'Chemical + Magnetic Equivalence' Problem

[Zipzap]

QUESTION:

Describe all the protons that are chemically equivalent to each other, and magnetically to each other. Then, give the Pople notation describing the proton spin system(s) in the molecule. Full marks for proper justification

http://imgur.com/VXhQl

 

ATTEMPT TO SOLVE THIS:

For the methyl that's 'dashed', I know that the 3 protons on it are chemically and magnetically equivalent. For the sake of naming it, I assigned it as A₃. However, I forgot 'why' it is that this is the case. Can somebody remind me of the reasoning behind this?

 

For the wedged proton, I figured that it just stood on its own. Since it's the only proton on that carbon, I called it B.

 

I seem to be having great trouble assigned the cis protons, though. While I know that there exists a symmetrical axis between the two, I also know that "cis"-coupling is close to ~11 Hz.

If I go by the symmetry argument, they are chemically equivalent. Following up on that, they would not be magnetically equivalent due to the different coupling paths they would take towards the wedged proton. If this was the case, I would assign them as MM'.

On the other hand, my gut (for a lack of better words) tells me that they are not chemically equivalent due to the fact that "coupling is never observed between chemical shift equivalent nuclei, be it from symmetry or by accident, not because the B_n field disappears but because spin transitions that would reveal the coupling are forbidden by symmetry" (taken from http://orgchem.color.../splitting.html).

^It's possible that my justification for either option is lacking, which is why I'm also posting this question here for feedback.

 

If described by Pople Notation, I either have two options: A₃BMM' if I use the argument by symmetry, or A₃BMX if I use the argument by coupling.

 

What are your thoughts, fellow forum members?

[hypervalent_iodine]

Sorry this took me so long to get to. I did get your PM, but I may or may not have forgotten about it later on when I had the time to reply.

 

I'm not sure that sentence you quoted really supports the argument that those two protons aren't equivalent. The compound has symmetry and since the -CH=CH- system is planar, each hydrogen attached to either side of the double bond will be in the exact same environment as the other. If you build a 3D model of the system, you should see how they are indistinguishable from one another.

 

Hope that helps.