Home / Notes / Class 12 / Chemistry / Aldehydes and ketones
Structure and naming of carbonyl compounds, preparation, physical properties, nucleophilic addition, oxidation and reduction, tests to distinguish aldehydes from ketones, aldol condensation and uses of methanal and propanone.
Aldehydes (R-CHO) and ketones (R-CO-R') both contain the carbonyl group C=O. In an aldehyde the carbonyl carbon is bonded to at least one hydrogen, so the group is at the end of the chain. In a ketone it is bonded to two carbon atoms. The carbonyl carbon is sp² hybridised, the group is planar with bond angles of about 120°, and the C=O bond is polar, with a partial positive charge on carbon and a partial negative charge on oxygen.
IUPAC names: aldehydes end in al (methanal HCHO, ethanal CH₃CHO) and ketones end in one (propanone CH₃COCH₃, butanone CH₃COC₂H₅). The carbonyl carbon of an aldehyde is number 1. In ketones the chain is numbered to give the C=O the lowest number, as in pentan-2-one. Common names are formaldehyde, acetaldehyde and acetone.
Preparation of aldehydes and ketones:
Aldehydes and ketones are polar, so their boiling points are higher than those of alkanes of similar mass. They are lower than those of alcohols because they cannot form hydrogen bonds with each other. The lower members dissolve in water, since the oxygen forms hydrogen bonds with water. Methanal is a gas, ethanal boils at about 21 °C, and propanone is a liquid that boils at 56 °C.
Nucleophilic addition is the typical reaction. The nucleophile attacks the partly positive carbon, the pi bond breaks and the oxygen gets a negative charge, then takes a proton. Aldehydes are more reactive than ketones, because they have less steric hindrance, and because alkyl groups in ketones push electrons toward carbon and reduce its positive charge.
Examples of nucleophilic addition:
Oxidation and reduction:
Tests that separate aldehydes from ketones:
Aldol condensation: aldehydes and ketones with an alpha hydrogen react in dilute alkali. Two molecules of ethanal join to form 3-hydroxybutanal (an aldol), which loses water on heating to give but-2-enal. Aldehydes with no alpha hydrogen, such as methanal and benzaldehyde, undergo the Cannizzaro reaction in concentrated alkali, where one molecule is oxidised to the acid salt and another is reduced to alcohol.
Uses: methanal as a 40% solution called formalin, used to preserve specimens and to make Bakelite and other resins. Ethanal is used to make ethanoic acid and other chemicals. Propanone is a common solvent (nail polish remover) for paints and plastics. Benzaldehyde is used in perfumes and flavours.
Study tip: Learn the three distinguishing tests with their exact colours, and know which compounds give iodoform. For mechanism questions, always draw the attack of the nucleophile on the carbonyl carbon and the movement of electrons to oxygen. Aldol and Cannizzaro are often asked together, so remember the alpha hydrogen rule that separates them.
Say the meaning first, then open the card to check.
The C=O functional group of aldehydes and ketones.
Addition of a nucleophile to the carbonyl carbon, breaking the pi bond.
Ammoniacal silver nitrate, which gives a silver mirror with aldehydes.
Alkaline copper(II) tartrate, reduced to red Cu₂O by aliphatic aldehydes.
Reaction of two carbonyl molecules in dilute alkali to give a hydroxy carbonyl compound.
Reaction of an aldehyde with no alpha hydrogen in strong alkali, giving an alcohol and an acid salt.
A hydrogen on the carbon next to the carbonyl group.
A 40% solution of methanal in water, used as a preservative.
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
In aldehydes the carbonyl carbon has one small H atom and only one electron-donating alkyl group, so it is more positive and less crowded. In ketones two alkyl groups reduce the positive charge and block the approach of the nucleophile.
Alcohol molecules form hydrogen bonds with each other through O-H. Carbonyl compounds have no hydrogen attached to oxygen, so they only have dipole forces, which are weaker.
Warm each compound with ammoniacal silver nitrate. Ethanal is oxidised to ethanoate and silver ions are reduced to metallic silver, which forms a shining mirror on the tube. Propanone is not oxidised and gives no mirror.
Propanone has a CH₃CO- group. With iodine and sodium hydroxide: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂O. A yellow precipitate of iodoform with an antiseptic smell shows a positive test.
In dilute NaOH, a base removes an alpha hydrogen from one ethanal and forms a carbanion. This adds to the carbonyl carbon of a second ethanal, giving 3-hydroxybutanal, CH₃CH(OH)CH₂CHO (aldol). On heating, the aldol loses water and gives but-2-enal, CH₃CH=CHCHO.
In the presence of a trace of base, HCN gives the cyanide ion CN⁻, a strong nucleophile. CN⁻ attacks the partly positive carbon of C=O and the pi electrons move onto oxygen, giving an alkoxide ion. The alkoxide takes H⁺ from HCN and forms the cyanohydrin (CH₃)₂C(OH)CN and regenerates CN⁻. The product has one carbon more than the ketone, and hydrolysis gives a hydroxy acid.
Answer: 6.0 g of propan-2-ol
Answer: 17.6 g of ethanal
Answer: 21.6 g of silver
Answer: C₃H₆O, propanal