What this chapter covers17.1

The carbonyl group, C=O, is the functional group at the heart of this topic. In aldehydes it sits at the end of the carbon chain; in ketones it sits between two carbon groups. The chemistry that follows — reduction, nucleophilic addition of cyanide, and the tests that tell aldehydes from ketones — all begins at the electrophilic carbon of the C=O bond. This page covers the synthesis, reactions and identification of aldehydes and ketones as required by topic 17.1.

The verbs tell you how much detail to give. 17.1.1 says recall: reagents, conditions and the distinction between distillation and reflux. 17.1.2 says describe: equations, reagents, conditions and what you would see. 17.1.3 says describe the mechanism: curly arrows showing electron movement at every step. 17.1.4 and 17.1.5 say describe and deduce: you must give the test and interpret the result. 17.1.6 says deduce: from the tri-iodomethane result, work back to the CH₃CO– structural feature.

The page is in two parts. The first covers aldehydes and ketones: how they are made, their reactions, and the tests that identify them. The second is a review section with a data summary and a self-test.

The carbonyl group17.1

A carbonyl group is a carbon atom double-bonded to an oxygen atom, C=O. In an aldehyde that carbon carries at least one hydrogen atom, so the group is always at the end of the chain: RCHO (or HCHO for methanal). In a ketone the carbonyl carbon is bonded to two other carbon atoms, so it sits inside the chain: RCOR′.

The naming rule: take the longest chain that contains the C=O, drop the –e of the parent alkane and add –al for an aldehyde or –one for a ketone. A locant is needed for the C=O in ketones with five or more carbons (pentan-2-one, pentan-3-one) but never for aldehydes, where the C=O is always at C1.

Physical properties

The C=O bond is strongly polar (oxygen is much more electronegative than carbon), giving the carbonyl group a permanent dipole. Carbonyl compounds can accept hydrogen bonds from water — oxygen has lone pairs — but they cannot donate them, because there is no O–H or N–H. So aldehydes and ketones have higher boiling points than alkanes of similar M_r (permanent dipole–dipole forces), but lower boiling points than the corresponding alcohols (no hydrogen bonding between molecules). The short-chain members are soluble in water because water molecules can hydrogen-bond to the carbonyl oxygen.

Making aldehydes and ketones by oxidation17.1.1

Both aldehydes and ketones are made by oxidising alcohols with an acidified oxidising agent — potassium dichromate(VI), K₂Cr₂O₇, or potassium manganate(VII), KMnO₄, acidified with dilute sulfuric acid. The oxidising agent removes two hydrogen atoms from the alcohol: one from the O–H and one from the C that carries the –OH.

Primary alcohols → aldehydes: distillation

A primary alcohol has two hydrogens on the –OH carbon, so the aldehyde produced can be oxidised again to give a carboxylic acid. To stop at the aldehyde:

• use the alcohol in excess, so there is not enough oxidising agent for the second step;
• distil the product off as it forms — the aldehyde has a much lower boiling point than the alcohol (no hydrogen bonding between aldehyde molecules), so gentle heating sends it out of the flask before it meets any more oxidising agent.

CH₃CH₂OH + [O] → CH₃CHO + H₂O    ethanol → ethanal

CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O    propan-1-ol → propanal

Secondary alcohols → ketones

A secondary alcohol has only one hydrogen on the –OH carbon, so after one oxidation step there are none left. The ketone cannot be oxidised further by acidified dichromate, so the conditions make no difference: distillation or reflux gives the same product. Reflux is normally used to ensure complete reaction.

CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O    propan-2-ol → propanone

CH₃CH₂CH(OH)CH₃ + [O] → CH₃CH₂COCH₃ + H₂O    butan-2-ol → butanone

The colour change is the same in every case: orange dichromate turns green (Cr³⁺), or purple manganate(VII) becomes colourless (Mn²⁺).

Reactivity of the carbonyl group17.1.2

The carbon of the C=O bond carries a significant δ+ charge, because oxygen is much more electronegative. That makes the carbonyl carbon electrophilic — it attracts nucleophiles. At the same time, the π bond is exposed (unlike the C–O single bond in an alcohol, which is shielded by the rest of the molecule). A nucleophile can therefore attack the carbonyl carbon and break the π bond, adding across the C=O in much the same way that an electrophile adds across C=C.

This pattern — nucleophilic addition — is the mechanism behind both of the reactions in 17.1.2: the addition of HCN and the reduction by NaBH₄ or LiAlH₄.

Reduction to alcohols17.1.2(a)

Aldehydes and ketones are reduced to alcohols by NaBH₄ (sodium tetrahydridoborate, in water or aqueous ethanol) or LiAlH₄ (lithium tetrahydridoaluminate, in dry ether). Both deliver a hydride ion, H⁻, to the δ+ carbonyl carbon. The oxygen then gains a proton from the solvent.

CH₃CHO + 2[H] → CH₃CH₂OH    ethanal → ethanol (primary alcohol)

CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃    propanone → propan-2-ol (secondary alcohol)

An aldehyde always gives a primary alcohol; a ketone always gives a secondary alcohol. No aldehyde or ketone can give a tertiary alcohol by this reduction, because the former carbonyl carbon ends up carrying the hydrogen it gained from the reducing agent.

Reaction with HCN: nucleophilic addition17.1.2(b)

When an aldehyde or ketone is heated with hydrogen cyanide, HCN, in the presence of a trace of potassium cyanide, KCN, as catalyst, the cyanide ion CN⁻ adds across the C=O to form a hydroxynitrile (also called a cyanohydrin).

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

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

The product has one more carbon atom than the starting material — the CN has added a carbon to the chain. This matters in synthesis, because hydrolysing the nitrile group gives a carboxylic acid (18.1.1(b)), so the two-step sequence aldehyde → hydroxynitrile → hydroxy acid is a way to extend the carbon chain by one.

The mechanism of nucleophilic addition17.1.3

The mechanism has two steps. Draw it with curly arrows showing where each electron pair moves.

Step 1 — Nucleophilic attack

The cyanide ion, CN⁻, is the nucleophile. Its lone pair on the carbon end attacks the δ+ carbonyl carbon, forming a new C–C bond. As the bond forms, the π electrons of the C=O shift onto the oxygen, which becomes negatively charged. The intermediate is an alkoxide ion.

Step 2 — Protonation

The alkoxide ion is a strong base. It takes a proton from HCN, forming the –OH group and regenerating CN⁻, which can catalyse another molecule. This is why only a trace of KCN is needed.

Detecting carbonyl compounds: 2,4-DNPH17.1.4

2,4-Dinitrophenylhydrazine (Brady's reagent, 2,4-DNPH) reacts with any compound that contains a C=O group — aldehyde, ketone, or a more complex molecule with either group present. The test detects the presence of a carbonyl group but does not say whether it is an aldehyde or a ketone.

Test: Add a few drops of the unknown compound to a solution of 2,4-DNPH in methanol and concentrated sulfuric acid.

Positive result: A bright orange or yellow crystalline precipitate forms. This is the 2,4-dinitrophenylhydrazone — a condensation product in which the C=O oxygen and two hydrogens have been lost as water.

Negative result: The solution stays clear. The compound does not contain a carbonyl group.

RCHO + 2,4-DNPH → orange/yellow precipitate + H₂O

RCOR′ + 2,4-DNPH → orange/yellow precipitate + H₂O

The precipitate can be filtered, recrystallised and its melting point measured. Each carbonyl compound gives a derivative with a unique melting point, so comparing it with a data table identifies the compound. This is how carbonyl compounds were identified before spectroscopy became routine.

Distinguishing aldehydes from ketones17.1.5

Aldehydes are easily oxidised to carboxylic acids; ketones are not. The two tests below exploit this difference: each uses a mild oxidising agent that an aldehyde can reduce but a ketone cannot.

Tollens' reagent (the silver mirror test)

Reagent: a solution of silver(I) ions in aqueous ammonia — [Ag(NH₃)₂]⁺, made by adding a few drops of dilute NaOH to silver nitrate solution and then adding dilute ammonia until the precipitate just redissolves.

Procedure: Warm the mixture gently in a clean glass tube (often in a water bath).

With an aldehyde: Ag⁺ is reduced to metallic silver, which deposits on the glass as a silver mirror. The aldehyde is oxidised to the carboxylate ion.

With a ketone: No reaction. The solution stays colourless.

RCHO + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → RCOO⁻ + 2Ag(s) + 4NH₃ + H₂O

Fehling's solution

Reagent: two solutions mixed just before use — Fehling's A (aqueous copper(II) sulfate, blue) and Fehling's B (sodium potassium tartrate in NaOH). The tartrate keeps the Cu²⁺ in solution at high pH.

Procedure: Add the unknown compound and heat.

With an aldehyde: the deep blue colour disappears and a brick-red precipitate of copper(I) oxide, Cu₂O, forms. The aldehyde is oxidised to the carboxylate ion.

With a ketone: No reaction. The solution stays blue.

RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O(s) + 3H₂O

The tri-iodomethane (iodoform) reaction17.1.6

The tri-iodomethane reaction detects a specific structural feature: a CH₃CO– group (a methyl group directly bonded to a carbonyl carbon). Ethanal, CH₃CHO, has this group. Propanone, CH₃COCH₃, has it. But propanal, CH₃CH₂CHO, does not — it has a CH₂ next to the C=O, not a CH₃.

Reagent: Alkaline aqueous iodine, I₂(aq) with NaOH(aq).

Procedure: Warm the compound with the reagent.

Positive result: A yellow precipitate of tri-iodomethane, CHI₃ (iodoform), which has an antiseptic smell. The organic product is the carboxylate ion, RCO₂⁻.

CH₃COR + 3I₂ + 4NaOH → CHI₃(s) + RCO₂⁻Na⁺ + 3NaI + 3H₂O

With ethanal (R = H): the carboxylate ion is HCO₂⁻ (methanoate).

With propanone (R = CH₃): the carboxylate ion is CH₃CO₂⁻ (ethanoate).

Data summary17.1

Self-test17.1

Twenty questions on topic 17.1. Each question has one correct answer and an explanation. Reset to start again.