 Today we plan to discuss one of the most intriguing debates that went on for about 10 years from 1997-98 to about 2008 in the field of excited state processes, excited state dynamics. The reason why this is very close to my heart is that it started when I was a PhD student and I was working in proton transfer so and we could see that something interesting is breaking out and the two groups involved one were of Tahitara who was in Institute of Molecular Sciences Japan at that time later on he moved to Rikken and the other group was Ahmed Zuel's group and this thing started in about 1995, 96, 97 when all of us could clearly see that Ahmed Zuel is going to get the Nobel Prize. So a potential Nobel laureate and on the other hand we have a 33 year old Japanese researcher lead group somebody who at that time did not was not confident in speaking English never went out of Japan. So that was almost a battle of David and Goliath and at the risk of giving a spoiler in the story in Bible Goliath one that is what often happens in real life as well but it is not really an issue of winning or losing this debate actually teaches us several interesting things maybe we will come to that at the end of this discussion. This discussion is expected to span two modules before getting into this debate of Seven as a Indole Dimer. Let me first present something else that also comes from the group of Tahitara more or less in late 90s. So all of us know Karsha's rule and all of us know that the reason why this emission spectrum is expected to be mirror image of absorption spectrum is when we draw Jablonski diagram this is what we always draw that this is S0 let us say this is S2 there are they can be many more of course and let us say this is S1 different vibrational states of S1. What we have learnt is that you excite no matter where you can excite to S2 you can excite to a higher vibrational level of S1 there is always ultrafast vibrational relaxation to V dashed equal to 0 the lowest vibrational state of S equal to 1 and that is where the emission takes place from. So this is the 0 dashed 0 transition you can have things like 0 dashed to 1 transition you can have something like 0 dashed to 2 and so on and so forth. And while excitation while exciting you can excite 0 0 dash and you can excite to higher levels as well. So we are very familiar with this kind of a picture where this is the absorption spectrum energy decreases this way so wavelength increases this way and this is the emission spectrum. And we as we have said earlier this is what we expect to see for well behaved molecules of course the entire business that we have is about molecules that are not well behaved because if all fluorescence spectra are mirror images absorption spectrum then there is not much you can do about it anyway. So when we have excited state processes we get this red shifted band somewhere here. So this is the mirror image spectrum and this is the spectrum from that arises from the excited state process I will just write ESP. Now the reason why you do not get emission from S2 usually is that if I draw not like this but in a little different way if I draw potential energy surfaces then generally this is the case this is S0 this is S1 this is S2 this is S3 and so on and so forth. Of course relative position of minima can be different what we are we want to highlight more is that there is an energy gap between S0 and S1 generally and generally S1 S2 S3 these three are not only close in energy but generally there is crossing of the potential energy surfaces that crossing can be adiabatic non adiabatic that is a different question. So since energy gap is small it is very easy for non radiative processes to happen among all these excited SN states and S1 is the lowest energy state there is a big energy gap so according to energy gap law V dashed equal to 0 of S1 that is where the molecule can reside for a while and that is when fluorescence gets a chance. This is the very premise of Karsha's rule and we have already seen how Karsha's rule can be violated if excited state processes are formed are there then you get redshifted spectrum. Now we go to the other end of the story is it possible in any way to get emission from S2 or S3 and this is very famous example that most of us might know and that is a azuline. So that used to be referred to for a long time as the azuline anomaly in azuline the major fluorescence is from S2 to S0 why that is because for azuline I will just write here the energy diagram is like this this is S0 S1 actually has an overlap with S0 and there is a big S1 S2 gap this is the conventional picture that has been known for 50, 60 years now. So that is why if you excite to S1 non radiative processes take over if you excite to S2 then only you can get some emission. So azuline has an emission band that is at a higher energy than the lowest energy absorption band because lowest energy absorption band of course is S0 to S1 emission in this case is S2 to S0 okay so this is something that is known classically. Now Tahara's group had tried to ask the question what happens in very short times is it that S2 is really non-fluorescent no fluorescence takes place or is it that the lifetime is so small that you do not see the fluorescence in the steady state spectrum. So the question is this is it possible and now I will draw only three lines to make things simple S0 S1 S2 so excite to S1 you get fluorescence and you get some non radiative processes understood excite to S2 what is believed is this non radiative process takes over okay and there can be some non radiative process between S2 and S0 as well but suppose for the sake of argument we say that there is emission for S2 from S2 as well. First of all where will that emission come higher energy or lower energy compared to the S1 emission naturally higher energy yeah now the problem is this we have discussed this multi exponential model of fitting fluorescence decays right. So this is the most popular model where you fit like this I at time t equal to I at time 0 sum over I AI e to the power minus t by tau I yeah and steady state intensity and now I will write lambda also at some particular lambda or better write frequency there is a conversion factor between frequency and lambda at some frequency is equal to integral of I nu at t dt yeah so this turns out to be sum over AI tau I and then what we know is that if there are several components then the contribution to steady state intensity of the ith component is given by AI tau I divided by sum over I AI tau I I believe this is all this we have discussed now see suppose I have a 2 exponential decay by exponential decay tau 1 is something like 0.1 picosecond and I will exaggerate just to bring out the fact tau 2 let us say is 10 nanosecond okay and let us say they contribute 50 50 AI is 0.5 sorry A1 is 0.5 and A2 is also 0.5 what will be the contribution of this tau 1 species associated with tau 1 what will be the contribution to the steady state intensity 0.5 multiplied by so I can write like this Is of 1 will be 0.5 multiplied by 0.1 if I write in picosecond divided by 0.5 multiplied by 0.1 plus 10 nanosecond means how much 10000 picosecond right so 10 to the power 4 so what will you do you are going to neglect this 0.1 with respect to 10 to the power 4 naturally so in the numerator you have 0.05 yeah in the denominator you have 0.5 into 10 to the power 4 what is the answer 10 to the power minus 5 really very small yeah so the point is if you have emission from a higher energy state that is short lived then you will not see it in steady state is the point made yeah where will it be seen is there any way of seeing it there is a way of seeing it and the only way of seeing it is by using ultrafast dynamics we have discussed already let us say this is your emission spectrum steady state yeah this is steady state x axis let us write is let us write frequency let us be consistent but then I should have perhaps written drawing the other way anyway y axis is intensity let us say I have two components like one like this one like this this is the 0.1 picosecond component and this is the 10 nanosecond component suppose I look at the steady the time resolved emission spectrum which we have discussed earlier and suppose we look at the spectrum at time t equal to something like 0.2 picosecond will you agree with me that that spectrum at initial times very short time after excitation will be dominated by this species that has a lifetime of 0.1 picosecond so that is one way in which you can actually see a spectrum that is elusive in steady state right so this is what mainly Takeuchi and Tahara had done prior to prior to this debate actually so what they had done is that taken some molecules and they had demonstrated that you can actually see the S2 spectrum I encourage you to find those papers and read them right in our presentation we are not going to talk about them but what Takeuchi and Tahara had done before the 7 as I indulge debate broke out was that they could actually demonstrate that you can see fluorescence from the S2 state at very small times after excitation okay that is the background we need before we can get into the discussion. So we are not saying Kasa's rule is wrong we are not saying that Jablonsky diagram we had drawn is wrong all we are saying is that you can hold strictly when only when you do steady state spectroscopy when you do time resolved spectroscopy then you can actually see things that are relaxing in very fast time scale and therefore are not observed in steady state spectroscopy similarly if you go down further to at a second which is now the state of the art then you can see things that we assume even now in femtosecond time scale to be instantaneous almost nothing is instantaneous anymore okay so the take home message from this is you can actually see fluorescence from a higher energy excited state if you look at small times post excitation now we get back to the debate and see how this concept turned out to be critical in this discussion okay. So we will talk about excited state reaction dynamics in nonpolar solvents manifested by excited state double proton transfer or 7 as I indulge dimer mainly we will discuss these papers there are many more as I am going to show you we are actually presenting almost only one side of the debate we are not presenting the other side so I encourage you to read everything it is an engrossing debate now I must thank Rosa Tahitara for sharing the slides with me that first of all you are almost if not here see it from the horses perspective and it also saved me a lot of time I did not have to prepare the slides myself so 7 as I indulge dimer is a widely popular model for DNA base pairs because if you just look at the dimer it will remind you of how ATGC are there in DNA okay. So you have 2 kinds of nitrogen here 1 in a 5 membered ring 1 in a 6 membered ring you have 2 kinds of nitrogen and the hydrogen is covalently bonded to the nitrogen in the 5 membered ring in both the cases okay. So this is where I used to get confused that why is it going from nitrogen to nitrogen please be very clear about that the 2 nitrogen are not equivalent 1 is in a 5 membered ring 1 is in a 6 membered ring their acidity is basically acidity is different so in ground state the proton is covalently bonded to this 5 membered ring nitrogen and hydrogen bonded to the 6 membered ring nitrogen of the neighbor of the partner and that is true in the other end as well what is the symmetry of the system what is the point group do we have a point of inversion point no point of inversion yes or no actually we do we do actually we do dimer I am talking about the dimer not the monomer point of inversion is there actually but that hardly matters when we are trying to determine the point group unless nothing else is there do we have a principal axis of symmetry in the dimer we have a C2 axis right perpendicular to the plane do we have a horizontal plane right so that makes it C2H do we have perpendicular C2 axis perpendicular C2 axis 1 is there no nothing is there there is no perpendicular C2 axis actually because you cannot draw C2 axis in maybe you are thinking of this nitrogen that nitrogen right that maybe true if you have a C2 axis through these 2 nitrogens but then these carbons will not find a partner and for those who might be wondering what we are talking about in case you need we are talking about symmetry of the molecule and we have a prior NPTEL course on symmetry in chemistry those lectures are freely available on YouTube in case you need help you can refer to them okay so this this has a C2H symmetry we will mention to the we will refer to the C2H symmetry in passing a little later so this is something that was known for a long long time almost as long as the 3 hydroxy flavone case that we discussed earlier that you have this dimer you excite it and then when you excite remember organic acids become stronger acids in excited state this NH is an organic acid this nitrogen in the 5 membered ring is an organic base bases become stronger bases and therefore there is a double proton transfer this was known very well I am going to show you the spectra it was manifested amply in the spectrum and the energy surfaces were also worked out and the energy surfaces is qualitatively same as that what we had for 3 hydroxy flavone asymmetric double well potentials in ground and excited state with the reverse asymmetry between them okay the state that is more stable in the ground in the ground state the well let us say the form that is more stable in the ground state is less stable in the excited state that is why proton transfer takes place in the excited state and since 2 protons are getting transferred it is called excited state intermolecular double proton transfer in our lab we have done some study on another molecule which we call BPOH2 there we have excited state intermolecular double proton transfer and it is manifested in the steady state spectra as I told you like this okay these are the absorption and emission spectra at different concentration of 7 as I told do not think that dimers are formed at very low concentration they are not in fact even for 10 to the power – 4 molar concentrations which is pretty high for fluorescence you do not get dimers okay in absorption you see there is a little bit of change this one the blue one is monomer and the other one which color is it green or red red the red one is for dimer so as concentration increases from 10 to the power – 4 molar to 10 to the power – 2 molar the absorption changes little bit this little bit of red shift and the structure also changes but you get more drastic changes in the emission spectrum this is the monomer emission spectrum between 300 and 400 nanometer and you see it is more or less mirror image of the absorption spectrum okay and these spectra are all I think normalized to this peak as concentration increases what you see is in this 400 nanometer to 700 nanometer region at 10 to the power – 4 molar there is hardly anything actually something is there what you see it only when you write when you draw a semi log plot and as the concentration increases at the look for the C2 axis later it is not there as concentration increases the stroke shifted band characteristic of excited state process keeps on increasing that is why the primary assignment of this to the excited state process happening in dimer was done right so this is something that was known and even reaching there actually was not easy the reason why this debate started in 1998 and not in 1978 is that people had to characterize people had to say with confidence that this is a dimer so several kinds of studies had to be done mass spectroscopy and so on and so forth they had to be confident that excited state double proton transfer is taking place a lot of studies were already there before the debate broke out what was the debate that is what we will take in the next module.