My question is apart from the C in the straight chain with $\ce{-OH}$ group, should $\ce{N}$ be counted as chiral center since it also has tetrahedral geometry and all different substituents?
There are more such examples.
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Also, check out these related questions: 1, 2, 3 and 4. Still leaving this open, because I’m not sure if one of these is an exact duplicate. – Jan Nov 07 '15 at 13:14
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Inversion at nitrogen chiral centres proceeds rapidly, so the answer is no. – orthocresol Nov 07 '15 at 16:08
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@orthocresol Would you explain, please? – Kartik Sharma Nov 08 '15 at 13:24
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Read those related questions. – orthocresol Nov 08 '15 at 14:15
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You can surely count that nitrogen atom as a "chiral center" because it has four different groups attached to it.
However, the net optical activity due to that chiral center is zero because the two optical isomers from that nitrogen atom are rapidly interconverting with each other ("Amine inversion"). They always exist as a racemic mixture, and hence are optically inactive.
Gaurang Tandon
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