Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: sundberg on October 28, 2005, 01:15:10 PM
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Hi!
I had an exam today in analytical organic chemistry which went pretty fine, but I hade one simple (!) NMR spectra that I just couldn't and still can't solve. It drives me mad, and I really should know this but I need your help. ;D
The question is which stereoisomer of C5H6O is present in the attached spectra (CDCl3, 25C, 60MHz). I've drawn a wide array of different molecules but nothing add up! My best guess for now is CH3-CH=C=CH2-COH but that isn't good either I think... (I'm a bit embarrased that I can't solve it, and it just drives me mad :o) What am I missing here?
http://img475.imageshack.us/img475/2082/nmr0025ej.jpg (http://img475.imageshack.us/img475/2082/nmr0025ej.jpg)
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C5H6O ---> 6H
CH3-CH=C=CH2-COH ---> 7H
C5H6O or C5H7O?
The ratio of the peak intensities?
C5H6O is a molecular formular or empirical formula?
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I accidently added a hydrogen too much of course. :)
I meant CH3-CH=C=CH-COH
And it's the molecular formula. No integration data of the peaks is supplied.
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Beside C5H6O and CDCl3, is there any compound?
I wonder the highest peak is come from another substance.
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in my eyes the substance is Pent-1-en-3-in-1-ol
CH3-C#C-CH=CH-OH (# is a triple bond)
you have an OH H at 6.30 which is strongly coupled with the next H and a little bit coupled with the one after
The next H is the one at 4.50 (that works for a vinyl H). It's peak is split up to a dupletic duplet because it is coupled with two other H atoms. Note: This is not a quadruplet: In a quadruplet the horizontal differences would be all the same and the two center peaks would go higher up according to 1:3:3:1. In this case, if you take a ruler, you notice the stronger coupling to OH (big gap in the middle) and the weaker coupling to the vinyl H (smaller gaps on the side).
The second vinyl H is the on at 3.05. You have weak coupling to the OH (just as it was the other way around of course) and stronger coupling to the other vinyl H. Maybe it's value of 3.05 is so low because of the M+ effect of the O (which leads to a negative charge there).
The three isolated methyl hydrogens lead to a big single peak at 3.80. It's a little bit higher than this kind of H should be. That might be for some kind of conjugation effects.
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But it is a kind of enol form which will tautomerize it self.
Something about chemical shifts:
CH3O- is 3.X (singlet)
-HC-HC=CH-O- are 3.0, 6.x and 4.x (all are doublet of doublet)
The chemical shifts are quite match to the spectrum . However, the structure cannot be solved.
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I accidently added a hydrogen too much of course. :)
I meant CH3-CH=C=CH-COH
And it's the molecular formula. No integration data of the peaks is supplied.
the ratio of peak intensities can be measured in cm (the integralsign in your spectrum)
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According to my experience, here is the answer:
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The intensity ratio should be 1:1:3:1, but your one is 1:2:2:1, and also, the multipicity of the peaks doesn't match.
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You are right!
So, what about this one (if I read between the lines, it is the one you suggested)?
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But it is a kind of enol form which will tautomerize it self.
Couldn't the enole be stable because of conjugational effects? It's just the same in Malone-acid-ester (Di-ethyl-propan-di-oat). I read that it's mostly enole and not a diketone because than you have conjugation.
Albert: The hydrogen behind the triple bond is totally isolated. It could have very little coupling with the other hydrogen two Cs away at the most. But it would not be a doublet of a doublet.
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The following spectrum was created using ACD/HNMR version 2.51.
The settings are those described by sundberg.
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ok, I guess that's as close as it gets. what are the differences, solution effects?
could you show me the spectrum of my substance too? I am just wondering. At least the coupling might be right. The chemical shifts aren't.
By the way: You probably have to buy that program. Or is there some kind of share ware version of it?
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Here is your molecule: the spectra are quite different, as you see.
For what concerns the programme I used, you have to pay for it, but there are other programmes like this one (maybe some of them might be shareware).