Identify the 'product' in the following reaction.
Pent - 3 - enenitrile $\xrightarrow[\large{\text{H}_3\text{O}^+}]{\large{\text{Al H (i-Bu)}_2}}$ Product

The given reactant is pent-3-enenitrile, which has the structure $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CN}$.

Diisobutylaluminium hydride (DIBAL-H), represented as $\text{AlH(i-Bu)}_2$, is a selective reducing agent. It reduces nitriles to imines at low temperatures, which upon subsequent acidic hydrolysis ($\text{H}_3\text{O}^+$) yield aldehydes.

DIBAL-H does not reduce isolated carbon-carbon double bonds.

The reaction proceeds as follows:
$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{C}\equiv\text{N} \xrightarrow{\text{AlH(i-Bu)}_2} \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}=\text{N}-\text{Al(i-Bu)}_2$

$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}=\text{N}-\text{Al(i-Bu)}_2 \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CHO}$

The final product is pent-3-enal.

Answer: Pent - 3 - enal