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Natural Science Forum / Chemistry / Organic Synthesis / April 2007proton NMR
I am having trouble understanding the H NMR for 5-norbornene-2,3- dicarboxylic anhydride (to see it, please see: http://www.aist.go.jp/RIODB/SDBS/cgi-bin/direct_frame_top.cgi?lang=eng). To begin with, I don't understand why H's D and E are labelled as non- identical hydrogens, when you only see one peak for D/E on the spectrum. I am also unclear as to why you can't see the difference between B and C on the spectrum (they are merged into one peak, when it seems that B is closer to the electronegative =O than C, and thus more deshielded). Any help would be appreciated. Thanks Dnia 4/11/2007 4:58 PM, sarahsbs@gmail.com napisa?(a): > http://www.aist.go.jp/RIODB/SDBS/cgi-bin/direct_frame_top.cgi?lang=eng). > > To begin with, I don't understand why H's D and E are labelled as non- > identical hydrogens, when you only see one peak for D/E on the > spectrum. Are you sure it is one signal? Why is it quartet then? :-) Actually, these are two overlapped triplets (or rather pseudo-triplets as the theoretical pattern is more complicated; on high-resolution spectrum you should probably see more signals) > I am also unclear as to why you can't see the difference > between B and C on the spectrum (they are merged into one peak, when > it seems that B is closer to the electronegative =O than C, and thus > more deshielded). Hint: cyclopentene ring. Michal On Apr 11, 10:58 am, sarah...@gmail.com wrote: > I am having trouble understanding the H NMR for 5-norbornene-2,3- > dicarboxylic anhydride (to see it, please see:http://www.aist.go.jp/RIODB/SDBS/cgi-bin/direct_frame_top.cgi?lang=eng). > To begin with, I don't understand why H's D and E are labelled as non- > identical hydrogens, when you only see one peak for D/E on the [quoted text clipped - 5 lines] > Any help would be appreciated. > Thanks Hey. I'm having a lot of the same questions as you. Have you found the answers to them. I don't think I got the cyclopentene ring hint. Also, the overlapping triplets - since D and E are in different electronic states, does that mean they split eachother. In which case D would be split by E and C? Have you made any progress with it? On 4/23/2007 6:07 PM, jmk2111@columbia.edu wrote: >> 5-norbornene-2,3-dicarboxylic anhydride >> http://www.aist.go.jp/RIODB/SDBS/cgi-bin/direct_frame_top.cgi?lang=eng). > > Hey. I'm having a lot of the same questions as you. Have you found > the answers to them. I don't think I got the cyclopentene ring hint. It was a superficial explanation, indeed. A double bond shifts signals of beta protons downfield, roughly to the same extent, irrespective of the atom in gamma position (systems >CH-CR=X, X = C, O, N, S). While one may expect stronger shift in the systems with more electronegative atoms, this is not always true. Thus, similar shift found for our compound should not be a surprise (however, the question "why" is always very good! :-) ). I don't know if there is one and trivial explanation of this phenomenon. The other factor is geometry of the compound. The SDBS spectrum refers apparently to the endo isomer (with anhydride residue below norbornene skeleton). For exo configuration, the difference between B and C is much larger, ca. 0.5 ppm. Maybe building models of these structures could be helpful. Protons involved in 5-member ring are shifted downfield additionally (this applies to both B and C protons). > Also, the overlapping triplets - since D and E are in different > electronic states, does that mean they split eachother. In which case > D would be split by E and C? Have you made any progress with it? Sorry, no triplets but doublets. My mistake. Only ^2J coupling (D - E) is visible at this low resolution. ^3J coupling (D - C, E - C) depends strongly on conformation od H-C-C-H dihedral angle and may vary from 0.5 ppm (angle 90o) to over 10 ppm (angle 0 or 180o). In this particular case it is ca. 1.5 Hz. On SDBS's spectrum you may notice only signal broadening due to this weak coupling. Michal |
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