Nuclear magnetic resonance spectroscopy (A-level only) — AQA A-Level Chemistry
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Nuclear magnetic resonance spectroscopy (A-level only) explained
No single instrument hands you a structure, which is why examiners set questions that give three at once.
Read the full explanation
Mass spectrometry gives the relative molecular mass from the molecular ion peak, plus clues from the fragments. Infrared spectroscopy names functional groups: a broad absorption around 3230 to 3550 cm⁻¹ says an alcohol O-H, one near 1680 to 1750 cm⁻¹ says a carbonyl. ¹³C NMR counts the carbon environments. ¹H NMR counts the hydrogen environments, and its integration and splitting say how many hydrogens sit in each and what is next door. Elemental analysis fixes the empirical formula. Put the molecular formula, the functional groups and the environments together and usually only one structure survives.
The use of the δ scale for recording chemical shift.
Chemical shift is quoted on the delta scale in parts per million. Zero on that scale is set by tetramethylsilane, added to the sample, and every other signal is reported as its distance from that peak divided by the operating frequency of the spectrometer. Dividing by the operating frequency is the important part, because it makes delta a ratio: a sample run on a low-field machine gives the same delta values as one run at high field, so the Data Booklet tables work for every instrument. Proton shifts run from roughly 0 to 12 ppm, carbon shifts from roughly 0 to 220 ppm. By convention the scale increases to the left, so a deshielded nucleus is described as downfield and tetramethylsilane sits at the right-hand edge.
Chemical shift depends on the molecular environment.
A nucleus never feels the whole applied field, because the electrons around it circulate and shield it. Anything that pulls electron density away, an oxygen, a halogen, a carbonyl group or a benzene ring, leaves the nucleus less shielded, so it resonates further left at a higher delta value. That is why a methyl group bonded only to carbon appears near 0.7 to 1.2 ppm, a CH₂ group attached to the single-bonded oxygen of an ester appears near 4.1 ppm, a methyl next to a carbonyl near 2.1 to 2.6 ppm, an aldehyde proton near 9.5 ppm and a carboxylic acid O-H near 11 ppm. Protons in identical surroundings give a single peak, so symmetry cuts the number of peaks you actually see.
Students should be able to: explain why TMS is a suitable substance to use as a standard 1 13 use H NMR and C NMR spectra and chemical shift data from the Chemistry Data Booklet to suggest possible structures or part structures for molecules use integration data from H NMR spectra to determine the relative numbers of equivalent protons in the molecule use the n+1 rule to deduce the spin–spin splitting patterns of adjacent, non-equivalent protons, limited to doublet, triplet and quartet formation in aliphatic compounds.
Tetramethylsilane works as the standard because its twelve protons are all equivalent and give one sharp peak; it is unreactive, non-toxic and soluble in organic solvents, it boils at about 27 °C so it is easily removed, and its peak lies to the right of nearly every other signal so nothing overlaps it. Integration gives the ratio of protons in each environment, never the absolute number, so a ratio of three to two to three needs the molecular formula before it becomes a structure. The n+1 rule says a signal is split into one more line than the number of non-equivalent protons on the neighbouring carbons: a quartet means three neighbours, a triplet two, a doublet one. An ethyl group therefore always shows a triplet and a quartet.
Your focus
- State what mass spectrometry, infrared spectroscopy, ¹³C NMR and ¹H NMR each contribute to a structure determination.
- Deduce a structure from combined infrared, ¹³C NMR and ¹H NMR data, quoting a value for every deduction.
- Test a proposed structure against each piece of data given and reject it when one piece disagrees.
Show all 16 objectives
- Given a ¹H NMR spectrum with no integration values and no resolved splitting, state which two deductions can no longer be made from it.
- State the reference compound for the delta scale and the shift assigned to it.
- Explain why chemical shift is quoted in parts per million rather than in hertz.
- Use a shift from a spectrum together with the Data Booklet to name the environment responsible.
- Predict whether two spectrometers of different field strength give the same delta value for a peak, and justify the answer.
- Explain shielding and deshielding in terms of the electron density around a nucleus.
- Predict which of two protons in a molecule appears at the higher chemical shift, and justify the choice.
- Assign a peak at a stated shift to an environment using the Data Booklet.
- Count the proton environments in a given structure and so predict the number of peaks.
- Explain three properties that make tetramethylsilane a suitable standard.
- Deduce the number of protons on a neighbouring carbon from a doublet, triplet or quartet.
- Combine an integration ratio such as 2:2:2:3:3 with the molecular formula to give the actual number of protons in each environment.
- Assemble shift, integration and splitting data into a part structure, then into a full structure that fits the molecular formula.
Nuclear magnetic resonance spectroscopy (A-level only) exam tips
Marking Points
- one mark for using infrared to identify a functional group, quoting the wavenumber range from the Data Booklet
- one mark for using the number of peaks in a ¹³C NMR spectrum to give the number of carbon environments
- one mark for using ¹H NMR integration to give the relative numbers of protons in each environment
- one mark for a final structure consistent with every piece of data quoted, not just with one spectrum
- one mark for stating that tetramethylsilane is the standard and is assigned a shift of zero
- one mark for chemical shift being measured in parts per million
- one mark for explaining that dividing by the spectrometer frequency makes the value independent of the field used
- one mark for reading a shift against the Data Booklet table and naming the environment it corresponds to
- one mark for shielding by surrounding electrons as the reason a nucleus resonates where it does
- one mark for stating that an electron-withdrawing neighbour deshields the nucleus and raises the shift
- one mark for using a quoted shift to name a specific environment, such as a proton on a carbon bonded to the single-bonded oxygen of an ester at about 4.1 ppm
- one mark for recognising equivalent protons as giving one peak rather than several
- one mark for tetramethylsilane giving a single peak because all of its protons are equivalent
- one mark for a further property of tetramethylsilane, such as being unreactive, non-toxic or easily removed because it is volatile
- one mark for using an integration ratio to give the relative numbers of equivalent protons
- one mark for applying the n+1 rule, for example a quartet because the neighbouring carbon carries three protons
- one mark for a part structure consistent with the shift, the integration and the splitting taken together
Examiner Tips
- 💡These are often levels-of-response questions, so cover a stage for each technique and let the reasoning run in order from data to structure.
- 💡Quote the actual value you are reading, the wavenumber or the shift, beside every deduction you make.
- 💡Finish with a drawn structure; a description in words rarely secures the top level.
- 💡Always write ppm after a shift; a bare number is not a chemical shift.
- 💡When a shift is given, find its row in the Data Booklet before writing anything, and quote the environment the table names.
- 💡Remember which end is which: a peak far to the left of the spectrum is strongly deshielded.
- 💡Name the environment in the Data Booklet's own wording when you justify a shift; the mark scheme is written from that table.
- 💡Read shift and splitting together, because each narrows the possibilities the other leaves open.
- 💡Check for symmetry first: two identical methyl groups are one environment and one peak.
- 💡Write the reasoning the mark scheme uses, quartet because the neighbouring carbon has three hydrogens, rather than just quartet.
- 💡Take each peak in the order shift, then integration, then splitting, and assemble the fragments at the end.
- 💡If a spectrum alone cannot settle the structure, say what is missing, for example that overlapping peaks hide the splitting.
Common Mistakes
- proposing a structure that fits the infrared spectrum but has the wrong number of carbon environments
- quoting a wavenumber with no functional group attached to it, or a functional group with no wavenumber
- using the fingerprint region to assign individual functional groups
- ignoring the molecular formula, and so suggesting a structure with the wrong number of atoms
- quoting a chemical shift in hertz rather than in parts per million
- saying delta values change when a stronger magnet is used
- treating tetramethylsilane as the solvent rather than as a reference added in small amount
- reading the proton table when the spectrum is a carbon spectrum, or the reverse
- assuming a higher shift means more hydrogens, which is what integration tells you and shift does not
- expecting a proton three or four bonds from an oxygen to be shifted as far downfield as one on the carbon bonded to it
- treating an O-H or N-H peak as having a fixed shift, when it is broad and variable
- counting every hydrogen as its own environment and predicting far too many peaks
- reading an integration value as the actual number of protons rather than as a ratio
- counting the protons on the signal's own carbon when applying the n+1 rule, instead of those on the neighbouring carbons
- splitting a signal with protons that are equivalent to it, which do not split each other
- expecting an O-H proton to split its neighbours, when it normally appears as a singlet
- saying tetramethylsilane is used because it is cheap, which is not a chemical reason the scheme credits