Friday, 11 January 2008

1H NMR Analysis: Common Myths and Misconceptions

The analysis of NMR spectra, in particular 1H-NMR, is certainly an exciting and challenging area of research and considerable effort has been made in the past 50 years or so to overcome many of the difficulties that this analysis faces. Actually, most of the work on this subject was done in the early 50’s and computer programs for the analysis of complex spin systems appeared in the 60’s and 70’s (see for example, references [1-4]). However, even though the analysis of 1H NMR spectra is for sure one of the most frequent tasks carried out by chemists, almost on a daily basis, I have found that there are several very popular misconceptions and fallacies amongst chemists which I would like to bring to your attention. In fact, some of these incorrect explanations can be found in books and articles. By doing this, I could run into the risk of introducing some inaccuracies in my comments (which is very likely). I’d like to take advantage of the interactive nature of this blog so, if you see any errors, please, just post them as a comment below.
This post will be somewhat lengthy, I tried to cover these points with all the detail I could think of, including illustrative figures and potential pitfalls, so grab your favourite caffeinated drink before you start reading!

Myth #1 All NMR signals are symmetric

When a chemist is first introduced into 1H NMR analysis, he/she is trained with the basis of the simple and well known first order multiplet patterns which, by definition, are always symmetric. It would be nice if all multiplets would behave following these simple rules. However, lines in any real-life spectrum are highly composite (more on this below), so no experimental line is "pure"; generally, experimental lines are a superposition of a large number of transitions. For example, let’s take the very simple case of the NMR spectrum of ethanol at 500 MHz:

The multiplets in this synthetic spectrum show indeed a large degree of symmetry and, in principle, as expected from the first order rules, we observe 4 lines for the CH2 group (triplet) and 3 lines for the CH3 group (assuming that coupling with OH does not occur).
This is certainly what one would observe in a modern high field spectrometer, but the truth is that there are 12 distinct main transitions in the CH2 group and 13 in the CH3 group, but all of them are never resolved. However, this more complex than expected structure of the CH3 and CH2 groups can be appreciated if the spectrum is recorded at lower field as illustrated in the CW experimental spectrum (from reference [2]).

This does not mean that all the 12+13 transitions only exist at low field. It’s just that the distance between the different transitions depends on the magnetic field in such a way that at higher fields they are too close to be observed and several transitions often contribute to the same peak. For example, in the following two figures, all these transitions are displayed as sticks superimposed on the observed multiplets which, apparently, look like a first order quadruplet and triplet respectively. As you can see, the inner structure is much more complex.

By the way, do not take too lightly the powerful capability of Mnova to classify and display graphically all the transitions corresponding to a given nucleus. If only first order rules were used, this would be a trivial task, however, when a rigorous quantum mechanics treatment is carried out to calculate all the transitions, this classification is far from being trivial. Combination lines, that is, transitions which correspond to the simultaneous change of spin of several nuclei, are particularly challenging on this classification system. They are forbidden in the first order limit but will have a small, non-zero intensity in the general case. These lines, when their intensity is large enough, are also taken into account in the classification system of Mnova. To the best of my knowledge, this capability is available in Mnova only, but if you know any other application which allows you to select one proton with the mouse and display the entire individual transitions corresponding to that nucleus (and the other way round, that is, selecting one transition in the spectrum and highlighting the nucleus causing such transition), please let me know.

Well, you may argue that even though the fine structure of the ethanol spectrum is somewhat complex, the truth is that the multiplets look symmetric and first order compliant (please note however that even in the simulated 500 MHz spectrum, the multiplets are STILL not symmetric, look at the sticks: The asymmetry is about 15% which is not little!) . But let us take, as a new example, a spectrum comprising an ABX spin system (as per Pople notation) with the couplings depicted in this figure:

In this spectrum we can observe the expected doublet for the X nucleus as a result of its coupling with B. However, if the chemical shift difference between A and B is reduced, the splitting pattern will become more complex. For example let’s now take a look at the same spin system when |A-B| = 15 Hz.

Interestingly, the X nucleus appears as a double doublet (disregarding the two small combination lines) even though it’s only coupled to nucleus B (JAX = 0). This kind of situation, where a nucleus appears to be coupled to another nucleus, even though the actual coupling constant is zero, is called Virtual Coupling. In general, if one nucleus is coupled to another nucleus which forms part of a strongly coupled group, the former nucleus will behave as if it were coupled to all the members of the spin system.
Do you still believe that 1H NMR spectra are always symmetric? Let’s make the chemical shift difference between A and B even smaller (5 Hz).

Clearly, an analysis of the X nucleus based on first order rules will yield incorrect values for the coupling constants. Obviously, the AB spin system can only by interpreted by means of a rigorous quantum mechanics treatment. From a symmetry standpoint, the X nucleus is symmetric (which could mislead one to assume it is a first order multiplet) but the AB multiplet is not.
Typical examples of ABX systems are pro-R and pro-S protons of methylenes in pro-chiral molecules.

So as a conclusion, in general multiplets are never really perfectly symmetrical, even disregarding the fact that they often overlap. In addition, relaxation effects in coupled systems are nontrivial and affect each transition in a different way, and there are often other effects such as exchange processes, co-presence of isomeric forms, impurities, etc.

Myth #2 High magnetic fields make the analysis of 1H NMR spectra by means of first order rules possible

Well, this is not completely wrong as it’s true that by increasing the spectrometer frequency, most of the 1H spin systems may become suitable for First-Order Analysis at least as a rough or grosso modo approach. For example, if we were to go back to the previous ABX spin system, and increase the magnetic field, the chemical shifts between A and B will become larger in such a way that we can arrive to a point in which the 3 individual multiplets can be perfectly handled with simple first order rules (see figure below).

However, there are many situations in which higher magnetic fields will be of little or no help. Let me present just a couple of examples:
  1. Saturated fatty acids
  2. AA’BB’ spin system

Consider the 1H NMR spectra of fatty acids with long saturated chains. In the figure below I have simulated the 1H NMR spectra of 3 fatty acids of different lengths. The first thing to notice is that the CH3 does not appear as a simple triplet. This is because it is (weakly) coupled to a CH2 group which is strongly coupled to the next CH2 group in the chain, leading the CH3 resonance to show virtual coupling with the CH2 protons in the chain. Even at higher fields, the pack of CH2 groups is very strongly coupled, though such higher fields may make it possible to resolve the resonances of some of the CH2 groups. However, as the number of methylene groups increases, the "newly resolved" methylenes will have quite small relative chemical shifts and they will always couple to their neighbors by J's of the same order of magnitude. So, whilst the CH2s at the extremes of the chain become 1st order when increasing the magnetic field, there are always others which are strongly coupled, and still others which are insufficiently resolved.

Another, more striking example, occurs in spin systems of the type AA’XX’ (or AA’BB’ if the chemical shifts are close) as is the case, for example, in spectra of o-dichlorobenzene (ODCB). This compound is often used to calibrate instrument resolution.
In an AA’BB’/AA’XX’ system there are 2 pairs of magnetically non-equivalent protons with the same chemical shift, that is, they are chemically equivalent. In general, when groups of chemically equivalent nuclei exist, second order effects are expected. As the chemical shift difference separating the nuclei in the molecule is always zero because of the symmetry, second order phenomena will always exist regardless of the magnetic field applied. If the magnetic field is increased, it will be possible to get a larger chemical shift difference between the AA’ and the BB’ groups, but not between A and A’ or B and B’, so that the highest simplification one can achieve by increasing the magnetic field is to move from an AA’BB’ group to an AA’XX’ group which is a second order spin system too.
There are a total of 12 transitions for each spin (the AA’ part or the BB’ part).
As an example, the figure below depicts the spectrum of ODCB at two magnetic field strengths, 60 MHz and 400 MHz.. At 60 MHz the inner lines are more intense than the outer lines and there is an evident lack of symmetry. These peculiarities are easily recognizable as second order effects and they are caused by the small chemical shift difference between AA’ and BB’. If the spectrometer field strength is increased (e.g. 400 MHz), these effects are reduced as the chemical shift difference between the two groups is now larger and we can now see two more symmetric multiplets. However, each multiplet does not follow the first order rules (e.g. they are not simple doublets of doublets of doublets) because A and A’ and B and B’ will always be strongly coupled regardless of the magnetic field strength. Once again, the only way to accurately analyze these systems is by means of quantum mechanics calculations.

Myth #3 Protons within a CH3 group do not couple with each other

I have found very frequently that chemists assume that CH3 protons are not coupled because only a singlet is observed in the spectrum. This is simply not true! The correct explanation is that couplings within a magnetically equivalent group such as a methyl group do not affect the spectrum appearance, but coupling definitely exists (it is known that geminal couplings are usually large and negative). In other words, if we synthesize a spectrum with and without couplings within a magnetically equivalent group, the spectrum will look exactly the same, so for calculation purposes, these couplings can be neglected.

That is all for now I hope you find these comments of some use and/or interest and if there are any points where I have failed to make myself dear, please do not hesitate to post a comment here. I’d also like to thank Stan Sykora for several useful discussion about this work.


[1] Pople J.A., Schneider W.G., Bernstein H.J., High-resolution Nuclear Magnetic Resonance, McGraw-Hill, New York 1959
[2] Roberts J.D., Nuclear Magnetic Resonance: Applications to Organic Chemistry, McGraw-Hill, New York 1959
[3] Early History of Nuclear Magnetic Resonance
[4] Automatic Analysis of NMR Spectra: An Alternative Approach Diehl P., Sykora S., Vogt J., J.Magn.Reson. 19, 67-82 (1975) [click here]


David Bradley said...

I'm fairly sure I didn't believe any of those myths before, but your explanations are very clear and put paid any vague doubts I had. Have you written an NMR textbook yet? "A Dummies Guide to NMR" would be perfect. ;-)


Carlos Cobas said...

Thanks David, I'm delighted that you found my explanations useful and clear.
As for your suggestion on the book, I'm very pleased that you think I could take this on, although my commitments at the moment make it unthinkable. It's something I'd like to do, I hope I can one day.

Enda said...

Good explanations. I agree with David that organic chemists in general badly need a "Dummies Guide" to many switch off when maths is introduced. Kind of like an instruction manual for a machine, rather than an explaination of how the machine works.

akhdar-mohit said...

Thanks for the detailed explanations--they really help to put everything into perspective. I'm just curious about why the coupling between protons on a methyl group doesn't show up? Could you explain this a little more?

Henry Rzepa said...

I know this thread is rather old, but the last question asked why couplings between protons in a methyl group do not show up.

In my own tutorials, I give the answer that yes, the couplings are present, but the transitions that would give rise to observable lines in the spectrum are simply of zero intensity. In other words, the 2nd order effects tend to reduce the intensity of the peaks which reflect the coupling to zero when the chemical shift difference becomes zero. One can see this most easily for the AB spin system as one reduces the Δν between the two spins, until when its zero, the outer two lines of the AB quartet have reduced their intensity to zero.

Carlos Cobas said...

Dear Henry,
Many thanks for your comment. Without entering into all the quantum mechanics details (which you can find in J.Chem.Phys. 52, 5949-5951 (1970)), the transitions of a CH3 group (A3-group more generally) are not zero. Instead, it appears that the A3-group consists of four transitions with non-zero intensities and actually, with different line widths!

So there are two points to consider here:
First, and as an example, if an A3 group is synthesized (= simulated) using quantum mechanics with a computer program (like the Spin Simulation module in Mnova), in order to take fully advantage of magnetic equivalence, this A3 particle is split into 2 pseudo-nuclides: one with spin 1/3 and one with spin 3/2.
It turns out that the spin 1/2 pseudo-nucleus has 2 levels and 1 transition, whereas the spin 3/2 has 4 levels and therefore 3 transitions, accounting for a total of 4 transitions. Now, all the 4 transitions have exactly the same frequency and we DO SEE all of them (more about this in a moment); they are just overlapped, but there is no transition with zero intensity.

Second, qualitatively speaking, the transition in the 1/2 spin isomer is always considerably narrower than the 3 transitions in the spin 3/2 isotopomer.
When I wrote that “We SEE all of them”, I mean that in a real spectrum we see only one peak, but if we could list the resulting transitions coming from a computer program, we will see all resulting numeric values. Furthermore, if we could, somehow analyze the shape, inhomogeneity permitting, we would see basically 3 components with different line widths. This phenomenon is more important than most people realize: The fact that a CH3 line is the convolution of several lines with different line widths explains why when one tries to fit this resonance with e.g. a Lorentzian line shape (e.g. using Mnova GSD – Global Spectral Deconvolution), one gets unexpected large residues. The reason is that the true lineshape of a methyl group is, because of the convolution of several lines with different line widths, no longer Lorentzian.
Finally, when the methyl is coupled to something and 2nd order starts to be visible a bit, the transitions start spreading out in a very predictable way.

Anonymous said...

Does anyone know why the descriptor 'apparent' is appended to first order multiplets -- is it necessary to use this term to denote chemical non-equivalence of the two coupling partners?