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 Question:
The problem presents an organic reaction where an alkylbenzene (ethylbenzene, $C_6H_5-CH_2-CH_3$) is subjected to strong oxidative conditions using alkaline potassium permanganate ($KMnO_4$) followed by an acidic workup. We must identify the final organic product.

Step 2: Detailed Explanation:
Alkaline $KMnO_4$ is an extremely powerful oxidizing agent.
When it acts upon an aromatic ring holding an alkyl side chain, a very specific and aggressive reaction occurs:
Regardless of how long or complex the alkyl side chain is (whether it is a methyl, ethyl, propyl, or isopropyl group), as long as there is at least one hydrogen atom attached directly to the benzylic carbon (the carbon immediately attached to the benzene ring), the entire side chain is aggressively cleaved and oxidized completely down to a single carboxylic acid group (-COOH).
1. The benzylic carbon undergoes severe oxidation to form a carboxylate salt initially (potassium benzoate) because the reaction is in an alkaline medium. The rest of the carbon chain is cleaved off as $CO_2$ and $H_2O$.
2. The subsequent acidic workup step ($\text{ii) } H_3O^+$) protonates the intermediate carboxylate salt to yield the final, stable carboxylic acid.
Thus, ethylbenzene ($C_6H_5-CH_2-CH_3$) is oxidized directly to benzoic acid ($C_6H_5-COOH$).

Step 3: Final Answer:
The product is benzoic acid, corresponding to option (d).