of potassium permanganate (\( KMnO_4 \)).

### Step-by-Step Solution:

1. **Identify the Compound**:
The compound given is \( CH_3C(CH_3)=CH-CH_3 \). This is an alkene due to the presence of a double bond between the carbon atoms.

2. **Understand the Reagents**:
- **Sodium Periodate (\( NaIO_4 \))**: This reagent is known for its oxidative cleavage capability, which means it can break double bonds and convert them into carbonyl compounds (aldehydes or ketones).
- **Potassium Permanganate (\( KMnO_4 \))**: This reagent is a strong oxidizing agent that can add hydroxyl groups (OH) across the double bond in a cis manner.

3. **Cis Addition of Hydroxyl Groups**:
When \( KMnO_4 \) is used, it will add hydroxyl groups to the double bond in a cis configuration. Therefore, the double bond in the compound will be converted into a 1,2-diol:
\[ CH_3C(CH_3)(OH)CH(OH)CH_3 \]
This results in a compound with two hydroxyl groups on adjacent carbon atoms.

4. **Oxidative Cleavage by Sodium Periodate**:
After the formation of the 1,2-diol, the \( NaIO_4 \) will cleave the bond between the two carbons that have the hydroxyl groups. This cleavage will yield:
- A ketone from the secondary alcohol (the carbon with two methyl groups).
- An aldehyde from the primary alcohol (the carbon with one methyl group).

The products from this cleavage will be:
- **Ketone**: \( CH_3C(=O)CH_3 \) (Acetone)
- **Aldehyde**: \( CH_3CHO \) (Acetaldehyde)

5. **Further Oxidation of Aldehyde**:
Since \( KMnO_4 \) is still present in the reaction mixture, the acetaldehyde (\( CH_3CHO \)) will undergo further oxidation to form a carboxylic acid:
\[ CH_3CHO \xrightarrow{KMnO_4} CH_3COOH \]
This results in the formation of acetic acid (ethanoic acid).

6. **Final Products**:
The final products of the reaction are:
- Acetone (\( CH_3C(=O)CH_3 \))
- Acetic acid (\( CH_3COOH \))

### Conclusion:
Thus, the final answer to the question is that the reaction of the compound \( CH_3C(CH_3)=CH-CH_3 \) with \( NaIO_4 \) in the presence of \( KMnO_4 \) gives acetone and acetic acid.