Carbonyls, carboxylic acids and chirality

The C=O group is the busiest functional group in organic chemistry. This topic covers aldehydes, ketones, carboxylic acids and the compounds made from them, the mirror-image molecules that some of their reactions produce, and the instrumental methods (NMR, mass spectrometry and chromatography) that chemists use to work out what they have made.

Naming the compounds in this topic

FamilyFunctional groupName endingExample
aldehyde–CHO (C=O at the end of the chain)-alCH3CH2CHO, propanal
ketoneC=O within the chain-oneCH3COCH2CH3, butanone; CH3COCH2CH2CH3, pentan-2-one
carboxylic acid–COOH-oic acidCH3CH2COOH, propanoic acid; HOOC–COOH, ethanedioic acid
ester–COO–alkyl …-oateCH3COOCH2CH3, ethyl ethanoate
acyl chloride–COCl-oyl chlorideCH3COCl, ethanoyl chloride
amide–CONH2-amideCH3CONH2, ethanamide; CH3CONHCH3, N-methylethanamide
nitrile–C≡N-nitrileCH3CH2CN, propanenitrile

Chirality and optical isomerism

A carbon atom joined to four different groups is a chiral centre (an asymmetric carbon). The four groups are arranged tetrahedrally, and there are two ways of doing that which are mirror images of each other. No amount of turning makes one fit exactly on top of the other: they are non-superimposable mirror images, like your left and right hands. The two forms are enantiomers, or optical isomers.

How reaction mechanisms decide optical activity

Whether a reaction gives one enantiomer or a racemic mixture depends on the shape of the particle that the attacking species meets.

Aldehydes and ketones

In the C=O group oxygen is much more electronegative than carbon, so the bond is polar, Cδ+=Oδ−. The electron-poor carbon is attacked by nucleophiles, unlike the C=C in alkenes, which attracts electrophiles. The C=O bond is also stronger than a C=C bond, so carbonyls do not take part in the electrophilic addition reactions of alkenes.

Oxidation: telling aldehydes from ketones

An aldehyde has a hydrogen on its carbonyl carbon and is easily oxidised to a carboxylic acid. A ketone has no such hydrogen, and mild oxidising agents leave it alone. Primary alcohols oxidise to aldehydes and then acids, secondary alcohols to ketones, and tertiary alcohols not at all.

Reagent and conditionsWith an aldehydeWith a ketone
potassium dichromate(VI) and dilute sulfuric acid, heat under refluxorange Cr2O72− turns green (Cr3+)stays orange
Fehling's or Benedict's solution (blue Cu2+ complex), warmbrick-red precipitate of copper(I) oxide, Cu2Ostays blue
Tollens' reagent ([Ag(NH3)2]+, made from silver nitrate and aqueous ammonia), warm in a water bathsilver mirror forms on the inside of the tubeno change

Full equations, with ethanal being oxidised to ethanoic acid in each case:

Reduction with lithium tetrahydridoaluminate

LiAlH4 in dry ether, at room temperature, reduces aldehydes to primary alcohols and ketones to secondary alcohols. Write the reducing agent as [H]:

Nucleophilic addition of hydrogen cyanide

HCN adds across the C=O bond to give a hydroxynitrile, adding one carbon to the chain. The reagent is HCN with some potassium cyanide, at room temperature; HCN itself barely dissociates, so the KCN supplies enough cyanide ions, the actual nucleophile, to start the reaction.

Oδ− Cδ+ H₃C H ⁻:CN :O⁻ C NC H₃C H H–CN OH C NC H₃C H + ⁻CN 2-hydroxypropanenitrile ethanal intermediate (alkoxide ion)
Nucleophilic addition. The lone pair on the cyanide ion attacks the δ+ carbon while the C=O π electrons move onto the oxygen. The negative oxygen of the intermediate then takes a proton from HCN, which regenerates a cyanide ion.

The carbonyl group is planar, so CN− can attack from either face. With ethanal, or any unsymmetrical ketone, the product carbon is chiral and the two faces give the two enantiomers equally: the product is a racemic mixture with no optical activity.

Two more tests

Carboxylic acids

Making carboxylic acids

Reactions of carboxylic acids

ReagentProductEquation
LiAlH4 in dry etherprimary alcoholCH3CH2COOH + 4[H] → CH3CH2CH2OH + H2O
reactive metalsalt + hydrogen2CH3COOH + 2Na → 2CH3COONa + H2
alkalisalt + waterCH3COOH + NaOH → CH3COONa + H2O
carbonate or hydrogencarbonatesalt + water + CO22CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
phosphorus(V) chloride, room temperatureacyl chlorideCH3COOH + PCl5 → CH3COCl + POCl3 + HCl
alcohol + concentrated H2SO4 catalyst, heatester + waterCH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

Acyl chlorides

Acyl chlorides are far more reactive than carboxylic acids. The carbonyl carbon carries two electronegative atoms, O and Cl, making it strongly δ+, and Cl− is a good leaving group. They react with nucleophiles at room temperature, quickly and completely, and every reaction gives off steamy fumes of HCl (or makes a chloride salt if a base is present).

Reacts withProductEquation
watercarboxylic acidCH3COCl + H2O → CH3COOH + HCl
alcoholesterCH3COCl + CH3CH2OH → CH3COOCH2CH3 + HCl
concentrated ammoniaprimary amideCH3COCl + 2NH3 → CH3CONH2 + NH4Cl
primary amineN-substituted amideCH3COCl + 2CH3NH2 → CH3CONHCH3 + CH3NH3+Cl−

With ammonia and amines, the HCl formed is neutralised by a second molecule of the base, so two are needed; with ammonia you see white smoke of ammonium chloride. The reaction with an alcohol is a better way of making esters than using the carboxylic acid, because it is fast and goes to completion.

Esters and polyesters

Condensation polymers: polyesters

In condensation polymerisation monomers join and a small molecule, usually water or HCl, is lost at each link. To build a chain, each monomer needs a reactive group at both ends. A polyester forms from:

Terylene (PET), used for clothing fibres and drinks bottles, is made from benzene-1,4-dicarboxylic acid and ethane-1,2-diol. Its repeat unit is –OC–C6H4–COO–CH2CH2–O–. To find the repeat unit of any polyester, take one of each monomer, remove the atoms lost as water or HCl, and join them through the ester link, with bonds left open at both ends.

Polyesters contain polar C=O and C–O bonds that can be attacked, so unlike poly(alkenes) they can be hydrolysed: by acid, giving back the dicarboxylic acid and the diol, or by alkali, giving the diol and the salt of the acid. This makes them biodegradable.

Spectroscopy: what radiation does to molecules

RadiationEffect on moleculesUsed for
radio waves (in a strong magnetic field)change the spin state of certain nuclei, such as 1H and 13CNMR spectroscopy and MRI body scanners
microwavesmake polar molecules rotate fasterheating food, which contains water
infraredmake bonds vibrate more stronglyIR spectroscopy to identify bonds
ultravioletcan break bonds, for example C–Cl or Cl–Cl, into radicalsstarting radical reactions

NMR spectroscopy

Nuclei with an odd number of protons or neutrons, such as 1H and 13C, behave like tiny magnets. In a strong magnetic field they can line up with or against the field, and radio waves of exactly the right frequency flip them between the two. The frequency needed depends on the chemical environment of each nucleus: electronegative atoms nearby pull electron density away and shift the signal. This shift is measured on the δ scale in parts per million (ppm), relative to a reference.

13C NMR

Each different carbon environment gives one peak, so the number of peaks is the number of carbon environments. Carbons near electronegative atoms or in C=O groups appear at higher δ. Symmetry matters: propanone has three carbons but only two environments, so two peaks.

Type of carbonApproximate δ / ppm
C–C (alkyl)5 to 40
C–O, C–Cl, C–N40 to 90
C=C, aromatic carbons90 to 150
C=O in acids and esters160 to 185
C=O in aldehydes and ketones190 to 220

High resolution 1H NMR

A proton NMR spectrum gives four pieces of information:

  1. Number of peaks = the number of different hydrogen environments. Hydrogens in the same environment are equivalent; ethanol, CH3CH2OH, has three.
  2. Chemical shift, δ, tells you what is near each hydrogen (table below).
  3. Relative peak area (the integration trace) is proportional to the number of hydrogens in each environment: 3 : 2 : 1 for ethanol.
  4. Splitting: each peak is split by the hydrogens on the neighbouring carbon atoms. A peak with n neighbouring hydrogens splits into n + 1 lines (the n + 1 rule). Equivalent hydrogens do not split each other, and O–H protons normally appear as a singlet that does not split its neighbours.
Type of hydrogenApproximate δ / ppm
R–CH30.7 to 1.2
R–CH2–R1.2 to 1.4
R–OH (alcohol)0.5 to 5.0, variable
H–C next to C=O (H–C–C=O)2.0 to 2.6
H–C next to O (H–C–O, alcohols, ethers, esters)3.3 to 4.3
H–C=C (alkenes)4.5 to 6.0
aldehyde, R–CHO9.0 to 10.0
carboxylic acid, R–COOH10.0 to 12.0
Neighbouring H atoms (n)01234
Lines in the peak (n + 1)1, singlet2, doublet3, triplet4, quartet5, quintet
Relative line heights11 : 11 : 2 : 11 : 3 : 3 : 11 : 4 : 6 : 4 : 1

The ethyl group, CH3CH2–, is the pattern to recognise: a triplet of area 3 (the CH3, next to a CH2) and a quartet of area 2 (the CH2, next to a CH3). A singlet means no hydrogens on the neighbouring carbons, for example a CH3 joined to a C=O or an O.

Mass spectrometry

In the mass spectrometer an electron is knocked off each molecule, forming the molecular ion, M+, which is both a positive ion and a radical:

M → [M]+• + e−

The molecular ion peak is the one with the highest m/z (ignoring the small M+1 peak from 13C), and it gives the relative molecular mass. Many molecular ions break up, or fragment: a covalent bond breaks to give a positive ion, which is detected, and a radical, which is not: [M]+• → X+ + Y•. The more stable the ion, the taller its peak; acylium ions, R–C≡O+, and secondary or tertiary carbocations are especially common.

m/zCommon fragment ion
15CH3+
29C2H5+ or CHO+
43CH3CO+ or C3H7+
57C2H5CO+ or C4H9+

Both are C5H10O, so both give the molecular ion at m/z 86.

Pentan-3-one, CH3CH2COCH2CH3, breaks next to the C=O to give [CH3CH2CO]+ at m/z 57 (and C2H5+ at 29): [CH3CH2COCH2CH3]+• → CH3CH2CO+ + •CH2CH3.

Pentan-2-one, CH3COCH2CH2CH3, gives a strong peak at m/z 43 from CH3CO+ and one at 71 from CH3CH2CH2CO+, but no peak at 57.

The same molecular ion but different fragments: fragmentation patterns tell isomers apart.

Chromatography

Chromatography separates the components of a mixture between a mobile phase, which moves, and a stationary phase, which does not. Each component is shared between the two phases to a different extent: one that dissolves better in the mobile phase, or is held less strongly by the stationary phase, travels faster. A solid stationary phase separates by adsorption; a liquid stationary phase separates by relative solubility.

Putting the techniques together

A compound contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass. Divide by Ar: C 54.5 ÷ 12 = 4.54, H 9.1 ÷ 1 = 9.1, O 36.4 ÷ 16 = 2.28. Divide by the smallest: C : H : O = 2 : 4 : 1, so the empirical formula is C2H4O (mass 44).

The mass spectrum has its molecular ion at m/z 88, so the molecular formula is C4H8O2.

The IR spectrum has a strong absorption at 1740 cm−1 (C=O) but no broad absorption at 2500 to 3300 cm−1, so there is no O–H: the compound is an ester, not a carboxylic acid.

The 1H NMR spectrum shows three peaks: δ 4.1, quartet, area 2; δ 2.0, singlet, area 3; δ 1.3, triplet, area 3.

The quartet and triplet together are an ethyl group, and the CH2 at δ 4.1 must be joined to oxygen (O–CH2CH3). The singlet of area 3 at δ 2.0 is a CH3 next to C=O with no neighbouring hydrogens (CH3CO–).

The compound is ethyl ethanoate, CH3COOCH2CH3. Check: its 13C NMR spectrum would show four peaks, including one near 171 ppm for the ester C=O.

Quick checks

Which of these molecules has a chiral centre?

In butan-2-ol, carbon 2 carries H, OH, CH3 and C2H5: four different groups. In propan-2-ol the two CH3 groups are the same.

A single enantiomer of 2-bromobutane reacts with OH− by an SN2 mechanism. What is the product?

SN2 attack comes from the side opposite the bromine in one step, inverting the centre. There is no carbocation; a racemate would point to SN1.

Which reagent gives a positive result with propanal but not with propanone?

Only the aldehyde is oxidised, reducing Ag+ to a silver mirror. 2,4-DNPH reacts with both; the iodoform test is positive for propanone (a methyl ketone) but not propanal.

Why is the product of HCN addition to ethanal optically inactive?

2-hydroxypropanenitrile does have a chiral centre, but both enantiomers form in equal amounts, so their rotations cancel.

Which reaction is the best way to make ethyl ethanoate quickly and completely?

Acyl chlorides react with alcohols at room temperature, quickly and irreversibly. Acid-catalysed esterification with the acid is slow and reaches equilibrium.

How many peaks are there in the 1H NMR spectrum of methyl propanoate, CH3CH2COOCH3, and what is the splitting of the CH3 joined to oxygen?

There are three environments: CH3 of the ethyl group (triplet), CH2 (quartet) and O–CH3. The O–CH3 is a singlet because its neighbour across the oxygen is the C=O carbon, which carries no hydrogen.

Which is NOT a reason for using TMS as the reference in NMR?

TMS is added in small amounts as a reference and gives a peak at δ 0. Dissolving the sample without adding peaks is the job of a solvent such as CDCl3.

On a TLC plate the solvent front moved 8.0 cm and a spot moved 2.6 cm. What is its Rf value?

Rf = 2.6 ÷ 8.0 = 0.325 ≈ 0.33. Rf is always less than 1; 3.1 is the ratio upside down.

Interactive diagrams

Turn the mirror image, then tick the box to lay it over the original. However you turn it, no more than two of the four groups ever line up.
Drag the slider to change the proportion of the two enantiomers and watch the plane of the emerging light turn. At 50 : 50, a racemic mixture, there is no rotation.
Click a peak, or a labelled group in the formula, to see which hydrogens cause it. Check that the number, position, area and splitting of the peaks fit the structure, then switch to the 13C spectrum.