Pyrrole undergoes electrophilic aromatic substitution more rea...

Question

Pyrrole undergoes electrophilic aromatic substitution more readily than benzene, and mild reagents and conditions are sufficient. These reactions normally occur at the 2-position rather than the 3-position, as shown in the following example.

b. Explain why pyrrole reacts more readily than benzene, and also why substitution occurs primarily at the 2-position rather than the 3-position.

Accepted Answer

Hey everyone and welcome back into this video. We are going to solve the following problem. The electro chromatic substitution reaction rate for thio finn is greater than that of benzene. It often reacts under mild reaction conditions, substitution and tie often occurs preferably at position to explain why thigh often is more reactive than benzene and why substitution preferably takes place At position two then position three. Now we're given our reaction. We essentially want to think about the sigma complex and the stability of sigma complex whenever the substitution takes place. Let's keep in mind that that because this is tailfin sulfur atom defines position number one. So we have two positions to analyze that be our position number two. In position number three. Now let's start by analyzing position number two. And what do we have when the attack takes place from position number two on electro file. So first of all we are going to draw our five member ring. Okay, so that's how five member going. We're going to add our pipe bombs. Let's not worry about the lone pair of sulfur yet. But of course we can add the two lone pairs. Now because we are thinking about position two, we're going to show an attack on the electro file in a generic Okay, is that B. E. Plus because we're adding it at position number two, we want to make sure that we draw our product correctly. So now we may simply add our five number grain and now let's attach The Electra file at position number two meaning there's a lack of electrons at position number three and we end up with a positive charge, that carbon. Right? So what we can see now, we can think of residents. First of all, there's an adjacent pi bond. This is our first residence structure. And we are going to de localize these by electrons getting a positive charge at the adjacent position. This corresponds to our second residence form. We may now to our five member ring. This is our newly created pi bond. And right here at the carbon atom, we have our positive charge. Now from here, we notice that we can still have some sort of the localization because we have lone pairs on sulfur and they can be de localized as well. We can use this lone parent sulfur form a pi bond and this would be sufficient for us to get an additional resonance form. This is our second residence structure And we are ready to draw the 3rd 1. So for the third one we will have a double bonded sulfur. Now sulfur has a double bond. And because it has a double bond, we are going to have a formal positive charge because there are bonds. Right? It's very similar to whenever you have oxygen being bonded. Now there's an additional pi bond. This is our electro file. We stopped drawing our residence forms because there's nothing else. We don't have any adjacent pi bond here to the localized lone pair. Or basically the only thing we can say is just break the pi bond, sulfur has already three bonds. It has an octet, right? So essentially we stop here, we got our resonance forms. Now for comparison, let's see what happens when we have our substitution at position number three. And we can do that in green for compared some. So whenever you're thinking about position number three, or just going to draw The five member ring, let's add our pi bonds. And then we are going to start with this bond. Show the attack on the electro file. And now we're going to add it at position number three. My apologies. Let's not add resonance arrow. This is just the part where we have an attack on the electro file. So let's try the result and resonance form. To begin with, we are going to attach the electro file at position Number three. And because we are breaking the spy bond, we must have a positive charge of this carbon. So this is our first resident structure. We can label it as structure number one. And then from here, we have to think of a possible residence. Let's keep in mind that in each case sulfur still has two lone pairs. And this is the only way to produce a resonance form. We can only use that lone pair in the formation of a pi bond. This produces us with the second residence form sulfur is now double bonded. There is still one lone pair remaining, there will be positive form a positive charge on sulfur. That's how we electrify. I'll And that's our pi bond. Let's make sure that we draw it clearly. So you understand that this pi bond is between two carbon atoms. Right? So this is our second residence form and there's nothing else we can do. Essentially, we cannot be localized that lone pair over here because silver has an arc dot white, essentially, you would be forming two double bonds on sulfur. That would cause ring strain. You cannot really do that for a five member ring. You cannot make sure make this part linear. Right? You cannot have two adjacent double bonds on sulfur because that would correspond to linear geometry. So we stop with our resident structures. We got two resonance structures. And now we have to compare. First of all, we can say that the structure that we have been analyzing so far, this is staff in it is more reactive, right? The reason why it's more reactive then benzene is because sulfur is donating its own pairs into the stabilization of the sigma complex. Right? So this is the reason basically we can say that lone pair on sulfur is de localized, meaning it is stabilizing the sigma complex. So that's the first and search our problem. And then the second part of our problem was to understand why substitution primarily takes place at position two instead of position three. Notice how we got to residents forms right? We have to residents forms number one and two. Going back when we think about position number three, we got three residence forms. Right? So at position number two, we have three residents forms at position. My apologies at position number two, Yes. At position number two, we have three residents forms at position number three. We have two residents forms because there is a greater number of residents forms whenever the substitution takes place at position number two, then at position number three. This implies the greater stability of the sigma complex and greater stability implies a greater reaction rate. Therefore, we expect this to be the major product, right? Because it has more resonance forms. And this pretty much explains why the substitution reaction is preferred at position number two instead of position number three. Based on our analysis, we can say that the correct answer to the multiple choice question is c often is more reactive because the sulfur atom increases the ring electron density. The electro file attack at position two is favored because the sigma complex has more resonance forms than it has when the ring is attacked at position three. Thank you for watching. And I hope to see in the next video

Key Concepts

Electrophilic Aromatic Substitution (EAS)

Aromaticity and Electron Density

Regioselectivity in Substitution Reactions

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