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Proton nuclear magnetic resonance (¹H NMR) spectroscopy is the most information-rich tool the synthetic organic chemist has for determining how the hydrogen atoms in a molecule are arranged. Every chemically distinct hydrogen "environment" produces its own peak; the position of that peak (chemical shift δ) reports the electronic neighbourhood; the area under the peak (integration) reports how many hydrogens share that environment; and the splitting pattern (multiplicity, governed by the n+1 rule) reports how many hydrogens sit on the immediately adjacent carbons. Together with the coupling constant J (Hz) — beyond the strict A-level syllabus but indispensable in real diagnosis — these four parameters let a chemist reconstruct a carbon skeleton on paper from a single trace. The whole experiment is referenced to tetramethylsilane (TMS, δ = 0) and run in a deuterated solvent (CDCl₃, D₂O, DMSO-d6) so the solvent itself does not swamp the spectrum. This lesson develops each parameter, then runs four worked unknowns of increasing complexity.
Spec mapping (AQA 7405): This lesson maps to §3.3.15 of the AQA A-Level Chemistry specification (NMR spectroscopy — proton NMR). It builds directly on §3.3.15 ¹³C NMR (lesson 3, carbon-13-nmr) and is preceded by §3.3.13 mass spectrometry (lesson 1, mass-spectrometry, anchored in L0 of this course) and §3.3.14 infrared spectroscopy (lesson 1 of the Analytical Techniques sequence, infrared-spectroscopy, anchored in L1). The synthesis lessons (L6 combined-spectroscopy and L7 unknown-identification) require students to chain MS → IR → ¹³C → ¹H to deduce full structures. Refer to the official AQA specification document for exact subsection wording.
Assessment objectives: Recall of the chemical shift table, the role of TMS as the reference, the requirement for deuterated solvents, and the statement of the n+1 rule are AO1 items. Interpreting a given ¹H spectrum (counting environments, reading off δ, computing integration ratios, and applying the n+1 rule to assign multiplicity) is AO2 — the bulk of the marks in any spectroscopy question. Deducing a full structure from a combined data set (molecular formula + ¹H spectrum + optional ¹³C / IR / MS / D₂O shake) is AO3 — the synoptic top-of-paper exercise that distinguishes A-grade from A* candidates.
The ¹H nucleus (a single proton) has an intrinsic property called nuclear spin, with quantum number I = ½. Place it in a strong external magnetic field B₀ and the spin can take one of two orientations: parallel to the field (lower energy, the α state, m_I = +½) or antiparallel (higher energy, the β state, m_I = −½). The energy gap between the two states is
ΔE = γℏB₀ = hν
where γ is the gyromagnetic ratio of the proton (γ/2π ≈ 42.58 MHz T⁻¹) and ν is the Larmor frequency. In a 400 MHz spectrometer (B₀ = 9.4 T) the proton Larmor frequency is, by definition, 400 MHz. Applying a radio-frequency (RF) pulse at exactly that frequency excites α → β transitions; the relaxation of nuclei back to α emits the same frequency, which the instrument detects as a free-induction decay (FID) and Fourier-transforms into the familiar spectrum.
The key point — and the reason NMR is a structure-determination technique at all — is that not every proton in a molecule resonates at exactly the bulk Larmor frequency. The local electron cloud around each ¹H nucleus generates a small opposing magnetic field that "shields" the proton from B₀, so the effective field it experiences is
B_eff = B₀ (1 − σ)
where σ is the (dimensionless) shielding constant. Different chemical environments have different electron densities and therefore different σ values; this is why a CH₃ adjacent to a CH₂OH gives a peak at a different position from the OCH₂ adjacent to it.
Because the absolute resonance frequency depends on instrument field strength (60, 100, 250, 400, 600, 800 MHz are all common), positions are reported as a chemical shift δ in parts per million (ppm) relative to a reference compound:
δ = (ν_sample − ν_reference)/ν_reference × 10⁶
The standard reference is tetramethylsilane, Si(CH₃)₄ (TMS), defined as δ = 0. TMS is chosen because:
Higher δ = lower shielding (the proton experiences more of the external field) = more electronegative or deshielding environment. Lower δ = higher shielding = more "electron-rich" environment.
Key Definition: Chemical shift δ is the dimensionless, instrument-independent position of an NMR absorption, reported in ppm relative to TMS (δ = 0). It encodes the electronic environment of the nucleus.
The following ranges must be familiar; AQA will provide a data booklet in the exam, but fluency means you read assignments off instantly rather than chasing the booklet line-by-line. Approximate values, in ppm:
| Hydrogen Environment | δ (ppm) | Example |
|---|---|---|
| Alkyl CH/CH₂/CH₃ (no nearby functional group) | 0.7–1.5 | CH₃ in propane (≈0.9) |
| Allylic C–H (next to C=C) | 1.7–2.5 | CH₃ in propene (≈1.7) |
| C–H adjacent to C=O (α to ketone, ester, aldehyde) | 2.0–2.7 | CH₃ in propanone (≈2.1) |
| C–H adjacent to OR (ether, alcohol, ester O–CH) | 3.3–4.0 | OCH₃ in methanol (≈3.4) |
| C–H adjacent to halogen | 3.0–4.5 | CH₃ in chloromethane (≈3.1) |
| Vinyl C–H (alkene =C–H) | 4.5–7.0 | CH₂ in ethene (≈5.3) |
| Aromatic C–H | 6.5–8.0 | C–H in benzene (7.3) |
| O–H, alcohol (broad, exchangeable) | 1–5 | OH in ethanol (variable) |
| O–H, carboxylic acid (broad, very deshielded) | 9–12 | OH in ethanoic acid (≈11.5) |
| N–H, amine (broad, exchangeable) | 0.5–4.0 | NH₂ in ethylamine (variable) |
| Aldehyde C–H (CHO) | 9.0–10.0 | H–C=O in ethanal (≈9.8) |
Exam Tip: The two broad, variable, exchangeable resonances (O–H, N–H) shift dramatically with concentration, solvent and hydrogen bonding. The carboxylic O–H sits very deshielded (9–12 ppm) because the C=O withdraws electron density from the O–H via the carboxyl group. Both classes disappear on a D₂O shake — see below.
Within a single class, the precise δ varies systematically with the electronegativity of attached atoms:
The effect decays sharply through bonds: in 1-chloropropane CH₃CH₂CH₂Cl, the C1 protons (δ ≈ 3.5) are heavily deshielded, the C2 protons (δ ≈ 1.7) modestly, and the C3 methyl (δ ≈ 1.0) hardly at all.
Two protons are in the same chemical environment if they can be interconverted by a symmetry operation of the molecule (a rotation, reflection, or improper rotation that maps the molecule onto itself). In practice, draw the structure and ask: which Hs are interchangeable by symmetry?
The number of peaks in an ideal spectrum equals the number of environments; degenerate accidental overlap of inequivalent peaks is a real-world annoyance but can be ignored at A-level.
The area under each peak (the integral) is strictly proportional to the number of protons in that environment, because the RF detector measures total absorbed energy, which scales with the number of resonating nuclei. Modern spectrometers display either a stepped integral trace or numerical integrals next to each peak.
For ethanol (CH₃CH₂OH):
Exam Tip: In exam questions the integral is often reported as a relative number (e.g. "1.0 : 0.67 : 0.33"). Multiply through to clear decimals and check the total matches the number of hydrogens in the molecular formula. If your ratio gives a sum that doesn't divide cleanly into the H-count in the formula, recheck — accidental peak overlap or miscounted environments is the usual cause.
Protons on adjacent carbons couple magnetically through the bonding electrons (a three-bond, H–C–C–H, through-bond interaction). The local field experienced by one proton is shifted up or down by a tiny amount depending on the spin orientation (α or β) of each neighbouring proton. The result: a peak that would otherwise be a singlet is split into a set of closely-spaced sub-peaks.
The n+1 rule states: if a proton has n equivalent neighbouring protons (on directly bonded adjacent carbons), its signal is split into n+1 lines, with intensities given by Pascal's triangle.
| Neighbours (n) | Pattern | Name | Intensities (Pascal) |
|---|---|---|---|
| 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 |
| 5 | 6 lines | sextet | 1 : 5 : 10 : 10 : 5 : 1 |
Three rules govern when the n+1 rule applies:
Adding a few drops of D₂O (deuterium oxide, "heavy water") to an NMR sample swaps the labile protons (those on O, N or S) for deuterium:
R–OH + D₂O ⇌ R–OD + HOD
Deuterium (²H, I = 1) has a different Larmor frequency from ¹H and is invisible in the ¹H spectrum. So after a D₂O shake, the OH, NH, COOH and other exchangeable peaks disappear (or shift to the HOD water peak at δ ≈ 4.7 ppm). Non-exchangeable C–H peaks are unaffected.
Exam Tip: "Which peak disappears on a D₂O shake?" is the standard examiner's probe for O–H or N–H assignment. The answer should always reference the rapid proton exchange between the labile site and the D₂O reagent — not just "because it's an alcohol."
Both are C₂H₆O (Mᵣ = 46), so MS alone cannot separate them.
The D₂O shake is the cleanest, fastest tie-breaker for that pair.
If the NMR sample were dissolved in ordinary (protiated) chloroform, the solvent itself would contribute roughly 10⁴ more protons to the spectrum than the dissolved analyte — its peak would saturate the receiver and bury the analyte signals. The fix is to use a deuterated solvent:
The residual protiated solvent peak is a useful internal δ check ("the CHCl₃ in this CDCl₃ is at 7.26 — calibration is correct").
A-level treats multiplicity qualitatively (a triplet, a doublet, a quartet), but in reality each multiplet has a measurable coupling constant J, in hertz (not ppm — J is field-independent). The spacing between adjacent lines within a multiplet is J. Typical values:
Two coupled protons share the same J. The clearest A-level use is distinguishing cis- from trans-alkenes by inspecting the J across the C=C: J ≈ 10 Hz → cis; J ≈ 17 Hz → trans. This is the standard A*-band synoptic move and one of the very few places coupling constants enter UK pre-undergraduate work.
A workable NMR sample requires:
Practical-skills box: A real spectrum has a baseline, peak labels in ppm, multiplicities and integrals — all of which can be redrawn on paper for an exam answer.
Four unknowns of increasing complexity. In each case the question is the same: deduce the structure from the molecular formula and the ¹H spectrum.
Spectrum:
Degree of unsaturation = (2·2 + 2 − 6)/2 = 0 (no ring or multiple bond). Three environments, integration 3:1:2 = 6H total (matches the formula). D₂O-exchangeable singlet at δ 2.6 → OH. Triplet (3H) coupled to a quartet (2H) → CH₃–CH₂– fragment. The CH₂ is at δ 3.7, consistent with attachment to oxygen. Assemble: CH₃CH₂OH (ethanol).
Spectrum:
Degree of unsaturation = (2·4 + 2 − 8)/2 = 1 (one C=O or ring). Three environments, integration 3:3:2 = 8H total. Singlet at δ 2.04, 3H → an isolated CH₃ with no neighbouring Hs, consistent with CH₃–C(=O)–. Triplet (3H) at δ 1.26 coupled to a quartet (2H) at δ 4.12 → CH₃–CH₂–O– (the high δ of the CH₂ shows it is bonded to an oxygen). Assemble: CH₃C(=O)OCH₂CH₃ (ethyl ethanoate / ethyl acetate).
Spectrum:
Degree of unsaturation = 1 (C=O implied). Three environments, integration 3:3:2 = 8H total. Singlet 3H at δ 2.14 → isolated CH₃ next to C=O (consistent with the methyl of CH₃C(=O)–). Triplet 3H at δ 1.05 coupled to quartet 2H at δ 2.46 → CH₃–CH₂– fragment, with the CH₂ deshielded into the C–C(=O) range (δ 2.0–2.7). Assemble: CH₃–C(=O)–CH₂–CH₃ (butan-2-one / butanone). Note how distinguishing butanone from ethyl ethanoate hinges on the shift of the CH₂ (2.46 vs 4.12) — the same multiplicity pattern, but the CH₂ next to a C=O is in a different shift window from a CH₂ next to an O-of-ester.
Spectrum:
Degree of unsaturation = (2·3 + 2 − 6)/2 = 1. Three environments, integration 3:2:1 = 6H total. The very deshielded broad singlet at δ 11.7 that exchanges on D₂O is the hallmark of a carboxylic acid O–H (range 9–12). The triplet/quartet at 1.16/2.36 ppm is a CH₃CH₂– fragment with the CH₂ in the "α to C=O" window (2.0–2.7). Assemble: CH₃CH₂COOH (propanoic acid).
These four worked examples cover the recurring exam moves: count environments, read the chemical shift table, apply the n+1 rule for multiplicity, use the D₂O shake to pin labile protons, and assemble fragments using the integration to balance the H-count.
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