Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: miss_molecule on May 03, 2009, 02:51:04 AM
-
Hi,
I am doing a synthesis of 1.1-diphenylbutane-1.3-diol. I am supposed to discuss the NMR-spectra, but it seems quit complicated!
This is what I have done so far:
chemical shift - multiplicity - hydrogens - coupling constants - assignment
1.2ppm - doublet - 3 - 6.4 - CH3...
2.3ppm - Quartet - 1 - 8.4 - ...
2.5ppm - quartet (or two doublets?) - 1 - 5.3 - ...
3.1ppm - doublet - 1 - 3.2 - ...
3..8-3.9ppm - multiplet- 1 - * - ...
4.8ppm - singlet - 1 - * - ...
7.2-7.5 - multiplet - 10 - * - H-Ar
+some impurities
I feel really lost here...
Why are the two hydroxyl groups so different? One is a singlet, and the other one is about 14 signals! I am supposed to draw a coupling tree for this one... And why does the CH2 result in two different peaks? Does it have something to do with cis/trans? And what about the coupling constants?? They really seem wrong...
I don't have to discuss the aromatic region though, and I am happy for that:)
thanks a lot if you choose to help me!
-
Hmmm, just a wild guess: are you sure you are dealing with one enantiomer? Because maybe if you are dealing with a racemate, some of the signals would actually be doubled. What is more, the CH2 protons might also appear separately (like in sugars).
-
Why are the two hydroxyl groups so different? One is a singlet, and the other one is about 14 signals!
Are you sure the one with 14 signals is an OH? I think it's something else.
And why does the CH2 result in two different peaks? Does it have something to do with cis/trans?
Those two H's are diastereotopic. I believe I linked a wiki page about topicity before, but here it is again. If there's anything in it that you need cleared up, please let us know:
http://en.wikipedia.org/wiki/Homotopic_groups
And what about the coupling constants?? They really seem wrong...
It's hard to say from what you've posted, because I'm not sure how you obtained them. Could you show how you obtained them?
Any chance you could scan in a copy of the NMR?
-
Hi again..
I will try to scan my spectra tomorrow, and post my coupling constant calculations.
I see now that it can't be the OH causing the multiplet, but then i really don't know what it is...:(
So the two Hydrogen atoms on the same carbon causing two different signals, are different in R/S not cis/trans? I remember having learned about this some years ago... And I believe that this was a part that I really understood. Thanks!
Is a racemate the mix of two enantiomers?
-
racemate is a 50:50 mixture of two enantiomers. (it's easy to check really, run an optical rotation experiment - racemates are optically inactive). However, NMR can't show you the difference between enantiomers (you'd have to add chiral shift reagents to see it). Diastereomeric protons, though, can be seen in the NMR as different peaks with different chemical shifts and coupling constants.
Scan the spectrum (if you have a 13C spectrum it would also help a bit) and post it.
Perhaps COSY spectrum would tell you more about the relationships between your mysterious peaks
-
So the two Hydrogen atoms on the same carbon causing two different signals, are different in R/S not cis/trans? I remember having learned about this some years ago... And I believe that this was a part that I really understood. Thanks!
Yes, the two hydrogen atoms on the same carbon cause two different signals b/c of the chiral center. It is similar to the case with cis/trans alkenes, where two hydrogens on the same carbon give different signals. The reasons are slightly different, though.
For alkenes, the lack of rotation around the C=C double bond causes the hydrogens to be in different chemical environments all the time. With a chiral molecule, although you may have free rotation around single bonds, the two hydrogens of the CH2 still never experience the same chemical environment.
The added complexity is that diastereotopic hydrogens on the same carbon (same carbon = geminal) not only cause different signals, they also split each other. They can also still be split by hydrogens one carbon away (one carbon away = vicinal).
Geminal splitting frequently gives different J-values than vicinal, and the N+1 rule only works if all the coupling constants are the same. If the J-values are different, you start getting doublets of doublets (dd), doublets of triplets (dt), doublets of quartets (dq) or other combinations (ddd, dddd, ddq, etc.).
I'll try to post something more visual later, when I get to my laptop.
-
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi19.photobucket.com%2Falbums%2Fb185%2Fworm193%2Fsplitting.gif&hash=bb99b91dc968170e641bbefb04fa36bd771c0020)
These are very limited examples. If you want to see pretty much every possibility that's worth analyzing, Google "Hoye splitting pattern".
-
This is my spectra (sorry about the quality!):
-
So, there's a lot of splitting in the signal @ 3.85 ppm. Which hydrogen would you expect to have the most splitting? (even if you used the N+1 rule).
If you look at one of the hydrogens of the CH2, how many neighboring hydrogens does it have? How many geminal and how many vicinal?
-
I have tried to work this through now...
I would expect a "sixlet" from the hydrogen on the carbon with the hydroxyl group. This is as far I can see the hydrogen with the most splitting. From the others I would expect three doublets and two singlets (for the OH's) (don't care about the aromatic region). I have an expansion of the region with the multiplet, and it could look like a "sixlet" of doublets. Could this be because it is a racemate?
I have never heard about geminal and vicinal before, but if I understand you correctly, one hydrogen on CH2 would have one geminal and one vicinal? I don't count the hydrogens on the OH do I? But what kind of signal would this make?
If I am expecting singlets from the OH's, this would be at 4.8 and ...? There is only one singlet! And how do I know which OH that is? I would guess it is the one furthest away from the rings, maybe ?
Sorry for my stupid questions...
-
Calculation of coupling constants
-CH3=((1.19-1.18)/106 * 400*10^6Hz = 6.4 Hz
****=((2.35-2.29)/3)/106 * 400*10^6 Hz = 8.4 Hz
****=((2.51-2.47)/3)/106 * 400*10^6 Hz = 5.3 Hz
****=((3.07-3.06)/106 * 400*10^6 Hz = 3.2 Hz
Does this make any sense at all??
-
They're not stupid questions at all. It's a challenging spectrum to analyze in detail.
One thing that's unusual...it's not common to see splitting for an OH, but one of your OH's IS split.
On the flipside, if that OH is being split, it's splitting another signal in return. You should post that expanded view of the signal at 3.85 ppm, b/c it's looking like a dddq (yikes! :)).
You're right about the number of geminal and vicinal hydrogens. Since the J values are different, though, the splitting pattern becomes (Ng + 1)x(Nv+1)= (1+1)x(1+1) = 2 x 2 (a doublet of doublets)= 4 peaks
The dd gives 4 peaks, but is not a quartet. From my earlier post, can you see the key difference between them?
Keep in mind that's just one of the hydrogens of the CH2
-
Here is the expansion.
I have worked on this 8 hours with no break now:) Time to eat! I will look at your comments later. Thanks a lot!
-
Maybe this will get you started:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi19.photobucket.com%2Falbums%2Fb185%2Fworm193%2Fsplitting2.gif&hash=e56b058e3a1cd13a3c8714f8e6e17573818b78d1)
-
Hi again!
This is my thoughts:
On the carbon for the methylene, there is a chiral centre. This will make the environment of the two hydrogens different. They both have one geminal and one vicinal hydrogen. The J-values will be different, and the splitting pattern becomes (Ng + 1)x(Nv+1)= (1+1)x(1+1) = 2 x 2 (a doublet of doublets)= 4 peaks. Two doublets of doublets are observed in the spectra, at 2.32 and 2.49ppm.
Hydroxyl usually do not split on a NMR spectra, but this obviously happened here, it is split into a doublet. If the OH is being split, it must split another signal in return, this means that the hydrogen atom on the hydroxyl will split the signal for the hydrogen on the neighboring carbon.
The singel hydrogen makes the most complex signal. With 5 neighboring hydrogens, a sixlet would be expected. The observed signal is a doublet of quartets of doublets. What I am guessing is that the OH is plitting to doublets, The CH3 to a quartet, and the methylene complicates further by doubling everything. Am i right??
And what about the coupling constants? none are the same, except those for the doublet of doublets... Can that be right??
-
Oh no >:(
I just saw that the OH can't be splitting the complex signal, because the expansion is treated with D2O. Then I seriously don't understand anything about this signal... ??? and I am running out of time :'(
-
I agree it doesn't look like the "multiplet" is being split by the OH. Was the sample only treated with D2O for the expansion? If so, that might explain why you see a doublet for the OH in the normal spectrum, but no corresponding splitting in the expansion.
Maybe D2O was added to the expansion so that you didn't have the extra, even more complicating splitting. I wouldn't call it a multiplet as is, but I might with that extra level of splitting.
Could you assign the splitting pattern for the"multiplet" (like "dd")? What is it split by? Hint there are three types of Nv.
-
Only the expansion is treated with D2O.
For me, the "multiplet" looks like dqd. it its split by the three hydrogens on the methyl, and the two on the methylene. I can't see anything else able to split it, when OH is removed?? So what is the third Nv? Or does the two hydrogens on methylene count as two different Nv's? I would still expect an "sixlet" though...
-
Yes, the two hydrogens of the methylene are different Nv's. If the OH were also causing splitting it would be a ddqd. 2 doublets for the methylene, a quartet for the methyl, and another doublet for the OH, not necessarily in that order. This was what my earlier yikes was about.
With D2O for the expansion, it simplifies to a ddq, which is still complicated but J-values can be determined.
-
Thanks a lot.
I'm going home now, have to think more about this later tonight ;) It is way to complicated for me! I will try to calcualte the J-values, and see if it makes more sense.
But what about the other coupling constants? none are the same, except those for the doublet of doublets... Can that be right?? (see table)
-
a slight correction to my last post...a dqd would be more correct than ddq for the "multiplet"
As for J-values, you should have one for each type of adjacent hydrogent. E.g. for a dd, there should be 2 J-values. For a dqd, there should be 3 J-values.
You should see J-values match up, though. That is, if the CH3 is split into a doublet by CH-O and has a J-value of X Hertz, the CH-O is split into a quartet by CH3 with a J-value of X Hertz.
-
I came up with the blue/red image several posts ago, b/c I recognized it as a dqd. If you have a dqd, there are a max of 2x4x2 (16) peaks. In your case, a couple of those peaks are overlapping.
Can you see the two quartets in the picture? (I'll post it again)
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi19.photobucket.com%2Falbums%2Fb185%2Fworm193%2Fsplitting2.gif&hash=e56b058e3a1cd13a3c8714f8e6e17573818b78d1)
-
I see what you mean, and the J- values fits perfectly! ;D
So the coupling tree for the dqd would look like this? (se picture)
But what about the coupling trees for the methylene hydrogens?
I just want to say thank you sooooo much! This page is amazing! I love that you don't just post the answear, but make me think instead. In a few year I will in fact be a chemistry teacher myself 8) But not at this level, haha ;)
-
There ya go! Very nicely done!!
Have you tried the methylene splitting since I mentioned a dd should have 2 J-values? The table above shows only one J-value for each.
Good luck with the teaching! ;D That's great to hear, b/c we need more good HS chemistry teachers!!