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Tuesday, March 18, 2008

Is My 1H NMR Spectrum Quantitative?

I often have students ask me, "Is my 1H NMR spectrum quantitative?". Of course we all know that the area under the signals of properly recorded 1H NMR spectra represent the number of protons responsible for the signal, however the answer to the question may be a bit more involved than a simple "yes". The following things must be taken into account.
Relaxation - In order for a spectrum to be quantitative, each of the NMR signals must be sufficiently relaxed to equilibrium before a pulse is applied. One should ideally have an idea what the T1 is for each of the signals and make sure that the interpulse delay (i.e. the sum of the acquisition time and the relaxation delay) is at least 5 times greater than the longest T1 if 90 degree pulses are used or at least twice the longest T1 if 30 degree pulses are used. This should provide data quantitative to about 95%. If more precise data is required, for example in kinetic isotope studies, the interpulse delay must be made even longer.
Baseline Roll - The reliability of integrals depends on the quality of the baseline in the spectrum. One should ensure a perfectly flat baseline is achieved by way of baseline correction routines.
Precise Phasing - Phasing errors in the spectrum (even minor ones) will result in imprecise integrals. Automatic phasing routines cannot always be relied upon. One should expand each signal vertically to ensure a proper absorptive phase using manual baseline correction.
Signal-to-Noise Ratio - Noisy spectra will give noisy integrals. The noise is directly related to the precision of the integral. The concentration of the sample or the number of scans should be chosen so that the signal-to-noise ratio is at least 100:1.
Excitation Profile - The pulses used to measure the spectrum must be sufficiently short to produce a flat excitation profile over the entire spectrum. This is potentially a problem for paramagnetic samples whose resonances span a very large chemical shift range. For more on excitation profiles click here.
Chemical Exchange - The signals from exchangeable protons may be very broad (therefore low signal-to-noise ratio) and will give imprecise integrals. If a deuterated solvent with exchangeable protons is used, then the exchangeable protons in the sample will exchange with the exchangeable deuterons of the solvent thus reducing or eliminating the exchangeable proton signals of the sample. This can also be used to your advantage as an assignment tool.
Solvent Suppression by Presaturation - When a solvent line is presaturated, the integrals of the signals close to the solvent signal will be diminished. If water suppression is used, some exchangeable signals from the sample could also be diminished due to exchange from the saturated protons of the solvent.
Edge Effects - Some NMR setups will produce baseline artifacts (smiles or frowns) at the edges of the observation window. One should choose a spectral width sufficiently large so that the signals will not be affected by this artifact.


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Anonymous said...

Does the 95% accuracy level hold for 19F NMR,, assuming relaxation times are also sufficiently long?

Glenn Facey said...


19F has a very large chemical shift range. In some cases, there may be problems with uniform excitation due to finite pulse width effects. if one peak is close to resonance while another is tens of kHz off resonance, the off-resonance peak may be under represented compared to near-resonance peak. Otherwise if the peaks of interest are close together and near-resonance, the data should be quantitative if the T1's are respected.


Amin said...

Hi Dr. Facey
I have a problem with comparing integration of a specific peak in different nmr spectrums.
My problem is that the values are relative to other peaks on that spectrum, not others.
Is there any way to obtain absolute values that can be compared inter-spectrums?

Glenn Facey said...

If you want to obtain absolute integrals for comparison between spectra, you should dope your samples with an external integration standard such that the concentration of the standard is known in your samples.

Anonymous said...

Dear Glenn,
thanks for your post.

I use for quantitative measurements a TD that results in a FIDRES of at least <= 0.2 Hz/pt.
What is your minimum FIDRES in quantitative measurements?



Glenn Facey said...

Hi Klaus,
Thank you for the question. To be honest I have never paid any attention to FIDRES (=SWH/TD). I make sure TD is sufficient such that my FID decays into the noise and then zero fill to at least 2*TD.