Question

Identify the product formed in the following reaction:
$C_6H_5 - CH_2 - CH_3 \xrightarrow[\text{ii) } H_3O^+]{\text{i) alk. } KMnO_4} \text{Product}$

• A. $C_6H_5 - CH_2 - COOH$
• B. $C_6H_5CH_2 - CH_2 - COOH$
• C. $C_6H_5 - OH$
• D. $C_6H_5 - COOH$

Hint:

Remember the "Side-Chain Oxidation Rule": No matter the length of the alkyl chain, strong oxidation of any alkylbenzene possessing benzylic hydrogens will always yield benzoic acid. Only a tert-butyl group (lacking benzylic hydrogens) will resist this oxidation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Vigorous oxidation of alkylbenzenes with alkaline $KMnO_4$ converts the entire side chain into a carboxylic acid group ($-COOH$).

Step 2: Formula Application:
$Ar-R \xrightarrow{KMnO_4} Ar-COOH$. This happens as long as there is at least one benzylic hydrogen.

Step 3: Explanation:
Ethylbenzene ($C_6H_5-CH_2-CH_3$) has two benzylic hydrogens on the first carbon. Regardless of the length of the alkyl chain (methyl, ethyl, propyl, etc.), the chain is "chopped off," and the benzylic carbon is oxidized to a carboxyl group, forming Benzoic acid.

Step 4: Final Answer:
The product is Benzoic acid ($C_6H_5-COOH$).