 Hello and welcome to this new segment of CD spectroscopy and MOSBUS spectroscopy for chemist. My name is Arnabdutta and I am an associate professor in the department of chemistry IIT Bombay. So, over the last few classes we are discussing about symmetry and how it is an integral part of the nature because one particular portion of the symmetry is chirality and nature uses chirality as one of the tools for molecular recognition and over here when you are talking about molecular recognition this chiral molecules in the forms of proteins in the forms of carbohydrates and nucleic acids they all become important because all these molecules which are actually an integral part of the biological segment is chiral in nature. For an example proteins when you talk about proteins proteins are made out of amino acids which are the building blocks for it and these are mostly L amino acids in nature whereas other hand when you talk about carbohydrate they are D carbohydrate in nature. So, these D and L are talking about their absolute configuration in the three-dimensional space on the other hand nucleic acids itself are not chiral but when they form this helical structure or the other secondary structure like beta sheet and all the systems the secondary structures are all chiral in nature and when biological system are talking among each other or actually interacting with other molecules this chiral environment becomes the key and that actually signifies which particular amino acids which is going to interact which particular carbohydrate is going to interact and how it recognize them. So, that particular system the molecular recognition and one of the important factor of that is this chiral environment and that is why we are interested to know about the symmetry and chirality and then when you are talking about chirality we found we can define it in the form of symmetry elements and also point groups which we have derived from mathematics and over there we found there are five symmetry elements can be found which can be E or identity operator C n rotational axis sigma reflection plane which can be sigma H sigma V or sigma D depending on its interaction its positioning with respect to the principal axis C n and improper acceleration S n and center of inversion I and among them all those things can be present in a molecule this particular symmetry element S n or improper axis of rotation is the key to find out whether my molecule will be chiral or not and over there we found that when you are talking about improper axis of rotation why it is important because by the definition of chirality or a chiral molecule or chiral object we say the mirror image is not super imposable or indistinguishable. So, over there what does it mean that I am doing a mirror image that means I am doing a sigma operation and then I am doing some rotation to find it out whether it is matching or not. So, I am doing sigma and C n together which is nothing but give me a S n axis of system and that is why we make it much more simpler when S n is equal to S 1 that means when n equal to 1 that is nothing but a sigma because I am doing a C 1 rotation which is nothing but an identity operator or living the object as it is. So, in S 1 is nothing but sigma. So, if you have a sigma you cannot have chirality if you are S 2 when you are talking about S n equal to S 2 it can be written as I center of inversion. So, if you have a center of inversion or S 2 you cannot be chiral if you have any other S n where n is greater than 2 even then the molecule cannot be chiral. So, that is the mathematical interpretation of chirality when a molecule can be chiral and if I try to bring that in the point group we have the C n and D n point groups where we found this particular point group does not contain any S n axis. So, they can be chiral in nature. So, in simplistic form to find out a molecule or an object whether it is chiral or not we have to simply just find out what is the point group of it if it belongs to C n or D n it will be chiral in nature. And then we try to find out which particular spectroscopic technique we can perform to find out whether the molecule is chiral or not and we found out there are multiple options and among them CD spectroscopy is one of the leading one we also have optical rotation along with it but CD spectroscopy is a crucial one where we found that we can observe its optical absorbance and if you have the optical absorbance band like this and if it the molecule is chiral and this particular band is chiral you are going to see some portion of it the chiral or it can be the opposite direction to which will define two different enantiomers. So, if say enantiomer A and enantiomer B actually belongs to the same molecule that means they are enantiomer of each other these two will show a CD spectroscopic curve where the x axis is the wavelength and y axis is the delta epsilon mole inverse centimeter inverse or we can also put that with ellipticity which is given in milli degree unit and we can find yes a molecule is chiral or not by CD spectroscopy if you get a signature in the CD spectroscopy then you can say yes my molecule is optically active and very interestingly this chiral molecule if you are seeing in a CD spectra that CD spectra should always have a precursor optical spectra unless you have optical active band you cannot have a CD active band. So, that is why first we perform an optical spectroscopy experiment figure it out which are the bands we are looking into and then if your molecule is chiral then you will find the CD active band among all this available optical active bands. So, that it is a thing we found and that is why CD spectroscopy becomes one of the important tools for finding out whether my molecule is chirality or not. Next we look into another system which is nothing but applications of CD spectroscopy. Next we look into applications of CD spectroscopy and over there so far we have covered three different examples. The first one we covered it is actually a cobalt based molecule which is bind with a peptide A, C, D, L, P, C, G and on the last class we have covered that or we found this molecule in presence of oxygen it goes sequential oxidation. First there is an oxidation of the thiolate ligand because over here when you talk about this system it is found that it is being bound with 2 nitrogen and 2 sulfurs and then the first molecule of oxygen reacts with it in the cobalt 2 state. The cobalt does not change its oxidation state the sulfur one of the sulfur becomes sulfenate or H O 2 and then further reaction with oxygen then the cobalt becomes cobalt 3 and along with it this sulfonate group comes into the plane of the 2 nitrogens and it goes to a ligand or solvent with the axi-ligation and it goes to a octahedral geometry. So, over there the first is oxidation of the thiolate and the second one is oxidation of cobalt 2 to cobalt 3 and that we have found over here and this particular sequential oxidation we found with respect to CD spectroscopy because CD spectra give us the first hint that there is a change in the molecular primary structure because all those things are actually part of this peptide which is chiral in nature and as I am going through sulfate to sulfonate sulfate to sulfonate that actually triggers a change CD spectroscopy can monitor that and then over here the cobalt 2 becomes cobalt 3 and it goes to an octahedral geometry and again CD spectroscopy can configure that because the presence of this chiral ligand scaffold and its interaction with the central metal. Over here I want to mention one more time the metal itself is not chiral in nature the metal active site however the presence of chiral ligand around it induces optical activity to this metal center especially metal centered optical absorbance band like D transition band or LMCT band. So those becomes chiral reactive the second example we figured out another sample with a cobalt saline ligand where this is actually bind with a nitrogen and oxygen molecule and over there we found this molecule is actually bound with the carboxylate ion and this is actually coordinating the cobalt especially in the cobalt 3 option state and stop it from binding to a another protons to produce hydrogen and this can only occur if I acidify it enough and the pH goes to below 3 only then this molecule actually opens up because nitrogen and oxygen remains as it is but this carboxylic acid group becomes COH so it does not bind and at the same time this protons are actually also acts as a proton release a proton can directly go and bind with the cobalt and it can trigger the proton to hydrogen transformation or hydrogen evolution reaction in short form we call them HGR. So that is possible to happen and we found it is happening over there because this is getting protonated and ligate off from this cobalt and again CD spectroscopy is actually allowed us to follow that when you record the CD spectroscopy at different PHS we found when I am crossing this barrier of pH 3 to lower than that we get a significant and visible change in the CD spectra which signifies there is a change in the primary coordination sphere of the metal because this particular band is an optically active band in the visible region not coming from the ligand itself but belongs to the metal center and that becomes CD active not only that it shows as a change this alteration allows us to follow what is happening in the system. Then the last one we followed with G quadruplex which actually shows that the four guanidine can form a nice planar structure which typically remain in the bottom of the chromosome in the telomeres region or telomere region and this particular G quadruplex can form in different forms parallel, antiparallel and hybrid we make sure of these two and again CD spectroscopy can tell us how this molecule is going to all these different kinds of possible orientation and this has particular signature and we can easily follow that to find out whether the molecule the G quadruplex is forming in the parallel, antiparallel and hybrid mode and also it will also allow us to find out whether I have a G quadruplex or not. Then today we are going to follow an another segment about amino acid resimerization. So, when we talk about an amino acid we already discussed that this molecule is tetrahedral in nature around the alpha carbon or the chiral carbon where if you put the hydrogen at the back carboxylic acid over here the R group over here R is the changeable group of an amino acid and this is the amine group over here we find out how it is oriented the carboxylic group R group and N group which is known as the corn rule and you can see it is oriented left hand wise. So, this is the L amino acid and there is also possibility of presence of another molecule where now I am putting R over here amine over here now you can see the corn rule still applies here but now I am moving in the opposite direction right hand side direction. So, this is the D amino acid. So, you have discussed earlier L amino acid is the most dominant form found in the nature when nature produces amino acid it is L amino acid in nature and this particular orientation why it is preferred there are multiple versions of that why it is happening and we have discussed even the possibility of an extraterrestrial material is actually influencing that which actually triggers the formation of amino acid of a particular geometry or chirality is the L amino acid. Now the question is what happens if I leave the L amino acid for a long time in presence of water it slowly converts to D amino acid. So, I should put it this way. So, it is reversible but it slowly convert to D amino acid. So, once what it is L amino acid it can be D amino acid in a while. Now I am going to give you several example of amino acid and it will give you a time scale in the terms of half life which gives us an idea how much time it requires to produce an L amino acid to D amino acid. So, if you start with 1 gram of L amino acid it slowly start producing D amino acid and at 1.2 time they will find only 0.5 gram of L amino acid is actually present the rest of them become D amino acid. So, if you follow this reaction it will be a faster reaction in nature and from the faster reaction you can find out the half life how much time it requires to get half of the amino acid to resimerize. And if we look into different amino acid we will take histidine as an example and the half life for L histidine to convert it to DMD histidine how much time it takes it takes 5500 years. So, it is a huge long time. Then if I take tyrosine the time scale is almost 8600 years it is longer. If I take alanine it is a polar amino acid it has a half life period of 32,000 years very long time. And then valine an amino acid with particular hydrocarbon chain and it is one of the slowest to convert to D amino acid at 180,000 years. So, you can see it is actually quite slow in nature. Now, the first question is why there is a difference over here between these two versus this hydrocarbon one a side chain and secondly how I can follow it up how I can follow this resimulation is happening. First we try to understand why the resimulation happens and how it is happening. So, for that we will start with the amino acid structure. So, again I am drawing an L amino acid this is the structure we draw but in reality in presence of water what happens they actually stay in their most stable form the carboxylic acid group becomes carboxylate because a neutral water around P7 the carboxylate is the most stable form. So, our group similarity is the nitrogen the amine groups become in is 3 plus because that is the most stable form of any primary amine group in water at P7. So, you can see at one side of the system is a positive ammonium group one side is negative carboxylate group and this particular form is known as zwitter ionic form. This zwitter ionic form is the most prevailing form for any amino acid present in nature and neutral condition and this is where all the reaction began when it start resimizing. So, the next thing happens this particular molecule react with some acid and this actually form the protonation form. So, this particular group gets protonated carboxylate the rest of it remain as it is. So, only the protonation of the most charge group over there it is a carboxylate it becomes carboxylic acid. Then comes the next step this proton over here CH bond holds out and create a CC double bond and if you want to create a CC double bond over here this particular group has to move and becomes O minus which actually takes this particular proton and if this is actually happening then what we are going to see is the following of a tautomeric form. Now this any C plus group come into the plane of the molecule. So, that is what is actually happening when this tautomizer happens and over here you can see now my alpha carbon was there previously it was tetral in nature. So, it is holding its chirality but now what happens it becomes a part of a CC double bond. Once it form the part of a CC double bond it becomes planar in nature and once it form a planar structure then it loses its chirality. So, it is not chiral because this actually contain a plane of symmetry. So, now what happens this becomes achiral and from here the molecule wants to turn back because this is not a very stable form. However, this is a possible structure it can go through as a tautomeric structure and how it is going to turn back it is going to return back the electrons over there this will come over here and this becomes carbon ion and this carbon ion is going to form over here and this carbon ion is going to abstract the proton and form the amino acid back and because it is a carbon ion derived from this C double bond C planar structure it can go either direction and there it forms either of the amino acid. So, this is one side and there is also possibility of the other side. So, this becomes NH3 plus this becomes R. So, you can see over here carboxylic R amine. So, this is the L amino acid and over here carboxylic R H. So, this is D amino acid and later on you can lose this extra proton over here and from carboxylate R group NH3 plus and H or carboxylate NH2 let us put R group over here NH3 plus and hydrogen and this becomes the D amino acid over here. So, that is how starting material and L amino acid has to go through a planar structure where it can lose its chirality and from this planar structure it can again regain the chirality, but over here if it is happening without any chiral environment in nature itself, but without any chiral environment it would produce both of them 50-50 and at one point time they will resume. So, that means I will see both the amino acid D and L present in the similar structure. So, now the question is why histidine and tyrosine are better? So, for that I am going to draw the side chains of them. So, side chains of histidine is CH2, 5-member ring. So, that is how the structure look like for histidine or tyrosine there is a structure. So, now over here you can see histidine and tyrosine this is actually a imidazole ring this is actually phenol ring. So, both of them are aromatic in nature. So, they actually have a natural drift of electrons towards them. So, through this electronic movement that the electron is going to come towards this electron dense aromatic rings. So, they can stabilize extra electron and this carbon if you remember it is bind with the alpha carbon which is actually forming the carbon ion. So, this planar structure then it goes to this carbon ion intermediate this carbon ion intermediate is actually getting stabilized by this aromatic ring and that is actually one of the factor which stabilizes the carbon ion intermediate with this particular group. And if it is stabilized more that means as we saw in the previous structure the possibility of coming from this part to this particular carbon and intermediate is high. So, you will end up at a higher rate in the resimulation region and that is why it is happening for histidine and tyrosine and one more factor also come into the picture that is due to the rapid proton relay. As you can see during this particular carbon and formation from the original structure there is exchange of protons between the system with the broad peak which is actually present in the vicinity. And over here this particular tyrosine, phenol or imidazole they actually have some groups which can exchange proton very fast. So, this actually create a rapid proton relay. So, when this carbon ion is forming a lot of proton has to be exchanged from this place to that place and this particular groups provided the source of the proton which is very near to it. So, that is why that is also another addition to this whole process. So, it is stabilized by the aromatic ring through minus i effect and rapid proton relay both of them affect the first formation of the carbon ion intermediate which is reflected into the fast resimulation compared to the other groups which do not have them. And that is why we find that the histidine and tyrosine is moving very fast whereas alanine and valine is pretty slow because they do not have a group which can stabilize the carbon ion nor they have protons to exchange its fast. Now, the question is how do I find it? How do I monitor that my L amino acid is converting to D amino acid? So, first of all we have to find out that when we start this reaction we have 100 percent L and 0 percent D and at the end of the reaction when the reaction will be end will be ending of 50 percent L 50 percent D. So, that is we are going to finally get if we allow this reaction to go for thousands of years. So, what we actually do we follow this reaction with respect to D and L ratio in the beginning of the reaction there is actually no D amino acid at all. So, it should be 0 and at the final stage because there is 50 percent L 50 percent D D by L ratio would be 1. So, that is what we are actually going from DL ratio 0 to DL ratio 1. So, how do you monitor that we can use again CD spectroscopy where we can follow CD spectroscopy of the amino acids where we start say this is the L amino acid and this is the D amino acid. So, when we start we will see only L amino acid and slowly with respect to time I start seeing this is actually going down and at one point time it will be 0 there will be no signature because D and L will remain in the similar concentration. So, that is I am going to follow. So, what I can follow is that particular wavelength over here what was the delta Epsilon and how it is changing with respect to time at this particular wavelength and with respect to time. So, this should slowly come down and that we can measure and extrapolate it to figure it out what will be the half life of this reaction and that is how it is being measured of 1000 years and more than 1000 years nobody is doing the experiment for 1000 years running. So, it is a extrapolation of the result and what is the advantage. So, as we know this L amino acid is solely going to maximize to a D amino acid. So, that means if I am doing this reaction even for a fossil the old stage sample I can look into what is the D and L comparison what is the absolute delta Epsilon value I know the 100 percent L 100 percent D value. So, from there I can back calculate how long the system was present in a biological form where L amino acid will be preserved 100 percent. Once it is died it is out of this biological chiral environment only then it start to racemize and go to 50 percent L 50 percent D. So, from there we can actually extrapolate that. So, that is why series spectroscopy can also be used in studies of old samples paleontology and all those things to figure it out what is the age of different fossils. So, with that we will like to conclude our segment over here where we found series spectroscopy can be useful tool to find out the racemization rate of L and D amino acid in HL phenomena. And by that we can follow the age and the reaction time of different old fossil fuel samples. Thank you, thank you very much.