I'm slightly confused about NMR.

I seem to remember at school that if a molecule is symmetrical in $\ce{^1H}$ NMR, that only one peak would be shown e.g. propane's terminal $\ce{-CH3}$.

Consider the two molecules shown below. enter image description here

My professor remarked that both these isomers would show a doublet and triplet in their proton-decoupled $\ce{^31P}$ NMR and I was struggling to work out why.

Consider the molecule on the left as molecule A and the right as molecule B.

My train of logic was that the axial $\ce{Ph3P}$ group on molecule A would be split by each $\ce{P}$ atom that is equatorial, so using the expression $2nI + 1$ where $I = \frac{1}{2}$ so our singlet would be split into a triplet.

Likewise, the equatorial $\ce{Ph3P}$ groups would both only see one axial $\ce{Ph3P}$ group, hence only a singlet would be observed.

I used the same logic on molecule B.

Is this the correct reasoning for the doublet and triplets observed in both molecules?


1 Answer 1


“Likewise the equatorial Ph3P groups would both only see one axial Ph3P group, hence only a singlet would be observed” - I think you’ve just made a little mistake. The equatorial groups can see one phosphorus other than themselves, so (2nI + 1) = 2, not 1!


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