December 22, 2024, 11:00:48 AM
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Topic: 13C NMR (hydrobenzoin)  (Read 8744 times)

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Offline smghz

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13C NMR (hydrobenzoin)
« on: February 26, 2018, 07:32:22 PM »
Would it be reasonable to conclude that the peaks beginning after 125ppm, and possibly including the small peak at 140, are aromatic? And thus, the carbons in the 70-80ppm range are those connected to O's? It seems so counter intuitive since those are only two carbons and there are visibly more peaks there than in the aromatics region. I'd love to know your insight. thanks

Offline smghz

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Proton deshielding in C versus O
« Reply #1 on: February 26, 2018, 07:48:37 PM »
For hydrobenzoin, the non-aromatic carbons are each attached to a hydroxyl. I was wondering, which proton would be more deshielded: the one in hydroxyl or the one attached to the carbon? It seems to me that it is the hydroxyl protons, but when I looked up the ppm range of O-H, it was in 1-4 ppm range, and that didn't match up with my observations since there is another peak right around 5 that is more deshielded.

Offline billnotgatez

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Re: 13C NMR (hydrobenzoin)
« Reply #2 on: February 26, 2018, 10:31:29 PM »
@smghz  I hope you do not mind but
I merged these 2 threads since both are talking about hydrobenzoin and can be addressed together.

Offline hypervalent_iodine

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Re: 13C NMR (hydrobenzoin)
« Reply #3 on: February 27, 2018, 01:27:10 AM »
Have you considered the presence of solvent peaks from the chloroform you've used? Based on the fact that you integrated the chloroform peak in the proton spectra, I'm guessing not.

Offline Babcock_Hall

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Re: 13C NMR (hydrobenzoin)
« Reply #4 on: May 01, 2018, 01:57:48 PM »
It looks as if there are two peaks in very close proximity in the H-1 spectrum; one is chloroform, and I don't see how one is going to get a completely clean integral under these circumstances.

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