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?
The two protons Hb and Hc are also chemically inequivalent, because they have a different chemical environment with the methyl group, i.e. they don´t have the same chemical shift.

In a chiral molecule, the protons in a methylene group are rarely really chemically equivalent. There is usually a preferred orientation.