2
$\begingroup$

Why do nuclei that have a smaller / faster correlation time have a higher / slower T1 / T2?

From my understanding:

  • Fast Brownian motion creates a wide range of B.local (local magnetic field created by dipole of spin X near spin Y) fluctuation frequencies
  • Slow motion gives rise to fewer frequencies but more intense fluctuations
  • Fluctuations at the right frequency cause transitions between energy levels / relaxation / re-establishment of equilibrium
  • The larger a molecule the slower it tumbles (correlation time is slower)
  • The slower the tumbling the faster the relaxation

To me, logically, it'd make sense that if a molecule tumbles faster then the relaxation should be faster, not the other way round. Can anyone explain this?

Thanks

$\endgroup$
  • $\begingroup$ Careful: It's not so much about the intensity of the fluctuations, but about how their influence integrates over time. $\endgroup$ – Karl Dec 13 '18 at 18:55
  • 1
    $\begingroup$ Fast Brownian motion does not usually create larger fluctuations, they just have a wider frequency spectrum. $\endgroup$ – Karl Dec 13 '18 at 18:57
2
$\begingroup$

First recall that the T1 and T2 times reflect different things in the NMR experiment. T1 is the decay of population back to equilibrium and T2 is the decay of the coherence between the spins so we expect them to behave differently in different solvents.

Because of the interaction with other nuclear spins in the molecule, and with electrons in the solvent molecules as they tumble around in solution, the individual magnetic moments making up the magnetisation experience slightly different local magnetic fields. These fields change in time (because the molecules are tumbling in solution) and induce random transitions between the NMR spin states and these random transitions must cause the magnetisation to decay back to equilibrium. The main causes are

(a) Rotational Diffusion. Molecules interact via dipole-dipole coupling which changes as the molecules rotate and/or diffuse. Rotational diffusion is a major cause of decay.

(b) Spin-Spin interaction. This is a direct effect caused by any paramagnetic species, eg. O$_2$ and can be dominant in aerated solutions. It occurs because the magnetogyric ratio ($\gamma$) for paramagnetic molecules and ions is approx 1000 times greater than for nuclei.

(c) Chemical Shift Anisotropy. The shielding observed in chemical shifts is due to the locally induced magnetic fields. As a molecule tumbles in solution, its position with respect to the external field $B_0$ changes and this acts to reduce the magnetisation because, from the molecule's viewpoint, the external magnetic field now appears to be randomly varying.

Solvent motion is characterised by a correlation time $\tau_c$ describing the angle that a solvent molecule has at any time $t$ compared to the angle it had at time zero, clearly this will depend on the nature and viscosity of the solvent. The correlation time can be considered as the tumbling (rotational diffusion) time of a molecule in solution, and is proportional to the solute size and solution viscosity.

Population decay T1 occurs fastest when motion of solvent is of same frequency as transition frequency, so is most effective close to transition (approx RF) frequency, say, 400MHz and so T1 times are short. (You can think of this, classically, as two oscillators of the same frequency becoming in resonance with one another and so are able to interact maximally).

In solvents with long correlation times motion is too slow to cause transitions and the solvent can be likened to an effective static average field and so interact only slightly and T1 times are long.

In solvents with short correlation times these can be so short that they present only an average field to the nuclear spin i.e. they move very fast compared to the solute's motion, and so interact only slightly and T1 is long. The effect this has is shown in the figure.

T2 dephasing time is unaffected by fast solvent for the same reasons as for T1 just described, but because T2 depends on dephasing between spins, which has a range of low frequencies, therefore T2 is sensitive to slow solvent motion (i.e. is in resonance with slow motion) and decreases as $\tau_c$ increases.

T1T2 decay times

Sketch of log T1 and log T2 times vs log correlation time in an NMR experiment.

$\endgroup$
4
$\begingroup$

Relaxation is often very counterintuitive and isn't a particularly easy subject to understand, so it's not always useful to argue based on what seems to be logical. I'm not sure where you're studying this from, but it seems to be slightly lacking in the detail needed to properly understand what's going on. Here is a brief overview of what's going on, but I strongly recommend consulting a textbook which will explain it in more detail. Chapter 9 of James Keeler's Understanding NMR Spectroscopy (2nd ed.) is a very good reference. You can find a pdf of the corresponding chapter in the previous edition here.

The link between rate of tumbling (correlation time, $\tau_c$) and the "amount of fluctuations at the right frequency" is given by the spectral density

$$j(\omega) = \frac{2\tau_c}{1 + \omega^2\tau_c^2}$$

This function tells you how much the random fields are fluctuating at each frequency $\omega$. For relaxation, the important frequency to look at is the Larmor frequency $\omega_0$, i.e. the central quantity is

$$j(\omega_0) = \frac{2\tau_c}{1 + \omega_0^2\tau_c^2}$$

By differentiating this with respect to $\tau_c$ you can find that this is at a maximum when $\tau_c = 1/\omega_0$. The rate of spin-lattice (longitudinal) relaxation is proportional to $j(\omega_0)$, so it is fastest when $\tau_c = 1/\omega_0$. That means that $T_1$ will be smallest when $\tau_c = 1/\omega_0$. When $\tau_c$ decreases below this, or increases above this, spin-lattice relaxation will be slowed down and $T_1$ will increase.

On the other hand, spin-spin (transverse) relaxation depends on the sum $j(\omega_0) + j(0)$. $j(0)$ is simply equal to $2\tau_c$, and you can show (again with calculus) as $\tau_c$ increases, the total sum $j(\omega_0) + j(0)$ increases continually. So $T_2$ decreases when $\tau_c$ increases.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.