I have measured fluoresence spectra of 4-fluoroindole and 5-fluoroindole in alcoholic medium at room temperature. The fluorescence intensity of 4-fluoroindole is reduced ~100 times than that of 5-fluoroindole .The spectral shape is also different which is due to La Lb factor that I know but what is the electronic factor behind this quenching?
Changes in the energetics of the system should be evident in the absorption spectra. Fluorescence efficiency is often dictated by the efficiency of the nonradiative decay processes, which can be affected by the differences of the virational structures of the two compounds.
Thanks Ivanova and Byrne. But there is no change in their absorption spectra which indicates that it is not the ground state phenomenon.So I want to know what is the electronic factor that guides here between position 4 and 5 in fluoroindoleindole (may be it is from organic chemistry point of view ).I guess there may be some electron density change by different amout in the benzene ring in position 4 and 5 that somehow affects the NH proton.
Could you show the absorption spectra? Spectral shapes in fluorescence can be influenced by so-called "self absorption", if they are not measured at low enough concentrations
Hi Byrne,
Thanks for ur comment. Actually i want to tell that both have same number of peak in their absorption spectra but the peak intensity and position are different so my words 'there is no change' is not correct,sorry for my mistake.And concentration of the solution was 100 micromolar range.
Well yes - the transition energies are very different, so you would expect the fluorescence to be very different.
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