Question

In the oxidation of C\(_6\)H\(_5\)-CH\(_2\)-CH\(_3\) by KMnO\(_4\) the product formed is

• A. C\(_6\)H\(_5\)-CH\(_2\)-CHO
• B. C\(_6\)H\(_5\)-CH\(_2\)-COOH
• C. C\(_6\)H\(_5\)-COOH
• D. C\(_6\)H\(_5\)-CH\(_2\)-OH

Hint:

Alkyl side chain on benzene is oxidized to -COOH by KMnO\(_4\).

Answer 1

The Correct Option is C

The oxidation of ethylbenzene (C\(_6\)H\(_5\)-CH\(_2\)-CH\(_3\)) by potassium permanganate (KMnO\(_4\)) is a common reaction in organic chemistry to convert alkyl side chains into carboxylic acids. Let's go through the solution step by step to understand why the product formed is benzoic acid (C\(_6\)H\(_5\)-COOH).

Understanding the Reaction: Potassium permanganate is a strong oxidizing agent. It oxidizes alkyl side chains attached to aromatic rings to carboxylic acids, regardless of the chain's length. This oxidation typically occurs through the formation of intermediate alcohols or aldehydes, but KMnO\(_4\) directly oxidizes these intermediates to carboxylic acids in the presence of sufficient oxidizing conditions.

Oxidation of Ethyl Group: The ethyl group (CH\(_2\)-CH\(_3\)) attached to the benzene ring is oxidized by KMnO\(_4\) to form a carboxylic acid group (COOH). The reaction can be simplified as:

1. \(\text{C}_6\text{H}_5\text{-CH}_2\text{-CH}_3 \xrightarrow{\text{KMnO}_4} \text{C}_6\text{H}_5\text{-COOH}\)

Justification of Product: Since KMnO\(_4\) is highly efficient at oxidizing the side chain to a carboxylic acid and because of the direct transformation in the presence of an acidic or neutral medium, the product formed is benzoic acid (C\(_6\)H\(_5\)-COOH).

Why Other Options are Incorrect:

• C\(_6\)H\(_5\)-CH\(_2\)-CHO: This compound is an aldehyde and is typically not stable under oxidizing conditions of KMnO\(_4\). In the presence of a strong oxidizing agent, aldehydes are further oxidized to carboxylic acids.
• C\(_6\)H\(_5\)-CH\(_2\)-COOH: This represents an intermediate step (if applicable), but these intermediates are generally unstable and proceed further to form benzoic acid.
• C\(_6\)H\(_5\)-CH\(_2\)-OH: Alcohols can indeed be intermediates but will be oxidized further to carboxylic acids in KMnO\(_4\) conditions.

Conclusion: The complete oxidation of C\(_6\)H\(_5\)-CH\(_2\)-CH\(_3\) by KMnO\(_4\) leads to the formation of the stable product C\(_6\)H\(_5\)-COOH, which is benzoic acid.