I have these H-NMR results for a 3-FEA batch (not my 3-FEA, not my NMR, I just have the screenshot of the results). I'd like to practice my NMR skills
on them but I'm having a little bit of a hard time.
This is the proton numbering scheme I'm going to use just so we're on the same page:
So I see the peaks between 6.75 and 7.25 that correspond pretty well to the hydrogens on the aromatic ring. In total, we should get an integral of
four there, but the NMR shows 5. I'm guessing the extra bit may be due to the CDCl3. So that's forgiven.
I'm seeing the peaks between 1.25 and 1.75 that correspond to the methyl groups, and integral looks nice.
Now between 2.75 and 3.75 I'm seeing an integral of five. This should be six, no? H14, H15, H4, H16, H17, and the amine H3. I'm guessing this is the
amine H3 not showing up. Why is that?
Now there's of course the suspicious peak at 9.75. The aldehyde range. Given the broadness, and the integral semi close to 1, could this be the amine
hydrogen? Maybe the fact that its 1 to 2 corresponds to the fact that this 3-FEA might have been a hydrochloride salt? But why is it so downfield? It
should be significantly more upfield?
Maybe the downfield hydrogen corresponds to the hydrogen in some 3-fluorobenzaldehyde impurity from the starting material? But thats huge, and we
don't see other huge peaks corresponding to that.
What's going on here?
1. Where's the amine hydrogen?
2. Whats that big peak downfield?
Thanks.thors.lab - 7-10-2020 at 13:31
Oh, could the peak at 9.75 correspond to HCl from the HCl salt of 3-FEA? In which case, there's not a stoichiometric amount of HCl. Right?DraconicAcid - 7-10-2020 at 17:18
The broad peak is definitely the one on N, quite possibly protonated.thors.lab - 7-10-2020 at 17:43
The broad peak is definitely the one on N, quite possibly protonated.
Why does it appear so far downfield? Shouldn't these amine hydrogens be closer to 3ish?
And why does its integral not match the expected?
[Edited on 8-10-2020 by thors.lab]DraconicAcid - 7-10-2020 at 18:23
I'm guessing that because the N is protonated, it will be strongly deshielded.
Hydrogens on N seldom integrate properly, due to the quadropole moment of N (at least I seem to recall that's the excuse I learned). That's also why
N-bound hydrogens always give broad peaks, even if not hydrogen-bonded.
