Question

Identify the 'product' in the following reaction.
Pent - 3 - enenitrile \(\rightarrow _{\text{H}_3\text{O}^+\,}^{\text{Al H (i-Bu)}_2\,}\) Product

• A. Pent - 3 - en - 1 - amine
• B. Pentanal
• C. Pent - 3 - enal
• D. Pent - 3 - en -1 - ol

Hint:

DIBAL-H reduces a nitrile to an imine, which hydrolyses to an aldehyde without touching the C=C.

Answer 1

The Correct Option is C

Step 1: Count the carbons:
The nitrile carbon is C1 of the name, so the chain has five carbons: $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{C}\equiv\text{N}$.

Step 2: Follow the reagent:
DIBAL-H delivers one hydride to the $\text{C}\equiv\text{N}$ carbon, making an aluminium imine. Aqueous acid then converts $\text{C=NH}$ to $\text{C=O}$ and releases ammonia.

Step 3: Result:
The $\text{C=C}$ at C3 is unaffected, so the aldehyde keeps it: pent-3-enal.
Over-reduction to the amine or alcohol does not happen with this reagent, and saturating the chain to pentanal would need hydrogenation conditions.

Final Answer:
Option (C), pent-3-enal. \[ \boxed{\text{Pent-3-enal (C)}} \]