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Related to this previous question about detection of small couplings between protons.


In a recent conversation with our labs NMR technician, it was stated that coupling constants decrease in the presence of electronegative atoms:

...and you have a lot of electronegative atoms around, so thats going to make those coupling constants drop off significantly...

Source: 'Personal communications', aka the labs NMR-tech

Thinking about it, this is something I've always known but possibly never thought about to any great deal, despite the fact it shows up in even the most basic organic chemistry textbooks:

...electronegative atoms, which decrease 2J and 3J couplings between protons

Source: Organic Chemistry, Clayden & Warren

Digging a little deeper, this idea comes up a lot in NMR textbooks too, and a few trends are discussed, for instance in Günther's NMR spectroscopy:

For substituted ethanes the relation between the change of electronegativity, caused by the replacement of a hydrogen atom with a group X, and the coupling constant is given by:

3J = 9.41 – 0.80 (EX – EH).

Source: NMR spectroscopy, Günther

What I can't find, in any great detail, is why the coupling constant depends on electronegativity the way it does.

From what I've read, it has something to do with the polarisability of the bond, but again, its maybe not completely intuitive how this effects coupling between two protons separated by 2/3 bonds.


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  • $\begingroup$ One possibility is that the bonds to electronegative atoms have more p character, so the remaining bonds are shorter and have more s character, than in the case of the unsubstituted compound. But intuitively, two atoms that are farther apart should have weaker coupling, so perhaps there is another factor at play. $\endgroup$ – iad22agp Jun 5 '17 at 10:39
  • $\begingroup$ I don't know where my Oxford Chemistry Primer NMR book is, but I usually consult it for these types of questions first... $\endgroup$ – Zhe Jun 6 '17 at 19:28
  • $\begingroup$ @Zhe. I feel slightly offended :P In all genuine seriousness, I've looked in a few books (see references in question) and haven't found anything that convinces me. There is an 'MO' type explanation involving bond polarisation, but its not terribly thorough in explaining why this affects the J value. $\endgroup$ – NotEvans. Jun 6 '17 at 19:30
  • $\begingroup$ :P I was trying to say (apparently failing at it) that the primer is an excellent reference that I've found useful. If I knew where my copy was and had it on hand, I would've looked it up for you (assuming it was there). Unfortunately, it's either packed away with the rest of the chem books in storage, or I donated it to MIT. $\endgroup$ – Zhe Jun 6 '17 at 20:11
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    $\begingroup$ @Zhe it isn't in there, unfortunately. I checked inside when I saw this question - surprisingly there is a lack of detail in many textbooks as to why this is observed... $\endgroup$ – orthocresol Jun 6 '17 at 20:30
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Let's set some starting parameters so we are hopefully working from the same page. Interestingly, the summary that an increase in electronegative substitution leads to a decrease in coupling constant is not strictly correct per se.

For geminal coupling, electronegative substitution at the alpha position leads to a more positive coupling constant. H-H Geminal couplings are negative, and hence we observe a coupling that becomes smaller in magnitude. (There are a few examples where they actually go the other way, but still become more positive). On the other hand, H-F geminal couplings are positive, and so an increase in 2JHF coupling is observed with an increase in electronegative substitution. Electronegative substitution at the beta position has the opposite effect, leading to a more negative coupling; this actually gives a coupling that is larger in magnitude. Effects from substituents at the beta position are usually smaller than for the alpha position, but the effect is nevertheless opposite.

For vicinal couplings, electronegative substitution leads to a reduction in magnitude of the coupling constant.

The magnitude of scalar spin coupling depends on a few parameters:

  • the angles between overlapping bonds between coupled nuclei
  • electronegativity of substituents
  • bond length

The traditional accounts for the mechanism of spin-spin coupling might go something like...

In the simplest terms, spin coupling comes about because the magnetic moment of nucleus A interacts with the electrons surrounding it, and these electrons produce a local magnetic field at the neighbouring nucleus B. Spin coupling is dominated by the Fermi contact over a short number of bonds. The Fermi contact describes the magnetic interaction of an electron within the nucleus. Only s-orbitals can contribute electron density at the nucleus (all other orbitals have nodes at the nucleus), and so bonding to entities that will promote s-character of molecular will strongly influence the coupling constants.

The more stable orientation of for the nucleus and electron are antiparallel (because they have opposite sign magnetogyric ratios), and for electrons in a bond also to be antiparallel. For the most stable arrangement we get something like (lower case = electron spin, upper case = nuclear spin, u=up, d=down, - are bonds):

Ud-uDu-dUd-uDu-.....

Essentially, the observable nuclear transitions depend on the molecular orbitals, and how well the molecular orbitals stabilise the antiparallel, or spin-polarised configuration. There are some comprehensive explanations of the molecular orbital model given in Gunther, as you have already encountered, and first published by Pople and Bothner-by. You will also find an excellent account, as always, on Reich's NMR page.

A coupling constant is positive, by definition, if it stabilises the anti-parallel electron spins, as shown above. So, a 1J will be positive, 2J negative (usually) and 3J positive. If the probability of finding an electron in the up state is the same as for the down state, the coupling constant will be zero. For a simple 2 nucleus, 2 electron bond system, if there is an up-electron at one nucleus, then there is a very high chance that there will be a down electron at the other nucleus, and so the coupling will be large. For 2 nuclei across multiple bonds, the spin state of an electron at nucleus A will have a negligible effect on the spin state of electron at nucleus B, and so the coupling will be very small.

...Pause for a breath....

So, to the crux of your question... If a substitution element can help stabilise the anti-parallel arrangement of the electron spins, by reducing the energy gap for the HOMO-LUMO of the molecular orbitals then the coupling constant will become more positive. Electron-withdrawing groups serve to reduce the energy gap between these orbitals (arguments based on symmetry-allowed transitions - read the above references). σ-acceptors will reduce the LUMO, while π-donors raise the HOMO; both have the same outcome of making the coupling more positive.

This explains the geminal trend, but what about vicinal? There are some theories about alternating polarisation along the bond pathway, so that stabilising the anti-parallel alignment in one molecular orbital will destabilise the adjoining orbitals, hence the trend towards reducing J. It all gets a bit vague about here, and not very convincing. There was some good work by Altona et al in the 80s/90s (see for example) extending the Karplus equation to account for electronegativities, and improved Karplus equations were published, but nothing universal. At this point, we begin to accept that the ability to describe the physical nature of what is going on here is not really going to work using an emperical model.

More recent work by Cremera and Gräfenstein, for example, pretty well put these old 'textbook' explanations back in their boxes, and describe the mechanisms for spin coupling as composites of 4 contributing factors: Fermi contact, spin dipole, diamagnetic spin orbit and paramagnetic spin orbit. It all gets very Hamiltonian early on, but essentially there are co-dependent interactions of all these terms in the electronics. Fermi contact coupling is propagated over multiple bonds and occurs through space via a molecular orbital vector involving the tails of the orbitals, and these tail interactions are strongly influenced by the substituents. An electronegative substituent on a carbon will increase the local Fermi contact, but lead to a decrease in the tail orbital overlap, leading to a reduction of vicinal coupling, for instance.

Some very useful references for you might include:

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