JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

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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
(C)2-butene✓ Correct
(D)4-methylpent-2-ene

Answer: (C)

Solution

Hex-4-ene-2-ol (

CH X_{3} CH (OH) CH X_{2} CH = CHCH X_{3}

) is oxidized by PCC to hex-4-en-2-one (A). Sodium hypoiodite (iodoform reaction) cleaves the methyl ketone to give pent-3-enoic acid (B). Heating with soda lime (decarboxylation) removes

CO X_{2}

to give but-2-ene (C), which is 2-butene.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option

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