37. Analytical techniques
- Syllabus
- 9701–2028–2029
- Section
- 37
- Level
- A2

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Topic 37.1
The stationary phase stays fixed (for example alumina on a solid support); the mobile phase is the solvent that moves through it. A sample travels from the baseline, while the solvent front marks the furthest solvent position.
A compound repeatedly partitions between phases according to polarity and intermolecular attraction. Its Rf is distance travelled by the spot divided by distance travelled by the solvent front.
If a spot moves 3.0 cm and the solvent front moves 6.0 cm, Rf = 0.50. Measure both distances from the baseline.
Rf has no units and should be between 0 and 1; do not measure from the edge of the paper or compare values from different solvent systems without care.
Rf is the distance travelled by a solute spot divided by the distance travelled by the solvent front, with both distances measured from the baseline.
Rf has no units and lies between 0 and 1. It helps compare a substance with a reference only when the stationary phase, solvent, temperature and measurement method are the same.
A spot that travels 2.4 cm while the solvent front travels 8.0 cm has Rf = 0.30. A second spot at 6.4 cm has Rf = 0.80, not 0.80 cm.
Do not measure from the paper edge or use the solvent-front distance as the denominator for one spot and the baseline distance for another.
A solute travels farther when it is relatively soluble in the mobile phase and interacts weakly with the stationary phase. Strong attraction to the stationary phase holds it back and lowers Rf.
Polarity, hydrogen bonding and the solvent composition control the balance. “More soluble” means relative to the other phase, not simply soluble in an absolute sense.
A polar compound may remain closer to polar alumina while a less strongly adsorbed compound moves with the solvent and gives a larger Rf.
A larger Rf does not mean a larger molecule or a higher boiling point; it describes phase interactions in that chromatographic system.
Topic 37.2
In gas chromatography the stationary phase is a high-boiling, non-polar liquid held on a solid support. The mobile phase is an unreactive carrier gas. Retention time is the time from injection to the detector signal.
Compounds repeatedly partition between the gas and stationary liquid. The instrument records when each compound emerges, not its Rf.
A compound with weak attraction to the stationary liquid spends more time in the carrier gas and reaches the detector sooner.
The carrier gas is not the sample and retention time is not a boiling point, although volatility can influence how long a compound remains in the column.
In a gas chromatogram, each resolved peak represents a component and its retention time helps identify it. The peak area is proportional to the amount detected, so relative areas estimate percentage composition when detector response factors are comparable.
Add the relevant peak areas, then divide each area by the total area and multiply by 100. Do not use peak height unless the method specifically validates it.
Peak areas 20, 30 and 50 give estimated compositions of 20%, 30% and 50%.
A larger retention time does not mean a larger percentage; time identifies the component, while area estimates amount.
Retention time is longer when a compound spends more time dissolved in or adsorbed by the stationary phase rather than travelling with the carrier gas.
Volatility and intermolecular interactions both matter. Compare retention times only under the same column, temperature program and carrier-gas conditions.
Two compounds with similar injection amounts can have different retention times because one has stronger attraction to the non-polar stationary liquid.
Retention time is not a universal identity constant: changing the column or temperature can shift it.
Topic 37.3
A carbon-13 NMR spectrum has one signal for each chemically distinct carbon environment, ignoring usually small isotope effects. Symmetry can make several carbons equivalent and reduce the signal count.
Count environments in the structure, then compare the expected number and approximate chemical shifts with the spectrum. Use shift regions as support, not as a substitute for connectivity.
Propane gives two carbon environments because the two terminal CH₃ groups are equivalent; propanone gives two environments for the methyl carbons and the carbonyl carbon.
The number of carbon atoms is not automatically the number of signals, and signal intensity is not a reliable direct carbon count in routine 13C spectra.
The number of carbon-13 NMR signals equals the number of chemically distinct carbon environments in the molecule. Symmetry and rapid equivalence reduce the count.
Label carbons, compare their attached groups and surroundings, then merge positions related by a symmetry operation. Do not count equivalent carbons twice.
Butan-2-one has four carbons but three environments: the terminal methyl groups are not equivalent because one is next to C=O and the other is not.
Signal count is a structural count, not a carbon atom count and not a direct measure of signal height.
Topic 37.4
A proton NMR spectrum reveals different proton environments. The integration trace gives relative numbers of protons, and splitting shows coupling to neighbouring non-equivalent protons.
Build a structure by combining three clues: number of signals, relative integrals and chemical-shift regions. Use the n+1 rule for simple adjacent proton sets, then check the total hydrogen count.
An ethyl group gives a triplet for CH₃ and a quartet for CH₂ with an integral ratio 3:2; an isolated OH may appear as a broad singlet.
A quartet does not mean four protons: it usually means one proton set is split by three neighbouring protons.
Chemical shift depends on the proton’s electronic environment: electronegative atoms and π systems usually deshield nearby protons. Splitting depends on neighbouring non-equivalent hydrogens.
Assign each proton set to a shift region, then count adjacent hydrogens and apply n+1 only when the sets are sufficiently equivalent for the simple model.
Protons next to an oxygen appear downfield from an ordinary alkyl group; a CH₂ next to CH₃ is commonly split into a quartet while the CH₃ is split into a triplet.
Do not predict splitting from the total number of hydrogens in the molecule or treat exchangeable OH/NH protons as fixed n+1 partners.
Tetramethylsilane (TMS) is used as the reference compound assigned δ = 0 ppm in NMR. Sample peaks are reported relative to this standard, making chemical shifts comparable.
TMS gives one sharp signal because its twelve protons (and four carbons in 13C NMR) are equivalent. It is chemically inert and easy to remove from the spectrum.
A proton signal at 2.1 ppm lies 2.1 ppm downfield from TMS under the same instrument conditions.
The TMS peak is not a sample impurity to interpret as an unknown structure; it defines the scale.
A deuterated solvent such as CDCl₃ replaces most solvent hydrogen atoms with deuterium, which is not detected in ordinary proton NMR in the same way. This prevents a huge solvent ¹H signal masking the sample.
The solvent must dissolve the sample and be sufficiently non-reactive. Small residual protonated-solvent peaks can still appear and should not be mistaken for the compound.
Dissolving an organic sample in CDCl₃ allows its proton signals to be observed while the deuterated solvent also provides a lock signal for the instrument.
Deuterated does not mean proton-free in an absolute sense; residual solvent peaks and exchangeable protons may remain visible.
Exchangeable O–H and N–H protons can be replaced by deuterium when D₂O is added. Because deuterium is not normally observed in a proton NMR spectrum, the corresponding signal disappears or weakens.
Compare spectra before and after adding D₂O. A disappearing broad signal supports an O–H or N–H assignment, but it does not by itself distinguish the two.
An alcohol spectrum with a broad signal that vanishes after D₂O treatment provides evidence for an O–H proton.
Do not expect ordinary C–H peaks to disappear; their hydrogens do not exchange rapidly with D₂O under the test conditions.