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.

a. Propose mechanisms for the bromination of furan at the 2-position and at the 3-position. Draw the resonance forms of each sigma complex, and compare their stabilities.

Accepted Answer

Hey everyone and welcome back the electra Philip aromatic substitution reaction. Wait for indoor pool 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. Proposing mechanism for the nomination of Indo. You do not have to show steps for generating the booming ion. You may begin with indo and possibly charged bro. Mean iron draw resonance forms for each sigma complex. Now, first of all, we are going to look at the scheme and we may notice that our starting material reacts with an electro file and we're adding electro file at position number three. Just as suggested by the problem in the ring contains position number one foreign hydrogen. And then if we go counterclockwise, that would be our position number two. Position number three. Indeed, position number three now contains our electro file that has replaced hydrogen. So let's see how that works. Based on the reaction scheme. We already know what we are going to expect. So we can simply redraw our starting material and now what we're going to do, we don't need to show how our electro file is generated. We have positively charged roaming. We are just going to show that because we're going to add positions. We need to make sure that we're using this pi bond as our attack step on the electro file and our electro file is positively charged roaming. So let's indicate that electro file and we are going to add it at position number three. Just as suggested by the reaction product. It says that this would be the major product. So we are going to clearly show how we are forming our intermediate or base lower sigma complex. Let's use residents arrows because we're forming our sigma complex. And now we are going to add our sixth member drain. Let's make sure that we keep the pi bonds in the same orientation as we started off with. For complete clarity. Now we can draw our five member ring and we understand that we are basically breaking this pi bon and we're adding romain at position three. So let's see that. And because we have a lack of electrons at this position, we are going to have our positive charge. That's because we have broken the pi bon. We have to think about the possible residence forms for this intermediate. And we may simply notice that there's only one residence form. We have a lone pair of nitrogen which can be the localized into the pi system. Right? We can essentially form a pipe on out of it. And if we do so night and we'll have four bonds and a formal positive charge. So let's draw the result in residence form. That's our first residence form. And then we are ready to draw the 2nd 1. So let's add our six member ring. That's our nitrogen. And we understand that we are forming a pi bond in this position. So we're going to add a formal positive charge on nitrogen because it has four bonds, we are going to add roaming and that's the only possible residence form that we can get noticed that we cannot really shift these pi electrons over here because Nigeria cannot have five bonds. It already has an octet, it has four bonds. So this is our these two actually are our only possible residence forms, the first residence form and the second one. And then we simply want to think about the formation of the final product. First of all, let's add and close our brackets because we are done with the residents forms and we can think of the second residence form. Of course, we need to remove that proton and we want to make sure that we are restoring that pipe on. You may notice how this position still has hydrogen and we are essentially going to use the electrons transfer from Iran booming bond to this hydrogen forming HBR But that's not really important because we're not worried about the formation of the sigma. My apologies, the formation of the electro file. So we didn't include the fact that we needed to use iron three bromide because it's not present in our mechanism. We can just show that these electrons would be essentially placed in a form of a pi bond. Of course we will first will have an attack on hydrogen. But then the next hour would indicate that we are forming a pipe on at this position. And because we're forming a pipe on over here, we're breaking the adjacent pi bon and transferring the electrons onto nitrogen. So nitrogen would have a lone pair. This would give us our final product. So let's draw it from the second residence form. We get our final product Which contains our six member ring. Now let's draw our five members ring. We understand that the pi electrons now appear in the form of a lone pair on nitrogen and we are forming a pi bond over here and we have roaming at position number three. Justice indicated by the problem. So, this is indeed our final product. We got our mechanism. Once again, the steps are the attack on the electro file by the pipe on. Keep in mind that we haven't shown the formation of the electro file step because we were not asked to do so, even though I run three bromide and die atomic romain, they are required for this reaction. So we skipped the first step and we are essentially showing that there's an attack on the electro file. We are essentially adding it at position number three over here because that's what the problem suggests. The reaction is much faster at position number three. So we have added our electro file and because we have a lack of electrons in this carbon, we have a positive charge. We drew the second residence form because the original one is the first one. Since we can de localize this lone pair of electrons on nitrogen. Right? So we are forming our new pi bond and we get a formal positive charge on nitrogen. And then we essentially have to remove our hydrogen from the structure. So we are transferring these carbon hydrogen electrons in the form of a pi bond, breaking the adjacent pi bond by transferring the pi electrons in the form of a lone pair on nitrogen. And this is how we get our final product. So we can now label this whole mechanism is our final answer and based on the mechanism concluded that the correct answer choice to the multiple choice question is B. However, if you have any questions, don't hesitate to leave a comment down below. Thank you for watching.

Key Concepts

Electrophilic Aromatic Substitution (EAS)

Resonance and Stability of Sigma Complexes

Position of Substitution in Furan

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