37. Analytical techniques
- Syllabus
- 9701–2028–2029
- Section
- 37
- Level
- A2
Exam analysis
37.1 Thin-layer chromatography
- Syllabus
- 9701–2028–2029
- Topic
- 37.1
- Level
- A2
A TLC plate separates spots between a fixed solid and a moving solvent
| Term | Meaning in TLC |
|---|---|
| stationary phase | a fixed polar solid such as aluminium oxide on a solid support; solutes adsorb to its surface |
| mobile phase | a polar or non-polar solvent that rises through the stationary phase and carries dissolved solutes |
| baseline | pencil line where sample/reference spots start; placed above the solvent level |
| solvent front | furthest position reached by the mobile phase, marked immediately when the plate is removed |
| Rf | distance travelled by a component from the baseline divided by distance travelled by the solvent front from the baseline |
Rf=distance from baseline to solvent frontdistance from baseline to centre of solute spot
Apply small sample and reference spots to the pencil baseline, allow them to dry, place the plate in a covered container with solvent below the baseline, develop it, then remove the plate and mark the solvent front before evaporation changes its position.
Use pencil because ink could dissolve and separate. If the baseline is submerged, the sample dissolves directly into the solvent reservoir rather than travelling from a defined start.
Calculate Rf, then interpret spots only against the same TLC system
Measure every distance from the same baseline. For a spot that travels 2.4 cm while the solvent front travels 8.0 cm, Rf = 2.4/8.0 = 0.30. Rf has no unit and normally lies from 0 to 1.
| Chromatogram evidence | Supported interpretation |
|---|---|
| sample spot aligns with a reference spot and has the same Rf | sample may contain that reference compound under those identical conditions |
| sample gives several separated spots | sample contains more than one detectable component |
| one expected pure substance gives an extra spot | impurity or an additional component may be present |
| same compound run with a different solvent/stationary phase | Rf may change; do not use the old value as an identity constant |
An Rf match is comparative evidence, not unique proof of identity. Use the same stationary phase, mobile-phase composition, temperature and measurement method, ideally running reference and sample on the same plate.
Do not write cm after an Rf, measure from the plate edge, or identify a compound by comparing values obtained in different chromatographic systems.
Rf reflects competition between stationary-phase attraction and mobile-phase solubility
A solute travels farther when it is relatively more soluble in the mobile phase and interacts more weakly with the stationary phase. Stronger adsorption or intermolecular attraction to the stationary phase holds it back and lowers Rf.
| Relative behaviour | Movement | Rf |
|---|---|---|
| stronger interaction with stationary phase | retained for longer | lower |
| greater relative solubility in mobile phase | carried farther with solvent | higher |
With polar alumina/silica and a relatively non-polar solvent, a solute capable of stronger polar or hydrogen-bonding interactions is often retained more strongly. Changing solvent polarity changes the competition and can change every Rf.
A larger Rf does not mean a larger molecule, higher boiling point or universal lower polarity. It reports the balance of interactions in that specific stationary/mobile phase system.
37.2 Gas / liquid chromatography
- Syllabus
- 9701–2028–2029
- Topic
- 37.2
- Level
- A2
GLC uses a high-boiling liquid stationary phase and an unreactive gas
| Term | Meaning in GLC |
|---|---|
| stationary phase | high-boiling-point, non-polar liquid coated on a solid support inside the column |
| mobile phase | unreactive carrier gas such as helium or nitrogen that moves vaporised sample through the column |
| retention time | time from sample injection until that component reaches the detector and produces its peak |
Volatile sample components distribute between travelling with the carrier gas and interacting with/dissolving in the stationary liquid. Their different residence times separate them before detection.
The chromatogram plots detector response against time. Peak position gives retention time; GLC does not use the TLC Rf ratio.
The carrier gas is the mobile phase, not the sample. Retention time is an experimental time under specified column and operating conditions, not the compound's boiling point.
Use relative peak areas to calculate percentage composition
Each resolved peak represents a detected component. Its retention time locates the component, while its integrated area represents the relative amount detected when response is treated as proportional under the stated method.
% component i=∑areas of all relevant peaksarea of peak i×100
If no integration values are supplied and peaks are approximated as triangles, estimate each area as 1/2 x base width x height. Use the areas consistently; peak height alone can mislead when widths differ.
For integrated areas 20, 30 and 50 arbitrary units, the total is 100 and the estimated percentage compositions are 20%, 30% and 50%.
A later retention time does not mean a larger percentage. Retention time concerns peak position/interaction; relative area concerns composition. Real detector response factors may require calibration, but use the supplied or integrated areas as the exam data specify.
Stronger stationary-phase interaction gives a longer retention time
A component with stronger attraction to or greater solubility in the stationary liquid spends a larger fraction of its journey retained in that phase. It moves with the carrier gas for less time and therefore reaches the detector later.
| Relative stationary-phase interaction | Fraction of time moving with gas | Retention time |
|---|---|---|
| weaker | greater | shorter |
| stronger | smaller | longer |
With the specified non-polar stationary liquid, intermolecular compatibility can make non-polar components interact more strongly. Compare compounds under the same column, temperature and carrier-gas flow because those conditions also shift measured times.
Retention time is not a universal identity constant and cannot be explained from polarity or boiling point alone. The required causal statement is the component's relative interaction with the stationary phase in that actual system.
37.3 Carbon-13 NMR spectroscopy
- Syllabus
- 9701–2028–2029
- Topic
- 37.3
- Level
- A2
Use carbon-13 peak count and shifts together to constrain a structure
To interpret a simple 13C NMR spectrum: 1) count sample peaks to find the number of chemically distinct carbon environments; 2) assign each chemical-shift region to plausible carbon environments; 3) combine those constraints with the molecular formula and other data; 4) draw candidate structures; 5) reject any candidate whose environment count or required shifts disagree.
| Approximate delta / ppm | Carbon environment indicated |
|---|---|
| 0-50 | saturated alkyl carbon |
| 30-70 | saturated carbon near C=O, halogen or O (use the narrower supplied data ranges) |
| 100-125 | nitrile carbon |
| 110-160 | alkene or arene carbon |
| 160-185 | carboxylic acid or ester carbonyl carbon |
| 190-220 | aldehyde or ketone carbonyl carbon |
Propanone gives two sample signals: one in the ketone C=O region for the carbonyl carbon and one in the saturated-carbon region for the two equivalent CH3 carbons. Two peaks therefore represent three carbon atoms in two environments.
Treat TMS at 0 ppm as the reference and ignore a labelled solvent peak such as CDCl3 near 77 ppm. Routine 13C peak heights are not directly proportional to the number of carbons in an environment.
Peak count alone rarely proves one structure, and a shift region alone does not determine connectivity. A proposed structure must satisfy both the number of distinct environments and every chemically plausible peak position.
Predict one carbon-13 peak for each chemically distinct carbon environment
Label every carbon, compare the atoms/groups reached in each direction, and merge only carbons related by genuine molecular symmetry or chemical equivalence. The number of remaining classes is the predicted number of 13C signals.
| Molecule | Carbon atoms | Distinct environments / predicted peaks | Reason |
|---|---|---|---|
| propane | 3 | 2 | the two terminal CH3 carbons are equivalent |
| propanone | 3 | 2 | the two CH3 carbons are equivalent; C=O is distinct |
| propan-2-ol | 3 | 2 | the two CH3 carbons are equivalent; central C-O carbon is distinct |
| butan-2-one | 4 | 4 | carbonyl C, adjacent CH3, CH2 and terminal CH3 all have different surroundings |
| benzene | 6 | 1 | all six ring carbons are symmetry-equivalent |
To explain a reduced peak count, state which carbons are equivalent and identify the symmetry that exchanges them without changing the molecule. Merely saying 'it is symmetrical' is incomplete if the equivalent positions are not identified.
Do not count carbon atoms, hydrogens or peak height. Butan-2-one is not a three-environment molecule: its two terminal CH3 groups are not equivalent, and its CH2 and carbonyl carbon add two further environments, giving four.
37.4 Proton NMR spectroscopy
- Syllabus
- 9701–2028–2029
- Topic
- 37.4
- Level
- A2
Combine shift, integration and splitting to deduce a proton NMR structure
For a simple 1H NMR spectrum: 1) count sample signals for distinct proton environments; 2) use chemical shifts to assign likely local groups; 3) reduce relative peak areas to the simplest proton ratio; 4) use each splitting pattern and n+1 to count equivalent protons on adjacent carbon atoms; 5) assemble candidate fragments; 6) reject any structure that fails the formula, total integral, shift or splitting evidence.
| Spectrum feature | Structural information |
|---|---|
| number of signals | number of chemically distinct proton environments |
| chemical shift, delta / ppm | electronic environment and nearby functional groups/pi systems |
| relative integrated area | relative number of protons in each environment |
| singlet/doublet/triplet/quartet/multiplet | adjacent non-equivalent proton count through n+1 within the simple model |
A 3H triplet paired with a 2H quartet supports CH3-CH2-: CH3 is split by two adjacent protons, while CH2 is split by three. Its shifts then show what group is attached to the CH2 end.
O-H and N-H signals are often broad and exchangeable, and they do not reliably participate in simple n+1 splitting. Confirm them by D2O exchange rather than forcing them into the neighbouring-proton pattern.
Multiplicity is not proton count: a quartet means splitting into four sub-peaks, commonly by three adjacent protons. No single clue proves the structure; all four evidence types must agree.
Predict each proton environment's shift and simple splitting pattern
Label chemically equivalent proton sets. For each set, use its nearest functional/electronic environment to estimate chemical shift, then count non-equivalent protons on directly adjacent carbon atom(s) and apply n+1 within the syllabus's simple splitting model.
| Approximate delta / ppm | Common proton environment |
|---|---|
| 0.9-1.7 | saturated alkyl C-H |
| 2.0-3.0 | C-H next to C=O or aryl/alkene system |
| 3.2-4.0 | C-H next to O or another electronegative atom |
| 4.5-6.0 | alkene H |
| 6.0-9.0 | arene H |
| 9.3-10.6 | aldehyde H |
| about 9-13 | carboxylic-acid O-H; broad/variable |
| Adjacent equivalent protons, n | n+1 result |
|---|---|
| 0 | singlet |
| 1 | doublet |
| 2 | triplet |
| 3 | quartet |
| more/overlapping simple neighbours | multiplet as appropriate |
Equivalent protons do not split each other. Do not count all hydrogens in the molecule, and do not treat exchangeable O-H/N-H protons as reliable fixed splitting partners.
TMS defines zero on the chemical-shift scale
Tetramethylsilane, Si(CH3)4, is added as the NMR reference and assigned delta = 0 ppm. Every sample signal is reported by its shift relative to this standard, allowing spectra recorded under different field strengths to use a common ppm scale.
| Property of TMS | Why it helps |
|---|---|
| all 12 protons are equivalent | gives one sharp, strong signal |
| highly shielded protons | signal lies at 0 ppm, away from most organic sample peaks |
| chemically inert | is unlikely to react with the sample |
| volatile | can be removed readily after measurement |
A sample signal at 2.1 ppm is 2.1 ppm downfield from the TMS reference. The TMS peak defines the axis and is not part of the unknown compound.
Do not assign the TMS peak as a sample environment or impurity. Its job is to provide the zero reference, not structural information about the analyte.
Deuterated solvents avoid a dominant solvent proton signal
An NMR sample must be dissolved, but an ordinary proton-containing solvent would produce a large 1H signal that could mask the sample. A deuterated solvent such as CDCl3 replaces most 1H with 2H, which resonates outside the ordinary proton NMR observation used here.
The solvent must dissolve the sample and not react with it. Deuterated solvent also supports the instrument lock, while small residual protonated-solvent peaks can remain.
Using CDCl3 instead of CHCl3 greatly suppresses the chloroform proton signal so the analyte's proton environments can be observed.
Deuterated does not mean absolutely proton-free. Recognise labelled residual-solvent peaks rather than assigning them to the unknown structure.
D2O exchange identifies O-H and N-H proton signals
Add D2O and compare the proton NMR spectrum before and after. Exchangeable O-H or N-H protons are replaced by deuterium, which is not observed at the same position in ordinary 1H NMR, so their signal disappears or becomes much weaker.
R−OH+DX2OR−OD+HODR−NHX2+DX2OR−NHD+HOD
A disappearing broad signal supports assignment to O-H or N-H. Combine this with chemical structure or other evidence because disappearance alone does not distinguish oxygen from nitrogen.
Ordinary C-H signals should remain: their protons do not exchange rapidly with D2O under the test conditions. Do not treat every changed baseline feature as an exchangeable proton without a before/after peak comparison.