Furan undergoes electrophilic aromatic substitution more readi...

Question

Furan undergoes electrophilic aromatic substitution more readily than benzene; mild reagents and conditions are sufficient. For example, furan reacts with bromine to give 2-bromofuran.

Furan structure with numbered positions, showing bromination reaction to form 2-bromofuran in dioxane.

b. Explain why furan undergoes bromination (and other electrophilic aromatic substitutions) primarily at the 2-position.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem. The electro chromatic substitution reaction rate for indo is since the power of 13 times greater than that of benzene, it often reacts under mild reaction conditions, substitution in Indo occurs preferably at position three. Explain why Indal is more reactive than benzene and why substitution preferably takes place at position three than position too. Okay, so we're given our reaction. We have our structure. Let's understand that this is our benzene ring which is fused with in the ring. We first will have to identify our positions. This is our nitrogen atoms. So this would correspond to position number one of the ring. Then we would have position number two over here at Jason tonight. And in position number three, we understand that the reaction takes place much faster at position number three. Notice how we are adding this electro file that substitutes the hydrogen atom. Right? We have an electro chromatic substitution at position three instead of position number two. So let's see why that happens. And to see why that happens. We are first of all going to analyze the substitution at position number two. So we're going to read through our structure and they were essentially adding electro file. Right? So we would have an attack on electro file that would be our generic electro file. E. Plus because we are first of all analyzing position to let's suppose that we have an attack on our electro file and then we are getting our intermediate. Now notice how we are adding our electro file at position number two. So let's at E. And then we would end up with a positive charge at this position because we had a loss of electrons that this carbon, so it has a lack of electrons. This is our nitrogen with a lone pair and hydrogen. So this is our result on carbon Karen and we can essentially think about the resin stabilization. Notice how first of all we have this adjacent pipe on and we can use that pipe bon and the localized the electrons to form a positive charge on the benzene ring. Or basically at this position because if we shift the pi electrons here, we would have the positive charge at the intersection, right at the common atom between the two rings. So then we draw our next residence form, that would be our benzene ring. And now we understand that we are forming a pipe on at this position, meaning we end up with a positive charge over here. So that's our next residence form. And we have to keep thinking about the possible residence forms. So from here, because the positive charge is at the intersection, we have two possibilities. We can either use the lone pair of nitrogen or the pi bond from here to get to different residence forms. So, first of all, let's suppose that we're thinking about the d localization of this lone pair of nitrogen and we understand that this is one way to get our next residence form. So for the green pathway, we end up with the following Now because we used our lone pair, we are going to create a pi bond. Andijan would have the formal positive charge because now it has four bonds. We are going to add our electro file as well as our pi bon at the very top. However, on the other hand, we can de localize these pi electrons and form a pi bond at this position of getting a positive charge at this carbon. So if you think about that black pathway, we get our next residence for him. Let's add our substitutes. Now there's still loan paranoid region according to the black pathway, let's look at the black arrow and we understand that we have our pipe on over here, we have a lack of electrons at this carbon, so it would have a positive charge. And now we are going to add that remaining pi bon. Now from here we can actually de localized these by electrons and get a positive charge at a different position. So that would be an additional residence form. Let's draw it out. That's our electoral file. Now we are forming a pipe on at this position and then we get our positive charge over here this carbon. Right? So we have got our residents forms, we noticed that we have 12345 residents forms, we can also label them, this is our first one, the second one, The 3rd 1, the 4th 1 and the 5th 1. The reason why we're drawing resonance forms is because we want to think about the stability of the sigma complex. We want to understand how stable the complexes. So yes, indeed. This looks like a very stable structure. Right? Essentially we have five resonance forms and the greater the number of significant resonance forms the better for the stability of the complex. And now let's think about the stability of the complex form when we perform electro aromatic substitution at position number three. So, first of all, we're going to draw our starting material. And now let's suppose that the generic electro file in the form of E plus is attacked by the pi bon. And then we are adding it at position number three. So that would give us the following structure. We are breaking the pi bon and because we're adding our electro file at positions, we we will have a positive charge here at Jason's nitrogen. You may notice now that here we only have one possible way for residents. So this is our first resident structure and then we can de localize this lone pair of electrons over here to form a positive charge on nitrogen and we can do that. So if we show that the localization, we end up with the following resonance form, that will be our six member ring. Notice how we haven't affected any of the pi bonds on the benzene ring. But now nitrogen contains four bonds 1234. Right? So we have a former positive charge on nitrogen and our electro file. Now there are no more resonance forms possible. Right? We cannot really use our pi bond. Let's actually redraw these pi bonds exactly the same as in structure number one. Because we don't want to affect them, We didn't really use them in our residence. And essentially this pipe bond cannot be used in the localization because nitrogen already has a double bond, we cannot have that ring strain of course, and nitrogen cannot have five bonds as well. Right, We cannot have that kind of scenario. So this these two are our only residents form. So we only have to. Now we have five residents forms for the electro file being added at position number two. Right, But we have only two residents forms For position # three. And that's a problem. Right, because we have more resonance forms for The substitution reaction at position # two. So what is the actual reason for a higher stability of this complex and a higher reaction rate at position number three, recall that generally we want to make sure that the the sigma complex is as stable as possible, meaning we want to have as many as possible significant resonance form. So this would imply that substitution at position number two would be more preferred because we have five verses to But in this case it's not so significant. Notice how in this one we didn't use Benzene for our resonance forms. And Benzene re change its aromatic city right? It still has six pi electrons. It's flat. So Benzene is still aromatic in each of the resonance forms. And recall that aromatic city implies stability. But if you look at the resonance forms of the previous one, notice how we actually used this pi bond right? In in the first residence form we use this pi bond to de localize it. And you may notice that here in particular we only end up with 12344 pi electrons. So Benzene loses automaticity meaning even though we have a greater number of residents forms, notice how Benzene loses automaticity. Once again, it doesn't have six pi electrons. It only has four same thing here and same thing here. So we only have one significant residence form which is stable. Right. Benzene here is still aromatic. So we only have one significant aromatic form. These ones are still significant but Benzene ring is not aromatic in any of them. And actually because there are four pi electrons on benzene, we can think of it as being anti aromatic. Right? Because whenever you have four and pi electrons, that implies. And sierra metis city and anti aromatic compounds are highly unstable. So these 1234 forms do not produce automaticity and their highlands stable. So we only have one stable form vs two stable forms. And that implies because benzene is aromatic in each case, Position number three is preferred over position number two. So we are going to label these residents forms indicating that this would be our major product. In addition to that, we can answer why indoor ring is more reactive than benzene ring. Notice how there's a lone pair of nitrogen and this is the fact that there's a lone pair of nitrogen, nitrogen atom increases the ring electron density. Whenever we managed to use a lone pair on a heater atom that can be de localized, essentially it increases the reactivity on the ring because it's increasing the electron density or basically it is activating. Right? So we can conclude that answer choice C. Is the correct answer to the multiple choice question in though is more reactive than Benzene because the nitrogen atom increases the electron density. Right? We we notice how this lone pair is the localizer. We are increasing the electron density on the ring and in addition the electra filic attack at position three is favored, right? Because it does not result in the loss of aromatic city of the fused benzene ring. Notice how the two residents forms are both aromatic and in the previous structure only one form was aromatic. All of the other four were not aromatic with respect to benzene. And that means this sigma complex is more stable because automaticity is retained an aromatic city implies stability. That would be it. Thank you for watching. And I hope to see you in the next video.

Key Concepts

Electrophilic Aromatic Substitution (EAS)

Reactivity of Furan

Position of Substitution in Furan

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