10
$\begingroup$

I am running a proton NMR on a liquid organic product. I understand that we wouldn't want to use anything that contains hydrogen atoms as a solvent, but I'm not sure why we need a solvent at all.

$\endgroup$
1
  • 4
    $\begingroup$ Well, solid-state NMR does exist, and in theory any liquid can be a ‘solvent’. $\endgroup$
    – Jan
    Oct 2, 2016 at 21:34

3 Answers 3

14
$\begingroup$

There is a minimal volume you need to fill into the NMR tube to be able to get a good shim, this is around 500 microliters with regular 5mm tubes. Not using a solvent would require a lot more of your actual sample just to fill the tube. In many cases you might not even have enough of your product to fill an NMR tube.

The solvent does have some effect on the spectra, it could be harder to compare spectra of pure product to the more typical spectra measured in a common solvent. You're measuring in a different environment, product in product is different than product in solvent.

Modern NMR spectrometers use the deuterium signal from the solvent for the lock, which corrects a drifting magnetic field. You could measure without the lock if your experiment is short or your signals are very broad anyway. But this is not something that every NMR facility has just set up for all users, you might have to manually set this up at the spectrometer.

I suspect that the automatic shimming methods typically used now would not be able to deal with a sample without solvent. But I never tried that.

$\endgroup$
2
  • $\begingroup$ The shifting effect of the solvent is usually not too bad, but can be confusing sometimes. It mostly depends on polarity. Water is the extreme, it comes at 0.4 in benzene, and 4.8 in water. Methylene group in ethanol drifts from 3.3 to 3.7, acetone from 1.5 to 2.2 $\endgroup$
    – Karl
    Oct 2, 2016 at 19:02
  • $\begingroup$ The issue with amount of substance is probably the primary reason. You usually need to fill the tube with 5 cm of solvent to get something that is relatively uniform. Otherwise, anisotropic effects make it hard to get a spectra that has clean sharp lines. $\endgroup$
    – Zhe
    Oct 2, 2016 at 20:02
10
$\begingroup$

Just to add some clarity to some already acceptable answers by others....

For running solution (liquid) state NMR, a solvent is not necessary. There are a number of applications where dilution with a solvent is unwarranted and counterproductive. So, why use a solvent for running liquid state NMR?

  • The main reason a solvent is used is to improve the resolution of signals.

And by main reason, I mean completely overriding factor if resolution is of any importance at all. All other factors are secondary. If resolution is of little importance (broad signals are expected or of no concern), then diluting with a solvent is unnecessary, and can simply be wasting valuable spectrometer time.

Diluting in a low viscosity solvent significantly increases the tumbling rate of molecules in solution, and decreases signal broadening through T1 mechanisms. This is why the most common NMR solvents are small molecules with low viscosity; chloroform, acetone, methanol, etc, and not something like glycerol. Most neat organic liquids that are made in the lab have very high viscosities, compared to common solvents, and the slow molecular tumbling will lead to efficient T1 relaxation, and hence very broad linewidths. In order to see the effects of solvent viscosity on linewidths of observed signals, compare the solvent linewidths of residual CDCl3, acetone-d6, dmso-d6.

-edit-

As an example - look at the following spectra of propylphenylketone in various solvents, and neat. Image 1 shows a standard spectrum of 1% PPK in acetone-d6. An expansion of the highlighted region is shown for various conditions in image 2.

enter image description here

For all samples below, standard 1H parameters were used; 1 scan, 64k data points over 20ppm, processed with 64k points without any apodization. All samples were shimmed using automatic gradient shimming routines.

enter image description here

As can be seen very clearly, the optimum resolution is obtained in the sample with the lowest viscosity - acetone. As the viscosity of the sample increases, the resolution decreases. In fact, the level of resolution obtained in acetone-d6 and that for acetone-d0 is almost identical. The spectrum below shows that excellent results for resolution can be obtained without lock and without deuterated solvent for a 1% sample.

enter image description here

The linewidth for the signals in non-deuterated acetone is about half that for neat PPK. This really drives home the importance of suitable solvents for running high resolution NMR. There is a balance that needs to be struck with NMR. It is an inherently insensitive technique, and it is important to get as much material in the coil volume as possible, but still critical that it is low in viscosity.

-end edit-

Never mind arguments about shimming and sample depth and sample amount. There are a squillion ways around these problems. Different size probes, different size tubes (running 3mm tubes inside a 5mm probe is a very common method of reducing the amount of required sample volume for instance), susceptibility plugs, shigemi tubes, sample depth adjustments are all standard applications for routine NMR.

The advantage of providing a spectrometer lock is secondary; an internal lock is not necessary for the vast majority of NMR experiments. In fact, most modern 2D experiments use a lock-hold switch which actually turns off the lock to allow gradient pulses to be used. For the vast majority of 1H experiments, simply turning off the field sweep will allow a stable enough field for a few minutes of data collection. Over the course of an hour or so, though, this would lead to significant broadening of signals for 1H - but perfectly fine for 13C experiments.

Even the argument of signal dampening by intense solvent peaks is not valid. Modern spectrometers can deal with suppression of multiple intense solvent peaks without any concern. Even the application of digital quadrature detection filtering allows for a very specific region of the spectrum to be observed without collecting signals for intense solvent peaks. Shaped pulses for selective excitation and even selective methods such as Hadamard are all well enough developed methods for getting around the problem of dynamic range conflict due to large solvent peaks. It is simply the case that, for the majority of the samples, we simply make use of the fact that we have dissolved our sample in a suitable solvent that excludes troublesome solvent peaks. In some cases, this is not so. In most protein work, deuterated solvent (D2O) for lock is added at a maximum of 10% to ensure exchange with labile amides is limited. These types of systems are typically analysed in 90%H2O/10%D2O. Metabolomic screening methods of blood plasma, urine, juice etc etc are typically run in non-deuterated conditions, or with minimal addition.

Let's debunk some of the statements on this page:

  • "I suspect that the automatic shimming methods typically used now would not be able to deal with a sample without solvent"

Not the case. There are multiple methods of automatic shimming. The most common, and most effective, uses a gradient pulse method to establish field homogeneity plot. This can be used on either 2H or 1H. All that is required is the shift of a well refined peak (singlet works best, but are not necessary). I use this method on samples of neat ionic liquids and it gives excellent shims above and beyond what I can ever achieve by hand.

  • "Theoretically, the way that NMR is commonly taught, you don't need a solvent, but it's very simplified because a lot of the technical details are quite complicated. In practice, the way that NMR is carried out, the instrument needs the solvent as a "lock" for the magnetic field"

That's a bit like saying 'because science'. The spectrometer does not need a lock to establish a stable field. Turning off the field sweep will keep the spectrometer field stable for long enough to run all but the highest resolution experiments.

  • "The issue with amount of substance is probably the primary reason"

A 1.7mm microprobe uses 30uL of sample. A 1mm microprobe uses 5uL of sample. Stopflow microcoil systems use as little as several nL of sample. Sample preparation for these still require sufficient dilution in a suitable solvent for optimised resolution. For 5mm probes (the most common), you can use 3mm tubes, or susceptibility-matched inserts to significantly reduce the amount of material required. However, if the sample is too concentrated, viscosity effects (and differences in susceptibility if using plugs) will lead to poor resolution.

  • "The solvent does have some effect on the spectra, it could be harder to compare spectra of pure product to the more typical spectra measured in a common solvent. You're measuring in a different environment, product in product is different than product in solvent."

which is why when you report your NMR data you should state the concentration of your samples. But who does that these days? It is generally assumed that samples are measured at high dilution when being characterised. At reasonable dilution, [solute]<<<[solvent], changes in chemical shift are minimal.

$\endgroup$
1
  • $\begingroup$ Oh well, to be honest, I hardly know much about it :) When I saw the question, I was hoping you would come along. Thanks for clarifying. $\endgroup$ Oct 12, 2016 at 10:31
1
$\begingroup$

First of all, are all your samples a liquid? Viscosity effects line width. Solids are run without a lock. Liquids can be run without a lock if you program the magnet drift into the pulse sequence. So basically, If you want high resolution, you have to compensate for magnet drift, hence, lock on to a strong, narrow, known peak.

$\endgroup$
1
  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. I am not an NMR expert, could you explain what you mean with lock? $\endgroup$ Oct 3, 2016 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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