Chemical Reactions of Aldehydes and Ketones

Hold two colourless liquids up to the light and they look identical, yet one turns a test tube into a silver mirror and the other does nothing at all. The difference is a single hydrogen atom sitting on the carbonyl carbon — present in every aldehyde, absent in every ketone — and it is the key to almost every reaction and every identification test in this topic. Aldehydes and ketones share one core reactivity, nucleophilic addition at an electron-poor carbonyl carbon, but aldehydes carry two extra tricks: they are easier to attack, and they are easy to oxidise. This article works through why nucleophilic addition happens, how the acidic alpha-hydrogen powers the aldol and Cannizzaro reactions, and how Tollens', Fehling's, and the iodoform test turn that one-hydrogen difference into a reliable diagnostic you can run in a test tube.

The core insight

The carbonyl carbon in aldehydes and ketones is electron-poor and open to nucleophilic addition, aldehydes react faster because they are less hindered and less electron-rich than ketones, and the acidic alpha-hydrogen in aldehydes powers both self-condensation (aldol) and self-redox (Cannizzaro), while Tollens', Fehling's, and iodoform tests exploit these same differences to identify a given carbonyl compound.

The core idea

Why the carbonyl carbon reacts at all. Oxygen is more electronegative than carbon, so the $\text{C=O}$ bond is polarised, leaving the carbon short of electron density and the oxygen rich in it. A nucleophile — an electron-rich species — is drawn straight to that carbon. The $\pi$ bond breaks as the new bond forms, pushing both electrons onto oxygen as a negative charge, which is picked up by a proton in a follow-up step.

\text{Nu}^- + \text{R}_2\text{C=O} \longrightarrow \text{R}_2\text{C(Nu)-O}^- \xrightarrow{\text{H}^+} \text{R}_2\text{C(Nu)-OH}

where: $\text{Nu}^-$ is the attacking nucleophile, $\text{R}_2\text{C=O}$ is the aldehyde ( $\text{R}$ = H and alkyl/aryl) or ketone ( $\text{R}$ = two alkyl/aryl groups), and the tetrahedral alkoxide intermediate is protonated to give the neutral addition product.

Why aldehydes beat ketones. Two independent effects both favour the aldehyde. Sterically, an aldehyde has only one bulky group next to the carbonyl carbon (the other position is a small hydrogen), so an incoming nucleophile has an easier approach than in a ketone, which is flanked by two bulky groups. Electronically, alkyl groups are weak electron donors (+I effect); a ketone has two such groups pushing electron density onto the carbonyl carbon, partly cancelling its electron deficiency, while an aldehyde has only one such group (or none, for methanal), leaving its carbon more electrophilic and more attractive to a nucleophile.

Why the alpha-hydrogen is special. A hydrogen on the carbon next to $\text{C=O}$ is unusually acidic, because removing it leaves a carbanion whose negative charge can delocalise onto the electronegative carbonyl oxygen by resonance — a stabilisation ordinary carbanions do not get. This single fact drives two named reactions of aldehydes possessing that hydrogen, and a third reaction for aldehydes that lack it entirely.

Figure

Comparing nucleophilic addition at an aldehyde and a ketone carbonyl carbon

Why it works

There is no single numeric formula governing these reactions — the syllabus fixes named reactions and named tests, each following from the same two structural facts: an electrophilic carbonyl carbon, and (in aldehydes with alpha-H) an acidic alpha-hydrogen. Aldol condensation happens because dilute base can remove that alpha-hydrogen from one molecule of aldehyde, generating a resonance-stabilised carbanion that is itself a good nucleophile; it then attacks the electrophilic carbonyl carbon of a second, unreacted aldehyde molecule exactly as in the general addition scheme above, giving a beta-hydroxy aldehyde that dehydrates on warming because eliminating water to extend conjugation with the carbonyl is thermodynamically favourable. Cannizzaro reaction happens only when an aldehyde has no alpha-hydrogen at all — with no acidic hydrogen to remove, concentrated hydroxide instead attacks the carbonyl carbon directly, and the resulting tetrahedral intermediate transfers a hydride ion to a second aldehyde molecule: one molecule is oxidised to the carboxylate, the other reduced to the alcohol, a disproportionation. Tollens' and Fehling's tests both rely on the aldehyde having a hydrogen directly on the carbonyl carbon, which lets a metal oxidant (silver ammine or alkaline copper tartrate) pull it off while oxidising the aldehyde to a carboxylate; ketones have no such hydrogen on that carbon and cannot be oxidised this way, so they give no reaction. The iodoform test works differently: it does not care whether the compound is an aldehyde or ketone, only whether it contains, or can be oxidised in situ to contain, a methyl ketone group $\text{CH}_3\text{CO-}$ , which iodine and base can halogenate and cleave to yield the yellow precipitate $\text{CHI}_3$ .

Worked example

Predict and explain the product when ethanal is treated with dilute NaOH followed by heat, and state whether Fehling's test would be positive for ethanal.

Step 1. Check for an alpha-hydrogen. Ethanal, $\text{CH}_3\text{CHO}$ , has three alpha-hydrogens on its methyl carbon, so it undergoes aldol condensation rather than Cannizzaro, which needs the absence of alpha-H.

Step 2. Form the carbanion and attack a second molecule. Dilute $\text{OH}^-$ removes an alpha-H to give the resonance-stabilised carbanion $^-\text{CH}_2\text{CHO}$ , which then attacks the carbonyl carbon of a second ethanal molecule, giving an alkoxide that is protonated by water to 3-hydroxybutanal:

\text{CH}_3\text{CHO} + {}^-\text{CH}_2\text{CHO} \rightarrow \text{CH}_3\text{CH(O}^-\text{)CH}_2\text{CHO} \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}

Step 3. Dehydrate on heating. The beta-hydroxy aldehyde loses water on warming, since elimination that extends conjugation with the carbonyl is favourable, giving the alpha,beta-unsaturated aldehyde crotonaldehyde:

\text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\Delta} \text{CH}_3\text{CH=CHCHO}

Step 4. Check Fehling's test separately. Ethanal is an aliphatic aldehyde with a hydrogen directly on its carbonyl carbon, so warming it with Fehling's reagent oxidises it to the carboxylate and reduces $\text{Cu}^{2+}$ to a brick-red $\text{Cu}_2\text{O}$ precipitate.

The product of the base-then-heat sequence is crotonaldehyde, $\text{CH}_3\text{CH=CHCHO}$ , and Fehling's test on ethanal is positive.

Competitive stretch

Distinguish propanal from propanone using two different named tests. Tollens' test: propanal ( $\text{CH}_3\text{CH}_2\text{CHO}$ ) has a carbonyl hydrogen and is oxidised, depositing a silver mirror, while propanone ( $\text{CH}_3\text{COCH}_3$ ) has no such hydrogen and gives no mirror. Iodoform test: propanal has no $\text{CH}_3\text{CO-}$ group and is not acetaldehyde, so it gives no yellow precipitate, while propanone is a methyl ketone and does give one. Running both tests, one positive-for-aldehyde and one positive-for-ketone, cross-checks the identification even if only one reagent set is available on the day.

Common traps

Assuming the iodoform test is specific to aldehydes — it is specific to the $\text{CH}_3\text{CO-}$ group or to acetaldehyde and related alcohols, so most aldehydes, including propanal, give a negative result.
Applying Cannizzaro conditions to an aldehyde that does have alpha-hydrogens — such an aldehyde will instead undergo aldol condensation; Cannizzaro applies only when no alpha-H is present.
Expecting Fehling's test to work on aromatic aldehydes such as benzaldehyde — it does not; Fehling's is diagnostic only for aliphatic aldehydes.
Forgetting that ketones simply cannot undergo Tollens' or Fehling's oxidation, since there is no hydrogen on the carbonyl carbon to remove — a negative result there rules out an aliphatic aldehyde but proves nothing else on its own (aromatic aldehydes such as benzaldehyde give a negative Fehling's yet are still aldehydes).

Ask the tutor: Want to walk through the full curved-arrow mechanism for aldol condensation or practise picking the right named test for a given unknown pair? Ask the AI tutor to run you through it.

If both aldehydes and ketones undergo nucleophilic addition, why do only aldehydes give positive Tollens and Fehling tests

Nucleophilic addition needs only an electrophilic carbonyl carbon, which both classes have. Tollens' and Fehling's tests are different: they are oxidations that specifically remove the hydrogen sitting directly on the carbonyl carbon. Aldehydes have that hydrogen; ketones, with two carbon substituents on the carbonyl carbon and no hydrogen there, simply have nothing for the oxidant to remove, so they cannot be oxidised this way and give no reaction.

Why does the aldol product need heating to lose water, and does that always happen

The initial aldol (a beta-hydroxy aldehyde) is a stable, isolable product on its own. Heating drives dehydration because removing water creates a carbon-carbon double bond conjugated with the carbonyl, which is thermodynamically more stable. In an exam, "aldol condensation" usually implies both addition and dehydration, giving the alpha,beta-unsaturated carbonyl compound, while "aldol addition" alone stops at the hydroxy-aldehyde stage — read the question carefully for which is asked.

Can a ketone ever give a positive iodoform test

Yes, if it is a methyl ketone, meaning it has the $\text{CH}_3\text{CO-}$ group attached to another carbon, such as propanone or butan-2-one. The iodoform test does not distinguish aldehydes from ketones at all — it only detects that specific methyl-ketone fragment, or acetaldehyde, or an alcohol that iodine and base can first oxidise into one of these. A ketone without that exact group, such as pentan-3-one, gives a negative iodoform test.