The answer is the very complicated spectrum B. The spectra were calculated with the following parameters: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.
3 comments:
Thanks for peaking my interest! Your post came up in a tweet on my twitter account feed, and I followed the link to your page. Having done NMR spectroscopy, both solution and solid state, in grad school and as a post-doc, it's been about three years since I have used my training on a concept experiment. THanks for reviving some of my old learnings :) I appreciate it!
Gregory Darnell, M.S./Ph.D
Chicago, IL
dear Glenn,
Thank you for your blog.
In butane, in each methylene groups, both protons are chemically but not magnetically equivalent, as their coupling constant with each proton of the other methylene has no reason to be the same, something easy to see in a Newman projection of the most stable conformation of butane.
Does this point can modify the spectrum even more ?
Gaspard Huber
Saclay, France
Gaspard,
I think thet spectrum would be more complicated as you suggest due to the methylene protons on the same carbon being magnetically inequivalent only if the molecule was completely rigid. The free bond rotations make the methyl protons on a single carbon and the methylene protons on a single carbon magnetically equivalent.
Glenn
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