Classify the following compounds as aromatic, antiaromatic, or...

Question

Classify the following compounds as aromatic, antiaromatic, or nonaromatic.

(a) Chemical structure of annulene, a cyclic compound with alternating double bonds, used for classification of aromaticity.

(b) Chemical structure of an annulene compound with labeled bonds for classification as aromatic, antiaromatic, or nonaromatic.

Accepted Answer

Hey everyone and welcome back identify each of the compounds shown below as non aromatic and T aromatic or aromatic. Provide a brief explanation for your answer. Now we're given two structures. We essentially just want to recall criteria for our metis city. Aromatic compounds must be playing our cyclic pi systems and number two they must have four and plus two by electrons as defined by the Hucles rule where N is an integer value, right? Such as +01234 and so on. Now anti aromatic structures will share the same first rule. They must be playing our cyclic pi systems and number two instead of having four. M plus two, we will just have four and I electrons where N is an integer. If the structure is not planer. In other words, the first grades here and is the most important one, then the structure cannot be aromatic. It cannot be anti aromatic. It must be non aromatic. So let's see how that works. First of all, whenever we are thinking about aromatic city, we have to look at our ring structure. Notice how we have our pi system, right? It's cycling. We have our ring. We just want to understand whether or not it's plainer and generally planer implies that all of the atoms are sp two hybridized. So notice how all of the double bonded atoms. If you recall R. S. P two hybridized. So they must be tribunal planer in arrangement. So we are going to label all of our double bonded carbon atoms and you may notice that we have all of them double bond. It's so that implies all of the carbon atoms must be sp two hybridized. You may also recall that sp two hybridized carbon atoms. According to the B. S. C. P. R. Theory produce tribunal plane arrangement which implies that they're playing all right. But let's be extra careful. Yes. Generally. And in most cases this is true. But let's take a look at several internal hydrogen. They are implicit. What about taking these two hydrogen notice that they're pretty puckered right there, essential adjacent to two carbon atoms. Okay, So they are adjacent to these two carbon atoms. And if you try to draw a tribunal plane arrangement, you will essentially end up with having bond angles not being equal to 120° right? So they're these bond angles are much lower. So basically because this hydrant is really squeezed, we cannot draw it in plain Instead is basically being directed out of the plane of the rain because it's puckered or basically here we have a crowded structure. So this hydrogen starts to point out of the plane of the ring G2 that craftiness. And similarly, if we think of another internal hydrogen such as this one, exactly the same story, right? We cannot really do it over here because the bond angle wouldn't be degrees just as defined by the triggering plane arrangement. So these two hydrogen must be must start pointing out of the plane to avoid being puckered, meaning this structure is not planners, we can say number one, not planer and this is where we have to be extra careful. It's not always sufficient to think about sp two hybridization. But also whether or not it's sensible to draw that hydrogen in a tribunal planer arrangement. Or could we actually avoid plainer arrangement by essentially increasing our bond angle or making sure that hydrogen points out of the plane and the crowd nous is decreased. So in this case, that's really what we want to keep in mind. And this concludes that the first structure is not plainer because it's not plain or it doesn't satisfy the first requirement of automaticity or ansi automaticity, meaning we can conclude that the first structure is non aromatic. And this is the reason for our brief explanation why the first structure is not aromatic. It's basically non aromatic. So let's label this is our answer. And for number two. First of all, let's check if the structure is plainer. Let's label our sp two carbon atoms. They are double bonded carbon atoms. And we noticed that we have one of them over here at this position, right. Which is not double bonded. And essentially it has two implicit hydrogen atoms. So let's draw them out. Let's count the number of electron regions around this carbon. So there'll be 1234 single bonds and four single bonds implied that it would be sp three hybridized because it's sp three hybridized. This carbon atom in particular in green, is tetra he'd roll in arrangement. And as a result, our cycling pi system is not planer gets the presence of this sp three hybridized carbon. So once again, The 2nd 1 is not plainer because there is an Sp three hybridized carbon atom in the ring, and therefore because it's not planer, it's not aromatic, it's not anti aromatic. It must be non aromatic as well, just as the previous structure. So we're going to say that it's also non aromatic. Let's label this is our answer to the second problem. And based on our answers, we may conclude that the correct answer to the multiple choice question is B. However, if you have any questions, don't hesitate to leave a comment down below and thank you for watching.

Classify the following compounds as aromatic, antiaromatic, or nonaromatic.
(a)
(b)

Key Concepts

Aromaticity

Antiaromaticity

Nonaromaticity

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