Cyanohydrin Formation

Acetone cyanohydrin remains a major industrial intermediate: rather than transport and handle volatile, extremely toxic anhydrous HCN gas directly, manufacturers generate and consume acetone cyanohydrin in situ (or ship it as a comparatively more manageable liquid) as the entry point to the acetone cyanohydrin (ACH) process for methyl methacrylate, the monomer of Plexiglas/PMMA — millions of tonnes of this cyanohydrin-derived plastic are produced annually. Mandelonitrile and related aromatic cyanohydrins are intermediates toward mandelic acid derivatives used in cosmetic chemical peels and as chiral resolving/building agents in pharmaceutical synthesis. Enantioselective cyanohydrin formation, catalyzed by (R)- and (S)-selective hydroxynitrile lyases (oxynitrilases) isolated from almond, cassava, and other plant sources, is run industrially to make single-enantiomer cyanohydrins that are converted to chiral pyrethroid insecticide side chains (e.g., in the synthesis of cyanopyrethroids like cypermethrin and deltamethrin) and to pharmaceutical building blocks such as chiral α-hydroxy acids and β-blocker precursors — a direct, large-scale descendant of the racemic-vs-enantioselective distinction developed here, and the natural bridge to <a class="wiki-link" href="/wiki/chem-orgo-cyanohydrins-in-synthesis" data-track="wiki_article.body.wiki_link">Cyanohydrins in Synthesis</a>. In biochemistry, the reverse reaction — cyanohydrin lyase-catalyzed release of HCN from cyanogenic glycosides such as amygdalin (found in almond and apricot pits) and linamarin (cassava) — is a plant chemical-defense strategy that depends on exactly the same reversible carbonyl/HCN equilibrium discussed here, now run in the thermodynamically "backward" direction by a dedicated enzyme.

Ask the tutor: How does the choice of chiral catalyst or oxynitrilase enzyme determine which enantiomer of a cyanohydrin is formed, and how is that selectivity used downstream in Cyanohydrins in Synthesis?

Cyanide ion is a small, linear, resonance-delocalized nucleophile, $\text{:C}\equiv\text{N}:^- \leftrightarrow {}^-\!:\!\text{N}\!=\!\text{C}:$, that attacks predominantly through carbon because carbon bears more of the negative charge and its orbital overlaps better with the incoming carbonyl $\pi^*$. When cyanide approaches an aldehyde or ketone carbonyl, its lone pair on carbon aligns antiperiplanar to the developing $\pi$ system exactly along the Bürgi–Dunitz trajectory (nucleophile–carbon–oxygen angle near 107°, not 90° or 180°), because this maximizes overlap between the nucleophile's HOMO and the carbonyl's LUMO ($\pi^*_{C=O}$) while the developing alkoxide lone pair swings into the plane once rehybridization to $sp^3$ is underway. The carbonyl carbon starts $sp^2$, trigonal planar, with the $\pi$ bond's node in the molecular plane and its two lobes above and below. Cyanide can attack from either face with equal probability if the carbonyl is prochiral (i.e., its two substituents R and R′ are different, or R = H), because the molecule has a plane of symmetry (the $sp^2$ plane itself) but no inherent preference for one face over the other. This is the microscopic origin of the racemic outcome: nothing in an achiral cyanohydrin-forming reaction distinguishes the Re face from the Si face, so the product is a 50:50 mixture of both new stereoisomers — unless something chiral (an enzyme, a chiral Lewis acid, a chiral catalyst H-bond donor) is present to make the two transition states diastereomeric in energy. Underlying all of this is a kinetic subtlety that makes cyanohydrin formation a favorite pH-rate case study: the actual nucleophile is CN⁻, not HCN, yet the reaction is normally run with an HCN/KCN buffer near pH 8–10 rather than with pure hydroxide-quenched cyanide at high pH. The rate law reflects a competition between having enough free cyanide ion (requires the medium not be too acidic, since HCN, $pK_a \approx 9.2$, is a weak acid and protonates readily) and having the carbonyl in an electrophilic, unprotonated state (requires the medium not be too basic, though for simple aldehydes/ketones — unlike acetals — over-basicity is not itself deactivating; the true bottleneck is simply cyanide's own protonation equilibrium and the fact that added hydroxide does not accelerate the intrinsically fast addition step further once cyanide is fully liberated).

The overall transformation is the addition of H–CN (delivered as CN⁻ plus a proton source) across the C=O π bond:

where: R, R′ = H, alkyl, or aryl; the product is the cyanohydrin, an α-hydroxynitrile; the double arrow signals genuine, measurable reversibility. The rate law for the productive (cyanide-mediated) pathway is second order overall, first order in carbonyl and first order in free cyanide ion — not in total cyanide added:

where: $k_2$ is the bimolecular rate constant for nucleophilic attack of cyanide on the carbonyl carbon; $[\text{CN}^-]$ is the free, unprotonated cyanide concentration set by the HCN/CN⁻ acid–base equilibrium at the working pH. Because HCN is a weak acid, the free cyanide fraction is fixed by the Henderson–Hasselbalch relationship:

where: $pK_a(\text{HCN}) \approx 9.2$ at 25 °C; this fraction is essentially 0 at pH 4 and essentially 1 at pH 13, with the steep transition centered exactly at pH 9.2. The position of the equilibrium itself — how far to the right the reaction sits — is captured by the standard carbonyl-addition equilibrium constant, exactly as for hydration:

where: a large $K_{eq}$ means the cyanohydrin is thermodynamically favored (typical of aldehydes and unhindered/activated ketones); a small $K_{eq}$ means the free carbonyl dominates at equilibrium (typical of hindered or electron-rich ketones).

Reading the two equations together: rate depends on $[\text{CN}^-]$, which increases monotonically with pH up to a plateau near pH ≈ 12; but $K_{eq}$, an equilibrium property, does not depend on pH at all (it is a thermodynamic ratio of neutral species concentrations). In practice, then, chemists buffer the reaction near pH 8–10: high enough to liberate a useful concentration of CN⁻ so the reaction proceeds at a convenient rate, but not so basic that handling becomes hazardous (very basic solutions of cyanide still contain volatile HCN in equilibrium and are exactly as toxic) or that side reactions (e.g., benzilic-acid-type rearrangements of very electrophilic diketones, or Cannizzaro competition for base-sensitive aldehydes) become competitive.

Mechanism 1 — Base/near-neutral (cyanide-ion) pathway

Goal: show, arrow by arrow, how free CN⁻ converts a ketone or aldehyde into the alkoxide form of the cyanohydrin, and how a subsequent proton transfer delivers the neutral product.

Step 1 (nucleophilic attack). The carbon lone pair of cyanide ion attacks the carbonyl carbon. A curved arrow runs from the C lone pair on :C≡N:⁻ into the carbonyl carbon; a second curved arrow runs from the C=O π bond onto oxygen, generating an alkoxide. The trajectory obeys the Bürgi–Dunitz angle (~107° from the C=O axis), which is the angle that lets the incoming nucleophile's filled orbital overlap cleanly with the empty $\pi^*_{C=O}$ while avoiding a destabilizing four-electron clash with the in-plane oxygen lone pairs. The carbon rehybridizes from $sp^2$ to $sp^3$ as the new $\sigma$ bond forms; the substituents (R, R′, and the developing C–CN and C–O⁻ bonds) pyramidalize away from the approaching nucleophile.

Step 2 (protonation of the alkoxide). The tetrahedral alkoxide is a strong base ($pK_a$ of the conjugate cyanohydrin O–H is roughly 13–14, similar to a simple secondary/tertiary alcohol) and is rapidly protonated by any available Brønsted acid in solution — water, HCN itself, or the conjugate acid of the buffer. A curved arrow runs from an O–H bond of the proton source into the alkoxide oxygen's incipient bond, while the arrow from the acid's O–H (or H–CN) bond breaks to place the electron pair back on that heteroatom.

Notice that Step 2 as written regenerates a cyanide ion, so HCN here is acting catalytically as the proton shuttle — this is exactly why only a catalytic amount of strong base (KOH, KCN, or a trace of Et₃N) is required in practice even though the reaction consumes one equivalent of "HCN" overall: the base's real job is to keep a small standing concentration of free CN⁻ available to attack, while bulk neutral HCN, present in large excess, serves as the terminal proton source.

Mechanism 2 — Acid-catalyzed / direct HCN pathway

Goal: show why, at lower pH, protonation of the carbonyl oxygen can precede nucleophilic attack by (a small amount of) surviving cyanide, and why this pathway is intrinsically slower for cyanohydrin formation than for simple hydration.

Step 1 (carbonyl protonation). Under mildly acidic conditions, the carbonyl oxygen's lone pair is protonated by H₃O⁺ (or by HCN itself acting as a weak Brønsted acid), a curved arrow running from the oxygen lone pair to the incoming proton, generating an oxocarbenium ion whose resonance structure places substantial positive charge on carbon.

Step 2 (nucleophilic attack by residual cyanide). The oxocarbenium carbon is now a far better electrophile, but at low pH the concentration of the actual nucleophile, free CN⁻, has collapsed (recall the Henderson–Hasselbalch curve above pKa 9.2). Whatever CN⁻ remains attacks the protonated carbonyl carbon along the same Bürgi–Dunitz trajectory as before; the π bond electrons shift fully onto the (already protonated) oxygen.

Step 3 (no further step needed — product is already neutral). Unlike Mechanism 1, the alkoxide never appears; the product cyanohydrin is obtained directly because the oxygen was protonated first. The net rate under acid conditions is throttled almost entirely by the vanishing $[\text{CN}^-]$ term (which falls off exponentially below pH 9.2) even though the electrophile is more reactive; this is why laboratory cyanohydrin syntheses are never run in strongly acidic solution — the productive nucleophile is starved. This is the same "nucleophile-limited vs. electrophile-limited" tension seen in hydrate formation, but the effect is far more dramatic here because HCN's $pK_a$ (9.2) sits so much higher than water's (15.7 as $pK_a$ of $\text{H}_3\text{O}^+$/effectively irrelevant) — meaning cyanide, unlike hydroxide/water, is substantially protonated across the entire practical working pH range, and the experimentalist must actively manage this balance rather than simply "add acid to catalyze."

Example 1 — Foundational: acetaldehyde + KCN/HCN buffer, full curved-arrow mechanism and equilibrium check. Acetaldehyde ($\text{CH}_3\text{CHO}$) is treated with KCN buffered by HCN at pH 9.0. Predict the product, sketch the mechanism, and estimate whether the cyanohydrin will be favored at equilibrium given that $K_{eq} \approx 1.3\times10^4\ \text{M}^{-1}$ for this substrate (a typical literature value reflecting an unhindered, weakly activated aldehyde). Step 1. At pH 9.0, using $pK_a(\text{HCN}) = 9.2$:

So about 39% of total cyanide is present as free CN⁻ — plenty to drive the addition at a convenient rate; this is why pH 9 buffers (rather than pH 4 or pH 13) are the standard laboratory choice. Step 2 (mechanism). Free CN⁻ attacks the aldehyde carbonyl carbon along the Bürgi–Dunitz trajectory: the carbon lone pair of cyanide forms a new C–C bond to the aldehyde carbon while the π electrons of C=O shift onto oxygen, generating the alkoxide $\text{CH}_3\text{CH(O}^-\text{)(CN)}$. This alkoxide is then protonated by HCN (regenerating catalytic CN⁻) to give the neutral cyanohydrin, $\text{CH}_3\text{CH(OH)(CN)}$ — lactonitrile. Step 3 (equilibrium). Starting from, say, $[\text{CH}_3\text{CHO}]_0 = 0.50\ \text{M}$ and $[\text{HCN}]_0 = 0.50\ \text{M}$ (chosen deliberately unequal-looking in the small-x approximation sense, though here total concentrations are set equal for a clean 1:1 stoichiometric mixture — let $x$ be the amount converted):

Because $K_{eq}$ is large, expect $x$ close to 0.50 M; let $y = 0.50 - x$ (the small residual free carbonyl):

Solving the quadratic:

So at equilibrium $[\text{CH}_3\text{CHO}] \approx 6.2\times10^{-3}\ \text{M}$, i.e., over 98.8% conversion to lactonitrile. This confirms the qualitative expectation: an unhindered aldehyde with a large $K_{eq}$ goes essentially to completion under a cyanide buffer. Limiting-case check: if instead $K_{eq}$ were vanishingly small (as for a very hindered ketone, see Example 3), the same algebra would give $y \to 0.50$ M, i.e., negligible conversion — confirming the formula correctly interpolates between "goes to completion" and "barely reacts" as $K_{eq}$ spans orders of magnitude.

Example 2 — Applied: cyanohydrin as a masked carboxylic acid (Kiliani-type one-carbon homologation logic). Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$) reacts with NaCN/H₂SO�4 (generating HCN in situ) to give mandelonitrile, $\text{C}_6\text{H}_5\text{CH(OH)(CN)}$, which upon aqueous acid hydrolysis gives mandelic acid, $\text{C}_6\text{H}_5\text{CH(OH)(COOH)}$. Walk through why the cyanohydrin step is chosen over direct nucleophilic addition of, say, a carboxylate equivalent. Step 1 (why CN⁻ and not, e.g., acetate or another carbon nucleophile). Cyanide is small, strongly nucleophilic through carbon (soft nucleophile, good HOMO–LUMO match with the carbonyl's $\pi^*$), and — crucially — the C≡N triple bond survives the addition step intact as a spectator group that can be later unmasked into –COOH, –CONH₂, or –CH₂NH₂. No other convenient one-carbon nucleophile installs a group with this much latent functional-group value in a single step. Step 2 (mechanism, condensed). CN⁻ attacks the aldehyde carbon (Bürgi–Dunitz trajectory, as in Mechanism 1); the alkoxide is protonated to give racemic mandelonitrile — racemic because benzaldehyde's carbonyl carbon is prochiral and achiral CN⁻ attacks both faces equally, exactly as depicted in the racemization figure above. Step 3 (hydrolysis, previewed mechanistically). Aqueous acid (typically hot HCl or H₂SO�4/H₂O) protonates the nitrile nitrogen, water attacks the resulting nitrilium carbon (a nucleophilic addition mechanistically parallel to cyanide's own attack on the original carbonyl — the nitrile is itself a π-electrophile once protonated), and the resulting imidic acid tautomerizes to an amide, which is further hydrolyzed through a second tetrahedral-intermediate addition–elimination sequence to the carboxylic acid. The stereocenter set in the cyanohydrin step (the benzylic C–OH carbon) is untouched by any of this chemistry, since none of the bonds to that carbon are broken; mandelic acid is obtained racemic, carrying forward exactly the stereochemical outcome fixed at the cyanohydrin stage. Numbers check. Mandelonitrile's $K_{eq}$ for cyanohydrin formation from benzaldehyde is smaller than acetaldehyde's (aromatic conjugation stabilizes the free carbonyl and slightly disfavors addition, an electronic effect exactly analogous to why aromatic aldehydes hydrate less than aliphatic ones) — literature values put $K_{eq}$ for benzaldehyde cyanohydrin around $2\times10^2\ \text{M}^{-1}$, roughly two orders of magnitude below acetaldehyde's. Using the same 0.50 M/0.50 M setup:

Equilibrium conversion is $(0.50-0.0476)/0.50 \approx 90.5\%$ — still favorable, but visibly less complete than acetaldehyde's >98.8%, exactly reflecting the smaller $K_{eq}$ and confirming that aromatic conjugation measurably (not just qualitatively) suppresses the addition equilibrium.

Example 3 — Exam/competition level: comparing $K_{eq}$ across a steric/electronic series and predicting a "does it react usefully" verdict. Rank the following four carbonyl compounds by expected cyanohydrin $K_{eq}$, from largest to smallest, and identify which one(s) would be judged synthetically impractical to convert to a cyanohydrin in useful yield: (a) chloral, $\text{Cl}_3\text{CCHO}$; (b) acetaldehyde, $\text{CH}_3\text{CHO}$; (c) acetone, $(\text{CH}_3)_2\text{C=O}$; (d) di-tert-butyl ketone, $(\text{CH}_3)_3\text{C–CO–C(CH}_3)_3$. Step 1 (electronic reasoning). Chloral's three chlorines are strongly electron-withdrawing by induction; they destabilize the developing positive character on the carbonyl carbon less relevantly than they stabilize the developing negative charge on the adjacent alkoxide oxygen in the tetrahedral product, and — as with hydrate formation — the inductive pull lowers the LUMO ($\pi^*_{C=O}$) energy, making the carbonyl carbon a much better electrophile even before any nucleophile arrives. Expect chloral's $K_{eq}$ to be the largest of the four, by a wide margin, mirroring its behavior in hydration (chloral hydrate is a stable, isolable solid). Step 2 (steric reasoning across the aldehyde/ketone divide). Moving from acetaldehyde (one methyl, one H on the carbonyl carbon) to acetone (two methyls) replaces a small hydrogen substituent with a second bulky methyl group in the immediate vicinity of the forming tetrahedral center; the transition state and product both suffer increased $A^{(1,3)}$-type steric strain as the carbon rehybridizes from planar $sp^2$ (120° angles, no eclipsing penalty) to tetrahedral $sp^3$ (109.5°, now four substituents crowding). This is the same steric argument used for hydrate equilibria: ketones have systematically smaller $K_{eq}$ than aldehydes with comparable electronics. Step 3 (the extreme steric case). Di-tert-butyl ketone carries two very bulky tert-butyl groups flanking the carbonyl carbon. The steric compression on going to the tetrahedral cyanohydrin is severe enough that $K_{eq}$ collapses to a very small number — this ketone is a classic example (alongside its notorious resistance to simple hydration and even to reaction with many hydride reagents) of a carbonyl too hindered for productive nucleophilic addition under normal conditions. Step 4 (ranking and verdict).

Practically: chloral, acetaldehyde, and acetone all give isolable, useful cyanohydrin yields (acetone cyanohydrin, in particular, is manufactured on an enormous industrial scale as a stable, storable, transportable source of anhydrous HCN, notably for methacrylate ester production). Di-tert-butyl ketone would be judged synthetically impractical: even with large excess KCN/HCN and long reaction times, the equilibrium concentration of cyanohydrin would remain too small to isolate in useful yield, exactly as predicted by the same steric argument that renders it resistant to hydration.

Cyanohydrin chemistry is inseparable from the career of Emil Fischer, the German chemist whose systematic use of cyanohydrin formation and subsequent hydrolysis/reduction let him lengthen sugar chains one carbon at a time and establish the configurations of the aldoses — the Kiliani–Fischer synthesis, developed jointly with Heinrich Kiliani, who first showed in 1886 that aldoses react with HCN to give cyanohydrins that hydrolyze to aldonic acids. Fischer's own genius lay in recognizing that the pair of diastereomeric cyanohydrins formed at the new stereocenter (since the sugar carbonyl carbon is prochiral but the rest of the molecule already carries defined stereocenters, making face-selectivity a diastereomeric rather than enantiomeric question) could be separated, correlated, and used to build up the entire family of sugar configurations by rigorous, careful degradation and comparison — essentially solving stereochemical assignment problems with 19th-century tools well before X-ray crystallography existed.