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

Question

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

(c)

(d)

Accepted Answer

Everyone. And welcome back in today's video, we are going to identify each of the compounds shown below as non aromatic and see aromatic or aromatic. Provide a brief explanation for your answer. Let's keep in mind that whenever a compound is aromatic, It must satisfy the following criteria. No one, that must be a planner cyclic pi system. Number two, there must be four and plus two by electrons where an is an integer. Right? That's where the Hucles rule states. On the other hand, if a structure is anti aromatic, the first requirement stays plainer and cyclic pi system. So that's exactly the same. The main difference is that there must be four and by electrons where an is an integer. If the structure is not a planner cyclic pi system, it's just non aromatic. That said write the first requirement. As long as it's not satisfied, the structure is non aromatic. That's how we're going to approach this problem. So, to begin with, if you will look at the first structure, we want to understand if it's plainer, we are going to label all of our double bonded carbon atoms within the cyclic structure. First of all. First of all, the cyclic, it's a pi system because it has pi bonds around it. Right? And then we want to understand whether or not every carbon atom is sp two hybridized. So all of the carbon atoms in red will be sp two hybridized because there are double bonded, they have three electron regions around them were basically three different bonds. One carbon carbon single bond, one carbon carbon double bond, one, carbon, hydrogen single bonds. Three electron regions implied that this is sp two hybridized. So that's 1234. Next one, all of these are double bonded. And because all of them are double bonded within the ring, that means the structure is plainer. So yes, number one is plain or cyclic pi system. So the requirement is satisfied. And number two, the question is, how many pi electrons do we have? Let's count the number of pi bonds starting with this 11234567. Right, So we got seven pi bonds, Seven pi bonds. Every pi bond has two electrons. So we're going to multiply by two electrons within one pi bond, two pi electrons. Or basically let's fix that. We want to make sure that our dimensional dimensional analysis is consistent and we noticed that we have a total of word scene by electrons. 14 pilots trans. Which equation does it satisfy? is 14 divisible by four. That's not an integer. Right? So we can essentially check if 14 can be equal to four am plus two. We subtract two from both sides. We divided both sides by four and equals three. That's an integer. So it's aromatic because it's plainer, it satisfies the second equation. There are four and plus two pi electrons because the number of electrons is 14. So we can say number +24 and plus two is satisfied, meaning it's aromatic. Now. For the 2nd 1 is the structure planner. That's our first thing to analyze, let's label all of our double bonded carbon atoms within our ring structure. Right. That's our cyclic structure. So we can say we are labeling all of our double bonded carbon atoms. Do we have any atom that is not double bonded? No. We have all of them double bonded. So that means the structure is cyclic and planner. Because all of these carbons are sp two hybridized. And we have our pi system, we have a lot of pi bonds. So the first requirement is satisfied. And what about the number of pi electrons? Let's count the number of pi bonds that B 12 from here. 345678. Now, here we have to be very careful. Yes, we did get eight pi bonds corresponding to eight times to 16 pi electrons. But we have an internal L keen. Right? This double bond in particular is not part of the pi system. In other words, we have to be extra careful whenever we have an internal elke notice how this is an internal elkin because we don't have any hydrogen is bonded to it. Right. We essentially have our carbon carbon bonds everywhere. While all the other double bonds, they essentially contain hydrogen atoms. Okay, so because we have our internal L king, just one of them over here. These two pi electrons are not included in our compound. Whenever we calculate the pi electrons to satisfy either the Hucles rule, which would be for employees to or the four and rule for our medic and anti aromatic compounds. So we are essentially not going to include these two electrons. This structure for more references called piranhas and it has been shown experimentally that the internal elkin part does not belong to our pi system. And we are not going to include these electrons in our calculation meaning even though we are seeing eight pi bonds, only seven pi bonds are included in the calculation. So the number of pi electrons would once again be 14. Keep in mind that there will be 14 by electrons in the in in our cyclic pi system, we are not going to include these extra two. And as we previously said, if it's 14, then n equals three, so it satisfies the four M plus to your rule. And therefore despite the fact that there are these two additional pi electrons, we're not including them because this is an internal l king and therefore the actual number that we're using is 14 pi electrons as a result because the structure satisfies the four employees to rule. This structure is also aromatic. Okay, so let's label this is our answers. The first structure is aromatic And the 2nd 1 is aromatic based on our answers. The correct answers to the multiple choice question is to be 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.
(c)
(d)

Key Concepts

Aromaticity

Antiaromaticity

Nonaromatic Compounds

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