`C_(6)H_(5)CH_(2)CH_(2)CH_(3)` is when oxidised in the presence of alk. `KMnO_(4)` the product obtained is

Options

A. `C_(6)H_(5)CHO`
B. `C_(6)H_(5)COOH`
C. `C_(6)H_(5)CH_(2)CH_(2)CHO`
D. `C_(6)H_(5)COCH_(3)`

Answer

To solve the question regarding the oxidation of the compound \( C_6H_5CH_2CH_2CH_3 \) in the presence of alkaline \( KMnO_4 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
The compound \( C_6H_5CH_2CH_2CH_3 \) can be recognized as phenylpropane. The phenyl group \( C_6H_5 \) is attached to a propyl chain \( CH_2CH_2CH_3 \).

2. **Understand the Oxidizing Agent**:
Alkaline \( KMnO_4 \) is a strong oxidizing agent that can oxidize alkyl chains. It is known to oxidize primary and secondary alcohols, aldehydes, and even aliphatic chains to carboxylic acids.

3. **Oxidation Process**:
In this case, the propyl chain (which is a part of the molecule) will undergo oxidation. The terminal carbon (the one attached to the \( CH_3 \)) will be oxidized to a carboxylic acid.

4. **Determine the Product**:
The oxidation of the propyl group will lead to the formation of a carboxylic acid. Specifically, the propyl group \( CH_2CH_2CH_3 \) will be oxidized to \( COOH \), resulting in the formation of benzoic acid \( C_6H_5COOH \).

5. **Final Product**:
Therefore, the final product of the reaction when \( C_6H_5CH_2CH_2CH_3 \) is oxidized with alkaline \( KMnO_4 \) is benzoic acid \( C_6H_5COOH \).

### Conclusion:
The answer to the question is that the product obtained is **benzoic acid** \( C_6H_5COOH \).