Explain why each compound is aromatic, antiaromatic, or nonaro...

Question

Explain why each compound is aromatic, antiaromatic, or nonaromatic.

(d)

(e)

(f)

Accepted Answer

Hey everyone and welcome back classify each of the following compounds says non aromatic and aromatic or aromatic provide a brief explanations for your answer were given three structures and we just want to have basic rates here for our metis city. So an aromatic compound would correspond to a planner cycling by system. And the second criterion would be the huckle school. There must be four and plus two by electrons where an is an integer valley such as 01234 and so on. Now if a structure is anti aromatic, you still want to make sure that it's plainer in cyclic high system. But the second grades here and it's a bit different. You must have for an pi electrons where N is an integer. Notice how it implies that if the structure is not planer, it cannot be aromatic and that's it. That will be our main criterion for non aromatic structures. Let's begin with the first one. And let's understand if it's plainer. We are going to begin with the criteria and of being a plainer structure. So whenever we are thinking about a plainer structure, we want to make sure that we identify our sp two hybridized carbon atoms. We have these two, then we have these two and this is our carbo Karen recalled that carbo Karen's are also planner because they have three electron regions around them or basically three single bonds and three electron regions imply that this atom is also sp two hybridized. So we have five atoms within our ring that are sp two hybridized. But this oxygen has four electron regions because it has four electron regions. We first of all once understand whether it's actually sp three hybridized or perhaps its lone pair is de localized and its hybridization is different. Right? So let's see if its lone pair can participate in residence. We can just draw our ring. We are going to add our double bonds, there's a carbon carry on. And you may notice that one of the lone pairs of oxygen can actually be the localized. You can create a pi bond over here because you have an adjacent pi bond and you're going to break that pi bond and transfer the electrons in the form of a lone pair on carbon. And this is how you get one of your resonance structures. This would actually be a very minor resident structure. Right, Because we would have a lot of charges. So let's do it a bit differently. Let's actually shift the pi bond immediately over here to stabilize our positive charge or basically neutralize the positive charge. That would be more accurate. So, we can essentially, through our residence form for this one. This would be our sixth member dring. Then we understand that we are creating a pi bond at this position which gives auction a formal positive charge, we still have a pipe on over here and now here we are shifting our pi bond at a different position and that would be our residence form. Notice how we clearly understand now that oxygen has three electron regions. One of the two lone pairs. One single bond, one double bond. So that implies oxygen is actually sp two hybridized. And because it's sp two hybridized, it's also within the plane or structure meaning all of the six atoms form a planar cyclic pi system. Well done. Right. We have all of our atoms being sp two hybridized and we have our pi electrons in the ring. So now we need to check the second rule, how many pi electrons do we have? We have T double bonds, right. Two double bonds times two pi electrons in each would give us four pi electrons. But in addition to that the lone pair of oxygen is also d localized into our pie system. So that means we need to add two more. So we have two pi electrons from oxygen. Those are our initial rain ones. And then if we add them we get a total of six pi electrons. And we can just see that if we equate 6 to 4. And and by six by four we don't get an integer. So let's try this. The first equation for the high cultural we can just equate six pi electrons equals four AM plays too. You may subtract two from both sides and you get four equals for an which implies that an equals one and this is an integer. So our equation for employers to the high cultural is satisfied linearity is satisfied. So we can say that both great syria are satisfied, meaning this structure is aromatic. Let's label that. So it's aromatic. Now let's look at the second structure. First of all, let's label our double bonded carbon atoms because we want to check whether it's plainer or not. So that would be 1234. This is double bonded as well. So once again we have a similar scenario. These are double bonded carbon atoms. So there S P two hybridized. And then we simply want to look at our ring and understand whether this lone pair is the localized or localized. So if we do that, we are going to draw a potential resonance form for number two. This is our original structure. Okay, let's add our pi bonds, our loan pairs. And then we have a metal group. So now there is indeed a possibility to de localize this lone pair because you may notice how we have this adjacent pipe bond. Right? And we can essentially break the pi bond form a new pi bond. A disposition by transferring these electrons here and then break the final pi bond by transferring the lone pair electrons onto oxygen. So this lone pair is actually de localized, meaning it's part of our high system. And if we draw the resultant residence form, we essentially get our double bond over here, auction has three bonds. So it must have a formal positive charge, one lone pair remaining, Then we are adding our pie pons. And now this oxygen would have a formal negative charge because it only has one bond. Right? And you may notice that similarly to what we had before this oxygen is also sp two hybridized. Because now you can clearly see how it has 123 election regions, one single bond, one double bond, one lone pair. And therefore we are going to include it among our sp two hybridized atoms. And you may clearly see that all of our rank has only sp two hybridized atoms, meaning it's playing our cyclic pi system. So the first criterion is satisfied. We can essentially say that yes, it's plainer and cyclic because all of the atoms are SPT and we already know that it has six by electrons. Two double bonds. Just as we said before, two times two gives us four pi electrons. Plus this d localized lone pair gives us additional two by electrons. If we add them we get six. And as we said, if it's six pi electrons, the integer would be an equals one. For the Hucles rule, meaning the second rule is also satisfied. It's actually aromatic. So it's also aromatic list labeled at are automatic. And now for the final one, we are going to check our criteria once again. So let's label our double bonded atoms. So those will be our double bonded carbon atoms. Okay, now we essentially have nitrogen with a lone pair. We can immediately understand that. It's the localized. We can take this lone pair shifted over here, right to create a pi bond And then break the adjacent pipe bond by transferring these electrons at that position. So essentially it would show you that because this lone pairs de localized, there will be 123 electron regions meaning nitrogen is also sp two hybridized. But there is a bit of a problem. This carbon atom, if you notice has t implicit hydrogen atoms. Okay, and because it has two implicit hydrogen atoms, we can count the number of electron regions at this carbon. So essentially there would be 1234 electron regions. And essentially it implies that this carbon atom is sp three hybridized because it's sp three hybridized, the geometrical arrangement around it etcetera. He'd roll. So we can say that number three is not playing or right, It's not playing her. And because the first grades here and it's not satisfied, it's not dramatic. It's not anti aromatic. It essentially implies that it must be non aromatic because the first grades here and it's not satisfied. So we can say that it's non are automatic. Okay, so let's label our answers. The first structure is aromatic. The second one is aromatic. The third one is non aromatic based on our answers. We may conclude that the correct answer to the multiple choice question is the However, if you have any questions don't hesitate to leave a comment down below. Thank you for watching.

Explain why each compound is aromatic, antiaromatic, or nonaromatic.
(d)
(e)
(f)

Key Concepts

Aromaticity

Antiaromaticity

Nonaromaticity

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