Identify product 'A' in the following reaction $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \frac{\mathrm{i})…
Identify product 'A' in the following reaction
$\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \frac{\mathrm{i}) \mathrm{DIBAl}-\mathrm{H}}{\mathrm{ii}) \mathrm{H}_{3} \mathrm{O}^{+}} \longrightarrow \mathrm{A}$
$\mathrm{R}-\mathrm{CONH}_{2}$
$\mathrm{R}-\mathrm{COOH}$
$\mathrm{R}-\mathrm{CH} \mathrm{O}$
$\mathrm{R}-\mathrm{CH}_{2} \mathrm{NH}_{2}$
Solution
To convert a nitrile ( $\mathrm{R}-\mathrm{C} \equiv \mathrm{N}$ ) to an aldehyde ( $\mathrm{R}-\mathrm{CHO}$ ), you can use DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis with H3O+. The reaction steps are:
1. Add DIBAL-H to the nitrile ( $\mathrm{R}-\mathrm{C} \equiv \mathrm{N}$ ).
2. Hydrolyze the intermediate with $\mathrm{H} 3 \mathrm{O}+$ to get the aldehyde ( $\mathrm{R}-\mathrm{CHO}$ ). This process selectively reduces the nitrile to an aldehyde, avoiding further reduction to an amine.