[kriggy]

Hi guys, as a part of one class we are supposed to interpret NMR of unknown compound.
Ive  been given chemical formula which C₂₂H₂₆N₅O₃F and few NMR experiments:
¹H ¹³3, ¹⁹F, cosy, hmbc and hsqc.
I understand the basics what those spectra can tell me but I cant find a way to start.
I was thinking to start with aromatic region but Im not sure here:
I have dublet of dublets at 8ppm, dublet of dublets at 7,67 and dublet of triplets at 7,27. I was thinking its trisubstitued pyridine ring with 3 hydrogens next to each other since I have only 5 carbons in aromatic area (98-125ppm).
The thing is, I dont understand the dublet of triplets. If there were only 3 hydrogens then it would be just triplet (or dublet of dublets) so it means there is another substituent that can split the signals. I guess it could be fluorine but it would then split the other hydrogen signals.

Thanks for any help

[Irlanur]

There is some sense in the fact that NMR Spectra are usually given as Spectra and not as sentences;) (altough in papers the peaks are usually given in tables, how annoying...)

a good start would be to try to find out how many protons are under each peak. can you distinguish all signals? write that down.
then add to each proton how the splitting looks like and what the chemical shift tells you.
then have a look at the hsqc to determine which protons are bound to which carbon and how many protons you have per carbon. build up subunits of the molecule by looking at the H-H-couplings in the cosy. you should be able to connect quite a bit of the molecule. use the hmbc for getting those pieces together.
always keep in mind the chemical formula and how many double bonds you should have.

[kriggy]

Oh yea, I forgot to upload the picture :-D There is detail of COSY spectra I got


There should be 1 proton under each peak (integrals 1,0 0,78 and 0,96)
Its the splitting that doesnt make much sense - to be more specific why the triplet is split into doublet. It means that there is something else to the 3 protons but why it doesnt split the other protons too?
I think the protons are in 1,2,3 positions since the one interaction seems very weak and 1,2,4 positions would give different splitting pattern

[OrgXemProf]

Perhaps it might be useful to compare your 1H and 13C chemical shift data with that of a model compound.

Consider the proton NMR spectrum of 2,3-dimethylpyridine; the aromatic protons in this molecule absorb in the region delta 7.0 - 8.3 ppm. The corresponding 13C NMR spectrum of this compound displays five aromatic carbon absorption signals in the range 121-157 ppm.

(reference to proton NMR spectrum of 2,3-dimethylpyridine: http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi, SDBS Number 694).

Also, you mentioned that the 19F NMR spectrum of your unknown compound is available. Do you observe spin-spin splitting pattern(s) in this spectrum?
 
Sorry - that's about all I can offer in the absence of actual 2D NMR spectra (COSY, HMBC, HSQD) of your unknown compound.

[OrgXemProf]

Aha - now I see the COSY spectrum.

Let's assume for the moment that your unknown compound contains a 2,3-disubstituted pyridine moiety, so that there indeed are three contiguous H-nuclei present in an aromatic ring. We then might refer to these H-nuclei as H(ortho), H(meta) and H(para), where the o-,m-,p- designations refer to their respective positions relative to ring nitrogen.

If the J-couplings between H(ortho)-H(meta) and H(meta)-H(para) are about the same, this could give rise to an apparent triplet for H(meta). Then, is it possible that long-range coupling of H(meta) to 19F might produce the doubled triplet located at ca. delta 7.25? (Please check your 19F NMR spectrum.)

Just a thought....

[kriggy]

Perhaps it might be useful to compare your 1H and 13C chemical shift data with that of a model compound.

Consider the proton NMR spectrum of 2,3-dimethylpyridine; the aromatic protons in this molecule absorb in the region delta 7.0 - 8.3 ppm. The corresponding 13C NMR spectrum of this compound displays five aromatic carbon absorption signals in the range 121-157 ppm.

(reference to proton NMR spectrum of 2,3-dimethylpyridine: http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi, SDBS Number 694).

Also, you mentioned that the 19F NMR spectrum of your unknown compound is available. Do you observe spin-spin splitting pattern(s) in this spectrum?
 
Sorry - that's about all I can offer in the absence of actual 2D NMR spectra (COSY, HMBC, HSQD) of your unknown compound.

1) Thanks for suggesting looking at model compound, the dimethylpyridine doesnt match but it seems very useful to do.
2) You mean splitting in 19F spectrum? There is only 1 peak at -109,96ppm. If you mean splitting by F which could be seen in 13C spectrum then Im not sure but it seems that there is none (very weird right? maybe bad quality of picture sicne i got only PDFs with the spectrum)
3) Im aware of that. I dont want you (or anyone else) to solve it for me since I want to be able to do it in the future (and th exam is actualy to explain how I did proceed even if I didnt get the whole structure)
hsqc
cosy
13C
1h
As I found out, there is something wrong with HMBC i got - crosspeaks doesnt fit so working on it atm

[OrgXemProf]

If the unknown compound exhibits long-range 19F coupling to, say, a pyridine ring proton, then I was thinking that you should see the corresponding splitting in the 19F spectrum. Does the one peak at ca. -110 ppm in the 19F NMR spectrum display any fine structure or other evidence of coupling to another nucleus?

It's difficult to evaluate the stated 19F chemical shift without knowing the reference. Frequently, FCC_l3 and F₃C-CO₂H and sometimes hexafluorobenzene are used as internal standards for this purpose.

I don't know any way to arrive at the structure of a C₂₂ compound other than to very patiently and systematically work through the various C-H and H-H connectivities via analysis of relevant 1D and 2D spectra (re: Irlanur's prior suggestion). It's worthwhile to perform cross-checks along the way to verify that your assignments are consistent with the rest of the spectral information.

The whole process reminds me of a Sudoku puzzle: If you make a mistake in a numerical assignment, by the time you realize that something is wrong it turns out that the mistake was made so far back in the sequence of events that it can be very difficult to retrace and then to correct. So let it be with NMR interpretation.

A further complication: The appearance of spurious ("false") crosspeaks in 2D spectra can be a real nuisance. I have found this to be especially true when dealing with NOESY spectra!

[kriggy]

It is indeed very much like sudoku but I solved many of those :-D
But never did NMR like this before.
I think there was no reference in those samples but they were corrected to residual solvent signal. The fluorine peak doesnt show any structure, its just singlet. This raises one qustion: are peaks in F NMR splitted by hydrogens like in 1H NMR
I asked the teacher and he told me that the not fitting peaks are not mistake but they are caused by something so I guess its the fluorine atom.

Thanks for help guys, I realy apreciate it

[Irlanur]

If you record a Fluorine spectrum with H decoupling, you won't see a splitting. But you will see the splitting due to the Fluorine in the 1H-Spectrum.

[kriggy]

but how can i know if the splitting is due to fluorine or hydrogen?

Anyway, I have few fragments assembled but one thing doesnt make sense to me, it he HSQC spectra if I look at carbon at ~43 ppm i see two spots which direct me to two hydrogen peaks at ~3,9 and 3,7. From what I know it means that those hydrogens are enantiotopic (is it correct?) but if they are, how is that they have different splitting? From the detail of 1H i see that one of those peaks is sextet and the other is heptet (split into 7 im not sure about the name) or could it be that the splitting is same but the visualization is just not detailed enough? 

and one more qustion:
i have a 1h singlet at 3,3 ppm with no integration in H spectra and in HMBC it shows cross peaks across all range of 13C. Any idea what it could be?