Question:

Predict the product (B) in the following sequence of reactions \[ \text{Ethylbenzene} \xrightarrow{KMnO_4 / KOH} A \xrightarrow{H_3O^+} B \]

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KMnO$_4$ converts any alkyl benzene → benzoic acid (if benzylic H present).
Updated On: May 2, 2026
  • Benzaldehyde
  • Benzophenone
  • Benzene
  • Acetophenone
  • Benzoic acid
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Solution and Explanation

Concept: Oxidation of alkyl side chain in benzene
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Step 1: Understand the reagent

KMnO$_4$ in alkaline medium is a strong oxidizing agent. Key rule:
• Any alkyl side chain attached to benzene (with at least one benzylic H) is oxidized to COOH. ---

Step 2: Identify substrate

Ethylbenzene = C$_6$H$_5$–CH$_2$CH$_3$ It has benzylic hydrogens → oxidation possible. ---

Step 3: Oxidation reaction

\[ C_6H_5-CH_2CH_3 \xrightarrow{KMnO_4/KOH} C_6H_5COOK \] Intermediate A = potassium benzoate ---

Step 4: Acidification

\[ C_6H_5COOK \xrightarrow{H_3O^+} C_6H_5COOH \] Product B = Benzoic acid --- Final Answer: \[ \boxed{\text{Benzoic acid}} \]
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