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So consider this molecule. We have to determine whether nitrogen atom is chiral or not.
enter image description here

MY IDEAS: Now going by all I have read, for chiral atoms only criteria is substituents to be different. Here two substituents are same but as we can see orientations of the substituents is different. So aren't the two substituents technically different because after all they aren't going to react the same way to plane polarized light?

I tried to break it into possibilities: Now in first case I think we could rotate around one of the sigma bonds attached to nitrogen and form a conformation with a plane of symmetry. So is in this case nitrogen not a chiral atom? enter image description here Now in the second case I do think we can't prove the two substituents to be same, no matter how much we rotate. So chiral?

CONCLUSION: I am a beginner in this topic and not really confident about my thought process

Can we really use conformers of molecules to support their symmetry because after all conformers are ever changing, not stuck in one orientation?

Also what's the logic behind checking the plane and centre of symmetry of molecules? How does that relate to rotation of PPL light?

->If two substituents of a sp3 hybridized atom are identical, we say not optically active. But both being identical will rotate PPL light in same direction, so a net rotation will be produced. ->So isn't it optically active?

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    $\begingroup$ So, where did you find this tetravalent nitrogen atom? ;-) $\endgroup$ – paracetamol Aug 14 '18 at 5:08
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No, it is not an asymmetric centre. To test this, you can draw the two potential stereoisomers:

isomers

Note that isomers A and B are identical (by C2 rotation), so the nitrogen cannot be an asymmetric centre. If it were one, then these isomers would by definition be distinct.

The attached groups are identical, from the point of view of the nitrogen - both have (R) stereochemistry. If one were (R) and the other (S), the nitrogen would constitute a pseudoasymmetric centre - https://goldbook.iupac.org/html/P/P04921.html

Hope this helps.

Also, don't get confused about the difference between the nitrogen being a sterogenic centre, and the molecule as a whole being chiral. The molecule is chiral, since it is non-superimposable on its mirror image. Chirality is a molecular property, and although frequently people will refer to a particular atom in a molecule as "chiral", this is a misnomer. A better description of an atom with 4 different groups attached is "sterogenic". It is entirely possible to have a molecule containing stereogenic centres but the molecule is achiral (e.g. meso compounds).

The logic behind checking the plane and centre of symmetry is that planes and centres of symmetry are conserved on reflection in a mirror, so checking for these symmetry elements is a shortcut to checking whether a molecule satisfies the definition of chirality.

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  • $\begingroup$ Please could you help with what's wrong with my ideas $\endgroup$ – tiffy Aug 14 '18 at 8:54
  • $\begingroup$ Part of your issue is in distinguishing the difference between configuration and conformation. The configuration of a molecule is the description of which atoms are bonded to which, and in what 3D arrangement around each atom. You can only change configuration by breaking or making bonds. Conformations are just different arrangements of atoms achieved by rotating around the single bonds in the molecule. Stereochemistry concerns configuration only, not conformation. $\endgroup$ – PCK Aug 14 '18 at 12:20
  • $\begingroup$ In the current case, look at the two alcohol stereocentres and define them as (R) or (S) - you should find that both are (R). So, from the point of view of the nitrogen atom, it is attached to X, Y and two identical, (R)-configured groups. Since two of the groups attached to the tetrahedral N are identical, it is not a stereogenic centre. $\endgroup$ – PCK Aug 14 '18 at 12:25
  • $\begingroup$ But remember this is only to tell whether the nitrogen atom is stereogenic, it says nothing about whether the molecule as a whole is chiral. The gold standard test for that is always to draw its mirror image and check whether it is possible to superimpose the two. $\endgroup$ – PCK Aug 14 '18 at 12:28
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Yes two substituents with different orientations are considered as two different groups to determine chiral centre. So here nitrogen is a chiral centre. If a compound posses plane or centre of symmetry it is said to be optically inactive. Optically inactive compounds don't rotate plane of polarized light. Though conformers rotate, but the relative position of the groups bonded to central atom remains same. So even if you draw conformers, you will not find the two substituents to be identical.

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    $\begingroup$ Are you sure the two substituents as pictured are not the same...? $\endgroup$ – orthocresol Aug 14 '18 at 7:11
  • $\begingroup$ Yes they are different $\endgroup$ – Lakshay Aug 14 '18 at 7:12
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    $\begingroup$ (-1) I suggest checking the CIP label, you will find they are the same, both (R). That atom in the middle is not a chiral centre (let's put aside the fact that it's a tetravalent nitrogen). Furthermore, molecules can possess symmetry elements but still be optically active. For example $\ce{[Co(en)3]^3+}$ (point group $D_3$) has threefold and twofold rotation axes, but is chiral. $\endgroup$ – orthocresol Aug 14 '18 at 8:18

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