The shape of the lesson
By the end of the lesson, learners can
- Shift, environments, areas: Electronegative atoms near a hydrogen pull electron density away from it, so its peak moves to higher δ (to the left of the spectrum).
- Splitting by the n+1 rule: Hydrogens on the next carbon along nudge the field a proton feels.
- Four spectra to know: Read each row from left to right: shift tells you the type of proton, area the number, splitting the neighbours.
- Number of peaks = number of H environments.
- Shift (Data Booklet) = type of H; TMS is 0.
- Area ratio = number of H in each environment.
- n+1 rule: n H on adjacent carbons give n+1 lines; O–H vanishes with D2O.
One picture every idea returns to
In a magnetic field, 1H nuclei absorb radio waves at frequencies that depend on their electron surroundings. The chemical shift, δ in ppm, is measured relative to TMS, Si(CH3)4, set at 0. The number of peaks equals the number of different proton environments, and the relative peak areas (integration) give the ratio of hydrogens in each. Each peak is split by the hydrogens on neighbouring carbons according to the n+1 rule: n neighbours give n+1 lines, so a singlet, doublet, triplet or quartet. Samples are dissolved in a deuterated solvent such as CDCl3, which gives no proton signal, and shaking with D2O makes O–H and N–H peaks disappear.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Count, weigh and split the protons.. A proton NMR spectrum answers four questions at once: how many hydrogen environments there are, what kind each one is, how many hydrogens are in each, and how many hydrogens sit on the neighbouring carbons. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Shift, environments, areas | Electronegative atoms near a hydrogen pull electron density away from it, so its peak moves to higher δ (to the left of the spectrum). Project tab 3 of the lab (Chromatography and NMR lab, proton NMR spectra) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Splitting by the n+1 rule | Hydrogens on the next carbon along nudge the field a proton feels. Project tab 4 of the lab (Chromatography and NMR lab, the n+1 rule) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Four spectra to know | Read each row from left to right: shift tells you the type of proton, area the number, splitting the neighbours. Project tab 3 of the lab (Chromatography and NMR lab, proton NMR spectra) and let a student drive it while the class predicts. | Jigsaw |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Four spectra to know, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Splitting comes from H on the adjacent carbon, not from the H in the group itself: a CH 3 next to a CH 2 is a triplet, not a quartet. | Peak area gives the number of H; the number of lines does not. The O–H peak is usually a singlet and is the one that disappears with D 2 O. TMS is at 0 and δ increases to the left. |
Proton NMR spectroscopy: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Give three reasons why tetramethylsilane, Si(CH3)4, is used as the reference in proton NMR.
Predict the high-resolution proton NMR spectrum of ethyl ethanoate, CH3COOCH2CH3: number of peaks, approximate δ, relative areas and splitting.
Explain why the spectrum is recorded in CDCl3 rather than CHCl3.
Compound Y, C3H7Cl, gives two peaks: a doublet at 1.5 (6H) and a seven-line peak at 4.1 (1H). Identify Y and explain the splitting.
Describe how proton NMR distinguishes propanone, CH3COCH3, from propanal, CH3CH2CHO.
The spectrum of an unknown compound has a broad singlet at 3.0. Describe a test, using NMR, to show whether it is an O–H proton.
Answer key and marking notes
- All 12 hydrogens are equivalent, so it gives a single strong peak, defined as δ = 0.
- Its peak is upfield of (to the right of) almost all organic protons, so it does not overlap them.
- It is inert, non-toxic and volatile, so it does not react with the sample and is easily removed.
- Three environments, so three peaks in the ratio 3 : 2 : 3.
- CH3–C=O: about 2.0, 3H, singlet (no H on the adjacent carbon).
- O–CH2: about 4.1, 2H, quartet (3 H on the adjacent CH3).
- CH3 of the ethyl group: about 1.3, 3H, triplet (2 H on the adjacent CH2).
- CHCl3 contains a hydrogen, which would give its own peak and could hide sample peaks.
- Deuterium does not absorb in the proton NMR region, so CDCl3 gives no signal.
- Y is 2-chloropropane, CH3CHClCH3: two equivalent CH3 groups (6H) and one CH (1H).
- Each CH3 is next to the CH with 1 H: 1 + 1 = 2 lines, a doublet.
- The CH is next to 6 equivalent H: 6 + 1 = 7 lines; it is shifted to 4.1 by the Cl on the same carbon.
- Propanone: one peak, a singlet at about 2.1 (6H).
- Propanal: three peaks in the ratio 3 : 2 : 1, including the –CHO proton at about 9.8.
- Propanal also shows splitting: CH3 triplet, CH3CH2CHO multiplet, CHO triplet.
- Add a few drops of D2O, shake, and record the spectrum again.
- If the peak at 3.0 disappears, it was an O–H (or N–H) proton, exchanged for D, which gives no proton signal.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
How many peaks are in the low-resolution 1H NMR spectrum of ethanol, CH3CH2OH?
- A3
- B2
- C6
- D1
In the high-resolution 1H NMR spectrum of chloroethane, CH3CH2Cl, how is the CH3 peak split?
- Ainto a triplet
- Binto a quartet
- Cinto a doublet
- Dit is a singlet
Why is the 1H NMR spectrum of an alcohol sometimes recorded a second time after shaking the sample with D2O?
- Athe O–H proton exchanges with deuterium, so its peak disappears and can be identified
- BD2O is the reference compound that defines δ = 0
- CD2O makes every peak split into a doublet
- DD2O dissolves the sample so that the C–H peaks become larger
Using the Data Booklet ranges, in which region does the proton of an aldehyde group, –CHO, appear?
- A9.3–10.5 ppm
- B2.0–2.5 ppm
- C0.9–1.7 ppm
- D6.0–9.0 ppm
Why are 1H NMR samples dissolved in CDCl3 rather than CHCl3?
- Adeuterium does not absorb in the proton region, so the solvent gives no peak
- BCDCl3 is used as the zero reference instead of TMS
- CCDCl3 removes O–H peaks from the spectrum
- DCDCl3 splits the sample peaks so they are easier to read
The 1H NMR spectrum of compound K, C4H8O, is shown. There are no peaks above 5 ppm, and the IR spectrum shows a strong C=O absorption. What is K?
- Abutanone, CH3COCH2CH3
- Bbutanal, CH3CH2CH2CHO
- C2-methylpropanal, (CH3)2CHCHO
- Dethoxyethene, CH2=CHOCH2CH3
What does the high-resolution 1H NMR spectrum of 2-chloropropane, CH3CHClCH3, show?
- Aa doublet (6H) and a seven-line peak (1H)
- Ba triplet (6H) and a quartet (1H)
- Ctwo triplets and a sextet in the ratio 3 : 3 : 1
- Da singlet (6H) and a singlet (1H)
Two esters, C4H8O2, both give three peaks in the ratio 3 : 2 : 3. Ester S has a singlet at 3.7 and a quartet at 2.3. Ester T has a singlet at 2.0 and a quartet at 4.1. Which pair of identities is correct?
- AS is methyl propanoate, CH3CH2COOCH3; T is ethyl ethanoate, CH3COOCH2CH3
- BS is ethyl ethanoate; T is methyl propanoate
- Cboth are ethyl ethanoate; the shifts depend only on the solvent
- DS is propyl methanoate; T is butanoic acid
Quiz key
Ethanol has three proton environments: CH3, CH2 and OH, so three peaks in the area ratio 3 : 2 : 1. Two would ignore the OH proton, which does give a peak (until D2O is added). Six counts atoms rather than environments. One would require all hydrogens to be equivalent, as in propanone.
The CH3 is next to a CH2 carrying 2 H, so by the n+1 rule it gives 2 + 1 = 3 lines, a triplet. A quartet is the splitting of the CH2 peak, caused by the 3 H of the CH3. A doublet would need just 1 neighbouring H, and a singlet none.
ROH + D2O → ROD + HDO. Deuterium gives no peak in a proton spectrum, so the O–H signal vanishes, identifying it (N–H protons behave in the same way). The reference at δ = 0 is TMS. D2O does not cause splitting, and it does not change the C–H peaks, which keep their areas.
The aldehyde proton is strongly deshielded by the C=O group and appears at 9.3–10.5 ppm, which makes aldehydes easy to spot. 2.0–2.5 is for H on a carbon next to C=O (as in CH3–C=O), 0.9–1.7 for ordinary alkyl H, and 6.0–9.0 for H on an aromatic ring.
CHCl3 contains a proton, which would give a large solvent peak that could hide sample peaks. CDCl3 has deuterium instead, which gives no signal. TMS, not the solvent, defines δ = 0. Removing O–H peaks is done by shaking with D2O, and the solvent does not cause splitting.
The singlet at 2.1 (3H) is a CH3 next to C=O with no neighbouring H; the quartet at 2.4 (2H) and triplet at 1.0 (3H) are an ethyl group, with its CH2 next to C=O. That is CH3COCH2CH3. Butanal and 2-methylpropanal are aldehydes and would show a –CHO peak at 9.3–10.5, which is absent. Ethoxyethene has no C=O and would show alkene H well above 5 ppm.
The two CH3 groups are equivalent (6H) and each is next to a CH with 1 H, so they give a doublet. The CH has 6 equivalent neighbours, giving 6 + 1 = 7 lines, shifted to about 4.1 by the Cl. A triplet and quartet would be an ethyl pattern. Treating the CH3 groups as different gives too many peaks. Singlets would need no hydrogens on adjacent carbons.
In S the singlet at 3.7 is an O–CH3 (no neighbouring H) and the quartet at 2.3 is a CH2 next to C=O, so S is methyl propanoate. In T the singlet at 2.0 is CH3–C=O and the quartet at 4.1 is O–CH2, so T is ethyl ethanoate. Swapping them puts each group in the wrong shift range. Chemical shifts are characteristic of structure, not just the solvent. Propyl methanoate would give four peaks including HCOO at about 8, and butanoic acid would show a COOH peak at 9.0–13.0.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.