Nitrogen Compounds: Question 2

Worked solution

Step 1: What reaction is occurring

Pentan-3-one is a ketone, $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$. Its carbonyl carbon is electron-poor (δ+), because oxygen is more electronegative than carbon. A trace of $\text{KCN}$ provides a small, catalytic concentration of $\text{CN}^-$ ions, which act as a nucleophile and attack this δ+ carbon in a nucleophilic addition reaction. As the new C–C bond forms, the $\text{C=O}$ pi bond breaks and its electrons move onto the oxygen, giving an alkoxide intermediate that is then protonated (by a further HCN molecule) to restore a neutral -OH group. Overall, one molecule of HCN adds across the C=O bond, giving a hydroxynitrile with both a new -OH group and a new $\text{CN}$ group on the former carbonyl carbon.

Step 2: Building the product

The former carbonyl carbon of pentan-3-one already carries two ethyl groups ($\text{CH}_3\text{CH}_2$– on each side). Addition of HCN adds an -OH and a -CN group to this same carbon, giving:

$\text{CH}_3\text{CH}_2\text{-C(OH)(CN)-CH}_2\text{CH}_3$

This is 2-ethyl-2-hydroxybutanenitrile. Note that it has six carbon atoms in total, the five carbons of pentan-3-one plus the one new carbon contributed by the incoming $\text{CN}^-$ ion.

Step 3: Why the other options are wrong

• A shows simple reduction of the ketone to an alcohol (pentan-3-ol), with no CN group at all, this ignores the whole point of using HCN.
• C shows two CN groups added and no -OH group; only one CN and one OH can add, since HCN adds once across a single C=O bond.
• D shows the CN group replacing a hydrogen atom on a carbon next to the (still intact) carbonyl group, rather than adding across the C=O bond itself. This is not how nucleophilic addition to a carbonyl compound works.

Final answer

B, $\text{CH}_3\text{CH}_2\text{C(OH)(CN)CH}_2\text{CH}_3$, the hydroxynitrile (cyanohydrin) formed by nucleophilic addition of HCN across the carbonyl group of pentan-3-one.