Organic Chemistry · Part 7 of 9

Carbonyl & Acyl Mechanisms

Reaction Mechanisms, Part 7 · 9 sections · about 21,209 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

The carbonyl group is the single most important functional group in organic chemistry, and it does three quite different things depending on what is attached to it. It adds nucleophiles. It substitutes them, if there is a leaving group. And it makes the hydrogens next door acidic. This part takes each in turn, and gives ester hydrolysis the full treatment it deserves — eight distinct mechanisms with the Ingold labels, and the evidence that distinguishes them. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 9 sections in Part 7

  • 1Nucleophilic addition to C=O — trajectory, electronics, sterics Free below
  • 2The addition reactions themselves — cyanohydrins, bisulfite, hydride, hemiacetals and acetals
  • 3Nitrogen nucleophiles — imines, oximes, hydrazones, enamines, and the pH–rate profile
  • 4The tetrahedral intermediate and the reactivity order of the acid derivatives
  • 5Ester hydrolysis in full — the Ingold classification and the evidence
  • 6Transesterification, amides, anhydrides and acid chlorides
  • 7Keto–enol tautomerism and the acidity of the α-hydrogen
  • 8α-Halogenation and alkylation
  • 9Aldol and Claisen — one mechanistic idea

Nucleophilic addition to C=O — trajectory, electronics, sterics

Free extract

Section G.1 of Part 7, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

Why the carbonyl carbon is attacked at all, from which direction, and what makes one carbonyl compound more reactive than another.

The polarity, and the orbital behind it

Oxygen is far more electronegative than carbon (Pauling values 3.44 against 2.55), so the C=O bond is strongly polarised. Two descriptions of the same fact are in circulation and you should be fluent in both:

  • Resonance / curly-arrow language. The carbonyl is a hybrid of the neutral form and a charge-separated form C⁺–O⁻. That minor contributor is why the carbon carries δ+ and the oxygen δ−.
  • Orbital language. The π bonding orbital is polarised towards oxygen; therefore the π* antibonding orbital — which is the LUMO — is polarised towards carbon. A nucleophile donating into the LUMO is donating mostly into a lobe sitting on carbon.

The orbital picture is the one that predicts things. It tells you not only where the nucleophile attacks but from what direction, and it explains why a carbonyl reacts with soft, weakly basic nucleophiles far faster than a simple electrostatic argument would allow.

Nucleophilic addition: a nucleophile bonds to the carbonyl carbon, the π electrons move onto oxygen, and the carbon changes from trigonal sp² to tetrahedral sp³. The product is an alkoxide (from an anionic nucleophile) or, after proton transfer, an alcohol.
Mechanism — nucleophilic addition of an anionic nucleophile
  1. Attack. The nucleophile’s lone pair attacks the carbonyl carbon along the Bürgi–Dunitz trajectory. Simultaneously the C=O π electrons move onto oxygen. One arrow from Nu⁻ to C; one arrow from the C=O bond to O.
  2. Tetrahedral alkoxide. The carbon is now sp³, the oxygen carries the negative charge. This species is often the rate-determining transition state’s immediate product and is frequently observable.
  3. Protonation on work-up. Adding dilute acid (or, in a protic solvent, the solvent itself) protonates the alkoxide to give the alcohol.

Notice the order. With an anionic nucleophile you attack first and protonate afterwards — if you protonated first you would simply destroy the nucleophile. With a neutral, weak nucleophile such as water or an alcohol the order is the other way round:

Mechanism — acid-catalysed addition of a neutral nucleophile
  1. Protonate the carbonyl oxygen. The oxygen lone pair takes a proton from the acid, giving a protonated carbonyl — equivalently, a carbon-centred cation stabilised by an oxygen lone pair.
  2. Attack. The neutral nucleophile (H₂O, ROH) attacks the now much more electrophilic carbon.
  3. Deprotonate. A base in solution — the conjugate base of the catalyst, or another molecule of solvent — removes the extra proton, returning the catalyst.
Note: The catalyst is returned in step 3. If your mechanism consumes acid without regenerating it, you have written a reagent, not a catalyst, and an examiner will notice. The same discipline applies to base catalysis throughout this Part.

Where does the nucleophile come from? The Bürgi–Dunitz trajectory

Draw a carbonyl group flat on the page. The nucleophile does not attack in the plane of the paper, and it does not attack straight down at 90° either. It comes in over one face of the carbonyl, tilted away from the oxygen, at an angle of about 107° to the C=O axis.

The Bürgi–Dunitz trajectory90° path: rejectedCORRNunucleophile approaches105–107°Bürgi–Dunitz angleπ* — the LUMO, fat on CC pyramidalises; O bends awayOverlap with π* argues for 90°; repulsion from the filled π bond pushes back.
Nucleophiles do not attack a carbonyl carbon perpendicular to the C=O plane. They come in over the face of the carbonyl at about 107° to the C=O axis — tilted away from the oxygen. The angle is a compromise: overlap with the large π* lobe on carbon pulls the nucleophile towards the perpendicular, while repulsion from the filled π bond and the oxygen lone pairs pushes it back. Everything about steric control of carbonyl addition follows from the fact that the incoming group leans over one face of the C=O plane.

Why 107° and not 90°? Two effects pull in opposite directions.

  1. Orbital overlap wants a perpendicular approach. The π* lobe on carbon points perpendicular to the molecular plane, so head-on donation into it argues for 90°.
  2. Electron–electron repulsion pushes the nucleophile back. At 90° the incoming lone pair sits directly on top of the filled π bond, which is a four-electron repulsion, and it is also uncomfortably close to the oxygen lone pairs. Tilting away from oxygen relieves both.

The observed compromise is roughly 105–110°. It is not far from the tetrahedral angle, which is a useful way to remember it: the nucleophile arrives already pointing at the position it will finally occupy. As the C–Nu bond forms, the carbon pyramidalises and the oxygen swings away from the incoming group; the geometry barely has to change.

The practical payoff is stereochemistry. Because the nucleophile leans over one face of the carbonyl and is tilted away from oxygen, whatever sits on that face — and whatever sits on the side away from oxygen — controls which face is attacked. Every facial-selectivity model you will meet (Cram, Felkin–Anh, axial attack on cyclohexanones) is an argument about which of the two Bürgi–Dunitz trajectories is less obstructed. If you draw the attack perpendicular, all of those models become impossible to reason about.

What makes one carbonyl more reactive than another

Two independent axes control the rate and the position of the equilibrium: electronics (how electron-poor is the carbon?) and sterics (how crowded will the tetrahedral product be?). Fortunately they usually point the same way, which is why the familiar reactivity order works so well.

FactorEffect on additionReason
Alkyl groups on the carbonyl carbonDecrease reactivityAlkyl is weakly electron-donating (hyperconjugation and +I), so it feeds electron density into the δ+ carbon; and it blocks the Bürgi–Dunitz trajectory
Electron-withdrawing groups (Cl, CF₃, NO₂, C=O)Increase reactivity strongly−I pulls density off the carbonyl carbon, and destabilises the sp² starting material more than the sp³ adduct
Conjugating groups (aryl, vinyl, OR, NR₂)Decrease reactivityDonation into the C=O π* raises the LUMO and stabilises the planar starting material; the stabilisation is lost on going to sp³
Bulk (branching, tert-butyl, ortho substituents)Decrease reactivity, sometimes to zeroThe tetrahedral adduct crowds four groups where three sat before; the transition state already feels it
Ring strain that is relieved on going sp² → sp³Increases reactivity dramaticallyA three- or four-membered ring is happier with a 109° carbon than a 120° one
The two columns almost always agree, which is why “aldehydes are more reactive than ketones” is a safe rule — it is being supported by both electronics and sterics at once.

The standard ladder, most reactive first:

H₂C=O > RCHO > ArCHO > R₂C=O > ArCOR > ArCOAr   —   and separately   F₃CCOCF₃ > Cl₃CCHO ≫ anything unsubstituted
Reactivity falls left to right. Methanal has no alkyl group at all; benzophenone has two aryl groups feeding the carbonyl and blocking both faces. Structures are RDKit-rendered from SMILES.

The hydration equilibrium — the cleanest measurement of all this

Dissolve a carbonyl compound in water and some of it adds water across the C=O to give a 1,1-diol, usually called a gem-diol or simply the hydrate. The reaction is a true equilibrium, it is fast, and the equilibrium constant is measurable — which makes it the standard yardstick for carbonyl electrophilicity.

Khyd = [R₂C(OH)₂] / [R₂C=O]   (water in large excess, so [H₂O] is absorbed into the constant)
Carbonyl compoundKhyd (water, ~25 °C)% hydrateWhy
Propanone (acetone) CH₃COCH₃~1.4 × 10⁻³~0.1%Two donating methyls; crowded tetrahedral product
Ethanal (acetaldehyde) CH₃CHO~1.06~50%One methyl only — halfway
Methanal (formaldehyde) H₂C=O~2.3 × 10³>99.9%No alkyl group at all; smallest possible steric demand
Trichloroethanal (chloral) Cl₃CCHO~2.8 × 10⁴>99.99%Three chlorines withdrawing hard
Hexafluoropropanone (CF₃)₂CO~10⁶essentially completeSix fluorines — the extreme case
Cyclopropanonevery largeessentially completeRelief of ring strain on going sp² → sp³
Trends first, numbers second. The span from acetone to hexafluoroacetone is about nine orders of magnitude for a change that touches only the substituents.
Mechanism — hydration — base-catalysed (the fast route above pH 7)
  1. Hydroxide attacks the carbonyl carbon: one arrow HO⁻ → C, one arrow C=O → O.
  2. Tetrahedral alkoxide R₂C(O⁻)OH forms.
  3. Protonation by water gives the neutral hydrate and regenerates hydroxide. Catalytic in HO⁻.
Mechanism — hydration — acid-catalysed (the fast route below pH 4)
  1. Protonate the carbonyl oxygen — the carbon becomes far more electrophilic.
  2. Water attacks the protonated carbonyl, giving a protonated hydrate R₂C(OH)(OH₂⁺).
  3. A second water removes the extra proton, giving the hydrate and regenerating H₃O⁺.
Note: Both routes reach the same equilibrium position. Catalysis changes how fast you get there, never where ‘there’ is. The pH–rate profile for hydration is therefore a V (fast in acid, fast in base, slowest near neutral) — do not confuse it with the bell curve of G.3, which arises from a completely different cause.

Why chloral and formaldehyde are the exceptions

These two get asked about again and again, and they are exceptions for different reasons. Saying “because they are more reactive” earns nothing.

Formaldehyde — a steric and hyperconjugative story. Methanal is the only carbonyl compound with no carbon substituent. It therefore loses nothing on pyramidalisation: there are no alkyl groups to be pushed together, and there are no alkyl groups donating electron density into the π* to stabilise the sp² form in the first place. Both terms favour the hydrate, and formaldehyde in water is essentially entirely methanediol. This is why “formalin” is not a solution of H₂C=O at all, and why formaldehyde solutions deposit polymeric paraformaldehyde on standing.

The two classic exceptions, and what they really are in aqueous solution.

Chloral — a purely electronic story. Three chlorine atoms on the adjacent carbon exert a powerful −I effect. That does two things at once, and it is worth separating them because examiners like the distinction:

  • It destabilises the starting material. The carbonyl carbon already carries δ+; putting a strongly electron-withdrawing group next to it makes an already electron-poor centre worse.
  • It stabilises the product only weakly, but crucially it removes the destabilisation: in the hydrate the carbon is no longer δ+ to anything like the same degree, so the −I penalty largely disappears.

The result is chloral hydrate, Cl₃CCH(OH)₂ — a stable, crystalline, isolable compound, historically the first synthetic sedative-hypnotic and still the textbook proof that a gem-diol can be a real substance rather than a book-keeping device. Hexafluoroacetone hydrate is the same argument taken further.

⚠ Common mistakes & exam traps

  • “gem-diols are unstable and cannot be isolated” is false. Chloral hydrate, ninhydrin and hexafluoroacetone hydrate are bottled solids. The correct statement is that gem-diols of simple aldehydes and ketones lie on the wrong side of an equilibrium.
  • Do not draw the nucleophile attacking perpendicular to C=O. It is not just cosmetically wrong — the entire family of facial-selectivity models depends on the 107° trajectory, and a 90° drawing makes them unanalysable.
  • Do not confuse the V-shaped pH–rate profile of hydration with the bell-shaped profile of imine formation. Hydration is fast at both extremes because acid and base each catalyse it. Imine formation is slow at both extremes because two different steps fail at the two ends (G.3).
  • A hydrate is not an oxidation product. Adding water across C=O changes no oxidation state. Students who write chloral hydrate as an oxidation of chloral have confused addition with redox.

Place these four in order of increasing Khyd, and justify each comparison: benzaldehyde, propanone, ethanal, 2,2,2-trichloroethanal. Medium

Propanone is the smallest. Two methyl groups donate into the carbonyl and crowd the tetrahedral carbon. K ≈ 10⁻³.
Benzaldehyde next? Careful — benzaldehyde is an aldehyde, so sterically it beats propanone, but the phenyl ring is conjugated to the C=O and that conjugation is lost on hydration. In practice benzaldehyde hydrates very little; it sits close to propanone and below ethanal. Ordering benzaldehyde below ethanal is the point of the question.
Ethanal. One methyl only, no conjugation. K ≈ 1, i.e. about half hydrate at equilibrium — the classic ‘middle’ case.
Trichloroethanal is by far the largest. Three −I chlorines, and no conjugation to lose. K ≈ 10⁴.
Answer: propanone < benzaldehyde < ethanal < trichloroethanal. State the reason in each comparison — steric vs conjugative vs inductive — because a bare ordering earns half the marks at best.
Advanced / reference layer

Where the 107° number actually comes from, how far it can be trusted, the thermodynamic decomposition of Khyd, and the cases where the simple picture fails.

Structure correlation: the experimental origin of the trajectory

The Bürgi–Dunitz angle was not calculated and it was not measured on a reacting system. It came from crystal structures of stable molecules, by an argument that is one of the most elegant in physical organic chemistry and is worth understanding rather than quoting.

Dunitz and Bürgi collected structures of amino-ketones in which a nitrogen lone pair sits at various fixed distances from a carbonyl carbon, held there by the molecular framework. Where the framework holds the nitrogen far away, the C=O is normal and planar. Where the framework forces the nitrogen closer, two things happen together: the C⋅⋅⋅N distance shortens, the carbonyl carbon pyramidalises, and the C=O bond lengthens. Plotting one against the other traces out a smooth path.

The structure-correlation principle: a set of static structures scattered along a deformation coordinate maps out the reaction path for that deformation. Nature samples the low-energy valley; a family of crystal structures is a series of snapshots along its floor.

The angle at which the approaching nitrogen sits, measured throughout the series, clusters near 105° and is remarkably insensitive to how far along the path the structure has travelled. That constancy is the real result: the trajectory is a line at fixed angle, not a curve that swings round as the bond forms.

Note: 105° or 107°? Both numbers are in circulation, and they come from two different pieces of work. The structure-correlation analysis of amino-ketone crystal structures — the experiment described above — gives 105° ± 5° (Bürgi, Dunitz and Shefter, J. Am. Chem. Soc. 1973, 95, 5065; Bürgi, Dunitz, Lehn and Wipff, Tetrahedron 1974, 30, 1563). The equally often quoted 107° comes from the calculated approach of hydride to a carbonyl carbon. Quote either, but do not attach the ±5° error bar to 107°: the error bar belongs to the crystallographic number. ‘About 105–107°’ is the safe form of words in an answer.

How reliable is the number?

Reliable as a teaching number, softer as a physical constant. Three qualifications matter at postgraduate level:

  1. The angle depends on the nucleophile. Small, hard nucleophiles (hydride) come in closer to the perpendicular; large, polarisable ones sit further back. Computed transition states span roughly 95–110° depending on the nucleophile and the level of theory.
  2. It depends on how early the transition state is. A highly reactive nucleophile attacking an unhindered aldehyde has a very early, reactant-like transition state in which the geometry barely deviates from planar; the angle is then poorly defined because the interaction is weak.

Decomposing Khyd: why addition is entropically expensive

Hydration takes two molecules and makes one. That costs entropy: a typical bimolecular association in solution carries ΔS° of order −120 to −170 J K⁻¹ mol⁻¹, which at 298 K is a −TΔS° penalty of roughly +35 to +50 kJ mol⁻¹. Any addition that proceeds at all must pay for that out of the enthalpy of the new σ bond.

ΔG° = ΔH° − TΔS°   and   ΔG° = −RT ln Khyd

This has three consequences that are examinable in disguise:

  • Intramolecular additions are enormously favoured relative to intermolecular ones, because the entropy has already been paid. This is why sugars exist almost entirely as cyclic hemiacetals (G.2) while acyclic hemiacetals of simple aldehydes barely form.
  • Cyclic acetals from diols are far more favourable than two separate alcohols — two molecules become two rather than three becoming two. That is the whole reason ethylene glycol is the standard carbonyl protecting reagent, and it is also why raising the temperature reverses additions.

Quantifying the substituent effect

The Taft equation is the aliphatic cousin of the Hammett treatment of Part 5. It separates a polar term from a steric term:

log(k / k₀) = ρ*σ* + δEs

For carbonyl hydration ρ* is large and positive — electron withdrawal accelerates — and δ is negative, confirming that bulk retards. It is the two columns of the table above, quantified.

Strain, and where the electronic model breaks down

Two families of compound hydrate far more than any electronic argument predicts, and both are worth knowing because they are the standard ‘explain this anomaly’ questions.

Small rings. An sp² carbon in a three-membered ring is forced to accept an internal angle of 60° when it would like 120°. Rehybridising to sp³ drops the preferred angle to 109.5°, which relieves a great deal of that strain. Cyclopropanone is therefore almost entirely hydrated in water, and cyclobutanone is substantially so. The same effect — usually called I-strain, internal strain — explains why cyclopentanone is more reactive towards addition than cyclohexanone, and why cyclohexanone is the least reactive of the common cyclic ketones.

Strain relief and flanking carbonyls both stabilise the sp³ adduct. Ninhydrin, the amino-acid stain, is supplied and used as the hydrate; its two flanking C=O groups destabilise the central carbonyl so severely that the gem-diol wins.

α-Dicarbonyls. Put two carbonyl groups next to each other and each destabilises the other — two adjacent δ+ carbons, and two competing demands on the same σ framework. Hydrating one of them removes the clash. Ninhydrin is the textbook case; glyoxal and pyruvic acid behave the same way.

Easy
Which member of each pair is attacked faster by a nucleophile, and why? (a) propanal or propanone; (b) ethanal or 2,2,2-trifluoroethanal; (c) benzaldehyde or benzophenone.
Show solution
(a) Propanal — one alkyl group instead of two, so less electron donation into the carbonyl and less crowding in the tetrahedral product. (b) 2,2,2-Trifluoroethanal — three fluorines withdraw electron density inductively, making the carbonyl carbon much more δ+. (c) Benzaldehyde — benzophenone has two aryl groups, each donating into the C=O by conjugation and each blocking a face. The rule: aldehyde > ketone, with electron withdrawal helping and donation or bulk hindering.
Med
Hexafluoroacetone forms a stable, isolable hydrate; acetone does not. Give two distinct reasons, one thermodynamic in origin and one steric.
Show solution
Electronic (thermodynamic). Six fluorines exert a very strong −I effect. This destabilises the carbonyl form, where the carbon is already δ+, far more than it destabilises the sp³ hydrate, where the carbon carries two oxygens and much less positive character. The equilibrium therefore shifts massively towards the hydrate — Khyd of order 10⁶. Steric/hyperconjugative. A CF₃ group is not a π-donor and is a very poor hyperconjugative donor, so unlike a methyl it gives the carbonyl form no stabilisation at all. Acetone, by contrast, has two methyls that do stabilise the sp² carbon and that crowd each other in the sp³ adduct. Note the structure of the answer: in every Khyd comparison you must say what happens to both sides of the equilibrium.
Med
Predict, with reasoning, whether the pH–rate profile for the hydration of acetaldehyde is bell-shaped, V-shaped, or flat.
Show solution
V-shaped. Hydration is catalysed independently by acid (protonating the carbonyl oxygen to activate the carbon) and by base (delivering the far better nucleophile HO⁻ instead of H₂O). Both catalysed pathways are fast; the uncatalysed water reaction in the middle is slow. So the rate is high at low pH, falls to a minimum around neutrality, and rises again at high pH — a V. Contrast G.3: imine formation is bell-shaped because there acid is needed for one step but destroys the reagent needed for another, so both extremes are bad. The habit worth forming is to ask, for each limb, which step is rate-determining there.

Sources for G.1 — all primary citations to be confirmed by an expert before publication.

  • Clayden, Greeves & Warren, Organic Chemistry, chapters on nucleophilic addition to the carbonyl group — the standard undergraduate treatment of the trajectory and of hydration equilibria.
  • Anslyn & Dougherty, Modern Physical Organic Chemistry — structure correlation, the Taft treatment, and the thermodynamics of addition.
  • H. B. Bürgi, J. D. Dunitz and E. Shefter, ‘Geometrical reaction coordinates. II. Nucleophilic addition to a carbonyl group’, J. Am. Chem. Soc. 1973, 95, 5065–5067, and H. B. Bürgi, J. D. Dunitz, J.-M. Lehn and G. Wipff, ‘Stereochemistry of reaction paths at carbonyl centres’, Tetrahedron 1974, 30, 1563–1572 — the structure-correlation origin of the trajectory.

Read the rest of Part 7

The remaining 8 sections of this part — The addition reactions themselves — cyanohydrins, bisulfite, hydride, hemiacetals and acetals, Nitrogen nucleophiles — imines, oximes, hydrazones, enamines, and the pH–rate profile, The… — and all nine parts of Reaction Mechanisms are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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