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Topic: How do I find the compound using H NMR and IR spectra?  (Read 844 times)

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Offline AWK

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Re: How do I find the compound using H NMR and IR spectra?
« Reply #15 on: May 14, 2020, 01:26:02 PM »
https://orgchemboulder.com/Spectroscopy/specttutor/arom.shtml
Did you read this link. Patterns for HNMR of disubstituted benzenes are shown there.
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Offline Navidad

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Re: How do I find the compound using H NMR and IR spectra?
« Reply #16 on: May 14, 2020, 01:43:49 PM »
https://orgchemboulder.com/Spectroscopy/specttutor/arom.shtml
Patterns for HNMR of disubstituted benzenes are shown there.
Sorry for the stupid question, but in the example of the disubstituted benzene, how does that peak show 4 protons? I am confused.

Offline AWK

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Re: How do I find the compound using H NMR and IR spectra?
« Reply #17 on: May 14, 2020, 01:58:20 PM »
Sometimes peaks can overlap. On proton spectra, integration helps to notice peak overlapping.
In your proton spectrum, you have twice two superimposed signals with very similar chemical shifts. And you don't have to consider any coupling between protons for this. In my opinion, on the C-13 spectrum, you also have two signals superimposed and additionally covered with a 120 ppm line. The benzene ring substitution pattern is also seen in the IR spectrum. Collecting all this information, it is possible to propose a meaningful structure of the studied compound.
But you have done it.
« Last Edit: May 14, 2020, 02:28:55 PM by AWK »
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Offline Babcock_Hall

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Re: How do I find the compound using H NMR and IR spectra?
« Reply #18 on: May 14, 2020, 03:24:42 PM »
Which one is sp2- hybridized?
You said the same thing twice.
Those above 100 ppm.  Those below 100 ppm are sp3-hybridized.  Carbon-13 spectra tend to spread things over a greater ppm range, which is helpful in this instance.

With respect to the degrees of unsaturation, there is one degree that is not yet accounted for.  This suggests that there is one more ring or one more double bond.  The IR spectrum and the C-13 NMR are quite helpful in identifying it.

One other point which can be helpful in interpreting C-13 NMR spectra.  Carbons with no directly attached hydrogen atoms often show lower intensity than those that have attached hydrogen atoms. 
« Last Edit: May 14, 2020, 05:19:55 PM by Babcock_Hall »

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