Need Help with Organic Spectroscopy & NMR?

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SUMMARY

This discussion focuses on interpreting NMR spectra for tri-methyl substituted benzenes. The key takeaway is that the number of distinct proton signals in the aromatic region of the H NMR spectrum indicates the substitution pattern: one signal corresponds to 1,3,5 substitution, two signals to 1,2,3 substitution, and three signals to 1,2,4 substitution. The aromatic region is confirmed to be around 7 ppm, and integration is recommended for further differentiation of the spectra when necessary.

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  • Understanding of NMR spectroscopy principles
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  • Knowledge of H NMR and C NMR signal interpretation
  • Ability to analyze integration and splitting in NMR spectra
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  • Learn about the different substitution patterns of aromatic compounds
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Chemistry students, organic chemists, and researchers working with NMR spectroscopy and aromatic compounds will benefit from this discussion.

suski
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anyone here is good at organic spectroscopy ??
I really need help with NMR spec > <

thanks a lot !
 
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I don't know if I'd call myself 'really good' with NMR spectra, but I've dealt with deciphering them before. I'm sure others here have too, so just tell us what your specific problem is rather than asking if we can solve it first...
 
the question i am trying to solve has 3 structures given, all three are benzenes substituted with 3 CH3- groups but each with different substitution locations
and the question asks to look at 3 spectrums which corresponds to each of these 3 tri-methyl benzenes and to determine which spectrum is for which tri-methyl benzene..
mm...does that make sense ?
is there a way to determine the positions of the methyl groups ? like by looking at the H NMR or C NMR ?

Thanks a lot!
 
Yes it should be easy. The only possible ways you could trisubstitute w/ methyls would be
in the

1,2,3

1,3,5

1,2,4 positions.

Just look for the number of distinct proton signals you see in the aromatic region in the HNMR. If you see just 1 signal it must correspond to 1,3,5 substitution, 2 signals to 1,2,3 substitution, and 3 signals to 1,2,4 substitution.

You don't even need to worry about splitting or integration at all.
 
thanks!
mm...there are two singlets for two of the HNMR spectrum and the other one has a doublet of doublet...
is the aromatic region around 7ppm ?
 
suski said:
thanks!
mm...there are two singlets for two of the HNMR spectrum and the other one has a doublet of doublet...
is the aromatic region around 7ppm ?

Yes the aromatic region is down around 7 ppm. If you can't figure out which one is which from just the number of signals, then go a step further, either pick integration or splitting to figure out which one is which. I'd pick integration since H on aromatic rings can be tricky sometimes because you can have long range coupling. From integration you should be able to figure out which one is which.
 
suski said:
thanks!
mm...there are two singlets for two of the HNMR spectrum and the other one has a doublet of doublet...
is the aromatic region around 7ppm ?

Yeah, that happens. The carbon signal would act like GNW indicated but the Proton spectrum will give singlets for the regular (regular = symmetrical) isomers. The odd isomer (nonsymmetrical) will give you a more fully coupled spectrum in proton. Try looking at the methyl groups for the two spectra that have singlets. Notice anything?
 

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