Nitrogen Compounds: Question 6

Worked solution

Step 1: What reduction does to a nitrile

A nitrile group, $-\text{C}{\equiv}\text{N}$, is reduced by an excess of $\text{LiAlH}_4$ in dry ether. Four hydrogen atoms add across the carbon-nitrogen triple bond, converting it fully into a $-\text{CH}_2\text{NH}_2$ group:

$\text{CH}_3\text{CH}_2\text{CN} + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$

(The $[\text{H}]$ represents hydrogen supplied by the reducing agent; dilute acid is added afterwards simply to work up, i.e. destroy, the excess $\text{LiAlH}_4$ and any aluminium complexes.)

Step 2: Counting the carbon atoms

Propanenitrile, $\text{CH}_3\text{CH}_2\text{CN}$, already has three carbon atoms (the nitrile carbon is counted as part of the chain in the name “propanenitrile”). Reduction does not add or remove any carbon atoms, it only adds hydrogen atoms across the triple bond, so the product, propylamine (propan-1-amine), also has three carbon atoms:

$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$

Step 3: Why the other options are wrong

• A shows reduction to an alcohol, as if the nitrogen atom had been lost entirely. LiAlH4 reduces the nitrile’s carbon-nitrogen bond, it does not replace nitrogen with oxygen.
• C shows only partial reduction, stopping at the imine intermediate; with an excess of $\text{LiAlH}_4$, the imine is reduced further, all the way to the saturated primary amine.
• D shows the amide formed by partial hydrolysis of the nitrile (addition of water, catalysed by acid or alkali) rather than reduction (addition of hydrogen), a different reagent and a different reaction altogether.

Final answer

B, $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$, propylamine (propan-1-amine), formed by full reduction of the nitrile group of propanenitrile with excess $\text{LiAlH}_4$.