HookThe silver mirror that only an aldehyde can make
Unscrew the top of a vacuum flask and look at the mirrored inner wall. That silver film was almost certainly laid down by an aldehyde. In 1835 the chemist Justus von Liebig found that an aldehyde will hand its electrons to silver ions dissolved in ammonia, plating out a flawless layer of metallic silver on the inside of the glass — the process that silvered mirrors, Christmas baubles and Thermos flasks for well over a century. Pour a ketone into exactly the same solution and nothing happens: no mirror, no reaction, no change.
That single difference in behaviour — one family reacts, its near-twin refuses — runs right through this section. Aldehydes and ketones both contain the carbonyl group, a carbon double-bonded to oxygen, and they behave almost identically when a nucleophile attacks that carbon. But an aldehyde carries a hydrogen atom on the carbonyl carbon and a ketone does not, and that lone hydrogen is the difference between a compound that is easily oxidised and one that shrugs off every mild oxidising agent you throw at it. Get the carbonyl group's shape and polarity clear in your head and the rest of the section — two diagnostic tests, one chain-lengthening addition, and one reduction — falls into place as variations on a single theme: something electron-rich attacking a carbon that is electron-poor.
ModelThe carbonyl group and the great divide
The carbonyl group is \(\text{C=O}\). Oxygen is far more electronegative than carbon, so it pulls the shared electrons towards itself: the carbon is left electron-deficient, \(\delta+\), and the oxygen electron-rich, \(\delta-\). That permanent dipole is the reason the whole family is attacked by nucleophiles — electron-rich species drawn to the \(\delta+\) carbon — and it is the engine of every reaction in this section.
An aldehyde has the carbonyl at the end of the chain, so the carbonyl carbon also carries a hydrogen: the group is \(-\text{CHO}\) and the names end in -al (ethanal, \(\text{CH}_3\text{CHO}\)). A ketone has the carbonyl in the middle of the chain, flanked by two carbon groups and no hydrogen: the group is \(\text{C=O}\) between two carbons and the names end in -one (propanone, \(\text{CH}_3\text{COCH}_3\)).
That structural detail decides everything about oxidation. To oxidise a carbonyl to a carboxylic acid you must break a bond to the carbonyl carbon and put an \(-\text{OH}\) there; an aldehyde has a C–H bond in exactly the right place to be replaced, so \(\text{RCHO} + [\text{O}] \rightarrow \text{RCOOH}\) happens readily. A ketone has only C–C bonds around the carbonyl and no hydrogen to lose, so mild oxidising agents cannot touch it. That is not a minor quirk — it is the basis of the two tests that tell the families apart.
MechanismThree tests that tell aldehyde from ketone
Because aldehydes oxidise and ketones do not, any mild oxidising agent that changes colour on being reduced becomes a diagnostic test. There are three you must know cold, each with its reagent, its condition and its observation.
Acidified potassium dichromate(VI), warmed gently: with an aldehyde the orange \(\text{Cr}_2\text{O}_7^{2-}\) is reduced to green \(\text{Cr}^{3+}\); with a ketone it stays orange. Tollens' reagent — ammoniacal silver nitrate, containing the \([\text{Ag(NH}_3)_2]^+\) ion — warmed in a water bath: an aldehyde reduces the silver(I) to a silver mirror on the tube wall, \(\text{Ag}^+ + e^- \rightarrow \text{Ag}\), while a ketone leaves it colourless and clear. Fehling's solution (or Benedict's), a blue complex of copper(II), warmed: an aldehyde reduces the copper(II) to a brick-red precipitate of copper(I) oxide, \(\text{Cu}_2\text{O}\), turning the blue solution orange-red; a ketone stays blue.
In every case the aldehyde is oxidised to a carboxylic acid (or, in the alkaline conditions of Tollens' and Fehling's, to the carboxylate ion) while the metal ion is reduced. The ketone gives a genuine negative — no mirror, no red precipitate, no colour change — and that absence of a reaction is itself the identification.
An unknown liquid X has the formula \(\text{C}_3\text{H}_6\text{O}\). It could be propanal, \(\text{CH}_3\text{CH}_2\text{CHO}\), or propanone, \(\text{CH}_3\text{COCH}_3\). Warmed with Fehling's solution, X turns the blue solution into a brick-red precipitate. Identify X and write the oxidation equation.
The brick-red precipitate is copper(I) oxide, formed only when the copper(II) is reduced — which only an aldehyde can do. So X is propanal, not propanone (propanone would have left the solution blue). The aldehyde is oxidised to propanoic acid:
\[\text{CH}_3\text{CH}_2\text{CHO} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\]
Using \([\text{O}]\) to represent the oxidising agent is the accepted A-level shorthand and keeps the equation balanced. To confirm the assignment you could repeat with Tollens' reagent: propanal would deposit a silver mirror, propanone would not.
MechanismNucleophilic addition of HCN — growing the chain by one carbon
Cyanide ion, generated from potassium cyanide with a little acid, is a strong nucleophile, and it adds across the carbonyl to make a 2-hydroxynitrile. This is a valuable synthetic step because it lengthens the carbon chain by one and installs two functional groups at once — a nitrile and a hydroxyl.
The mechanism is nucleophilic addition, and AQA expects the curly arrows exactly. The lone pair on the \(:\text{CN}^-\) ion attacks the \(\delta+\) carbonyl carbon, forming a new C–C bond; at the same time the \(\pi\) bond of \(\text{C=O}\) breaks and both its electrons move onto the oxygen, giving a negatively charged alkoxide intermediate. That oxygen then takes a proton (from HCN or water) to give the neutral hydroxynitrile. One curly arrow from the cyanide lone pair to the carbon, one from the double bond to the oxygen, then protonation — three arrows, in that order.
Because the carbonyl carbon is planar to begin with, the cyanide can attack from either face with equal probability. When the product has a new chiral centre this produces equal amounts of the two enantiomers — a racemic mixture that is optically inactive, exactly the reasoning met in the previous section on optical isomerism.
Give the product and the mechanism outcome when hydrogen cyanide adds to ethanal, \(\text{CH}_3\text{CHO}\).
The cyanide ion attacks the carbonyl carbon; the oxygen becomes an alkoxide and is then protonated. The product is 2-hydroxypropanenitrile:
\[\text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN}\]
Count the carbons: ethanal has two, the product has three — the chain has grown by one, which is the synthetic point of the reaction. Now check the product carbon that used to be the carbonyl: it carries \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\) and \(-\text{CN}\), four different groups, so it is a chiral centre. Because cyanide attacked the flat carbonyl from both faces equally, the two enantiomers form in equal amounts and the product is a racemate — optically inactive despite every molecule being chiral.
MechanismReduction to alcohols with NaBH4
Aldehydes and ketones are reduced to alcohols by sodium tetrahydridoborate(III), \(\text{NaBH}_4\) (also called sodium borohydride), usually in aqueous or alcoholic solution. In A-level equations the reducing agent is written as \([\text{H}]\): an aldehyde gains two hydrogens to give a primary alcohol, \(\text{RCHO} + 2[\text{H}] \rightarrow \text{RCH}_2\text{OH}\), and a ketone gives a secondary alcohol, \(\text{RCOR'} + 2[\text{H}] \rightarrow \text{RCH(OH)R'}\).
Mechanistically this is another nucleophilic addition, but the nucleophile is now a hydride ion, \(:\text{H}^-\), delivered from the \(\text{NaBH}_4\). The hydride attacks the \(\delta+\) carbonyl carbon, the \(\pi\) bond breaks onto the oxygen to give an alkoxide, and the oxygen is then protonated by water to give the alcohol — the same three-step shape as the cyanide addition, with \(\text{H}^-\) playing the role of the nucleophile.
A point AQA likes to probe: \(\text{NaBH}_4\) is selective. It reduces the polar \(\text{C=O}\) bond but leaves a non-polar \(\text{C=C}\) alkene bond untouched, because the alkene has no \(\delta+\) carbon to attract the hydride. So reducing an unsaturated aldehyde such as prop-2-enal with \(\text{NaBH}_4\) converts the carbonyl to an alcohol while the carbon–carbon double bond survives intact.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Every distinguishing-test question is marked on three things together: the reagent, the condition and the observation. 'Fehling's solution' alone is not enough — write 'warm with Fehling's solution: blue solution turns to a brick-red precipitate with an aldehyde, stays blue with a ketone'. Dropping the 'warm', or the colour change, drops marks. Quote the metal-ion change where you can (Ag⁺ reduced to Ag; copper(II) reduced to copper(I) oxide) to show you understand it is a redox test.
For oxidation equations use \([\text{O}]\) and balance them; \(\text{RCHO} + [\text{O}] \rightarrow \text{RCOOH}\). For reductions use \(2[\text{H}]\) and state whether a primary or secondary alcohol results.
Mechanism marks are the strictest on the paper. A curly arrow must start from a bond or a lone pair and end precisely where the new bond forms. For HCN addition, show the arrow from the cyanide lone pair to the \(\delta+\) carbon, the arrow from the \(\text{C=O}\) bond onto the oxygen, the negatively charged intermediate, and the final protonation — and mark the \(\delta+\) and \(\delta-\) on the carbonyl. Name the mechanism as nucleophilic addition and name the nucleophile (cyanide ion, or hydride ion for the \(\text{NaBH}_4\) reduction). If asked why the product is optically inactive, explain that the planar carbonyl is attacked equally from both faces to give a racemate — the reasoning, not just the word.