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Topic: NMR Help  (Read 27752 times)

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Garneck

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NMR Help
« on: May 20, 2005, 05:19:46 PM »
Ok, I have a problem. My girlfriend has a 1H NMR spectrum of "a probably aromatic ketone" (at least that's what she said). From what I suspect, this compund is benzophenone, you know - two benzene rings connected by a C=O group. All the shifts are over 7 ppm, so that suggests those are aromatic protons.

I can't confirm this on SDBS, because their 90 MHz spectra suck sometimes (yeah, the resolution is crappy when you have too many signals). So, could someone please confrim my suspicion?

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Offline Borek

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Re:NMR Help
« Reply #1 on: May 20, 2005, 05:43:32 PM »
How??
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Garneck

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Re:NMR Help
« Reply #2 on: May 20, 2005, 05:53:37 PM »
I don't understand what you mean, Borek.

Don't you see the image?
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Offline Borek

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Re:NMR Help
« Reply #3 on: May 20, 2005, 06:05:19 PM »
I see now, something was wrong at first. No idea why.
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Offline movies

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Re:NMR Help
« Reply #4 on: May 20, 2005, 08:11:41 PM »
How many MHz is that spectrum at?  If it's 500 MHz then it looks right on for benzophenone.  Even the coupling constants work out.

All the peaks are shifted by about 0.15 ppm from the data I have, but that could be from a different solvent or perhaps the spectrum isn't referenced properly to TMS.

I got the data from this book, 3rd ed., p. 219.
http://www.amazon.com/exec/obidos/tg/detail/-/3540678158/
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Garneck

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Re:NMR Help
« Reply #5 on: May 21, 2005, 02:50:29 AM »
Quote from: movies on May 20, 2005, 08:11:41 PM
How many MHz is that spectrum at?  If it's 500 MHz then it looks right on for benzophenone.  Even the coupling constants work out.

All the peaks are shifted by about 0.15 ppm from the data I have, but that could be from a different solvent or perhaps the spectrum isn't referenced properly to TMS.

I got the data from this book, 3rd ed., p. 219.
http://www.amazon.com/exec/obidos/tg/detail/-/3540678158/

I don't know, movies. My girlfriend didn't tell me that, besides, she wouldn't even know ;)

Ok, now if it's benzophenone, then I don't understand one thing - I'll draw it. I've pointed to the "same" (I don't remember how to call them properly in english) protons. Now I have a question.
The red protons couple with 2 blue ones and one green. Why isn't there a quartet signal from those two protons? It's been a while since I interpreted spectras.

Thanks for your answers.
« Last Edit: May 21, 2005, 02:51:01 AM by Garneck »
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Offline Winga

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Re:NMR Help
« Reply #6 on: May 21, 2005, 06:54:29 AM »
"same" => "equivalent"

The two meta-protons (red) are chemically equivalent, and you only need to study either one of them.

Just concentrate on one of the meta-protons.
From its proton NMR signal, the multiplicity is triplet (2 nearby protons) and the chemical shift is 7.4979.
Then, you study another meta proton, it gives you the same signal pattern and position.
The two signals are overlapped to each other.

Another example:
3-ethylpentane
What is the multiplicity of methylene(-CH2-) protons?
« Last Edit: May 21, 2005, 06:55:04 AM by Winga »
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Garneck

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Re:NMR Help
« Reply #7 on: May 21, 2005, 07:34:10 AM »
Quote from: Winga on May 21, 2005, 06:54:29 AM
"same" => "equivalent"

Thanks!

Quote from: Winga on May 21, 2005, 06:54:29 AM
The two signals are overlapped to each other.

Another example:
3-ethylpentane
What is the multiplicity of methylene(-CH2-) protons?

I get it.

I'll have a quartet from those groups, when they couple with the three protons from CH3- and 3 signals will be overlapped, right?
« Last Edit: May 21, 2005, 07:39:38 AM by Garneck »
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Offline Winga

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Re:NMR Help
« Reply #8 on: May 21, 2005, 07:40:40 AM »
Quote
I get it.

I'll have a quartet from those groups.

I think you got it, but the answer is wrong! ;D
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Garneck

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Re:NMR Help
« Reply #9 on: May 21, 2005, 07:43:09 AM »
Quote from: Winga on May 21, 2005, 07:40:40 AM
I think you got it, but the answer is wrong! ;D

See my corrected answer
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Offline Winga

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Re:NMR Help
« Reply #10 on: May 21, 2005, 07:46:58 AM »
Quote
I'll have a quartet from those groups, when they couple with the three protons from CH3- and 3 signals will be overlapped, right?
I think using the term "overlap" to describe the signal is not enough, because this overlapping will also lead to "increase in peak intensity".
« Last Edit: May 21, 2005, 07:49:07 AM by Winga »
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Offline Winga

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Re:NMR Help
« Reply #11 on: May 21, 2005, 07:50:13 AM »
I mean the multiplicity is wrong.
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Garneck

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Re:NMR Help
« Reply #12 on: May 21, 2005, 08:27:48 AM »
 ???

So what will it look like then? I mean, the whole spectrum.
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Garneck

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Re:NMR Help
« Reply #13 on: May 21, 2005, 08:29:41 AM »
Duh! I know now. I forgot that the 3 carbon has a proton attached to it. Forget my question.
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Offline Winga

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Re:NMR Help
« Reply #14 on: May 21, 2005, 10:41:51 AM »
Chemical shift: ortho-H > para-H > meta-H
It is because the carbonyl group is electron withdrawing.
Just figure out the resonance structures and you will know why.
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