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Topic: Counting NMR Peaks  (Read 8113 times)

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Offline rozarria

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Counting NMR Peaks
« on: November 25, 2008, 02:50:06 PM »
Would this compound have 9 different peaks on an NMR chart?



At first, I thought it would be 6, but my professor said that because there are 2 different hydrogens (1 is cis, the other is trans) attached to each cyclic carbon, those are actually 2 different peaks. So, I counted 9 instead. Is this correct?

Offline azmanam

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Re: Counting NMR Peaks
« Reply #1 on: November 25, 2008, 03:04:16 PM »
Yeah.  I would have said 6 first, too.

They are diastereotopic protons.  They probably will give different signals.  If you want to impress your instructor, let him/her know the equatorial ones will be more deshielded and will show up more downfield.

I think 11 is the right answer - the methylene protons on the ethylbromide chain are also diastereotopic.***

http://cobalt.rocky.edu/~barbaroj/equivalent_hydrogens.pdf
http://www.chemistry.ccsu.edu/glagovich/teaching/316/nmr/couplingtopic.html
http://www.cem.msu.edu/~reusch/VirtTxtJml/Spectrpy/nmr/nmr1.htm#nmr1
(scroll down to the end of the pi-electron functions section)




***Wait.

That's not a stereocenter.  The molecule is not chiral.

The axial/equatorial protons will still give different signals - that is, the ring protons ARE diastereotopic - but the methylene protons on the side chain are enantiotopic, not diastereotopic.

My mistake.  9 is probably right.

(the links are still good, though)

edited to correct mistake
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Offline azmanam

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Re: Counting NMR Peaks
« Reply #2 on: November 25, 2008, 03:14:12 PM »
That's actually a really interesting example.

Some of the protons are diastereotopic and some are enantiotopic.

I can't think of another example like that.
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Offline azmanam

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Re: Counting NMR Peaks
« Reply #3 on: November 25, 2008, 03:16:18 PM »
It's also really interesting, because the protons opposite to ethyl group are interesting.  Changing one of those to deuterium would NOT create chiral centers.  The molecule will still not have chiral centers.

But the chair would favor the conformation with the ethyl group equatorial - so the protons opposite the ring ARE defined as one favoring axial and one favoring equatorial, so they would still give different signals because they are still in different environments.

Very interesting question.  Thanks for asking.
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Offline azmanam

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Re: Counting NMR Peaks
« Reply #4 on: November 25, 2008, 03:18:30 PM »
Maybe butanal diethyl acetal would be another example.

The protons on the butane chain would be enantiotopic, but the methylene protons in the ethyl groups would be diastereotopic.

Sorry.  I don't know why this question fascinates me so much, but it does...  I'm done now.
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Offline macman104

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Re: Counting NMR Peaks
« Reply #5 on: November 25, 2008, 03:18:54 PM »
I agree, it looks like it should be 9.

Offline rozarria

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Re: Counting NMR Peaks
« Reply #6 on: November 25, 2008, 04:49:02 PM »
Hahaha, thanks for the help, azmanam. By the way, the hydrogens on the side chain substituent (the part branching out of the cyclic ring) are all on the same plane, right? They can't be pointing towards or away from you? Which is why they only count for one peak collectively?

Offline azmanam

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Re: Counting NMR Peaks
« Reply #7 on: November 25, 2008, 05:29:05 PM »
Quote
the hydrogens on the side chain substituent (the part branching out of the cyclic ring) are all on the same plane, right

no.  the carbon atoms are still tetrahedral and sp3 hybridized, thus the protons are still oriented in 3-dimensions.  There is free rotation about both of the C-C bonds in the side chain.  So if you arbitrarily position the two carbon atoms in the plane, then the two protons on each carbon atom are either in or out of the plane.

There are two reasons they coalesce into one signal.  First, there is free rotation about the C-C bonds.  Thus, the protons are not locked into one chemical environment the way the axial/equatorial protons are in the cyclohexane ring are.

Second, even though there is free rotation, if the protons are diastereotopic, they will give slightly different chemical signals.  The classical way for determining if a set of protons are diastereotopic or not is to exchange one proton for a deuteron (a deuterium atom).  If that forms a new stereocenter, then the two protons are stereotopic.  If you exchange the other proton for a deuteron, you will either form the enantiomer of the the first deuterium-substituted analog, or a diastereomer of the first analog.  If you make enantiomers, the protons are enantiotopic and give the same NMR signal.  If they are diastereomers, they protons are diastereotopic and give different NMR signals.

In this case, the side chain protons on the first carbon atom can be substituted with deuterium to give enantiomers (the carbon center on the ring bearing the side chain is still not a stereocenter, so exchange with deuterium on the side chain only gives one stereocenter).  Because exchange gives a pair of enantiomers, those protons are enantiotopic and will only give one signal.

If you exchange protons on the same carbon within the ring, you generate a stereocenter at the carbon bearing the protons being exchanged for deuterium... AND the ring carbon bearing the leaving group is now a stereocenter.  There are now two stereocenters in the molecule where one of the ring protons is substituted for deuterium.  Exchaning the other proton on that center gives a diastereomer of the first compound, thus the protons are diastereotopic and will give different signals in the NMR.
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