The compound `Me_(3)C-NH_(2)` on oxidation with acidic `KMnO_(4)` gives:

Options

A. a.
B. b.
C. c.
D. d.

Explanation

To solve the question regarding the oxidation of the compound `Me3C-NH2` (tert-butylamine) with acidic `KMnO4`, we can follow these steps:

### Step 1: Identify the structure of the compound
The compound `Me3C-NH2` is tert-butylamine, which has the following structure:

```
CH3
|
CH3 - C - NH2
|
CH3
```

### Step 2: Understand the role of acidic KMnO4
Acidic `KMnO4` is a strong oxidizing agent. It is known to oxidize amines, particularly converting primary and secondary amines to their corresponding nitro compounds. In this case, we have a tertiary amine.

### Step 3: Oxidation of the amine group
When tert-butylamine is treated with acidic `KMnO4`, the `NH2` group (amine) will be oxidized to a nitro group (`NO2`). The reaction can be summarized as follows:

```
Me3C-NH2 → Me3C-NO2
```

### Step 4: Write the final product
After the oxidation, the final product will be tert-butyl nitro compound, which can be represented as:

```
CH3
|
CH3 - C - NO2
|
CH3
```

### Conclusion
Thus, the compound `Me3C-NH2` on oxidation with acidic `KMnO4` gives `Me3C-NO2`.

### Final Answer
The correct answer is `Me3C-NO2`.

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