April 16, 2024, 02:19:39 PM
Forum Rules: Read This Before Posting


Topic: excited state questions  (Read 8040 times)

0 Members and 1 Guest are viewing this topic.

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
excited state questions
« on: July 12, 2015, 04:59:16 AM »
i have some fundamental excited state organic chemistry questions would like to seek for some comments and explanations.

1) consider uv-vis absorbing aromatic compounds, with spin allowed pi to pi* transition, different compounds have different molar absorptivity. So i am just curious what properties of the aromatic compounds determine the molar absorptivity(or extinction coefficient)

2) quite few class of compounds like rhodamine and coumarin are know to give intense fluorescence while a large number of the rest are nearly non fluorescent. Is there any general rules for a compound to have strong fluorescent? (I know that there should be a good electron acceptor so it can stabilize the excited state and prevent form undergo TICT; and also have a rigid structure for non radioactive decay. But is there any way when i look at a molecule i know that whether it is likely to have strong fluorescence?)

Here are some questions on fluorescence quenching:
3) I do some search on internet and it said that simple heterocyclics like pyridine and furan dont exhibit fluorescence since the excited singlet state will quickly converted to triplet state to quench the fluorescence. So my questions is why it only occur on simple heterocyclics while those fused ring one can prevent intersystem crossing? Is there any way or rules that may easily distinguish whether ISC occur?

4) It is known that when say 7-Hydroxycoumarin is alkylated on the hydroxy position, the fluorescence will quenched(and the quenching is enhanced by electron withdrawing groups but why?) and quite a number of google fig explain the quenching is done by PET. They describe the electron is transfer from the alkyl group to the coumarin. So i wonder how this is possible as i think the HOMO of the alkyl group should be lower in energy than the "hole" for the electron transfer to occur. (So my question is: why the methylated coumarin is quenched while the free alkoxy coumarin is strongly fluorescence; is the quenching done by PET? and how is the PET occur?)

5) I previously read a post saying that the porphyrin is fluorescence and when one of the hydrogen substituted by bromine, the fluorescence is quenched due to rapid ISC and this is known as heavy atom effects. I am just curious why heavy atom like bromine facilitate ISC?
and i also saw that there is one class of fluorescence compounds called eosin which contain bromide substituents but it is still exhibit fluorescence. Why ISC is not occuring in this case?

Final question, not too related to fluorescence...
6) Diazirine compounds are know to undergo photolysis by UV to give nitrogen and carbene, and as i know, this is a concerted reaction. How can this photolysis explained by MO theory? (ie: UV will excite electron to which orbitals to cause the fragmentation?)

Sorry for bring so many question

Thanks :D
« Last Edit: July 12, 2015, 05:15:53 AM by khwcm »

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4041
  • Mole Snacks: +304/-59
Re: excited state questions
« Reply #1 on: July 12, 2015, 07:06:36 AM »
1) Spin? I understand one electron jumps from one level to an other.

π to π* is for one isolated double bond. Aromatics have more electron pairs spread around more nuclei, so that more electronic levels are available and occupied, needed more names than π and π*. It's already the case for conjugated dienes.

I'd say: the volume available (polycyclic, conjugated enes) for π electrons has a big effect, then the affixed atoms and groups. Having several independent aromatic rings per molecule must also increase the absorptivity per mole. Data there
http://satellite.mpic.de/spectral_atlas
Not only the wavelength of the absorption peaks change, their amplitude too.

2) Radioactive? Radiative?
« Last Edit: July 12, 2015, 09:23:44 AM by Enthalpy »

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3480
  • Mole Snacks: +528/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: excited state questions
« Reply #2 on: July 12, 2015, 11:37:09 AM »
All questions with no simple answers.
1) consider uv-vis absorbing aromatic compounds, with spin allowed pi to pi* transition, different compounds have different molar absorptivity. So i am just curious what properties of the aromatic compounds determine the molar absorptivity(or extinction coefficient)
For pi-pi* transitions only, the biggest factor is probably molecular symmetry. Configuration interaction causes additive and subtractive mixing of transition probabilities. E.g., consider the weak and strong pi-pi* transitions of porphine. 

Quote
2) quite few class of compounds like rhodamine and coumarin are know to give intense fluorescence while a large number of the rest are nearly non fluorescent. Is there any general rules for a compound to have strong fluorescent? (I know that there should be a good electron acceptor so it can stabilize the excited state and prevent form undergo TICT; and also have a rigid structure for non radioactive decay. But is there any way when i look at a molecule i know that whether it is likely to have strong fluorescence?)
From a starting point of "all molecules fluoresce", the reason some (indeed, most) fluoresce less is because the rate constant for non-radiative decay of the excited state is faster than the rate constant for fluorescence. So, the degree of fluorescence requires understanding what impacts the rate constant for fluorescence and the rate constant for non-radiative deactivation. Actually, the latter tends to be generally more important - or at least easier to qualitatively predict. For most molecules, the rate of non-raditative decay is dominated by vibronic coupling between the ground electronic state and the emissive excited electronic state. We can roughly say this coupling is related to a Franck-Condon type interaction, and therefore summarize as follows: molecules with larger energy gaps have slower non-radiative rate constants, and molecules that are more rigid have slower non-radiative rate constants. This is why there are more blue-emitting organic molecules than far-red emitting organic molecules, and it is also why most fluorophores tend to have rigid, highly conjugated organic frameworks.

This is a fairly simplistic treatise of the subject, as there are many other factors at play - solvent polarity, the presence of heavy atoms, and so forth - but ultimately all of them do come down to differences in non-radiative or ratiative rate constants.

Quote
3) I do some search on internet and it said that simple heterocyclics like pyridine and furan dont exhibit fluorescence since the excited singlet state will quickly converted to triplet state to quench the fluorescence. So my questions is why it only occur on simple heterocyclics while those fused ring one can prevent intersystem crossing? Is there any way or rules that may easily distinguish whether ISC occur?
Yes. The most obvious is the presence of a heavy atom, which will almost always ensure a fast rate of intersystem crossing, and hence low fluorescence yield.  The influence of heteroatoms is less well understood, but the predominant theory is that heteroatoms have non-bonding electrons, and as such have low-lying n :rarrow:π* transitions that effectively short-circuit the excite state before fluorescence can occur (again, high rate of non-radiative decay). This low-lying state is formally a triplet state, but it is not a pure pi-pi* triplet state. (For the same reason, the presence of open-shell metal ions can quench fluorescence of, say, porphyrins).

Quote
4) It is known that when say 7-Hydroxycoumarin is alkylated on the hydroxy position, the fluorescence will quenched(and the quenching is enhanced by electron withdrawing groups but why?) and quite a number of google fig explain the quenching is done by PET. They describe the electron is transfer from the alkyl group to the coumarin. So i wonder how this is possible as i think the HOMO of the alkyl group should be lower in energy than the "hole" for the electron transfer to occur. (So my question is: why the methylated coumarin is quenched while the free alkoxy coumarin is strongly fluorescence; is the quenching done by PET? and how is the PET occur?)
Generally, relative polar fluorophores like the coumarins and xanthines are highly sensitive to substitutions because of the potential for vibronic deactivation (as described above), the formation of intermolecular charge transfer states, and deactivation due to solvent polarity effects. But I'll have to give this specific question some thought before I answer.

Quote
5) I previously read a post saying that the porphyrin is fluorescence and when one of the hydrogen substituted by bromine, the fluorescence is quenched due to rapid ISC and this is known as heavy atom effects. I am just curious why heavy atom like bromine facilitate ISC?
It is understood that "heavy atoms" (although note, there's no definitive point at which an atom becomes "heavy") have increased spin-orbit coupling constants, which mediate the transition between singlet and triplet excited states. The effect is not subtle, but does depend on the "relative heaviness" of the atoms involved. I've seen a lot of attempts to explain why heavy atoms have larger spin orbit coupling constants.
Turro offers a classical physics explanation, and I'm paraphrasing here since it's been a few years since I read it and I don't have a copy of my book handy: heavy atoms have a larger core charge, therefore electrons accelerate quite a bit more in the vicinity of heavy nuclei than lighter nuclei, and thus have a larger probability of flipping their spins. I'm not sure how compelling this argument is - there's a larger degree of "quantum funniness" going on, and as you know explanations using classical physics are typically insufficient when it comes to describing quantum phenomena. Unless you feel compelled to dig further, I'd leave it at "the spin-orbit coupling coefficient of an atom depends is related to its nuclear mass, and larger spin-orbit coupling coefficients increase the rate constant for intersystem crossing - i.e., effectively increase the probability of an electron crossing from a singlet to triplet energy surface, a process that is nominally forbidden. 

Quote
and i also saw that there is one class of fluorescence compounds called eosin which contain bromide substituents but it is still exhibit fluorescence. Why ISC is not occurring in this case?
I don't have an exact answer to this question, but I can speculate: As above, the fluorescence yield is related to a delicate balance between all rate constants for radiative and non-radiative deactivation of the excited state. A heavy atom increases the rate constant for non-radiative decay (specifically, the rate constant for intersystem crossing). However the magnitude of this effect is going to be sensitive to a lot of things - particularly, the geometrical nature of the excited state and its proximity to the bromines. Which is to say, the bromines have to be located near where electrons are in the fluorescent state in order to interact with them and bring about a singlet-to-triplet conversion. In the limit that there is no electron density on the carbons to which the bromines are attached, there will be in principle no effect of having a bromine attached to the molecule. (Although, keep in mind there are always through-space interactions; the heavy atoms do not even have to be attached to the molecule to exert an effect - iodinated solvents can reduce the luminescence yield through aheavy atom effect, although the magnitude of the effect is less than the presence of a heavy atom directly on the fluorophore). I don't know off-hand what carbon positions in xanthine dyes have the most electron density, but this does have a great effect on how effective attached heavy atoms are at quenching fluorescence. Also do note that xanthine dyes have very high quantum yields to start with, and thus even with moderate degrees of quenching, observing fluorescence is still possible (this isn't an all-or-none proposition). Also note that erythrosyn B, the iodinated analogue of eosin, exhibits a fluorescence yield of only 12%, compared to 65% for eosin (compared to ~100% for fluorescein, the non-halogenated parent molecule). The excited state lifetime for erythrosyn B is about 600 ps, compared to 3 ns for eosin. So you see, there is nothing fundamentally different going on in the eosin family of molecules - the halogens are still quenching the fluorescence. It's mostly a difference in scale, and the fact that the parent fluorescein is such a fluorescent molecule to start with.

I'd have to do a little research to answer your last question, sorry.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
Re: excited state questions
« Reply #3 on: July 12, 2015, 01:42:11 PM »
Thanks for Enthalpy and Corribus explanations.

I would like to have some follow up questions if you dont mind

1) I dont have very strong physical chemistry backgroud(though i really love physical chemistry LOL) so i dont really understand what do you mean configuration interaction:
"Configuration interaction causes additive and subtractive mixing of transition probabilities"
The weak and strong pi-pi* transitions of porphine i think should be govern by the change in orbital angular momentum?

I was having this question cuz i was told that partial hydrogenation of porphyrin to give chlorin may have the molar absorptivity increase by 8 times, which suprize me alot. Do you have any explanation for this observation?

2) I understand that the pi to pi* excitation are governed by frank condon principle but i dont understand why "the rate of non-raditative decay is dominated by vibronic coupling between the ground electronic state and the emissive excited electronic state". Would you mind elaborate abit more?

3) For simple heterocyclics like pyridine and furan dont exhibit fluorescence, you comment that "the predominant theory is that heteroatoms have non-bonding electrons, and as such have low-lying n :rarrow: π* transitions that effectively short-circuit the excite state before fluorescence can occur molecules with larger energy gaps have slower non-radiative rate constants"
Do you mean that the quenching is done by something similar to FRET? (ie: the pi to pi* electron in excited state singlet state relax to ground state and excite the electron from n to pi* triplet?)
So do you mean that open-shell metal ions can quench fluorescence by d-d transition instead of n to pi* in porphyrins?

btw, why is that fused ring fluorescence heterocycles wont have this quenching problem?

4) Really looking forward to your answer ;D 
I am also really suprized that like the fluorescein diacetate are fluorescent quenched aside from the Hydroxycoumarin and naphthalimide one.
So there is nothing to do with photon induce electron transfer?

I just came up with 2 more questions, i hope it is not too much >.<

7) For organic molecule pi to pi* transition (in uv-vis spectrum), the major absorption peak is quite broad usually (across 100 nm). so my question is why the absorption peak isnt a sharp peak instead?

8 ) A lot of organic compounds may undergo rapid ISC to triplet state and have the fluorescence quenched. So i am just curious to know why it is not common for them to undergo phosphorescence after ISC?
Meanwhile, Ru(bpy)32+ make use of spin orbit coupling to give triplet excited state and eventually have phosphorescence. so why Ru-bpy type molecule could achieve such luminescence while lots of molecule couldnt?

really thanks for your kind explanation. I really love your illustration and comments on eosin  :)
« Last Edit: July 12, 2015, 02:29:45 PM by khwcm »

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3480
  • Mole Snacks: +528/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: excited state questions
« Reply #4 on: July 12, 2015, 08:40:30 PM »
Quote
1) I dont have very strong physical chemistry backgroud(though i really love physical chemistry LOL) so i dont really understand what do you mean configuration interaction:
"Configuration interaction causes additive and subtractive mixing of transition probabilities"
The weak and strong pi-pi* transitions of porphine i think should be govern by the change in orbital angular momentum?
Bear in mind, configuration interaction is a theoretical construct. But, more or less it refers to the fact that electronic wavefunctions can mix if they have the appropriate symmetry, and when they do so, you get linear combinations (i.e., additive or subtractive). In the case of porphyrin, you have two sets of two nearly degenerate transitions between the two homo and two lumo orbitals. One of these pairs is polarized in the x-direction and the other in the y-direction. So what you get is basically two x-polarized transitions of practically the same energy that are fully allowed. However if we allow for transitions/states of appropriate symmetry to mix, you get basically additive and subtractive combinations of these states. The additive combination becomes even more allowed and is blue-shifted, and the subtractive combination becomes partially forbidden and is red-shifted. You'd have to play around with the transition moment integral to see why that's the case, but effectively this is why porphyrin optical spectra exhibit a low energy, partially forbidden set of pi-pi* bands (often called the Q-bands) and a higher energy, fully allowed transition (called the B- or Soret-band). Any substitution to the porphyrin macrocycle breaks the degeneracy of the HOMO and LUMO orbitals, resulting in a poorer degree of mixing, and therefore an increase in the intensity of the Q-band. This mixing is what we can casually refer to as "configuration interaction" - the interaction between two potential electron configurations. It is certainly a peculiarity of quantum mechanics. In a classical world, the electron configuration would have to be one or the other. In the quantum world, it's both at the same time. In fact, it is not so dissimilar from the quantum double slit experiment, where an electron basically interferes with itself.

Anyway, this type of CI manifests in most very symmetric pi-conjugated molecules. Which is why for the most symmetric molecules - e.g., benzene - the lowest energy optical transition actually has very low extinction coefficient, despite being what would appear to be a fully allowed pi-pi* transition. 

A lot of theoretical research in the 1960s and beyond was on ways to calculate the effects of configuration interaction to better predict energy levels of organic molecules.

Quote
I was having this question cuz i was told that partial hydrogenation of porphyrin to give chlorin may have the molar absorptivity increase by 8 times, which suprize me alot. Do you have any explanation for this observation?
I am not familiar with the electronic absorption spectra of chlorin, but given that it has less symmetry than porphine, I could see have it may have stronger molar absorptivity.

Quote
2) I understand that the pi to pi* excitation are governed by frank condon principle but i dont understand why "the rate of non-raditative decay is dominated by vibronic coupling between the ground electronic state and the emissive excited electronic state". Would you mind elaborate abit more?
If you don't mind, I will postpone answering this until tomorrow. I have a nice figure that illustrates it better than I can describe with words alone.

Quote
3) For simple heterocyclics like pyridine and furan dont exhibit fluorescence, you comment that "the predominant theory is that heteroatoms have non-bonding electrons, and as such have low-lying n :rarrow: π* transitions that effectively short-circuit the excite state before fluorescence can occur molecules with larger energy gaps have slower non-radiative rate constants"
Do you mean that the quenching is done by something similar to FRET? (ie: the pi to pi* electron in excited state singlet state relax to ground state and excite the electron from n to pi* triplet?)
So do you mean that open-shell metal ions can quench fluorescence by d-d transition instead of n to pi* in porphyrins?
Again, I think the figure I will show you tomorrow will help. But basically any time you introduce lower energy molecular states that provide an avenue for excited-state relaxation, this will tend to reduce the excited state lifetime, and hence the fluorescence yield. It's kind of like this: imagine you are a part of a hiking group that has walked to a cliff, and the tour guide tells you all that you've got to get to the bottom to finish the tour. But: there's only one way down: jumping. You stand up there and look over the edge, there's a pond at the bottom that your guide promises you is deep enough to make jumping safe, but even so maybe it takes the average person quite a long time to work up the nerve to just throw themselves out into free fall. The total amount of time it takes for the group to get to the bottom is long. Now imagine the same scenario but there's a couple of smaller ledges between you and the ground below, such that you can hop down gradually. Don't you imagine that the total time to get to the bottom now will be a much less? Sure, some thrill seekers may still jump, but most people will opt for the safer route.

I hope this analogy is clear. d-d states, n-pi states, and so forth, are all little ledges that allow electrons to quickly get back down to the ground state, but they don't emit photons when they relax this way. So, they take away from the fluorescence yield of the molecule.
 
Quote
btw, why is that fused ring fluorescence heterocycles wont have this quenching problem?
My guess would be that whereas pi-pi* transitions are highly delocalized, n-pi* transitions are less so. The coupling between states is to some degree dependent on geometry. So the coupling between pi-pi* transition and n-pi* transition in pyridine is good, it is less good as the pi system becomes larger. Porphine is also a heteroatom system, but it has a reasonable fluorescence yield. There is also the matter of whether there is electron density on the heteroatom. In simple symmetric systems the electron density on each position is identical, but this is not the case in more complicated ring systems.

(This is also, by the way, one of the reasons why the triplet yield goes down in conjugated polymers, because singlet states tend to be highly delocalized, whereas triplet states are not. So, you get poor spatial coupling between the triplet and singlet wavefunctions as the number of conjugated repeat units increases.)

Quote
7) For organic molecule pi to pi* transition (in uv-vis spectrum), the major absorption peak is quite broad usually (across 100 nm). so my question is why the absorption peak isnt a sharp peak instead?
Lifetime broadening is one part of it. But, organic molecules also have a lot of conformational freedom - even fairly rigid ones, which broadens out transitions. Finally, most of this work is done in solution phase. The possible solvent interactions with the fluorophore also contribute to broadening mechanisms. If you do low-temperature (77K) experiments, you will see these transitions narrow up quite a bit, because less favorable conformations are removed from the equation.

Quote
8 ) A lot of organic compounds may undergo rapid ISC to triplet state and have the fluorescence quenched. So i am just curious to know why it is not common for them to undergo phosphorescence after ISC?
Meanwhile, Ru(bpy)32+ make use of spin orbit coupling to give triplet excited state and eventually have phosphorescence. so why Ru-bpy type molecule could achieve such luminescence while lots of molecule couldnt?
Metal polypyridyl complexes are sort of hybrid species. The low-lying electron states are not pi-pi* transitions. They are metal-to-ligand charge transfer transitions. Formally we call them singlets and triplets, but this terminology is really most meaningful when the two electrons have the same spatial delocalization. In the case of an MLCT transition, one of the electrons has been transferred away from the metal center and is located on a ligand. There is still some interaction between these electrons, particularly in ru-bpy, because of their close proximity, but I don't think it's appropriate to call it a true triplet state, either. The lifetimes of true triplet states are generally in the hundreds of microseconds or even longer. The MLCT states of ru(bpy)3 are only in the low microsecond range.

As to why phosphorescence isn't common. Well, if the intersystem crossing rate from excited singlet to excited triplet is fast, generally so to is the intersystem crossing rate from the excited triplet back to the ground singlet. Also, oxygen (a ground state triplet) does a good job of quenching triplet states of organic molecules because it turns this into a spin-allowed process. The diffusion limited process is on the order of a microsecond at room temperature, meaning that unless the triplet state is lower in energy than that of singlet oxygen, or the radiative rate is faster than a microsecond, you're going to quench your phosphorescence pretty efficiently. So, you often don't observe phosphorescence unless your solution is deoxygenated, and even then you may not see it because competitive nonradiative decay is fast.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3480
  • Mole Snacks: +528/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: excited state questions
« Reply #5 on: July 13, 2015, 03:59:19 PM »
So, I located the image I wanted to upload but despite these good intentions I forgot to bring it home with me. But I did find on my hard drive here an incomplete version of it, which I've attached below.

For some generic fluorophore you have a ground singlet (blue parabola) and excited singlet (red parabola). The y-axis is energy and the x-axis is some generic nuclear coordinate. In reality these potential surfaces would be multi-dimensional but so simplicity we can represent them as 1 agglomerate dimension. Each of these electron states has associated vibrational states as well - for the blue surface they are shown as hashes along the right hand side. For the red surface, only the lowest energy one is shown.

Assuming you prepare an excited singlet state of a single molecule, the first thing that happens more or less is that there is fast (sub picosecond) vibrational relaxation to the lowest lying vibrational state of the excited singlet surface (V0 of S1). Now, eventually this excited S1 state will relax, which means that electrons will transfer to the S0 surface, and specifically some vibrational level of the S0 surface. The probability of any such transition occurring is proportional to the Franck-Condon factor, which is basically the overlap integral between the vibrational wavefunction of the initial state and that of the final state.

For fluorescence to occur, a transition has to happen directly between the S1 state and a low-lying vibrational level of the S0 surface, emitting a photon in the process. However, a competitive non-radiatve pathway is that the V0,S1 state changes into a {Vn,S0} state, where n is some vibrational level of the blue surface that has approximately the same energy as V1 of the red surface. This is called a "hot ground state" and relaxes back down to the V0 state of the blue surface very rapidly, and without emitting a photon. This process is called internal conversion and is one of the most significant non-radiative relaxation pathways.

Now, you will see the figure shows two possible scenarios, one for a visible-emitting fluorophore, where the S1-S0 energy gap is large, and one for a NIR-emitting fluorophore, where the {S1-S0} energy gap is small. The probability of going from {V0,S1} to {Vn,S0} is proportional to the Franck-Condon factor between the {V0,S1} wavefunction and {Vn,S0} wavefunction. As illustrated in the figure, it should be easy to see that the degree of overlap in the visible fluorphore is smaller than that in the NIR fluorophore, because in the former, most of the probability density of the {Vn,S0} wavefunction is on the "sides" of the blue well, but that of the {V0,S1} wavefunction is in the "middle". In the case of the NIR fluorophore, the overlap is better, and hence smaller energy gaps tend to result in faster rates of nonradiative relaxation, and hence smaller fluorescence yields.

One thing this figure neglects to show is the effect of the nuclear coordinate. My better version has a third scenario shown, which is a visible fluorophore but where the red curve is shifted somewhat to the right, reflecting a large geometric change between the ground and excited electronic states - i.e., a floppy molecule. You should be able to imagine that this scenario gives rise to good overlap of the two vibration wavefunctions, and hence faster rates of non-radiative decay. This is why rigid molecules - those with very little change in equilibrium geometry between excited and ground states - tend to have the largest fluorescence yields.

In reality there's a bit more to it; there are more than one donor and acceptor wavefunction you really have to worry about if you want to be quantitative, for instance. But, this is the gist of it. Hope all that was clear.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
Re: excited state questions
« Reply #6 on: July 15, 2015, 10:01:27 AM »
Thanks for the answering and the figure is really clear especially the wavefunction was excgarated. I dont realize the amplitude is greatest when i studied quantum mechanics.

I still have some questions in my mind hope you dont mind answering them if possible  :)

For Q3
you said that the pi to pi* transitions are highly delocalized while the n to pi* transitions are less;
Are you refering to the delocalization in the excited state? What is the reasons contributing to the above statement?

And i would also ask why singlet states tend to be highly delocalized, whereas triplet states are not?

Thanks really much

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
Re: excited state questions
« Reply #7 on: July 19, 2015, 12:30:23 PM »
just some questions pop up to my mind:
9) i saw that when a radical-linked fluorophores is fluorescently quenched, while after pairing up with another radical will exhibit the fluorescent:
http://turroserver.chem.columbia.edu/images/publications/NJT952.gif
Is that the radical is usually high energy enough to short-circuit the singlet to give triplet state like those open shell metals?

10) I had recently come a cross to a table from here(http://www.nature.com/nmeth/journal/v12/n3/pdf/nmeth.3256.pdf) and i crop a part of it as shown in the attachment.
do you have any comment on why the rhodamine with -N(CH3)2 have such a low quantum yield compare with the primary amine -NH2 derivative?
Is there anything to do with the -NH2 have a similar excited state energy level as the TICT intermediate, and migh be in equilibrium so this result to a higher quantum yield?

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3480
  • Mole Snacks: +528/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: excited state questions
« Reply #8 on: July 19, 2015, 09:09:12 PM »
Hi, sorry I didn't reply to your question.  Have been out of town.  I'll try to write something tomorrow.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline MrTeo

  • Chemist
  • Full Member
  • *
  • Posts: 312
  • Mole Snacks: +31/-9
  • Gender: Male
Re: excited state questions
« Reply #9 on: July 20, 2015, 01:26:09 PM »
I think I can help with the eosin question. If I remember correctly my Photochemistry classes eosin (which is just halogenated fluorescein, as Corribus already mentioned) shows a phenomenon called "delayed fluorescence" (type E, from eosin, there's also another type, P, much more complex that I can't really recall). Basically there is a good ISC transition constant but also an equally good "reverse" ISC constant that causes the eosin to fluoresce (even though with a lower quantum yield) with a much longer lifetime as the electrons spend some time in the triplet state where some decay, with a lower yield, via phosphorescence and ISC to the fundamental state, while others go back to the singlet state and undergo radiative decay.
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: excited state questions
« Reply #10 on: July 20, 2015, 03:21:18 PM »
10) I had recently come a cross to a table from here(http://www.nature.com/nmeth/journal/v12/n3/pdf/nmeth.3256.pdf) and i crop a part of it as shown in the attachment.
do you have any comment on why the rhodamine with -N(CH3)2 have such a low quantum yield compare with the primary amine -NH2 derivative?
Is there anything to do with the -NH2 have a similar excited state energy level as the TICT intermediate, and migh be in equilibrium so this result to a higher quantum yield?

The authors of the Nature Methods paper address this in the text:
Quote
A plausible explanation for the lower quantum efficiency of N,N,N′,N′-tetraalkylrhodamines such as 2 is the formation of a twisted internal charge transfer (TICT) state. In this process fluorophore 2 absorbs a photon to give an excited state (2*; Fig. 1b), which is followed by electron transfer from the nitrogen atom to the xanthene ring system with concomitant twisting of the Caryl-N bond (2TICT). TICT is energetically favorable in tetraalkylrhodamine dyes owing to the lower ionization potential of N,N-dialkylanilines versus less substituted anilines25. The TICT form relaxes without emission of a photon, leading to rapid nonradiative decay of the excited state25, 26, 27. The 2TICT diradical intermediate may also undergo irreversible bleaching reactions27. Thus, rhodamine derivatives in which TICT is disfavored should exhibit increased quantum efficiency, longer fluorescence lifetimes and higher photostability.

You can probably find more information by reading ref 25.

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
Re: excited state questions
« Reply #11 on: July 21, 2015, 11:36:39 AM »
Thanks for everyone's reply :D just still some puzzle in my mind....
Quote
Basically there is a good ISC transition constant but also an equally good "reverse" ISC constant that causes the eosin to fluoresce (even though with a lower quantum yield) with a much longer lifetime as the electrons spend some time in the triplet state where some decay, with a lower yield, via phosphorescence and ISC to the fundamental state, while others go back to the singlet state and undergo radiative decay.
Wow this was the first time for me to know something reverse ISC. Is there any criteria for it to occur(or anything may facilitate it to occur?

Quote
You can probably find more information by reading ref 25.
I had looked briefly but i am more curious if TICT is the only factor affect the quantum yield, since observing fluorescein diacetate is fuorescent quenched while fluorescein is highly fluorescent, so i am wondering whether electron density on rhodamine will participitate in any fluorescence quenching process.

Meanwhile, anyone may explain how fluorescein diacetate can be fuorescent quenched?

Thanks everyone helping me

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3480
  • Mole Snacks: +528/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: excited state questions
« Reply #12 on: July 21, 2015, 12:05:28 PM »
@khwcm

Quote
you said that the pi to pi* transitions are highly delocalized while the n to pi* transitions are less;
Are you refering to the delocalization in the excited state? What is the reasons contributing to the above statement?

What I was getting at is this: in a pi-pi* state, both the HOMO and LUMO have a large extent of delocalization. In an n-pi* transition, the LUMO may be delocalized but the HOMO isn't - it's located solely (with good approximation) on the non-bonding atomic orbital of the heteroatom. Therefore as the pi system expands, the spatial disparity between the pi and pi* molecular orbitals on the one hand, and the nonbonding atomic orbitals on the other, grows.  Another way to put this might be that as the pi system expands, electrons spend, on average, their time farther away from the nonbonding orbital - which might translate into a lower probability of quenching by the heteroatom as the pi-system expands. Other thing is that, unlike in simple ring systems like benzene, in fused ring systems the probability of finding an electron in the vicinity of a nuclear (carbon) center is not the same for each carbon atom in the system. Electrons naturally accumulate near some carbon nuclei more than others. My thinking was that this might translate into differences in fluorescence quenching by the heteroatom depending on where it's located - if it's in a position where electrons naturally spend more time, it might be more efficient at quenching the excited state. There is relatively recent theoretical work to show that the position of the heteroatom does significantly impact the molecular orbital (HOMO and LUMO) characteristics for fused ring systems: see Chen et al, JPCA, 2014, 118, 5700. Orbitals don't always correlate exactly with where electrons are located in states, but it was at least a starting point.

Based on this, I formulated two hypotheses:

1. As the size of the ring system expands, the efficiency of heteroatom quenching decreases and therefore the fluorescence yield should increase. (Or, the disparity in quantum yield between the heteroatom-containing fluorophore and that of the non-heteroatom-containing control gets smaller.)

2. The fluorescence yield will change depending on where the heteroatom is, with higher yields being more likely to be observed when the heteroatom is located at a position where electron density in the 1st excited state is largest.

For hypothesis one, I tried to find quantum yield data for benzene/pyridine, naphthalene/quinoline, and anthracene/acridine. If my hypothesis was right, the difference in QY between naphthalene/quinoline should exceed that between anthracene/acridine. Benzene is a bad one to start with since it has a low quantum yield itself. Moving on. Quinoline has a QY of about 0.001, whereas naphthalene is roughly 25%, although the range of data in the literature is quite large - another problem of trying to learn anything about quantum yields. Surprisingly, acridine - which is basically anthracene (three fused rings) with a single nitrogen in place of one of the carbons - also appears to be really low, practically nondetectable. (See http://onlinelibrary.wiley.com/doi/10.1002/bbpc.19810850212/abstract). The QY of anthracene is about 36% or so. So far, it doesn't appear that the first hypothesis bears out, but more on that in a minute.

For hypothesis two, my immediate thought was to compare the fluorescence yields of quinoline and isoquinoline, which differ only in the placement of the heteroatom (nitrogen) on the naphthalene core. It was far harder to find good quantum yield data than I expected, but Li and Lim (JCP 1972 57 605) report quantum yield values of ~0.01 and ~0.001, respectively. It's possible this could be taken as support of my hypothesis, but later work on the fluorescence yields of these molecules (Anton and Moomow, JCP 1977 66 1808) showed that reported yields are extremely sensitive to even trace levels of hydrogen bonding solvent molecules (e.g., water) - this being because hydrogen bonding to the heteroatom raises the energy level of the n-pi* state so that it lies above the pi-pi* state, and so qunching isn't observed. We might take something from the fact that isoquinoline appears to be more sensitive to this than quinoline, but without reliable data it's hard to make conclusions one can be confident in. Nevertheless, it's fairly clear that differences in electron density at the heteroatom position would play a minor role, if any, in the degree of quenching of the excited state compared to the presence of a heteroatom in the first place. At least, this is so for naphthalene derivatives - considering that the QY of naphthalene is ~25%, far higher than either quinoline or isoquinoline. So, no data to definitively support hypothesis 2 either.

That said, I think the hypotheses still have merit. For instance, because the nonbonding orbital of nitrogen won't change no matter how large the ring system is, it is reasonabe to expect that the pi-pi* energy gap will shrink more rapidly than the n-pi* energy gap as the size of the system expands. At some point, there should be a cross-over point, and at that point the n-pi* state should be unable to quench the pi-pi* state. I looked to see if I could find a N-containing tetracene but I wasn't able to find an an experimental report of this moleucle in my (brief) literature search, so I guess this hypothesis will go unproven for now. :)

Quote
And i would also ask why singlet states tend to be highly delocalized, whereas triplet states are not?
This was still a matter of intense research several years ago. I haven't kept up with the latest in molecular photophysics for the last few years, particularly on conjugated polymers, so maybe there are better explanations for it now than what I was using in the past. So, I wouldn't feel comfortable trying to answer this without doing some reading first.

Quote
9) i saw that when a radical-linked fluorophores is fluorescently quenched, while after pairing up with another

radical will exhibit the fluorescent:
http://turroserver.chem.columbia.edu/images/publications/NJT952.gif
Is that the radical is usually high energy enough to short-circuit the singlet to give triplet state like those open shell

metals?
I think I'd prefer to see the whole paper before commenting, but yes radicals often quench fluorescence, either due to photochemistry or because they facilitate things like low energy intramolecular charge transfer states that provide convenient avenues for fast nonradiative excited state relaxation.

Quote
10) I had recently come a cross to a table from here(http://www.nature.com/nmeth/journal/v12/n3/pdf/nmeth.3256.pdf) and i crop a part of it as shown in the attachment.
do you have any comment on why the rhodamine with -N(CH3)2 have such a low quantum yield compare with the primary amine -NH2 derivative?
Is there anything to do with the -NH2 have a similar excited state energy level as the TICT intermediate, and migh be in equilibrium so this result to a higher quantum yield?
Adding to the previous post, the flexibility of the alkyl chains also makes a difference, as you can see from your table. The solvent polarity as well, as polar solvents stabilize the charge transfer state. The more rigid structures are less sensitive in this, since rotation is hindered.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline khwcm

  • Regular Member
  • ***
  • Posts: 47
  • Mole Snacks: +0/-1
Re: excited state questions
« Reply #13 on: July 24, 2015, 01:07:05 PM »
Thanks for your clear detail explanations :D

Sponsored Links