Alkenes
An alkene X undergoes catalytic hydrogenation (H₂/Ni) to give n-butane. The same alkene X decolorises bromine water and also decolorises cold, dilute KMnO₄ solution. When X is treated with ozone followed by oxidative workup (H₂O₂), it gives only propanoic acid as the organic product. What is the IUPAC name of X?
Select the correct option:
Solution
But-2-ene
Step 1 — Hydrogenation gives n-butane (C₄H₁₀), so X must be a butene (C₄H₈). Step 2 — It decolorises Br₂/H₂O (confirms C=C present) and cold dilute KMnO₄ (confirms alkene, not aromatic). Step 3 — Ozonolysis with oxidative workup converts each carbon of the C=C to a carboxylic acid (if it bears at least one H) or a ketone (if it bears no H). Only propanoic acid (CH₃CH₂COOH) is produced, meaning both fragments are identical 3-carbon units: CH₃CH₂– on each carbon of the double bond. This is consistent with but-2-ene (CH₃–CH=CH–CH₃), where each C of the double bond carries one CH₃ and one H → each gives CH₃CH₂COOH upon oxidative ozonolysis. But-1-ene would give methanal + propanal (different fragments). Therefore X is but-2-ene.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More alkenes Practice Questions
Among but-1-ene, cis-but-2-ene, trans-but-2-ene, and 2-methylpropene, the most thermodynamically sta...
Among but-1-ene, cis-but-2-ene, trans-but-2-ene, and 2-methylpropene, the most thermodynamically sta...
Restricted rotation about a carbon-carbon double bond can give rise to geometrical isomers, and the ...
Restricted rotation about a carbon-carbon double bond can give rise to geometrical isomers, and the ...
Hydrogen bromide is added to propene in the absence of any peroxide, and the major organic product n...
Hydrogen bromide is added to propene in the absence of any peroxide, and the major organic product n...
An unknown symmetrical alkene on reductive ozonolysis yields only propanal as the carbonyl product, ...
An unknown symmetrical alkene on reductive ozonolysis yields only propanal as the carbonyl product, ...
Markovnikov's rule predicts that the addition of HBr to propene (CH₃–CH=CH₂) will predominantly yiel...
Markovnikov's rule predicts that the addition of HBr to propene (CH₃–CH=CH₂) will predominantly yiel...
About This Question
- Subject
- chemistry
- Chapter
- hydrocarbons
- Topic
- alkenes
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
But-2-ene
Step 1 — Hydrogenation gives n-butane (C₄H₁₀), so X must be a butene (C₄H₈). Step 2 — It decolorises Br₂/H₂O (confirms C=C present) and cold dilute KMnO₄ (confirms alkene, not aromatic). Step 3 — Ozonolysis with oxidative workup converts each carbon of the C=C to a carboxylic acid (if it bears at least one H) or a ketone (if it bears no H). Only propanoic acid (CH₃CH₂COOH) is produced, meaning both fragments are identical 3-carbon units: CH₃CH₂– on each carbon of the double bond. This is consistent with but-2-ene (CH₃–CH=CH–CH₃), where each C of the double bond carries one CH₃ and one H → each gives CH₃CH₂COOH upon oxidative ozonolysis. But-1-ene would give methanal + propanal (different fragments). Therefore X is but-2-ene.
This hard difficulty chemistry question is from the chapter hydrocarbons, covering the topic of alkenes. It appeared in the 2025 exam.
Looking for more practice? Explore all chemistry questions or browse hydrocarbons questions on RankGuru.