Consider the following reaction sequence:
The product 'B' is :

Solution

Chemistry ma’am · 4 steps

DIBAL-H stops the nitrile at the aldehyde, and that aldehyde then does a crossed aldol with CH₃CHO.

1. Find A. AlH(i-Bu)₂ (DIBAL-H) adds only one hydride to -C≡N, giving an imine–aluminium complex. Water hydrolyses it to -CHO. So A is 4-aminobenzaldehyde, H₂N-C₆H₄-CHO, and the -NH₂ group is untouched.
2. Who makes the enolate. A has no α-hydrogen, so it cannot form an enolate. Only CH₃CHO can. Dilute NaOH turns it into ⁻CH₂CHO, which attacks the carbonyl carbon of A. This crossed aldol is the Claisen–Schmidt reaction.

   H₂N-C₆H₄-CHO + CH₃CHO →(dil NaOH) H₂N-C₆H₄-CH(OH)-CH₂CHO →(Δ, -H₂O) H₂N-C₆H₄-CH=CH-CHO

3. Heat dehydrates. Heating removes water from the β-hydroxy aldehyde. The new C=C is conjugated with both the ring and the C=O, so the dehydrated product is strongly favoured.
4. So. B is 4-aminocinnamaldehyde, H₂N-C₆H₄-CH=CH-CHO, which is option (b).

The idea to keep. DIBAL-H takes nitriles and esters only to the aldehyde. Then an aldehyde with no α-H plus one that has an α-H, in dil NaOH with heat, gives the conjugated -CH=CH-CHO product.

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