Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: alansary on September 22, 2012, 10:45:27 PM
-
Hi chemists ,
i've got four question and i've answered them all. i just want to double check if my answer is correct or not.. and if its not please explain to me.
I've put the answers in separate pic to be clear and readable.
also requested to show the correct stereochemistry where appropriate , but don't know how to do it. need help with that.
thanks in advanced.
-
I'm sorry but I cannot make out your scribblings on these pictures.
Please re-scan them and make them more readable. Thanks.
-
I have rescan the questions , and this is my answers. what do u think ?
-
Molecule number 4 looks OK, but what is the double bond stereochemistry, cis or trans.
Molecule number 6, where is the aldehyde signal? It should be around 9.5. Also the signal at 3.9 is of a carbon directly attached to an oxygen atom, for example a methoxy. I think you will need to think again here.
Molecule 7, form the spectra it does not seem to have a carbonyl group in it. These signals are usually around 160-200 ppm in the 13C spectra.
-
ok, Ill have another look at it . but what about no.5 ?
-
It does not look correct to me, I miss the 4H singlet for the 2 CH2Cl groups.
Your suggestion shows a symmetrical molecule so for your structure you might expect only 2 signals in the 1H-NMR.
You need to re-examine this one.
-
Molecule 4 could be cis since the J value is between 6 to 9. in trance the coupling constant has value (15)
I have attached Molecule 6.
Ill attache the rest tomorrow
thanks for your help.
-
hi again,
here Molecule number 5
-
molecule 6 must be a para disubstituted aromatic ring.
molecule 5 as well must be para disubstitued.
-
I must be missing something in 4, but I seem to see a signal at 2.1 ppm that integrates to 2H. What portion of the molecule is that?
-
Solvent?
The structure looks OK
-
Solvent?
The structure looks OK
I disagree. The splitting patterns are not consistent with but-2-enal. The terminal Me group is a triplet, so it must be adjacent to a CH2 - and that CH2 is the peak at 2.1
Regarding double bond geometry - the coupling constants cannot be calculated from the ppm difference without being given the frequency of the spectrometer.
molecule 6 must be a para disubstituted aromatic ring.
molecule 5 as well must be para disubstitued.
Agreed.
For 7, this is definitely not 2-nonanone (lacking Me group triplet).
-
Then surely the molecular formula given must be wrong. It needs to be C4H8O.
-
I believe so. The spectrum given matches the spectrum for pent-2-enal on the Sigma-Aldrich website (http://www.sigmaaldrich.com/spectra/fnmr/FNMR004958.PDF) exactly.
-
There seem to be a lot of errors recently in printed material handed out students. Or is this just coincidence?
-
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??
-
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?
-
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
-
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.
-
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.
-
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.
-
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.
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.
-
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?
-
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.