Q43 / 150 · MHT CET 2026 (15 April Shift 1) · Chemistry
Identify the compound formed when But-2-ene is treated with $\text{KMnO}_4$ in dilute $\text{H}_2\text{SO}_4$.
- AButanoic acid
- Correct answerBAcetic acid
- CButanal
- DButanol
Solution
But-2-ene has the structure $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$.
When treated with acidic $\text{KMnO}_4$ ($\text{KMnO}_4$ in dilute $\text{H}_2\text{SO}_4$), it undergoes oxidative cleavage. The strong oxidizing agent completely breaks the carbon-carbon double bond, oxidizing each of the resulting fragments into a carboxylic acid.
$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \xrightarrow{\text{KMnO}_4, \text{H}^+} 2 \text{CH}_3\text{COOH}$
The product formed is acetic acid (ethanoic acid).
Answer: Acetic acid
When treated with acidic $\text{KMnO}_4$ ($\text{KMnO}_4$ in dilute $\text{H}_2\text{SO}_4$), it undergoes oxidative cleavage. The strong oxidizing agent completely breaks the carbon-carbon double bond, oxidizing each of the resulting fragments into a carboxylic acid.
$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \xrightarrow{\text{KMnO}_4, \text{H}^+} 2 \text{CH}_3\text{COOH}$
The product formed is acetic acid (ethanoic acid).
Answer: Acetic acid
Hence, the correct answer is option: 'B'.