Question

Summarize the relationship between the positive/negative charges in resonance structures and proton chemical shift. Solved Problem Analysis: Below is a 1H-NMR analysis for 4-nitro-N-vinylaniline. Read the analysis then critique the presented solution. Determine if the solution is correct or not. Write a paragraph on how you would convey the same information (include corrections or suggestions to improve the analysis presented). Tip: you may want to build the compound in ChemDraw and use the 1H-NMR predictor to help with your critique. Analysis: The vinyl group (C=C double bond group) has no resonance structures possible with the surrounding groups so all of the protons on the alkene will all be in the 6.5 ppm region. The proton on the nitrogen of the amine falls on and off easily so it will not couple to any other protons (resulting in a singlet peak) and will have a fairly broad chemical shift region where it can appear. The two protons ortho to the nitroso groups will appear as two doublets, more upfield than what you would typically see for an aromatic proton (this is because they are more shielded). The two protons ortho to the amine group will appear as two doublets, more downfield than what you would typically see for an aromatic proton (this is because they are more deshielded).

Answer 1

For the 1H NMR spectrum of 4-nitro-N-vinylaniline. The ortho protons to the -NO2 group are more shielded (lower ppm shift value) due to resonance with the -NO2 group at the 4- position of the ring. -NO2 is an electon withdrawing group. This gives a +ve charge on carbon and a negative charge on oxygen of the -NO2 group. the lone pair of nitrogen also plays a major role in shieding these protons. On the other side, the ortho protons to aniline nitrogen are mored deshielded and thus appears much mor downfield (higher ppm shift value). This is attributed to the resonance effect with the -nitrogen of the -NO2 group.