Why does the color of iodine differ when it forms a charge transfer complex with benzene (pi donor) - pink and ethanol (electron pair donor) - brown? Does the HOMO-LUMO energy differ, if so how?
Answer
The formation of charge transfer complexes between iodine and aromatic hydrocarbons according to
\[\ce{Ar + I2 <=> Ar\cdot I2}\] is a long known phenomenon and has been described by Benesi and Hildebrand in A Spectrophotometric Investigation of the Interaction of Iodine with Aromatic Hydrocarbons. The following data for the absorption maxima of iodine in the VIS and UV range (wavelength in nm) in different solvents are taken from that article.
\begin{array}{lrrr} \mathbf{solvent} & \lambda_{max}(\textrm{VIS}) & \lambda_{max}(\textrm{UV})\\ \hline \textrm{benzene} & 500 & 297\\ \textrm{toluene} & 497 & 306\\ \textrm{p-xylene} & 495 & 315\\ \textrm{mesitylene} & 490 & 333\\ \hline \textrm{trifluormethylbenzene} & 512\\ \textrm{n-heptane} & 520 & \\ \end{array}
Note that the effect is stronger in the UV than in the VIS range and the shift correlates with the (combined) $+I$ effects of the methyl groups, while a $\ce{CF3}$ substituent disfavours then formation of a charge transfer complex.
Ethanol, on the other hand, is a protic and polar solvent, nice to stabilize charged species.
One might argue that under these conditions, other species than in the case of aromatic solvents exist - the solvation of triiodide $\ce{I3-}$ is certainly possible here.
In addition to the coordination of iodine by ethanol - we cannot exclude interaction with several molecules of ethanol:
\[\ce{n\ EtOH +I2 <=> I2\cdot(EtOH)_{n}}\]
it is conceivable that disproportionation of iodine and solvation of the resulting charged species might play a role here:
$$\ce{ I2\cdot(EtOH)_{n} + I2 <=>[\ce{EtOH}] I+\cdot(EtOH)_{n} + I3-} $$
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