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?