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Nuclear magnetic resonance (NMR) spectroscopy is arguably the most powerful analytical technique available for determining the structure of organic molecules. While mass spectrometry tells you the molecular mass and IR reveals functional groups, NMR tells you about the carbon and hydrogen framework — how many distinct environments exist, how many atoms are in each, and how they are connected.
This lesson focuses on the principles of proton (¹H) NMR, the most commonly used form of NMR spectroscopy.
Certain atomic nuclei possess a property called nuclear spin. Nuclei with an odd number of protons, an odd number of neutrons, or both, have non-zero spin and are described as "NMR-active." The nuclei most relevant to organic chemistry are:
The common isotope ¹²C has zero spin and is NMR-inactive. This is why ¹³C NMR relies on the small natural abundance (1.1%) of the ¹³C isotope.
When placed in a strong external magnetic field (B₀), nuclei with spin = ½ can adopt one of two orientations:
The energy difference (ΔE) between these two states depends on the strength of the external magnetic field and the type of nucleus. When radiofrequency (RF) radiation of exactly the right frequency is applied, nuclei in the lower energy state absorb the energy and "flip" to the higher energy state. This is resonance.
The frequency of radiation needed is typically in the radiofrequency range (hundreds of MHz), which is why NMR spectrometers use powerful magnets and RF transmitters.
Different hydrogen atoms in a molecule experience slightly different local magnetic fields because the electrons surrounding them shield the nucleus from the external field. Electron-rich environments provide more shielding, reducing the effective magnetic field felt by the nucleus, while electron-poor environments provide less shielding (deshielding).
The result is that different hydrogens resonate at slightly different frequencies. Rather than recording these as absolute frequencies (which depend on the instrument's magnetic field strength), we report them as chemical shifts (δ) measured in parts per million (ppm).
The chemical shift scale is calibrated using tetramethylsilane, (CH₃)₄Si (TMS), which is defined as δ = 0.00 ppm. TMS is used because:
All chemical shifts in a ¹H NMR spectrum are measured relative to TMS. Most organic hydrogen signals appear between δ = 0 and δ = 12 ppm.
The following table summarises the chemical shift ranges you need to know. Learn these values — they are essential for interpreting spectra:
| Type of proton | δ / ppm | Typical example |
|---|---|---|
| R–CH₃ (alkyl, no nearby electroneg. groups) | 0.7–1.2 | CH₃ in ethane (δ ≈ 0.9) |
| R–CH₂–R (alkyl) | 1.2–1.4 | CH₂ in propane (δ ≈ 1.3) |
| R₃CH (alkyl) | 1.4–1.7 | CH in 2-methylpropane |
| CH₃–C=O or CH₃–C=C | 1.9–2.5 | CH₃COCH₃ (δ ≈ 2.1) |
| N–CH₃ or N–CH₂ | 2.2–2.8 | N(CH₃)₃ (δ ≈ 2.3) |
| O–CH₃ or O–CH₂ | 3.3–4.3 | CH₃OH (δ ≈ 3.4) |
| Cl–CH or Br–CH | 3.0–4.2 | CH₃Cl (δ ≈ 3.1) |
| C=C–H (alkene) | 4.5–6.0 | CH₂=CH₂ (δ ≈ 5.3) |
| Ar–H (aromatic) | 6.5–8.0 | C₆H₆ (δ ≈ 7.3) |
| R–CHO (aldehyde) | 9.0–10.0 | CH₃CHO (δ ≈ 9.8) |
| R–COOH (carboxylic acid) | 10.0–12.0 | CH₃COOH (δ ≈ 11.4) |
| R–OH (alcohol) | 0.5–5.0 (variable) | Depends on H-bonding |
| R–NH₂ (amine) | 1.0–4.5 (variable) | Depends on H-bonding |
A central concept in NMR is identifying which hydrogen atoms are in the same chemical environment (equivalent) and which are in different environments (non-equivalent).
Equivalent protons are hydrogen atoms that are in identical chemical environments. They give rise to the same NMR signal. To determine equivalence, ask: "If I replaced this hydrogen with a different atom (say deuterium), would the resulting molecule be identical to the one formed by replacing the other hydrogen?"
Ethane (CH₃CH₃): All 6 hydrogens are equivalent — one peak in the ¹H NMR.
Ethanol (CH₃CH₂OH): Three environments — the CH₃ hydrogens (3H), the CH₂ hydrogens (2H), and the OH hydrogen (1H). This gives three peaks.
Propanone (CH₃COCH₃): All 6 hydrogens are equivalent (both methyl groups are in identical environments) — one peak.
Propan-1-ol (CH₃CH₂CH₂OH): Four environments — CH₃ (3H), central CH₂ (2H), CH₂ next to OH (2H), and OH (1H). Four peaks.
The number of peaks in a ¹H NMR spectrum equals the number of distinct hydrogen environments in the molecule.
How many distinct ¹H environments are there in each of the following?
| Compound | Formula | Environments | Number of peaks |
|---|---|---|---|
| Methane | CH₄ | All H equivalent | 1 |
| Chloromethane | CH₃Cl | All 3 H equivalent | 1 |
| Ethanol | CH₃CH₂OH | CH₃, CH₂, OH | 3 |
| Propanone | CH₃COCH₃ | Both CH₃ equivalent | 1 |
| Para-xylene | 1,4-C₆H₄(CH₃)₂ | Ring H (4H), CH₃ (6H) | 2 |
| Ethyl ethanoate | CH₃COOCH₂CH₃ | 3 environments | 3 |
The area under each peak (the integral) is proportional to the number of hydrogen atoms in that environment. Modern spectrometers display integration as a stepped line superimposed on the spectrum, or as numerical values.
For example, if a spectrum shows two peaks with an integration ratio of 3:2, there are 3 hydrogens in one environment and 2 in the other (or 6:4, or 9:6 — the ratio is what matters).
Integration does not give the absolute number of hydrogens directly. You need additional information (such as the molecular formula from mass spectrometry) to determine the actual numbers. If the molecular formula tells you there are 10 hydrogens total and the integration ratio is 3:2, then the actual numbers are 6H and 4H.
NMR samples are usually dissolved in a deuterated solvent such as CDCl₃ (deuterated chloroform) or D₂O (deuterium oxide). Deuterium (²H) has a different resonance frequency from ¹H, so the solvent does not produce interfering peaks in the ¹H NMR spectrum.
However, CDCl₃ always contains a tiny amount of CHCl₃ (with regular ¹H), which appears as a small peak at δ = 7.26 ppm. This is sometimes visible in spectra and should not be confused with a sample peak.
The hydroxyl (O–H) proton in alcohols and carboxylic acids is often broad and can appear at variable chemical shift positions depending on concentration, temperature, and solvent. This is because:
The D₂O shake test exploits this: if you add a few drops of D₂O to the NMR sample and the peak disappears, it was an exchangeable O–H (or N–H) proton. This is a standard technique for identifying O–H and N–H peaks.
Mistake 1: Confusing the number of peaks with the number of hydrogen atoms. The number of peaks equals the number of hydrogen environments, not the total number of H atoms. A CH₃ group gives one peak integrating for 3H.
Mistake 2: Forgetting molecular symmetry. Propanone (CH₃COCH₃) has 6 hydrogens but only one peak because the two CH₃ groups are equivalent by symmetry.
Mistake 3: Assuming an OH peak is always present and always at the same position. The OH chemical shift is variable (δ = 0.5–5.0 for alcohols, δ = 10–12 for acids) and the peak may be broadened beyond detection. Use the D₂O shake test to identify it.
Mistake 4: Not using the molecular formula to determine absolute hydrogen numbers from integration ratios. Integration gives ratios only — you need the total number of H from the molecular formula to convert ratios to actual numbers.
¹H NMR spectroscopy reveals:
Combined with molecular formula information from mass spectrometry and functional group data from IR, NMR provides the detailed structural framework needed to identify unknown compounds. In the next lesson, you will learn how to use chemical shift values and splitting patterns to determine connectivity.
Edexcel 9CH0 specification, Topic 19 — Modern Analytical Techniques II, sub-strand 19.1 introduces ¹H (proton) nuclear magnetic resonance (NMR) spectroscopy: the principle that ¹H nuclei in different chemical environments resonate at different frequencies in a magnetic field; the use of tetramethylsilane (TMS, Si(CH₃)₄) as the chemical-shift reference at δ = 0 ppm; the relationship between number of peaks and number of distinct ¹H environments; the use of integration (peak area) to determine the ratio of hydrogens in each environment; and the use of CDCl₃ or other deuterated solvents (refer to the official specification document for exact wording). Examined on Paper 2 (Core Organic and Physical Chemistry) and synoptically on Paper 3 (General and Practical Principles). The data booklet provides chemical-shift ranges for ¹H environments.
Question (7 marks):
Consider the compounds below:
(a) For each compound, state the number of ¹H environments and the integration ratio you would expect to observe in the ¹H NMR spectrum. (4)
(b) Explain why the OH proton in ethanol typically appears as a broad singlet at variable chemical shift. (2)
(c) State why deuterated chloroform (CDCl₃) is used as the solvent rather than CHCl₃. (1)
Solution with mark scheme:
(a)
(b) The OH proton exchanges rapidly between molecules via hydrogen bonding (proton exchange), and the rate of exchange depends on concentration, solvent, and trace water. M1. Because the proton spends a fraction of time as part of different hydrogen-bonded states, its chemical shift varies and the peak appears broad and at variable position; for the same reason, OH protons usually do not show the splitting expected from coupling to neighbouring CH protons. A1.
(c) CHCl₃ would give a strong ¹H NMR peak at ~7.26 ppm from the solvent proton, swamping the spectrum. CDCl₃ has ²H (deuterium) instead of ¹H and so contributes no significant ¹H peak (the residual CHCl₃ in CDCl₃ gives only a tiny peak at 7.26, used as a secondary reference). A1.
Total: 7 marks (M5 A2).
Question (6 marks): Predict the number of ¹H environments and the relative integration for each of the following:
(a) methylbenzene (toluene) C₆H₅CH₃ — (2) (b) methylpropane (CH₃)₃CH — (2) (c) propan-1-ol CH₃CH₂CH₂OH — (2)
Mark scheme decomposition by AO:
(a)
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