p-Xylene undergoes nitration much faster than benzene...

Question

p-Xylene undergoes nitration much faster than benzene. Use resonance forms of the sigma complex to explain this accelerated rate

Accepted Answer

Welcome back everyone in today's video, we are going to solve this problem, fennel is ruminated under much milder conditions than Benzene draw residence forms of the sigma complex that is formed after the electro filic attack to describe why fennel reacts faster than benzene. Now, first of all, we're going to think about this scenario and of course we just need to draw final itself. Let's recall that final essentially implies the presence of benzene that contains an alcohol group. Oh age. And because we are interested in an electra filic attack of course this example analyzes rumination in particular. But we are just going to look at this mechanism and the residence forms for any electro file. E. So in the first step of an electro filic aromatic substitution, we're essentially using the attack at the electro file. And because wages an Ortho or power director, let's just think about the power position. For simplicity. It doesn't really matter which position you analyze because we're just interested in resonance. But let's suppose that we're interested in the power position. So the pipe on attacks the electro file. And as we said in the general case that electro file can be labeled as E. Plus, right? And then when we have that attack, as we said, we are going to analyze the power position So we can troll the result on structure here during our reaction. We are essentially getting the following structure which involves the original ring. And then because we are breaking this pi bond, let's not draw it right. We're essentially going to add our electro file at this position. Keep in mind that there is still that implicit hygiene atom but we can expand it and then here we have our electro file added. And since now this carbon legs electrons, it has a positive charge because we have broken the pi bond and we only created a bond at this position but we have no additional bond here. So we have a positive charge. Okay, well done. And now we are ready to think about residents first will notice that we can de localize that positive charge by transferring the pi electrons right essentially what we can see, we can transfer the pi electrons to this position and then we will have a positive charge over here. Right? Because we will have a lack of electrons. So this will be our second residence form. Let's keep in mind that this is the first one, then we control residence arrows and we control the result on second residence form. Okay, so we noticed that first of all we have created a pi bond at a different position. Now we are going to add hydrogen and our electro file, then the O. H. Group and then we have a positive charge at this position because we simply have a lack of electrons after breaking the pi bon. And now from here we are going to think a very similar way what else can we do in order to stabilize that charge, we can be localized that once again notice how there is a lone parent oxygen which can be used or basically de localized right in the creation of the pi bond. And this will allow us to get rid of this positive charge here, but we will have a positive charge and oxygen. So this will give us our third resident structure because this one is the second one and we can simply choice. So that will be our six member ring, our double bonds. Now we have an additional double bond at this position. And then here we have oh age with a formal positive charge on oxygen. Now let's add Haijun and our electro file. And from here we are ready to think about our next residence structure. What else can we do to to stabilize the structure by an additional residence form. So if we think about the benzene ring, let's label the previous structure as residence form number three. Now you may notice that you may simply the localized these pi electrons. And the reason why you can see so is because you have a contract it system, right, you're going to be localize these elections over here because by doing so you can break this pi bon and get rid of the positive charge and oxygen. So this gives us an additional resonance structures that will be our ohh with a single bond. Now, because we have broken the pi bond, we can draw the ring and then we may notice that we have our double bond, our newly created double bond over here and because this carbon lacks electrons, we are going to have our positive charge at this position as well as hygiene and our electro file. Okay, so that will be our structure number four. Let's label this is our structure number four. And now simply let's think about the residents of Benzene in a similar scenario. If we choose a different color in particular green, let's suppose that the starting material is benzene instead. So in the first step, we have an attack on the electro file. We are forming our sigma complex. Right? If we add an electro file at this position then we would have a positive charge here. And now we are ready to start drawing the resonance forms. That would be the first one because the original one is the residence form itself. Now, if we shift these electrons, we get a positive charge here. So let's draw the result. And Residents form that would be our 2nd 1. Let's label it as our 2nd 1. And now we have an additional possibility. We can essentially shift these by electrons to form a positive charge here and there's nothing else we can see so far. You may notice that we have covered our cycle. We got three residents forms. Right? We cannot move forward. We can only go backwards right by bringing these by electrons back. So that would correspond to the previous structure meaning we are done. And now we noticed that we understand the context. Now, what is the reason that fennel is more reactive to an electra Felix substitution? Keep in mind that in this problem it was referred to domination. And we can say that in this case e is equivalent to be roaming right for complete consistency. And we may notice that percival for benzene we got three resonance forms, three residents forms and for Fennel We got four residence forms. Recall that an additional residence form can stabilize our sigma complex more or basically to a greater extent, right? And this is indeed the case for fennel. We may label our starting reaction and all of the residence forms required for it. So let's do that. These are our three ones. And then if you notice at the bottom we have the fourth one. So these are the required forms for fennel. And specifically notice how going from 2 to 3 involves the presence of oxygen. Right? We can actually de localized the lone parent oxygen itself. And Benzene doesn't have that oxygen. So the additional residence form comes from the fact that there is oxygen present and it allows us to achieve that extra resonance form. An extra resonance form implies that there is stabilization a greater stabilization of the sigma complex and you may understand that the greater disability, the foster the reaction rate and this explains why funeral reacts faster than Benzene because it has higher stability. Whenever we are thinking about the sigma complex, it has four residence forms versus just three residence forms for Benzene, and that would be the main reason. The greater stability implies the Foster reaction rate based on our residence forms drawn. The correct answer to this multiple choice question is B. However, if you have any questions, don't hesitate, Slaver comment down below and thank you for watching.

p-Xylene undergoes nitration much faster than benzene. Use resonance forms of the sigma complex to explain this accelerated rate

Key Concepts

Nitration of Aromatic Compounds

Resonance and Stability of Sigma Complex

Electron-Donating Effects of Alkyl Groups

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