Draw resonance forms to show the charge distribution on the py...

Question

Draw resonance forms to show the charge distribution on the pyrrole structure.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to use residents forms to illustrate the distribution of charge on the pier. Older relative structure shown below, let's recall that residents essentially implies that there must be the localization of a lone pair of electrons across the pi system. That means the most important part to understand is that we have a lone pair on nitrogen. So that's the only lone pair that we see. And there are two adjacent pi bonds which allow us to de localize that lone pair into the pi system. So let's see how that works. First of all, let's recall that the welfare structure is a residence form itself. So this is our first resident structure. Now we are going to draw the second one by first of all, including the residents arrow. And then we need to think about the concept of residence. So the way it works is that you're going to start with the lone pair and you're going to de localize it by forming a pi bond out of it. But because you are forming a pi bond out of it, you have to break the adjacent pi bond and you have to transfer these by electrons onto carbon atom in the form of a lone pair. So we meet now through our second residence form, that would be neither chan we can start by drawing our five member ring and then we will distribute the pi bonds appropriately. Now we are forming a pi bond disposition, we are breaking this pi bond and because we're transferring a lone pair of electrons onto this carbon, we're going to draw a lone pair. And in addition to that, carbon would have a negative charge because we have bonds, there's an implicit hygiene and a lone pair. So that would give us a sum of five or minus five, gives us negative one. Based on the formal charge formula. And then we still have our original pi bond at this position, 19 has four bonds, which according to the formal charge formula, gives it a formal positive charge. Right? That would be 5 -4 bonds. This gives us positive one. We're using five because nitrogen has five valence electrons and that will be our second residence structure. We will just keep going Now, if we start with this lone pair formed, we are going to create a pi bon out of it and break the adjacent pi bon and transfer these by elections in the form of a lone pair. So let's draw our five member ring, This is our double bond. Nigerian still has a formal positive charge. We are forming a pi bond at a different position over here and we are transferring these pi electrons in the form of a lone pair. So let's add a formal negative charge and a lone pair. So that's our resident structure, number three. Well done. And now we are going to think about the 4th 1. So because we have our lone pair of electrons. Once again, we are going to start at the lone pair and we're going to use that lone pair in the formation of a pi bond at this position. And here's our Jason pi bond that we're going to break and we're going to transfer these by electrons onto this carbon. This will give us the 4th residence form, nitrogen. Let's draw our five member ring. Now we notice that we have our original pi bond, we are forming a new pi bond at this position and we get our lone pair at this carbon nitrogen still has a formal positive charge there for bonds. And now we need to think whether this structure number four can undergo residence further or do e o get the same structure. So if we look at a possible residence form, we can use the Sloan pair now and de localize it, forming a new pipe bon and breaking the adjacent pipe. On looking at the structure formed, we control Our five member Green. Once again, we noticed that we have our pipe on over here, nitrogen with a formal positive charge. Now we are transferring a lone pair of electrons onto this carbon, meaning there's a lone pair with a negative charge on that carbon. And we are forming a new pi bon in this position. So this is our fifth form so far, It's different from any other form. And now we need to think what happens next. What if we shift this lone pair over here so that we get our pi bond over here? The original pi bond looks the same as in the original structure. And then because we are going to break this pipe bond when we shift this lone pair over here and then break the pi bond, we are going to transfer these pi electrons onto nitrogen in the form of a lone pair. So that's exactly same structure as structure number one, meaning we are done. There are no more possibilities to draw additional resonance forms. Now looking at all of these forms, notice how each carbon had a negative charge and recall that carbon is less electro negative than nitrogen, nitrogen would not prefer to have a positive charge whenever we have a competition with carbon, we call that a more electro negative atom prefers to have a negative charge. But because nitrogen is more electro negative than carbon and carbon has a negative charge instead, these are minor resonance forms. And this implies that the first form is the most significant one. All of the atoms have an octet, we have formal charges of zero in these structures. We essentially contain och tats, but the formal negative charge is on a less electro negative atom and this is not preferred. So that said, the first form is the most significant form, we are ready to label our answers these five resident structures and looking at them, we can conclude that the correct answer to the multiple choice question is B. However, if you have any questions, don't have this, leave your comment down below. Thank you for watching.

Draw resonance forms to show the charge distribution on the pyrrole structure.

Key Concepts

Resonance Structures

Aromaticity

Charge Distribution

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