The linewidth of peaks when taking NMR spectra of quadrupolar nuclei (11B or 14N for example) is often very broad, making their interpretation challenging.

Recently, when acquiring 11B-NMR spectra of some boronic esters I'd made, the NMR technician pointed out that the linewidth is much improved in highly symmetrical environments (which my boronic esters were not) – she proved this by running a sample of $\ce{LiBH4}$, but couldn't really offer an explanation of why this was the case, other than to point out that its a relaxation effect.

Claridge[1] confirms this point about symmetry:

14N linewidth of $\ce{N(Me)4+}$ is less than 1 Hz whereas that for $\ce{NMe3}$ is nearer to 80 Hz.

Arguably, both $\ce{N(Me)4+}$ and $\ce{NMe3}$ are fairly symmetrical, being Td and pseudo-C3v respectively, which leads me to think that what the molecule needs is some kind of spherical symmetry, which only Td (and Oh, arguably less useful in synthetic organic chemistry) can provide.

Some further reading through Claridge[1] reveals that this is something known as quadrupolar relaxation, with Reich[2] confirming, though not really expanding beyond an equation describing the relaxation process and a comment that there is "friction" between the nucleus and the surrounding electron.

All of the explanations I've found have focused a lot on explaining this mathematically (Levitt describes it quite nicely, but slightly beyond my understanding). Is there an intuitive explanation for the relaxation taking place in quadrupolar nuclei, and why high symmetry environments lead to narrow peak shape?

[1]: Claridge, T. D. W. High Resolution NMR Techniques In Organic Chemistry; Elsevier:Amsterdam, 2016

[2]: https://www.chem.wisc.edu/areas/reich/nmr/08-tech-01-relax.htm, accessed 20/08/2017.

  • 3
    $\begingroup$ A large electric field gradient at the nucleus facilitates quadrupolar relaxation, in a highly symmetric environment the field gradient is (near) zero and quadrupolar relaxation becomes much slower. $\endgroup$ – ron Aug 20 '17 at 13:50

I know that you asked for an intuitive explanation, but I'm afraid that if one wants to go beyond ron's comment, the technical aspects are somewhat necessary. It isn't incredibly important to understand the maths, but I have included it as I want to enable interested readers to go further, and it provides a basis for the last section, so I make no apology for it. I have also assumed some basic knowledge of NMR theory.1

Linewidths and relaxation

Following a pulse sequence, the spectrometer detects transverse magnetisation (i.e. magnetisation in the xy-plane). The signal that is detected is generally of the form $s(t) \propto \exp(\mathrm i \Omega t)\exp(t/T_2)$, where $T_2$ is the transverse or spin-spin relaxation time.

If one takes the Fourier transform of $s(t)$, one gets the spectrum $S(\Omega)$, and it can be shown that the widths of the Lorentzian peaks thus obtained are proportional to $1/T_2$. To be precise, the linewidth at half-height $\Delta \nu_{1/2}$ is given by

$$\Delta \nu_{1/2} = \frac{1}{\pi T_2}$$

Therefore, fast transverse relaxation (which corresponds to small $T_2$) leads to broad lines, and slow relaxation leads to narrow lines.

Quadrupoles and the electric field gradient

For quadrupolar nuclei ($I > 1/2$), a major source of relaxation arises due to interaction of the nuclear electric quadrupole moment with the electric field gradient at the nucleus.

An introduction to the above two concepts is as follows: a quadrupole moment reflects an asymmetry in the charge distribution of the nucleus, and can be thought of as two opposing dipoles placed next to each other. The electric field gradient refers to the derivative of the electric field $E$, which is itself the derivative of the electric potential $V$ (so, the field gradient is the second derivative of the potential). These derivatives are evaluated at the spatial position of the nucleus.

The electric field gradient therefore most generally has nine components, since two derivatives, each with respect to 3 axes, are taken. These are represented by $q_{ij}$ (where $i,j$ are one of $x,y,z$):

$$q_{ij} \equiv \left.\frac{\partial^2V}{\partial i\,\partial j}\right|_\text{at nucleus}$$

However, by a suitable transformation of coordinates, the cross derivatives $q_{ij}$ ($i \neq j$) can be set to zero (this essentially involves diagonalising the matrix $\mathbf{q}$). The remaining quantities $q_{xx}$, $q_{yy}$, $q_{zz}$ are referred to as the principal values, and without loss of generality we can adopt the convention $|q_{zz}| \geq |q_{yy}| \geq |q_{xx}|$. Since the electric potential obeys Laplace's equation $\nabla^2 V = 0$, this also means that $q_{xx} + q_{yy} + q_{zz}$ is necessarily equal to 0 (this is often referred to as "traceless", since the trace of the matrix $\mathbf{q}$ is 0). Therefore, there are only two independent parameters. These parameters are chosen to be $q_{zz}$, and $\eta$, a quantity called the biaxiality and defined as

$$\eta = \frac{q_{yy} - q_{xx}}{q_{zz}}.$$

Quadrupolar relaxation2

The interaction of the quadrupole moment with the electric field gradient contributes to the energy of the nucleus (exactly analogous to how an electric dipole moment interacts with an electric field).3 This interaction is modulated by molecular tumbling. That is to say, as the molecule rotates in solution, the magnitude of this interaction changes as well.4

If this interaction varies at the frequency corresponding to the transition between two spin states, then it is possible to induce this transition, thereby leading to spin relaxation. (More strictly speaking, it has to contain a component that oscillates at the transition frequency.) In this aspect, it is analogous to the more well-known case of how magnetic dipolar interactions between two nuclei can lead to relaxation, which is described quite thoroughly in Keeler's book.1 For spin-1/2 nuclei in solution-state NMR, dipolar interactions are the primary method of relaxation since these nuclei do not have quadrupole moments. For quadrupolar nuclei, though, quadrupolar relaxation is the primary method of relaxation, simply because this interaction term tends to have a large magnitude.

For our purposes, it is sufficient to use a derived result, which you seem to have already come across. I know that I am glossing over a lot, but as far as I can tell, the derivation of this is very involved:5

$$\frac{1}{T_1} = \frac{1}{T_2} = \frac{3\pi^2}{10}\frac{(2I+3)}{I^2(2I-1)} \left(\frac{e^2q_{zz}Q}{h}\right)^2 \left(1 + \frac{1}{3}\eta^2 \right) \tau_\mathrm{c}$$


  • $T_1$ and $T_2$ are the spin-lattice (longitudinal) and spin-spin (transverse) relaxation times respectively; these only account for relaxation via the above mechanism and not for other possible mechanisms
  • $I$ is the spin of the nucleus
  • $e$ is the elementary charge, $\pu{1.602 \times 10^-19 C}$
  • $q_{zz}$ is the main component of the electric field gradient, as explained above
  • $Q$ is the quadrupole moment of the nucleus
  • $h$ is Planck's constant, $\pu{6.626 \times 10^-34 J s}$
  • $\eta$ is the biaxiality, defined above
  • $\tau_\mathrm{c}$ is the correlation time, which is a constant that essentially measures the rate of molecular tumbling (see books in ref 1 for more information).

Please, no more maths

OK - the only remaining maths is to analyse the equation above and identify the factors which contribute to relaxation, and hence, linewidths. For the actual question, the relevant portion is $e^2q_{zz}Q/h$, which is often written as $\chi$ and termed the nuclear quadrupole coupling constant. Some representative values of $\chi$ are given in ref 2.

In particular, the value of $q_{zz}$ is mainly determined by the distribution of electrons close to the nucleus, and this provides an obvious link to the molecular geometry of the quadrupolar nucleus. For nuclei in tetrahedral, octahedral, cubic, or spherical environments, $q_{zz}$ is, by symmetry, equal to zero. Therefore, quadrupolar relaxation by the above mechanism does not take place, $T_2$ is long, and linewidths are small. One sees very narrow linewidths for the central atoms in, for example, $\ce{Cl- (aq)}$ (neglecting solvation effects, spherical symmetry); $\ce{NMe4+}$ and $\ce{SO4^2-}$ (tetrahedral), and $\ce{[Co(CN)6]^3-}$ (octahedral). On the other hand, if such symmetry is lacking, then $q_{zz} \neq 0$, there is usually fast quadrupolar relaxation by the above mechanism, and linewidths are large.

$q_{zz}$ is directly tied to the distribution of electrons and not the distribution of bonds. Although there is nearly always a correlation, there are some interesting cases where this is not true; in $\ce{Et2N-NO2}$, for example, the $\ce{NO2}$ nitrogen has an extremely sharp peak, despite the local molecular symmetry not intuitively suggesting that it would be the case.2 (The $\ce{Et2N}$ nitrogen, as expected, has a very broad peak.)

Since we went the full distance to obtain the above expression, we can also extract a bit more information from it. The rate of relaxation is also proportional to the nuclear quadrupole moment $Q$; some quadrupole moments are tabulated in ref 2. Some nuclei which have small quadrupole moments (namely deuterium, lithium-6, and cesium-133) have smaller linewidths and are much more amenable to NMR studies. Also, the factor of $I^2$ in the denominator also means that higher spins tend to give sharper lines.

A bonus: coupling to quadrupolar nuclei

$\ce{^13C}$ NMR experiments are often run with proton decoupling - this means that when the carbon signal is collected, the protons are irradiated with a radiofrequency field. This has the effect of inducing rapid transitions between the two spin states of the protons, and each carbon-13 nucleus effectively "sees" an average of the two spin states, causing the C–H couplings to be lost.

Since quadrupolar relaxation also induces rapid transitions between spin states of quadrupolar nuclei, it leads to what is effectively a "self-decoupling" of all quadrupolar nuclei. However, the couplings can still be seen when the quadrupolar nuclei are in highly symmetrical environments: for example, the $\ce{^19F}$ spectrum of $\ce{[BrF6]+}$ is shown here.6 Both $\ce{^79Br}$ and $\ce{^81Br}$ have roughly equal abundance (~50%) and a nuclear spin of $3/2$, so we expect to see two 1:1:1:1 quartets of equal intensity, one arising from $\ce{[^79BrF6]+}$ and one from $\ce{[^81BrF6]+}$.

Fluorine NMR spectrum of BrF6+

A few other examples are presented on Hans Reich's page. Of course, these can be understood from the discussion above about how molecular symmetry affects quadrupolar relaxation.

In fact, because deuterium has a small quadrupole moment and undergoes slow quadrupolar relaxation, coupling to deuterium can often be seen. All organic chemists should recall that $\ce{CDCl3}$ appears as a 1:1:1 triplet - this is why!

Notes and references

  1. For an exposition of basic NMR theory see either Keeler, Understanding NMR Spectroscopy, 2nd ed. (Wiley) or Hore, Nuclear Magnetic Resonance, 2nd ed. (OUP).

  2. A necessarily mathematical, but still fairly accessible, introduction to the topic is provided in: Gerothanassis, I. P.; Tsanaktsidis, C. G. Nuclear electric quadrupole relaxation. Concepts Magn. Reson. 1996, 8 (1), 63–74. DOI: 10.1002/(SICI)1099-0534(1996)8:1<63::AID-CMR5>3.0.CO;2-N.

  3. The full Hamiltonian is provided in Appendix A.7 of Levitt, Spin Dynamics, 2nd ed. (Wiley).

  4. Abragam (ref 5) writes that this is best understood in terms of "a fluctuating electric field gradient acting on the quadrupole moment of the nucleus", and refers to: Bloembergen, N.; Purcell, E. M.; Pound, R. V. Relaxation Effects in Nuclear Magnetic Resonance Absorption. Phys. Rev. 1948, 73 (7), 679–712. DOI: 10.1103/PhysRev.73.679. (Non-paywall version from Harvard available here.)

    However, as far as I can tell, Bloembergen et al. simply describe the case for dipolar relaxation of spin-1/2 nuclei, and then write that "the interaction of the electric quadrupole moment of the deuteron with a fluctuating inhomogeneous electric field can bring about thermal relaxation. We omit the analysis of this process, which parallels closely the treatment of dipole-dipole interaction".

  5. This equation is given without explanation in Gunther, NMR Spectroscopy, 3rd ed. (Wiley). An explanation can be found in Abragam, The Principles of Nuclear Magnetism (OUP), including the conditions under which this equation holds true. Quadrupolar relaxation is dealt with on p 313, but going through the earlier sections is necessary to understand it. (I didn't do that.)

  6. Gillespie, R. J.; Schrobilgen, G. J. Hexafluorobromine(VII) cation, $\ce{BrF6+}$. Preparation of hexafluorobromine(1+) hexafluoroarsenate(1-) and hexafluorobromine(1+)-fluorodecafluorodinabinonate(1-) and characterization by fluorine-19 nuclear magnetic resonance and Raman spectroscopy. Inorg. Chem. 1974, 13 (5), 1230–1235. DOI: 10.1021/ic50135a043.


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.