I think it's two, but I am not able to access/find any information on the net that supports my claim or justifies why it's wrong.

One chiral center is the obvious carbon attached to $\text{-(COOH)}$ group. I think another one is the secondary Nitrogen because it has three different groups attached to it + a lone pair. Moreover, it's also a part of a cyclic ring structure so that should slow down Nitrogen Inversion.

What's the actual answer to the above mentioned question and why?

  • 4
    $\begingroup$ I think you would need stronger obstacle to affect nitrogen inversion. $\endgroup$
    – Poutnik
    Jan 2 at 8:32

1 Answer 1


From the conceptual perspective applying (CIP)-rules, you are right; beside the carbon atom both $\alpha$ to nitrogen and the carboxylate, the nitrogen may be either up or down the plane of the cycle -- and given the pattern of substitution (including the free electron pair on $\ce{N}$), there could be diastereomers of proline.

However in praxi there are not. Embedded in a conformational flexible five membered cycle (with little cyclic strain, compared to cycles of four or three atoms) with single bonds only, the two forms of the free electron pair up or down interconvert into each other so rapidly, that this averages out like there were no stereogenic centre present. (Build molecular models and try this manually.) This is reason the Wikipedia article states as general rule

«For a compound that would otherwise be chiral due to a nitrogen stereocenter, nitrogen inversion provides a low energy pathway for racemization, usually making chiral resolution impossible.»


(For free ammonia at room temperature, the same page states 30 billion flips per second.) Thus for proline, count only one stereogenic centre.

Note, rigid cycles may prevent this flips to occur (e.g., Tröger bases).


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