[DoctorDomo]

Lets say you have this compound:

I know that 19F atoms can couple with each other over much greater distances, so while protons couldn't couple across the whole benzene ring, 19F atoms on the other hand can. Firstly, what peaks would you expect to see from that compound? I'd expect to see 3 peaks, but I'm baffled by what splitting would occur.

Firstly lets start with 3 atoms together, lets say they're at positions 1,2 and 3 starting from the let (so the lone 19F atom is at position 5). Atoms 1 and 3 are equivalent, so they wont couple, but they'll both couple with atom 2, forming a regular doublet I presume. But then theres atom 5. That will couple with them too won't it, so in this case will it split the doublet into a doublet of doublets because atom 5 is in such a different electronic environment?

Then atom 2, it will couple with its two neighbours forming a triplet, but can it couple with atom 5 (located para to it) or is that too far away for it to couple? Someone told me that the presence of electron withdrawing groups affects this. So lets change the compound to this:

Now the ring is more electron deficient, will that increase the range of 19F-19F coupling, making it more possible for flourine 2 and flourine 5 to couple?

Then what about this compound:

Should be an easy one, there are 2 unique 19F atoms there, so you get two peaks but how does the splitting work? I would have expected flourines 1 and 4 to couple with flourines 2 and 3, forming a triplet, then the same thing happening for fluorines 3 and 4, but the real life spectrum I saw, didn't have two triplets. It has 2 peaks, but they're not triplets. Heres one of them:

The phase is corrected. What the hell is this multiplet? Both peaks (they're pretty far apart) have these weird multiplets.  The spectrum is proton decoupled. Why are there so many peaks points, I expected to see triplets.

[discodermolide]

Looks like two overlapping quartets.

[DoctorDomo]

I'm confused about how that could happen. On the molecule you have two pairs of spin equivalent protons, so they shouldn't couple, true? If thats true, then why would they be split by 2 protons into quartets, rather than triplets? And there appears another multiplet way downfield with the same pattern. Why would there be 4 unique 19F atoms on this molecule?

Lets call them the protonated atoms 1 and 2, and the 19F atoms 3,4,5,6. Atoms 4 an 5 are ortho to each other and are spin equivalent. Atoms 3 and 6 are para to each other and are also spin equivalent (are they not?). I assumed that atoms 4 and 5, woudl couple to atoms 3 and 6 to form a triplet. But if I think about it, atom 4 is located meta to atom 6, so that would mean atom 6 is in a different electronic environment (with respect to atom 4) than say atom 3 which is right next door to atom 4. Is that whats going on here? Sorry, I confused myself with that question.

[Corribus]

I didn't read all this very closely, but have you considered the protons (and carbons) couple very strongly to fluorine nuclei? Getting strange splitting patterns when you have a lot of fluorines/protons together is not uncommon.

http://chem.ch.huji.ac.il/nmr/whatisnmr/hetcoup.htm#hcf
http://faculty.missouri.edu/~glaserr/8160f09/fluoroacetone_NMR.pdf

EDIT: Unless your spectrum is decoupled, in which case, never mind.

[DoctorDomo]

The spectrum is proton decoupled. Plus, there are only 2 protons so the splitting wouldn't be of equal intensity for each peak, the protons would split the nearest neighbours more effectively than the ones on the opposite side of the ring. The molecule for that spectrum I showed is actually ortho-acetyl-tetrafluorobenzene, so proton coupling definitely isn't the cause of this. 17O isn't abundant to influence the spectrum. Maybe its something to do with resonance structures? The NMR was done at room temperature, but I don't know if resonance structures can influence splitting in fluorine spectra or not.

[Babcock_Hall]

The compound with four fluorines in a row may be AA'XX' or AA'BB'.  In other words for each pair of fluorines with the same chemical shift, the two nuclei are not magnetically equivalent.  Ortho-dichlorobenzene is another such compound, and it is covered in Silverstein's book.  Thus there are more peaks in this sort of system than are predicted by simple rules.  Simulation programs are good at handling these sorts of situations in my limited experience.