Question:
Vigorous oxidation of n-propylbenzene with alkaline potassium permanganate ($\mathrm{KMnO_4/KOH/\Delta}$) followed by hydrolysis gives
Vigorous oxidation of n-propylbenzene with alkaline potassium permanganate ($\mathrm{KMnO_4/KOH/\Delta}$) followed by hydrolysis gives
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Strong oxidation rule:
- Any alkyl benzene $\rightarrow$ benzoic acid (if benzylic H is present)
Updated On: Apr 30, 2026
- phenol
- phenyl acetic acid
- phenyl propionic acid
- benzaldehyde
- benzoic acid
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The Correct Option is
Solution and Explanation
Concept:
Strong oxidizing agents like $\mathrm{KMnO_4}$ oxidize any alkyl side chain attached to a benzene ring to a carboxylic acid group, provided there is at least one benzylic hydrogen.
Step 1: Identify substrate.
n-propylbenzene has a side chain with benzylic hydrogen.
Step 2: Oxidation effect.
Entire side chain is oxidized to $\mathrm{-COOH}$ regardless of length: \[ \mathrm{C_6H_5CH_2CH_2CH_3 \xrightarrow{KMnO_4/KOH,\ \Delta} C_6H_5COOH} \]
Step 3: Product formed.
\[ \text{Benzoic acid} \]
Step 1: Identify substrate.
n-propylbenzene has a side chain with benzylic hydrogen.
Step 2: Oxidation effect.
Entire side chain is oxidized to $\mathrm{-COOH}$ regardless of length: \[ \mathrm{C_6H_5CH_2CH_2CH_3 \xrightarrow{KMnO_4/KOH,\ \Delta} C_6H_5COOH} \]
Step 3: Product formed.
\[ \text{Benzoic acid} \]
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