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Author Topic: EPR Spectrum  (Read 6588 times)

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Sonntag

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EPR Spectrum
« on: August 22, 2014, 07:31:20 AM »

Hi guys,

can you predict the EPR spectrum of this molecule?

I would say, the unpaired electron is at the C-atom between the -CH2 and -CH3. So, because of the protons (S = 1/2), we will get 3 x 4 signals. The double bonding leads to a stronger triplet than quartet. So, I would expect a triplet of quartets.

What would you say?

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rwiew

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Re: EPR Spectrum
« Reply #1 on: August 22, 2014, 11:55:27 AM »

I don't see a radical in your structure?
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Sonntag

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Re: EPR Spectrum
« Reply #2 on: August 22, 2014, 12:05:48 PM »

Yeah, you're right. It's just the molecule used in the polymerization. Anything else I need to assume (and belongs to my question). ;)
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rwiew

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Re: EPR Spectrum
« Reply #3 on: August 22, 2014, 07:41:49 PM »

So, you want a radical on the carbon that's currently in the double bond and at the same time you want to keep the double bond (that's your argument for stronger triplet)? That's a pentavalent carbon, does not make sense. Please can you draw a structure with the radical in question indicated, the molecule that you showed does not show any EPR response as it does not contain unpaired electron density.
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Sonntag

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Re: EPR Spectrum
« Reply #4 on: August 23, 2014, 04:24:43 AM »

After starting the polymerization I assume the growing chain looks like this. What do you think?
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Sonntag

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Re: EPR Spectrum
« Reply #5 on: August 23, 2014, 04:26:24 AM »

-
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rwiew

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Re: EPR Spectrum
« Reply #6 on: August 23, 2014, 11:38:19 AM »

Cool, I see what you mean now. So we have 3 options - triplet of quartets, quarter of triplets or sextet. I wouldn't be surprised if it is a sextet (the hyperfine coupling constants with both groups of protons essentially equivalent). But if I were to speculate about the two being inequivalent, then I'd consider hyperconjugation between the C-H/C-C bonds and the radical - you could say a C-H bond has a slightly higher donor ability and hence the hyperconjugation with the -CH3 group will be slightly larger than with CH2R and hence larger coupling with the methyl. This would give you a quarter of triplets. Of course your argument with the double bonding does not work - the double bonding is not present anymore in the radical, you see that?
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Sonntag

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Re: EPR Spectrum
« Reply #7 on: August 24, 2014, 02:36:14 AM »

Thank you, for your help, especially the pro and cons for two inequivalent coupling constants. :)
And yeah, the double bond argumentation was a too quick thought.
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