Styrene on oxidation with $KMnO_{4}$ gives compound 'A', its IUPAC name is :

Question

Accepted Answer

Styrene, which is also known as vinylbenzene, is an aromatic hydrocarbon with the chemical formula C₆H₅CH=CH₂. When styrene undergoes oxidation with potassium permanganate (KMnO₄), a strong oxidizing agent, it is converted into benzoic acid. Here’s why Option 4 is the correct answer:

1. **Explanation for Option 4 (Benzoic acid):**
   - During the oxidation process, the double bond in styrene is cleaved, and the side chain is oxidized to form a carboxylic acid group. The phenyl group (C₆H₅-) remains intact, and the ethylene part (CH=CH₂) is converted into a carboxylic acid (COOH). This results in the formation of benzoic acid (C₆H₅COOH).
   - The reaction can be represented as:
     $$
     \text{C}_6\text{H}_5\text{CH=CH}_2 + \text{KMnO}_4 \rightarrow \text{C}_6\text{H}_5\text{COOH}
     $$
   - Therefore, the IUPAC name of the compound formed is benzoic acid, which corresponds to Option 4.

2. **Explanation for why other options are incorrect:**

- **Option 1 (Phenol):**
     - Phenol (C₆H₅OH) is an aromatic compound where a hydroxyl group is directly attached to the benzene ring. Oxidation of styrene with KMnO₄ does not introduce a hydroxyl group to form phenol. Therefore, this option is incorrect.

- **Option 2 (Benzaldehyde):**
     - Benzaldehyde (C₆H₅CHO) is an aromatic aldehyde. The oxidation of styrene with KMnO₄ does not result in the formation of an aldehyde group. Instead, it forms a carboxylic acid group. Hence, this option is incorrect.

- **Option 3 (2-Phenyl ethanoic acid):**
     - 2-Phenyl ethanoic acid (C₆H₅CH₂COOH) is a different compound where the carboxylic acid group is attached to a methylene group (CH₂) linked to the phenyl ring. The oxidation of styrene does not lead to this structure; it directly forms benzoic acid. Thus, this option is incorrect.

In conclusion, the oxidation of styrene with KMnO₄ results in the formation of benzoic acid, making Option 4 the correct answer.