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This question already has an answer here:

Is it necessary that any atoms which is chiral must have 4 different atoms around the central chiral atom(maybe it can be carbon if it's organic molecule) bonded to it? More specifically: I mean can't be just three or two different groups or atoms attached to central molecule make a molecule chiral? And if no then please explain

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marked as duplicate by Mithoron, andselisk, Todd Minehardt, Jon Custer, a-cyclohexane-molecule Jan 5 '18 at 19:59

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A nitrogen atom connected to three different groups is enough to make the molecule chiral. In this case, the lone pairs "serve the role" of the fourth group needed to make it chiral. Quote from http://research.cm.utexas.edu/nbauld/CHAPTER%2021.htm:

It is interesting to note that, since the nitrogen atom of amines is tetrahedral, such a nitrogen can be a stereocenter if it has three different R groups attached. By definition, the fourth group is an electron pair, so that all four groups are different.

However, it is observed that when chiral amines are generated, they very rapidly undergo an umbrella-like inversion to generate the corresponding enantiomer, quickly racemizing the amine. Certain amines, for which this inversion is especially difficult, can be prepared and are relatively stable as a single enantiomer.

An atom connected to three different atoms with all the four atoms in the same plane is achiral because the plane of the molecule is a mirror plane hence the molecule and its mirror can overlap.

Please also note there are cases where the chirality is not caused by a simple chirality center, such as helicenes.

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    $\begingroup$ Only your last sentence actually addresses the OPs question. $\endgroup$ – Karl Jan 5 '18 at 9:31
  • $\begingroup$ The question wasn't about whether non-carbon atoms can be chiral or not it was about whether you need 4 different atoms. And amines are a bad example to choose as they are often not chiral because of fast inversion at the nitrogen. $\endgroup$ – matt_black Jan 5 '18 at 11:47
  • $\begingroup$ The question is about whether you need 4 different groups connected to the same atom, and the answer is "no, three is enough". Amine is an example. I know about the fast inversion and have quoted them. So why do you think it does not address the OP's question? @Karl $\endgroup$ – Weijun Zhou Jan 5 '18 at 11:57
  • $\begingroup$ Didn't Werner devise a cobalt complex where the cobalt was attached to only one different ligand? The way the ethylenediamine ligands were arranged made a configuration that could not be matched with its mirror image, and IIRC he even resolved the enantiomers! $\endgroup$ – Oscar Lanzi Jan 6 '18 at 2:13
  • $\begingroup$ Because at low enough temperature to avoid inversion, the free electron pair is the fourth substituent, and the geometry is very nearly a tetrahedron. $\endgroup$ – Karl Jan 6 '18 at 18:28

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