Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: Sach on November 12, 2017, 05:07:39 AM
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Hi guys, I did direct arylation to get the following molecule (dimethyl 4,4'-(3-(((benzyloxy)carbonyl)amino)thiophene-2,5-diyl)dibenzoate), see attachment for the molecule. I took an H-nmr spectrum that you can find in the word document (total spectrum and zoomed in). I am relatively new to H-nmr although i did study it to be able to interpret it.
To start the interpratation, from my understanding of the H-nmr, there should be 8 types of hydrogens but on the spectrum I can only see 6 type of hydrogens. i am not sure if I missed anything so far.
But is there any chance this spectrum could belong to the molecule I posted in the attachment?
Thank you in advance
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I do not beleive its the wanted product, have you compared it with the NMR och the Cbz-amino thiophene? The aromatic area looks like this startingmaterial.
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Yes I did and it doesn't really look like the starting material. The tlc also confirmed that there was no Cbz-protected thiophene after the direct arylation.
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OK, I think the benzylic CH2-protons should be downfield compared to the methylester-protons? You have almost seven prototons at 3,9ppm and I think it would be strange if the benzylic and methylester-protons came at this same shift. The aromatic protons could be OK though. The peaks you have at 1-1,5ppm is probably impurity.
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I checked now and the methylester-protons should come lower than i expected so I think your compound is OK.
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But shouldn't I have 8 types of hydrogens on the spectrum because my molecule has 8 different types of hydrogens? Tomorrow, I have to tell my professor which peak belongs to which hydrogen but I still couldn't figure it out.
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Sorry Sach, what is missing is the benzyloxycarbonyl group, both the benzyl-CH2 and from the phenyl-ring.
The benzyl-CH2 should be att 5ppm and not at 4ppm if I am not mistaking. I have no explanation for this.
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No problem, I still can't thank you enough.
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According to the H-nmr, is there any chance that this is my molecule with protecting group removed? So thiophene with both benzene rings and NH.
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Yes, I have been thinking that too, do you have a mass-spectrometer in the lab?
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My professor was also thinking that this could be the molecule with the protecting group removed and that would be actually good news. No we don't have a mass spectrometer but I can take a C-nmr on another campus.
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Why don't you do a TLC stain or assay to see if you have a free amine? Such as this trick (https://www.thermofisher.com/order/catalog/product/28997)
Perhaps you could even see if it will go back and forth between aqueous and organic phases using dilute acid and base washes
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I will see if it's possible with a compound we have available because otherwise we probably won't buy it.
But thnx a lot.
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I got the following result from C-nmr. The number of carbon environment more or less checks out but I can't see an ester in this spectrum (chemical shift 150-200). You can find the spectrum in the attachment. This is not my unprotected molecule like I thought unless I missed something. Is there someone who can confirm this?
Thnx in advance
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Personally, I would rerun that. The signal to noise ratio isn't great, so you're quite possibly not seeing peaks because they're lost to the background.
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Pretty sure this isn't what you're looking for. The 1H spectrum should have 12 unique environments by my count (since the amine breaks the symmetry of the thiophene so the aryl groups are non-equivalent) and the 13C should have 22 peaks I think, which yours has nothing like. Whatever you've got seems to be much smaller or more symmetrical.
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I also thought that the amine would break the symmetry but I wasn't sure if it would really show on C-nmr. I have recently been studying C-nmr to be able to interpret it better (I studied it during graduation which is more than 2y ago now). Thank you all for your input, I appreciate it.
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I am actually thinking that this C-nmr indicates my molecule without the benzyl carbamate (amine group and protecting group removed).
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yeah the sample conc. seems low. I suggest using more sample and maybe more scans as well.
btw if you want to be sure that you have the amine group, the easiest way is to use ninhydrine stain. Its cheap and most likely you have some lying randomly in cabinet
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Ok thank you
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In the attachment, you can find the infrared spectrum I obtained. I want to see whether or not it is the IR spectrum of the molecule you can find in the attachment (unprotected amine because I suspect that the protecting group is removed according the C-nmr). According to me, there is a peak (2 peaks) at 3300-3400 that indicate a primary amine. There also seems to be ester (peaks around 1750). I was thinking that the peaks around 3000 should stand for aromatic hydrocarbons but they should be above 3000 while in this case, they are below 3000.
Now I have the H-nmr, C-nmr and the IR but I am still not completely able to relate it to a molecule. Can someone help me further with this?
Thnx in advance guys