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
- A2 -pentene
- Bproponaldehyde
- C2 -butene
- 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.