Question on C13 NMR

  • Thread starter Bladibla
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  • #1
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Main Question or Discussion Point

I currently have a problem with having 2 small peaks, normally from what I would recall would correspond to 2 fully substituted carbons. (The molecule I have predicted is ethyl-3-bromobenzoate) With one carbon fully substitued at the '3 position of the benzene, the carbon of the carbonyl group, and the carbon of '1 position of benzene fully substituted with the carbon group, wouldn't there be 3 small peaks?

The calculation I do for the individual resonances come out perfectly; however, one of the problems is that the '1 carbon of benzene is shown as a peak with high intensity. My question is is it possible for a carbon fully substitued in a benzene ring to have a high resonance?

Thanks.
 

Answers and Replies

  • #3
chemisttree
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It was this molecule, by the way. http://www.sigmaaldrich.com/catalog/search/ProductDetail/ALDRICH/335819

Why is there only 2 small peaks for this molecule for the C13?
No se... you should get six peaks in the aromatic region, one for the carboxyl and two for the ethyl ester plus those of the solvent(s). Did you lose lock during the acquisition? How concentrated was your sample? Was the proton decoupler on during acquisition? How many acquisitions in your spectra? What does the S/N look like? Does the FID look normal or is it noisy?

Lots to think about...

Are you sure that you aren't looking at a bad proton spectrum?
 
  • #4
chemisttree
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Never mind my previous post. I misunderstood what you were asking... I thought you only saw 2 peaks in your spectrum. Yes, you should see three carbons that lack NOE enhancement and appear smaller. Sometimes the carbonyl carbon is a little harder to see but you should see it (well-removed from the aromatic region).
 

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