Hello everyone!
I have a question about H-NMR, especially -NH2 peak.
First, the material is (R)-Morpholin-3-ylmethanol hydrochloride (CAS 1212377-10-0).
I analyzed H-NMR for several lots (different vendor) and there is a difference in -NH2 peak.
(-NH2 (9-10ppm): one is from the structure of -NH, and the other is from the ionic bond from HCl salt)
In some cases, the -NH2 peak split clearly into two, and in other cases, the -NH2 peak is merged into one. What makes the -NH2 peak different?
P.S: There is also a difference in -OH peak (5.4 ppm; sharpness?). Whate makes it different?
Please reply T.T
Many many thanks!
Xuelang Gao
I used same solvents (DMSO-d6) and same NMR machine at the same time (Analysis was conducted on the same day in a row). They are different batches from different vendors, but same materials.
Guobing Xiang The sample which shows merged -NH2 has 0.20% of water content and the split one has 2.1 % of water content by volumetric Karl Fischer. So I added several water drop to the sample which showed merged -NH2 and re-analyzed the NMR. However, -NH2 peak was not changed..It still merged.
Add water will make NH, NH2, OH peak dis-appearing for sure. Drop of water is way too much, I believe!
We added D2O to do NMR again for finding out NH, NH2, OH before.
Xuelang Gao
Guobing Xiang Rossen Buyukliev I already conducted 2D NMR and all peaks were assigned. And whether adding water or not, no change occured except the water peak (~3.3 ppm) increased in the spectrum. If you think that water makes the -NH2 peak split or merged, can anyone explain the related principle?
Note To: Xuelang Gao : I believe that the chemical structure you drew in your NMR prediction answer is not the correct isomer. Morpholine has a 1,4- arrangement of the nitrogen and oxygen in the 6-membered ring, not the 1,3-arrangement, so I’d be interested in seeing that NMR prediction for the 1,4-isomer with the CH2OH group next to the nitrogen atom.
To: Soyoung Na: Could it be that one of your batches was mislabeled and is actually the racemic mixture instead of the pure R-isomer? Perhaps there is some sort of interaction between the D- and L- isomers in DMSO that is less-dominant in the pure R-isomer which allows for some slowing of the exchange rate of the ammonium protons and CH2OH proton? That would be rather bizarre, but not impossible. Measuring the optical activity would give you a quick answer.
Ed Brown I analyzed chiral purity of both two batches and the (S) form was < 0.1 % in both cases. First case (merged one) showed 0.03 % and the second case (split one) showed 0.08 % of (S)-form. But I don't think only 0.05 % difference makes these phenomenon..
Yes, I would agree that such a tiny difference in enantiomeric purities would probably not make a large difference in the 1H-NMR spectrum. If you have checked the optical purities and the 13C-NMR spectra are identical, too, then it is unlikely that these differences in the 1H-NMR spectra will make any difference to the chemistry that follows.
Ed Brown Yes, I agree. Isn't it related to hydrogen bonds in substances including HCl salts? -OH peak (5-6 ppm) is also a big difference in sharpness including H2O peak (3.3 ppm). It is hard to understand that there is such a big difference when it was taken in the same solvent at the same concentration on the same day T.T.
It is difficult to say what is going on for sure without more experimentation. Clearly, in one sample, the axial and equitorial hydrogens on the amine nitrogen atoms in the molecules are fixed as axial and equitorial hydrogens on the NMR time scale and therefore give the two different resonance peaks at about 9.4 and 9.7 ppm; in that sample, they do not exchange quickly with solvent or with other molecules and each other, so give two different peaks for the axial and equitorial ammonium hydrogen atoms. In the other sample, the exchange rate between the axial and equitorial hydrogen atoms on the amines is very much faster and so they give a merged, broad peak half-way between 9.4 and 9.7 ppm. Perhaps, in one sample, there is a trace amount of unprotonated amine present in excess of the HCl, which would allow the lone pair of the nitrogen to invert in the ring and also exchange with the hydrogens in other molecules quickly-- which would lead to the exchange of axial and equitorial protons and averaging of those resonances. What happens if you add a tiny amount of NaOH or NH4OH to your sample with the resonances at 9.4 and 9.7 ppm that has a tiny amount of H2O present? I’m curious to know the answer.
Ed Brown I'm sorry for late response! I added a tiny amount of NaOH and the result showed that -OH peak (5-6 ppm) and water peak broadened like figure 2, but -NH2 peak didn't split. It also broadened more.. But it's interesting to see the change of -OH peak depends on the HCl content.
That is an interesting result about the -NH2+ peak getting broader when OH- is present. So, perhaps if you do the opposite and add H+ Cl- to an original sample in d6-DMSO where the -NH2+ peak was broad to begin with, perhaps the extra acid will lock the axial and equitorial protons on nitrogen and slow the exchange on the NMR time scale? Perhaps it will split into the two separate resonances for each different H on nitrogen, like the other sample did? I’m not sure if this would happen, but it might be an interesting exploration that wouldn’t be too difficult or expensive to try. And then it would relate your two different batch samples to one-another. Or, a 13C NMR spectral comparison would probably work, too. The protons are decoupled from the carbon atoms usually in 13C NMR, so you should end up with the same spectra for each batch sample in D6-DMSO. If you didn’t get the same 13C resonances as singlets in identical positions in each different spectrum, you would know that the compounds are different in structure.
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