Isobutyraldehyde on reaction with formaldehyd | vedclass

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(C) $1$. Isobutyraldehyde $((CH_{3})_{2}CHCHO)$ reacts with formaldehyde $(HCHO)$ in the presence of $K_{2}CO_{3}$ (aldol condensation followed by Can

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(C) $1$. Isobutyraldehyde $((CH_{3})_{2}CHCHO)$ reacts with formaldehyde $(HCHO)$ in the presence of $K_{2}CO_{3}$ (aldol condensation followed by Cannizzaro-like step or simply aldol addition) to form compound '$A$',which is $3$-hydroxy-$2,2$-dimethylpropanal $(HO-CH_{2}-C(CH_{3})_{2}-CHO)$.$2$. Compound '$A$' reacts with $KCN$ (cyanohydrin formation) to yield compound '$B$',which is $2,4$-dihydroxy-$3,3$-dimethylbutanenitrile $(HO-CH_{2}-C(CH_{3})_{2}-CH(OH)-CN)$.$3$. Hydrolysis of '$B$' gives the corresponding carboxylic acid,$2,4$-dihydroxy-$3,3$-dimethylbutanoic acid $(HO-CH_{2}-C(CH_{3})_{2}-CH(OH)-COOH)$.$4$. This hydroxy acid undergoes intramolecular cyclization (lactonization) to form a stable lactone,which is $3$-hydroxy-$4,4$-dimethyltetrahydrofuran-$2$-one (as shown in option $C$).

Isobutyraldehyde on reaction with formaldehyde and $K_{2}CO_{3}$ gives compound '$A$.' Compound '$A$' reacts with $KCN$ and yields compound '$B$,' which on hydrolysis gives a stable compound '$C$.' The compound '$C$' is .... .

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