2
$\begingroup$

If an object is chiral, its mirror image and itself are non-superimposable and represent two distinct versions of the same object. A hand, for instance, can be either the right hand or the left hand. Or a helix, which can be winding clockwise or counter-clockwise.

With molecules is the same. A molecule for which its mirror image is non-superimposable with itself is chiral, whether or not it contains any carbon. Whenever that is the case, the molecule and its mirror image form an enantiomeric pair and they are each other's enantiomer.

I have been playing with the principal axes of inertia as tools in attempts at superimposing two different molecules when I asked myself what would be the relationship between the axes of inertia of each molecule in an enantiomeric pair. I was thinking that the length of each vector might be the same since the relative distances of atoms and their angles are the same, but maybe the axes themselves are distinct since the spatial disposition changed, but nevertheless they have a certain relationship between one another?

So my question is: Is there a known relationship between the principal axes of inertia of molecules in an enantiomeric pair?

$\endgroup$
3
  • 3
    $\begingroup$ Chiral pairs are mirror symmetric, so are all their (vectorial) properties. Now what is your question? $\endgroup$
    – Karl
    Commented Dec 3, 2021 at 7:58
  • $\begingroup$ @Karl What do you mean they are mirror symmetric? I thought the entire point of being chiral was that the object does not have any $S_n$ axis of symmetry (which would include inversion centers as they are equivalent to $S_1$ and regular reflections, which are the same as $S_2$). $\endgroup$
    – urquiza
    Commented Dec 7, 2021 at 15:18
  • $\begingroup$ Mirror symmetric to each other, of course. $\endgroup$
    – Karl
    Commented Dec 7, 2021 at 22:15

1 Answer 1

2
$\begingroup$

Your reasoning concerning the same bond lengths and angles giving the same moments of inertia and therefore the same principal axes is correct. Chiral enantiomers have indistinguishable rotational constants and moments of inertia. However, the projection of the dipole moment on these axes is different. This feature was exploited in this paper to distinguish between the different enantiomers using three-wave mixing with microwaves.

$\endgroup$
5
  • 1
    $\begingroup$ Wow, I had heard of this paper as an incredibly complex form of microwave spectroscopy to distinguish enantiomers. An easier and recent way is chiral tagging in microwave spectroscopy. A small chiral alcohol is injected along with the analyte. It forms a diastereoisomer and diasterisomers have a different center of mass hence distinguishable spectra. $\endgroup$
    – ACR
    Commented Dec 3, 2021 at 6:40
  • 1
    $\begingroup$ It's probably one of the coolest applications of MW spectroscopy, but indeed not for the faint of heart 😉 The method you refer to is indeed more practical to distinguish enantiomers (although I believe that the author's are/were working on a commercialization of the technique), I added the link mostly as an illustration of the effect asked for by the OP. $\endgroup$
    – Paul
    Commented Dec 3, 2021 at 6:55
  • $\begingroup$ I agree this paper is not for the faint of heart, which includes myself. $\endgroup$
    – ACR
    Commented Dec 3, 2021 at 7:53
  • $\begingroup$ The paper is brilliant. However, is there a simple computation of a system of vectors (akin to the moments of inertia) that I could perhaps make that would behave like the dipole moments? $\endgroup$
    – urquiza
    Commented Dec 3, 2021 at 17:42
  • $\begingroup$ The principal axes don't provide information about the positive or negative direction, but I suppose you could define a random vector in your molecule, mirror it and look at it's projections on the principal axes in both enantiomers (similar to Fig. 1 of the paper). This quantity does not have any physical meaning, but could perhaps help identify the different enantiomers. $\endgroup$
    – Paul
    Commented Dec 3, 2021 at 20:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.