Isobutyraldehyde on reaction with formaldehyde and K₂CO₃ 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

Chemistry ma’am · 4 steps

Build the chain one step at a time, then let the γ-hydroxy acid close into a five-membered lactone.

1. Crossed aldol. Isobutyraldehyde, (CH₃)₂CH-CHO, has just one α-H. With mild base its carbanion attacks HCHO, which has no α-H and is the more reactive carbonyl. This gives A = HOCH₂-C(CH₃)₂-CHO, and with no α-H left it cannot dehydrate.
2. Cyanohydrin. CN⁻ adds to the CHO carbon of A, giving B = HOCH₂-C(CH₃)₂-CH(OH)-CN.
3. Hydrolysis. Hydrolysis turns -CN into -COOH: HOCH₂-C(CH₃)₂-CH(OH)-COOH. Counting from the COOH carbon, the CH₂OH group sits on the γ-carbon, so this is a γ-hydroxy acid.

   HOCH₂-C(CH₃)₂-CH(OH)-COOH →(-H₂O) five-membered γ-lactone

4. So. The γ-hydroxy acid closes into its stable lactone, C. Going round the ring: ring O, CH₂, C(CH₃)₂, CH(OH), C=O. The OH is on the carbon next to C=O, and the gem-dimethyl carbon is next to that. This is option (c).

The idea to keep. Number the carbons from the COOH. A hydroxy group on the γ or δ carbon closes the acid into a five- or six-membered lactone, and that lactone is the stable product.

Answer from the NTA final answer key. Solution by PrepWiser.