NMR Basics — The Logic Behind Chemical Shift
Students usually meet chemical shift as a table to be memorised. It is far better understood as a consequence of one simple fact: the magnetic field a nucleus experiences is not the field you applied. Electrons circulate, generate their own small field, and the nucleus feels the difference. Everything in a proton NMR spectrum — why an aldehyde sits at 9.8 ppm, why an alkyne proton sits near 2 ppm despite being on an sp carbon — follows from that.
Shielding, in one equation
σ is the shielding constant. Larger σ → smaller effective field → lower resonance frequency → the signal appears further upfield (smaller δ).
Electron density around a nucleus opposes the applied field, so electron-rich protons are shielded and resonate at low δ, while protons next to electronegative atoms are stripped of electron density, deshielded, and pushed downfield. That single trend explains most of the aliphatic region of a spectrum:
| Proton environment | Typical δ / ppm | Dominant reason |
|---|---|---|
| Si(CH₃)₄ (TMS reference) | 0.00 | Si is electropositive — unusually shielded |
| R–CH₃ | 0.9 | Baseline alkyl |
| R–CH₂–R | 1.3 | Slight deshielding by extra carbon |
| C≡C–H (terminal alkyne) | 1.8–3.1 | Anisotropic shielding by the triple bond |
| CH₃ next to C=O | 2.1–2.6 | Inductive + anisotropy of the carbonyl |
| CH₃–O | 3.3–4.0 | Strong inductive withdrawal by oxygen |
| Vinyl C=C–H | 4.5–6.5 | sp² carbon + π anisotropy (deshielding) |
| Aromatic Ar–H | 6.5–8.0 | Ring current (deshielding outside the ring) |
| R–CHO | 9.5–10.5 | Inductive + carbonyl anisotropy, both deshielding |
| R–COOH | 10–13 | Deshielding plus hydrogen bonding |
Why the scale is in ppm and not in hertz
The resonance frequency scales with the magnet, so the same proton resonates at a different frequency on every spectrometer. Dividing by the spectrometer frequency removes that dependence entirely:
Working form: δ = Δν (Hz) ÷ spectrometer frequency (MHz)
Q. On a 400 MHz spectrometer a signal appears 1240 Hz downfield of TMS. Find δ, and predict where it appears on a 100 MHz instrument.
Working. δ = 1240 Hz ÷ 400 MHz = 3.10 ppm.
On a 100 MHz instrument the same proton is displaced by 3.10 × 100 = 310 Hz — but δ is still 3.10 ppm. That invariance is the entire point of the scale: a shift quoted in ppm is a property of the molecule, while a shift quoted in Hz is a property of the molecule and the magnet.
The one quantity that does not scale: J
Q. A doublet has J = 7.2 Hz on a 400 MHz instrument. What is J at 100 MHz, and what is the separation in ppm in each case?
Working. Coupling arises through bonds, not through the applied field, so J = 7.2 Hz at both fields.
Separation at 400 MHz: 7.2/400 = 0.018 ppm.
Separation at 100 MHz: 7.2/100 = 0.072 ppm.
So on a higher-field magnet multiplets shrink in ppm while chemical shifts stay put — which is exactly why 400 and 600 MHz instruments resolve crowded spectra that a 60 MHz machine turns into an unreadable lump. It also gives you a diagnostic: if a splitting measured in Hz changes when you change magnets, it was never a coupling — it was two different chemical shifts.
Anisotropy — the part electronegativity cannot explain
Inductive reasoning alone predicts that an alkyne proton, on an sp carbon with 50 % s character, should be strongly deshielded — more so than a vinyl proton. Experimentally it appears around 2 ppm, well upfield of the vinyl protons at 5–6 ppm. The reason is the shape of the induced electron circulation.
In an alkyne the cylindrical π system circulates about the molecular axis. The proton lies along that axis, in the region where the induced field opposes B₀, so it is shielded. In benzene the π electrons circulate in the plane of the ring; a proton attached to the ring sits outside the loop where the induced field reinforces B₀, so it is deshielded to 7.3 ppm. Same phenomenon, opposite sign, decided purely by geometry.
The clinching evidence is [18]annulene, an aromatic ring large enough to have protons both outside and inside. Its outer protons appear near δ 9.3 and its inner protons near δ −3.0 — a spread of more than twelve ppm within one molecule, with no difference in electronegativity anywhere. Nothing but a ring current can produce that.
A full worked assignment
Q. Predict the ¹H spectrum of ethyl acetate, CH₃–CO–O–CH₂–CH₃.
Three environments, so three signals, with intensity ratio 3 : 2 : 3.
- δ ≈ 1.26, triplet, 3H, J ≈ 7.1 Hz — the CH₃ of the ethyl group. It has two neighbours, so n + 1 = 3 lines. It is a plain alkyl methyl, hence low δ.
- δ ≈ 2.04, singlet, 3H — the acetyl CH₃. No neighbouring protons, so no splitting; deshielded to about 2 ppm by the adjacent carbonyl.
- δ ≈ 4.12, quartet, 2H, J ≈ 7.1 Hz — the OCH₂. Three neighbours give four lines, and attachment to ester oxygen deshields it strongly.
Check the coupling. The triplet and the quartet must share the same J, because they are coupled to each other. If a printed spectrum shows 7.1 Hz for one and 6.2 Hz for the other, the assignment is wrong — coupling is strictly mutual.
Check the integrals. 3 : 2 : 3 matches C₄H₈O₂. Integration gives ratios, never absolute counts, so you must anchor them against the molecular formula.
Common mistakes
- Explaining every shift by electronegativity. The alkyne proton at 2 ppm and the [18]annulene inner protons at −3 ppm are anisotropy effects, not inductive ones.
- Quoting a coupling constant in ppm. J is always in Hz. Reporting it in ppm hides the field dependence that proves it is a coupling at all.
- Treating integration as an absolute proton count. It is a ratio; scale it using the molecular formula.
- Expecting a sharp OH or NH peak in a predictable place. Exchangeable protons move with concentration, solvent and temperature, and are often broad. Confirm them with a D₂O shake, which removes the signal.
- Applying "n + 1" to non-equivalent neighbours. The rule holds only when all n neighbours are equivalent; otherwise you get a doublet of doublets, not a triplet.
- Forgetting that equivalent protons do not split each other. The six protons of acetone are one singlet.
Summary
| Quantity | Depends on magnet? | Units | What it tells you |
|---|---|---|---|
| Chemical shift δ | No | ppm | Electronic environment of the nucleus |
| Frequency offset Δν | Yes | Hz | δ × spectrometer frequency in MHz |
| Coupling constant J | No | Hz | Number and geometry of neighbours |
| Integration | No | relative | Ratio of protons per environment |
| Multiplicity | No | — | n + 1 for n equivalent neighbours |
Do the arithmetic without slips. Converting between Hz and ppm, scaling integrals and checking molecular formulas are exactly the small calculations that cost marks when rushed. The ABC Chemistry Calculator Suite has the molar mass, unit-conversion and scientific tools you need alongside your spectra.
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