Carbonyl Compounds

0. What Examiners Want

Where this topic appears. Recent AS papers have asked for: the products of cold dilute and of hot concentrated KMnO₄ with an alkene and whether each reacts with 2,4-DNPH; which compounds give different results with alkaline I₂(aq); which reagent gives a positive result with propanone; the observations when a ketone and an aldehyde (made by cleaving a diol with an oxidising agent) are each tested with Na₂CO₃, alkaline iodine, 2,4-DNPH and Tollens’ reagent, and an equation for reduction with NaBH₄ using [H]; the number of carbonyl isomers of C₅H₁₀O (counting stereoisomers); two isomeric butanols identified by Fehling’s and alkaline iodine; and the identification of straight-chain isomers of C₅H₁₀O from a table of 2,4-DNPH, Tollens’ and alkaline iodine results. Infrared spectra of carbonyl compounds are built with Topic 22.

1. Structure, Naming and Properties

Aldehydes and ketones both contain the carbonyl group, C=O. The carbon atom is sp² hybridised with a planar arrangement of bonds (bond angle about 120°), and the C=O consists of one σ bond and one π bond (Topic 13).

Naming. An aldehyde takes the ending -al. The CHO carbon is always carbon 1, so no number is needed. A ketone takes the ending -one, and from pentanone upwards a number shows the position of the C=O (number from the end that gives the smaller number). Both families have the general formula C_nH_2nO, so an aldehyde and a ketone with the same number of carbons are functional group isomers (propanal and propanone are both C₃H₆O).

Polarity and properties. Oxygen is more electronegative than carbon, so the C=O bond is polar: the carbon is δ+ and the oxygen is δ−. Molecules therefore attract each other by permanent dipole–dipole forces, which are stronger than the induced dipole forces in alkanes but weaker than hydrogen bonds, because there is no O–H bond. Propanal boils at about 49°C, between butane (about −1°C) and propan-1-ol (about 97°C). The smaller carbonyl compounds (methanal, ethanal, propanone) mix with water because the lone pairs on oxygen accept hydrogen bonds from water; solubility falls as the hydrocarbon chain gets longer.

Worked example (counting isomers). How many carbonyl compounds have the formula C₅H₁₀O? Aldehydes: pentanal, 3-methylbutanal, 2-methylbutanal and 2,2-dimethylpropanal (four structures). Ketones: pentan-2-one, pentan-3-one and 3-methylbutan-2-one (three structures). That is 7 structural isomers. In 2-methylbutanal, CH₃CH₂CH(CH₃)CHO, the second carbon is attached to four different groups (H, CH₃, C₂H₅ and CHO), so it has a pair of optical isomers. The total including stereoisomers is 8.

2. Making Aldehydes and Ketones

Oxidation of alcohols (see Topic 16). Warming with acidified potassium dichromate(VI) turns the orange solution green as Cr₂O₇²⁻ is reduced to Cr³⁺. In equations the oxidising agent is written as [O].

• Primary alcohol to aldehyde: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O. Add the oxidising agent slowly to the warm alcohol and distil off the aldehyde as it forms. The aldehyde boils at a lower temperature than the alcohol (no hydrogen bonding between its molecules), so it leaves the mixture before it can be oxidised further. If the aldehyde stays in the flask with excess oxidant, it is oxidised to the carboxylic acid: CH₃CHO + [O] → CH₃COOH.
• Secondary alcohol to ketone: CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O. Heat the alcohol with the oxidising agent. The ketone cannot be oxidised further by these reagents, even under reflux with excess oxidant, so it does not need to be distilled off.

Cleavage of a C=C bond. Hot concentrated acidified KMnO₄ breaks the alkene double bond and oxidises each fragment. A carbon that carried two alkyl groups becomes a ketone, and a carbon that carried one hydrogen becomes an aldehyde which is oxidised straight on to a carboxylic acid. Examples: (CH₃)₂C=CHCH₃ + 3[O] → (CH₃)₂CO + CH₃COOH; and CH₃CH=CHCH₃ + 4[O] → 2CH₃COOH. (Cold dilute KMnO₄ only adds two OH groups to give a diol and does not break the C–C bond.)

Cleavage of a 1,2-diol. An oxidising agent such as sodium bismuthate(V), NaBiO₃, breaks the C–C bond between the two carbons that carry OH, and each of those carbons becomes C=O. (CH₃)₂C(OH)CH(OH)CH₃ + [O] → CH₃COCH₃ + CH₃CHO + H₂O. A question may give you an unfamiliar reagent like this; apply the same logic (each C–OH carbon becomes C=O) and use the structures in the question.

3. Reduction

Reducing a carbonyl compound adds hydrogen across the C=O bond to form an alcohol. The reducing agent is written as [H].

• Aldehyde to primary alcohol: CH₃CHO + 2[H] → CH₃CH₂OH, and CH₃CH₂CHO + 2[H] → CH₃CH₂CH₂OH.
• Ketone to secondary alcohol: CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃, and pentan-3-one gives pentan-3-ol.
• NaBH₄: an aqueous alkaline solution, with warming. It is the milder reagent.
• LiAlH₄: dissolved in dry ether (such as diethyl ether) at room temperature. It is a stronger reducing agent and reacts violently with water, so the solvent must be dry. Unlike NaBH₄, it also reduces carboxylic acids (Topic 16).

Mechanism. NaBH₄ and LiAlH₄ act as sources of the hydride ion, H⁻, which is a nucleophile. The reaction follows the same two steps as the addition of HCN (next section): H⁻ attacks the δ+ carbon, the C=O π electrons move onto oxygen to give an alkoxide intermediate, and the O⁻ then takes H⁺ from water to form the alcohol. The type of reaction is nucleophilic addition.

4. Nucleophilic Addition of HCN

Alkenes are attacked by electrophiles because the C=C is a region of high electron density. The C=O carbon is electron-poor (δ+), so it is attacked by nucleophiles, which are species with a lone pair that can form a new bond to carbon. Addition across C=O by a nucleophile first is called nucleophilic addition.

Aldehydes and ketones react with hydrogen cyanide, HCN, to form a 2-hydroxynitrile (the OH and CN are on the same carbon). HCN is a weak acid and gives very few CN⁻ ions, so a little KCN is added as the source of the nucleophile. HCN is volatile (b.p. about 26°C) and extremely poisonous, so it is made in the mixture from KCN and dilute acid, in a fume cupboard: KCN + H₂SO₄ → HCN + KHSO₄. The Cambridge syllabus lists the conditions as HCN with KCN as catalyst, with heating, so quote ‘HCN and KCN, heat’ in an exam (in the laboratory the addition also goes at or near room temperature, which is why the gas is kept in the mixture rather than driven off).

CH₃CHO + HCN → CH₃CH(OH)CN (2-hydroxypropanenitrile)

CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN (2-hydroxy-2-methylpropanenitrile)

Mechanism in words. (1) The lone pair on the carbon of CN⁻ attacks the δ+ carbon of C=O. At the same time the π bond of C=O breaks and both electrons go to oxygen, giving a negatively charged intermediate. (2) The O⁻ of the intermediate takes a proton from HCN (or from water or H⁺), forming the O–H bond of the hydroxynitrile and releasing a new CN⁻.

Using the product. The nitrile group can be hydrolysed to a carboxylic acid by heating under reflux with dilute hydrochloric (or sulfuric) acid: CH₃CH(OH)CN + HCl + 2H₂O → CH₃CH(OH)COOH + NH₄Cl. With dilute NaOH(aq) the sodium salt forms, and acid is then added to release the carboxylic acid. Because the carbon of the CN is new, the sequence adds one carbon to the chain.

Link to A Level. The product from ethanal, CH₃CH(OH)CN, has four different groups on one carbon, so it is chiral. The C=O is planar and CN⁻ can attack from either face with equal probability, so the product is a racemic mixture with no optical activity (Topic 29).

5. Tests for Carbonyl Compounds

2,4-Dinitrophenylhydrazine (2,4-DNPH). Add the unknown to a solution of 2,4-DNPH. An orange (yellow-orange) precipitate shows an aldehyde or a ketone. The reaction is a condensation reaction: the two molecules join and a small molecule, water, is lost. For ethanal: C₆H₃(NO₂)₂NHNH₂ + CH₃CHO → C₆H₃(NO₂)₂NHN=CHCH₃ + H₂O. Other compounds that contain C=O, such as carboxylic acids and esters, give no precipitate. The precipitate can be purified by recrystallisation, and its melting point, compared with data tables, identifies the original carbonyl compound.

Tollens’ reagent is aqueous silver nitrate in excess ammonia (ammoniacal silver nitrate), a mild oxidising agent containing Ag⁺. Warm the unknown with it. An aldehyde is oxidised and Ag⁺ is reduced to silver, which coats the tube as a silver mirror. A ketone gives no change. Because the mixture is alkaline, the aldehyde is oxidised to the carboxylate ion, RCOO⁻, not the free acid.

• Reduction half-equation: Ag⁺ + e⁻ → Ag
• Overall: CH₃CHO + 2Ag⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag + 2H₂O

Fehling’s solution is an alkaline solution of copper(II) ions (deep blue). Warm the unknown with it. An aldehyde is oxidised and the blue Cu²⁺ is reduced to Cu⁺, which appears as a red-orange precipitate of copper(I) oxide, Cu₂O. A ketone leaves the solution blue.

• Reduction half-equation: 2Cu²⁺ + 2OH⁻ + 2e⁻ → Cu₂O + H₂O (or simply Cu²⁺ + e⁻ → Cu⁺)
• Overall: CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O + 3H₂O

Both reagents work because an aldehyde has a hydrogen on the carbonyl carbon and is therefore easy to oxidise; a ketone would have to break a C–C bond to be oxidised, which mild reagents cannot do.

6. The Tri-iodomethane Test

Warm the unknown with alkaline aqueous iodine (I₂ with NaOH(aq)). A pale yellow precipitate of tri-iodomethane, CHI₃ (antiseptic smell, melting point 119°C) shows the group CH₃CO–. It happens in two stages: (1) the three hydrogens of the CH₃ group next to the C=O are replaced by iodine, giving RCOCI₃; (2) hydroxide ions hydrolyse this, releasing CHI₃ and the carboxylate ion RCOO⁻.

• Propanone: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂O
• Butanone: CH₃COCH₂CH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃CH₂COONa + 3NaI + 3H₂O
• Ethanal: CH₃CHO + 3I₂ + 4NaOH → CHI₃ + HCOONa + 3NaI + 3H₂O

Positive: ethanal, propanone, butanone, pentan-2-one and any other methyl ketone. Negative: methanal, propanal and all other aldehydes, pentan-3-one, and carboxylic acids and esters. The test also detects alcohols with CH₃CH(OH)– (Topic 16): the iodine first oxidises the alcohol to CH₃CO–.

7. Identifying Unknowns and Linking Routes

Worked identification. D, E and F are straight-chain isomers of C₅H₁₀O. All three give an orange precipitate with 2,4-DNPH, so each contains C=O and each is an aldehyde or ketone. Only D gives a silver mirror with Tollens’ reagent, so D is the aldehyde, pentanal (straight chain). E and F are the straight-chain ketones, pentan-2-one and pentan-3-one. Only E gives a yellow precipitate with alkaline iodine, so E contains CH₃CO– and is pentan-2-one; F is pentan-3-one.

Using 2,4-DNPH melting points. The derivatives of butanal and propanone both melt at about 126°C, so a melting point alone cannot separate them. If the unknown also gives a red precipitate with Fehling’s solution, it is the aldehyde, butanal, and the melting point and the Fehling’s result together identify it.

Linking reactions into routes. Carbonyl compounds sit at the centre of many two- and three-step syntheses. For example, ethene to 2-hydroxypropanoic acid:

Propan-2-ol and propanone can be interconverted: oxidation with acidified K₂Cr₂O₇ goes one way and reduction with NaBH₄ the other. Always state the reagent, the conditions and the intermediate for each step (Topic 21 builds on this).

8. Quick Sheet and Checklist

Before you leave the question, check:

• The oxidation says distil (aldehyde) or reflux with excess (acid), and names the colour change.
• Reduction equations use 2[H] and give a primary alcohol from an aldehyde, a secondary alcohol from a ketone.
• The HCN mechanism has arrows from the carbon lone pair of CN⁻ and from the C=O bond, plus partial charges and the intermediate.
• Every test gives the observation for both classes (silver mirror or no change; red ppt or stays blue).
• 2,4-DNPH is used to show C=O, and Tollens’ or Fehling’s to separate aldehyde from ketone.
• Iodoform answers name CHI₃ and look for CH₃CO– next to the C=O; ethanal is the only aldehyde that passes.