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JEE Main Chemistry question

ChemistryOrganic Oxidation & Reduction ReactionsJEE Main

On ozonolysis, an alkene (X) forms two carbonyl compounds, namely butan-2-one and methanal. The total number of methyl groups present in alkene (X) is .

Answer: 2

Solution

For ozonolysis to cleave cleanly to methanal (H₂C=O) on one side and butan-2-one (CH₃–CO–C₂H₅) on the other, the alkene must be CH₂=C(CH₃)(C₂H₅), i.e., 2-methylbut-1-ene: the terminal =CH₂ gives methanal, and the

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Question ID VK-150155