Analytical techniques

All four subtopics here are printed under "A Level subject content" in the 9701 syllabus and every objective carries the tier "A Level". None of it is AS, and it is a separate topic from topic 22 of the same name, whose objectives are all tiered AS and cover infrared spectroscopy and mass spectrometry. If you are working at AS, topic 22 is yours and this one is not. This one is examined on Paper 4, and Paper 1 is the AS multiple-choice paper, so no practice on this site is tagged to this topic.

This is the last topic in the syllabus, and NMR is the largest part of it by examination weight.

Syllabus points

37.1 Thin-layer chromatography

• Describe and understand the terms: stationary phase, for example aluminium oxide on a solid support; mobile phase, a polar or non-polar solvent; Rf value; solvent front and baseline.
• Interpret Rf values.
• Explain the differences in Rf values in terms of interaction with the stationary phase and of relative solubility in the mobile phase.

37.2 Gas/liquid chromatography

• Describe and understand the terms: stationary phase, a high boiling point non-polar liquid on a solid support; mobile phase, an unreactive gas; retention time.
• Interpret gas/liquid chromatograms in terms of the percentage composition of a mixture.
• Explain retention times in terms of interaction with the stationary phase.

37.3 Carbon-13 NMR spectroscopy

• Analyse and interpret a carbon-13 NMR spectrum of a simple molecule to deduce the different environments of the carbon atoms present, and the possible structures for the molecule.
• Predict or explain the number of peaks in a carbon-13 NMR spectrum for a given molecule.

37.4 Proton (¹H) NMR spectroscopy

• Analyse and interpret a proton (¹H) NMR spectrum of a simple molecule to deduce: the different environments of protons present using chemical shift values; the relative numbers of each type of proton present from relative peak areas; the number of equivalent protons on the carbon atom adjacent to the one to which the given proton is attached, from the splitting pattern, using the n+1 rule (limited to singlet, doublet, triplet, quartet and multiplet); the possible structures for the molecule.
• Predict the chemical shifts and splitting patterns of the protons in a given molecule.
• Describe the use of tetramethylsilane, TMS, as the standard for chemical shift measurements.
• State the need for deuterated solvents, for example CDCl₃, when obtaining a proton NMR spectrum.
• Describe the identification of O-H and N-H protons by proton exchange using D₂O.

Thin-layer chromatography

Every form of chromatography separates a mixture by the same principle: the components distribute themselves differently between a stationary phase and a mobile phase that moves over it, so they travel at different speeds.

• Stationary phase: a thin layer of aluminium oxide or silica on a glass or plastic support.
• Mobile phase: a solvent, polar or non-polar, which rises up the plate by capillary action.
• Baseline: the pencil line near the bottom on which the samples are spotted. Drawn in pencil, because ink would itself separate and run.
• Solvent front: the furthest point the solvent reaches before the plate is removed. It must be marked at once, before the solvent evaporates.

The spot must start above the level of the solvent in the tank, otherwise the sample dissolves off the plate into the solvent and nothing separates.

Rf values

Rf = distance moved by the spot / distance moved by the solvent front

both measured from the baseline, and to the centre of the spot.

Worked example. The solvent front travels 8.0 cm from the baseline; a spot travels 3.4 cm.

Rf = 3.4 / 8.0 = 0.425

so Rf is 0.43.

Rf is always between 0 and 1, so an answer greater than 1 means the two distances have been divided the wrong way round. It is a constant for a given compound in a given solvent on a given stationary phase, so a component is identified by running a known reference alongside it and comparing, or by comparing with a tabulated value obtained under the same conditions.

Explaining a difference in Rf

Two effects compete, and both must be named:

• Adsorption onto the stationary phase. A substance that is more strongly attracted to the polar stationary phase spends more time held on it and moves less, giving a lower Rf.
• Solubility in the mobile phase. A substance that is more soluble in the solvent spends more time moving with it, so it travels further and has a higher Rf.

For a polar stationary phase such as aluminium oxide, a polar compound, able to hydrogen bond to it, is strongly adsorbed and has a low Rf. A non-polar compound is weakly adsorbed and, in a non-polar solvent, highly soluble, so it has a high Rf.

A single-sided answer that mentions only solubility, or only adsorption, is a partial answer.

Gas/liquid chromatography

Used for volatile compounds, and capable of separating and quantifying a mixture in one run.

• Stationary phase: a high boiling point, non-polar liquid coated onto an inert solid support, packed into a long coiled column in an oven.
• Mobile phase, also called the carrier gas: an unreactive gas such as nitrogen or helium.
• Retention time: the time between injection and the appearance of the peak at the detector, and it is characteristic of a compound under a given set of conditions.

The sample is injected, vaporised, and carried through the column by the gas. Each component partitions between the gas and the liquid stationary phase, and the detector produces a peak as each emerges.

Explaining retention time

A component with a greater attraction to the stationary phase, or a higher boiling point so that it spends less time in the vapour, dissolves more in the liquid and is held back, giving a longer retention time. A component that is more volatile or less soluble in the stationary phase spends more time in the gas stream and emerges sooner.

Because the stationary phase is non-polar, non-polar components are retained longest, which is the reverse of the TLC case and worth noting deliberately so the two do not merge.

Percentage composition

The area under each peak is proportional to the amount of that component. So:

percentage of a component = (area of its peak / total area of all peaks) × 100

Worked example. A chromatogram shows three peaks with areas 24, 60 and 36 arbitrary units.

Total area:

24 + 60 + 36 = 120

Percentages:

24 / 120 × 100 = 20

60 / 120 × 100 = 50

36 / 120 × 100 = 30

so the mixture is 20 per cent, 50 per cent and 30 per cent of the three components.

Two limitations are worth having ready. Retention time alone can be ambiguous, since two compounds may happen to have similar values, which is why GLC is often coupled to a mass spectrometer. And the detector's response is not identical for every compound, so peak areas give exact percentages only after calibration.

NMR: the shared idea

Nuclei with an odd mass number, including ¹H and ¹³C, behave as tiny magnets. In a strong magnetic field they align either with or against it, and radio waves of the right frequency flip them from the lower to the higher energy state. The frequency absorbed depends on the electron environment around the nucleus, because surrounding electrons shield it from the applied field.

So each chemically distinct environment gives its own peak, at its own chemical shift, δ, measured in ppm against a standard.

Tetramethylsilane, TMS, is that standard, and it is defined as δ = 0. Four reasons it is chosen, and questions ask for them:

• All twelve of its protons are equivalent, so it gives a single sharp peak.
• That peak is well away from the region in which almost all organic protons absorb, so it does not overlap the sample's peaks.
• It is chemically inert and does not react with the sample.
• It is volatile, with a low boiling point, so it is easily removed and the sample recovered.

Deuterated solvents such as CDCl₃ are used because an ordinary solvent's own hydrogen atoms would give large peaks of their own, obscuring the sample. Deuterium, ²H, does not absorb in the proton NMR region, so a deuterated solvent is effectively invisible.

Carbon-13 NMR

Simpler than proton NMR, because there is no splitting to interpret and no integration to read. There is one thing to do: count the environments.

The number of peaks equals the number of chemically different carbon environments. Two carbons are in the same environment if they are equivalent by the symmetry of the molecule.

Worked examples.

• Ethanol, CH₃CH₂OH: the two carbons are different, so 2 peaks.
• Propanone, CH₃COCH₃: the two methyl carbons are equivalent by symmetry, and the carbonyl carbon is different, so 2 peaks, not 3.
• Propan-1-ol, CH₃CH₂CH₂OH: three different carbons, so 3 peaks.
• Benzene, C₆H₆: all six carbons are equivalent, so 1 peak.
• Methylbenzene, C₆H₅CH₃: the methyl carbon, then the ring carbons in four environments (the one bearing the methyl, and the 2, 3 and 4 positions, with 5 and 6 equivalent to 3 and 2), so 5 peaks.
• 1,4-dimethylbenzene: the methyl carbons are equivalent, and the ring has only two environments, so 3 peaks.

Counting the environments in a substituted benzene ring is the step that separates a correct answer from a plausible one, so work round the ring position by position and pair up the ones related by symmetry.

The chemical shift then narrows the identification: a carbon in a C=O group appears far downfield, around 190 to 220 ppm; aromatic carbons around 110 to 160; a carbon attached to oxygen around 50 to 90; and an ordinary alkyl carbon below about 50. Values are given in the Data Booklet, so use it rather than memorising.

Proton NMR

Four features to read, and a full interpretation uses all four.

1. Number of peaks

Equals the number of proton environments, counted by symmetry exactly as for carbon. All three protons of a methyl group are equivalent to one another, so a CH₃ group is always one environment.

2. Chemical shift

Tells you what each set of protons is attached to. Approximate values, but always check the Data Booklet:

3. Relative peak areas, the integration

The area under each peak is proportional to the number of protons in that environment. The trace is usually shown as a stepped line, and the heights of the steps give the ratio.

The ratio is a ratio, not the absolute number. A 2:3 ratio could be 2 and 3 protons, or 4 and 6. Use the molecular formula to decide which.

4. Splitting, and the n+1 rule

A peak is split by the protons on the adjacent carbon atom. If there are n equivalent protons on the neighbouring carbon, the peak is split into n+1 lines.

Read the rule backwards when interpreting a spectrum, which is how it is used: a triplet means there are two protons on the neighbouring carbon, and a quartet means three.

Note that protons in the same environment do not split each other, so the three protons of a methyl group produce a single peak whose splitting is decided entirely by the neighbouring carbon.

O-H and N-H protons do not split, and are not split, because they exchange rapidly between molecules. They appear as singlets, often broad.

The D₂O test

Shake the sample with a few drops of D₂O, heavy water, and run the spectrum again. The O-H and N-H protons exchange with deuterium:

R-OH + D₂O ⇌ R-OD + HOD

Deuterium does not absorb in the proton NMR region, so the peak due to that proton disappears. Any peak that vanishes after adding D₂O was therefore an O-H or N-H proton, and this is the standard way of identifying it, since its chemical shift alone is too variable to be reliable.

Worked interpretation

A compound of formula C₃H₆O₂ gives a proton NMR spectrum with three peaks:

Reading them in turn:

• Three peaks means three proton environments, and the areas 3 : 2 : 1 account for all six hydrogens.
• The peak at 11.5 is far downfield and disappears with D₂O, so it is the O-H of a carboxylic acid. That fits the two oxygens in the formula.
• The peak at 1.2, area 3, is a triplet, so it is a CH₃ with two protons on the adjacent carbon: a CH₃ next to a CH₂.
• The peak at 2.3, area 2, is a quartet, so it is a CH₂ with three protons on the adjacent carbon: a CH₂ next to a CH₃. Its shift also places it next to a C=O.

Putting the fragments together, CH₃-CH₂-COOH, which is propanoic acid. Check the formula: C₃H₆O₂, as given.

That triplet-and-quartet pair, in a 3 : 2 area ratio, is the signature of an ethyl group and appears constantly, so learn to recognise it on sight.

Common mistakes

• Drawing the chromatography baseline in ink, or starting the plate with the spots below the solvent level.
• Dividing the wrong way round in an Rf calculation, giving a value above 1.
• Measuring to the edge of a spot rather than to its centre.
• Explaining a difference in Rf by solubility alone, without mentioning adsorption onto the stationary phase.
• Saying a component with a longer retention time in GLC is more soluble in the mobile phase. It is more soluble in the stationary phase.
• Using peak heights rather than peak areas for percentage composition.
• Counting equivalent carbons or protons as separate environments, so propanone is given three carbon peaks instead of two.
• Miscounting environments in a substituted benzene ring by ignoring its symmetry.
• Applying the n+1 rule to the protons in the peak itself rather than to those on the adjacent carbon.
• Reading a quartet as four neighbouring protons rather than three.
• Expecting an O-H peak to be split, or to appear at a predictable shift.
• Treating an integration ratio as the absolute number of protons without checking against the molecular formula.
• Saying TMS is used because it is cheap, rather than for its single sharp peak outside the normal range, its inertness and its volatility.
• Forgetting that the solvent must be deuterated, so the solvent's own protons would swamp the spectrum.