# NMR of trigonal bipyramidal molecules

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. 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

“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!