`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 problem, we need to determine the product formed when the compound \( C_6H_5CH_2CH_2CH_3 \) (which can be represented as a phenylpropyl group) is oxidized in the presence of alkaline \( KMnO_4 \).

### Step-by-Step Solution:

1. **Identify the Structure of the Compound**:
The compound \( C_6H_5CH_2CH_2CH_3 \) consists of a benzene ring (C6H5) attached to a propyl group (CH2CH2CH3). This means it has a total of 6 carbons in the benzene and 3 in the propyl group, making a total of 9 carbons.

2. **Understand the Oxidation Process**:
Alkaline \( KMnO_4 \) is a strong oxidizing agent. It can oxidize alkyl side chains attached to aromatic rings. In this case, it will oxidize the propyl group (the alkyl chain) completely.

3. **Complete Oxidation of the Alkyl Chain**:
The propyl group (CH2CH2CH3) will be oxidized to form a carboxylic acid. Since the propyl group has 3 carbons, the complete oxidation will convert each carbon in the alkyl chain into a carboxylic acid group (-COOH).

4. **Formation of the Product**:
The oxidation of the propyl group will yield:
- The terminal carbon (C1) will be oxidized to a carboxylic acid (-COOH).
- The other two carbons (C2 and C3) will also be oxidized to form a carboxylic acid.

Therefore, the entire propyl group will be converted into a carboxylic acid, resulting in the formation of benzoic acid (C6H5COOH).

5. **Final Product**:
The final product obtained after the oxidation of \( C_6H_5CH_2CH_2CH_3 \) in the presence of alkaline \( KMnO_4 \) is benzoic acid (C6H5COOH).

### Conclusion:
The product obtained from the oxidation of \( C_6H_5CH_2CH_2CH_3 \) with alkaline \( KMnO_4 \) is **benzoic acid**.

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