 Okay friends, so in this session we are going to discuss few reactions which is based on aromaticity, okay. So far we have discussed about aromaticity, examples also we have discussed how do we assign, how do we calculate the number of pi electrons and how do we assign aromaticity non-aromaticity or anti-aromaticity to any compounds, right. So now in this session we are going to discuss about reactions which involves aromaticity, okay. So I will write down the heading here, reactions, reaction involves aromaticity, okay. First reaction we have is hydrogenation, okay. So now if I write down one example here, suppose the molecule is this and hydrogenation reaction is taking place here, right. So what are the possible product we have? The possible product can be any one of these. The double bond will break and we will get a hydrogen attached with it. This is one of the product when both the hydrogen atom will attach over here. Another possibility is what? We can have this also or we can have this also or we can have this also, okay. But in all these possible product you see the product which is aromatic is nothing but this. This molecule has 6 pi electron and it is aromatic in nature. We have seen this example separately also. The larger molecule is this, right, which is non-aromatic. So we will see the smaller one. A smaller one is what? The benzene ring itself which is aromatic, hence the compound is aromatic. So because of this aromatic nature of this compound, because of the aromatic nature of this compound, this is the major product we have. We will not get any one of these product in this reaction, okay. So this is the hydrogenation reaction we have. Another reaction also we can show you which is this hydrogenation, hydrogenation, this molecule, okay, hydrogenation. Here the product will be what? Aromatic behavior will be maintained. So the product will be this, hydrogen will attach here, right. Here the product will be, this pi bond is in conjugation with this, will not attach hydrogen over here, okay. Now here what happens? The hydrogen will attach on here, will break this double bond. So the product we will get here is this. This is the product we get, okay. And why this product? Because in this product we have three aromatic rings, which is nothing but one, two and three. Three aromatic rings forms here and that is why this is the most preferred product we get here, okay. Now you will see the another examples of the aromatic reactions and one more thing since we are breaking three bonds here. So we will take here three equivalents of this hydrogen, okay. That is also important to mention. We are taking one equivalents of this hydrogen. Then only we will break only one bond, okay. Here also we will take one equivalent. What happens if we take two equivalent? Then this bond will also break, right. We will get this aromatic compounds, okay. The next type of reaction you see is reaction in presence of metal. The first reaction is hydrogenation. The second one in presence of metal. For example you see if I take cyclobutene, cyclobutadiene and in presence of sodium, right, in presence of sodium. So this sodium what happens? It will lose two electron because I am taking two moles of this. So it forms two Na plus, plus two electron, okay. And because of this what happens? This homolysis cleavage will be there, okay. One electron will have here, right, unpaired electron. And this unpaired electron will get paired by one electron each from this sodium metal. And finally we get what? We get this double bond. So we have one pair of electron here, one pair of electron here. Negative charge Na plus will be here, Na plus will be here, right. Here also we have the negative charge. So here the one electron of this carbon we have. One electron from this metal, one electron from this metal, okay. And now why this reaction takes place here? Because you see this molecule now it has 6 pi electron which is aromatic in nature. And further the reaction proceeds. The cleavage takes place like this, cleavage of bond. Similarly if I take this example in presence of metal, okay. Then what happens? Again two moles of this will take, similarly we will let the same compound, one pair of electron, one pair of electron, negative charge, negative charge, Na plus, Na plus, right. So we have 6, 10 electron, this is also aromatic, 10 pi electrons aromatic, okay. Another example you see, this is 4 pi electron, anti-aromatic, right. And when this reaction takes place in presence of 2 Fe 3 plus, it forms this. It takes both the electron from here and forms 2 Fe 2 plus, okay. So one of the metal can provide electron also and other iron, metal iron, right, which is electron deficient can take electron from this carbon atom and forms positive charge here. So because of this what happens? It has 2 pi electron now and which is aromatic, okay. So this kind of reaction takes place in presence of metal, okay, which is you know very important reaction we have. One more reaction I will write down here, which is the case of non-aromatic compound. Like you see this one, this is we have hydrogen present here, here and it has, it is non-aromatic compound, sp3 hybridized carbon, non-aromatic compounds. So in this case, what happens? It converts into double bond, double bond, negative charge plus half of H2. One hydrogen atom goes out and since the negative charge on the top most carbon atom, you see it becomes what? It becomes aromatic in nature, 6 pi electron aromatic in nature, we use sodium here. So this is how the second reaction we have in which the aromaticity is involved and then the further reaction proceeds. So we will get the product in all this kind of reaction according to the stable intermediate product which is nothing but the aromatic product we have, okay. So if you have non-aromatic compound, you can convert into aromatic, anti-aromatic and convert into aromatic if it is possible and then we will write down the product, okay. So similarly in this we have third type of reaction also which is the Lewis acid base reaction. It is the important one. The third type is the most important out of the three, Lewis acid base reaction if you have. For example, you see you have RCl alkyl halide and this RCl is heated in presence of AlCl3, okay, AlCl3. Then what happens? We will get an intermediate here RCl, this Al partial negative, partial positive charge. This AlCl3 is a Lewis acid behaves as Lewis acid, aluminum has d orbital and it has only six electrons here so it has tendency to accept electron, right. So the lone pair of chlorine takes part in the reaction and it has tendency to lose this electron delta positive delta negative charge. Similarly what happens? This bond breaks and we get here R positive plus AlCl4 negative, okay. Now this cation if it is aromatic then the reaction proceeds with a very faster rate, okay. So this cation only positive charge only may provides aromaticity in few cases. For example, you see if I write down this reaction and when this is heated in presence of AlCl3 then what happens? The product we will get is this positive charge here double bond, this has 2 pi electron and it is aromatic in nature, right, aromatic in nature. Again you see if you have this molecule and it is heated in presence of AgCl sorry AgNO3 then what happens? We will get this ion plus we will get the precipitate of AgCl, right because this is aromatic, aromatic in nature resonance stabilized, okay. What happens if you have this molecule? This compound is what? This compound is non-aromatic we are going towards aromaticity from non-aromatic. This compound is also non-aromatic and we are going towards aromatic compound. Now this compound is also non-aromatic, right when it combines with AgNO3 you see we are getting what? Positive charge double bond double bond plus AgCl precipitate if possible, right. But this compound is what? It has 4 pi electron so it is anti-aromatic, right. So you see non-aromatic since it is more stable than anti-aromatic then this reaction will not proceed at all will not get this product in this reaction because we are not we cannot get the reaction will not proceed towards the lesser stable state, okay. So lesser stable compound will not form here the reaction will not proceed, okay. So reaction always goes towards from anti-aromatic to non-aromatic possible from non-aromatic to aromatic possible, okay. Reverse way it is not possible, okay. These are the few reactions where the aromaticity are involved, okay. One question on this has been asked in J exam, right. I will give you that question and then we will solve it, right. When this molecule reacts with or I will write down the question like this. A, B and C which of this molecule, molecule reacts with AgNO3 with fastest rate with fastest rate, okay. So you see first of all if this molecule reacts with AgNO3 it forms this ion positive charge. If this molecule reacts it forms positive charge here and if this molecule reacts it forms, okay. Now you look at all this molecule this one is aromatic 2 pi electron, this one is non-aromatic sp3 hybridized and this one is anti-aromatic. So the reaction which reacts with the fastest rate will the product the molecule reacts with the fastest rate is nothing but this because it is forming aromatic compound or the order of the reaction will be this aromatic then non-aromatic and then anti-aromatic, okay. So this type of reactions questions they may ask based on aromaticity you must memorize all these concepts, okay. Thank you.