共鳴 Magnetic Resonance · Laboratory experiment

Proton NMR of simple organic liquids: chemical shift and spin–spin coupling

Aim: to acquire the free-induction decay of a liquid sample in a pulsed FT-NMR spectrometer, Fourier-transform and phase it, measure the chemical shifts δ relative to TMS, the coupling constants J and the integrals, and show that δ (ppm) and J (Hz) do not depend on the field while the multiplet separations in Hz do.

FT-NMR spectrometer

superconducting magnet · probe · console · Bloch sphere

Free-induction decay (real channel)

¹H spectrum

Click a line to pick its exact position; double-click to zoom out. The δ axis runs right to left, as is conventional. The integral trace is shown in orange.

Table A: picked lines

Table B: integrals

Analysis

Theory

A proton (I = ½) in a field B₀ precesses at the Larmor frequency \(\nu = \gamma B_0/2\pi\) (42.577 MHz/T). Electrons around the nucleus shield it, \(B_{loc} = B_0(1-\sigma)\). Shifts are quoted relative to tetramethylsilane (TMS) and are field independent in ppm:

$$ \delta = \frac{\nu - \nu_{TMS}}{\nu_0}\times10^6\ \text{ppm} .$$

Neighbouring spins couple through the bonding electrons with \(H_J = J\,\mathbf I_1\cdot\mathbf I_2\). When \(\Delta\nu \gg J\) (first order), n equivalent neighbours split a signal into n + 1 lines with binomial intensities (Pascal's triangle), spaced by J Hz, which is independent of the field.

Exact simulation

This spectrometer does not use the n + 1 rule. For every coupled group it builds the full spin Hamiltonian

$$ \hat H = \sum_i \nu_i \hat I_{z,i} + \sum_{i<j} J_{ij}\,\hat{\mathbf I}_i\cdot\hat{\mathbf I}_j ,$$

diagonalises it exactly (block by block in Mz) and computes the transition intensities \(|\langle b|\hat I^-|a\rangle|^2\). At low field (60 MHz), when Δν approaches J, you therefore see real second-order effects: roofing (inner lines grow) and extra lines in the aromatic region of toluene. At 600 MHz the spectrum becomes first order.

The spectrum is obtained like a real one. The complex FID \(s(t)=\sum_k a_k e^{i2\pi(\nu_k-\nu_{ref})t}e^{-t/T_2^*}\) plus noise (S/N ∝ √NS) is apodised, zero-filled, Fourier-transformed and phased.

Procedure

  1. Choose ethanol at 300 MHz and acquire 8 scans. Auto-phase the spectrum.
  2. Click every line of each multiplet. Label the lines in a multiplet with the same group name (e.g. CH₂). The analysis gives δ as the multiplet centre and J as the mean line spacing.
  3. Switch to integrate mode and drag across each multiplet. Integrals are normalised to the smallest one, which gives the proton ratios (3 : 2 : 1 for CH₃ : CH₂ : OH).
  4. Repeat at 60 MHz: δ and J are unchanged, but the multiplets move closer together in Hz.
  5. Try toluene at 60 MHz and at 600 MHz to see second-order effects disappear.

Reference data (CDCl₃)

  • Ethanol: CH₃ 1.25 (t), CH₂ 3.72 (q), OH ≈ 1.3 (s, exchange-broadened); J = 7.0 Hz
  • Ethyl acetate: CH₃CO 2.05 (s), OCH₂ 4.12 (q), CH₃ 1.26 (t); J = 7.1 Hz
  • 2-Propanol: CH₃ 1.22 (d), CH 4.04 (septet); J = 6.1 Hz
  • Acetaldehyde: CHO 9.80 (q), CH₃ 2.21 (d); J = 2.9 Hz
  • Toluene: CH₃ 2.36 (s), aromatic 7.16–7.25 (m)

Viva questions

Why does the OH proton not show coupling?

It exchanges between molecules faster than 1/J (catalysed by traces of acid or water), so its coupling averages to zero and the line broadens.

Why is TMS the reference?

It has 12 equivalent, strongly shielded protons that give one sharp line at the high-field end. It is inert and volatile.

Why do equivalent protons not split each other?

For magnetically equivalent spins, the coupling term commutes with the total-spin operators that cause the transitions, so it has no effect on the observed spectrum.