Understanding Benzene: Visible Light and Excited Electrons

[gracy]

know benzene is colourless...but i can't seem to get my head around it at the moment! As benzene has delocalised electrons, it should require less energy to excite the electrons. And Visible light is a lower frequency to UV? So according to the equation: E=hv Benzene should be coloured as it absorbs visible light? I know this is wrong and it is colourless, but i can't see how it works? Thanks for any help in advance!

 

[Borek]

it should require less energy to excite the electrons

Less than what?

 

[DrDu]

There is not just one UV but UV is a very broad range covering many octaves (in contrast to visible light which covers only about one octave). So e.g. a saturated hydrocarbon has some electronic transitions in the far UV (approx. 130 nm wavelength), while an isolated CC double bond absorbs at about 165 nm and the conjugated pi bond in benzene give rise to absorption at about 245 nm. For comparison, the visible range (violet) starts at about 400 nm. See any introductory text about UV- spectroscopy.

 

[Altered State]

Less than what?

This is pretty much it.

For a molecule to absorb light of a given wavelength, there must be two molecular orbitals (one filled and one with room for an electron) between which an electronic transition is allowed. The energy difference between these two orbitals, equals the energy of the light that will be absorbed in that transition or excitation process.

Pi conjugation usually makes possible transitions between pi and n molecular orbitals closer in energy than in a situation where this conjugation is missing. Briefly speaking, if the conjugation is enough to make these orbitals close enough to absorb light from the visible spectrum, the compound will have color. Otherwise, like in the case of benzene, the conjugation is not enough to absorb visible light, but UV light will indeed be absorbed.