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Normally, an aldehyde carbon is the target of an attack: oxygen pulls electron density away, leaving it partly positive. The benzoin condensation flips this rule for a moment. A cyanide ion turns that carbon into the attacker, and two benzaldehyde molecules join into one bigger molecule, benzoin.
The reaction is old. Friedrich Wöhler and Justus Liebig first reported the condensation in their 1832 study of the benzoyl radical, Untersuchungen über das Radikal der Benzoesäure; according to Wikipedia's history of the reaction, the catalytic cyanide version was developed by Nikolay Zinin in the late 1830s. The mechanism you learn today came 71 years after the first report, from Arthur Lapworth's 1903 paper on reactions involving the addition of hydrogen cyanide to carbon compounds.
What happens in the benzoin condensation, step by step?
According to the Organic Chemistry Portal, adding cyanide to make a cyanohydrin reverses the usual charge preference of the carbonyl, and the aldehyde carbon becomes nucleophilic once it is deprotonated. Here is the sequence in four moves:
Cyanide adds. CN⁻ attacks the carbonyl carbon of the first benzaldehyde, giving an alkoxide, Ph–CH(O⁻)–CN.
A proton moves. The H on that carbon is now acidic, because the nitrile next to it and the benzene ring both spread out a negative charge. The proton shifts to oxygen, leaving Ph–C⁻(OH)–CN. This carbanion is the key species: the old aldehyde carbon is now electron-rich.
The carbanion attacks. It adds to the carbonyl carbon of a second benzaldehyde, forming a new C–C bond.
Cyanide leaves. After another proton transfer, the first carbon becomes a C=O again and pushes CN⁻ out. The product is benzoin, Ph–CO–CH(OH)–Ph.
PubChem lists benzoin as 2-hydroxy-1,2-diphenylethanone, formula C14H12O2. It is an α-hydroxy ketone: a C=O and an –OH on neighbouring carbons.
Why does cyanide work, and not just any base?
Cyanide has to do three jobs, and few small ions can do all of them:
A good nucleophile, so it adds quickly to the carbonyl in step 1.
A good electron sink, so the nitrile group stabilises the carbanion in step 2.
A good leaving group, so it can be pushed out in step 4 and start the cycle again.
A hydroxide ion can add to a carbonyl, but it cannot stabilise a carbanion on the same carbon the way a nitrile does. That is why the same benzaldehyde with concentrated alkali goes a different way, as you can see in our post on the crossed Cannizzaro reaction.
Cyanide is not the only option. The Organic Chemistry Portal notes that a thiazolium salt can also catalyse the reaction. Thiamine (vitamin B1) has a thiazolium ring, which is how living cells carry out similar acyloin chemistry. Ronald Breslow studied how thiamine does this in his 1958 paper On the Mechanism of Thiamine Action IV.
Be careful in the lab: cyanide is highly toxic. The US CDC fact sheet states that cyanide prevents the body's cells from getting oxygen, and lists sodium cyanide and potassium cyanide among its forms. Learn it on paper.
How do you check the answer with a mass count?
PubChem gives the molar mass of benzaldehyde, C7H6O, as 106.12 g/mol and of benzoin, C14H12O2, as 212.24 g/mol.
Two benzaldehydes: 2 × C7H6O = C14H12O2.
Two molar masses: 2 × 106.12 = 212.24 g/mol, exactly benzoin.
So no small molecule is lost, which tells you this "condensation" loses no water, unlike an aldol condensation. Every atom of both aldehydes ends up in the product.
How do you spot it in a question, and what are the traps?
Use this checklist when you see an aldehyde and a reagent:
Reagent: KCN or NaCN, usually in aqueous ethanol, with only a small amount of cyanide. Think benzoin.
Substrate: an aromatic aldehyde with no α-hydrogen, such as benzaldehyde. With α-hydrogens, the base-driven aldol condensation competes.
Product: an α-hydroxy ketone, with the two aryl groups on adjacent carbons.
The common mistakes:
Writing a cyanohydrin as the final answer. The cyanohydrin anion is only step 1; with benzaldehyde and catalytic cyanide, the reaction continues. Our post on why a crowded ketone resists cyanohydrin formation looks at cyanohydrin formation on its own.
Putting CN in the product. It leaves in step 4.
Mixing it up with Cannizzaro. Concentrated NaOH gives an alcohol and a carboxylate salt; cyanide gives one joined molecule.
Drawing the curved arrow from the aldehyde carbon in step 3 without first showing the proton shift in step 2. That shift is the step where the carbon becomes nucleophilic, so do not skip it.
If mechanisms like this still feel like strings of arrows, look up the idea of nucleophiles and electrophiles in the Fermi Concept Library, then try writing out a few named reactions in the Fermi Problem Library, where a step-by-step solver explains the reasoning rather than just giving you the product.
