Home / A-Level Chemistry 9701 / Topic 37 / Revision notes
Download PDFQuestion bank →Topic 37 lessons

Analytical techniques (A Level)

A-Level Chemistry 9701 · Topic 37 revision notes

Everything from the 3 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.

  1. Chromatography
  2. Proton NMR spectroscopy
  3. Carbon-13 NMR spectroscopy

1Chromatography

Syllabus 37.1

All chromatography separates a mixture between a stationary phase, which stays put, and a mobile phase, which moves through it. A component that interacts more strongly with the stationary phase moves more slowly. In thin-layer chromatography (TLC) the stationary phase is silica or alumina on a plate and separation is mainly by adsorption; each spot has an Rf value = distance moved by the spot ÷ distance moved by the solvent front. In gas-liquid chromatography (GLC) an inert carrier gas carries the vapourised sample through a column coated with a non-volatile liquid, and separation is by partition. Each component has a characteristic retention time, and the area under each peak is proportional to the amount of that component, so relative peak areas give the percentage composition.

TLC plate: solvent front 8.0 cm above the baseline baseline (pencil) solvent front A B mix std 8.0 5.6 2.0 Rf = spot ÷ front A: 2.0 ÷ 8.0 = 0.250 B: 5.6 ÷ 8.0 = 0.700 B has the same Rf as the standard
Fig. 1: measuring Rf values on a developed TLC plate

Two phases, two numbers.

  • Rf = spot distance ÷ solvent front distance, both from the baseline.
  • TLC: adsorption onto silica or alumina; GLC: partition into a liquid.
  • Retention time identifies a GLC peak, under fixed conditions.
  • Peak area ÷ total area × 100 gives percentage composition.

Watch out!

Rf is spot distance ÷ solvent front distance, never ÷ the length of the plate, and both are measured from the baseline. In GLC the amount of a component is given by the peak area, not its height, and the percentage uses the total of all peak areas. A long retention time means strong interaction with the stationary phase, not a large amount of substance.

2Proton NMR spectroscopy

Syllabus 37.2

In a magnetic field, 1H nuclei absorb radio waves at frequencies that depend on their electron surroundings. The chemical shift, δ in ppm, is measured relative to TMS, Si(CH3)4, set at 0. The number of peaks equals the number of different proton environments, and the relative peak areas (integration) give the ratio of hydrogens in each. Each peak is split by the hydrogens on neighbouring carbons according to the n+1 rule: n neighbours give n+1 lines, so a singlet, doublet, triplet or quartet. Samples are dissolved in a deuterated solvent such as CDCl3, which gives no proton signal, and shaking with D2O makes O–H and N–H peaks disappear.

¹H NMR spectrum of ethanol, CH₃CH₂OH 0 1 2 3 4 5 6 7 8 9 10 chemical shift, δ / ppm (TMS = 0) CH₂ 2H, q OH 1H, s CH₃ 3H, t TMS s = singlet, t = triplet, q = quartet areas 2 : 1 : 3; OH peak vanishes with D₂O
Fig. 1: the high-resolution proton NMR spectrum of ethanol

Peaks, shift, area, splitting.

  • Number of peaks = number of H environments.
  • Shift (Data Booklet) = type of H; TMS is 0.
  • Area ratio = number of H in each environment.
  • n+1 rule: n H on adjacent carbons give n+1 lines; O–H vanishes with D2O.

Watch out!

Splitting comes from H on the adjacent carbon, not from the H in the group itself: a CH3 next to a CH2 is a triplet, not a quartet. Peak area gives the number of H; the number of lines does not. The O–H peak is usually a singlet and is the one that disappears with D2O. TMS is at 0 and δ increases to the left.

3Carbon-13 NMR spectroscopy

Syllabus 37.2

Only about 1.1% of carbon atoms are 13C, but they absorb radio waves in a magnetic field just as protons do. The number of peaks equals the number of different carbon environments, so symmetry matters: propan-2-ol gives two peaks, propan-1-ol three. The chemical shift range is wide, about 0 to 220 ppm, and tells you the type of carbon: alkyl 5–55, C–O 50–90, alkene and aromatic 110–160, C=O 160–220. Routine spectra show no splitting, so every peak is a single line, and peak heights do not reliably count carbons. Combined with IR (functional groups) and mass spectra (Mr), it pins down a structure.

¹³C NMR spectrum of ethyl ethanoate CH₃COOCH₂CH₃: four carbon environments, four peaks 0 20 40 60 80 100 120 140 160 180 200 chemical shift, δ / ppm C=O (ester) 171 C–O 60 21 14 alkyl carbons no splitting: one line per environment
Fig. 1: the carbon-13 NMR spectrum of ethyl ethanoate

Environments, ranges, then combine.

  • Peaks = carbon environments; use symmetry.
  • Ranges: alkyl 5–55, C–O 50–90, C=C and aromatic 110–160, C=O 160–220.
  • No splitting, and heights do not count carbons.
  • Combine: MS for Mr, IR for groups, NMR for the skeleton.

Watch out!

Count environments, not carbon atoms: the two CH3 groups of propanone give one peak. 13C peaks are not split and their heights do not give the number of carbons, unlike proton peak areas. A carbon bonded to O sits at 50–90, while a C=O carbon is far further left at 160–220; do not confuse the two.

Back to A-Level Chemistry 9701 contents · (c) 2026 TheLucidSTEM · thelucidstem.com