Revision Notes
OxfordAQA A-Level Chemistry: Nuclear magnetic resonance spectroscopy (9620) – Revision Notes
Condensed OxfordAQA 9620 NMR revision notes: chemical shift, TMS, solvents, ¹³C and ¹H spectra, integration, n+1 splitting and a quick self-test.
- Subject
- Chemistry
- Level
- A LEVEL
- Topic
- Nuclear magnetic resonance spectroscopy
- Author
- Marlbridge Academic Team
- Updated
Aligned to OxfordAQA A Level Chemistry (9620), Version 5.3 (first teaching 2019, first AS and A-level exams 2020). Official specification .
Syllabus page (what it covers and how it is assessed): OxfordAQA A Level Chemistry.
Syllabus points this page covers
9620 (A Level)
- 3.3.15.1 Nuclear magnetic resonance spectroscopy
- 33 Nuclear magnetic resonance spectroscopy (whole topic)
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Short notes for the last weeks before the exam. For full explanations and five worked examples, use the NMR study guide.
Coverage: section 3.3.15, Nuclear magnetic resonance spectroscopy (3.3.15.1), in the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. It is International A-level only content, examined in Unit 4; Unit 5 (practical and synoptic) may also include it. Practise with the NMR practice questions, revise infrared and mass spectrometry in Organic analysis, and use the course hub, the printable checklist and the free diagnostics to track the rest of the course.
3.3.15.1 in ten lines
- Analytical techniques are used together to confirm the structure of a new compound.
- NMR locates ¹³C or ¹H atoms within a molecule.
- ¹³C spectra are simpler than ¹H spectra.
- Chemical shift is recorded on the δ scale (ppm).
- Chemical shift depends on the molecular environment.
- An integrated ¹H spectrum shows the ratio of ¹H atoms across environments.
- ¹H samples are dissolved in deuterated solvents or CCl₄.
- TMS is the standard, at δ = 0.
- You suggest structures from ¹H and ¹³C spectra plus data-booklet chemical shifts.
- The n+1 rule covers doublets, triplets and quartets in aliphatic compounds only.
No practical activity is listed for this section.
Terms to define precisely
| Term | Definition to write |
|---|---|
| Environment | The surroundings of a nucleus; nuclei in the same environment give one signal |
| Equivalent | Made identical by the symmetry of the molecule |
| Chemical shift, δ | Signal position in ppm, relative to TMS at 0 |
| Integration | Relative signal area in a ¹H spectrum, proportional to the number of H |
| Splitting | A ¹H signal divided into n + 1 peaks by n non-equivalent H on adjacent carbons |
| Deuterated solvent | A solvent with ¹H replaced by ²H (D), e.g. CDCl₃ |
Shift ranges (approximate)
Use the Chemistry data booklet in the exam. These rounded ranges are written for these notes.
| ¹H near… | δ / ppm | ¹³C type | δ / ppm |
|---|---|---|---|
| Alkyl only | 0.7–1.7 | Alkyl | 5–45 |
| C=O | 2.0–3.0 | Next to C=O | 20–55 |
| Cl or Br on the same C | 3.0–4.5 | Bonded to Cl or Br | 30–75 |
| O on the same C | 3.1–4.3 | Bonded to O | 50–90 |
| C=C (H on the double bond) | 4.5–6.0 | C=C | 100–150 |
| Aldehyde CHO | 9.0–10.0 | C=O, acid or ester | 160–185 |
| Acid COOH | 10.0–12.0 | C=O, aldehyde or ketone | 190–220 |
The trend: an electronegative atom or a C=O group close to the nucleus means a larger δ. Two electronegative atoms on the same carbon push the signal further still.
TMS in four points
Tetramethylsilane, Si(CH₃)₄:
- one intense signal: 12 equivalent H and 4 equivalent C;
- signal at the low-δ end, clear of nearly all organic signals;
- inert: no reaction with the sample;
- volatile: evaporates off, so the sample can be recovered.
Solvent choice
| Solvent | Suitable for ¹H NMR? | Reason |
|---|---|---|
| CDCl₃ | Yes | D (²H) gives no ¹H signal |
| CCl₄ | Yes | Contains no H |
| CHCl₃ | No | Its ¹H gives a huge signal that swamps the sample’s |
Deuterium still has a nucleus, but it absorbs at a completely different frequency from ¹H, so nothing from a CDCl₃ molecule appears in the ¹H spectrum.
¹³C spectra
- Lines = carbon environments. Look for symmetry first.
- δ says what kind of carbon each one is.
- No splitting and a wide δ range (about 0–220 ppm), so one clean line per environment. That is why ¹³C spectra are simpler.
Worked reminder. Butanedione, CH₃COCOCH₃, is symmetrical: two equivalent C=O carbons (190–220) and two equivalent CH₃ carbons. Answer: 2 lines.
¹H spectra: five-step method
- Count signals: that is the number of H environments.
- Integration: divide trace heights by the smallest, make whole numbers, scale to the formula.
- Shift: match each δ to the table to locate the group.
- Splitting: n + 1 peaks means n H on the adjacent carbons.
- Assemble and check: draw the structure and tick off every signal, ratio and splitting.
Worked reminder. 1,2-dimethoxyethane, CH₃OCH₂CH₂OCH₃ (C₄H₁₀O₂), gives two singlets with trace heights 21 mm and 14 mm. Ratio 21 : 14 = 3 : 2. The formula has 10 H, so scale by 2: 6 H and 4 H. The two CH₂ groups are next to each other, but they are equivalent, so they do not split each other and give a singlet.
The n+1 rule
| H on adjacent carbons | Pattern | Fragment it suggests |
|---|---|---|
| 0 | singlet | next to O, C=O, or a C with no H |
| 1 | doublet | a CH next door |
| 2 | triplet | a CH₂ next door |
| 3 | quartet | a CH₃ next door |
Fragments worth recognising on sight:
- 2H quartet + 3H triplet: an ethyl group, CH₃CH₂–.
- 1H quartet + 3H doublet: CH₃CH< with no other H neighbours.
- 1H triplet + 2H doublet: –CH–CH₂– where neither carbon has other H neighbours.
- 2H triplet + 2H triplet: –CH₂CH₂– with different groups at each end.
- 9H singlet: (CH₃)₃C–.
O–H protons usually appear as singlets in the spectra you meet, and their δ varies from sample to sample.
Combining techniques
| Data | Tells you |
|---|---|
| Molecular ion (mass spectrum) | Mr; a precise mass fixes the formula |
| IR absorptions | Bonds and functional groups |
| ¹³C lines and shifts | Number and type of carbon environments |
| ¹H signals, integration, splitting | H environments, how many H in each, neighbours |
Work in that order: formula, functional group, carbon skeleton, then H detail.
Must-know distinctions
- Shift vs splitting: δ reflects nearby atoms (O, Cl, C=O); splitting counts H on the carbons next door.
- Ratio vs number: integration gives a ratio until you scale it to the formula.
- ¹³C vs ¹H: ¹³C shows carbon environments with no splitting; ¹H gives integration and splitting.
- Ester vs acid: an ester has a ¹³C line at 50–90 (C–O) and no ¹H signal at 10–12.
- CH₃ next to O vs CH₃ next to C=O: both singlets; only δ tells them apart.
Quick self-test
- How many ¹³C lines does butanedione give, and how many ¹H signals?
- How many ¹³C lines does 3-methylpentane, CH₃CH₂CH(CH₃)CH₂CH₃, give?
- Integration trace heights are 8 mm, 12 mm and 20 mm. Give the simplest ratio.
- A signal is a doublet. How many H are on the adjacent carbons?
- Why can CCl₄ be used as a ¹H NMR solvent?
- Which property of TMS stops its signal overlapping the sample’s?
- A ¹³C line appears at δ 172. What type of carbon does it suggest?
- Predict the ¹H spectrum of 2,2-dichloropropane, CH₃CCl₂CH₃.
- Two 2H triplets appear in one spectrum. What fragment does this suggest?
- A ¹H signal appears at δ 3.5. Suggest what the H atoms are near.
- Give two features that make a ¹³C spectrum easier to interpret than a ¹H spectrum.
Answers
- 2 lines (C=O and CH₃); 1 signal, a singlet, because all 6 H are equivalent.
- 4 (the two end CH₃, the two CH₂, the CH, and the branch CH₃).
- Divide by 4: 2 : 3 : 5.
- 1.
- It contains no hydrogen, so it gives no ¹H signal.
- It absorbs upfield of nearly all organic H, so its signal stays clear of theirs.
- A C=O in a carboxylic acid or ester (160–185).
- One singlet (6H): the two CH₃ groups are equivalent and the central carbon has no H.
- –CH₂CH₂– with a different group at each end.
- An O (or a Cl or Br) on the same carbon.
- No splitting, so one line per environment; and a wider δ range, so lines rarely overlap.
Where marks are usually lost
- Missing symmetry, e.g. counting two lines for the equivalent C=O carbons in butanedione.
- Using n+1 on the H in the group itself instead of on the adjacent carbons.
- Splitting H on a carbon that is next to O or C=O, which should be a singlet.
- Giving equivalent neighbours a splitting: equivalent H do not split each other.
- Quoting the simplest integration ratio as the answer when the formula needs it scaled.
- Calling deuterated solvents “unreactive” instead of saying they give no ¹H signal.
- Listing “signal at zero” as a reason TMS is chosen.
- A structure that fits the splitting but puts a group on the wrong side of an O atom, against the δ data.
- Not checking the final structure against the molecular formula.
Official syllabus
OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Coverage here: section 3.3.15 Nuclear magnetic resonance spectroscopy (3.3.15.1), International A-level only.
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Study guides
OxfordAQA A-Level Chemistry: Nuclear magnetic resonance spectroscopy (9620)
OxfordAQA 9620 study guide to NMR: the δ scale, TMS, deuterated solvents, ¹³C and ¹H spectra, integration and the n+1 rule, with five worked examples.
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Practice questions
OxfordAQA A-Level Chemistry: Nuclear magnetic resonance spectroscopy (9620) – Practice Questions
Original OxfordAQA 9620 NMR practice questions with marked worked answers: TMS, solvents, ¹³C counts, integration, n+1 splitting and combined data.
Chemistry · OxfordAQA · A Level
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OxfordAQA A-Level Chemistry: Optical isomerism (9620)
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