NMR Spectroscopy: Proton and Carbon-13
NMR Spectroscopy: Proton and Carbon-13
Nuclear Magnetic Resonance (NMR) spectroscopy exploits the fact that certain nuclei (¹H and ¹³C) behave as tiny magnets. When placed in a strong external magnetic field, these nuclei can absorb radiofrequency radiation and flip between energy states. The frequency of absorption depends on the chemical environment of the nucleus.
¹H NMR (Proton NMR)
Proton NMR detects hydrogen atoms (¹H) in different chemical environments.
Chemical Shift (δ)
The chemical shift is the position of a peak on the spectrum, measured in parts per million (ppm) relative to the reference standard tetramethylsilane (TMS), which is assigned δ = 0.
Why TMS?
- It gives a single sharp peak (all 12 H atoms are equivalent)
- The peak is at a higher field than most organic signals (so it does not overlap)
- It is inert, volatile, and easy to remove
- It is soluble in most organic solvents
Typical Chemical Shift Values
| Environment | δ / ppm | Example |
|---|---|---|
| R–CH₃ (alkyl) | 0.7–1.6 | CH₃ in ethane |
| R–CH₂–R | 1.2–1.8 | CH₂ in propane |
| R–CO–CH₃ | 2.0–2.5 | CH₃ next to C=O |
| R–O–CH₃ | 3.3–3.9 | CH₃ in methyl ester |
| R–CHO | 9.0–10.0 | Aldehyde H |
| Ar–H | 6.5–8.0 | Benzene ring H |
| R–OH | 0.5–5.0 (variable) | Alcohol OH |
| R–COOH | 10.0–12.0 | Carboxylic acid OH |
| R–NH₂ | 1.0–5.0 (variable) | Amine NH |
Key principle: The greater the electron withdrawal (deshielding) around a proton, the higher the chemical shift (further downfield / to the left).
Number of Peaks
The number of peaks in a ¹H NMR spectrum equals the number of chemically distinct hydrogen environments in the molecule.
Example: Ethanol (CH₃CH₂OH) has three peaks:
1. CH₃ (3 equivalent H atoms)
2. CH₂ (2 equivalent H atoms)
3. OH (1 H atom)
Integration (Relative Peak Areas)
The area under each peak is proportional to the number of hydrogen atoms in that environment. NMR spectrometers display an integration trace (step curve) or print numerical ratios.
If the integration ratio is 3:2:1, this matches 3H : 2H : 1H (as in ethanol above).
Splitting Patterns (Spin-Spin Coupling)
Adjacent non-equivalent hydrogens cause peaks to split according to the (n+1) rule:
If a hydrogen has n equivalent hydrogen neighbours on adjacent carbons, its signal is split into (n + 1) lines.
| n (neighbours) | Splitting | Pattern name | Relative intensities |
|---|---|---|---|
| 0 | 1 line | Singlet (s) | 1 |
| 1 | 2 lines | Doublet (d) | 1:1 |
| 2 | 3 lines | Triplet (t) | 1:2:1 |
| 3 | 4 lines | Quartet (q) | 1:3:3:1 |
| 4 | 5 lines | Quintet | 1:4:6:4:1 |
Key: OH and NH protons usually appear as singlets (they do not split or get split) because rapid exchange with the solvent averages out coupling. A D₂O shake (adding deuterium oxide) replaces OH/NH protons with deuterium — these peaks disappear, confirming their identity.
Worked Example: Propanal (CH₃CH₂CHO)
Three hydrogen environments:
1. CH₃ (δ ≈ 1.0): split by 2 neighbouring CH₂ protons → triplet, integration 3H
2. CH₂ (δ ≈ 2.4): split by 3 CH₃ + 1 CHO neighbours. In practice, coupling to CHO and CH₃ may give a complex multiplet, but at A-Level, consider separately: coupled to CH₃ (3 neighbours → quartet region)
3. CHO (δ ≈ 9.8): split by 2 CH₂ neighbours → triplet, integration 1H
¹³C NMR (Carbon-13 NMR)
¹³C NMR detects the ¹³C isotope (1.1% natural abundance). It provides information about the number and type of carbon environments.
Key differences from ¹H NMR:
- Each chemically distinct carbon gives one peak (no splitting in routine proton-decoupled spectra)
- ¹³C is much less sensitive than ¹H (lower abundance + lower gyromagnetic ratio)
- Peak heights are NOT proportional to the number of carbons (do not use integration)
- The chemical shift range is much wider (0–220 ppm), making it easier to resolve different environments
Typical ¹³C Chemical Shifts
| Environment | δ / ppm |
|---|---|
| R–CH₃, R–CH₂–R, R₃CH | 5–55 |
| C–O (alcohols, ethers) | 50–90 |
| C=C (alkenes) | 100–150 |
| Aromatic C | 110–160 |
| C=O (aldehydes, ketones) | 190–220 |
| C=O (carboxylic acids, esters) | 160–185 |
| C≡N | 110–125 |
Worked Example: ¹³C NMR of Propanone (CH₃COCH₃)
Propanone has two carbon environments:
1. Two equivalent CH₃ carbons (δ ≈ 30 ppm)
2. One C=O carbon (δ ≈ 205 ppm)
The spectrum shows two peaks — confirming the molecule's symmetry.
DEPT Spectra (Extension)
DEPT (Distortionless Enhancement by Polarisation Transfer) is a ¹³C technique that distinguishes CH₃, CH₂, CH, and quaternary C by running three sub-experiments. At A-Level, you may simply need to know that ¹³C NMR plus DEPT can identify how many hydrogens are attached to each carbon.
Solving Structure Problems Using NMR
1. Molecular formula — determine from mass spectrometry or given data
2. Degrees of unsaturation — (2C + 2 + N − H − X) / 2 — tells you the number of double bonds and/or rings
3. ¹³C NMR — how many distinct carbon environments?
4. ¹H NMR — how many distinct H environments, in what ratio (integration), with what splitting pattern?
5. Chemical shifts — match environments to functional groups using data tables
6. Piece together the structure that is consistent with all the data
Exam Tips
- Always quote chemical shift values from the data booklet — do not memorise exact numbers, but know the general ranges
- For splitting, remember: the peak is split by neighbours on adjacent carbons, not by H on the same carbon
- If OH or NH peaks are suspected, describe the D₂O shake test
- In ¹³C NMR, count the number of peaks = number of unique carbon environments (not necessarily the number of carbons)