Wavelength difference is a big deal, I know. It can solely change the whole interaction between the chiral molecule & the light. But I am not sure what's the mechanism by which light of different wavelengths produces different angles of rotation.
So, why and how wavelength matters?
Answer
The rotation of plane polarised light by a solution of, say, sucrose depend on the ability of the oscillating magnetic filed of the light to induce an electric dipole moment in the molecule and the ability of the oscillating electric field of the light to induce a magnetic dipole moment.
For these interaction to have any magnitude it is supposed that the electrons in a molecule move in a helical path, or, alternatively that there are two linear dipoles generated by electron motion which are in planes that are at some angle to one another. Although this model is clearly artificial it emphasises the fact that movement of charge must follow somewhat crooked pathways under the influence of the radiation. The theory of this is very complicated but the result from a quantum calculation is relatively straightforwards and is that the molecular rotation M at wavelength $\lambda$ is given by
$$M_\lambda = a\sum_i \frac{\lambda_{0i}^2X}{\lambda^2-\lambda_{0i}^2}$$
where i represents all the electronic states of the molecule and the wavelength $\lambda_{0i} = c/\omega_{0i}$ where $\omega_{0i} = (E_i-E_0)/\hbar$ is the frequency of the $i^{th}$ excited state at energy $E_i$. The parameter $X=\mathrm{Im}(\mu_{m_{0i}}\cdot\mu_{e_{i0}})$ is the value imaginary part of the complex dot product of the induced magnetic and electric dipole moments and a are a set of constants independent of wavelength.
From this formula it is seen that the polarisation rotation depends on the wavelength. (The dependence of rotation angle on wavelength has been called optical rotatory dispersion.) We can also conclude
that experiments should be performed where the molecule has little absorbance but that the rotation will be larger as an absorption band is approached. (in an absorption band elliptically polarised light can be formed)
The strength of any optical transition is not important as the dot product X depends on induced moments and the angle between them. This also means that weak optical transitions can be as important as strong one in rotation the polarisation. If the induced dipoles are perpendicular then the rotation vanishes as the dot product is zero.
That many excited states i can be involved and their effects can cancel to some extent so the value of the rotation is hard to predict.
The polarisation rotation of two mirror image molecules are equal and opposite in size and if a molecule is identical with its mirror image polarisation rotation must be zero and the term X is zero.
When the wavelength $\lambda < \lambda_{0i} $ i.e. the wavelength crosses a transition then the signal changes sign and this is called the Cotton effect.
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