Question

Cyclohexene on oxidation by \(\text{KMnO}_4\) in dilute \(\text{H}_2\text{SO}_4\) produces

• A. benzene-1,4-dicarboxylic acid
• B. pent-2-enoic acid
• C. hexane-1,6-dioic acid
• D. benzene-1,2-dicarboxylic acid

Hint:

Acidified KMnO4 cleaves the C=C bond; in a ring this opens the ring to a dicarboxylic acid.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Hot acidified \(\text{KMnO}_4\) is a strong oxidising agent that breaks the carbon-carbon double bond and converts each carbon of the C=C into a carboxylic acid group (or a ketone if it is fully substituted).

Step 2: Apply to cyclohexene:
Cyclohexene has a double bond between two ring carbons. When it breaks, the ring opens. Each of the two former alkene carbons becomes a -COOH group.

Step 3: Name the product:
The six carbons are now in a straight chain with -COOH at both ends: \(\text{HOOC-(CH}_2)_4\text{-COOH}\). This is hexane-1,6-dioic acid (adipic acid).

Step 4: Why the other options are wrong.
Benzene dicarboxylic acids (A and D) contain a benzene ring, which cannot form from cyclohexene by oxidation. Pent-2-enoic acid (B) has only 5 carbons and still has a double bond, but all 6 carbons must be retained.

Final Answer:
Cyclohexene gives hexane-1,6-dioic acid on oxidation. \[ \boxed{\text{(C) }\text{hexane-1,6-dioic acid}} \]