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If a compound has a carbon atom with four different groups covalently bonded to it, it is called asymmetric and enantiomers of the compound can exist.

But imagine if one has a different central atom, such as a nitrogen or a sulfur where one of the four "groups" is an electron pair (see examples below).

An explosive chloramine and a sulfonium ion

Would such a thing still behave like a "normal" asymmetric carbon? If not, how else does it behave, i.e. are there enantiomers? Is my assumption that those will still have $\mathrm{sp^3}$ hybrid orbitals wrong? I would also enjoy pointers to literature, I was unable to find any.

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Generally, amine nitrogens will not behave like a normal asymmetric carbon. Simple amines are roughly $\mathrm{sp^3}$ hybridiized and the molecules you use as examples do have 4 (we include the lone pair of electrons as a substituent) different substituents around the central nitrogen atom. So in principle me might consider it asymmetric or chiral. But simple amines can undergo a process called nitrogen inversion, which essentially coverts one enantiomer into the other.

enter image description here

enter image description here

(image source)

However, if one can find a way to slow down or eliminate the nitrogen inversion process, then chiral amines can be isolated. One way to achieve this is to incorporate the amine nitrogen into a 3-membered ring (an aziridine). Now to achieve the planar state necessary for nitrogen inversion requires the bond angle in the 3-membered ring to open from 60° to 120°, an impossible task. Aziridines containing a chiral nitrogen atom have been isolated and characterized.

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  • $\begingroup$ The source for your image covers the exact system I was curious about. But it says it doesn't undergo nitrogen inversion. Is that true, or is the inversion simply of on consequence because the carbons are locked? $\endgroup$ – SendersReagent Apr 24 '16 at 18:21
  • $\begingroup$ Actually you mentioned already that the planar state is the issue, so it probably doesn't invert. Not because the final energy of the inversion is quite high, but because the transition state energy is extremely high. $\endgroup$ – SendersReagent Apr 24 '16 at 18:21
  • $\begingroup$ Thats's correct, the quinuclidine nitrogen being constrained in a rather rigid ring system also does not invert. $\endgroup$ – ron Apr 24 '16 at 18:24
  • $\begingroup$ Strictly speaking, an aziridine can invert without opening the bond angle in the ring to $120°$. The ligand attached to nitrogen could do all the moving. But then you would have to distort the bond angles to the ligand. $\endgroup$ – Oscar Lanzi Aug 16 '17 at 10:14
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Two cases are possible:

  1. The nitrogen case with inversion frequently happening. Here, the compounds inverts via a planar transition state going from $\mathrm{sp^3}$ hybridisation to $\mathrm{sp^2+p}$ and back to $\mathrm{sp^3}$. Because of this, asymmetric nitrogen atoms that are not forced into place due to steric strain (i.e. as bridgehead atoms in 1,4-diazabicyclo[2.2.2]octane) are impossible.

  2. The sulfoxide case. Sulfoxides $\ce{{R^1}-SO-R^2}$ have a free electron pair and a trigonal pyramidal configuration on the sulphur atom. The lone pair is in an orbital with a high s-contribution meaning that the nitrogen inversion highlighted above is rather complicated. Thus, sulfoxides can be assumed to be chiral on sulphur (although in practice they are often generated racemicly).

The case I called sulfoxide case (because that is generally the most well-known example) generally applies to all atoms with a single lone pair period three and up as the s-contribution to bonding orbitals decreases rapidly. Another common example of a heteroatom which can be a chiral centre is phosphorus, e.g. in phosphines $\ce{PHR^1R^2}$. Thus, the nitrogen inversion is actually more of a special case than a general rule.

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In the case of N, the compound is chiral but not isolable as it inverts. (compare quinuclidine). The diastereomors are not resolvable.

In the case of sulfoxides, these are isolable and chiral.

In the case of Carbon.Carbanion intermediates are nonresolvable but neutral ground state carbon does not invert spontaneously.

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    $\begingroup$ There are some exceptions, but simple carbanions generally do undergo rapid pyramidal inversion leading to racemisation. $\endgroup$ – orthocresol Sep 7 '16 at 15:21
  • $\begingroup$ i wrote that in haste $\endgroup$ – DrAzulene Sep 8 '16 at 20:05

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