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Offline ostudent

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fluorescence
« on: January 14, 2019, 01:41:33 AM »
See attachment for problem.

In this question, I am confused about the effect of rigidity on effects of UV radiation.
I agree that C is the answer because it is the most highly conjugated. However, why does C being "the most rigid due to its fused aromatic ring system" make it more likely to have intense yellow fluorescence. Can this be explained?

Offline Irlanur

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Re: fluorescence
« Reply #1 on: January 14, 2019, 03:25:26 AM »
The argument is a bit hand-waving, but one can argue than in order to be fluoresecent, the light energy that was taken up also has to be released via a radiative process. If a molecule has a lot of vibrational degrees of freedom, i.e. if it can wiggle a lot, most energy will be released as heat, not as visible light.

Offline rolnor

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Re: fluorescence
« Reply #2 on: January 14, 2019, 08:40:46 AM »
I think A and B have to little conjugation to be fluorescent?

Offline zarhym

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Re: fluorescence
« Reply #3 on: January 17, 2019, 08:43:27 PM »
We can change the question into this way.
Which one has the lowest excited singlet(S1) state?

Offline Enthalpy

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Re: fluorescence
« Reply #4 on: January 18, 2019, 05:01:55 AM »
We can change the question into this way.
Which one has the lowest excited singlet(S1) state?
And how would that relate with vibrational degrees of freedom?

Offline zarhym

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Re: fluorescence
« Reply #5 on: January 18, 2019, 10:27:29 AM »
I was thinking about the homo lumo gap. Better conjugation means lower S1/T1 state, which leads to less energy between the T1 and S0 state. Thus, the fluorescence will start to shift from UV to yellow with the growth of the conjugation.  Since C has the highest number of Pi electrons stack together, I would guess it should more likely to give yellow fluorescence among the first three choice.

I may be wrong about this.
Actually, I didn't think about vibrational degrees of freedom.
My first instinct is to use the jablonski diagram to explain this question.

Offline clarkstill

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Re: fluorescence
« Reply #6 on: January 22, 2019, 03:22:04 AM »
The argument is a bit hand-waving, but one can argue than in order to be fluoresecent, the light energy that was taken up also has to be released via a radiative process. If a molecule has a lot of vibrational degrees of freedom, i.e. if it can wiggle a lot, most energy will be released as heat, not as visible light.

Its going way WAY back for me, but isn't it that the electronic transitions are formally spin-forbidden, but when coupled with a vibrational change the process becomes allowed (so the electronic state relaxes more quickly)?

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