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Why is a tertiary amine with three different substituents not chiral? Doesn't the lone pair represent a fourth, different substituent?

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    $\begingroup$ See this en.wikipedia.org/wiki/Nitrogen_inversion $\endgroup$ – Soumik Das Jun 15 '18 at 17:05
  • $\begingroup$ @SoumikDas Also addressed in this question...probably close enough to be duplicate, actually. $\endgroup$ – hBy2Py Jun 15 '18 at 17:06
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    $\begingroup$ The other question is not a fully dupe of this question... For one thing, the statement of this question is wrong. Tertiary amines can definitely be chiral. It's just that the epimerize quickly, but they are definitely chiral when there are 3 different substituents. In addition, you can easily create an tertiary amine that does not invert by tying back the substituents. $\endgroup$ – Zhe Jun 15 '18 at 17:42
  • $\begingroup$ @Zhe Sir, plz. explain a bit " you can easily create an tertiary amine that does not invert by tying back the substituents " what is"tying back the substituents" $\endgroup$ – Hercules Jun 15 '18 at 17:48
  • $\begingroup$ Constraining the substituents within a rigid structure, possibly a cyclic species $\endgroup$ – Waylander Jun 15 '18 at 18:10
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The short answer to your question

Under conditions that inhibit inversion, an amine that has three different groups attached is chiral.

The above applies not just to tertiary but also to secondary amines.


Now, why inhibit inversion? Why does it make a difference?

As mentioned by Zhe (and mentioned in Newton's answer) in the comments:

Tertiary amines can definitely be chiral. It's just that they epimerize quickly, but they are definitely chiral when there are 3 different substituents. In addition, you can easily create a tertiary amine that does not invert by tying back the substituents.

If you notice, they are chiral for very short periods of time due to the inversion that takes place in amines. In regards to the time scale – Ammonia ($\ce{NH3}$) flips $4 \times 10^{10}$ times every second. Converting that into the time for one inversion that translates into an inversion every $\pu{2.5 \times10^{-11} s}$ which is $\pu{25 picoseconds}$. Such a scale of time cannot be discerned by any of our current methods to identify stereochemistry.

From http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch07/ch7-3.html

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In this case, even though a chirality center is present in each molecule, the sample is optically inactive since the optical activity of the two extremes of inversion averages out because they are enantiomeric.

For why umbrella inversion occurs, see this answer by SendersReagent


Now, what if we restrict this inversion by attaching free-rotation restricting groups? What happens then?

This restriction brings (or rather removes) a fact into the mix. There is no more inversion possible. This is because rearrangement into $\mathrm {sp}^2$ cannot take place because of the groups added since it becomes sterically inhibited in becoming planar. An example of this would be quinuclidine which has no way of forming a planar structure.

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    $\begingroup$ I'm not sure that one can go so far as to say what the "net observed" molecule is... $\endgroup$ – Zhe Sep 11 at 16:48
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    $\begingroup$ There is actually another way to get a tertiary or secondary amine to be chiral. Put it into a chiral environment, such as a chiral solvent. You will get nonequivalent interactions with the two enantiomers of the amines, so the equilibrium between the two enantiomers is nudged off symmetry. We see this experimentally as a change in optical activity with addition of the amine to the system. $\endgroup$ – Oscar Lanzi Sep 11 at 19:53
  • $\begingroup$ @OscarLanzi Is this a method to find chirality in amines? Also, why is there a difference in interaction? $\endgroup$ – Safdar Sep 11 at 19:54
  • $\begingroup$ To follow-up on @OscarLanzi's comment, there are Europium based chiral shift reagents frequently used to cause enatiomeric peaks to shift (by creating disastereomeric complexes). $\endgroup$ – Zhe Sep 16 at 19:26
  • $\begingroup$ @Safdar Is the planar structure a transition state? Also why in the electrostatic maps there is electron density above the nitrogen and not below it if the two structures transform into each other? $\endgroup$ – ado sar Oct 13 at 8:24

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