JEE Main 2022 · 28 Jun · Shift 2 · Chemistry Q76

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ChemistrySingle correct+4 / -1Question 76 of 90

Isobutyraldehyde on reaction with formaldehyde and $\mathrm{K_2CO_3}$ gives compound 'A'. Compound 'A' reacts with KCN and yields compound 'B', which on hydrolysis gives a stable compound 'C'. The compound 'C' is

\mathrm{HO{-}CH_2{-}C(CH_3)_2{-}CH(OH){-}COOH}

\mathrm{HO{-}CH_2CH_2{-}CH(CH_3){-}CH(OH){-}COOH}

Open chain HO-CH2CH2-CH(CH3)-CH(OH)-COOH.

γ-Lactone with OH on the carbon next to C=O and two CH3 on the following carbon (pantolactone)

Five-membered lactone: C=O, then CH(OH), then C(CH3)2, then CH2-O closing the ring.

γ-Lactone with two CH3 on the carbon next to C=O and OH on the following carbon

Five-membered lactone: C=O, then C(CH3)2, then CH(OH), then CH2-O closing the ring.

Answer and worked solution

Official answer: C

A. Crossed aldol: $\mathrm{HOCH_2C(CH_3)_2CHO}$ . B. Cyanohydrin. C. Hydrolysis gives $\mathrm{HOCH_2C(CH_3)_2CH(OH)COOH}$ , which closes to the stable γ-lactone (pantolactone).

(C)

Topics:Aldol and related condensations Nucleophilic addition

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JEE Main 2022 · 28 Jun · Shift 2, Chemistry question 76