When you stored your sample, what did you store it in (anything other than D2O and TSP) i.e. did you have any other reagents in your spectra from either the reaction or solvents ?

Also, what temperature did you store it in ?

The disapearance of the signal belonging to the protons of that methyl group could be due to exchange (H substitution by Deuterium) resulting from storage in D2O.

So this is my guess.

The Nitrogen of the isoquinoline, is slightly basic and might initiate the generation of a short-lived tautomeric structure which has a double bonded CH2 instead of the CH3. When this reverts back to the original structure it is possible that instead of an H it incorporates a D from the D2O. What you are seeing probably is a CHD2 structure, in which case however it should have a quintet like shape instead of a singlet for CH3.

The question does not provide adequate information about storage. Apart from methyl signal(s) from any other moiety missing ?

I guess your methyl group may have exchanged of with deuterium from your solvent (D2O).

It's very very difficult to exchange proton from methyl group. Actually methyl group has a very low Ka (acidity constant) but why not. You can try to do mass spectrometry to verify the MW and in fine the integrity of your compound.

hmmm....two questions/suggestions:

1) are you sure the methyl was part of your molecule or might it be a evaporating solvent

2) To clarify whether it is still there in deuterated form you could take a 13C (direct of HMBC type) or 2H Spectrum provided you have enough material....

Regards

Alfred

Always check the storage condition and the stability of compounds, because they usually affects the reproducibility of the spectrum.

Methyl group protons exchange with deuterium from D2O in your aplha-methyl heterocycle, especially in prolonged storage.

PS. It is well known that protons of alpha-Me group in six-membered heterocycles are weakly acidic, so that type of exchange may occur.

Good luck with your research!

I think your compound is unstable. If it is not, change the solvent or check T1 e T2. Maybe T1 is high... This help undestand the integral equal to 1.3. If this is true, so to use delay (d1) equal to 10 s or more.

Since you have mentioned that you are carrying out aggregation studies, longer storage times could have resulted in an aggregated state resulting in broad signals and probably disappearance. You can raise the temperature of recording to see if this aggregation is broken. If so, it will result in a slightly sharper signal.

Shape of a signal is determined by spin-spin relaxation time T2. T2 is proportional to Spectral width at half the height of the signal (designated by Nu 1/2.

When aggregation occurs, the relative motions of various groups in the molecule are affected i.e reduced. A reduced motion results in reduction of T2 and hence larger Nu1/2 - a broad signal which could appear as a disappeared signal if the aggregation is too much.

Higher temperatures and reagents which break down this aggregation could make this signal visible.

Thank you all for the informative suggestions...I already know the structure of the compound and it has a methyl group. I mainly want to understand the aggregation as i want to study the interaction of this compound with DNA sequences further. I have carried out ROESY (200ms -mixing time), HSQC and HMBC for the same sample. The cross peak which am supposed to get between the methyl group and other protons are coming in the roesy spectra. whereas am not having any peak of methyl in HMBC and HSQC. Moreover the 13C spectra of my sample is not proper even after 8000 scans.concentration of my sample is 3mM.

I suggest that the methyl group was transformed into CH2D, CHD2 and CD3 due to isotope exchange reaction.

There is a possibility for the methylhydrogens to exchange with the deuterated solvent.

Confirm the change in the mass due to exchange by mass spectrometry.

If the conditions of the spectra acquisition are the same before and after the storage, the described observations favor instability of the compound. Please, compare the spectra in more detail. Maybe other changes are also seen?

If you have any deuterium on the methyl, you can check it by performing 1H and/or 13C and/or HSQC experiments. With integral 1.3 there is still some H atom on your methyl but approximately 2 H are exchanged for D - so you have -CHD2.

1H: you will see quintet with intensities (1:2:3:2:1). The middle line of the signal should be close to the previous location of methyl group. The coupling is due to 2 deuterium atoms; coupling constant should be around 2-4Hz (1J(HD)).

13C: with H docoupling you will see quintet with intensities (1:2:3:2:1). The middle line of the signal should be close to the previous location of methyl group. The coupling is due to 2 deuterium atoms; coupling constant should be around 30Hz (1J(CD)). The intensity of the highest peak should be 1/3 of previous singlet so if you want to observe the multiplet you have to measure for at least 9 times longer time.

HSQC is much faster experiment. You might see slight broadening of the signal in H dimension but you should see a quintet in 13C dimension. This technique is faster than the 13C. In 400 MHz instrument the peaks of multiplet should be separated by 0.3 ppm (30 Hz).

Of course, you can also measure 2H (deuterium) spectrum directly. BUT in this case you either don't use deuterated solvent or you would need to use more advanced technique like deuterium watergate for example. You should have already good instrument as well. You would need to change the LOCK nucleus to different from deuterium. Ideally, to some H far from methyl H atom - perhaps some aromatic from the isoquinoline. All this will be already more complicated than previous experiments.

BTW if you evaporate the solvent (D2O) and dissolve the compound in normal water, you should be able to achieve back the methyl. It might take a little bit more time due to isotope effect in kinetics of the reaction. You could make it faster by adding of some base (e.g. NH3 is easy to remove) - but your compound must be stable in this condition.

It may be consequence of one-electron oxidation of the compound and subsequent electron exchange between radical cation (if it stable enough). See for example our article: A. N. Suslov, A. S. Morkovnik, G. S. Borodkin, PMR study of the degenerate electron transfer between 1,2,3-trimethyl-2-phenylbenzimidazoline and its cation radical, Theoretical and Experimental Chemistry (1991) Online version of article: http://link.springer.com/article/10.1007/BF00536425

Then, you do not have a 1-methyl isoquinoline derivative. Check if the heterocycle has not been opened (look at a singlet at 2.4-2.6 ppm or vinylic protons.