Proton NMRAQA A-Level Chemistry: Revision notes
Section 1
What a proton NMR spectrum shows
A ¹H NMR spectrum gives four pieces of information about the hydrogen atoms (protons) in a molecule:
- the number of peaks: one for each different proton environment
- the chemical shift, δ: the type of environment (use the Data Booklet)
- the integration: the relative number of protons in each environment
- the splitting pattern: the number of adjacent protons
Protons in identical environments are equivalent and give one peak. Unlike ¹³C NMR, ¹H spectra show splitting, which gives extra detail.
Go through the four pieces of information in order for every spectrum question.
Section 2
Solvents and the standard
Samples must not add peaks of their own. ¹H NMR spectra are obtained using samples dissolved in deuterated solvents (such as CDCl₃) or in CCl₄, which contain no ¹H atoms.
A solvent such as CHCl₃ would give its own ¹H peak, and one with many hydrogens would swamp the sample peaks. TMS is added as the standard, with δ = 0.
CHCl₃ is not suitable: it contains hydrogen and gives its own peak.
Section 3
Integration
The area under each peak is proportional to the number of protons in that environment. An integration trace shows this as steps whose heights are in the same ratio.
Worked example: peak areas of 6.0, 4.0 and 6.0 give the ratio 3 : 2 : 3. If the formula has 8 H atoms, the peaks are caused by 3, 2 and 3 protons.
Always check that the numbers add up to the number of H atoms in the molecular formula.
Divide by the smallest area first, then scale to match the total number of hydrogen atoms in the formula.
Section 4
Spin-spin splitting and the n + 1 rule
A peak is split by protons on adjacent carbon atoms that are not equivalent to the protons giving the peak. The n + 1 rule: n adjacent protons give n + 1 peaks.
- n = 0: singlet
- n = 1: doublet
- n = 2: triplet
- n = 3: quartet
In ethyl groups, –CH₂CH₃, the CH₂ is a quartet (next to three H) and the CH₃ is a triplet (next to two H). At A Level you are limited to doublets, triplets and quartets in aliphatic compounds.
Protons are split by their neighbours, never by the protons in their own group.
Section 5
Putting it together: structure from data
Typical ¹H shifts (δ / ppm): R–CH₃ 0.7–1.2; CH₃ or CH₂ next to C=O 2.1–2.6; R–O–CH₂ 3.1–3.9; ester O–CH₂ 3.7–4.1; C=C–H 4.5–6.0; aromatic H 6.0–9.0; R–CHO 9.4–10.0; R–COOH 10.0–12.0.
Worked example: δ = 1.2 (triplet, 3 H), 2.1 (singlet, 3 H), 4.1 (quartet, 2 H), C₄H₈O₂. The triplet and quartet form a CH₃CH₂– unit, with the quartet at 4.1 showing it is bonded to the O of an ester. The singlet at 2.1 is a CH₃ next to C=O. The compound is CH₃COOCH₂CH₃, ethyl ethanoate.
Link each peak to a group: shift gives the group, area gives the number of H, splitting gives the neighbours.
Must Know
- Number of peaks = number of proton environments
- Chemical shift tells you the type of environment; use the Data Booklet
- Integration gives the relative number of equivalent protons
- n + 1 rule: n adjacent non-equivalent protons give n + 1 peaks; singlet, doublet, triplet, quartet
- Use deuterated solvents or CCl₄, with TMS as the standard
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Proton NMR
- A chemist records ¹H NMR spectra of organic compounds to find how many hydrogen atoms are in each environment. The sample for each spectrum is dissolved in a solvent, and tetramethylsilane (TMS) is added as the standard.The integration values for the three peaks in the spectrum of an ester, C₄H₈O₂, are 6.0, 4.0 and 6.0 (arbitrary units). Explain how these data are used to work out the relative numbers of equivalent protons.2 marks
- The ¹H NMR spectra of aliphatic compounds show splitting of peaks by hydrogen atoms on adjacent carbon atoms. A student is predicting the splitting patterns for 1,1-dichloroethane, CH₃CHCl₂, using the n + 1 rule.Use the n + 1 rule to explain why the signal from the CH₃ protons in 1,1-dichloroethane is split in the way you chose in part (a).2 marks
- Compound R has the molecular formula C₄H₈O₂. Its ¹H NMR spectrum has three peaks: δ = 1.2 (triplet, relative area 3), δ = 2.1 (singlet, relative area 3) and δ = 4.1 (quartet, relative area 2). Shift data (δ / ppm): R–CH₃ 0.7–1.2; CH₃ next to C=O 2.1–2.6; CH₂ bonded to the O of an ester 3.7–4.1.Explain how the splitting patterns in the spectrum of R show that it contains an ethyl group, –CH₂CH₃, attached to an atom with no hydrogen atoms.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).