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The carbonyl group — a carbon double-bonded to an oxygen — is the single most important functional group in organic chemistry. It appears in aldehydes, ketones, carboxylic acids, esters, acyl chlorides, amides, urea, sugars and almost every drug molecule. This lesson looks at the two simplest carbonyl families — aldehydes and ketones — and develops the structure, bonding, polarisation, nomenclature, physical properties, and the central nucleophilic-addition mechanism that drives all of next lesson's chemistry. Master this and the next four lessons (carbonyl tests, carboxylic acids, esters, acyl chlorides) become a single connected story rather than five isolated topics.
This lesson maps to the OCR A-Level Chemistry A (H432) specification at Module 6.1.2 (a)–(b) — the structure of carbonyl compounds, their nomenclature and physical properties (refer to the official OCR H432 specification document for exact wording). The lesson sequence intentionally leads with the polarisation argument (Section 1) because every reaction in Lesson 2 follows directly from it: aldehyde and ketone carbons are δ+, so they are attacked by nucleophiles, full stop. If you understand why C=O is polarised, you understand four lessons' worth of mechanisms before you ever draw an arrow.
Key Definition — Carbonyl compound: an organic molecule containing a >C=O group in which the carbon is bonded only to H and/or other carbons (aldehydes and ketones). When the carbonyl carbon is bonded to a heteroatom (–OH, –OR, –Cl, –NR₂) the compound is a carbonyl derivative (carboxylic acid, ester, acyl chloride, amide). All share the same δ+C=Oδ⁻ polarisation but differ in reactivity because of the second substituent.
The carbonyl group is the C=O double bond. In aldehydes and ketones the carbonyl carbon is bonded only to H and/or carbon, never directly to a heteroatom such as O, N or Cl (those would be carboxylic acids, amides or acyl chlorides).
Key Definition — Carbonyl group: The functional group >C=O, where a carbon atom is double-bonded to an oxygen atom.
The C=O double bond consists of one sigma (σ) bond and one pi (π) bond. The carbonyl carbon is sp² hybridised and the three atoms attached to it lie in a trigonal planar arrangement with bond angles close to 120°.
Because oxygen is much more electronegative than carbon (3.44 vs 2.55), the C=O bond is strongly polarised:
δ+C=Oδ−
This polarisation is the single most important thing to remember about carbonyls. It makes the carbonyl carbon an electrophile — a site of attack by nucleophiles — and underlies nearly every reaction you will meet in the next few lessons.
The bond enthalpy of C=O is approximately 745 kJ mol⁻¹, considerably greater than C=C (~611) or even N=N (~418). Most of this strength comes from the σ component, but the π component (~280 kJ mol⁻¹ of the total) is the chemically interesting half: it is the π bond that breaks when a nucleophile attacks. The π electrons are held closer to oxygen than to carbon because oxygen's 2p orbitals are lower in energy than carbon's 2p orbitals (oxygen's nuclear charge is +8 versus carbon's +6). The asymmetry of the π molecular orbital is what makes the carbonyl carbon δ+.
The C=O dipole moment is ~2.4 D (debye), roughly twice that of a C–Cl bond (~1.5 D) and far larger than C=C (~0 D). This is why propanone (μ = 2.88 D for the whole molecule) is a polar solvent that dissolves both polar solutes (water-miscible) and non-polar solutes (alkane-miscible) — the molecule has a permanent dipole big enough to interact with water, but the methyl groups still solubilise non-polar species.
The C=O group is bifunctional within itself:
The combination of these two sites is what makes carbonyls so versatile in synthesis — nucleophiles attack the carbon, but the oxygen can be activated by acid to speed up that attack. You will meet both halves of this picture in the next four lessons.
graph LR
A[Carbonyl C=O] --> B[Sigma + pi bond]
A --> C[sp2 C, trigonal planar 120 deg]
A --> D[Polarised: C delta + / O delta -]
D --> E[Carbon is electrophilic]
D --> G[Oxygen has 2 lone pairs - weak base]
E --> F[Attacked by nucleophiles]
G --> H[Protonated under acidic conditions to speed up attack]
The distinction between an aldehyde and a ketone rests on what is attached to the carbonyl carbon.
| Feature | Aldehyde | Ketone |
|---|---|---|
| Carbonyl C is bonded to… | At least one H | Two carbon atoms (no H on C=O carbon) |
| General formula | R–CHO | R–CO–R' |
| Position of carbonyl | End of chain | Middle of chain |
| Example | Ethanal CH₃CHO | Propanone CH₃COCH₃ |
The presence of that C–H bond on the carbonyl carbon of aldehydes makes aldehydes more reactive and more easily oxidised than ketones. This difference underpins every chemical test for aldehydes vs ketones that you will meet in Lesson 2.
graph TD
A[Carbonyl R-CO-X] --> B{What is X?}
B -->|H| C["Aldehyde R-CHO<br/>Carbonyl at end of chain<br/>Easily oxidised"]
B -->|Carbon R'| D["Ketone R-CO-R’<br/>Carbonyl in middle of chain<br/>Resists oxidation"]
| Compound | Structure | Class |
|---|---|---|
| Methanal (formaldehyde) | HCHO | Aldehyde |
| Ethanal (acetaldehyde) | CH₃CHO | Aldehyde |
| Propanal | CH₃CH₂CHO | Aldehyde |
| Benzaldehyde | C₆H₅CHO | Aromatic aldehyde |
| Propanone (acetone) | CH₃COCH₃ | Ketone |
| Butanone | CH₃COCH₂CH₃ | Ketone |
| Pentan-2-one | CH₃COCH₂CH₂CH₃ | Ketone |
| Pentan-3-one | CH₃CH₂COCH₂CH₃ | Ketone |
OCR expects you to name and draw aldehydes and ketones up to about 6 carbons, including branched chains. The naming follows the standard IUPAC machinery introduced in Lesson 1 of the Basic Organic course: identify the longest chain that contains the C=O group; number to give the C=O the lowest possible locant; apply the suffix (-al for aldehydes, -one for ketones); add substituents as prefixes with their own locants.
Aldehydes take the suffix -al. The carbonyl is always at carbon 1 — you do not need a locant for it because no other position is possible.
Rules:
Examples:
Ketones take the suffix -one. Because the C=O can be anywhere in the middle of the chain, you usually need a locant to say where.
Rules:
Examples:
Exam Tip: For 4-carbon ketones (butanone) and 3-carbon ketones (propanone) you do not write a locant. For 5 carbons and longer, you must — OCR will penalise you for writing "pentanone" because it is ambiguous.
When an aldehyde or ketone coexists with other functional groups (alcohols, halides, alkenes), the priority order — highest priority gets the suffix; the others become prefixes — runs:
−COOH>−COOR>−COCl>−CONH2>−CN>−CHO>−CO−>−OH>−NH2>−C=C−
So in 4-hydroxypentan-2-one, CH₃-CO-CH₂-CH(OH)-CH₃, the ketone gets the -one suffix and the alcohol becomes the hydroxy- prefix (rather than the -ol suffix). The numbering starts from the end that gives the C=O (the higher-priority group) the lower locant.
For aldehydes containing rings or branches:
Carbonyl compounds are polar because of the δ+C=Oδ- dipole. They cannot form hydrogen bonds with themselves (they lack an O–H or N–H), so their intermolecular forces are:
Compare some similar-sized molecules:
| Compound | Mᵣ | Boiling point (°C) | Strongest IMF |
|---|---|---|---|
| Butane C₄H₁₀ | 58 | –1 | London |
| Propanal C₂H₅CHO | 58 | 48 | Permanent dipole |
| Propanone CH₃COCH₃ | 58 | 56 | Permanent dipole |
| Propan-1-ol C₃H₇OH | 60 | 97 | Hydrogen bonding |
Propanal and propanone boil ~50 °C higher than butane because of their dipole-dipole forces. They boil ~50 °C lower than propan-1-ol because they cannot H-bond with themselves (propan-1-ol can).
Small carbonyls (C1–C4) are miscible with water because the lone pairs on the carbonyl oxygen can accept hydrogen bonds from water's O–H. As the carbon chain grows, the hydrophobic tail dominates and solubility drops. By C7 or so, most carbonyls are insoluble in water but soluble in non-polar solvents.
Key Insight: Carbonyl compounds can accept hydrogen bonds from water but cannot donate them to each other. This explains both their relatively low boiling points and their small-molecule water solubility.
The polarisation of C=O means two reactive sites coexist in a single functional group:
This combination makes carbonyls the central "hub" of organic synthesis. Every major transformation you will study in the next few lessons — reductions, additions, condensations, esterifications, nucleophilic substitutions — goes through (or begins at) a C=O group.
graph TD
A[Carbonyl C=O] --> B[Reductions: NaBH4 -> alcohol]
A --> C[Additions: HCN -> hydroxynitrile]
A --> D[Oxidation of aldehydes -> carboxylic acid]
A --> E[Tests: Tollens, Fehlings, 2,4-DNP]
A --> F[Combines with alcohol + acid -> ester]
The shared mechanism for nearly every carbonyl reaction is nucleophilic addition. In one sentence: a nucleophile attacks the δ+ carbon, the π bond breaks heterolytically with both electrons going to oxygen, and the resulting alkoxide is protonated to give the neutral product. Three curly arrows total. You will draw this in Lesson 2 for HCN and for NaBH₄; in Lesson 4 for esterification (acid-catalysed variant); in Lesson 5 for acyl chlorides (with an extra elimination step); and in Lesson 6 for the addition of amines to ketones.
Nu−+R2C=O⟶R2C(Nu)−O−H+R2C(Nu)−OH
The product is always a tetrahedral carbon carrying the new Nu group and an OH (or, after elimination, the C=O reforms with a new substituent in place of the original). Knowing this template is worth four lessons of revision.
Butan-2-one is CH₃-CO-CH₂CH₃, an unsymmetrical ketone. HCN provides CN⁻ as the nucleophile. The CN⁻ attacks the δ+ carbonyl carbon (the one between the methyl and the ethyl). The C=O π bond opens, the alkoxide is then protonated by H⁺. Product: CH₃-C(OH)(CN)-CH₂CH₃ — 2-hydroxy-2-methylbutanenitrile. This is a hydroxynitrile (cyanohydrin) and contains a brand-new chiral centre at the original carbonyl carbon, so the product is a 50:50 mixture of (R)- and (S)- enantiomers (a racemate). We meet this in detail next lesson.
When NaBH₄ reduces ethanal (CH₃CHO → CH₃CH₂OH), the carbonyl carbon's oxidation state changes from +1 (one C-H, one C=O equivalent to two C-O) to −1 (two C-H, one C-O). Net change: −2. Hydrogen's oxidation state stays at −1 (one new C-H bond from H⁻). Net electrons added: 2, gained by the carbon. This confirms the reaction is a 2-electron reduction. (Recall: H⁻ supplies one H plus 2 electrons in total across the two new bonds — one on C, one on O.)
Synoptic Links — Connects to:
ocr-alevel-chemistry-basic-organic / functional-groups-iupac(Lesson 1 — IUPAC nomenclature for -al and -one suffixes, locants, principal-group priority).ocr-alevel-chemistry-basic-organic / reaction-mechanisms(Lesson 4 — curly-arrow conventions, nucleophiles vs electrophiles, heterolytic vs homolytic bond fission; the template applied here for the first time to a polar π bond).ocr-alevel-chemistry-alcohols-haloalkanes / combustion-and-oxidation-of-alcohols(controlled oxidation of primary alcohols makes aldehydes; oxidation of secondary alcohols makes ketones — this is the synthetic on-ramp into Lesson 2).ocr-alevel-chemistry-carbonyls-polymers-spectroscopy / carbonyls-reactions-and-tests(Lesson 2 — direct application of the polarisation and bonding picture developed here).ocr-alevel-chemistry-carbonyls-polymers-spectroscopy / carbon-13-nmrandproton-nmr-combined-techniques(the C=O carbon resonates at δ 190–210 ppm in ¹³C NMR — a diagnostic signal; aldehyde H appears at δ ~9.5 ppm in ¹H NMR — also diagnostic).
Practical Activity Group anchor: PAG 7 — Qualitative analysis of organic compounds. The standard PAG 7 task is to identify an unknown carbonyl using 2,4-DNP, Tollens' and Fehling's tests; the structural framework developed in this lesson is the prerequisite for interpreting those test-tube observations.
Question (8 marks): Compound X is a carbonyl compound with molecular formula C₄H₈O. (a) Draw the displayed formulae of all four possible structures of X consistent with the formula and the carbonyl class. (b) The three aldehyde-class structures share a common property that distinguishes them from the ketone. State that property and explain it. (c) Predict and explain the relative boiling points of butan-2-one (M_r 72) and butan-1-ol (M_r 74).
| Mark | AO | Awarded for |
|---|---|---|
| 1 | AO1 | Butanal CH₃CH₂CH₂CHO drawn correctly |
| 2 | AO1 | 2-methylpropanal (CH₃)₂CHCHO drawn correctly |
| 3 | AO1 | Butan-2-one CH₃COCH₂CH₃ drawn correctly |
| 4 | AO2 | Recognising the three aldehydes / one ketone structural split |
| 5 | AO2 | Stating that aldehydes are easily oxidised to carboxylic acids whereas ketones are not |
| 6 | AO2 | Stating that butan-2-one has a lower b.p. than butan-1-ol |
| 7 | AO3 | Explaining: butan-2-one has dipole-dipole only; butan-1-ol has O–H hydrogen bonds |
| 8 | AO3 | Hydrogen bonds are stronger than dipole-dipole, so more energy is required |
AO split: AO1 = 3, AO2 = 3, AO3 = 2.
(a) Three carbonyl isomers of C₄H₈O: butanal CH₃CH₂CH₂CHO, 2-methylpropanal (CH₃)₂CHCHO, butan-2-one CH₃COCH₂CH₃.
(b) The aldehydes can be oxidised to carboxylic acids; the ketone cannot. This is because aldehydes have an H on the C=O carbon that can be replaced by OH, whereas ketones do not.
(c) Butan-2-one boils lower than butan-1-ol because butan-1-ol can hydrogen bond and butan-2-one cannot.
Examiner-style commentary: Marks for (a) and (b) covered; (c) missing the explicit comparison of bond strengths. Around 5/8.
(a) Three carbonyl structures of C₄H₈O:
(b) The two aldehydes are easily oxidised by acidified K₂Cr₂O₇ to butanoic acid and 2-methylpropanoic acid respectively (orange to green colour change). Butan-2-one cannot be oxidised under the same conditions because the C=O carbon has no H — there is no H available to be replaced by OH. This is the diagnostic property distinguishing the aldehydes from the ketone.
(c) Butan-2-one (b.p. 80 °C) has only permanent dipole-dipole interactions (and London forces) between its molecules because the H atoms are all bonded to C, not to O, so there is no O–H to participate in hydrogen bonding. Butan-1-ol (b.p. 118 °C) has a free O–H bond which forms hydrogen bonds between molecules. Hydrogen bonds are stronger than ordinary dipole-dipole interactions, so more energy is needed to separate butan-1-ol molecules — its boiling point is ~38 °C higher.
Examiner-style commentary: 7/8 — strong explanation, would benefit from quoting that the aldehyde C–H is the "replaceable" hydrogen converted to C–OH on oxidation.
(a) The carbonyl isomers of C₄H₈O are:
(b) Aldehydes are easily oxidised; ketones resist oxidation under standard lab conditions. Mechanistically, the aldehyde C=O carbon has a C–H bond that can be replaced by C–OH during oxidation — the C–H bond is broken in the oxidation by Cr(VI) and a new C–OH bond forms, giving the carboxylic acid R–COOH. Ketones have no such C–H bond on the carbonyl carbon (both attached atoms are C), so no oxidation occurs without breaking the carbon skeleton — which requires much harsher conditions (e.g. hot concentrated KMnO₄, splitting the molecule into two carboxylic acids).
(c) Both butan-2-one (M_r 72, b.p. 80 °C) and butan-1-ol (M_r 74, b.p. 118 °C) are polar molecules of similar mass and similar shape. Butan-2-one has the C=O dipole; butan-1-ol has the O–H dipole. The key difference: butan-1-ol has an O–H bond that can hydrogen-bond to another butan-1-ol molecule's lone-pair oxygen (donor + acceptor in every molecule). Butan-2-one has lone-pair oxygens that can accept H-bonds from other O–H-containing molecules (e.g. water), but its own H atoms are all bonded to C, so it cannot donate H-bonds — its molecules engage only in permanent dipole-dipole interactions between C=O groups (and London forces from C₄). A typical permanent dipole-dipole interaction is ~5–25 kJ mol⁻¹ at intermolecular distances; a typical O–H⋯O hydrogen bond is ~20–25 kJ mol⁻¹ but acts more directionally with stronger orbital overlap, so the net intermolecular cohesion in butan-1-ol is higher. The b.p. difference of ~38 °C reflects this — more energy needed to overcome H-bonds + dipole-dipole compared with dipole-dipole alone.
Examiner-style commentary: Full 8/8. Top-band discriminator: mechanistic justification for why aldehydes oxidise, explicit comparison of bond-energy magnitudes, recognition that butan-2-one is an H-bond acceptor to water but not a donor to itself.
Pedagogical observations — not fabricated statistics:
The history of the carbonyl group is the history of organic chemistry itself. August Kekulé drew the first explicit structural formulae of aldehydes and ketones in the 1860s as part of his programme to give organic compounds determinate structures. Adolf Baeyer (Nobel Prize 1905) developed the field of aldol and condensation reactions in the 1870s-80s, building carbohydrate chemistry from carbonyl building blocks. Robert Burns Woodward (Nobel Prize 1965) used carbonyl chemistry as the central tool of his total syntheses (cholesterol, strychnine, chlorophyll, vitamin B12); modern organic-chemistry textbooks still treat the carbonyl group as the workhorse of organic synthesis.
Beyond A-Level, the orbital picture becomes richer. The C=O π* orbital — the empty antibonding orbital lying just above the π — is the orbital that accepts electrons from incoming nucleophiles. Frontier molecular orbital (FMO) theory rationalises why HCN, NaBH₄ and even electron-rich aromatics (in the case of acyl chlorides + Friedel-Crafts) all attack the same carbon: the nucleophile's HOMO overlaps with the carbonyl's LUMO (which is largest at the carbon end). The Bürgi-Dunitz angle (~107°) describes the geometry of nucleophilic attack — the nucleophile approaches at ~107° to the C=O axis, not perpendicular. This is testable by X-ray crystallography on enzyme-bound carbonyl substrates.
In biology, the carbonyl group is everywhere. Glucose (and every other monosaccharide) is a polyhydroxy-aldehyde or polyhydroxy-ketone. Ribose, the sugar in RNA, is a 5-carbon polyhydroxy-aldehyde. Ketone bodies (acetoacetate, β-hydroxybutyrate) are the brain's backup fuel during starvation. Recommended reading: Clayden, Greeves and Warren, Organic Chemistry, Chapter 6 ("Nucleophilic addition to the carbonyl group"). Oxbridge interview prompt: "Why does formaldehyde (HCHO) exist as a stable trimer (paraformaldehyde) at room temperature but propanone does not?" — answer involves steric strain at the C=O carbon and the entropy/enthalpy balance of ring formation.
The errors that distinguish A from A*:
Carbonyl compounds — aldehydes (R–CHO) and ketones (R–CO–R') — are characterised by the C=O double bond between a carbon (sp², trigonal planar) and an oxygen. The bond is strongly polar (dipole moment ~2.4 D) because of oxygen's higher electronegativity, making the carbon δ+ and the oxygen δ−. This polarisation is the single dominant feature of carbonyl chemistry and the basis of every nucleophilic-addition reaction in the next four lessons. Aldehydes have at least one H on the C=O carbon and so can be oxidised to carboxylic acids; ketones cannot. Naming uses the suffix -al for aldehydes (C=O at C1) and -one for ketones (locant required for C₅+ chains). Boiling points sit between non-polar alkanes (no permanent dipole) and hydrogen-bonded alcohols (O–H donor + acceptor), reflecting the presence of dipole-dipole forces but the absence of self-hydrogen-bonding. Short-chain carbonyls are water-soluble because the C=O oxygen accepts H-bonds from water, even though carbonyls cannot donate H-bonds. The same polarisation drives the chemistry of carboxylic acids (Lesson 3), esters (Lesson 4) and acyl chlorides (Lesson 5).
Reference: OCR A-Level Chemistry A (H432) Module 6.1.2 (a)–(b) — structure, nomenclature and physical properties of carbonyl compounds (refer to the official OCR H432 specification document for exact wording).