Suppose you have,
NH2......N in a molecule and they are neighbour from one bond.
And an proton exchange is happening as following
-NH2........N
-N=H......NH
The exchange is slow on the NMR time cale so you observe both peaks.
In that case can we trust the peak integration values ? I mean, do they give the correct ratio of hydrogens ? For example as for, -N=H......NH, is it 1:1.
your question is obscure and obsolete! Why?
Please define if you mean R-NH2 and what is R?
On other part, what a kind of NMR solvent are you using? Please don 't use CDCl3 because it has HCl traces after the purification.
What a temperature gradient.
What a kind of NMR spectrometer and NMR tubes are you using)
JRG
Dear Grezes, let me say for two different tautomers and there exixst two exchangable protons, and lets say everything (solvent, NMR spectrometer, types of the NMR tubes, pressure, spinning rate, pulse paremeters, temperature etc.) is ok to happen an exchange. Even lets say an exchange is happenig. In that case can we trust the peak integration values since they reflect the relation between number of protons. Let me go more into detail please dont think about the valency number assume that is ok as well..
NH.......NH peak integration (1:1)
now asusume proton exchange is happening;
NH2*........N and N....NH2 vice versa (if I integrate the peaks shown with asterix will I see a value which indicates two protons when compared to the rest of the protons in the molecule in case of proton exchange)
Briefly, Can I trust ingration values of protons in my molecule even a proton exhange is happinin during the measurement of 1H NMR.
I think now its clear.
If it is undergoing slow exchange, and you can see the two exchanging peaks separately, then the integration should be good and you can trust it.
A similar discussion can be found here: http://chemistry.stackexchange.com/questions/27020/1h-nmr-broad-peaks
Hope that helps.
Dear Özgür,
thank you for your fast answer. It is for me not clear . Why?
when you write:
Suppose you have,
NH2......N in a molecule and they are neighbour from one bond.
And an proton exchange is happening as following
-NH2........N
-N=H......NH
NH2*........N and N....NH2 vice versa (if I integrate the peaks shown with asterix will I see a value which indicates two protons when compared to the rest of the protons in the molecule in case of proton exchange)...
What a kind of molecule? Is it aromatic? Polymer? Biopolymer...
You can answer via Message. It is confidential and only you and me can read the answer
A good question must show the molecule. See p.e the following link:
http://pubs.acs.org/doi/abs/10.1021/bi00074a013
JRG
It doesn't matter what the compound is - proton NMR will detect all of the protons in whichever environment they are. The chemical shift and appearance will be determined by the physical conditions (temperature, pH, concentration, etc.). Your total exchangeable proton integral will reflect the number of protons however they sit.
Exchangeable protons can give unreliable integrals for a number of reasons. If you have acidic protons in your solution you may protonate one of the nitrogens and this would lead to an increase of apparent integral. Conversely, if you have and D2O, this will decrease the apparent integral. As mentioned earlier, CDCl3 may contain small amounts of HCl or DCl either of which will affect the integral. Some batches of d6-DMSO contain D2O.
Last of all, integrating broad signals is inherently difficult as the signal extends out a long way and is often lost in the noise. In these circumstances, peak-fitting is the best way of determining the integral.
Lots of things to think about but the main thing is that NMR is a great atom counter!
Dear John, you words "It doesn't matter what the compound is - proton NMR will detect all of the protons in whichever environment they are. " and "Exchangeable protons can give unreliable integrals for a number of reasons. " is exactly what I want to hear. Best wishes.
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