Q16 / 90 · JEE Main 2022 (24 Jun Shift 2) · Chemistry

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

  • A
    2-pentene
  • B
    proponaldehyde
  • Correct answer
    C
    2-butene
  • D
    4-methylpent-2-ene

Solution

Figure

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. 

Hence, the correct answer is option: 'C'.

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