Magnetic resonance and molecular information

A magnetic field splits the states of a suitable nucleus

Nuclear magnetic resonance is isotope-specific. A nucleus with spin quantum number I has 2I+1 magnetic substates; an I=0 nucleus does not supply an ordinary nuclear-spin resonance. For an uncoupled spin-1/2 nucleus, a static field produces two levels and an appropriately tuned radiofrequency magnetic field can induce transitions. Nuclear spin is a quantum property, not a literal small spinning charged ball.

For a positive angular magnetogyric ratio γ in an unshielded model, the gap is ħγB₀ and ordinary frequency is ν = γB₀/(2π). In an isotropic chemical environment, an averaged dimensionless shielding σ modifies the simple expression to ν = γ(1−σ)B₀/(2π). Field, isotope, reference, solvent, temperature and interactions matter for interpreting an actual sample.

A chemical shift is a referenced frequency ratio

Define δ = (ν_sample − ν_ref)/ν_ref, with matched frequencies for the same nucleus and reference convention. The dimensionless fraction multiplied by 10⁶ gives the numerical shift in ppm. A positive δ means the sample frequency is higher than the reference frequency in this convention. Chemical shift is not the same quantity as absolute shielding.

An illustrative offset of 1200 Hz from a 400 MHz reference gives δ = 3.00 ppm. Under comparable isotropic chemical conditions and the same referencing convention, that relative shift gives an 1800 Hz offset at a 600 MHz reference. Use actual reference frequencies when precision matters; a nominal instrument label is not automatically the exact ν_ref.

A simple multiplet requires a simple coupling model

In a weak first-order model, coupling an observed spin-1/2 nucleus to n equivalent spin-1/2 neighbours with equal coupling J gives n+1 components with binomial relative areas. The relevant frequency separation must be much larger than J, and unresolved other couplings, exchange or inequivalent neighbours can change the pattern. Chemical equivalence alone is not always magnetic equivalence. The simple binomial areas also assume approximately equal populations of the neighbour spin configurations.

For a specified three-neighbour model with J=7.00 Hz, a quartet has relative component areas 1:3:3:1 at offsets −3J/2, −J/2, +J/2 and +3J/2 from its centre. These are ideal component areas, not universal peak heights; line widths and acquisition can affect what is displayed.

A small population excess is not a tiny lower-level population

For an ideal free-proton model at stipulated B₀=9.40 T, the current free-proton γ/(2π) gives approximately 400.228 MHz before rounding the field input. At 300 K, nearly half the nuclei occupy each level; the lower-state excess over the upper state is only about 3.20 × 10⁻⁵ of the total. This is not the exact chemically shielded resonance of a proton in a molecule.

Resolved proton integrations can give represented-number ratios under suitable relaxation, acquisition, baseline and separation conditions. Absolute counts also need a scale or other information. Routine decoupled carbon-13 areas are not automatically proportional to carbon counts. NMR patterns can constrain structures without proving a unique structure alone. EPR or ESR instead probes electronic magnetic moments, commonly through microwave transitions of systems with unpaired electrons.

Worked examples

Try it yourself

1. At a 600 MHz reference, what offset corresponds to +2.00 ppm under the stated relative-frequency convention?

   Show the answer and reasoning

   +1.20 × 10³ Hz.

   Multiply the reference frequency by the dimensionless shift 2.00 × 10⁻⁶.

2. How many first-order components follow from one equivalent spin-1/2 neighbour with the stated ordinary coupling assumptions?

   Show the answer and reasoning

   Two.

   The n+1 rule gives a doublet in this restricted model.

3. Does a small thermal lower-minus-upper excess imply only a small fraction occupies the lower state?

   Show the answer and reasoning

   No.

   Almost half can occupy each level while their difference is small.

4. Are routine decoupled carbon-13 signal areas automatically quantitative carbon counts?

   Show the answer and reasoning

   No.

   Relaxation and acquisition effects require appropriate quantitative conditions.

What to carry forward

• Specify isotope, field and reference.
• Keep ordinary frequency distinct from angular γ.
• Use n+1 only with its weak-coupling equivalence assumptions.
• Population excess, component areas and quantitative atom counts are different quantities.