[nirhananya]

Hello. My compound has a double bond attached to an aromatic ring.
I've done photooxygenation on this double bond (to get the 1,2-dioxetane four membered ring, doesn't matter). It's a simple reaction that I've done many times before (but not on this specific molecule).
Here is what happened: two of my aromatic hydrogens just disappeared! I can't see them in the H-NMR spectrum, but everything else in the molecule seems to be just fine.
So my question is: Is there any chance that the expected reaction did happened, but still I can't see the aromatic hydrogens for another reason? Does anybody have ever seen anything like that?
[I've already tried different solvents and also long relaxation time (d1). didn't help.]
Thanks

[kriggy]

Could you show us the structure and spectra?

[nirhananya]

I attached here a picture of the 1,2-dioxetane, while in the substrate there is no O-O four membered ring but a double bond instead. I also attached the spectra of the substrate and the product, and an enlargement of the aromatic zone of the product (you can ignore the small picks which are contamination).
You can see there those "hills". I thought that maybe they can explain the mystery. What do you think?

And another thing - I've just got a MALDI spectrum that confirms that this is the product. So where are my aromatic hydrogens???

[Irlanur]

Usually flat peaks ("hills") are indicative of exchange processes or fast relaxation. I don't know answer but I can't think of a way where you would have exchangeable aromatic Protons... anyone?

Another thought I have is about oxygen. it is a paramagnetic compound and if I think about oxygen and light it seems likely that you formed some kind of radical maybe? this could explain the fast relaxation of nearby protons. But I can't think of a mechanism or the exact physics behind the NMR spectrum...

[Irlanur]

I would suggest to

1) measure an NMR again.
2) put it in an EPR Spectrometer, if easily available.

[Dan]

Usually flat peaks ("hills") are indicative of exchange processes or fast relaxation. I don't know answer but I can't think of a way where you would have exchangeable aromatic Protons... anyone?

Seems unlikely, but could it be a bisphenol?

Nirhananya, have you got MS data?

[nirhananya]

Thank you everybody.
Well, the MS spectrum was hard to achieve (the compound is unstable), but it confirms that I've got the right product.
I also have a functional proof that confirms it - this compound emits light upon substitution of the boron by oxygen with H2O2 followed by hydrolysis (chemiluminsence), so I guess it's enough.
I know that aromatic hydrogens are not exchangable, but maybe the slow rotation around the Ar-C bond make the picks wide.
thanks again
Nir

[Babcock_Hall]

Are the hills at the expected chemical shift of the hydrogen atoms in question?  Just thinking out loud but two of the three aromatic H's might show three-bond coupling to boron, and the major isotopes of boron are both quadrupolar.  I am not sure whether or not this will have an effect, although if you could obtain a spectrum while decoupling boron, that might be helpful.
http://www.bruker-nmr.de/guide/eNMR/chem/B.html
http://u-of-o-nmr-facility.blogspot.com/2008/04/1-h-nmr-with-11-b-decoupling.html

[Irlanur]

good point. but you would also see it it in the reactant I guess...?

[Babcock_Hall]

My incomplete understanding is that if the T₁ of the boron changed between reactant and product, then the appearance of the H-1 spectrum might not be the same with respect to how much broadening one sees.  However, checking the starting material sounds like a good idea.  http://www.chem.wisc.edu/areas/reich/chem605/