I was presented today with the following molecule:
and asked the splitting pattern of $\ce{H_a}$ given that up to $\mathrm{^3J}$ and $\mathrm{^4J}$ coupling is allowed.
I thought that $\ce{H_a}$ couples to $\ce{H_b}$, $\ce{H_c}$ and $\ce{H_d}$ given that
- there is a $\mathrm{^3J}$ coupling to $\ce{H_b}$ (Clockwise)
- there is a $\mathrm{^4J}$ coupling to $\ce{H_c}$ (Clockwise)
- there is a $\mathrm{^4J}$ coupling to $\ce{H_d}$ (Anticlockwise)
Hence the splitting would be ddd.
However, the answer is apparently dd for the splitting of the $\ce{H_a}$ signal.
My thought was that anticlockwise, the $\mathrm{^3J}$ coupling is not possible due to the absence of a π bond in that direction down to $\ce{H_d}$ hence the coupling through the π system would not be possible.
Am I correct or is there another reason for no $\mathrm{^3J}$ coupling to $\ce{H_d}$?
\ce{…}should only be used for chemical compounds and reactions. Also, I added the compound name as image metadata (it's better to fill in something meaningful in place of "Enter image description here"), and I suggest you do the same in your future post as it makes finding your post much easier. $\endgroup$