Question:
Alkenes on oxidation with \(\text{KMnO}_4\) in dil \(\text{H}_2\text{SO}_4\) forms
Alkenes on oxidation with \(\text{KMnO}_4\) in dil \(\text{H}_2\text{SO}_4\) forms
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Acidified KMnO\(_4\) cleaves the C=C bond and oxidises the fragments to carboxylic acids.
Updated On: Oct 1, 2026
- Alkanol
- Alkanal
- Alkanone
- Alkanoic acid
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Acidified KMnO\(_4\) is a strong oxidant. It breaks the double bond completely, and any aldehyde formed is oxidised further.
Step 2: Detailed Explanation
The alkene splits into two fragments at the double bond. A \(=\text{CH}_2\) end becomes CO\(_2\), and a \(=\text{CHR}\) end becomes RCOOH.
Since aldehydes cannot survive the oxidising medium, they go on to the carboxylic acid.
The class of product asked in the question is alkanoic acid.
Final Answer:
Alkenes give carboxylic acids (alkanoic acids), option (D). \[ \boxed{\text{Alkanoic acid (D)}} \]
Acidified KMnO\(_4\) is a strong oxidant. It breaks the double bond completely, and any aldehyde formed is oxidised further.
Step 2: Detailed Explanation
The alkene splits into two fragments at the double bond. A \(=\text{CH}_2\) end becomes CO\(_2\), and a \(=\text{CHR}\) end becomes RCOOH.
Since aldehydes cannot survive the oxidising medium, they go on to the carboxylic acid.
The class of product asked in the question is alkanoic acid.
Final Answer:
Alkenes give carboxylic acids (alkanoic acids), option (D). \[ \boxed{\text{Alkanoic acid (D)}} \]
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