 Hello and welcome to this segment of optical activity. My name is Arnabdutta and I am an assistant professor in department of chemistry IIT Bombay. Optical activity is mostly found for the chiral molecules. So the chiral molecules can rotate the plane of the polarization of linearly polarized light to a particular direction and this particular phenomena is known as the optical activity. Now the question is why does an chiral molecule rotates the plane polarized light? And before there what is a plane polarized or linearly polarized light? So we will come into that first. So when we talk about a light, the light generally travels in the form of a wave because it is an electromagnetic wave. So now this is how the light looks like when it is traveling if you are looking from the sideways. But if I try to look into that from the front over here how does the light would look like? The light will look like as follows whenever it is at the center it will look like this then it is going down. So it will be like this then it is coming back and going back to zero again then it is reaching the maximum and so on and so forth. So that is how it looks like. So if I put that all in the same place this will look like nothing but a line vibrating like this. So sometime it is going to be in this way, sometime is in the middle, sometime it is on the top. So if we all combine them together this is how we say this is a light is coming through. So this is what one particular plane. So when the white light actually comes through it can show this behavior at any possible direction. So this is known as the unpolarized light. However if we pass this unpolarized light through some special crystal which absorb all the different orientations of the light except the one. It will release the light which is vibrating in one particular plane and that is known as the plane polarized or linearly polarized light. So now this plane polarized light if it passes through a molecule which is chiral in nature. So say this is my plane polarized light where it started from. I passes through a sample where I have a chiral molecule and at the end when I am getting the final portion. So instead of being present in the original position we see it actually rotates. So this particular rotation if I put that as an alpha is known as optical rotation. So that is known as the rotation of the plane polarized light by a chiral molecule and this is the optical rotation. Now the question is how do we analyze this rotation, this optical rotation. So for that we have to go a little bit on details where we will see this plane polarized light we are seeing is nothing but it is a combination of two circularly polarized light. One switch rotates right hand side, one switch rotates on the left hand side. So what do I mean by circularly polarized light? Circularly polarized light means when I start I have my direction of the electrical field in this particular direction then it starts rotates on the right hand side and it continues to do so. So the same electrical field it is going to do this rotation at the same place. So instead of showing them in different places I can show them in one place. So what we can see that this light is rotating in the same direction. So it will be like this, this, this and it will come back to the original position. So that is is the rotation but the rotation is happening on the right hand side. So we will say it is the right circularly polarized light. On the other hand there will be another component which rotates on the left hand side like this, this, this and so on and so forth which can again shown over here in the form of like this. So now this one will be called the left hand circularly polarized light. Now this right hand circularly polarized light and left hand circularly polarized light they actually come together. So we call them in short RCP and this is called the LCP. They combine together and their combination or superposition is actually produces the linearly polarized light. So how to understand that? So over here on the top I am going to draw the RCP. Then below that I am going to draw the LCP and at the end over there I am going to show what is the superposition of those two systems. So now let us start over there. Both of these are in the same direction. So I should see a line like that if I super impose both of them. So this is more of like a vector addition of this RCP and LCP. Then RCP start moving on the right hand side and the LCP moves the same amount but in the opposite direction. So now if I combine these two how it will look like? It will be at the same position but with lower magnitude. Why? Because you can just imagine one is this way, one is that way we get a vector over here. So that is coming over here. The next one this is rotated at 90 degree. This one also rotated at 90 degree. Now these two if I combine them they will cancel each other out. So I should get nothing. Then it start moving again and now I will see one but it is on the opposite direction because they also start rotating on the opposite direction. So this is right hand, this is the left hand circular polarized light and then they can go to the opposite end after a full 180 degree rotation and this is what I get. So now if we see the superpositions we are getting for each position it is nothing but it is a plane polarized light in reality. So it is nothing but a plane polarized light which is getting combined from this right hand and left hand circular polarized light. So that is why it is said that the plane polarized light is a superposition of circular polarized light. By circular polarized light we mean the RCP the right hand circular polarized light and LCP the left hand circular polarized light. Now when a chiral molecule interacts with this plane polarized light it is actually interacting with both right hand circular polarized light and left hand circular polarized light. And chiral molecule because of the asymmetry present in the electronic environment that can detect the difference between RCP and LCP the right hand and left hand circular polarized light and that is actually reflected in the form of optical rotation. Let us connect those dots. So say I have my chiral molecule present over here and a plane polarized light is going through there. So we can see actually it is not a simple plane polarized light but it is actually a combination of right hand circular polarized light and left hand circular polarized light which are at the same phase. So that is why I am always getting the plane polarized light over here. Now what happens once they both goes through the sample one of them say the right hand circular polarized light is moving faster than the left hand circular polarized light. So previously they were in the same angle of rotation from the beginning. Now they are not. So now say it has moved say 175 degree to the right hand side in the left hand they rotated only 155 degree. So now what happens there so now what happens the right hand circular polarized light and left hand circular polarized light their correspondent system is going to rotate from the original position where they are present. So over here they are going to rotate on the right hand side because right hand circular polarized light is moving more compared to the left hand side. So coming back one more time previously the right hand circular polarized light and the left hand circular polarized light started moving in the same direction in the same magnitude. So that is why when you look into the resultant of there they are always in the same position. However if one of them started moving more compared to the other so say the right hand circular polarized light is moving more than the left hand circular polarized light what I am going to get is the resultant is now shifted and that shifts from the original position and that is nothing but the optical rotation. So this optical rotation happens because the chiral molecule allows one of the circular polarized light to move faster compared to the other. Now movement of light through a sample depends on the interaction of the light that means the electromagnetic wave the electrical field of it and the electrons present in the molecule itself the polarizability so they actually interacts and that interaction is given by the refractive index. So what we can say the refractive index is given by N of the right hand circular polarized light and the left hand circular polarized light is not the same and that is why the difference in the movement and that is why the resultant moves from the original position and we see an optical rotation. So that is why the optical rotation is coming and that is coming because of the difference in the left hand circular polarized light refractive index and right hand circular polarized light refractive index and optical rotation is a function of their difference and this particular phenomena is known as circular birefringence because it is creating a difference in refractive index so that is why birefringence and it is showing that on circularly polarized light so that is why circular birefringence. Now when we talk about this optical rotation and we understand why the optical rotation is happening now this alpha value the optical rotation will depend on how much concentration of the sample I have how much sample is actually allowed to interaction with the light and the property of the light itself because light is giving the electromagnetic field which is actually interacting. So over here all these factors are actually combined over here in this equation which says if we are taking optical rotation of a particular sample and say the measured optical rotation is lambda observed and if we divided that by the path length of the system. So when we shine this plane polarized light so this is the plane polarized light which is nothing but a combination of two circularly polarized light RCP and LCP right hand and left hand circular polarized light it is going through a sample where my chiral compound is so it depends on how much path length I am allowing to get this interaction between this light and this sample this is my chiral sample and that is given by this particular term L the path length so that is taken by care of this factor how much is allowed to interact with light. So what is the concentration that will also depend more the concentration of the sample more will be the number of molecule present over here and that will have better interaction with the light and that is taken care by this concentration term put over here. Now we measure this optical rotation at a particular condition so generally it is at 20 degree centigrade so that is why this 20 term is given over here and this D defined by sodium D line which is actually a line coming from a sodium lamp at a particular wavelength. So we measure that at a particular wavelength at a particular temperature and then if we have a observed optical rotation and I divided that by the path length and the concentration we get a value which will be a constant which is given over here and generally we write the temperature and the lambda or the light which light we are using over here. So this value is a particular constant value and for any particular chiral molecule it is can be a signature value. So now as we know any chiral molecule has two enantiomers which are just mirror image of each other so they will have alpha values exactly the same so it is x degree they are rotating but for those two enantiomers for one enantiomer it will be plus x the other one will be exactly same value but the opposite minus x degree. So that is why by taking a sample and measuring the optical rotation and if we know the concentration and path length we can figure it out what is the standard or specific optical rotation value and not only that if we have a mixture of the samples we can also find out what is the concentration of either of the enantiomer and we can find out the enantiomeric excess which is an important term when we discuss about producing an optically active or chiral compound and finding out their priority. So we can do this experiment to find out how much is actually present with respect to one enantiomer. So now we talk about this optical rotation and as we just defined the optical rotation defines that how much is the plane porous light rotates from its original position and it can be in the direction of the right hand side which is taken as the plus sign which is also defined as the D or dex rotatory over there look into that carefully the D is written in this particular way and over there the rotation on the left hand side is written as minus or L as levo rotatory levo means left hand side dextro means right hand side. So those are right written over here in the lower case system which defines that it is a actual rotation of the plane porous light when a particular chiral molecule is exposed with to the plane porous light. Now this plus or minus sign has nothing to do with the actual configuration of the molecule the actual configuration of the molecule means how the molecule is actually oriented in three dimension. So for an example the absolute configuration again means the spatial orientation of different groups present in the molecule and over here I am showing you two molecules one is the lactic acid this is the lactic acid present over here I am drawn that in this particular configuration where the carboxylic acid present over here OH group over here CH3 over here H over here and this is found to rotate the plane porous light in the right hand direction. So it is denoted as plus or D molecule over here so it is written as plus lactic acid. Now with the same configuration keeping all the specific orientation specific relative orientation same I just change the carboxylate group to ester group the methyl ester of it and this molecule change the direction of the optical rotation from plus to minus although they are actual absolute configuration in the same way. So that means it says which particular direction the molecule is going to rotate is not directly correlated with the absolute configuration. So just by doing an experiment and finding out that whether it is a plus or minus I cannot really comment on the actual structure of the molecule so that is why we need a different nomenclature to designate the overall configuration or the absolute configuration of the molecule rather than just the direction of the optical rotation because they are not really correlated. For defining an absolute configuration there is a nomenclature method which has been developed by these three scientists Kahn in gold and Prelog and we call them as CIP rule by taking the first letter of each of their names which is known as the CIP rule. So what is the CIP rule? So in CIP rule we are going to prioritize the different groups bound to a chiral center for an example a carbon center with four different groups bound to it. So what the first thing we are going to do we are going to prioritize all those four different groups and for that Kahn in gold and Prelog defined some rules. The first rule is that we are going to define them with respect to their atomic number. So for an example I have a molecule over here so say I have a molecule over here where I have a hydrogen, I have a OH group, I have a CH3 group and I have a bromide group. So all is present. So if I want to know if this particular structure and if I take the mirror image of this system which will be the other enantiomer of it. So the absolute configuration of these two molecules are actually not the same they are different absolute configuration but can I define actually the absolute configuration and nomenclature that. For that I need to find out what is the priority. The priority is done with respect to the atomic number and the atomic number higher the atomic number higher will be the priority that is the simple rule. So out of these four groups we can find the bromide is coming from bromine has atomic number of 35 OH the oxygen is directly connected to the carbon. So we look into the atoms which are directly coordinated to the carbon or the chiral center. So out of this O and A is the oxygen is connected so we will take oxygen which has the atomic number of 8 then it bonds the methyl group and the carbon is the directly connected. So that is why the carbon has atomic number of 6 and the end it is the hydrogen 1. So the priority will be 35 greater than 8 greater than 6 greater than 1 or the bromide greater than hydroxyl greater than the methyl greater than the hydrogen. If there are two differently groups connected to it and both of them has the similar atoms bound first we should go to the next atom. So what does it mean now say I have a molecule where I have a CH3 group and a CH2 OH group present over there. Now directly it is connected to both carbons and both carbons has atomic number of 6. So there is not possible to prioritize them in this position. So for that we will go to the next one. So the CH3 molecule so let me just draw this molecule a little bit further away is connected to 3 hydrogen whereas this molecule is connected to 2 hydrogens and 1 OH group. So over there you can see the next ones are all hydrogen over there next two are hydrogen and one OH group. So this there is a carbon-carbon oxygen. So this oxygen will get the preference. So we go to the next one. So this is the chiral center. So first one was unable to differentiate the prioritization the next one actually allowed to do that. So over here the OH group will be prioritized. So the CH2 OH group have the highest priority because it has atomic number of 8. Then we draw the molecule in tetrahedral form and at the end we are going to draw the molecule such a way that we are able to distinguish the difference. For an example let me take the example of that original molecule we have discussed about earlier. So say I am talking about this particular molecule whereas a carbon is connected to CH3 and OH a beer and H. So over there what I am going to do is first draw this molecule such a way that the lowest priority group always go to the back. So over there when you draw the lines that means they are on the plane this wage bond means it is above the plane and this dotted wage bond means it is below the plane. So out of these four groups hydrogen is always at the back. The rest of the group we will just draw as it is. So first the beer then the OH then say the CH3 and then we numbered the prioritized. So bromine is the number one priority higher atomic number then the OH the next one and the CH3 and hydrogen is forced so it is on the backside. Now this 1, 2, 3 how they are connected to each other you can see if I want to connect them by a line I have to rotate clockwise. So if I have to rotate clockwise then the nomenclature gives by R the R terms come from the rectus which is nothing but right in Greek. Now if I draw the mirror image of this molecule say the bromide the OH now comes on this side the CH3 on this side the hydrogen still remains on the back again the prioritization is 1, 2 and 3 as per their atomic numbers which is directly connected to the chiral subbrune center. So now you look 1, 2, 3 if I want to connect that I have to rotate anti clockwise or a left hand side direction. So that is going to nomenclature as S or sinister which means left in Greek. So this is the R this is the S. So again what we need to do to find the absolute configuration first find out all the groups connected to the chiral carbon with respect to the priority and the priority defines by the higher atomic number. If the atoms linked to the chiral center is the same move to the next atom until you find the difference then draw the molecule in the tetradal orientation like this put the number 4 or least priority group at the back and then connect the 1, 2 and 3s if it is a clockwise connection it is the R if it is the left hand connection it is the S. So by that we can define the absolute configuration of the molecule how they are all each connected to each other. Now going to the next section we also look into the confirmation of different molecule what is the confirmation? Confirmation means how a molecule is connected through different bonds and if I rotate only a single bond I go to a different orientation of the same molecule without changing the absolute configuration and that is known as the confirmation. So take an example we take ethane. So ethane is a molecule like this so over here ethane is such a molecule so this is the hydrogen over here this is a hydrogen over here this is a hydrogen over there on the backside there are the hydrogens on the other carbon. So now say if I look through the molecule through this carbon-carbon bond I will try to see this molecule in this way the dot over here shows the first carbon the circle on the back shows the second carbon and that is how the molecule looks like so they looks like this kind of trigonal structure if I look from the front and that is how it is looks like and on the back that is also look like a trigonal structure. So this is known as the Newman projection and now over here if I rotate this carbon-carbon bond without changing anything else so what I can achieve? So now say I am rotating it I am leaving the first carbon as it is and by rotating 800 and 20 degrees I will go to the same place so now say I rotate only 60 degree and I will rotate to this place where each of this hydrogen and each of this CH bond are actually almost on top of each other. So these are kind of we say eclipse to each other so that is known as the eclipse form whereas this one is known as the staggered form and what is the difference between these two molecules they are same molecule ethane C2H6 but the difference is how the each of the bonds are oriented with respect to this carbon-carbon bond rotation and depending on that I can have these two particular orientation one is called the staggered one is called the eclipse. Now if we try to find out what is the energy of those systems now you can see because of this eclipse interaction there is a possible steric hindrance coming over here because through the space the hydrogens are coming very close to each other and they can face some steric hindrance so that is why eclipse is going to be destabilized and staggered form is the most stable form the hydrogens are specially as far as possible you can get into this structure 60 degree orientation and then if I put this different orientation and their energy with respect to the angular rotation of this cc bond I am going to get this kind of graph so this is where the staggered form eclipse from I begin with so it is higher energy then I rotate this bond and I following that with respect to this red circled hydrogen the red colored hydrogens which is keeping the first one same the back one I am rotating so I rotate 60 degree and you can see now the energy is going down because they are going far away from each other so it is getting stabilized now I am writing further and this red one is writing further coming to close to this hydrogen over here and again it is facing a eclipse orientation so it is going to hide in energy and so on and so forth it goes back and forth so finding out this kind of energy with respect to different conformer while I am writing the cc rotation this is known as the conformation and analysis of a particular molecule so this is one of the example how we can find out different conformers and how we can predict their energies so over there you can see the eclipse form because of this close interaction steric interaction it has 3 kilo kcal per mole higher energy compared to the staggered one now coming back to the last part of our discussion on optical activity about the optical isomerism in transition metal complex so generally when you talk about optical activity or chiral complex we generally talk about the organic molecule where we have a carbon with four different groups but transition metal complexes can also show chirality and the most common example of that when we have a molecule in the form of a metal bound with xx ligand where xx defines a bidented ligand that means a ligand which can bind the molecule with two different sides and if three of them are present so all together it has six coordination and each of the ligand is giving two coordination each so the examples of such bidented ligands are given over here so one of them can be the biperiodine which are two pyridine molecules connected to each other like in this way where these two nitrogen can come and bind with the metal so it can give a bidented ligand so this is known as the biperiodine shot from bpy and generally we write it as this way which shows that it is a bidented ligand then comes ethylene diamine so which is nothing but two amine groups connected with ethylene so this can also bind metal which is also again a bidented ligand its short form is n which is shown over there and this is also a bidented ligand and we generally also show that two nitrogen binding with a connected form so it is another bidented ligand form then we can have acetyl acetonate where we have a strong resonance coming over here with negative charge and this is can bind with another metal so this is also a bidented ligand and we show them generally in this particular direction so it is again a bidented ligand then we can have oxalates which can be shown over here similar way bidented ligand and acetyl acetonate is short form is ACAC so these are the most common bidented ligand so it can bind a metal with two different sides so that is known as the bidented ligand. Now a metal binding to a bidented ligand can have this particular structure so over here when we have formed this bidented ligand this is now a chiral molecule if I draw the mirror image of this molecule this will be the mirror image of this molecule and we can see if we try to overlap these two molecules they are not indistinguishable and superimposable so that means they are actually optically active compound so that means we need a nomenclature so that we can distinguish between those two enantiomers so for that we have a definition coming as delta and lambda isomer how to do that so now each particular bidented ligand we can see there are two connections over here and one of them is actually towards the observer like us and one of them is on the backhand side so this is very straightforward for this particular group because this is wage bond above the page this is dotted bond below the page so that means this is towards you this is backhand side. Now this one over here this is on the plane of the paper this is back hand side that means relatively this is towards us this is on the backhand side and between these two this is on the plane of the paper this is above so this is again towards us this is on the backhand side now once we figure it out which one one is towards that and which one is backward among each of the set of the bidented ligand. Now you have to connect them, connect them from the top to bottom. So among these two, this is on top, this is at bottom. So if we connect them, we have to go this direction. Similarly, between these two, this is top, this is bottom, connect them, we have to go this direction. Similarly for this, we have to go from this to this direction. So over here you can see from the top to bottom if I go from each of the bidented ligand, I have to rotate right hand side altogether. So this is known as the delta isomer. When I have to rotate right for going from the top to the bottom part of the same bidented ligand, what happens to its mirror image? In the mirror image you can see if I take this one from top to bottom, I am moving left hand side. This is top, this is bottom. So again I am moving left hand side, again moving left hand side. So this is altogether I am moving opposite side and which is expected because this is the mirror image of each other. So this is known as the lambda isomer. So how to do that? Again look into each of the bidented ligand, find out which one is on the top, which one is the bottom, connect them and find out for all three of them and see which direction I am moving. If I am moving right hand rotation, this delta, if I am moving left hand side, it is lambda. So that is how we can distinguish between these two isomers present over here. So delta and lambda. So you can get an idea about this absolute configuration. What do we get for the organic compound through R as nomenclature? So in the summary, in this optical activity discussion we found how to define an optical rotation and why it is happening. We found that optical rotation is connected with circular birefringence where the circularly polarized lights are actually have different refractive indexes. So one of them move faster than the other and that is why we see a difference in the plane polarized light which is reflected by the optical rotation. We can find a specific optical rotation which is going to be a constant for a particular molecule measured at a particular temperature with a particular wavelength of light. We can also find out the absolute configuration of a molecule because which is not directly correlated with the angle of rotation or the direction of the rotation. The absolute nomenclature we can find by prioritizing the four different groups connected with the carbon which is a chiral in nature. And from there, using the CaN in gold pre-log rules or CIP rules, we can nomenclature them through R or S molecule. We also discussed the confirmation analysis which is nothing but confirmation of the different molecules which is nothing but a single bond rotation and we can find a different orientation of the molecule and we can find how they are behaving with respect to the energy. And if we plot their energy versus different conformation, we call them the confirmation analysis. As we have discussed with the example of Ethan, then we looked into the transitional complex whether they can be chiral or not and we discussed one of the examples when a metal showing an octahedral coordination symmetry but bound with three different bidented ligand and over there they can show two different enantiomers delta and lambda and that we define with respect to how we can connect the frontier side and the back side of the same bidented ligand. If it is a right hand rotation for each of them, it is delta. If it is a anticloar case rotation or left hand rotation, it is the lambda isomer. So with respect to that, we would like to conclude this section for the discussion about the optical activity. Thank you.