Question
Predict the product when acetaldehyde (CHCHO) reacts with HCN. Show the mechanism and identify the type of reaction.
Solution — Step by Step
HCN adds across the C=O of an aldehyde. This is a nucleophilic addition — the cyanide ion (CN) is the nucleophile attacking the electrophilic carbonyl carbon.
HCN partially dissociates: HCN H + CN. The CN ion attacks the carbonyl C of CHCHO, breaking the -bond and pushing the electrons onto O.
Intermediate: CH-C(CN)(H)-O (alkoxide).
The alkoxide picks up a proton from the medium (or HCN), giving the final product:
CH-CH(OH)(CN) — known as 2-hydroxypropanenitrile (lactonitrile).
The C=O bond is polarized C and O. CN attacks the carbon. After protonation, the OH and CN end up on the same carbon — the original carbonyl carbon.
Final answer: product is CH-CH(OH)(CN), a cyanohydrin. Reaction: nucleophilic addition.
Why This Works
Carbonyl compounds are perfect substrates for nucleophiles because the C=O carbon is electron-poor. Aldehydes are more reactive than ketones because (a) less steric hindrance and (b) only one alkyl group donating electron density to the C.
The cyanohydrin product has a hydroxyl group and a nitrile, both useful for further synthesis — hydrolysing the nitrile to COOH gives an -hydroxy acid (lactic acid in this case).
Alternative Method
Direct memorisation: aldehyde + HCN → cyanohydrin (general formula RCH(OH)CN). Pattern recognition saves you from drawing the mechanism on every question.
Remember the reactivity order for nucleophilic addition: HCHO > RCHO > RCO. Formaldehyde reacts fastest; ketones slowest.
Common Mistake
Drawing the OH and CN on different carbons. Both end up on the original carbonyl carbon (which was the C of C=O). The mechanism doesn’t move the carbon backbone.
Confusing this with the haloform reaction (CHCHO + halogen + base → carboxylic acid). HCN addition gives cyanohydrin; haloform gives a carbon acid. Different reagents → different products.