Please help analyze the proton NMR spectra found in the link below. These are sp
ID: 494702 • Letter: P
Question
Please help analyze the proton NMR spectra found in the link below. These are spectra from a green wittig reaction involving 4-nitrobenzaldehyde. There is two NMR: the 1st picture is the crude one, the next 2 pictures are for the pure one. Please analyze and explain (multiplicities, type of H, splitting, coupling, integration, etc).
Thanks! THE LINK: https://drive.google.com/open?id=0B6xndDj0iCpISzYzLURLNDdaT0E
Can you just give me the peaks and their multiplicities and splitting? I need to fill in my data sheet. NItrobenzealdehyde was supposed to be removed cause it was limiting reagent. The other reactant was 4-(N.N-dimethylamino)benzyltriphenylphosphonium. And my prof said we will have mixture of cis and trans prod in crude
Explanation / Answer
The reaction of 4-nitrobenzaldehyde with 4-(N,N-dimethylamino)benzyltriphenylphosphonium(bromide) should result in the synthesis of 4-(N,N-dimethylamino)-4'nitrostilbene.
The NMR of the crude suggests the successful completion of reaction. There is no peak at around 10.0 ppm suggestive of complete removal of nitrobenzaldehyde which was the limiting reagent. Further, 2 peaks at 2.5 ppm show the presence of two methy groups (CH3) bonded to N. Also, there are sufficient peaks in the aromatic region from 6.2 ppm to 8.2 ppm to account for the aromatic and double alkene protons (both the isomers as your professor said). However, excess of peaks and the remaining aliphatic peaks are undesired and are due to impurities.
Unfortunately, the 2 pictures for the pure one does not contain the characteristic peaks (two CH3, singlets for 3 protons each and two alkene protons for each isomer) which suggests the absence of desired compound. Infact, the reactants too seem to be missing in these spectra. There are no peaks in the aromatic region, which reveals the absence of both 4-(N,N-dimethylamino)benzyltriphenylphosphonium and 4-nitrobenzaldehyde.