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JEE Main202224 Jun 2022Evening ShiftChemistryAlcohols Phenols and EthersActual

Hex- 4 -ene- 2 -ol on treatment with PCC gives ' A '. ' A ' on reaction with sodium hypoiodite gives ' B ', which on further heating with soda lime gives ' C '. The compound ' C ' is

Options

  1. A2 -pentene
  2. Bproponaldehyde
  3. C2 -butene
  4. D4 -methylpent- 2 -ene

Correct answer

C. 2 -butene

Step-by-step solution

Hex- 4 -ene- 2 -ol on reaction with PCC forms Hex- 4 -ene- 2 -one. Hex- 4 -ene- 2 -one on iodoform reaction gives pent- 2 enoic acid and iodoform. Pent- 2 enoic acid on decarboxylation with soda lime forms but- 2 -ene.

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