# Why can a solution show optical rotation?

Why can a solution show optical rotation? A solution, as a liquid, is rotationally isotropic, right? So, even if the molecules are chiral, because of the random orientation of the molecules, shouldn't the effect cancel out?

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Your claim: So, even if the molecules are chiral, because of the random orientation of the molecules, should not the effect cancel out? is not correct.

One should think in a way that molecules have different dispersion for left and right circularly polarized light. Hence one of the circular polarization will be retarded with respect to another which will result in rotated linear polarization.

If molecules also have different absorption the exiting light will be elliptic

EDIT: To visualize the effect take a spring and rotate it - you will see - no matter from which side you look it will be still left (or right), in a sense of clock or -counterclockwise rotation of the spring. Thus if you have only left springs - you will have more retardation for the left light.

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I don't see how this addresses the question; you don't explain why the random orientation cannot cancel out the asymmetric effect of the molecules on the different polarizations of light, you are just being pedantic about exactly what that effect is. – zwol Mar 25 at 15:20
To zwol - I hope edit is now explanatory. – Madan Ivan Mar 25 at 15:40

A solution of chiral molecules is rotationally isometric, yes, but rotation in 3D space cannot cancel out 3D mirror asymmetry. As usual with these things, it may help to imagine the 2D analogue. You cannot turn p into q by rotating it in the plane, even though you can by rotating it in the third dimension.

In the three-dimensional space we live in, planar molecules are never chiral, precisely because you can rotate them into their mirror images through the third dimension. But a chiral molecule like CHFClBr is mirror-asymmetric in three dimensions; to turn (D)-CHFClBr into (L)-CHFClBr by rotation. you would have to rotate it through a fourth spatial dimension. As there is no such dimension, a solution of pure (D)- or (L)-CHFClBr is mirror-asymmetric, even though it is rotationally isometric. And therefore it can have an asymmetric effect on other things, such as light.

(Fun historical fact: this is one of the pieces of evidence that persuaded 19th century chemists that atoms and molecules were physically real, not just a mathematical model for predicting the results of chemical reactions. A crystal of (D)-tartaric acid is obviously asymmetric, and optically active. Dissolve it in water and the solution is still optically active; therefore there must be something that can be asymmetric even when dissolved.)

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In a solution of achiral molecules, the effect of each molecule on polarized light will be cancelled by a molecule having the exact opposite (mirror-like) orientation. However, for chiral molecules, this opposite orientation can only be achieved by the other enantiomer of the molecule, which is not present in the solution (unless you have a mixture of the two enantiomers, which is optically inactive when the ratio is 1:1, known as a racemic mixture).

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Light whose wavelength is much larger compared to the size of individual molecules, sees the solution as a homogeneous medium. Optical rotation is due to phase lag between ordinary and extraordinary components of incident light. Read: 'Optical Electronics' by A. Ghatak and K. Thyagarajan.

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I don't see how this addresses the question; you don't explain why the random orientation cannot cancel out the asymmetric effect of the molecules on the different polarizations of light, you are just being pedantic about exactly what that effect is. – zwol Mar 25 at 15:21

Note that the combined system of light and the optical medium cannot be isotropic because the wave vector $\vec{k}$ singles out a direction. Isotropy of the medium alone does not forbid the rotation of the polarization vector $\hat{\epsilon}$ about $\vec{k}$, i.e., optical rotation; the isotropy merely demands that the sense and magnitude of such rotation be independent of the direction of $\vec{k}$.

Optical rotation can be either right-handed or left-handed with respect to $\vec{k}$. Now, consider the reflection of the entire system about the polarization plane (i.e., the plane formed by $\hat{\epsilon}$ and $\vec{k}$), which switches the handedness of the optical rotation. Notice that this reflection leaves $\hat{\epsilon}$ and $\vec{k}$ invariant. If, in addition, the properties of the medium are invariant under the reflection, the optical rotation should also stay the same (because everything it depends on remains unchanged). Hence, in this case, the optical rotation vanishes because it must stay the same while changing its handedness.

We have argued that the optical rotation must vanish if the medium is identical to its mirror image about the polarization plane. Conversely, to have a nonzero optical rotation, the polarization plane should not coincide with a plane of mirror symmetry of the medium. In case the medium is isotropic, reflection about the polarization plane is equivalent to reflection about any plane. Therefore, for an isotropic medium, a nonzero optical rotation is possible only if the medium possesses no plane of mirror symmetry. This is exactly the case of a solution containing chiral molecules.

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## protected by Qmechanic♦Mar 25 at 11:39

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