Principles of NMR and carbon-13 NMRAQA A-Level Chemistry: Revision notes
Section 1
What NMR tells you
Nuclear magnetic resonance (NMR) spectroscopy gives information about the position of ¹³C or ¹H atoms in a molecule. Nuclei placed in a strong magnetic field absorb radio-frequency radiation, and the frequency absorbed depends on the environment of the nucleus.
No single technique proves a structure. Scientists have developed a range of analytical techniques (mass spectrometry, infrared spectroscopy and NMR) that together enable the structures of new compounds to be confirmed.
¹³C NMR gives simpler spectra than ¹H NMR: each peak is a single line (no splitting) and there are fewer peaks.
Say that techniques are used together: NMR on its own rarely proves a structure.
Section 2
Chemical shift and the delta scale
The position of a peak is its chemical shift, δ, recorded on the delta scale in parts per million (ppm) relative to a standard.
The chemical shift depends on the molecular environment: electronegative atoms such as oxygen, and groups such as C=O, pull electron density from nearby carbon atoms and move their peaks to a higher δ.
The standard is tetramethylsilane, TMS, Si(CH₃)₄, whose peak is defined as δ = 0.
Chemical shift depends on the environment around the carbon, not on the number of carbon atoms in the molecule.
Section 3
Why TMS is a suitable standard
TMS is chosen because:
- all four carbon atoms are equivalent, so it gives one sharp peak
- its peak is at the low δ end, away from most sample peaks
- it is chemically inert, so it does not react with the sample
- it is volatile (boiling point 27 °C), so it is easily removed from the sample
- it is non-toxic
Link each TMS property to a consequence: four equivalent carbons mean one peak.
Section 4
Carbon environments and the number of peaks
In a ¹³C NMR spectrum, each different carbon environment gives one peak. Carbon atoms that are in identical environments, for example because of symmetry, are equivalent and give the same peak.
Examples:
- propanone, CH₃COCH₃: 2 peaks (two equivalent CH₃, one C=O)
- propanal, CH₃CH₂CHO: 3 peaks
- 2-methylpropan-2-ol, (CH₃)₃COH: 2 peaks
- ethyl ethanoate, CH₃COOCH₂CH₃: 4 peaks
Count the peaks, then ask which carbons could be equivalent.
Do not count every carbon atom: equivalent carbons give one peak between them.
Section 5
Using shift data to suggest structures
Use the Chemistry Data Booklet. Approximate ¹³C shift ranges (δ / ppm):
- C–C: 5–40
- C–Cl or C–Br: 10–70
- C–N: 25–60
- C–O (alcohols, ethers, esters): 50–90
- C=C: 90–150
- aromatic carbon: 110–160
- C=O in esters and acids: 160–185
- C=O in aldehydes and ketones: 190–220
Method: count the peaks, match each shift to a carbon type, then use the formula to build the structure. Other evidence (such as a test with Tollens' reagent) can separate aldehydes from ketones, as both absorb at 190–220.
A peak at 190-220 is aldehyde or ketone, 160-185 is acid or ester. Learn to read the booklet quickly.
Must Know
- NMR gives the positions of ¹³C or ¹H atoms; techniques are used together to confirm structures
- Chemical shift δ in ppm, relative to TMS at δ = 0; it depends on the molecular environment
- TMS: one sharp peak (four equivalent carbons), inert, volatile, peak away from the sample
- One peak per carbon environment; equivalent carbons share a peak
- ¹³C spectra are simpler than ¹H spectra
- Use the Data Booklet shift ranges to assign peaks
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Principles of NMR and carbon-13 NMR
- Nuclear magnetic resonance (NMR) spectroscopy is used together with other analytical techniques to confirm the structures of new compounds. A chemist records ¹³C NMR spectra of samples dissolved in a suitable solvent, with tetramethylsilane (TMS), Si(CH₃)₄, added as a standard.Explain what is meant by chemical shift and why the chemical shift of a carbon atom depends on its molecular environment.2 marks
- Compound Y has the molecular formula C₄H₈O₂. Its ¹³C NMR spectrum has four peaks, at δ = 14, 21, 60 and 171 ppm. Data booklet shift ranges (δ / ppm): C–C 5–40; C–O 50–90; C=C 90–150; C=O (esters, carboxylic acids) 160–185; C=O (aldehydes, ketones) 190–220.State the type of carbon atom responsible for the peak at δ = 171 ppm and for the peak at δ = 14 ppm in the spectrum of Y, using the shift ranges given.2 marks
- Compound Z has the molecular formula C₃H₆O. Its ¹³C NMR spectrum has two peaks, at δ = 30 and δ = 206 ppm. Data booklet shift ranges (δ / ppm): C–C 5–40; C–O 50–90; C=C 90–150; C=O (esters, carboxylic acids) 160–185; C=O (aldehydes, ketones) 190–220.Deduce the structure of Z. Explain your reasoning using the spectrum.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).