 I shall welcome you all once again to MSB lecture series on Interpretive Spectroscopy. In my last lecture, I started talking about UV visible spectroscopy. So let me continue from where I had stopped in my previous lecture. We started talking about electronic transitions. Of course, electronic transitions can be broadly classified into four types, sigma, sigma star, n to sigma star, n to pi star and pi to pi star. And sigma to sigma star are shown by saturated hydrocarbons in which all valential electrons are involved in the formation of sigma bond. That means no electrons are left unutilized for bond formation in the valential. So these transitions require very high energy and occur at lesser wavelengths, less than 200 nanometer and fall in vacuum UV region. And if we consider n to sigma star transition, compounds having non-bonding electrons on hetero atoms such as oxygen, nitrogen, sulfur or halogens can show such type of transitions. The energy required for these transitions decrease with the decrease in electron activity of hetero atom and therefore the wavelength of absorption increases. And then the third type is n to pi star. Compounds having double bond between hetero atoms, for example, C double bond O, C double bond ES, n double bond Y, etcetera show these kind of n to pi star transitions. These transitions requires only small amount of energy and takes place within the range of ordinary ultraviolet region. However, the intensity of absorption is generally very low and these transitions have an epsilon value of less than 10 raised to 4. Of course here when you look into overall the energy difference between these levels n to pi star is the least energetic. That is the reason it requires very low energy for the promotion of the electrons. Pi pi star compounds having double bonds show these transitions. The intensity of transition is very high for these transitions. These transitions have epsilon value in the range of 10 to the power of 4 or more and they are also allowed transitions. All these electronics spectroscopy if you look into it, all these compounds having different region of usable range. So that means definitely we have substituent effects. So what are those? A functional group capable of having a characteristic electron transition is called a chromophore. We should remember the chromophore is responsible for electronic transition. The attachment of a substituent group other than hydrogen can shift the energy of the transition. So that means substituents that increase the intensity and often wavelength of an absorption are called axochromes. And then substituents may have any one of the four effects on chromophore. What is that one? We call it as bathochromic shift that is also known as red shift. What happens in this one? Shift to longer wavelength. When it goes to longer wavelength means probably we are thinking of decreasing the gap between HOMO and LUMO. So then it requires low energy that is the reason it is called bathochromic shift or red shift or shifting to longer wavelength or it is a low energy transition. Then the second one is hypsochromic shift that is also called blue shift that means shifting to shorter wavelength and higher energy. That means when the gap increases we need to give more energy that is the reason it is called blue shift and also shift to shorter wavelength and also high energy transition. What is hypochromic effect? An increase in the intensity of the transition then hypochromic effect is a decrease in intensity. So these four terms we come across and these four comes as a consequence of the influence of substituent on chromophore. You can see here in this plot you can see if you say here it is a 200 nanometer and here is 700 nanometer and this is epsilon value. You can see hypochromic and hypochromic this is hyperchromic and this is hypochromic and this is bathochromic and hypsochromic. The moment you see the shift from left to right it has to be hypsochromic blue shift and right to left means it is bathochromic red shift and as the intensity is going high it is called hyperchromic and intensity is dropping it is called hypochromic. So these are the influence of substituents on chromophore which is responsible for electronic transition. So now here I have given a list of chromophores and with a specific example in each case and also the solvent in which we have looked into the transition and the lambda maximum and epsilon and also the type of transition very useful table this is. For example if you look into alkene we know that we have a double bond so immediately we can say it should be a pi pi transition and once it is a pi pi transition you can anticipate little higher epsilon value this is in the 13000 and lambda maximum is 177 nanometer the solvent consider is n heptane. Then if you take alkene we have a triple bond and again solvent user is n heptane it can show lambda maximum in the range of 178 to 1, 225 and then here epsilon value varies 10000, 2000 or 160 and here it is pi pi star transition other transitions can also be seen. And then if you consider carbonyl carbonyl group make aldehyde acid or a ketone. So in this case for example if you consider aldehyde n hexane is there and here you can anticipate 186 as well as 280 and for 186 epsilon value is little higher compared to 280 it is low and then for this acetone n hexane 180 and 293 and in case of 180 large epsilon value will be there in 293 it is low and here in the large one is due to n to sigma star and the lower one is due to n to pi star. We know that n to pi star is of weak intensity and same thing here whenever weak intensity is there you can conclude that it is not n to sigma star it is not even pi to pi star it is n to pi star. And then carboxylic group if you consider acetic acid an example recorded in the ethanol it shows 204 and then 41 so immediately when you look into the epsilon value one can conclude that certain it is due to n to pi star transition. And in case of amide group for example if you take acetylamide water is taken as a solvent because soluble in water this is 214 and then 60 again same thing it is n to pi star and then with azo compounds are there n double bond in ethanol and 339 and then 5 very low so again you can conclude that it is n to pi star. In case of nitro iso octane is a solvent taken and values 280 and epsilon value is again low 22 so this also again due to n pi star transition. If you take nitroso group recorded in ethyl ether 300 base or 100 where in case of 665 20 it is again n to pi star and then in case nitrate it is dioxane is the solvent chosen and then lambda maximum is 270 and then epsilon is very low again. So what we can see clearly is pi to pi star has higher epsilon value and n to sigma star also has reasonably higher epsilon value but wherever the epsilon absorptive coefficient is very low without any hesitation we can tell this is due to n to pi star transition. This gives some idea about the different type of chromo force and their characteristic lambda maximum and epsilon maximum and also the type of transition we come across. Then looking to substituent effect let us look into solvent effect. When we talk about the polarity comes into the picture increasing the polarity of the solvents shifts n to pi star transition to shorter wavelength that means addition of polar solvent results in blue shift n to pi star goes to shorter wavelength or higher energy that means gap is increasing. So this blue shift is mainly due to the greater stabilization of the ground state than the excited state through dipole-dipole interaction that means polar solvent induces dipole-dipole interaction as a result what happens the ground state will be dropped in energy than the excited state and hence the energy gap increases and the wavelength decreases and the energy required increases in frequency. So this is a typical polar solvent for example when you take non-polar solvent this is a situation when you add a polar solvent the ground state is more stabilized means it drops in energy as a result n pi star gap homo-homo gap increases and hence it shifts to blue region it is called blue shift. That means if a group is more polar in the ground state than in the excited state increasing the polarity of the solvent shifts the absorption to shorter wavelength. So this we should remember we come across hypsochromic shift blue shift. On the other hand increasing polarity of the solvent shifts pi to pi star transition to longer wavelength then it is called red shift. So the red shift is mainly due to the greater stabilization of the excited state than the ground state through dipole-dipole interaction formation of hydrogen bonds or solvent effect with the polar solvent is responsible for this kind of rearrangement. So that means increasing the polarity of the solvent brings down the pi-pi star gap to the lower level as a result what happens longer wavelength radiation is required for electronic transition this is called red shift. Red shift is mainly due to the greater stabilization of the excited state than the ground state. In case of earlier one we saw the blue shift is mainly due to the greater stabilization of ground state than the excited state opposite is true here. The red shift is mainly due to the greater stabilization of the excited state than the ground state through dipole-dipole interaction this is probably due to hydrogen bonding or solvation with the polar solvent. So you can see the situation in non-polar solvent this is the one and in the polar solvent this gap decreases and it goes to red shift that means longer wavelength that means if a group is more polar in the excited state than in the ground state the increasing the polarity of the solvent shifts the absorption to longer wavelength. This is a second type of solvent effect on pi-pi star transition previous one was n to pi star transition then what would happen if there is an extended conjugation an increase in conjugation decreases the energy required for electronic excitation that means extensive conjugation brings the gap between pi-pi star if the pi is homo and pi star is lumo this gap is considerably brought down if we have extended conjugation. From molecular orbital diagram shown here two atomic p orbitals form two sets of sp2 hybrid carbons here and here so form pi-pi star in ethylene so here you can see this is simple ethylene but if we compare this one to butadiene here mixing of 4p orbitals results in energetically symmetrical distribution of 4 MOs compared to ethylene due to which homo-lomo gap is reduced and hence transition energy goes to longer wavelength. So that means you can compare here this is what typical ethylene and this first butadiene here the gap is considerably lower here compared to what we see in case of ethylene. So that means delta e or energy required for homo-lomo transition is reduced here that means conjugation conjugation decreases the homo-lomo gap and hence it will be a redshift. So conjugation effect can be seen for different transitions here with the different systems the longer conjugated systems decreases the energy gap between homo and lumo and it progressively it becomes smaller and absorption shifts to longer wavelengths so you can see here ethylene to butadiene to hexatrine to octatetroine so that means you start from one double bond two double bonds conjugated three double bonds conjugated and four double bonds conjugated you can see very nicely the energy gap is diminishing and shifting to mostly longer wavelength so this is called conjugation effect. So now let us look into important rules such as Woodward Fischer rules for calculating lambda maximum for a given compound by considering all functional groups present in it. If you consider homo annular diene a cyclic diene with conjugated double bonds in the same ring they are called homo annular diene if you have conjugation or conjugated double bonds in the same ring this is called cyclic diene and then we look into hetero annular diene a cyclic diene with conjugated double bonds present in different rings it can be something like this or it can be something like this. So hetero annular diene is one in which we have a typical cyclic diene with conjugation are two double bonds present in different rings and endocyclic double bond a double bond present in a ring is called endocyclic double bond exocyclic double bond a double bond outside the ring with one of the doubly bonded atom is a part of a ring so this is something like this and here we have both endocyclic double bond and exocyclic double bond this is a typical example here where you have both so this how some of these unsaturated compounds are defined with terms homo annular diene hetero annular diene endocyclic double bond and exocyclic double bond with example in each case is given here. What are the Woodward Fischer rules for conjugated dienes trienes, Pauline etc let us look into it. Let us consider a parent value so by identifying the type of cyclic system we have to give a parent value what are those parent value a cyclic conjugated diene the value is 270 nanometer for homo annular conjugated diene it is 253 nanometer in case of hetero annular conjugated diene it is 214 nanometer we should remember now we should look into the increment provided we add something else to it for example each alkyze substituent or a ring residue add 5 nanometer and each exocyclic double bond we should add another 5 nanometer and then double bond extending conjugation 30 nanometer if we have oxochrome such as war 6 nanometer and SR 30 nanometer and if you have OCOH3 carboxylic group then there's no need to add anything and if you have chlorine or bromine add 5 nanometer and if you have an amide group Nr2 add 60 nanometer so one typical example is given here and calculated value here comes 234 nanometers by considering the hetero annular diene here and then the three ring residue is there so add one ring is 5 so three are there 15 and one exocyclic double bond is there another 5 increment so it comes around calculated lambda maximum comes around 234 and if you actually measure it would come very close to this value given here then if you see homo annular diene conjugation within the ring the value designated is 253 and two ring residue so 10 nanometer two alkyze substituents 10 nanometer it comes around 273 nanometer so this how one can effectively use Woodward-Fischer rules for conjugated dienes to predict the lambda maximum value okay now further continue if you have alpha beta unsaturated compounds also Woodward-Fischer rules are available parent values are given here alpha beta unsaturated acyclic or 6-membered ring ketone the value is 250 nanometer and alpha beta unsaturated 5-membered ring ketone is 202 and alpha beta unsaturated aldehyde is 207 nanometer is the parent value and then increments follows this order at alpha position if you have alkyze substituent or ring residue then add 10 nanometer at alpha position at beta position add 12 nanometer and at gamma and higher position add 18 nanometer and for every exocyclic double bond add 5 nanometer and double bond exocyclic to two rings simultaneously there's a 10 nanometer so that means we have to watch carefully the structure before we give this increments for the parent value and then double bond extending conjugation give 30 nanometer and homo annular conjugated dienes at 39 nanometer so one such example is considered here parent value is 215 here because this is alpha beta unsaturated 6-membered ring here and then one alpha ring is there 10 nanometer and one delta ring is there 18 nanometer one exocyclic double bond is there 5 nanometer one double bond extended conjugation 30 nanometer one homo annular conjugated diene is there 39 nanometer so calculated lambda maximum value for this compound considering the increments added to the parent value it comes around 317 nanometer so one can conveniently use this method to predict the lambda maximum value and then that can always compare with the experimental value so here some axochromes values are given if alpha position for hydroxy group 35 beta 30 and gamma 50 and similarly for over group alkoxy or iloxy group 35 30 37 respectively for alpha beta and gamma positions and if you have SR value SR is there function group so in this case what happens alpha nil beta it's 85 and gamma nil and if you have carboxylic group alpha position 6 beta also 6 and gamma whether it's in alpha beta or gamma 6 increment has to be given nanometer and for chlorine alpha position 15 beta position 12 and nothing need to be added if it's in gamma position and then similarly for bromine we add 25 and 30 so this is how we can carefully look into the increments by identifying the groups and their position and adding the increments for various oxochromes or other fragments to the parent value very conveniently very nicely very precisely we can calculate the lambda maximum value now let's look into the type of electronic transitions responsible for color of compound especially coordination compounds so that means in complexes instead of looking into the several transitions I showed you most of this transition whatever I showed you sigma to sigma star pi to pi star n to sigma star this are all mostly confined to organic moieties nevertheless we can also see in the metal complexes originating from ligands or metals to an extent and we get one more extra transition that's called DD transition in metal complexes for example if you consider 3d 4d or 5d series they have anywhere from D1 to D10 electrons in their valence shell and then in case of octahedral compounds we know from crystal field theory that they are split into t2g and eg and in case of tetrahedral complexes e and t2 opposite of that one and in case of square planar complexes we have dxy and homo and dx my square as lumo and between them electronic transition takes place that we call it as DD transitions so what are the conditions for DD transitions do really DD transitions are allowed or not all those things we shall look into detail in my next lecture now just I have given one example here hexa aqua titanium 3 plus a D1 system and here you can see when the electron is there and this electron after giving the energy it is promoted here this is called DD transition similarly if you take cobalt tetrachlorocobaltate 2 minus it is tetrahedral compound we have four electrons it's a high spin complex we have four electrons in e and three electrons in t2 so here which electron is responsible so either e this electron can go or this electron can go that's what we have seen so here we can see the transition so let us look into more details of DD transitions in my next lecture so until then see you all have a nice time