JEE Main 2022 · 24 Jun · Shift 2 · Chemistry Q76

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ChemistrySingle correct+4 / -1Question 76 of 90

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

2-pentene

proponaldehyde

2-butene

4-methylpent-2-ene

Answer and worked solution

Official answer: C

A. PCC oxidises the secondary alcohol: $\mathrm{CH_3COCH_2CH=CHCH_3}$ (hex-4-en-2-one).

B. The methyl ketone gives the haloform reaction with NaOI: $\mathrm{CH_3CH=CHCH_2COONa}$ (+ $\mathrm{CHI_3}$ ).

C. Soda lime decarboxylates it: $\mathrm{CH_3CH=CHCH_3}$ , 2-butene.

(C)

Topics:Named reactions Carboxylic acids Alcohols

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JEE Main 2022 · 24 Jun · Shift 2, Chemistry question 76