Tests For Unsaturation
A colourless organic liquid decolourises cold dilute alkaline potassium permanganate solution rapidly, and the structural feature responsible for this observation must be identified.
Select the correct option:
Solution
Presence of a carbon-carbon double bond
Cold dilute alkaline potassium permanganate, known as Baeyer's reagent, is a classic test for unsaturation. When an alkene is present, the carbon-carbon double bond is oxidized to a vicinal diol (glycol), and in the process the purple permanganate is reduced to brown manganese dioxide, so the solution loses its pink-purple colour. Therefore rapid decolourisation indicates a carbon-carbon double bond. A compound with only single bonds, an alkane, does not react with Baeyer's reagent and would not decolourise it, so that option is wrong. An aromatic benzene ring is unusually stable due to aromaticity and resists the mild oxidation by cold dilute permanganate, so it would not cause rapid decolourisation, making that option incorrect. A fully saturated cyclic ring such as cyclohexane likewise has no reactive pi bond and does not react, so it does not apply. This is the NCERT-described chemical test distinguishing unsaturated from saturated hydrocarbons. A plausibility check: only systems with reactive pi electrons reduce permanganate quickly, confirming the double bond.
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About This Question
- Subject
- chemistry
- Chapter
- hydrocarbons
- Topic
- tests for unsaturation
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
Presence of a carbon-carbon double bond
Cold dilute alkaline potassium permanganate, known as Baeyer's reagent, is a classic test for unsaturation. When an alkene is present, the carbon-carbon double bond is oxidized to a vicinal diol (glycol), and in the process the purple permanganate is reduced to brown manganese dioxide, so the solution loses its pink-purple colour. Therefore rapid decolourisation indicates a carbon-carbon double bond. A compound with only single bonds, an alkane, does not react with Baeyer's reagent and would not decolourise it, so that option is wrong. An aromatic benzene ring is unusually stable due to aromaticity and resists the mild oxidation by cold dilute permanganate, so it would not cause rapid decolourisation, making that option incorrect. A fully saturated cyclic ring such as cyclohexane likewise has no reactive pi bond and does not react, so it does not apply. This is the NCERT-described chemical test distinguishing unsaturated from saturated hydrocarbons. A plausibility check: only systems with reactive pi electrons reduce permanganate quickly, confirming the double bond.
This easy difficulty chemistry question is from the chapter hydrocarbons, covering the topic of tests for unsaturation. It appeared in the 2025 exam.
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