 Welcome to this new segment of CD spectroscopy and MOSBUS spectroscopy for chemist. My name is Arnab Dutta and I am an associate professor in the department of chemistry at IIT Bombay. So, in the previous segment we are discussing about the relationship between chirality and symmetry. So, let me just refresh your memory a little bit chirality and point group and the take home message from this particular segment was a molecule can be chiral if it does not have any SN axis. So, the presence of SN axis is directly connected to chirality why? Because when you define chirality we say we are taking a mirror image that means a reflection that means a reflection around a plane basically we are doing a sigma operation. Then we take this mirror image and try to fit whether it fits with the original molecule. Now, over here we are doing nothing but simple rotation around an axis. So, we are basically doing a sigma operation and CN operation without really understanding it. So, basically we are nothing but doing a SN operation. So, if the molecule does not fit on its original position or original structure that means you do not have an SN axis that means you are going to have a chiral molecule. So, that we have found and then if you also find out this can be simplified that means S1 means sigma plane. So, that means if you do not have a sigma plane there is a chance that you are going to be chiral and S2 is center of symmetry that means there is also a chance you have a chiral molecule if you do not have a center of symmetry but just simple absence of this particular symmetry element does not guarantee that you are going to have a chiral molecule. If you have this particular symmetry elements then it is guaranteed that your molecule will not be chiral that will be a chiral. On the other hand we have tried to find out what are the point groups are present there that they do not have any particular SN axis and we figure it out that is belongs to two groups CN and DN. The simple axis rotation group nothing other than CN axis and a diadol group nothing other than CN and N number of cities perpendicular to the CN. And I want to mention C1 is a special case of CN where you have nothing but only an identity operator available for a particular structure of the molecule. So, with that covered then today we are going to cover three different structures of molecules and find out how they are becoming chiral. So, let us go ahead with that. So, the first example we are going to take is the following an iron octahedral complex and over here I am going to have bidented ligands bind to it. What do I mean by bidented ligands? So, this structure I am drawing like this it is nothing but carboxylate acid groups. So, over here in a carboxylate binding group you can see these are the two centers where it will bind to the metal and over here it can bind with two different coordination side from the same ligand. So, that is why they are called the bidented ligand. And over here this iron is binding three such bidented ligand. Now the question is whether this molecule is chiral or not and as we learned earlier what you have to do just figure it out what is the point group of this molecule. So, let us figure it out and we have learned that earlier is this molecule linear as it is known is this molecule belong to octahedral or tetrahedral point group that means any cubic group. So, the answer is still no although you will say that this molecule is binding in an octahedral geometry that is true this molecule is binding in an octahedral geometry but that does not mean that the molecule is octahedral symmetry because to have an octahedral symmetry the molecule should possess a center of symmetry and for center of symmetry that means if you go from one part of the molecule to the other part through the center of the molecule which is iron in this case we will have to find a similar molecule or segment present. So, for an example if you take this oxygen go through this I can find an oxygen. So, it is valid oxygen to oxygen valid oxygen to oxygen valid but there is also some molecular portion present between this oxygen this is this carbon carbon bond if you go through this you do not find any if you go through this you do not find it same thing. So, that means this molecule does not belong to octahedral point group. Now, the question comes what is the CN point group present over here. So, this molecule actually has a C3. So, where is that C3? So, to do that I am going to draw this molecule in this particular format the front one separately compared to the background one. So, you can see these are the different ways the molecule can be showed. So, the back one is actually the blue for each of the segment and it should be this kind of wedge bond. So, these are the ones. So, each of the baryanted ligand you can see there is two part of it one is the front part one is the backward part. For an example over that this particular segment this dot this line this line actually says that this molecule is actually on the plane of the paper this bond whereas this dotted wedge line says that it is below the plane. So, that means among these two this is above this is below same thing over here this is above this is below this is above this is below relative terms and over there they are connected through carbon-carbon system and over there you can see they are not finding any other segment on the other side. So, they are not really octahedral symmetry. But if I now rotate it over here just connecting the front parts for all this molecule along with the backward parts for all this molecule you can see you are seeing two triangles and that means that if I rotate this molecule through this iron 120 degree I am going to get all the backward oxygen exchange places and the front one exchange place. But the molecular structure will remain same as the original one superimposable and indistinguishable with this particular rotation and this relation is nothing but a C3. So, that is where the C3 exist in this particular molecule. Now the next question whether I have n number of C2 perpendicular to Cn that means do I have three C2's perpendicular to the C3 or not if I can find one the others will be there. So, for that again I am drawing this particular molecule which is the oxygen and then do the connection and this molecule you can see if you rotate through this carbon bound system over here this kind of the oxygen and over here if you rotate 180 degree over here or C2 this oxygen goes over here this oxygen goes over there and their connection will also pass on and these two oxygen states it is on their place iron is playing say remaining on the same place. So, with that we can say this is actually having a C2 present over there in the molecule and if you have found one C2 because it is perpendicular to C3 you are also going to find the other two in a similar position that means through this oxygen oxygen carbon carbon bond of a oxalate ligand. So, that is present over there. So, yes we have three C2 perpendicular to C3 next question do I have a sigma H sigma H there is the C3 will be on the perpendicular plane and you can see it is obviously not there because the front and back is not going to exchange. So, that is a no sigma H is the question is do you have three sigma D's that means the sigma D will be present in between the C2's and if you try to put a sigma plane over here you can see obviously it is not going to work. So, that is why so somewhere in between this C2 it is not going to work because oxygen will come on the other side there is nothing there. So, that is why it does not have also sigma D's. So, what I am looking with no sigma D no sigma H 3 C2 perpendicular to C3 only Cn. So, this molecule belongs to the point group of D3. So, that means this molecule belongs to point group of D3 and as we learned earlier this molecule will be chiral in nature. Now the question is what will be the twin ensumers for that I am going to draw this molecule one more time and what you need to do find out what is the front part of a bidented ligand what is the background part. So, this is I am putting as O 1 O 1 or I should say O A O A O B O B O C O C that is because of our understanding where the molecule goes after a particular rotation or not they are not like different oxygen atoms they are all very similar and equivalent. So, now what we do for each particular bidented ligand from the front to back we find out which side I am actually rotating and for that what I am going to do for this bidented ligand over here this is the front side this is the back side. So, front to back if I want to rotate I have to rotate in this particular direction. Similarly, for this part this is the front part this is the back part I have to rotate in this direction again for this one the C1 this is the front part this is the back part rotate this direction. So, you can see the rotation remain in the similar direction it is like this a left hand side rotation and this particular side rotation is called a lambda isomer that means what happens to the other portion of it that let us take a look into it. So, so far we have find how to find the different isomers of this kind of D3 molecule and what is the nomenclature is going to come we found that is actually from front to back for each set of bidented ligand every bidented ligand have a two sides front and back go from front to back front to back front to back you see the similar rotation on this direction that means it is a lambda isomer. Now, let us take a mirror image of it because although we know it is a D3 point group molecules it should be chiral but let us double check it. So, over there this octagonal geometry is going to remain same. So, first let us draw that and then try to fill the molecule. So, this mirror image so this particular portion over here will come closer to the mirror that is closer to the mirror in the beginning. So, that is there OB on the top OC on the top is going to remain as it is this OC over here is going to come on this side it will be the OB and then the connections are going to be like this. So, now you can take a look into it and try to figure it out how it is going to work and you can see front to back for this side is this one front to back front to back. So, you can see all of a sudden my rotational system is changed and this is known as a delta isomer. So, these are the two different isomers we can actually find in this particular set of the molecule. And now we are going to look into a different variants of this particular molecule that will be our number two examples where we are going to use glycinate as a ligand. So, what is glycinate ligand? It is nothing but coming from a natural amino acid glycine that will cover in the next few segments. So, this molecule glycine this is the structure CH2 and then they showed him carboxylate and ammonia. So, this molecule typically bind with this ammonia and carboxylate that means it can also act as a bidented ligand but over here a subtle difference not both the atoms are same one of the coordination atom is nitrogen one of them is oxygen. So, let us take a look into it. So, let us take similar iron complex and over here I am not mentioning any oxygen state or anything just looking into the coordination geometry at this moment. So, over here I can say that nitrogen and oxygen can be like that nitrogen and oxygen like that. So, these are oxygen sorry and there is a other nitrogen and oxygen. So, that is how it is and I just want to put is this one as oxygen this one as nitrogen. So, nitrogen in the front part oxygen in the backward part for each of this ligand. So, over here you can see also nitrogen and oxygen is going like this. So, it is actually a left hand rotation but before going to what is the nomenclature of this particular enantiomer first find out whether it is enantiomer or not for that we need to find the point group. So, again we ask the question are you linear answer is no are you belong to a cubic group let me state the head or octahedral the answer is no but over here be very careful because when we talk about this tetrahedral and octahedral we are talking about the symmetry not about the coordination geometry coordination geometry is still octahedral. But it is symmetry which is a more important thing in this particular aspect that is not perfectly octahedral because over here you can see all the coordination atoms are also not same you have 3 sets of nitrogen 3 sets of oxygen. So, that is why it is not octahedral molecule next question is what is your CN and for that if you look very carefully all the nitrogen over here they are all in the forefront. So, now if you put over there and rotate a 120 degree they are going to remain on its own position similar things are following with the oxygen remain on the same place. So, that is why with that respect we can say yes we actually contain a CN and that is C3 very much similar to the previous case. Now the question is do I still have the 3 C2's perpendicular to the C3 present over there or not previously we found that it was actually present in this particular direction. So, if you rotate the C2 oxygens are exchanging places but now because I have nitrogen versus oxygen so they will exchange places so that will not be symmetric. So, there is no C2 present in this molecule so this is answer is no. Sigma H which should be perpendicular to the plane of rotation and I also do not see it 3 number of sigma V's sigma D's that we also cannot see. So, with respect to that this molecule only has C3 nothing else. So, this belongs to a point group of C3. So, if you have a C3 molecule it is going to be kyder and that is why this molecule will be one of the enantiomer and as you see we are rotating front to back portion for this molecule nitrogen to oxygen nitrogen to oxygen nitrogen to oxygen is rotating this way left hand side. So, it is a lambda isomer then the question is what is the mirror image of that. So, let us draw that. So, if I am doing mirror image over here let us consider this particular structure the molecule is the original structure. As soon as it comes here this nitrogen will come on the other side this nitrogen remain as the same position this oxygen remains as it is and what we are left with one of the other oxygen this and over here these are the connections and from here we can say nitrogen to oxygen nitrogen to oxygen oxygen to oxygen. So, it is actually waiting other ways. So, it is here delta isomer. So, this is what is actually happening with the glycinate ligand. Glycinate versus oxalate the difference is because you have now two different atoms as coordination site for the glycinate it is actually going to a little bit lower symmetry. So, that is it is C3 on the other hand if you look back to the iron oxalate that belongs to a D3 point group. So, which is actually higher in symmetry you have three extra C2s perpendicular C3 which is absent in the case of glycinate. Now, the last portion we will discuss about this particular molecule you all heard of H2O2 hydrogen peroxide. So, what is the structure of hydrogen peroxide? So, as you know it has oxygen oxygen bond and two hydrogen hydrogen bonds but how that particular thing is actually oriented they are not planar. So, what is the structure? The structure of the H2O2 molecule was described as an open book structure. So, this is the base of the book and is open such a way that you can see both this particular side. So, this is the front side. So, this is how it looks like and over there the oxygen oxygen bond over the sheets over there for H2O2 one hydrogen in one of the page one hydrogen the other page. So, this is the actual structure of H2O2 and over here although you said earlier is open book structure now you understand why because on the border of these two planes oxygen oxygen bond states and OH bonds goes to each of the other planes. So, this is what is a open book and nowadays you understand better say like open laptop kind of structure. Now, what is the point group of this molecule? To figure it out what we do we try to look into through this oxygen oxygen bond the front oxygen I am drawing as a line a dot and there is the hydrogen these are H1 and say this is the H2O2. The backward oxygen I am showing you as a circle. So, this one is actually a little bit tilted so somewhere around here. So, this is the structure of the molecule when you are looking through this oxygen oxygen bond why I need that because now if I want to find out the point group I say whether the molecule is linear answer is no tetrahedral or octahedral or any cube group no CN do you have a CN. So, for that we are doing this particular set of the molecule and over here you can see this is the front is the back and over here if I rotate C2 or 180 degree what will happen front oxygen will go to back oxygen come to the front and this hydrogen will come over here this hydrogen will go over there there is a motion you are going to see. So, that is why we have a CN and that is C2. Next question you do you have 2 C2's for particular C2 and answer is no there is no perpendicular C2 if it is there it has to be somewhere there or somewhere there and there is nothing. Next question it is sigma H containing molecule or not sigma H molecule has to be somewhere perpendicular to it and you can clearly see it is not. 2 sigma V's does it present now sigma V are 2 planes which are going to bisect this molecule and the only plane we can see that is bisecting the molecule is none. So, there is no such sigma plane present in this particular molecule. So, neither sigma H nor sigma V. So, if this is the case then we say the point group of molecule is actually C2. So, that is the point group of the molecule. So, if you have a C2 then the molecule should be chiral in nature. So, H2O2 molecule should be chiral in nature but when you go and try to find H2O2 we found this molecule is actually not chiral at all. Now the question is why it is not chiral? So, for that what we are going to do draw the mirror image of this particular molecule. So, over here this is the front carbon this is the front oxygen back oxygen is the front hydrogen this is the back hydrogen. So, obviously they are not fitting on each other now if I want to go from this structure to this one how I should do and that is doable if there is a bond angle rotation. So, let me draw this open book structure one more time. So, this is where we started with and after the mirror image that is how it is now. Now to bring them in the similar way this oxygen-oxygen band has to rotate so that this one come over there this hydrogen goes on the top that is the structure. So, that means there needs to be a rotation this hydrogen comes over here and this hydrogen goes on there and that is possible with the oxygen-oxygen bond rotation and this oxygen-oxygen bond rotation happens because it actually requires very low amount of energy and this particular bond rotation or as we say fluxionality in the molecule ensures that this particular molecule of the enantiomer will look very similar to the original one because in the previous cases when you are taking the mirror image we are not allowing to do any particular change or fluxionality over here we are allowing to change them the bond angle bond length all those things and if this is actually happening we create a fluxionality and that actually kills the chirality because now the enantiomer is not going to be staying as it is it is going to change to the original structure and over here we say this fluxionality is actually the important factor which ensures that the chirality is not seen so that is actually killing the system. So, with that we would like to close over here so today we go through three different examples one is the iron oxalate one is the iron glycinate where they are D3 and C3 point group each other and we found both of them are chiral we also know the nomenclature delta n lambda isomer. Then we come to this curious case of hydrogen peroxide very common molecule and we find out that this is actually chiral molecule but due to this oxinoxane bond rotation this molecule lose its chirality because it goes back to this original position and with that we would like to conclude over here we will continue with respect to the other applications of cd spectroscopy in the next class along with the molecular origin of cd spectroscopy or circular dichroism spectroscopy. Thank you, thank you all.