Question 46

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

Solution

Chemical Mechanism & Solution Profile

Reaction Summary Pathway:
Isobutyraldehyde + HCHO → Compound 'A' (Aldol Product)
Compound 'A' + KCN → Compound 'B' (Cyanohydrin Intermediate)
Compound 'B' + Hydrolysis (H3O+) → Compound 'C' (γ-Lactone / Option C)

1Cross-Aldol Condensation (Formation of 'A')

Isobutyraldehyde, (CH3)2CH-CHO, has exactly one acidic α-hydrogen. In the presence of a mild base (K2CO3), it forms an enolate which selectively attacks the more reactive, non-enolizable formaldehyde (HCHO).

(CH3)2C¯-CHO + H2C=O → HO-CH2-C(CH3)2-CHO Compound 'A' (3-Hydroxy-2,2-dimethylpropanal)

2Cyanohydrin Formation (Formation of 'B')

The aldehyde group of Compound 'A' is targeted by nucleophilic cyanide ions (CN¯) from KCN, producing a cyanohydrin intermediate containing both a nitrile moiety and hydroxyl components.

HO-CH2-C(CH3)2-CHO + CN¯ → HO-CH2-C(CH3)2-CH(OH)-CN Compound 'B'

3Acid Hydrolysis & Lactonization (Formation of 'C')

During aqueous acid treatment, two shifts happen in rapid succession:

The nitrile group (-CN) is completely hydrolyzed into a carboxylic acid group (-COOH).
Because a carboxylic acid and a hydroxyl group (-OH) exist in a 1,4-relationship within the same molecule, they undergo spontaneous intramolecular nucleophilic acyl substitution (esterification), releasing an H2O molecule to form a stable 5-membered cyclic ester ring (a γ-lactone).

HO-CH2-C(CH3)2-CH(OH)-COOH Intramolecular Attack (-H2O) → O O HO H3C H3C Compound 'C' (Option C)

Conclusion Matrix

Because the hydroxyl attachment generated on the original isobutyraldehyde core ends up next to the gem-dimethyl group while the initial primary alcohol site closing the ring system targets the carbonyl terminus, the connectivity perfectly maps out to match the configuration shown in Option C.

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