1H NMR of 4,4' azobenzene dicarboxylate

Hi all,

I've attached a screenshot of the 1H NMR spectrum for clarity.

The molecule in question is:

NaOOC-Ph-N=N-Ph-COONa

First problem: Cis-trans forms.

The trans form (as expected at RTP due to sterics) would surely display 4 different proton environments, however only two are observed.

The cis form would display 2 different proton environments due to symmetry - however this form is not expected at RTP due to sterics.

I've said in my analysis that due to resonance throughout the whole molecule, rotation about the N=N is possible and so, in effect, only two proton environments are present.

Second problem: Additional splitting.

As you'll be able to see, each part of each doublet is again split into a triplet.
The J values for the two doublets are ~8,6, so delta over J is less than one, indicating that complicated splitting patterns will be observed.

i've been advised that this splitting will either be down to 14N coupling or second order coupling (as the spltting is relatively large in terms of intensity).]

Trouble is, I'm not quite sure how to explain this!

Any thoughts/suggestions?

Thanks in advance!