Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: jamesrb on May 05, 2016, 10:25:07 PM
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Sorry for the ridiculous size of the image, I wanted everything to be readable.
I have the MS, IR, 1HNMR, and 13CNMR for two compounds and I was looking for some help. For the first one I have attached the four spectrums to this post below.
I have been told there is some degree of conjugation present (UV-VIS active).
MS:
The M+ peak is at 303 thus it must have an odd number of nitrogens.
IR:
I see a really small... something... around 3500. Not sure what that is. A very weak N-H stretch maybe? Perhaps this is an amine.
Also the C-H stretching that you normally see in this region is super weak.
I don't see C=O around 1700. I see a SHARP peak at 1680. It doesn't look like usual C=O peaks I see but maybe that is what it is.
I'm thinking there is a benzene ring present.
There is so much in the fingerprint region it's overwhelming.
HNMR:
The cluster of signals around 6.5 to 7.5 I think is due to aromatic protons and perhaps amide protons.
There is a singlet at ~3.7. This could be due to an amino group.
CNMR:
There is a quartet at 55. This is ether, ester, alcohol territory.
The cluster of singlets and doublets in the 125-145 range is probably the benzene carbons.
There is a singlet at 161. This is ester, amide, carboxylic acid territory.
There is a singlet at 170. This is ester, amide, carboxylic acid territory.
This is obviously a heavily substituted benzene ring or a multiple ring system with nitrogen and maybe even oxygen containing substituents. I'm thinking the sharp peak at 1680 in the IR must be C=O because so many of these HNMR and CNMR signals point to this being an amide. It is just so spikey.
(https://i.imgsafe.org/cc51523.png)
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Here is another one I think I am close to finishing but have hit a roadblock. The M+ peak is at 109 so it has an odd number of nitrogen atoms. The M+/M+2/M+4 peaks show a 9:6:1 ratio so there are two chlorine atoms. So a possible formula could be C2Cl2NH. The DOU of that would be 2. From the HNMR we can see there is only 1 type of proton environment (obviously). From the CNMR we see there is a CH and a C. How to get all that together is tricky. Is there was only 1 degree of unsaturation I would throw a carbon carbon double bond in there and call it a day. But there are two. I don't know how to get 2 chlorine atoms, 1 nitrogen, and 1 hydrogen onto 2 carbons with 2 double bonds, 1 triple bond, or 1 ring.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi68.tinypic.com%2F161xs2h.png&hash=c3dad3232d3c82809942a2514c29f167fbf284a7)
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you need to do some more thinking about the MS spectra; you are on the right track.
MW 109, so N = odd; 9:6:1 ratio so Cl2; C2H1N1Cl2....are correct.
What are the multiplet signals that show 9:6:1 ratio ?
What are the mass losses from M+. to the Cl-containing ions ? What combination of atoms will give these mass losses ?
Generally the best way to interpret a mass spectrum is as follows, setting up a Table of m/z vs possible composition, so
m/z Loss Composition ??
109 0 C2H1N1Cl2
82 ? ?
74 ? ?
47 ? ?
Go to it !
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MW 303; N = odd...correct
What can you learn from the ratio of the all 12Cn peak intensity to the 13C112Cn-1 of any ion signal ??
You should do this with m/z 303 (less accurate, because of low intensity signal) and m/z 135 (more accurate, because of high signal intensity).
Also, construct table as indicated in my reply to the MW 109 compound.