Aldehydes and ketonesEdexcel International A Level Chemistry: Revision notes
Section 1
Naming and structure
Aldehydes and ketones both contain the carbonyl group, C=O. In an aldehyde the carbonyl carbon is at the end of the chain, bonded to at least one hydrogen (–CHO), and the name ends -al: ethanal, CH₃CHO. In a ketone the carbonyl carbon is bonded to two carbon atoms and the name ends -one, with a number once the chain is long enough to need one: propanone, CH₃COCH₃; pentan-3-one, CH₃CH₂COCH₂CH₃.
In an aldehyde the carbonyl carbon is always carbon 1, so no number is needed. Draw them as structural, displayed or skeletal formulae; in skeletal form the C=O is a double line going to an O at the end or in the chain.
Write aldehydes as –CHO and not –COH, which looks like an alcohol.
Section 2
Polarity, boiling temperature and solubility
The C=O bond is polar because oxygen is more electronegative than carbon: the carbon is δ+ and the oxygen δ−. Carbonyl molecules therefore have permanent dipole–permanent dipole attractions as well as London forces, so they boil at higher temperatures than alkanes of similar Mr.
Aldehydes and ketones have no hydrogen bonded to oxygen, so they cannot form hydrogen bonds with each other. Their boiling temperatures are therefore lower than those of alcohols of similar Mr, which do form hydrogen bonds.
The oxygen has lone pairs, so carbonyl compounds can form hydrogen bonds with water (to the δ+ hydrogen of water). Short-chain aldehydes and ketones such as ethanal and propanone are miscible with water. Solubility falls as the hydrocarbon chain lengthens, because the non-polar chain cannot interact well with water.
Saying carbonyl compounds hydrogen bond with themselves. They only hydrogen bond with water (or other molecules that have an O–H or N–H).
Section 3
Distinguishing aldehydes from ketones by oxidation
Aldehydes are easily oxidised to carboxylic acids; ketones are not. This gives three tests (all warmed gently):
- Tollens' reagent (ammoniacal silver nitrate, [Ag(NH₃)₂]⁺): aldehyde gives a silver mirror as Ag⁺ is reduced to Ag. Ketone: no change.
- Fehling's or Benedict's solution (blue copper(II) complex in alkali): aldehyde gives a brick-red precipitate of copper(I) oxide. Ketone: stays blue.
- Acidified potassium dichromate(VI): aldehyde turns orange to green as Cr₂O₇²⁻ is reduced to Cr³⁺. Ketone: stays orange.
Use [O] in equations: CH₃CHO + [O] → CH₃COOH.
Writing the aldehyde as the oxidising agent. The aldehyde is oxidised, so it is the reducing agent; the metal ion is reduced.
Section 4
Reduction with LiAlH₄
Lithium tetrahydridoaluminate(III), LiAlH₄, in dry ether reduces aldehydes to primary alcohols and ketones to secondary alcohols. It acts as a source of hydride ions, H⁻. Use [H] in equations:
CH₃CHO + 2[H] → CH₃CH₂OH
CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃
The ether must be dry because LiAlH₄ reacts violently with water. A dilute acid is added afterwards to protonate the alkoxide ion.
Section 5
Nucleophilic addition of HCN
Aldehydes and ketones react with hydrogen cyanide in the presence of KCN to form hydroxynitriles. HCN is a weak acid, so KCN supplies CN⁻, the nucleophile.
- The carbonyl carbon is δ+. A lone pair on the carbon of CN⁻ attacks it (curly arrow from the lone pair to carbon).
- The C=O π bond breaks (curly arrow from the bond to oxygen), giving an intermediate with O⁻.
- O⁻ takes H⁺ from HCN (or water), giving the hydroxynitrile and regenerating CN⁻.
Ethanal gives 2-hydroxypropanenitrile, CH₃CH(OH)CN. The carbonyl group is planar, so CN⁻ can attack from either side with equal probability. The product has a chiral carbon, so the two optical isomers form in equal amounts: a racemic mixture with no optical activity. This is evidence for the planar carbonyl group and the mechanism.
Start the curly arrow at the lone pair on the carbon of CN⁻ (not on the negative sign) and end it at the carbonyl carbon.
Section 6
Tests for carbonyl compounds
2,4-dinitrophenylhydrazine (2,4-DNPH) is a test for the carbonyl group: aldehydes and ketones give a yellow or orange precipitate. The precipitate is filtered off, purified by recrystallisation, dried and its melting temperature measured. Comparing with data values for derivatives identifies the carbonyl compound. (The equation is not required.)
The iodoform test uses iodine and sodium hydroxide solution. Compounds containing a CH₃C=O group (ethanal and methyl ketones such as propanone and butanone; also CH₃CH(OH)– alcohols such as ethanol) give a pale yellow precipitate of triiodomethane, CHI₃.
To identify an unknown carbonyl: DNPH positive means aldehyde or ketone; Tollens'/Fehling's positive means aldehyde; iodoform positive means it contains CH₃C=O.
Saying a positive 2,4-DNPH test shows an aldehyde. It shows only a carbonyl group, aldehyde or ketone.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Aldehydes and ketones
- A chemist compares three compounds of similar relative molecular mass: butane (Mr 58, boiling temperature −1 °C), propanal (Mr 58, boiling temperature 49 °C) and propan-1-ol (Mr 60, boiling temperature 97 °C).Explain why propanal has a higher boiling temperature than butane, even though their relative molecular masses are the same.2 marks
- A technician has two unlabelled liquids, one is propanal and the other is propanone. She tests a sample of each with Tollens' reagent, with Fehling's solution and with acidified potassium dichromate(VI) solution, warming gently where necessary.Write an equation, using [O] to represent the oxidising agent, for the reaction of propanal with acidified potassium dichromate(VI), and state the colour change seen for propanal but not for propanone.2 marks
- Compound X has the molecular formula C₄H₈O. It gives an orange precipitate with 2,4-dinitrophenylhydrazine solution. It gives no reaction when warmed with Tollens' reagent. It gives a pale yellow precipitate when warmed with iodine and sodium hydroxide solution.Deduce the structure and name of compound X, explaining each observation.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).