 Hello everyone, welcome to the module on stereochemistry, we are still trying to discuss the arrangement of atoms in space and look at how does that give rise to different properties. In this lecture, we will go into a bit in detail about what are the symmetry elements of a molecule and how that dictates the chirality or the lack of it in a molecule. Before we get there, let us just quickly recap what we learnt in the last lecture. In the last lecture, we had seen different kinds of isomers that is constitutional and stereo isomers. By constitutional, we learnt that here to go from one form of isomer to another, one has to break a bond and rearrange the atoms and that would give rise to a constitutional isomers. That is the connectivity for a constitutional isomers are completely different. A classic example we had seen was between a cycloexin and few forms of hexene, one hexene, three hexene or others. On the other hand, if you come to the stereo isomers, here the main distinction is the connectivity remains the same but the way they are arranged in the space is what distinguishes the two isomers. And for this, we had looked at two classes. One is called as Cistan's isomers and we had looked at few examples of those and another is what are called as isomers based on a chiral centre or an asymmetric centre. So in this lecture, we are going to go into a bit more detail on that and see what are the different properties of that system. We had also learnt that there are two different kinds called as conformations and configurations. In conformations, you have different isomers which actually can rapidly or can be easily interconverted between one among one between the two without actually breaking the bond. However, in case of configurations, one has to break a bond to go from one configuration to the other configuration and in the example we saw where the two chiral forms of a particular molecule called two methyl butanol and we had also discussed in a bit detail about Cistan's isomers. So with this background, let us now get into looking at a bit more closely on stereo isomers which have a chiral centre or an asymmetric centre. So these are the two examples which we had looked at in the previous class. On the left hand side, you have a two methyl butanol derivative in which the methyl group is protruding out of the plane of the screen. On the right, you have the methyl actually going back and I am sure you are all aware of what is called as a CaN in-gold prilogue method of naming this or giving them the absolute configuration that is the R and S. So just to refresh your memories, let us just look at that. If you go by the CaN in-gold prilogue method, so then if you know we are looking at this particular carbon atom here and so then what you have is you are going to try and look at the priority of the carbon atoms. So we have one, this would be the first priority because that has the oxygen attached to it. Then you have the second priority is this which is having two carbon atoms and then the third priority would be to the methyl and you would also have something which is going back which is the hydrogen. So if I put it here like this, so in the CaN in-gold prilogue, we have the hydrogen always pointing away from us and that is indeed the case in this particular isomer. So if you now put it like this, we have one, we have two and three arranged in space. So then this would actually go as a clockwise and this would give rise to R enantiomer or R form and that is what we have here. Similarly, one can elucidate for the one on the right hand side and this we would call it a S enantiomer or the S configuration, correct. And this is what we typically saw as in a molecular model we saw as this where a central carbon atom is attached with four different units and this was also the definition or the proposition given by vanthoff to begin with. Now let us look at one particular statement and see if this particular definition that a carbon attached to four different substituents is chiral. Is this true or is this universal or are there any limitations on this particular statement? To understand that, let us go back and look at a very classic example of a system which contains two centers or two chiral centers. Here is an example of what is called as a tartaric acid. This was the first derivative which was studied by Pasture in 1850s around 1850s and what he observed is that there are two forms. One is called as the one form which rotates the plane of polarized light along one direction and that is why we call it as optically active because it rotates the plane of polarized light and similarly there is another form which actually rotates on the opposite direction. So these were the two forms which Pasture understood or Pasture isolated. However, there was also another form which was later realized that this was actually optically inactive. That means if you shine a polarized light on it, it would not absorb or it would not change the direction of the plane polarized light. So this was a bit puzzling. If you now look at this particular molecule which is shown here, in the light of what we just discussed above, what you indeed see is that it has what we call as a chiral centers. It has one and two chiral centers but still it is actually optically inactive. So how do we go about rationalizing or understanding this? And this is not just one particular molecule. There are actually many, many such systems where you could have chiral centers in the one, two positions which would be optically inactive and there are also different kinds of molecules called as allines which do not even contain a chiral center and they are still optically active. So how does one rationalize or understand this? So we need a much more broader perspective on this. In order to evolve a broader perspective, what we need to do is we need to go back and look at symmetry and chirality in relationship with each other. So far what we said was we just said that if you have a carbon atom attached to four different centers, then we called it as a chiral center or a molecule as chiral. However, when one has more than one carbon atoms or more than one chiral centers, then the whole system has to be considered in totality. So for that we have to actually understand what are the symmetry elements or a symmetry operations which are applicable in a molecule and how that leads to chirality or the lack of it. There are four major or four classes of symmetry operations which are possible on a molecule. One is called as a proper axis and proper axis of symmetry and other one is called a reflection or a symmetry plane and third is called the inversion center and the fourth is called as a improper axis. So now what we will do is we shall look at each one of them in a little more detail and try to understand them with a few examples and finally we will come back to what is the relationship between these elements and chirality of a given molecule or a system. Let us look in by looking at what is called as a proper axis of symmetry which is denoted by this C with underscore n and this is simply nothing but the axis along which if I rotate the molecule it should come back to the same position and the number of the turns which it would take is what is denoted as n here. It would become clear if I take you a simple example of a molecule of water. If I take a molecule of water which is shown here with the red oxygen and the two white as the hydrogens and if in order to get back to the same molecule or an indistinguishable molecule what I have to do is along this direction along the direction perpendicular to the oxygen I would rotate it by 180 degrees. I would go and rotate it by 180 degrees that would lead me to the same molecule again. And here the what n stands for is 2 because I need to rotate the molecule along this axis by 180 degrees to get back the same molecule. So if you put n is equal to 2 in the formula you would see that I need a 180 degree rotation and that is indeed what we did here. We went from a molecule like this we just rotated by 180 degrees to get the same molecule back. I hope this is gives you an idea of what I mean by proper axis of symmetry. To give you another simpler example so I have taken here an ammonia molecule n is 3 where the black ball represents the nitrogen and the green ones represent the hydrogen in this case. If I have to get back this molecule I need to do a 120 degree rotation that is this is the angle by which I need to rotate which is the 120 degrees. So if I rotate then I would get back the same molecule and I can keep on doing this to get the same same molecule. And if you now have divide the 360 degrees by 120 degrees that is angle needed to get back to an indistinguishable molecule you will now see that we have a C3 symmetry that is what is written here. Now let us go ahead and take a look at a slightly difficult but still a simple molecule which is called as an ethane which is something which we looked at in the first two lectures. So here I am showing you a staggered configuration of an ethane molecule where you see the 3 hydrogens which are actually coming towards you and a carbon and at the back we have one carbon and the 3 hydrogens correct. So I guess this would be again very apparent if you want to look at what are the symmetry elements present. So the one the one way to do is to hold this carbon and rotate the entire molecule by 120 degrees and see if I get a similar molecule back. So let us do that I am going to hold this carbon and I shall just rotate it till I get the get it back onto this position that is a 120. So you see that I have got exactly the same molecule back if I hold any one of them and rotate I would get back the same molecule. This is actually very similar to what you saw for the ammonia case that because we these three are in a predisposition along the tetrahedral of a carbon you will have a C3 symmetry in this case which is I guess is little more obvious. However what is not obvious is the C2 axis. So to understand the C2 axis let us do one thing let us now hold the molecule like this where what you see is that I hope you all can see that there is one particular plane which is going through this hydrogen the carbon the carbon and the hydrogen. So the plane which is actually going through these four atoms is one plane correct and if I have this particular plane what I can do is I am just going to hold this and I can come with an axis which is perpendicular to this and I will try to do a 180 degree rotation here. So around this if I do a 180 degree rotation I actually end up on the same molecule. I hope you see that there and I can just do that again. So the hydrogen is on the bottom is coming towards you and hydrogen at the back is coming towards me and if I do again a 180 degree rotation or a C2 you would see that the same molecule is retained that is the bottom hydrogen is again coming out of the plane and towards you and the one here at the top carbon is coming towards me. So this is what would give rise to a C2 symmetry and since I have I can create one plane I have one C2 symmetry which I am showing you currently but I can since there are I can actually similarly make three total planes where two hydrogens and two carbons are present like for example this is one one plane which I have currently now and what I can also do is I can also take these atoms which is again another plane and similarly you can also come up with another which is this. So it may not be very apparent now because all the substituents are exactly the same so it will be hard to see all the three planes and the consequent C2s but I hope you at least can see one of the planes and the C2 axis which is perpendicular to it or the orthogonal to it. Alright now let us go ahead and look at a slightly more symmetric or a complicated system. So here what we have is we have a benzene as the particular system and I guess you this is a little more easier to look at because the molecule is planar so you can actually just look at it in the plane of the board itself and here what you have is you have a different elements of symmetry one is you can do what is if you actually go orthogonal to the bold that is perpendicular to the bold have a I will just try and so if I draw the benzene again here assume it is symmetrical or the bond lengths are the same I can actually come up with an axis which is or the axis is perpendicular to the plane of the screen. So if I come up with this then what I can do is so if I rotate it by 60 degrees that is one carbon to 2 so then I would end up in the same molecule again right so the degree of rotation I am doing is 60 degrees so that means 360 by 60 that I have a C6 axis of symmetry in this case correct so that is one of the elements which is shown here that is the C6. Similarly I can also do a 120 degrees that is if I if I draw this so then I would do 120 degree rotation so if I now do a 120 degree rotation with the axis perpendicular to the plane of the benzene then I would again get the same molecule and if you now divide 360 by 120 you would get a C3 symmetry correct because even now the N is 3 so that is what we are looking at here is C3 so we understood hopefully C6 as well as C3 and now coming to the so these were actually where the axis was perpendicular to the screen of the or the plane of the benzene. So now let us come back and look at what is called as the C2 so in this particular case you can so you can try and do one thing which is now you are going to try and look at within the plane of the benzene so for that I will draw again once again so that it is a bit more clear so now what you can do is you can actually have a plane which is passing through this if you now along this plane which is within the plane of the benzene if you now flip it you would get back the same molecule right because the molecule is symmetric so now you get one C2 similarly I guess you can easily understand that I can get this C2 and I can also get this C2 so I already have 3 C2s now and so this is when you go from one atom to the other atom that is this to this or similarly to the other atoms now what happens if I go between the atoms so let us look at that in a so since the bond lengths are same although I am drawing double bonds here but this is actually alternating double bonds so the bond lengths are exactly the same so given that what I can do is I can actually go in between the two bonds as well so this is where I am actually going through the not going through atoms but I am going through bonds and this would also give me one C2 axis similarly I can have two more C2 accesses so what this tells you is that you can have symmetry elements of the symmetry operations going through not just bonds you can also have them going through atoms as well as through the bonds all right so now we looked at what is called as a proper access of symmetry and now let us go ahead and look at something what is called as a plane of symmetry or a reflection plane so in this what we will do is we will actually take the molecule and look at the reflection if there is any reflection in the molecule and on reflection does it give a same molecule or a different molecule so I understand that these symmetry operations which I am trying to show now are actually not that easy to understand if you do not have these kind of molecular models with you but what you can certainly do is you can go to this very nice website which is given here symmetry at the rate order order bind so this has very visual ways of looking at symmetry elements and symmetry operations in these simple molecules and even in some of the more complicated systems I think that would give you much more better feel of what these symmetry elements are so I would encourage you to go back and play around on this website. Coming back to symmetry plane so now let us again go back to a simple case which we looked at that is the molecule of water so if I take this molecule so is there any reflection plane or a plane which is which cuts the molecule equally equally so there are actually two symmetry planes here one if I take this molecule and if I cut it in this or if I put a plane across this I see that the atoms on either side are exactly the same that is if I put a plane here I have a hydrogen here and I have hydrogen on the other side so that means I have one plane of symmetry or one sigma plane here and what could be the other sigma plane here because this is a little obvious the one which I showed you is obvious so the other one is actually I could even cut through the atoms right I can actually in the plane of the molecule itself in the plane of water molecule if I cut or if I have a plane which is going through the atoms then that would also give me a sigma plane and so in totality I have two sigma planes for a simple water molecule I hope that is understandable and now let us look at the case of the ammonia which we had previously seen so if you now take this molecule again I am viewing it from this from the front on view I could have one sigma plane which is actually going through the carbon and one of the hydrogens and here what you see is that if I have a sigma plane going through it has both hydrogens on the other sides so that means it is a symmetric system so this would be one sigma plane and similarly I can have it going through the other two as well so in total in ammonia I would have three sigma planes and now coming to a more symmetric benzene molecule so you would have imagined or you would have already guessed that here also similar thing is possible so let us just look at that briefly so if I draw again and one more benzene for the to make it clear so I could have a plane which is actually we could have a sigma plane which is going through this and perpendicular to the perpendicular to the plane of the benzene and that would be one sigma plane and similarly I could draw two more here one and another one so I have now three sigma planes which are actually going perpendicular and through the atoms right and similarly as we did for the proper axis we can also do for the for the sigma plane that I can draw now between the atoms and through the bonds so yeah I can actually have one which is going through this and similarly between these two opposite bonds I can have one other going through and here again I have one more going through this so this is three sigmas through the bonds and three sigmas were through the atoms right and these are little more sort of intuitive whereas another one which people typically miss is that if I have a benzene I can also have it I can also slice through it or I can also have a plane going across the atoms and that would also give me a sigma so in total I will have seven that is one is the horizontal and the other six are vertical among the six three are going through the atoms and three are going through the bonds so you could actually play around with this you can draw the molecule on a sheet of paper or go through this website and play around and I am sure that will become a little more clearer to you as you go through that website now let us go ahead and try to look at what is called as an inversion symmetry so an inversion symmetry is one in which every atom should have a corresponding atom at exactly the opposite distance on the in the molecule so this might look a bit mathematical and abstract so to clarify that let us look at a case of the benzene again if we take the benzene and what we have here is that we have a center of inversion here and here I will draw the hydrogens just to be point out and now if I want to go from here to this carbon I am going to draw here in this in this direction and to find to see whether this is indeed the center of inversion I I need to come back equal distance and I indeed find another carbon similarly if I want to go towards an hydrogen I can go towards this hydrogen and to find out whether it is again a center of inversion I need to come back and find out I find another hydrogen here and I have an hydrogen here as well so you can do the same exercise for all different kinds of atoms present in the molecule and that would lead you to an in the center of inversion in benzene which is at the centroid of this of this hexagon and another let us look at another example which is this 1 2 dichloro dibromoethane so here I have a model of this 1 2 dichloro dibromoethane for you so what I have is the black ones are the carbons and the green ones are the chlorines and orange ones are the bromines and the white ones are the hydrogens correct so what I am trying to do is I am going to hold it like this let us I am going to hold it like this and let us say I have a center of inversion right in the middle of this bond and if I have that then let us go from here to the carbon atom or to the hydrogen atom I see that there is an I can find a hydrogen atom and if I actually go back from here to the on the other on the reverse direction I indeed see hydrogen atom here right so I have both hydrogens on this positive as well as a negative side similarly if you want to do for the chlorine you do see that from here going from center I can go and find the chlorine atom here and if I come back I do find a chlorine atom and the same thing holds for the bromine atom as well so this would tell you that the molecule is has a center of inversion which is again another important symmetry element and finally we will come to an important thing which is called as improper axis of symmetry which is nothing but rotation by 360 degrees by n which is the proper axis of symmetry which we saw previously followed by a reflection or a sigma plane which is perpendicular to the rotation axis let us say if this is my rotation axis in a molecule the sigma plane should be perpendicular to it that would define you an improper axis which is denoted by Sn let us look at a simple molecule which is the benzene again so here we know that we have a C6 that is I have a plane of benzene here and I know that I have a C6 which is perpendicular to this but I once I have the C6 in order to see whether I have an Sn or not I should put a reflection plane or I should put a sigma plane here and if I put a sigma plane which is cutting through the atoms of the benzene I indeed find that there is a sigma plane so that means I have a C6 followed by a sigma plane which is orthogonal to it so in the end I will have an S6 for benzene and similarly if you want to convince yourself of a slightly different atom what we can do is we can take this cyclohexane derivative with X being a substituent and what I will try and show you is that if I have this particular molecule I am going to rotate by 90 degrees okay so for sake of convenience let us number the atoms as 1, 2, 3, 4 and I am going to write down similar so I have my 1 here so the substituent X would be here 2, 3 and 4 so if I go to 2 the substituent is down and if I go to 3 the substituent is again up so I will write 3 here and if I come to 4 the substituent is down correct so now I have done a 90 degree rotation here so now what I will do is I will put a mirror plane here so if you put a mirror plane what you would end up and there is something like this so now this would become and this will come here so I hope you can see that the one which I have drawn here initially let us call this A and the one which we have got by operating a 90 degree see a 90 degree proper rotation followed by a mirror plane mirror plane or sigma and this is C4 I hope you can all see that and followed by sigma I have ended up with exactly the same molecule here that is on C1 I have the X which is pointing up and C2 the X is coming down and so on and so forth so in essence what I have is an S4 element of symmetry for this particular molecule so you must be wondering what is this got to do with chirality or why is he trying to tell us about symmetry rather than about chirality so the reason why I am trying to talk to you about chirality and the symmetry is that a molecule is said to be chiral not only when it has 4 different substituents but a molecule is said to be chiral in a more rigorous term if it has no SN or improper axis or improper elements of symmetry so only and only then a molecule is said to be chiral but please do note that a chiral molecule can have a proper axis of symmetry that does not forbid it from having any CNs but that is not a necessary condition it may or it may not have so if you just want to convince yourself what I would encourage you to do is to actually go back and try to test out these two hypothesis which I have shown you that S1 or improper axis with S1 improper axis of the first symmetry has it actually equal to plane of symmetry and S2 is nothing but an inversion symmetry so you can take a very simple molecule like ethane or the dichlorodibromoethane I showed you in the previous slide to convince yourself of these two these two aspects which actually embed both the plane of symmetry as well as inversion into the improper axis improper symmetry element as a result only one symmetry element that is improper is good enough to define whether a given molecule is chiral or not with this we shall stop here in the next lecture we shall go ahead and look at what are called as enantiomers and diastereomers thank you