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Topic: NMR Spectroscopy problems  (Read 10595 times)

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Offline alansary

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Re: NMR Spectroscopy problems
« Reply #15 on: September 23, 2012, 02:16:28 PM »
Hi again,

wich Molecule are you discussed about?! cant follow you

here the last one, please tell me if i answered it correctly or no??


Offline discodermolide

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Re: NMR Spectroscopy problems
« Reply #16 on: September 23, 2012, 02:20:59 PM »
We were discussing molecule number 4, which apparently has a wrong formula. One which does not fit the spectrum.

By the way your last suggestion is not correct. If it was you would expect olefinic signals to be present in the 1HNMR and they are not. Please rethink this one as well.
Also remember my suggestions for the 1,3-disubstituted benzene ring for the other two compounds.
p,s, this last one may be a ring structure?
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Offline alansary

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Re: NMR Spectroscopy problems
« Reply #17 on: September 23, 2012, 03:24:17 PM »
mmmm, is there 3 methyl groups and 1 hydroxyl group attached to the ring? because it seems to me there are 3 singlet in the 1HNMR, but the hydroxyl group confused me!!!!

so, it could be also this one ( see attached pic)

thank you

Offline Dan

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Re: NMR Spectroscopy problems
« Reply #18 on: September 23, 2012, 03:52:05 PM »
Can you explain your logic?

While I am almost certain that there is a cyclohexane ring due to the typical axial-equatorial and axial-axial couplings in the triple triplet around 4 ppm, I do not understand how you came up with cyclohexyl isopropyl ether.
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Offline alansary

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Re: NMR Spectroscopy problems
« Reply #19 on: September 23, 2012, 07:11:37 PM »
actually you right, there is no evidence to say (cyclohexyl isopropyl ether) is the answer.

what i have :
a cyclohexyl
and from 13C NMR there are 2 signals which mean 2 carbon atoms attache to O atom
-CH2-O-CH3 this will attached to the cyclohexyl ( the proton in CH2 shown as a singlet in 2.8 ppm)
the last C will also attache to the cyclohexyl in methyl group.

looking for your reply.













 

Offline orgopete

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Re: NMR Spectroscopy problems
« Reply #20 on: September 23, 2012, 09:03:36 PM »
Just my opinion, this would have been better if posted as separate problems when comments include, "which problem?".

Just as a manner of going about solving these problems. I Always prefer the Tinker Toy method. For example, in problem ??, the MF is C8H6Cl4. The integral of the NMR has 1:1:1, which I believe means the actual ratio is 2:2:2. What pieces can give this ratio? Be sure they are consistent with the chemical shifts, that is, how many aromatic and non-aromatic hydrogens are there? How many ways can they be put together? I believe it is a small number.
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Offline Dan

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Re: NMR Spectroscopy problems
« Reply #21 on: September 24, 2012, 04:49:13 AM »
actually you right, there is no evidence to say (cyclohexyl isopropyl ether) is the answer.

Then why suggest it as the answer? There is certainly evidence to suggest that cyclohexyl isopropyl ether is not the answer - the most compelling being that cyclohexyl isopropyl ether only has 6 unique C environments and is definitely not consistent with the 13C spectrum given.

Quote
what i have :
a cyclohexyl
and from 13C NMR there are 2 signals which mean 2 carbon atoms attache to O atom
-CH2-O-CH3 this will attached to the cyclohexyl ( the proton in CH2 shown as a singlet in 2.8 ppm)
the last C will also attache to the cyclohexyl in methyl group.

I also think there is a cyclohexane ring present. However I do not think it is an ether, I think it is probably an alcohol.

I am not sure exactly what the structure is.
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Offline orgopete

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trimethylcyclohexanol
« Reply #22 on: September 24, 2012, 04:29:24 PM »
Re #4

I think I know what the compound is, but I sort of cheated in figuring it out. That shouldn't be a great mystery, but let’s look at what we might know. There are nine peaks in the 13C-NMR, so there isn't any symmetry. In the 1H-NMR, there are three CH3 groups, two of which do not have an adjacent hydrogen, C(CH3)2. There is a CH3 doublet, so a CH3CH. The low field peak is a CHOH. C9 - C6 = C3, H18 - H12 = H6. The remaining six hydrogens are on three carbons, 3 x CH2. You could decide the C(CH3)2 is not adjacent to the CHOH. It seems likely two CH2 groups are adjacent to the CHOH. That leaves four compounds, can you draw the four possible structures?
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Offline Dan

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Re: NMR Spectroscopy problems
« Reply #23 on: September 24, 2012, 05:33:49 PM »
We are on the same page then, that's good. I can't see how to decide which one it is without access to the other coupling constants though.
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