I have been told that structurally symmetric molecules are optically inactive irrespective of the atoms that are being used. However I see optical activity as the force that turns plain polarised light by some angle and I think factors like mass, electrons, etc. should influence that. Any explanation would be great

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    $\begingroup$ Possible duplicate of Molecular chirality and optical rotation $\endgroup$ – Mithoron Jun 1 '16 at 11:03
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    $\begingroup$ Mass, electrons, and whatever other factors are already considered by symmetry; if the molecule has certain symmetry, then all these factors try to turn the plane clockwise and counterclockwise with exactly the same force, and so end up not turning it at all. Also, welcome to Chem.SE. $\endgroup$ – Ivan Neretin Jun 1 '16 at 11:04
  • $\begingroup$ Thanks but that is the result and I know it, I want to know how we come to that result $\endgroup$ – Prabhdeep Singh Jun 1 '16 at 11:11
  • $\begingroup$ physics.stackexchange.com/questions/15503/… $\endgroup$ – bon Jun 1 '16 at 11:21
  • $\begingroup$ The article is about how light rotates and I understand I want to know how symmetry cancels the factors like mass of atom, charge etc. $\endgroup$ – Prabhdeep Singh Jun 1 '16 at 11:36

Symmetry is strict. We only consider a molecule symmetric if the plane of symmetry transforms identical atoms onto identical atoms. The atoms transformed onto each other must have the same bonding partners, the same geometry, the same oxidation state etc.

There are cases where it is not immediately obvious. $\ce{NO2}$, for example, has a plane of symmetry traversing the nitrogen atom which transforms the oxygen onto each other. In simple figures, one would draw the molecule as $\ce{O=N-O.}$. However realise, that this is only one possible mesomeric depiction, another one being $\ce{.O-N=O}$ with the oxygens reversed. Even though they don’t look it, they are symmetrically identical. Therefore, we need no longer be concerned with charge or electronics.

How this plane of symmetry works: Remember that we have a macroscopic solution, usually in the range of $50~\mathrm{mmol/l}$ to measure optical activity. That is some $10^{21}$ molecules, all randomly oriented. The polarised light beam will interact with every molecule it meets, and either be polarised to the right or to the left. However, since the molecules are symmetric and numerous, for almost every molecule the beam interacts with there will be another one oriented in exactly the opposite way (as in mirror image), reversing the former’s effect for an overall zero-effect.

Thereby, things such as conformational variants are cancelled out by sheer numbers being averaged.

One final aspect remaining is mass. Many elements come in different isotopes that differ by mass and of course, if one of two symmetry-equivalent atoms happens to be a different isotope than the other that will make that molecule ‘chiral’. But remember again that we have sheer numbers we can average and that the isotope distribution is generally random. Thus, even different isotopes being present in different molecules do not change the overall macroscopic picture of an achiral molecule.

If you somehow manage to stereoselectively destroy a molecule’s symmetry by, e.g. having the re-methyl group $\ce{^{12}C}$ and the si one $\ce{^{13}C}$ then that molecule should be (weakly) chiral.

  • $\begingroup$ Different atoms would have different electron clouds(with diff atoms), so a tri-molecular symmetric structure should be active and that is exactly my poin $\endgroup$ – Prabhdeep Singh Jun 1 '16 at 13:01
  • $\begingroup$ @PrabhdeepSingh You really have to clarify what you mean there. Electron clouds (i.e. orbitals) are always symmetric or antisymmetric with respect to every single $\sigma$ a molecule has. (And light cannot distinguish between symmetric and antisymmetric.) $\endgroup$ – Jan Jun 1 '16 at 16:57

Molecules interact with plane polarized light through their electron clouds, not by the mass of their molecules or their charge.

So if you have a bulk solution of symmetric molecules, when the plane polarized light passes through there is no 'bias' for the light and it continues along its way.

This is similar in the case of a racemic mixture, where we have equal amounts of enantiomers because the little nudging to the left and right of the individual molecules are cancelled out in the bulk solution.

However, when you have an optically pure compound in bulk solution, when the light strikes it although the optically pure compound in bulk solution is randomly arranged, there is a very clear bias which causes a net rotation of the light which is observable.

You can imagine this by taking the example of arranging multiple fan blades fixed in a water pipe randomly. The mass of the fan blades themselves do not cause the rotation of the water - at most it causes turbulence. However, the orientation of the fans (which have a certain asymmetry) causes the flow of water to spiral in a certain direction. Regardless of whether the fan blade is facing forward, backward, perpendicular, to the flow of the water, the blades will cause the water to rotate in a certain way.

It's a bit of a challenge to visualize, but I hope this helps.

  • $\begingroup$ Different atoms would have different electron clouds, so a tri-molecular symmetric structure should be active and that is exactly my point $\endgroup$ – Prabhdeep Singh Jun 1 '16 at 12:56
  • $\begingroup$ But that makes no sense because the (1) electron cloud of the molecule is a property of the molecule, no longer of the individual atoms, and (2) if it is symmetric, there is no way for the light to distinguish left or right-handedness so there is no 'bias' for the light to rotate towards. $\endgroup$ – IT Tsoi Jun 1 '16 at 13:01
  • $\begingroup$ @PrabhdeepSingh this is a long shot but you can imagine that if light strikes a symmetric molecule, then it doesn't matter which way it rotates (left or right) because it takes the same amount of energy either way. However, if it strikes an asymmetric molecule, it is easier for it to rotate in one direction rather than the other, so more of the light rotates left than right, and then you have observable rotation to the left. $\endgroup$ – IT Tsoi Jun 1 '16 at 13:04

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