Carbonyl Chemistry — The Five Nucleophilic Addition Patterns
Almost every reaction of a carbonyl compound is one of five patterns wearing a different costume. Once you can name the pattern, you can predict the product of a reaction you have never seen — which is exactly what CSIR-NET and GATE questions are designed to test. This article sets out the electronic reason the C=O group is attacked at carbon, the geometry of that attack, the five patterns, and the quantitative arguments (equilibrium constants, real stoichiometry) that separate a good answer from a vague one.
Why the carbonyl carbon is electrophilic
Two effects point the same way. The σ and π bonds are both polarised towards the more electronegative oxygen, giving a permanent δ+ on carbon; and, more usefully for predicting reactivity, the LUMO is the π*C=O orbital, whose larger coefficient sits on carbon. A nucleophile's HOMO overlaps best where that coefficient is biggest. Both the electrostatic and the orbital argument therefore agree: the nucleophile attacks carbon, and the electrons of the π bond are pushed onto oxygen to give an alkoxide.
sp2, trigonal planar carbon → sp3, tetrahedral carbon. Every pattern below starts with this step.
The nucleophile does not approach perpendicular to the C=O bond, and it does not approach along it. It comes in at the Bürgi–Dunitz angle of roughly 107° to the C=O axis, on the side away from the oxygen, because that trajectory maximises overlap with the π* lobe while minimising repulsion from the filled π orbital. This angle matters: it is why a substituent that blocks one face of the carbonyl controls which diastereomer forms.
Reactivity order — and the reason for it
| Compound | Relative electrophilicity | Why |
|---|---|---|
| HCHO | Highest of the simple carbonyls | No alkyl group to donate electron density; no steric shield |
| RCHO (aldehyde) | High | One donating alkyl group, one small H |
| R2CO (ketone) | Lower | Two donating alkyl groups reduce δ+; two groups crowd the trajectory |
| RCOCl > (RCO)2O > RCO2R′ > RCONR′2 | Decreasing along the row | Lone-pair donation from the attached heteroatom into π* — strongest from N, weakest from Cl — plus leaving-group ability |
| ArCHO, ArCOR | Reduced vs the aliphatic analogue | The ring conjugates with C=O and delocalises the positive charge |
| CF3COR, CCl3CHO | Greatly increased | Strong inductive withdrawal raises δ+ and destabilises the starting carbonyl relative to the tetrahedral adduct |
The hydration equilibrium — a reactivity scale you can calculate with
Water adds reversibly to a carbonyl to give a gem-diol (hydrate). Because the equilibrium constant is measurable, hydration is the cleanest quantitative measure of carbonyl electrophilicity available.
Fraction present as the hydrate = Khyd / (1 + Khyd)
Worked example 1 — converting Khyd into a percentage.
Using the values usually tabulated for dilute aqueous solution:
Formaldehyde, Khyd ≈ 2.28 × 103:
fraction = 2280 ⁄ (1 + 2280) = 2280 ⁄ 2281 = 0.99956 → 99.96% hydrate.
"Formalin" is essentially a solution of the gem-diol and its oligomers, not of
HCHO.
Acetaldehyde, Khyd ≈ 1.06:
fraction = 1.06 ⁄ 2.06 = 0.5146 → 51.5% hydrate — very nearly a 50:50
mixture.
Acetone, Khyd ≈ 2.0 × 10−3:
fraction = 0.0020 ⁄ 1.0020 = 0.001996 → 0.20% hydrate. Acetone in water is,
for practical purposes, still acetone.
Two alkyl groups instead of none change the position of the same equilibrium by a factor of
about 106. (Exact K values differ slightly between textbooks and with temperature
and ionic strength — quote them as approximate.)
Worked example 2 — turning that ratio into an energy.
Khyd(HCHO) ⁄ Khyd(acetone) = 2280 ⁄ 0.0020 = 1.14 × 106.
The corresponding free-energy difference at 298 K is
ΔΔG° = −RT ln(ratio) with R = 8.314 J K−1 mol−1:
RT = 8.314 × 298 = 2477.6 J mol−1
ln(1.14 × 106) = ln 1.14 + ln 106 = 0.1310 + 13.8155 = 13.9465
ΔΔG° = 2477.6 × 13.9465 = 34 555 J mol−1 = 34.6 kJ mol−1.
So the two methyl groups of acetone are worth roughly 35 kJ mol−1 against
hydration — a genuinely large number, and a much better answer than "acetone is less
reactive".
Pattern 1 — reversible addition (weak, neutral or stabilised nucleophiles)
Water, alcohols, thiols, cyanide and bisulfite all add reversibly. The adduct is close in energy to the starting materials, so the position of equilibrium — not the rate — decides what you isolate.
- Alcohols → hemiacetal → acetal. The first addition needs only acid or base catalysis; going on to the acetal requires acid, because the hemiacetal OH must be protonated and lost as water to give an oxocarbenium ion, which the second alcohol then traps. Acetals are stable to base and to nucleophiles but hydrolyse in aqueous acid, which is precisely why they are the standard protecting group for aldehydes and ketones.
- HCN → cyanohydrin. The nucleophile is CN−, not HCN, so a little base is needed; too much acid shuts the reaction down by removing the cyanide.
- Bisulfite adducts are crystalline and water-soluble, and the reaction reverses on treatment with acid or base — a classical purification trick that works well for aldehydes and methyl ketones and poorly for hindered ketones.
Pattern 2 — irreversible addition (strong carbanion and hydride nucleophiles)
Grignard reagents, organolithiums, acetylides and metal hydrides give an alkoxide that has no low-energy way back. The reaction goes to completion, so rate and chemoselectivity matter instead of equilibrium.
| Reagent | Reduces aldehydes/ketones? | Reduces esters? | Reduces carboxylic acids / amides? | Practical note |
|---|---|---|---|---|
| NaBH4 | Yes | No (very slow) | No | Works in methanol or ethanol; the standard chemoselective choice |
| LiAlH4 | Yes | Yes → primary alcohol | Yes | Must be used in dry ether or THF; reacts violently with water |
| DIBAL-H, 1 equiv, −78 °C | Yes | Yes, but stops at the aldehyde | Nitrile → aldehyde | Low temperature is what traps the tetrahedral intermediate |
| RMgX / RLi | Yes → secondary or tertiary alcohol | Yes, twice → tertiary alcohol | Deprotonate the acid instead | Basic as well as nucleophilic; no acidic O–H or N–H may be present |
Worked example 3 — real stoichiometry of a borohydride reduction.
How much NaBH4 is needed, in principle, to reduce 0.10 mol of acetophenone?
Each BH4− can in principle deliver all four hydrides,
so moles of NaBH4 = 0.10 ⁄ 4 = 0.025 mol.
Molar mass of NaBH4 = Na 22.990 + B 10.81 + 4 × H 1.008
= 22.990 + 10.81 + 4.032 = 37.832 g mol−1
Mass required = 0.025 × 37.832 = 0.9458 g ≈ 0.95 g.
In practice a laboratory procedure uses an excess — often one full equivalent of
NaBH4 per carbonyl — because the protic solvent consumes hydride as well. State the
theoretical figure and then say why the real quantity is larger; an examiner is looking for
both halves.
Pattern 3 — addition then elimination of water (nitrogen nucleophiles)
Primary amines, hydroxylamine, hydrazines and semicarbazide add to give a carbinolamine, which then loses water to form a C=N compound: imine, oxime, hydrazone, semicarbazone. The rate–pH profile is bell-shaped with a maximum typically around pH 4–5, and the reason is worth memorising because it is a favourite short-answer question:
- At high pH there is not enough acid to protonate the carbinolamine OH, so the dehydration step is slow.
- At low pH the amine is protonated to RNH3+, which has no lone pair, so the addition step has no nucleophile.
- The optimum is the pH that leaves enough free amine and enough acid — mildly acidic conditions.
Secondary amines cannot form a neutral imine (no N–H left to lose), so with an α-hydrogen-bearing carbonyl they give an enamine instead.
Pattern 4 — addition–elimination at carboxylic acid derivatives
Here the tetrahedral intermediate has a leaving group to expel, so the carbonyl is regenerated and the net result is substitution at the acyl carbon rather than addition. The reactivity order acyl chloride > anhydride > ester > amide follows both the leaving-group ability (Cl− good, R2N− hopeless) and the degree of lone-pair donation into π* (weak from chlorine, strong from nitrogen). The practical consequence: you can always convert a more reactive derivative into a less reactive one, and never the other way round without activating the substrate first.
Pattern 5 — conjugate (1,4) versus direct (1,2) addition
An α,β-unsaturated carbonyl has two electrophilic sites: the carbonyl carbon (1,2 attack) and the β carbon (1,4 or conjugate attack). Which one wins is a textbook illustration of hard and soft reagent behaviour.
| Nucleophile | Character | Preferred site | Product from cyclohex-2-enone |
|---|---|---|---|
| CH3Li, RMgX | Hard, charge-controlled | 1,2 | 1-methylcyclohex-2-en-1-ol |
| (CH3)2CuLi (Gilman) | Soft, orbital-controlled | 1,4 | 3-methylcyclohexan-1-one |
| CN−, RS−, R2NH, stabilised enolates | Soft | 1,4 | β-substituted ketone (Michael addition) |
| NaBH4 alone | Mixed — often gives substantial 1,4 | 1,4 and 1,2 | Mixture; poor synthetic control |
| NaBH4 + CeCl3 (Luche) | Hardened by the lanthanide | 1,2 | Cyclohex-2-en-1-ol (allylic alcohol) |
The pattern generalises: hard, highly basic, strongly charged nucleophiles go for the site with the largest positive charge (the carbonyl carbon); soft, polarisable nucleophiles go for the site with the largest LUMO coefficient (the β carbon). Note also that 1,2-addition is often the faster reaction while 1,4-addition gives the more stable product, so the outcome can also be shifted by making the first step reversible.
Facial selectivity — the Felkin–Anh model
When the carbon next to the carbonyl is already a stereocentre, the two faces of the C=O group are no longer equivalent and one diastereomer dominates. The Felkin–Anh analysis puts the largest α substituent perpendicular to the C=O plane and anti to the incoming nucleophile, then brings the nucleophile in past the smallest group along the Bürgi–Dunitz trajectory. If the α substituent is an electronegative atom, it — rather than the bulkiest group — takes the anti position, because that arrangement best stabilises the developing negative charge. This is the standard route to predicting the major product of a Grignard or hydride addition to a chiral aldehyde.
Mistakes that cost marks
- Drawing the nucleophile attacking oxygen. Oxygen is the electron-rich end. It is protonated or coordinated to a Lewis acid — never attacked by a nucleophile.
- Explaining aldehyde > ketone reactivity by sterics alone. Both the electronic effect (alkyl donation reducing δ+) and the steric effect operate, and a full-mark answer names both.
- Saying acetals form under basic conditions. A hemiacetal will form with acid or base, but the second step requires an oxocarbenium ion and therefore acid. This is exactly why acetals survive Grignard reagents and LiAlH4.
- Getting the imine pH argument only half right. "Acid catalyses it" is incomplete — you must also say that too much acid destroys the nucleophile. The bell shape is the answer.
- Using LiAlH4 in methanol. It destroys the reagent instantly. NaBH4 tolerates alcohols; LiAlH4 needs dry ether or THF.
- Assuming a Grignard reagent gives 1,4-addition on an enone. Organolithiums and Grignards are hard nucleophiles and normally attack 1,2. Use a cuprate for conjugate addition.
- Forgetting that NaBH4 supplies four hydrides. Dividing by four is the difference between a correct stoichiometric answer and a fourfold error.
- Confusing addition with the α-carbon chemistry. Aldol, halogenation and alkylation proceed through the enol or enolate, not by attack at the carbonyl carbon. Read the question and decide which side of the molecule is reacting.
Where this appears in the exam
| Exam | Typical demand |
|---|---|
| CSIR-NET Chemical Sciences | Predict the product and stereochemistry of an addition; order carbonyls by Khyd or by reactivity; choose the reagent that gives 1,2 or 1,4; explain the bell-shaped pH profile |
| GATE Chemistry | Mechanism with correct curly arrows; chemoselective reduction of a polyfunctional substrate; Felkin–Anh prediction |
| IIT-JAM / CUET-PG | Aldehyde vs ketone reactivity with reasons; acetal protection and deprotection; cyanohydrin and bisulfite adducts |
| MSc coursework | Bürgi–Dunitz trajectory and stereoelectronics; hard–soft control of 1,2 vs 1,4; quantitative treatment of hydration equilibria |
Do the numbers, not just the arrows. Equilibrium fractions, ΔG° from an equilibrium constant, and reagent masses from molar masses all turn a descriptive answer into a quantitative one — and all three appear in the same question more often than students expect. The ABC Chemistry Calculator Suite keeps a scientific calculator, a molar-mass tool and the standard constants side by side.
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