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Notes/Chemistry/Paper 1/Carbonyl Compounds
CAIEAS Level9701§17

Carbonyl Compounds

Aldehydes and ketones: making them from alcohols, reducing them back, adding HCN (with the curly-arrow mechanism), and the 2,4-DNPH, Fehling's/Tollens' and tri-iodomethane tests that let you work out a structure.

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In the AS Hydroxy Compounds note you oxidised alcohols and made aldehydes and ketones. This note is about those two families. Both contain the carbonyl group, C=O\text{C}{=}\text{O}. Oxygen pulls the bond's electrons towards itself, so the carbon is slightly positive, and that one fact explains most of the chemistry here.

You will first make aldehydes and ketones from alcohols and reduce them back to alcohols. Then you will add HCN and draw its mechanism with curly arrows. Last come three tests: 2,4-DNPH for any aldehyde or ketone, Fehling's and Tollens' reagents to tell an aldehyde from a ketone, and alkaline iodine for the CH3CO−\text{CH}_3\text{CO}{-} group. By the end you can work out a structure from a table of test results.

Before you start you should be able to

  • Oxidation of primary and secondary alcohols with acidified K₂Cr₂O₇ or KMnO₄, and distilling versus heating under reflux (AS Hydroxy Compounds)

  • The [O] and [H] shorthand for oxidation and reduction equations (AS Hydroxy Compounds)

  • Curly arrows, lone pairs, dipoles and nucleophiles (AS Introduction to Organic Chemistry)

  • Electrophilic addition to alkenes, used here as the contrast case (AS Hydrocarbons)

  • Chiral centres and optical isomers (AS Introduction to Organic Chemistry)

  • The tri-iodomethane test on CH₃CH(OH)– alcohols (AS Hydroxy Compounds)

By the end of this page you can

  • Recall how aldehydes and ketones are produced: oxidation of primary alcohols with acidified K₂Cr₂O₇ or KMnO₄ and distillation to stop at the aldehyde, and of secondary alcohols with distillation to give ketones

  • Describe reduction of aldehydes and ketones with NaBH₄ or LiAlH₄ to produce alcohols, writing equations in [H] notation

  • Describe the reaction of aldehydes and ketones with HCN, KCN as catalyst and heat, producing hydroxynitriles exemplified by ethanal and propanone

  • Describe the mechanism of nucleophilic addition of HCN: dipoles, lone pairs, curly arrows from :CN⁻ to the δ+ carbon, the charged intermediate, and protonation; explain why ethanal gives a racemic mixture

  • Describe the use of 2,4-DNPH reagent to detect the presence of carbonyl compounds

  • Deduce the nature (aldehyde or ketone) of an unknown carbonyl compound from test results: Fehling's and Tollens' reagents, ease of oxidation

  • Deduce the presence of a CH₃CO– group from reaction with alkaline I₂(aq): yellow precipitate of tri-iodomethane plus RCO₂⁻

  • Combine the results of several tests to deduce the structure of an unknown compound

01

Meet the carbonyl group

One double bond, two family names

A carbonyl group is a carbon doubly bonded to an oxygen, C=O\text{C}{=}\text{O}. Two families carry it, and the difference is tiny to draw and huge in behaviour:

  • an aldehyde has the C=O\text{C}{=}\text{O} at the end of a chain, so the carbonyl carbon also carries a hydrogen: R–CHO\text{R–CHO}. Ethanal CH3CHO\text{CH}_3\text{CHO}, propanal CH3CH2CHO\text{CH}_3\text{CH}_2\text{CHO}, butanal CH3CH2CH2CHO\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}. Suffix -al; no locant is ever needed because the group can only sit at C1.
  • a ketone has the C=O\text{C}{=}\text{O} inside the chain, flanked by two carbons: R–CO–R′\text{R–CO–R}'. Propanone (CH3)2CO(\text{CH}_3)_2\text{CO}, butanone CH3CH2COCH3\text{CH}_3\text{CH}_2\text{COCH}_3, pentan-2-one. Suffix -one with a locant once the chain is long enough to need one.

Both families share the general formula CnH2nO\text{C}_n\text{H}_{2n}\text{O} (for open-chain members with one oxygen). Propanal and propanone are both C3H6O\text{C}_3\text{H}_6\text{O}, so a molecular formula alone cannot tell you which one you have — that is why the chemical tests later in this note matter. Notice also what is missing: unlike an alcohol, there is no O–H here. The chemistry of this topic all happens at the C=O\text{C}{=}\text{O} double bond.

Why everything here is nucleophilic

Oxygen is more electronegative than carbon (3.53.5 vs 2.52.5), so the electron pair of each C=O bond — including the π\pi pair — is pulled towards oxygen. The group is polarised:

Cδ+=Oδ−\text{C}^{\delta+}{=}\text{O}^{\delta-}

Compare the alkene's C=C\text{C}{=}\text{C}: both atoms are carbon, so the bond has no dipole and nothing attracts a nucleophile — alkenes react with electrophiles instead. The carbonyl is the opposite. Its carbon is electron-poor, so any species with a lone pair (a nucleophile) is attracted to it, and its oxygen is electron-rich, ready to pick up a proton at the end. The two reactions in this note follow the same pattern:

  • a nucleophile gives its lone pair to Cδ+\text{C}^{\delta+} (a hydride ion in reduction, :CN−:\text{CN}^- in HCN addition);
  • the π\pi bond breaks and its electron pair moves onto oxygen, which now carries the negative charge;
  • the oxygen picks up H+\text{H}^+, which finishes the reaction.

The tests later in the note depend on the same structure: 2,4-DNPH reacts with the C=O\text{C}{=}\text{O} group itself, while Fehling's and Tollens' reagents need the H on an aldehyde's carbonyl carbon.

ONE POLAR BOND RUNS THE WHOLE TOPICCOδ+δ−O is more electronegative (3.5 vs 2.5):both bonding pairs — the π pair too —sit nearer the oxygen.carbon = electron-poor (δ+)oxygen = electron-rich (δ−)NUCLEOPHILES AIM HERE:Nu−COδ+δ−lone pair → Cδ+ ; the π bondbreaks onto O, charge parks there.e.g. hydride (H⁻) or :CN⁻THE FAMILY SPLIT: ONE ATOMaldehydeRCOHan H on thecarbonyl CketoneRCOR′two carbons —no H to oxidise

Anatomy of the group: how electronegativity polarises C=O, why the carbon is the attack site, and the single structural feature — H or R on the carbonyl carbon — that splits the family.

Alkene vs carbonyl: same double bond, opposite attacker

Both families add things across their double bond, but the first step differs. An alkene's π\pi pair attacks an electrophile (the alkene is the nucleophile). A carbonyl's Cδ+\text{C}^{\delta+} is attacked BY a nucleophile (the carbonyl is the electrophile). If you catch yourself drawing an electrophile attacking the C=O carbon, stop: that carbon is short of electrons and has none to give.

Naming, classifying and locating the δ+ carbon

For each compound, give the systematic name, state whether it is an aldehyde, a ketone or neither, and mark any Cδ+\text{C}^{\delta+} of a carbonyl group:

(i) CH3CH2COCH2CH3\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3 (ii) CH3CH(CH3)CH2CHO\text{CH}_3\text{CH(CH}_3)\text{CH}_2\text{CHO} (iii) CH2=CHCH2OH\text{CH}_2{=}\text{CHCH}_2\text{OH} (iv) HOCH2CH2COOH\text{HOCH}_2\text{CH}_2\text{COOH}

Show full working
  1. 1

    (i) Five-carbon chain, C=O inside it on C3 → suffix -one: pentan-3-one, a ketone. Carbonyl carbon = C3, flanked by ethyl groups both sides; that C is Cδ+\text{C}^{\delta+}.

    Count from whichever end gives the C=O the lowest locant — 3 either way here. A ketone's carbonyl carbon is bonded to TWO carbons.

  2. 2

    (ii) Longest chain containing the CHO carbon: four carbons (the branch is a methyl), numbered from the aldehyde end → 3-methylbutanal, an aldehyde. Its carbonyl carbon carries H and exactly one carbon neighbour — the aldehyde pattern.

    The CHO carbon is always C1, so the branch is numbered from that end. Numbering from the other end would give the wrong name, 2-methylbutanal, which is a different compound.

  3. 3

    (iii) CH2=CHCH2OH\text{CH}_2{=}\text{CHCH}_2\text{OH} has a C=C and an –OH but no C=O: neither family. Name: prop-2-en-1-ol. Its double-bond carbons are NOT Cδ+\text{C}^{\delta+} — a bond between two carbon atoms has no permanent dipole.

    A δ+ comes only from a bond between atoms of different electronegativity. C=C is between two identical atoms, so it is not polarised.

  4. 4

    (iv) The −COOH-\text{COOH} carbon does carry a C=O\text{C}{=}\text{O}, but it also carries an –OH. That makes it a carboxylic acid (a later AS note), not an aldehyde or ketone, and it does not give the reactions of this note. For an aldehyde or ketone, the carbonyl carbon is bonded only to C and/or H. So: neither. Name: 3-hydroxypropanoic acid.

    A common MCQ trap: carboxylic acids and esters contain C=O but are NOT carbonyl compounds in this topic's sense. 2,4-DNPH, for example, reacts only with aldehydes and ketones.

Answer

(i) pentan-3-one, ketone, C3 is δ+ (ii) 3-methylbutanal, aldehyde, CHO carbon δ+ (iii) prop-2-en-1-ol, neither (iv) 3-hydroxypropanoic acid, neither — COOH is not this topic's carbonyl

Your turn

  1. 14 marks

    (a) Draw the full structural formulae of all aldehydes and ketones with molecular formula C5H10O\text{C}_5\text{H}_{10}\text{O} that have a single carbonyl group and no other unsaturation, and name each.

    (b) On your drawings, mark every carbon that carries a δ+\delta+ charge because of a C=O\text{C}{=}\text{O} dipole.

    Stuck? Show hint

    (a) Fix the CHO end first for the aldehydes, then move the ketone C=O along a five-carbon chain. (b) only ONE carbon per molecule earns the δ+.

    Show solution
    1. 1

      (a) Aldehydes — the CHO pins C1, so vary the skeleton instead: pentanal CH3CH2CH2CH2CHO\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}; 2-methylbutanal CH3CH2CH(CH3)CHO\text{CH}_3\text{CH}_2\text{CH(CH}_3)\text{CHO}; 3-methylbutanal CH3CH(CH3)CH2CHO\text{CH}_3\text{CH(CH}_3)\text{CH}_2\text{CHO}; 2,2-dimethylpropanal (CH3)3CCHO(\text{CH}_3)_3\text{CCHO}. Four aldehydes.

      The CHO group uses up one carbon at the end of the chain, so you are really listing the four C₄ skeletons that can carry it: straight chain, and branches at C2, C3 and a double branch at C2.

    2. 2

      (a) Ketones — slide the C=O along the chain: pentan-2-one CH3COCH2CH2CH3\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3; pentan-3-one CH3CH2COCH2CH3\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3; then the branched skeleton 3-methylbutan-2-one CH3COCH(CH3)2\text{CH}_3\text{COCH(CH}_3)_2. Pentan-4-one is pentan-2-one renumbered, not new; 2-methylbutan-3-one renumbers to 3-methylbutan-2-one. Three ketones: seven structures total.

      Systematic enumeration beats guessing: fix what the functional group forces (CHO at C1), then enumerate carbon skeletons. Always finish by checking each candidate against renumbering duplicates.

    3. 3

      (b) Exactly the carbonyl carbon in each structure — nowhere else. The adjacent carbons carry no partial charge at all: the dipole is localised on the C=O bond itself.

      A common slip is putting δ+ on the neighbouring carbons as well. The dipole belongs to the C=O bond only: O is δ−, the carbonyl C is δ+.

    Answer

    4 aldehydes (pentanal; 2- and 3-methylbutanal; 2,2-dimethylpropanal) + 3 ketones (pentan-2-one, pentan-3-one, 3-methylbutan-2-one); δ+ on the carbonyl C only

  2. 24 marks

    Four liquids, known to be butan-1-ol, butanal, butanone and butanoic acid, have lost their labels.

    (a) Which TWO of the four are aldehydes or ketones?

    (b) A friend says butanoic acid should count as well, "because it contains C=O". In one sentence, explain the flaw.

    (c) Butanal is heated under reflux with acidified potassium dichromate(VI). Name the organic product. Which ONE other liquid of the four gives the same product under the same conditions?

    Stuck? Show hint

    (b) ask WHAT ELSE is bonded to the carbonyl carbon in –CO₂H. (c) think back to how far a primary alcohol is oxidised under reflux.

    Show solution
    1. 1

      (a) Only the aldehyde and the ketone: butanal and butanone.

      In this topic, 'carbonyl compound' means an aldehyde or a ketone, even though other groups also contain C=O.

    2. 2

      (b) In butanoic acid the C=O carbon also carries an −OH-\text{OH}: it is a carboxyl group, which reacts as an acid, not as an aldehyde or ketone.

      'Contains C=O' is not enough: the –OH on the same carbon changes the chemistry completely.

    3. 3

      (c) Under reflux the aldehyde is oxidised to butanoic acid, CH3CH2CH2COOH\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}.

      Reflux keeps the aldehyde in contact with the oxidising agent, so it is oxidised all the way to the acid.

    4. 4

      (c) continued: butan-1-ol is a primary alcohol. Under reflux it is oxidised to butanal and then on to butanoic acid, so it gives the same product. (Butanone cannot be oxidised further, and butanoic acid is already the acid.)

      A primary alcohol and its aldehyde both end at the same carboxylic acid under reflux, because the aldehyde is just the halfway stage.

    Answer

    (a) butanal and butanone (b) the carbonyl carbon in –COOH also carries –OH, so it reacts as an acid (c) butanoic acid; butan-1-ol

Paper 1 · MCQReal past-paper questions · Reduction; HCN addition and mechanismPaper 2 · TheoryReal past-paper questions · Reduction; HCN addition and mechanism

The rest of this note

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Can you do all of these?

  • I can name and draw aldehydes and ketones up to six carbons, explain the Cδ+=Oδ− polarisation from electronegativity, and say why nucleophiles — not electrophiles — start carbonyl reactions

  • I can produce aldehydes and ketones from the right alcohols with acidified K₂Cr₂O₇/KMnO₄ AND DISTILLATION, work backwards from product to alcohol, and explain why refluxing instead over-oxidises the aldehyde

  • I can write reduction equations in [H] notation for NaBH₄/LiAlH₄, predict primary vs secondary alcohol products, and state which reagent reduces C=O while leaving C=C alone

  • I can draw the complete HCN/KCN nucleophilic addition mechanism — dipole, :CN⁻ arrow to Cδ+, π pair to O, charged intermediate, protonation from HCN with CN⁻ regenerated — name hydroxynitrile products including their +1 carbon count, and explain why ethanal gives a racemic mixture

  • I can state the 2,4-DNPH observation (orange/red/yellow precipitate), what it proves and what it does not, and use it for 'same functional group' answers

  • I can use Fehling's (blue → brick-red ppt) and Tollens' (silver mirror) results to deduce aldehyde vs ketone, write the oxidation equation to the acid, and explain WHY ketones do not react

  • I can apply the tri-iodomethane test: name CHI₃ as the yellow precipitate AND the RCO₂⁻ ion as co-product, track the lost carbon, screen molecules for CH₃CO– or CH₃CH(OH)– units, and identify ethanol/ethanal as special cases

  • I can solve multi-test deduction tables and counting MCQs by running the three-test funnel in order and enumerating candidates systematically

Paper 1 · MCQ
Real past-paper questions, sorted by difficulty, with mark schemes
Paper 2 · Theory
Real past-paper questions, sorted by difficulty, with mark schemes