Question

Which of the following alkenes on oxidation by \(\text{KMnO}_4\) in dil \(\text{H}_2\text{SO}_4\) forms adipic acid ?

• A. Hex-1-ene
• B. Hex-2-ene
• C. Hex-3-ene
• D. Cyclohexene

Hint:

Oxidative cleavage of cyclohexene opens the ring to a six-carbon dicarboxylic acid.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Hot acidic \(\text{KMnO}_4\) cleaves a \(\text{C=C}\) bond completely. Each carbon of the double bond becomes a carboxylic acid group (or a ketone if it was fully substituted).

Step 2: Adipic acid:
Adipic acid is hexanedioic acid, \(\text{HOOC-(CH}_2)_4\text{-COOH}\), with six carbons and two \(-\text{COOH}\) groups at the two ends.

Step 3: Detailed Explanation:
Cyclohexene is a six-carbon ring with one double bond. Cleaving the double bond opens the ring and turns both former alkene carbons into \(-\text{COOH}\), leaving four \(\text{CH}_2\) groups in between:
\[ \text{Cyclohexene} \xrightarrow{\text{KMnO}_4/\text{dil H}_2\text{SO}_4} \text{HOOC-(CH}_2)_4\text{-COOH} \]

Step 4: Why the other options are wrong.
Hex-1-ene (A) breaks into pentanoic acid and \(\text{CO}_2\). Hex-2-ene (B) gives ethanoic acid and butanoic acid. Hex-3-ene (C) gives two molecules of propanoic acid. None of these gives a dicarboxylic acid because the chains are open.

Final Answer:
Cyclohexene gives adipic acid, option (D). \[ \boxed{\text{Cyclohexene (D)}} \]