# NMR splitting of protons in a ring

I was presented with the following compound and asked to comment on the splitting of the proton $$\ce{H_a}$$ in $$\ce{^1H}$$ NMR.

The question presented today was:

What is the chemical shift (below) and splitting pattern for $$\ce{H_a}$$ up to and including $$\ce{^4J}$$?

I relabelled the molecule as shown:

but I was left stuck here. I know that $$\ce{H_a}$$ will be split by $$\ce{H_d}$$ and $$\ce{H_e}$$ to give a doublet of doublets (dd) but I was wondering, since the molecule is asymmetric. Now

are the protons $$\ce{H_c}$$ and $$\ce{H_d}$$ chemically equivalent but magnetically inequivalent

to yield the final splitting pattern as dddd?

• And I find the wording of the question by your teacher a bit weird.The interesting point is wether a 4J coupling is large enough to become observable!
– Karl
Dec 30 '19 at 17:07
• @Karl that was how the question was worded to me! I was just wondering if Hb and Hc are magnetically inequivalent, so they split the Ha signal with two different coupling constants? Dec 30 '19 at 17:56
• @Karl done. I've shortened the question down to grasp the main point and clarified any grammatical errors. Dec 30 '19 at 18:02
• Cool. ;) Ahem, a hint: you notice the methyl group at the bottom. I contains a lot of electrons. How do Hb and Hc feel about that?
– Karl
Dec 30 '19 at 18:11
• @Karl Hb would feel a stronger magnetic field strength of the methyl then Hc as the methyl group and Hb are in the same plane so they would be at different chemical shifts as a result - That's what I would say is my answer to your question. Is that right? Dec 30 '19 at 19:40