Edexcel International A Level Chemistry

Topic 15: Organic Chemistry: Carbonyls, Carboxylic Acids and Chirality

Mirror-image molecules, the C=O group and everything built on it, and the spectra and chromatography that prove what you have made.

The one idea

A left hand and a right hand have the same parts but do not match. Some molecules are the same: they have the same atoms and bonds, but one is the mirror image of the other. They behave identically in most reactions, yet they rotate plane-polarised light in opposite directions. That is chirality, and it opens this topic.

The rest of the topic follows one functional group, C=O. In an aldehyde or ketone it is attacked by nucleophiles, and the flat C=O group is why some products come out as a mixture of mirror images. Add an –OH to the carbon and you have a carboxylic acid. Swap that –OH for –Cl, –OR or –NH₂ and you have the derivatives: acyl chlorides, esters and amides. The topic ends with the tools for proving a structure: accurate mass, carbon-13 and proton NMR, and chromatography.

Three ideas carry most of the marks:

1. Chirality and optical activity as evidence. Whether a product keeps its optical activity tells you which mechanism ran (S_N1, S_N2, or addition to C=O).
2. The reaction web. From the alcohol or nitrile to the acid, and from the acid to its derivatives, with the reagent and condition for every step.
3. Putting spectra together. Use accurate mass, carbon-13 NMR and proton NMR together to deduce a structure.

The five parts

When you have finished, use Topic 15 on a page for a last read.

What you need first

• Nucleophilic substitution, S_N1 and S_N2, and the oxidation of alcohols, from the halogenoalkanes and alcohols topic.
• Mass spectra and infrared spectroscopy from Topic 10, which this topic extends.
• Intermolecular forces and hydrogen bonding, and curly arrows for movement of electron pairs.
• Equilibrium and Le Chatelier's principle, for esterification.

Values used in these notes

Check the shift ranges against the data booklet your exam gives you. They are quoted here as typical values.

Pages

1. Chiral centres

   A carbon atom with four different groups attached is a chiral centre. Its molecule can exist as two mirror-image forms that cannot be placed on top of each other.

   12 min 15.115.2

2. Drawing enantiomers

   A 3D diagram uses plain lines, a wedge and a dashed line around the chiral carbon. The mirror image swaps left and right while the wedge stays a wedge and the dash stays a dash.

   12 min 15.2

3. Optical activity and racemic mixtures

   A single enantiomer rotates the plane of plane-polarised light. Its mirror image rotates it by the same angle the other way, so an equal mixture of the two has no effect.

   13 min 15.315.4

4. Optical activity as evidence for a mechanism

   Start with a single enantiomer and check whether the product is optically active. A reaction that keeps the activity works by SN2, and one that loses it passes through a flat intermediate, as in SN1.

   14 min 15.5

5. Naming aldehydes and ketones

   Aldehydes end in -al with the C=O on the end carbon. Ketones end in -one with the C=O inside the chain. You need to draw each as structural, displayed and skeletal formulae.

   14 min 15.6

6. Physical properties of carbonyl compounds

   Aldehydes and ketones have no O–H bond, so they cannot hydrogen-bond to each other and boil lower than alcohols. Their oxygen atom can accept a hydrogen bond from water, so small ones dissolve.

   12 min 15.7(i)15.7(ii)

7. Oxidation tests for aldehydes

   Aldehydes are easily oxidised to carboxylic acids and ketones are not. Tollens' reagent, Fehling's or Benedict's solution and acidified dichromate(VI) all react with aldehydes and not with ketones.

   14 min 15.8(i)

8. Reducing carbonyl compounds with LiAlH₄

   Lithium tetrahydridoaluminate(III) in dry ether reduces an aldehyde to a primary alcohol and a ketone to a secondary alcohol. A ketone such as butanone gives a racemic product, because the flat C=O is attacked from both sides.

   12 min 15.8(ii)15.5

9. Addition of hydrogen cyanide

   Cyanide ion attacks the δ+ carbon of a C=O group in nucleophilic addition, making a hydroxynitrile. Because the C=O group is flat, an aldehyde gives a racemic mixture.

   16 min 15.8(iii)15.5

10. 2,4-DNPH and the iodoform test

    2,4-dinitrophenylhydrazine gives an orange precipitate with any aldehyde or ketone, and its melting temperature identifies which one. Iodine and alkali give a pale yellow precipitate of triiodomethane with a CH₃CO group.

    15 min 15.8(iv)15.8(v)

11. Practical, identifying carbonyl compounds

    A sequence of test-tube tests with 2,4-DNPH, Tollens' reagent and iodine and alkali separates aldehydes, methyl ketones, other ketones and alcohols. The melting temperature of the derivative then names the compound.

    15 min 15.8(i)15.8(iv)15.8(v)

12. Naming carboxylic acids

    A carboxylic acid contains the –COOH group and its name ends in -oic acid. The carboxyl carbon is always carbon 1, so a number is never needed for it.

    11 min 15.9

13. Hydrogen bonding in carboxylic acids

    A carboxylic acid has both a C=O oxygen to accept and an O–H hydrogen to give hydrogen bonds. This gives high boiling temperatures and, for small acids, complete solubility in water.

    14 min 15.10

14. Making carboxylic acids

    Carboxylic acids are made by oxidising primary alcohols or aldehydes under reflux with acidified dichromate(VI), or by hydrolysing a nitrile with acid or alkali. The nitrile route lengthens the carbon chain by one carbon.

    16 min 15.11

15. Reactions of carboxylic acids

    Carboxylic acids are reduced to primary alcohols by lithium tetrahydridoaluminate(III), neutralised by bases to give salts, and converted to acyl chlorides by phosphorus(V) chloride.

    15 min 15.12(i)15.12(ii)15.12(iii)

16. Esterification

    A carboxylic acid and an alcohol, heated with a little concentrated sulfuric acid as catalyst, form an ester and water in an equilibrium. Esters are named from the alcohol first and the acid second.

    14 min 15.12(iv)

17. Naming acyl chlorides and esters

    Acyl chlorides end in -oyl chloride and contain –COCl. Esters are named alkyl alkanoate, with the part from the alcohol first and the part from the acid second.

    12 min 15.13

18. Reactions of acyl chlorides

    Ethanoyl chloride reacts quickly at room temperature with water, alcohols, concentrated ammonia and amines. Each reaction swaps the Cl for –OH, –OR, –NH₂ or –NHR and gives off hydrogen chloride or its salt.

    15 min 15.14(i)15.14(ii)15.14(iii)15.14(iv)

19. Hydrolysis of esters

    Heating an ester under reflux with dilute acid reverses esterification and gives an equilibrium. Heating it with aqueous sodium hydroxide goes to completion and gives the sodium salt of the acid and the alcohol.

    13 min 15.15

20. Polyesters

    A polyester forms when a molecule with two –COOH groups reacts with a molecule with two –OH groups. A water molecule is lost at each ester link. Terylene is made from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

    12 min 15.16

21. Accurate relative molecular mass

    A high-resolution mass spectrometer measures the molecular ion peak to four decimal places. Atoms do not have whole-number masses, so the exact value tells you the molecular formula.

    14 min 15.17(i)15.17(ii)

22. Carbon-13 NMR

    Each carbon atom in a different environment gives one peak in a carbon-13 NMR spectrum. The chemical shift of the peak shows what the carbon is bonded to, and the number of peaks shows how many different environments there are.

    15 min 15.1815.19(i)15.19(ii)

23. Proton NMR, environments and peak areas

    A proton NMR spectrum has one peak for each set of equivalent hydrogen atoms. The chemical shift shows the type of hydrogen and the area under each peak is in the ratio of the number of hydrogens in that environment.

    15 min 15.20(i)15.20(ii)

24. Splitting and the (n + 1) rule

    At high resolution a peak splits into a pattern. A hydrogen with n hydrogens on the adjacent carbon atoms gives n + 1 lines, with heights in the ratio of Pascal's triangle.

    14 min 15.20(iii)

25. Predicting and deducing from proton NMR

    To predict a spectrum, find each hydrogen environment and give its shift, area and splitting. To deduce a structure, reverse the process and build pieces that fit every peak, then check them against the formula.

    16 min 15.20(iv)15.20(iii)

26. Chromatography and Rf values

    Chromatography separates a mixture because its components spend different amounts of time in a moving mobile phase and a fixed stationary phase. In paper and thin-layer chromatography each substance has its own Rf value.

    14 min 15.2115.22

27. Column chromatography

    High-performance liquid chromatography and gas chromatography pass a mixture through a column. Components leave at different retention times, and the detector, often a mass spectrometer, identifies and measures each.

    13 min 15.23

28. Putting the data together

    Use accurate mass for the molecular formula, carbon-13 NMR for the number of carbon environments and the functional groups, and proton NMR for the environments, ratio and neighbours. Then assemble a structure that fits every piece.

    16 min 15.17(i)15.19(i)15.20(iv)

Also in this topic: Topic 15 on a page

Spec checklist

Every point you can be examined on in this topic, and where it is explained. Tick them off as you go.