[riboswitch]

Charge transfer transition:

So a charge transfer transition is unlike the charge delocalization? You're right, because it appears that no relocation of charge density happens in carboxylate anions. The "negative charge" is only delocalized between the two electronegative oxygen atoms in a resonance structure. Intramolecular charge transfer, in contrast, refers to the process that changes the overall charge distribution in a molecule.

You said that charge transfer transition state offers a convenient relaxation pathway that short-circuits triplet formation/inter-system crossing. Of course, there are many reasons why inter-system crossing is not favoured, since singlet-to-triplet transitions are usually forbidden. Any part of a molecular structure that facilitates charge transfer transition will negatively influence triplet yield.

Well, you stated the example of coumarin laser dyes. It seems that twisted intramolecular charge transfer is a particular form of bond rotation around a carbon-amino nitrogen bond. If my interpretation is correct, the "twisting motion" is in fact a form of fluorescence quenching, a preferable non-radiative relaxation pathway that is dependent upon the solvent in which the molecule is dissolved. Solvents that stabilize the charge transfer state reduces the fluorescence yield of the excited molecule. An example of such solvent is water (which is easily observed in your very colourful and enticing photograph).

Fluorescence yield and triplet yield:

So a low fluorescence yield does not always correspond to high triplet yield? Since you said that there are numerous non-radiative pathways that can short-circuit triplet formation, low fluorescent yields can also correspond to low triplet yield. You already explained it in your last post; now I'm starting to understand things correctly.


Wrapping things up....

• In our problem, the highest occupied level is supposed to be m=±5 and the lowest unoccupied level is supposed to be m=±6
• In reality, particle-on-a-ring treatments of porphyrins using a 20 electron-model is incorrect, because it would imply that porphyrin is paramagnetic in its ground state. That's absolutely wrong because virtually all organic molecules are diamagnetic in their ground state.
• An 18 electron-model of porphyrin is more appropriate if I'm going to use particle-on-a-ring approximations. Such model will imply that the highest occupied levels do not contain two unpaired electrons in the ground state (the molecule, as a result, is diamagnetic, which is consistent with real-life observations).
• A superficial analysis of the free electron model of porphyrin will lead me to believe that there is a single 4-fold degenerate transition that is happening when I excite the molecule. The porphyrin spectrum tells me otherwise. There are actually two 2-fold degenerate transitions: a lower energy transition (which should be forbidden) represented by the Q-band of the spectrum, and the higher energy transition represented by the B-band of the porphyrin spectrum. Of course, the free electron model can account for this, but we need complicated mathematics to do that.
• Dioxygen molecule is a triplet in its ground state. It means the molecule is paramagnetic when it is not excited. If porphyrin were also paramagnetic in its ground state, a spin-allowed reaction would take place between ground-state dioxygen and ground-state porphyrin. This doesn't happen in real-life situations.
• A porphyrin can be excited to a triplet state. In order to do this, favourable conditions must be met in order to increase the probability of inter-system crossing. For example, the vibrational levels of the singlet excited state and the triplet excited state must overlap. The presence of heavy atoms near the conjugated system may also increase triplet yields in porphyrins.
• Interaction between triplet-excited state porphyrin and triplet ground-state dioxygen will produce singlet ground state porphyrin and singlet oxygen. The latter has interesting properties which are exploited in numerous applications, such as photodynamic therapy.
• As you've said earlier, to generate singlet oxygen, the excited triplet state of the photosensitizer has to have a higher energy than that of singlet oxygen. That means that identifying good photosensitizers is not an easy task.

Let me know if I said anything wrong. Thanks for helping me out. 

[Corribus]

So a charge transfer transition is unlike the charge delocalization?

In common parlance, charge transfer is a process, whereas charge delocalization is a state. I.e., the stability of the phenoxide anion is often attributed to the ability of the negative charge to be delocalized over the phenyl ring. On the other hand, charge transfer can result in charge delocalization. For instance, it is well known that when bipheny, is reduced, the two phenyl rings become coplanar, resulting in delocalization of the transfered charge over a larger area. Which is to say, the two phenyl groups in neutral biphenyl are largely independent, but in the anion (or cation) radical (or, for that matter, in the photoexcited neutral molecule), they are well-conjugated. (Ref: Almennigen, et al, J. Mol. Struct. 1985, 128, 59-76.) To put that in a photophysical context, we can turn again to the metal polypyridyl literature. [Ru(bpy)₃]²⁺ is a metal polypyridyl complex in which three bipyridine ligands are coordianted to a ruthenium dication. When photoexcited, an electron is transfered from the ruthenium metal to one of the three bipyridine ligands, whereupon the charge is delocalized over the entire ligand. This metal-to-ligand-charge-transfer (or MLCT) state gradually relaxes predominantly by luminescence back to the ground state on a microsecond timescale. Now what happens if we functionalize  the bipyridine ligands with phenyl groups? In the ligand, the phenyl group is canted with respect to the bipyridine plane due to steric-hindrance of the adjacent hydrogens, much like the two phenyl groups in biphenyl. When [Ru(4,4'-diphenyl-bpy)₃]²⁺ is photoexcited, an electron is again transfered to the ligand. Initially this transfered charge is only delocalized over the bipyridine ring. Over a time scale of ~2 picoseconds, however, the phenyl groups rotate, allowing the charge to become further delocalized. Evidently, the energy gained by delocalizing the charge is larger than the energy cost of having those hydrogens bumping up against each other. You can read more about this at the following reference, if you wish: Damrauer and McCusker, J. Phys. Chem. A, 1999, 103, 8440-6.

Any part of a molecular structure that facilitates charge transfer transition will negatively influence triplet yield.

With the usual acknowledgement that there are probably exceptions out there, and as long as we are understanding the triplet yield refers to a triplet pi-pi* state, yes.

In reality, particle-on-a-ring treatments of porphyrins using a 20 electron-model is incorrect, because it would imply that porphyrin is paramagnetic in its ground state. That's absolutely wrong because virtually all organic molecules are diamagnetic in their ground state.

I would soften the language. I wouldn't call it incorrect - I'd call it a poorer model than the 18 electron model. Also, I wouldn't say "virtually all", I'd say a "vast majority". In chemistry, you learn quickly to always leave room for exceptions. As soon as you say "all" or "none", someone will point out a counter example in about 5 seconds.

we need complicated mathematics to do that.

The mathematics are not so complicated to write down, but exact answers do not exist, so many approximations have to be made to find workable solutions. We can easily understand why there is a splitting, but quantitating them with any accuracy usually requires computers and sophisticated modelling programs. Semiempirical methods (parameter models based on experimental data) have can also be used with some success.

Dioxygen molecule is a triplet in its ground state. It means the molecule is paramagnetic when it is not excited. If porphyrin were also paramagnetic in its ground state, a spin-allowed reaction would take place between ground-state dioxygen and ground-state porphyrin. This doesn't happen in real-life situations.

Well, it does. The hemoglobin in your body binds oxygen, doesn't it? I guess to be pedantic, this is oxygen reacting with the open-shell iron center, not the porphyrin itself, but the porphyrin does play a role. Beyond that, it's better to speak in terms of fast rates and slow rates (or, alternatively, reaction probabilities) rather than what does or does not happen. Singlets can and do react with triplets - the reactions are just slow. "

This point about softening your language also carries through to your other statements as well. Generally your summary is good - you just have to be careful about speaking in absolutes.

[riboswitch]

Charge transfer

So biphenyl itself isn't planar. One ring is slightly canted in relation to the other. Its most stable conformation is achieved when the molecule has a dihedral angle of about 45°. If I haven't consulted the Internet, I would have said that the stable conformer has a 90° dihedral angle. The conformers of biphenyl reminds me so much of butane conformations that I have to review my Organic Chemistry notes just to refresh myself on the topic.

Anyways, you said that a charge transfer process happens when our ruthenium polypyridyl complex is photoexcited. But the molecule in its excited state eventually relaxes back to the ground state via luminescence. I'll just assume that the molecule undergoes fluorescence (radiative relaxation pathway) to return to its ground state. If one of the bipyridine ligands contains a phenyl group, then a "twisted intramolecular charge transfer" occurs: the phenyl-bipyridine bond is "twisted", charge delocalization results and the two molecules are now co-planar. Fluorescence is also quenched when the phenyl-bipyridine bond is "twisted". Of course, as said earlier, the presence of solvents that stabilize the charge transfer state (polar solvents) further reduces the fluorescence yield of our molecule.

What do you think? Did I explain it correctly?


Paramagnetism

Also, I wouldn't say "virtually all", I'd say a "vast majority". In chemistry, you learn quickly to always leave room for exceptions. As soon as you say "all" or "none", someone will point out a counter example in about 5 seconds.

Questions out of curiosity: are you suggesting that there are paramagnetic organic compounds? Let me guess, are they something like intermediate metabolites from biochemical reactions (like fructose 1,6-biphosphate)? Does paramagnetism caused by the presence of functional groups like the hydroxyl, amine or phosphate groups? Or maybe you're talking about organometallic compounds, like vitamin B₁₂?

...or probably magnetism is a spectrum and not an absolute characteristic? That would mean that organic molecules have varying "degrees" of paramagnetism...


Heavy atom?

That said, the presence of a heavy atom in the proximity of the conjugated system (what is "heavy" is arbitrary; there's no hard cut-off, but usually 3rd row or lower is heavy enough to be significant) generally will lead to high triplet yields because of the heavy atom's impact on spin-orbit coupling.

Another curiosity-driven question (if you don't mind): does the chelated metal in naturally occurring porphyrins count as the "heavy atom" that influences triplet yields?

[Corribus]

So biphenyl itself isn't planar. One ring is slightly canted in relation to the other. Its most stable conformation is achieved when the molecule has a dihedral angle of about 45°. If I haven't consulted the Internet, I would have said that the stable conformer has a 90° dihedral angle.

The average structure is a compromise between the energy gain due to delocalization in the coplanar structure and the energy gain due to minimization of steric interactions between the hydrogens.

Anyways, you said that a charge transfer process happens when our ruthenium polypyridyl complex is photoexcited. But the molecule in its excited state eventually relaxes back to the ground state via luminescence. I'll just assume that the molecule undergoes fluorescence (radiative relaxation pathway) to return to its ground state.

Strictly speaking, it's not really fluorescence because the excited state isn't really a singlet state. Formally it's a triplet state but it's not even really that, because of its charge transfer character. So we usually just called it luminescence and be done with it.

If one of the bipyridine ligands contains a phenyl group, then a "twisted intramolecular charge transfer" occurs: the phenyl-bipyridine bond is "twisted", charge delocalization results and the two molecules are now co-planar. Fluorescence is also quenched when the phenyl-bipyridine bond is "twisted". Of course, as said earlier, the presence of solvents that stabilize the charge transfer state (polar solvents) further reduces the fluorescence yield of our molecule.

What do you think? Did I explain it correctly?

I think you get the general idea. TICT as a formal designation is usually reserved for amino-functionalized dyes (for some reason), probably because of its dramatic effect on the luminescence properties of the dyes. Not only is fluorescence quenched, but it's also usually red-shifted as well (because the excited state is stabilized, or lowered in energy).

 

Questions out of curiosity: are you suggesting that there are paramagnetic organic compounds?

There are plenty of organic compounds with unpaired electrons, but not too many of them are stable. Organic compounds with a ground-state triplet would react readily with oxygen. Probably the most well known class of purely organic compounds with a ground-state triplet are the carbenes. Some of them are persistent, meaning they are quasi-stable. The triplet-state classes of course would need to be kept in inert atmosphere, though, and even then, they often readily dimerize (in a matter of seconds, long enough to be observed and characterized but not stored on your shelf). Persistent carbenes usually require the use of bulky substituents, which slow down the dimerization process - meaning the stability is usually kinetic rather than thermodynamic.

http://en.wikipedia.org/wiki/Persistent_carbene

Another curiosity-driven question (if you don't mind): does the chelated metal in naturally occurring porphyrins count as the "heavy atom" that influences triplet yields?

Most naturally occurring porphyrins and porphyrin derivatives are bound to 1st row transition metals, such as iron (heme), magnesium (chlorophyll), cobalt (vitamin b12), etc. The most common synthetic porphyrin is bound to zinc. Zinc is "heavy enough" to see a spin-orbit coupling effect, so the other transition metals would be, too. However the other transition metals are open shell, and the empty d-orbitals usually provide a rapid pathway for excited-state deactivation. For example, the excited-state lifetime of most iron porphyrins is on the order of picoseconds*, far too short to see fluorescence or intersystem crossing to the π-π* triplet-state. So, while the intersystem timescale might be affected by these metals, you'd never be able to measure it.

* Although, I managed to create one that was much longer. I cringe to think on it though, because it was perhaps the most annoyingly fickle compound I ever had the displeasure of working with. Making hemes sucks**, and whoever tells you otherwise is lying. Actually, I have quite a bit of respect for nature, which not only managed to get the iron atom to stick around, but actually was able to get it to do something bloody*** useful.

** You can quote me on that if you want.

*** I feel the need to obnoxiously point out my incredibly witty pun, in case you missed it.

[riboswitch]

Strictly speaking, it's not really fluorescence because the excited state isn't really a singlet state. Formally it's a triplet state but it's not even really that, because of its charge transfer character. So we usually just called it luminescence and be done with it.

You're saying that it's still a radiative relaxation pathway, but it's not technically fluorescence. I learned that fluorescence requires photo-excitation of singlet ground state electrons of an atom (or molecule) to higher energy singlet states, followed by a transition from the excited state back to ground state via emission of lower energy photons. The emitted photons have lower energies because of energy dissipations that occur during the excited state lifetime. Since our metal-polypyridine complex doesn't follow the general rules of fluorescence, then I'll just do what you just did: classify it as luminescence and be done with it! 

I think you get the general idea. TICT as a formal designation is usually reserved for amino-functionalized dyes (for some reason), probably because of its dramatic effect on the luminescence properties of the dyes. Not only is fluorescence quenched, but it's also usually red-shifted as well (because the excited state is stabilized, or lowered in energy).

Red-shifted? Like Stokes shift? 

There are plenty of organic compounds with unpaired electrons, but not too many of them are stable. Organic compounds with a ground-state triplet would react readily with oxygen. Probably the most well known class of purely organic compounds with a ground-state triplet are the carbenes.

So even persistent carbenes react readily with oxygen. No wonder they are usually seen only under laboratory-controlled conditions. I'll guess it's safe to say that there are no paramagnetic organic compounds detectable in biological conditions (like inside a cell). It's because the ubiquitous dioxygen molecule plays a very important role in the energy metabolism of living organisms and having biomolecules that are paramagnetic might disrupt key biological functions because of their instability in the presence of molecular oxygen.

* Although, I managed to create one that was much longer. I cringe to think on it though, because it was perhaps the most annoyingly fickle compound I ever had the displeasure of working with. Making hemes sucks**, and whoever tells you otherwise is lying. Actually, I have quite a bit of respect for nature, which not only managed to get the iron atom to stick around, but actually was able to get it to do something bloody*** useful.

I don't know if this might console you, but hemes lose their usefulness when carbon monoxide is present. So Nature's compounds have their own little flaws too. 

BTW, I never thought making hemes suck. But you worked with hemes anyway.

[Corribus]

Red-shifted? Like Stokes shift? 

Yes. Most of the Stokes shift occurs because relaxation prior to fluorescence results in a loss of some quantity of the excitation energy. In polar solvents, reorientation of solvent molecules around the polarized excited-state reduces the energy of the state on timescales faster than fluorescence - hence the larger Stokes shifts in polar solvents (usually). Note that in very rigid, nonpolar fluorophores, the Stokes shift is usually very small in nonpolar solvents. Not coincidentally, these fluorophores also tend to be the brightest.