 Now take a molecular complex because at the end it is an inorganic chemistry. So, if we go to an inorganic molecule like this one, a metal in an octahedral coordination geometry that I have three carbonyl groups over there, carbon oxides and three bromide groups over here. Now the geometries octahedral but is the point group or symmetries octahedral. If you look into it, your answer will be no. This molecule is actually not octahedral symmetry because for octahedral symmetry all the ligands has to be same. Now all the ligands are not same. This is which is known as a meridonial isomer and if you try to find out what is the point group, you will find out this molecule also belongs to C2V point group. So, here you have that C2 axis. If you wrote it 180 degree, the COA and BR axial axis, these three atoms remain as it is. This bromine goes to 180 degree, this carbon goes to 180 degree and same and vice versa. So, this molecule remains same. So, there is a C2. Two sigma V's are there along with this BR, M, BR trans and CO, M, CO planes. So, these are the two sigma V's. So, this molecule is also C2V molecule. Now the question comes is this molecule is active or not? Again, what I need to do is actually look into the character table of C2V and what I typically try to find out whether X and RX can be activated together. The answer is no. Y and Ry no. Z and RZ no. So, this molecule cannot be optically active. So, C2V molecule it is not chiral, it cannot be optically active. Now say I have a molecule, another metal complex, this one which we have discussed earlier, a bidentate ligand, NN is nothing but say ethylene diamine. In short form it is generally written as N. So, it is a MN3 complex and we have discussed earlier this molecule what is the point group and you can figure it out. If you look into this particular phase I have a C3 present over here and there are three C2 perpendicular to it going through each of this bidentate ligand. So, it has a C3, it has three C2 perpendicular to C3 and that's why it belongs to point group of D3. Now is this molecule chiral or not? So, for that again just look into the character table of D3. So, this is the character table of D3 again these are all available online. So, you can just check it later. So, now over here try to find out if X, RX and Ry and RZ can be activated together. So, over here you can see X and Y and RX and Ry have been combined together. So, that means they are active in two different dimensions and that is what this term E means. That means it is just W degenerate or it can have two different dimensions. So, this means this can be activated together with this E symmetry you can activate X, RX, Y, Ry together. Similarly, you can see over here Z and RZ in the same symmetry A2. So, they can be activated together. So, in this case mu E and mu M can be activated together and your molecule can be optically active. So, that is how it is done. And then the most important part is that do I have to find out the point group and look into the character table of each molecule? No, because we already know there are only two point groups, two sets of point groups possible which can have X, RX, Y, Ry and Z, RZ in the same symmetry only two sets of point groups. One is the CN, one is the DN. And N can be starting from one to anything for DN you have to have at least N equal to 2 because you cannot have a D1 molecule because you have to have at least N equal to 2 C2 so that you can have other C2s perpendicular to it with only a C1 you cannot have another C1 perpendicular to it. So, that is why in a molecule this can be N greater than equal to 1 for DN it has to be greater than equal to 2 or at any integer value. So, these are the only two point groups which will be optically active. So, from now on if you want to find out whether a molecule is optically active or not, you do not always need to find out the mirror image and try to match it up or try to find out a SM axis or try to find out any sigma or i just simply find out the point group and see if the point group belongs to CN or DN. If it is then it is optically active if not it is not. And now you know why it is connected because it has to have this particular property that X, RX, Y, Ry, Z, RZ has to be activated together so that your mu E and mu M the dipole moment due to electrical field, dipole moment due to magnetic field can be activated together so that you can have both this motion happening together so that you can create a helical motion and once you have the helical motion you have the property to detect the difference between RCP and LCP and that is the principal reason that you can differentiate a optical active molecule. Okay, up to here any questions are required please feel free to ask. Don't think that it is going to be a very stupid question or something like that there is no question is stupid every question is good so if you have any question please go ahead and ask okay no responses I assume that everybody is understanding everything so now we go to the next part. So once we have the understanding the origin of a chirality now the question is a chirality we can define or we can monitor in two different ways one is through the optical rotation and the other one is the ellipticity. Now this optical rotation we already discussed about that we call that ORD optical rotatory distortion where we actually try to find out what is the change in optical rotation in a range of wavelength and from there if you remember correctly we have discussed about the positive cotton effect negative cotton effect and plane curve so the extent of optical rotation varies with respect to the wavelength and it is very much sensitive to it. On the other hand ellipticity we can figure it out by cd or circular dichroism data which is directly dependent on the optical absorbance because ellipticity is originated from the difference in the absorbance between the right hand circular lipolyslite and left hand circular lipolyslite and over there what happens that unless and until you have absorbance in the first place you cannot have a cd spectrum so that is why we are quite sure that where I should look for if it is cd spectrum although it is also possible that not all the peaks are actually chirally active we'll come into that in a little bit later what do I mean by that. Now you have two options either you can run ORD where you actually find out the change of the optical rotation with respect to lambda where phi is a function of nl minus n bar or you can have a data of psi versus lambda where psi is a function of al minus ar and the positive and negative is just showing whether it is the value of al is higher or ar is higher similar to nl or nr in that ORD so this is the ORD data that is how you're going to look into this is the cd data you're going to look into now the question is which of them is actually will be more reliable data or much more easier data to connect the data with respect to its structure. Now mostly we look into optical rotation so far but that is actually not the best way to practically monitor an optically active compound because first of all you have to find out how this phi differentiates with respect to lambda so how it actually going with respect to that so something like this or it's exactly the opposite enantiomers something like that so you have to find it out now generally when you look into a molecule first we're probably going to do the optical spectrum first and with optical spectrum we know that wherever you have a maxima you are going to get a crossing point a zero crossing point because of the cotton effect and that is why you are going to see a huge change over here with respect to the optical rotation and the wavelength so that means if you want to pinpoint at which wavelength you should measure your optical rotation that is very tricky because even with a very small change you can have a huge difference so over here the change is very sensitive and you have to first figure it out exactly where it is happening you have to optimize it so it is an extra set of experiment you have to do you have to do it every and each and wavelength and then figure it out or you can find out the best possible condition is somewhere where no absorbance is there in a plane craft situation but in the beginning you cannot just comment on that because you don't know exactly how it is happening so that is why if you want to do a wordy data you have to measure the optical rotation at each and every wavelength and you don't know at which range you will get the best possible result because at the condition where it is absorbing you can have a huge difference which is very sensitive and it may happen at a wavelength where there is no absorbance so you have no heads up call from the molecular or any optical data to exactly where to look into so that is why wordy data is not that easy to find out once you get it then you can use it but getting to a wordy data of a new molecule is very much challenging and secondly you are looking into a difference between nl and nr circular birefringence the change in the refractive index of the left hand rotary and right hand rotary circularly porous light now as we know this refractive index is very much sensitive in a lot of things what is the temperature what is the density of the solution what is the viscosity of the solution all these different parameters comes and can affect the refractive index so that means you imagine if you want to find out a nl and nr difference with respect to wordy how many different things you have to optimize first the final example over here I am doing the same molecule at one molar concentration somewhere else in other part of the planet someone is doing at five molar concentration we can end up having different effect of the wordy because the density is changing and it is have going to have a direct effect on the refractive index changes so that is why it is also sensitive on the which is known as the physical chemical conditions so what do I mean by physical physical conditions that means the temperature that means the pH if I change the pH there is a slight change in the overall proteic environment around the molecule it changes the overall arrangement of ionic molecule around it change the hydrogen bonding environment that is going to have a direct effect on the nl and nr so that is why wordy will be dependent on all of them on the other hand if you look into the cd cd spectra is directly dependent on the absorbance now the absorbance is also dependent on temperature if you take the absorbance data at different temperature you might see some changes especially at lower temperature you will see the absorbance band is much more narrow at higher temperature is much more broad but to see such extent of difference you have to change the temperature at a much more larger range say from a liquid nitrogen temperature 77 Kelvin to room temperature then you're probably going to see a very good change if you're changing it from 298 Kelvin to 350 versus 273 you are not probably going to get a huge difference so absorbance all that is temperature dependent its sensitivity is pretty low and obviously the cd spectra we are looking into the difference between al and ar if absorbance itself is a property which is less sensitive the difference of al and ar will be also less sensitive so that is why these are much more less sensitive sample so as we just discussed a word will be very much dependent on the physicochemical condition absorbance is not so the cd spectra is actually going to have a less amount of effect of the physicochemical conditions on the result additionally for a cd spectra you exactly know what to look into how because first you are going to measure a optical spectra and from the optical spectra you exactly know where should we look into because unless you have optical spectra you cannot have a cd spectra so if you take a measurement of the optical spectra of a molecule first you find there are four peaks only that region you're going to look for the cd spectra if you are seeing any difference that means it is chiral if you're not seeing any difference it is not chiral a chiral molecule as simple as that so that is why for these two important reasons cd spectroscopy is actually a more reliable and practical experiment that can be used to find out the chirality of a molecule and that is what is actually used by white