Hi
I wanted to learn proton NMR shift by substitution group.
As you can see this picture, the main structure is like that
The first one is substituted methyl group(C1 proton)
and the second one is substitued fluorine(Also C1 proton). The control NMR peak is thrid one.
Firstly I thought the hydroxyl and imine group proton of first and second structure should be devided into two peaks. However only the structure which has fluorine only present divided peaks.
I don't know why. Actually the electron environment is different both side.
Is there any reason?? Please let me know. Thanks
you need to provide integrals for the spectra to make sense.
I assume you are talking about the two left most group of signals.
The left most groups has split there are 2 singlets.
The 2nd left most group appear to be only 1 singlet for the Me substituent.
Methyl is only a weak electron donating group, the difference may not be significant enough that the spectrometer can detect. A higher frequency NMR machine may be able to resolve the peak.
J coupling is mutual, i.e. JAB = JBA always. Thus there is never just one nucleus which shows J splitting -
there must be two, and they must have the same splitting constant J. However, both nuclei need not be protons -
fluorine (19F) and phosphorus (31P) are two other common nuclei that have spin ½ and 100% abundance, so they
will couple to all nearby protons. If these nuclei are present in a molecule, there are likely to be splittings which are
present in only one proton multiplet (i.e. not shared by two multiplets).
https://faculty.missouri.edu/~glaserr/8160f09/fluoroacetone_NMR.pdf
http://chem.ch.huji.ac.il/nmr/whatisnmr/hetcoup.htm#hcf
I assume this is the downfield region - aromatic protons to the right, phenol (OH) and imine (N=CH) protons to the left. A methyl substitution won't have as large an effect as a fluorine substituent, so this isn't particularly surprising. I would expect the phenolic OH to be most sensitive to electronic effects, which alter the basicity of the N to which they are H-bonded. You can verify that the downfield signals are the phenols by D2O exchange.
You need to provide spectra with signal integrals and also the chemical shift scale, and it would also be helpful if you clearly state with trace you assign to which compound. Without this information one can only speculate.
I agree with all comments above, but in special with doctor Baumann that there is a lack in the information that are - besides the chemical shift and integral - solvent used, spectrometer (300,400 MHz) and temperature in which the spectra were obtained. I would also like to know if they are real spectra or simulations of the proposed structure. Best regards
Thank you all!
I missed some information like integral value and solvent.
Anyway, as James said, the methyl group is weak electron donating group so the peak is not divided that much. I'll keep try to assign them! Thanks a lot.
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