Which resonance structure, A or B, is most contributing?

Question

Resonance structures A and B are shown, with a question about which is the most contributing structure.

Accepted Answer

Welcome back everyone. Which resonance structure contributes more to the resonance hybrid of South Cyclo hea 25 die in one iodine cyclo heta, 135 tri in A or B. We're given our structure and into resonance forms. And what we notice is that we have our positive charge and structure A for the six numbered ring. And for structure B, we have our positive charge for the seven numbered ring. And let's remember that carbon ions are less stable than carbon CS. Now, the reason is very simple, that's because carbon cion experience a positive conductive effect coming from the carbon carbon bonds, carbon ions. However, get an increase in the negative charge when electrons are supplied and they are less stable. So we actually want to focus on those negative charges and consider the resonance forms when it comes to that negative charge. Starting with structure A, we can essentially redraw the first Rison's form, right? And now what we're going to do, we're going to focus on that negative chart can essentially delocalized that long pair of electrons into our system. So we can control our second resonance form. Let's go ahead and do that So now, based on our arrow, we are simply saying, OK, now let's place that long pair of electrons on a different carbon atom. So first of all, we are forming our pi bond and we are adding our long pair at a different position. Now, we can delocalized that long pair. Once again, let's draw the result in resonance form. So now we said we're getting rid of the negative charge, getting rid of the pie Bo in essential. We can say that. Now we have our pi bond at a different position and we're going to add our result, a negative charge. We can still keep doing this because we have an adjacent pie bond, right? And we're going to form a negative charge once again on a different carbon atom. So let's see what we get from here. Let's remove the negative charge once again and the adjacent pie bond. So now we can clearly see that we get a pi bond at a different position, our negative charge. Now, from here, we can continue, right? Because we can form a pi bond at a different position. We haven't used it yet. So we're going to break the pie Bo and we're not going to worry about the adjacent ring. Essentially, we cannot delocalized our electrons yet. So we're just breaking our pi bond forming a long pair at a different position once again. So let's draw the fifth re form, right? Once again, we're removing the negative charge, removing the Python. Now, what we notice is that we have our negative charge on carbon, the pipe bond gets a new position. And now what else can we do? Can we get a new resins form? Well, essentially, yes, we can. So we can form a new pi bond even though we had it before. But now we will get a negative charge at a different carbon, which means that this is an additional resin structure. So let's draw the sixth one, right? So there are quite a lot we're going to add to result in pi bonds in that negative charge. And now we have to be careful. We need to think if we can get a new one. So yes, we can even though we had to achieve some pie bonds, but now that we will get something slightly different, we will get our negative charge on a carbon atom that um never had a negative charge yet, right? So that means we are getting an additional resin structure. Let's go ahead and draw the seventh one. So now we're getting our pie bond and we're getting our negative charge. Now, we have to think if there's any possibility for an additional one. So if we form our P bond, we will get a negative charge at the same position where we started, right? So we can say that OK, now we have our resin structures, there are seven of them. So this is what we get for A but now what about B, well, essentially for B we're going to redraw A and we are just going to change the positions of the positive and negative charges. So we're not really constrain our positive charge, right? And the resonance in that ring because carbon ions are less stable and we're interested in stabilizing carbon ions more than carbon CS. So what we're going to do now we're going to change our positive and negative charge N two negative and positive. So we have negative within our six membered ring and positive within our seven membered ring. What we notice is that we can form a pie bond and break the adjacent pie bond in one way. So this will give us our second resonance structure. Let's use our arrows and identify the resultant bonds. So now we're going to add that negative charge and the pi bond, we cannot move further in a counter clockwise direction because there's no adjacent pi bond. What we can do is just go in a different direction from the first resonance form. And we can essentially say that if we go in a clockwise direction and use the other adjacent pi bond, we will get our final risen's form, which will be the third one. So essentially, we can say that if we use another color, that negative charge can travel in a clockwise direction. So we will end up with an additional resonance form, right. So now we are breaking our pi bond on the bottom of the ring and we are forming another pi bond getting a negative charge. And we noticed that we only have three resonance forms versus seven and therefore A is more stable. It contributes more the re hybrid. Why? Well, basically because A has seven resins forms versus three resin forms for b the greater number of residence forms implies the higher stability of the negative charge. And that will be our final answer. Thank you for watching.

Which resonance structure, A or B, is most contributing?

Key Concepts

Resonance Structures

Stability of Resonance Structures

Charge Distribution

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