Spectrum B
Spectrometer frequency = 500 MHz
δ1 = δ4 = 0.333 ppm
δ2 = δ3 = 1.271 ppm
3J12 = 3J34 = 7.11 Hz
4J13 = 4J24 = -0.07 Hz
3J23 = 6.77 Hz
LB = 0.5 Hz
Spectrum A
Spectrometer frequency = 500 MHz
δ1 = δ4 = 0.333 ppm
δ2 = δ3 = 1.271 ppm
3J12 = 3J34 = 7.11 Hz
4J13 = 4J24 = -0.07 Hz
3J23 = 0 Hz
δ1 = δ4 = 0.333 ppm
δ2 = δ3 = 1.271 ppm
3J12 = 3J34 = 7.11 Hz
4J13 = 4J24 = -0.07 Hz
3J23 = 0 Hz
LB = 0.5 Hz
The only difference between the simulations is that in spectrum B a coupling of 6.77 Hz was assumed between the two methylene groups whereas in spectrum A the same coupling was taken to be zero. The reason spectrum spectrum B is so complicated is that despite the fact that both the methyl groups and both the methylene groups are chemically equivalent, they are not magnetically equivalent. This is true for both spectrum A and spectrum B however, in spectrum A the second order effects are small based on the parameters used in the simulation.
Thank you to Adrian Dingle for inspiring me to create this post.
The only difference between the simulations is that in spectrum B a coupling of 6.77 Hz was assumed between the two methylene groups whereas in spectrum A the same coupling was taken to be zero. The reason spectrum spectrum B is so complicated is that despite the fact that both the methyl groups and both the methylene groups are chemically equivalent, they are not magnetically equivalent. This is true for both spectrum A and spectrum B however, in spectrum A the second order effects are small based on the parameters used in the simulation.
Thank you to Adrian Dingle for inspiring me to create this post.