← Unit 4: How are organic compounds categorised, analysed and used?
How can organic compounds be analysed?
the principles and interpretation of proton (^1H) and carbon-13 (^13C) NMR spectroscopy (chemical shift, integration, n+1 splitting and number of carbon environments) and high performance liquid chromatography (HPLC, retention time and calibration curves) for the identification and quantification of organic compounds
A focused VCE Chemistry Unit 4 answer on proton and carbon-13 NMR, and HPLC. Covers TMS reference and chemical shift, number of environments, the n+1 splitting rule with examples, integration, ^13C NMR for counting carbon environments, and HPLC retention time with quantitative calibration curves.
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What this dot point is asking
VCAA wants you to interpret ^1H NMR (number of environments, chemical shift, integration, n+1 splitting) and ^13C NMR (number of carbon environments), and to describe how HPLC identifies components by retention time and quantifies them by using calibration curves.
The answer
Proton (^1H) NMR: what it tells you
A proton NMR spectrum gives four pieces of information for each set of equivalent hydrogens (each "environment") in the molecule:
1. The number of signals = the number of unique hydrogen environments. Hydrogens are equivalent (one environment) if you can interchange them by symmetry of the molecule. CH3CH2OH has 3 environments (the 3 CH3 protons, the 2 CH2 protons, the 1 OH proton).
2. The chemical shift (delta, in ppm) reflects the local electronic environment. More electronegative neighbours (O, N, halogens, C=O) shift the signal downfield (higher ppm). Reference compound is TMS (tetramethylsilane), set at 0 ppm.
Approximate chemical shift ranges to know:
| Environment | delta (ppm) |
|---|---|
| -CH3 (alkane) | 0.9 |
| -CH2- (alkane) | 1.3 |
| -CH3-CO- (alpha to C=O) | 2.1 |
| -CH2-Cl | 3.5 to 4.0 |
| -O-CH2- | 3.7 |
| -O-CH3 | 3.4 to 3.9 |
| -CHO (aldehyde proton) | 9 to 10 |
| -COOH (carboxylic acid proton) | 10 to 12 |
| -OH (alcohol) | 1 to 5, variable, often broad |
| -NH (amine) | 0.5 to 4, variable |
| Aromatic -CH= | 6 to 8 |
| Alkene =CH- | 5 to 6 |
3. Integration (peak area) is proportional to the number of hydrogens in that environment. A 3:2 integration ratio in an ethyl group corresponds to CH3 : CH2.
4. Splitting (multiplicity) follows the n+1 rule: a hydrogen environment with n equivalent neighbouring hydrogens on adjacent carbons appears as an (n+1)-multiplet. So:
- 0 neighbours: singlet.
- 1 neighbour: doublet.
- 2 neighbours: triplet.
- 3 neighbours: quartet.
- 4 neighbours: pentet, etc.
Hydrogens on the same carbon do not split each other if they are equivalent. Hydrogens on -OH and -NH typically do not split (exchange happens quickly) and appear as broad singlets.
Carbon-13 (^13C) NMR
Detects the natural ^13C isotope (1.1% abundance) instead of ^1H. The output is simpler:
- The number of signals = number of carbon environments. CH3CH2OH has 2 carbon environments.
- Chemical shifts span 0 to 220 ppm. Useful ranges: alkane C around 10 to 40, alcohol C-O around 50 to 80, aromatic C around 110 to 150, carbonyl C around 165 to 215 (ester/acid 165 to 180, ketone/aldehyde 190 to 215).
- Splitting is not used in routine ^13C NMR (protons are decoupled out; each carbon appears as a singlet).
- Integration is unreliable and usually not used.
The main use of ^13C NMR in VCE is counting carbon environments to distinguish isomers.
How to interpret a ^1H NMR spectrum (workflow)
- Count the number of signals = number of H environments.
- Read each chemical shift and look it up in your table to suggest the type of H.
- Read the integration to get the ratio of Hs in each environment. Scale to match the molecular formula.
- Read the splitting and use n+1 to count neighbours.
- Combine all of the above to propose a structure consistent with the data and with the molecular formula.
A useful starting trick: the molecular formula plus the count of environments narrows the candidates fast.
High performance liquid chromatography (HPLC)
HPLC separates components of a liquid mixture for both identification and quantification.
Principle. A liquid (the mobile phase) is pumped at high pressure (5 to 400 bar) through a column packed with a solid (the stationary phase). The sample is injected, and components travel through the column at rates that depend on their partition between the two phases. More polar species in a non-polar (reversed-phase) column move quickly; less polar species spend more time on the stationary phase and move slowly.
A detector (UV-vis absorbance, fluorescence, or mass spectrometry) at the column exit gives a chromatogram: detector response (y) versus time (x). Each component appears as a peak at its characteristic retention time Rt.
Use 1: identification. Compare the retention time of an unknown peak to that of a known standard run under the same conditions. Matching Rt is consistent with the same compound (but not proof; many compounds may share an Rt).
Use 2: quantification. The area under the peak is proportional to the amount of that component injected. A calibration curve of peak area against known concentrations of pure standard converts the unknown's peak area into a concentration.
Calibration curve procedure.
- Prepare a series of standards of pure analyte at known concentrations.
- Run each through HPLC under identical conditions and record the peak area at the analyte's Rt.
- Plot peak area (y) versus concentration (x); expect a linear relationship at low concentration.
- Run the unknown sample; measure the peak area at the same Rt.
- Read the concentration from the calibration line.
Strengths: separation and quantification of complex mixtures, can detect small amounts, works for non-volatile species that gas chromatography cannot handle.
Limitations: similar polarity compounds may co-elute, sample preparation can be involved, expensive instrument.
Worked example
A pure organic compound C3H6O gives a ^1H NMR spectrum with two signals:
- Signal A at 2.1 ppm, singlet, integration 6.
- Signal B at 9.6 ppm... wait, there is only ONE signal in this spectrum at 2.1 ppm, integration 6 H.
Actually let us re-do this. C3H6O with only one signal in NMR at 2.1 ppm, singlet, integration 6 means all 6 protons are equivalent and adjacent to a carbonyl. The only C3H6O molecule with all 6 H equivalent in this position is propan-2-one (acetone, CH3COCH3): the two methyl groups are equivalent by symmetry, each shifted to 2.1 ppm by the adjacent C=O, and both are singlets because the only "neighbour" is across the C=O carbon (which has no H of its own).
^13C NMR confirmation: propan-2-one has 2 carbon environments (the two equivalent methyl carbons and the carbonyl carbon), so we expect 2 signals: one around 30 ppm (methyl C) and one around 207 ppm (C=O).
Compare to propanal (CH3CH2CHO, also C3H6O): expects 3 ^1H environments (CHO around 9.6, CH2 around 2.4, CH3 around 1.1) and 3 ^13C environments. Not consistent with the one-signal spectrum.
Common traps
Confusing number of H atoms with number of environments. Ethanol has 6 hydrogens but only 3 environments (CH3, CH2, OH).
Forgetting OH and NH do not always split neighbours. They usually appear as broad singlets and do not split adjacent CH protons (rapid exchange).
Reading peak heights instead of peak areas. Always use the integration (area) for the H count, not the height.
Mis-applying the n+1 rule. The "n" is the number of H on adjacent carbons, not on the same carbon. CH2 protons on the same carbon are equivalent and do not split each other.
Forgetting to scale the integration to molecular formula. Integration ratios are relative; multiply by the total H count to get the actual numbers.
Calling HPLC qualitative only. It is both qualitative (Rt identifies) and quantitative (peak area, via calibration, measures concentration).
Saying Rt alone proves identity. Many compounds can share Rt under one set of conditions. Confirm by spiking with the standard or by an orthogonal method (MS, NMR).
In one sentence
Proton NMR identifies each unique hydrogen environment by its chemical shift, integration and n+1 splitting; ^13C NMR counts the number of carbon environments; and HPLC separates a liquid mixture so each component elutes at its own retention time with peak area proportional to amount, used together with a calibration curve to quantify a specific analyte.
Past exam questions, worked
Real questions from past VCAA papers on this dot point, with our answer explainer.
2024 VCE4 marksAn unknown ester (C4H8O2) gives a ^1H NMR spectrum with three signals. (a) Predict the chemical shifts and splitting patterns for ethyl ethanoate, CH3COOCH2CH3. (b) Use your answer to confirm whether the spectrum is consistent with ethyl ethanoate or methyl propanoate (CH3CH2COOCH3).Show worked answer →
A 4-mark answer needs three correctly assigned environments per structure and an n+1 splitting analysis.
(a) Ethyl ethanoate, CH3COOCH2CH3 has 3 hydrogen environments:
- CH3-CO- (a methyl next to a carbonyl): chemical shift ~2.0 ppm, singlet (no adjacent H), integration 3.
- -OCH2- (a methylene next to O and CH3): chemical shift ~4.1 ppm, quartet (3 neighbouring H on the adjacent CH3, n+1 = 4), integration 2.
- -CH2-CH3 (the terminal methyl of the ethyl): chemical shift ~1.3 ppm, triplet (2 neighbouring H on CH2, n+1 = 3), integration 3.
(b) Methyl propanoate, CH3CH2COOCH3 has 3 environments:
- CH3-CH2- (the methyl of the ethyl): ~1.1 ppm, triplet, integration 3.
- -CH2-CO- (a methylene next to C=O): ~2.3 ppm, quartet, integration 2.
- -OCH3 (a methoxy next to O): ~3.7 ppm, singlet, integration 3.
The spectrum will distinguish them by: ethyl ethanoate has a singlet at ~2.0 and a quartet at ~4.1; methyl propanoate has a singlet at ~3.7 and a quartet at ~2.3. Whichever singlet appears at the higher chemical shift identifies the isomer (~3.7 for methyl propanoate; ~2.0 for ethyl ethanoate).
2025 VCE3 marks(a) Outline how HPLC separates compounds in a mixture. (b) Describe how a calibration curve is used to determine the concentration of caffeine in a soft-drink sample.Show worked answer →
A 3-mark answer needs the separation principle, the role of retention time, and the calibration step.
(a) HPLC separation: a liquid sample is injected into a stream of solvent (mobile phase) that is pumped under high pressure through a column packed with a solid stationary phase. Different components have different affinities for the stationary phase, so they pass through at different rates. Each component elutes at its characteristic retention time Rt. A detector (often UV absorbance) at the column exit gives a peak for each component, with the area under the peak proportional to the amount.
(b) Calibration curve: prepare a series of standards of pure caffeine in the same solvent at known concentrations spanning the expected range. Run each through HPLC under the same conditions; record the peak area of caffeine for each. Plot peak area (y) against concentration (x). Run the soft-drink sample, measure the caffeine peak area, and read the corresponding concentration off the calibration line.
Markers also accept that the retention time identifies caffeine and the peak area quantifies it; the two functions are separate.
Related dot points
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