 We are now discussing pi molecular systems. We have completed our discussion of polyatomic molecules with sigma bonds only like methane and we have discussed the first molecule which has pi bonds that is ethylene. Today we extend the discussion to butadiene. So butadiene of course is little more complicated than ethylene you can think it is like two ethylene moieties that are joined together. So we will see how we get there and how we can build a Huckel theoretical description of a molecule like this. So before that just a small recap in ethylene we have worked out wave functions and energy we had expressed the molecular orbital pi molecular orbital as a linear combination of the two p orbitals on the two carbon atoms and we have obtained this kind of expressions psi plus the bonding orbital was 1 divided by root over 2 into 1 plus s multiplied by chi 1 plus chi 2 and psi minus the antibonding orbital turned out to be 1 divided by root over 2 into 1 minus s multiplied by chi 1 minus chi 2 and the associated energies where E plus equal to alpha plus beta divided by 1 plus s and E minus is alpha minus beta divided by 1 minus s and here the interesting thing is that you do not need to know the Hamiltonian for a Huckel theory. You actually work it out in terms of the integrals where Hamiltonian is just written as H. What do we do there? For equivalent carbon atoms we remember that H11 has to be equal to H22 we call this the Coulomb integral alpha and then we set it to 0 right because all measurements are with respect to alpha and H12 and H21 we said they are equal to each other and they are called the resonance integral this is what gives us the stabilization because of pi bond formation. Of course, S12 and S21 are our familiar overlap integral and S11 and S21 are just 1 because their atomic orbitals are normalized. So, we have set alpha to be equal to 0 because that essentially is the energy of a pz orbital in the sigma framework of the ethylene. So, that is your starting point and beta as you said is delocalization energy and we had told you the value is minus 75 kilojoule per mole no need to remember this value what is important to remember is beta is a negative quantity and we discussed very briefly what we do with S with that background let us go over to Buterain we have just added 2 carbon atoms and we have brought in 2 more p orbitals perpendicular to this molecular plane. What happens now is that everything remains the same is just that the secular determinant becomes larger earlier you had a 2 by 2 determinant now you have a 4 by 4 determinant. So, since the problem is a little more complicated it would help if you could bring in some more simplifying factors and that is provided by Huckel theory. We have said already that we set Coulomb integral to 0 what we now say is that this Hij equal to Hji we said that to be resonance integral beta only for adjacent atoms and we consider them to be 0 for others because resonance integral basically gives us the energy for delocalization of the electron over 2 atoms. Well the electrons are actually delocalized over the entire atom but remember these integrals have 2 wave functions right integral psi i H psi j that kind of an integral what we are saying is that if i and j are adjacent to each other then that integral has some value beta if you take 1 and 3 for example then the value is as good as 0 we are neglecting it because after all these integrals are all worked out numerically. So, in locations where say chi 1 has not too much of value it does not matter what the value of h chi 3 is you are going to get a 0 and vice versa. So, only for adjacent atoms we said that this Hij and Hji are equal to beta for everything else it is just 0 right it is an approximation but it is an approximation that works to a very large extent and there is some logic to it the logic is that we have to consider we do not have to consider atoms that are too far apart from each other when we build these pair wise terms like this where we have 2 orbitals 2 atomic orbitals from 2 atoms and the Hamiltonian operator operating on one of them. So, that is the justification put in very very simple terms. Now, let us see what we have another thing that we need to worry about is overlap integral even overlap integral in Huckel theory it is considered that by similar logic overlap integral can have some nonzero value only for adjacent atoms chi 1 and chi 3 is not going to have too much of overlap anyway because the distance is already fixed by the sigma bonding network they cannot come any closer. So, as we have studied for say H2 or H2 plus when they come close together then only the overlap integral increases from 0 to some determinable value. So, since 1 and 3 are far apart from each other anyway we can set that overlap integral to be equal to 0. In fact, in the simplest version of Huckel theory that we are going to use we are going to set all overlap integrals to be equal to 0. This might sound to be very arbitrary but then let us think what kind of overlap we are talking about we are talking about pi overlap like this not sigma overlap like this. So, the distances between the 2 at centers is already fixed they cannot come closer than this and p orbital are sort of held like this. So, how much will the overlap not much that is why it is approximately said to be equal to 0. Of course, if you want more precise results you cannot set them to be equal to 0 at least for nearest neighbors but for our purpose it is okay if you go with the simplest version of Huckel theory set overlap integral to be 0 as well Coulomb integral of course is set to 0 as we said earlier. So, now this determinant becomes very simple. So, for all these H11, H22, H33, H44 I can write alpha and then I can happily set them to be equal to 0. H11, H12 all that is happily equal to 0 we do not have to worry about them. So, what am I left with? I am left with wherever we have beta. So, H12 would be equal to beta right and H12 occurs in 2 places H21 is equal to H21 as we said earlier. What about H13? H13 is going to be 0 because 1 and 3 are not adjacent to each other do not have to bother about it. Similarly, H14 the distance is even bigger there is no way we can we have to worry about those but what about say H23? H23 is not equal to 0. H23 actually is equal to beta because 2 and 3 are adjacent to each other and H23 appears in 2 terms. Similarly, H34 also appears in 2 terms and they can be replaced as beta. So, what do I have now? What is the first term here? 0 minus 0 second term is beta then 0 0 then we have beta well this is beta minus ES12 right. So, that is equal to 0 ES11 remember this is not equal to 0 I think I made a mistake there ES11 is just S that is equal to 1. So, here I have minus E then I have this is beta. So, this way we can keep on writing and this is what we are left with alpha minus E beta 0 0 beta alpha minus E beta 0 0 beta alpha minus E beta 0 0 beta alpha minus E right. This is where we are started from we said that we are going to set H11, H22, H33, H44 those are actually alpha we are going to set it to 0 later on then we said S11, S22, S33, S44 these are all equal to 1. So, the first the diagonal terms become something like alpha minus E everywhere then H12 is beta and this minus ES12 that becomes 0 because S12 is equal to 0. So, here H13 minus ES13 that is 0 anyway. So, we get alpha minus E beta 0 0 beta alpha minus E beta 0 0 beta alpha minus E beta 0 0 beta alpha minus E. So, this determinant is equal to 0 all right. Now, we will simplify this a little further before going ahead. So, how can I simplify what happens if I divide say alpha minus E divided by B if I take some quantity X where X is alpha minus E divided by beta not B I keep on saying B for beta inadvertently please do not get confused when I see B in this context I actually mean beta. So, alpha minus E divided by beta I put it as X and now we have a nice determinant here X 1 0 0 1 X 1 1 0 1 X 1 0 0 1 X that determinant equal to 0 what is the next step we should expand it when we expand I leave it to you to do the expansion by yourself it is not difficult just show you the answer X to the power 4 minus 3 X square plus 1 equal to 0. Now, please do not get daunted seeing X to the power 4 X to the power 4 is just X square square. So, this is really a quadratic equation in X square is not it. So, we know how to solve quadratic equations we are going to proceed in the same way we just going to write this as X square square of that minus 3 multiplied by X square plus 1 equal to 0. So, what is X square going to be then X square will be minus B plus minus root over B square minus 4ac B square is 9 minus 4ac is 5 plus minus root over B square minus 4ac divided by 2 alpha that is 2 this is your answer X square is equal to 3 plus minus root over 5 by 2 and fortunately root over 5 is less than 3. So, we get all positive numbers here it does not matter whether you take the plus combination or minus combination you get all positive numbers and that is good because if X square is negative then we end up getting something that is imaginary that is unphysical. So, if you take square root of this now you are going to get 4 roots for X and they are I leave it to you to calculate it plus minus 1.61804 plus minus 0.61804 I find this to be particularly amusing because the characteristics is the only thing that changes 1 and 0 after decimal point it is 61804 in both the when all four terms actually that is what it is we have found the values of X. What is X by the way X essentially is this if I said alpha to be equal to 0 X is what minus C divided by beta. So, can I say that X is E in units of beta? So, if it is 1.61804 that is the energy in units of beta keeping in mind that beta is a negative quantity. Now, with that understanding we can draw the energy levels you have not drawn the wave functions yet we will get there but we have drawn the energy levels the center here is alpha equal to 0 we start calculations from here 0.61804 is really this separation do not forget because after all the expression does have alpha minus E kind of thing. So, and then this is 1.61804 even though I said it several times please remember plus 1.61804 beta actually means a negative quantity because beta is negative that is why it is lower. Now these are the energy levels what can I do now I can fill in the electrons like we always do what how many pi electrons are there in butadiene 4 right. So, I can just put in like this I will draw in the conventional manner not worrying too much about those spin wave functions just draw like this this is the configuration see for these 2 orbitals lower energy orbitals the energy is actually negative these are bonding orbitals for the higher 2 these are the anti-bonding orbitals their energies are positive. So, what we see here is that for butadiene the pi electrons reside in bonding orbitals only right and you can calculate what the bond order is pi bond order is 2 and if this was a valence bond theoretical treatment you would be able to draw these resonating structures very easily right I will draw 1 because it is easy 1 resonating structure is something like this and this is the structure that I can draw without much hassle let us see if I can draw the second one I think this is how you push the arrows. So, this is your second picture here the double bond is between atoms 2 and 3 carbon atoms 2 and 3 here I have minus sign here I have plus sign and if you go exactly the same way then the other thing that you could draw is just push the arrows in the opposite direction you can have minus plus please do not get confused we are discussing molecular orbital theory and now what I am drawing is really a valence bond description with resonance built in okay I am just trying to do a comparison we are going to come to this very soon from the molecular orbital picture. So, see which one of these 3 is going to contribute more naturally whenever there is charge separation we know that contribution is less. So, these 2 structures where you have the double bond in the middle that they are going to contribute less and the central one where you have double bond between 1 and 2 and 2 and 3 that is going to be the major contributor remember this and let us see whether we get something similar at least qualitatively from the molecular orbital picture. But before we do that we need to talk about wave functions then only can we think about insights that we can get from the wave functions.