The shape of the lesson
By the end of the lesson, learners can
- Counting carbon environments: Draw the full structure, then look for symmetry.
- Shift ranges for carbon: The more electronegative the atoms bonded to a carbon, and the more multiple bonds it has, the further left its peak appears.
- IR, MS and NMR together: Each technique answers a different question.
- Peaks = carbon environments; use symmetry.
- Ranges: alkyl 5–55, C–O 50–90, C=C and aromatic 110–160, C=O 160–220.
- No splitting, and heights do not count carbons.
- Combine: MS for Mr, IR for groups, NMR for the skeleton.
One picture every idea returns to
Only about 1.1% of carbon atoms are 13C, but they absorb radio waves in a magnetic field just as protons do. The number of peaks equals the number of different carbon environments, so symmetry matters: propan-2-ol gives two peaks, propan-1-ol three. The chemical shift range is wide, about 0 to 220 ppm, and tells you the type of carbon: alkyl 5–55, C–O 50–90, alkene and aromatic 110–160, C=O 160–220. Routine spectra show no splitting, so every peak is a single line, and peak heights do not reliably count carbons. Combined with IR (functional groups) and mass spectra (Mr), it pins down a structure.
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, One line for every carbon environment.. Carbon-13 NMR looks at the carbon skeleton directly. 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: Counting carbon environments | Draw the full structure, then look for symmetry. Project tab 5 of the lab (Chromatography and NMR lab, carbon-13 NMR) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Shift ranges for carbon | The more electronegative the atoms bonded to a carbon, and the more multiple bonds it has, the further left its peak appears. Project tab 5 of the lab (Chromatography and NMR lab, carbon-13 NMR) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: IR, MS and NMR together | Each technique answers a different question. Project tab 6 of the lab (Chromatography and NMR lab, identify the unknown) and let a student drive it while the class predicts. | Corners |
| 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, 15 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 IR, MS and NMR together, 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.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
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 |
|---|---|
| Count environments, not carbon atoms: the two CH 3 groups of propanone give one peak. 13 C peaks are not split and their heights do not give the number of carbons, unlike proton peak areas. | A carbon bonded to O sits at 50–90, while a C=O carbon is far further left at 160–220; do not confuse the two. |
Carbon-13 NMR spectroscopy: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
State the number of peaks in the 13C NMR spectra of (a) propan-1-ol and (b) propan-2-ol, and explain the difference.
Predict the 13C NMR spectrum of ethyl ethanoate, CH3COOCH2CH3, giving the approximate shift of each peak.
Give two differences between a routine 13C NMR spectrum and a high-resolution 1H NMR spectrum of the same compound.
Compound Z, C3H6O, shows: IR strong absorption at 1715 cm−1 and no broad band at 3200–3600 cm−1; 13C NMR peaks at 206 and 30; 1H NMR one singlet. Deduce the structure of Z.
In the mass spectrum of a hydrocarbon the M+ peak has height 24.0 and the M+1 peak has height 1.06. Calculate the number of carbon atoms in the molecule.
Pentan-2-one, CH3COCH2CH2CH3, and pentan-3-one, CH3CH2COCH2CH3, give almost identical IR spectra. Explain how 13C NMR distinguishes them.
Answer key and marking notes
- (a) 3 peaks: all three carbons are in different environments.
- (b) 2 peaks: the molecule is symmetrical, so the two CH3 carbons are equivalent.
- Four carbon environments, so four peaks.
- C=O of the ester about 171 (160–185); O–CH2 about 60 (50–90).
- Two alkyl CH3 carbons at about 21 and 14 (5–55).
- 13C peaks are single lines with no splitting; 1H peaks are split by the n+1 rule.
- 13C peak heights do not show the number of atoms, while 1H peak areas give the ratio of H atoms (also, the 13C shift range is much wider, about 0–220 ppm against 0–13).
- IR: C=O present and no O–H, so Z is an aldehyde or ketone.
- Two carbon environments with the C=O at 206 (ketone range); one proton environment with no neighbouring H.
- Z is propanone, CH3COCH3.
- n = 100 × (M+1 height) ÷ (1.1 × M+ height).
- n = 100 × 1.06 ÷ (1.1 × 24.0) = 4.02, so 4 carbon atoms.
- Both are ketones, so both absorb near 1715 cm−1 in IR and both have a C=O peak at 190–220 in 13C NMR.
- Pentan-2-one has no symmetry: five carbon environments, so five peaks.
- Pentan-3-one is symmetrical about the C=O: the two CH2 carbons are equivalent and so are the two CH3 carbons, giving only three peaks.
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 13C NMR spectrum of propanone, CH3COCH3?
- A2
- B3
- C1
- D6
Using the Data Booklet, in which range does a carbon bonded to a single oxygen atom, as in CH3OH, appear in 13C NMR?
- A50–90 ppm
- B160–220 ppm
- C110–160 ppm
- D5–55 ppm
Which statement about a routine 13C NMR spectrum is correct?
- Aeach carbon environment gives a single line, with no splitting
- Beach peak is split by the n+1 rule using neighbouring carbons
- Cthe peak heights give the number of carbon atoms in each environment
- Dthe chemical shift range is 0 to 13 ppm
How many peaks are in the 13C NMR spectrum of methylbenzene, C6H5CH3?
- A5
- B7
- C2
- D4
An unknown compound is analysed by mass spectrometry, infrared spectroscopy and 13C NMR. Which piece of information comes from the mass spectrum?
- Athe relative molecular mass, from the molecular ion peak
- Bthe functional groups present, from absorptions near 1700 and 3300 cm−1
- Cthe number of carbon environments
- Dthe number of hydrogens on adjacent carbons
Compound L, C3H8O, gives the 13C NMR spectrum shown. Its IR spectrum has a broad absorption at 3200–3600 cm−1. What is L?
- Apropan-2-ol, CH3CH(OH)CH3
- Bpropan-1-ol, CH3CH2CH2OH
- Cmethoxyethane, CH3OCH2CH3
- Dpropanone, CH3COCH3
An ester C4H8O2 has IR absorption at 1740 cm−1 and no broad O–H band. Its 13C NMR spectrum has four peaks, at 171, 60, 21 and 14, and its 1H NMR spectrum includes a singlet at 2.0. What is the ester?
- Aethyl ethanoate, CH3COOCH2CH3
- Bmethyl propanoate, CH3CH2COOCH3
- Cbutanoic acid, CH3CH2CH2COOH
- Dpropyl methanoate, HCOOCH2CH2CH3
An aromatic hydrocarbon C8H10 gives only three peaks in its 13C NMR spectrum. Which isomer is it?
- A1,4-dimethylbenzene
- B1,2-dimethylbenzene
- C1,3-dimethylbenzene
- Dethylbenzene
Quiz key
The two CH3 carbons are equivalent, and the C=O carbon is different, so there are two peaks (about 30 and 206). Three counts every carbon atom, ignoring symmetry. One would mean all carbons are equivalent. Six is the number of hydrogens, which 13C NMR does not count.
A carbon singly bonded to O is moved downfield to 50–90 ppm. 160–220 is the carbonyl (C=O) range, 110–160 is for alkene and aromatic carbons, and 5–55 is for alkyl carbons bonded only to C and H.
Neighbouring 13C atoms are very rare and routine spectra are run so that protons do not split the carbon signals, so every environment gives one line. The n+1 rule applies to 1H spectra. 13C peak heights are not proportional to the number of carbons, unlike proton peak areas. The 13C range is about 0–220 ppm; 0–13 is roughly the proton range.
The ring has four environments: the carbon bearing CH3, the two ortho carbons, the two meta carbons and the para carbon. Add the CH3 carbon: five peaks. Seven counts every carbon atom. Two treats the ring as one environment, as in benzene. Four forgets the CH3 carbon.
The M+ peak gives Mr, and the M+1 peak gives the number of carbons. Functional groups come from IR absorptions such as C=O and O–H. Carbon environments come from 13C NMR, and neighbouring hydrogens from splitting in 1H NMR.
The IR band shows an alcohol O–H. Two peaks means two carbon environments: 64 is a C–O carbon (50–90) and 25 is alkyl. Propan-2-ol has two equivalent CH3 groups and one CH(OH), so it fits. Propan-1-ol has three different carbons, so three peaks. Methoxyethane also has three carbon environments and no O–H. Propanone is C3H6O, not C3H8O, and its C=O peak would be near 206.
The 171 peak is an ester C=O, 60 is O–CH2, and 21 and 14 are two alkyl CH3 carbons. The 1H singlet at 2.0 is CH3–C=O, which fits ethyl ethanoate. Methyl propanoate also has four carbon environments, but its O–CH3 singlet would be near 3.7. Butanoic acid would show a broad O–H in IR and a COOH proton at 9.0–13.0. Propyl methanoate has no CH3–C=O and would show an HCOO proton near 8.
1,4-dimethylbenzene has two equivalent CH3 carbons, two equivalent ring carbons bearing CH3, and four equivalent ring CH carbons: three peaks. 1,2-dimethylbenzene gives four (CH3, C–CH3 and two kinds of ring CH). 1,3-dimethylbenzene gives five, because the ring CH carbons fall into three environments. Ethylbenzene gives six: CH3, CH2 and four ring environments.
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.