Absence of some signals on 1H and 13C spectra of polyaromatic carboximidate

i like this post (click again to cancel)

i dont like this post (click again to cancel)

remove favorite mark from this question (click again to restore mark)

Hi,
I have encountered a strange behaviour of some signals (they are absent) in a series of polycyclic aromatic carboximidates spectra. We have X-Ray or mass spectroscopy data for these compounds, so we know how they look like. All signals in 1H and 13C spectra match up to structures, but some protons and carbons don't produce peaks at all.

alt text

(1) Here we have this problematic fragment. Red atoms are not visible on corresponding spectra. Imine proton can be found near 14 ppm (CDCl3), but its intensity is like 0,05 in comparison to single aromatic proton's signal. Probably this peak belongs to 15N bonded protons. Red carbon signal is absent in 13C spectrum.

alt text

(2) Analogical Coumarin was synthesised. In case of this compound all signals can be found and described on respective spectra.

alt text

(3) Trifluoromethanesulfonyl amide of compund (1) behaves like Coumarin (2). All signals can be found and described.

alt text

(4) This compound is similar to (1) but instead of phenyl ring (twisted 90 deg in relation to the rest of molecule plane) we used here the ferrocenyl as an off-planar substituent (it's twisted approximately 40 deg). The second ferrocenyl is also incorporated in this structure, but its position is distant from this problematic region of molecule.

1H Spectrum: Imine proton is visible at 14 ppm with proper intensity (integral = 1). Ferrocenyl (red one on illustration) signals are not visible at all. Signals from the second ferrocenyl (not included on ilustration) are present with proper intensities.

13C Spectrum: Imine carbon can be found at 189 ppm (CDCl3). 1st ferrocenyl signals (all !) are absent, 2nd ferrocenyl signals are present.

Does anyone know what is going here? Maybe this is an effect of strongly quadrupolar (14N) vicinity and/or CSA of C=N bond? Any idea, hint or literature would be a great help.

Best regards,
Arek

asked Mar 14 '15 at 13:33

Arkadiusz Leniak
238

updated Mar 14 '15 at 14:28

3 Answers:

oldest newest most voted

i like this answer (click again to cancel)

i dont like this answer (click again to cancel)

Hi Arek! I haven't got ready-to-use explanation, but I can think of 4 reasons: 1. very long T1 2. or very short T2 relaxation times 3. chemical exchange 4. hindered rotation around single bond connecting these two rings.

Cases 3. and 4. probably can be rejected as you say nothing about doubling of signals or their broadening. Which is also suggesting that in case of (1) the problem is v. long T1 (quaternary carbon). The T1-problem may be tested by UDEFT sequence, which is available in Bruker's standard pulse program directory. I strongly encourage you to test it link text

In case of (4) problem may be more complex and somehow related to N lone pair of electrons.

Regards Luke

link

answered Mar 18 '15 at 00:15

zablociak
21

Luke,
I'm truly grateful to you for your answer,

In reference to the 1st compound (one can name it iminocoumarin, iminolactone or iminoether), I thought that the invisibility of this particular carbon signal could be caused (just like you wrote) by long relaxation times (I've experienced that kind of phenomenon in case of a few formyl carbons). I haven't occassion to utilize this UDEFT sequence, but according to your recomendation I will try to use it and check how it'll work in this peculiar case.

Talking about exchange - it's hard to say anything about it. There's no proton signal (one bonded to nitrogen) to measure at spectrum (dry CDCl3 deacidified a priori to measurments). Other solvents are out of question - hopeless solubility, decomposition or isomerization. DMSO results in opening of this 6-membered ring and outcomes with two isomers. Spectra in DMF looks just like the the one in CDCl3.
There is probably also some kind of hydrogen bond -O---H-N- (vestigial =N-H signal at 14 ppm).

alt text

In the case of 1st compound the culprit it's probably a quadrupole coupling (freak short T1 / T2 times). This statement can be proved by comparison to 2nd compound (oxygen instead of nitrogen -> everything goes smoothly).
3rd compound (suphamide) also could be some kind of proof. Sulphamide drastically changes electron density distribution across molecule and disturbs quadrupole properties of nitrogen nucleus.

Some singularity arises in case of 4th compound (ferrocenyl substituents). You mentioned about hindered rotation around single bond connecting these two rings. Could you be some kind and say something more about it? I've never met this kind of issue. How can I check if it is a matter of my case? Spectra in higher temperatures? (I've made spectra for 1st compoud up to 373K in DMSO - two isomers comes up to one set of signals, and DMF - basically only chemical shifts change a little. I didn't make low temperatures measurments).

I've checked also CSA. I've made 1H (1st compound) spectra on different spectrometers (200MHz, 400MHz, 500MHz, 600MHz). Only one signal shows broadenig in higher fields. The rest of signals are quite sharp.

alt text

Just like you, I also think that the lone pair at a nitrogen atom plays some major part in this incident, but I just can't understand and explain how (especially for 4th case). This must be some kind of quantum phenomenon.

Arek

link

answered Mar 20 '15 at 19:08

Arkadiusz Leniak
238

Arek, in case of comp. 1-3 you have (most probably) already answered your own questions. It only bothers me if one can test this on NMR more directly. Maybe this will help you: link text

The hindered rotation around C-C bond will be visible as doubling of signals in both 1H and 13C spectrum at low temperatures, then "collapse" and eventually one set of signals will be visible at higher temperatures. This is caused by the phenomenon of "rotamers", that are visible on NMR-timescale. So a big bulky ferrocenyl group could have energy barrier between both rotamers. The very good example (from my field of interest) are flavonoids called C-glycosides: link text In case of v. short T2s you'd observe simple broadening of signals at low temperature. Higher temperature would make them sharper (or visible). According to Glenn Facey: link text

The line width at half height for an NMR resonance for a dissolved substance (in the absence of exchange or quadrupolar effects) is 1/(piT2). A plot of log T2 vs correlation time first decreases and then comes to a plateau at longer correlation times. The correlation time is defined as the time a molecule takes to rotate by one radian. At higher temperatures the correlation time is shorter and at lower temperatures the correlation time is longer. Therefore at as the temperature is decreased one expects the correlation time to increase and the T2 to decrease. Since the line width at half height is 1/(piT2), the line width increases as temperature is decreased.

In fact I've tested one compound lately, in which short T2 problem occurred through intramolecular dipolar interaction. Imagine two rings in one molecule that are connected throug glycosidic bond and close to each other, but not fixed in their positions. Therefore, through-space interaction is causing them to relax quickly. I've acquired spectra at 25 and 40 degrC in CD3OD, and the latter was much sharper. This means, that in higher temperature molecular tumbling was much faster and therefore efficient relaxation was not possible. Anyway - you should try to make some temperature tests (maybe in DMF?). It may clear up the situation.

Regards Luke

link

answered Mar 23 '15 at 12:11

zablociak
21

Here (once you log in) you will be able to sign up for the periodic email updates about this question.

Your answer

Please start posting your answer anonymously - your answer will be saved within the current session and published after you log in or create a new account. Please try to give a good answer, for discussions, please use comments and please do remember to vote (login to vote)

toggle preview

Tags:

1H-NMR ×26
relaxation ×18
13C-NMR ×12
missing ×4
signal ×3

Asked: Mar 14 '15 at 13:33

Seen: 7,462 times

Last updated: Mar 23 '15 at 12:11

Related questions