When I asked my teacher whether a nitrogen bonded to 3 unique groups counted as a chiral center, they said that it did not because nitrogen undergoes rapid inversion at room temperature. What causes this? And what other atoms are capable of doing this?

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    $\begingroup$ There might be quantum tunnelling effects as well. It's possible that the smaller mass of N as compared to P etc. makes the tunnelling more likely $\endgroup$ Mar 3, 2016 at 10:55
  • $\begingroup$ @orthocresol Along the same lines is the higher electronegativity of N vs P, which makes the quantum tunnelling more favorable. To be honest, though, I don't even know what that means. I'm just regurgitating the Anslyn and Dougherty book's ideas. $\endgroup$ Mar 3, 2016 at 16:03
  • $\begingroup$ related chemistry.stackexchange.com/questions/464/… $\endgroup$
    – Mithoron
    Mar 3, 2016 at 18:52

1 Answer 1


In the second row of the periodic table, elements have relatively small differences between the size their $\mathrm s$- and $\mathrm p$-orbitals. Therefore, the orbitals of $\ce{NR3}$ can go from $\mathrm {sp}^3$ to $\mathrm {sp}^2$ with relatively little energy increase, so an amine can become planar and then reorient with the inverted stereochemistry. The same occurs with carbanions. I'm not certain, but I'm guessing oxonium cations do this, as well, and I would even guess that it would happen more quickly due to oxygen's smaller size.

The reason we don't see this with $\ce{NF3}$ is because orbitals with more $\mathrm p$ character are more affected by induction, so the highly electronegative fluorine atoms would lose electron density during inversion, which would be energetically unfavorable (to the point that it doesn't occur readily).

Chiral phosphines do not invert because the size difference between $\mathrm s$- and $\mathrm p$-orbitals for larger atoms is greater, so converting $\ce{PR3}$ $\mathrm{sp}^3$ orbitals to $\mathrm {sp}^2$ would require much contraction and, therefore, more energy than for the analogous amine.


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