Explain why each compound or ion should be aromatic, antiaroma...

Question

Explain why each compound or ion should be aromatic, antiaromatic, or nonaromatic.

(d)

(e)

(f) the [20]annulene dication

Accepted Answer

Hey everyone and welcome back in today's video, we are going to identify each of the structures shown below as non aromatic, anti aromatic or aromatic provide a brief explanation for your answer. In this problem, we are seeing three structures and we need to define criteria for us to answer this problem easily. So we have to recall that in order for the compound to be aromatic, It has to satisfy two criteria. First of all, it must be a planner and cyclic pi system. Right. Essentially it implies that you have a ring structure which only contains let's sp two hybridized carbon atoms and that essentially would correspond to adjacent double bonds. So the system would be conjugated, right and number two, if that compound is aromatic. According to the hotel's rule, there would be four M plus two pi electrons. Where an is an integer. In other words, you may recall that as long as you're substituting different integers here from an equal zero up to any number that you wish. In terms of integer, you may get the number of pi electrons that correspond to an aromatic compound. For example, if n equals zero, then you get to buy electrons. If N equals when you get six pi electrons, if n equals two, then you get eight plus two, right? That will be 10 pi electrons. And so on. Whenever you get your electrons, you can essentially solve for an and you can state whether it corresponds to the aromatic compound as long as you get an as an integer. Now the compound is not anti aromatic. If it satisfies the first condition, so if it's a planner and cyclic pi system in the second condition Is that we have four M high electrons. Where an is an integer. In other words, either see it either four pi electrons, right? Because we cannot have 04 times 00 pi electrons, we definitely have to have some sort of pi electrons. So and would be an integer greater than or equal to one. That would be a multiple of four, such as 48, 12, 16 and so on. Now, the compound is not aromatic. If it doesn't satisfy the first grade Syrian, right? If it's not a plane or cycling by system, it cannot be aromatic. So let's see how that works. If you look at the first structure, what we're going to do, we're going to think whether or not the first great Syrian is satisfied. Is it plainer? So we're going to identify all of the double bonded carbon atoms because the double bonded carbon atoms are sp two hybridized. We're going to skip this one for now and we're only going to circle the double bonded carbon atoms in blue. So we are done. Right. We have labeled all of our double bonded carbon atoms and now let's think about the remaining carbon atoms which are carbon ions because the lone pair in each case is the localized. In other words, you may simply draw a residence form. For example, if you take this lone pair, you can actually create a double bond out of it by breaking the adjacent double bond, so you're de localizing this lone pair, meaning this carbon can also have a double bond and therefore it implies that this carbon atoms also sp two hybridized and same goes to the other carbon atom. Right. We can essentially notice that this lone pair is the localized. It's where he easy to draw the residents form. We can create a double bond at this position and break the adjacent pi bond. Right, So both of these lone pairs belong to our system right there. The carbon atoms are also sp two hybridized. So the first criteria and is satisfied. We can say that criteria, number one is satisfied because all of the carbon atoms are sp two hybridized. And this implies that the whole ring structure is plainer and cyclic pi system because we have pi electrons over here within our double bonds. And also the lone pairs are de localized. Now let's think about the second grade Syrian, How many electrons do we have? First of all, we can essentially calculate the number of double bonds. So that would be 12345678. Right, So we have eight double bonds. Every double bond has two electrons. We can say that eight times two pi electrons. And we need to take into account these electrons because they're also the localized into our pie system. Right? They participate in restaurants, meaning we need to add two lone pairs and we can say plus two lone pairs times two electrons or simply four. So if you do the math, you notice that you get eight times two plus four, that would be equal to 20. Now the question is okay if the first criterion is satisfied, it's either aromatic or anti aromatic. But now, which one of the two you will sell? Based on the second criterion? If we get 20 pi electrons, it corresponds to the second equation for N. Right? Because If for n equals 20, then dividing both sides by four gives you n of five. And this is indeed an integer, meaning it satisfies the equation which has for an rather than for M plus two and therefore it corresponds to the anti aromatic compound. We can say that this compound actually not really compound. It's an iron, right? Is anti aromatic and see are automatic. Okay, so we are done with this one. Let's look at the second structure. We may start by looking at our cyclic system. It indeed has double bonded carbon atoms. So we can label them. We immediately noticed that all of the double bonded carbon atoms have been labeled but we actually have carbon atoms that are sp three hybridized. Notice how this carbon atom has an implicit hydrogen atom. So this would be an sp three hybridized carbon atom. Similarly, this one has an implicit hydrogen atom. Right? Because we see three bonds only every carbon atom has four bonds. Right? So this carbon atom is also sp three hybridized and we have two more over here and another one also the double bonds are not conjugated. Right? They're not forming a link. So the first criterion is not satisfied. We may say that the first great Syrian is not satisfied, meaning we can immediately tell that this structure is not aromatic at all. Right. Because we did not satisfy the primary crate Syrian for either being aromatic one or anti aromatic. And simply that's because we don't have that plane or cyclic by system, we have several or actually 1234 sp three hybridized carbon atoms. And therefore this ring structure is not planer and it's not a cyclic by system. Right? So if it's not planer it's not aromatic. And we can conclude that the second one is non aromatic. So let's write it down below non aromatic. Now let's look at the third one. So, the first question is, is it plainer cyclic pi system we see a lot of by electrons here. So let's start by labeling our double bonded carbon atoms and let's see if there are any remaining ones that are not sp two hybridized. So we are labeling our double bonded carbon atoms in red. And now let's think about the remaining ones. Notice how we have all of them labeled except from to these to have positive charges. And now the question is, can we have double bonds at these positions? The answer is yes, we can for example take this double bond and shift it over here. Getting a positive charge here. So the positive charges de localized, meaning this carbon atom is also sp three hybridized and therefore we are ready to include it in our structure. And now thinking about this one same thing, we see that the adjacent double bond can shift over here to produce a positive charge here. So the positive charge is also the localized meaning this carbon atom can have a double bond and therefore it's also sp two hybridized. So we included and we now see that all the atoms are included within the ring. They're all sp two hybridized. And therefore yes, we do have a planer cyclic structure. So we are going to label that the first grade Shirin is satisfied. And now our goal is to count the number of pi electrons to decide between aromatic and anti aromatic. So how many double bonds do we have? Let's start with this double bond,  aN:aN:000NaN 123456789. Right, So we have nine double bonds and every double bond has two pi electrons nine times two Gives us 18 pi electrons. Now let's think about the equation that it can correspond to. Let's start with four. 18 is not divisible by four. Right, It will be 4.5. And we need to make sure that N is an integer. Meaning we're going to look at the second equation If we set equals for M plus two and subtract two from both sides. We get 16 equals for N and divide both sides by four. If we divide both sides by four we get n equals four. And yes, this is an integer. It satisfies the equation for aromatic compounds, meaning we can simply conclude that the the third structure is aromatic. Okay, so now we see that the first structure is anti aromatic. The second one is non aromatic and the third one is aromatic. Let's label. This is our answers to clearly see how we assigned automaticity or non automaticity for our structures and looking at the answer choices because the first one is anti aromatic. The second one is non aromatic and the third one is aromatic. We can conclude that the that the correct answer to this multiple choice question is B. However, if you have any questions, don't have states, leave your comment down below in the comment section and thank you for watching

Explain why each compound or ion should be aromatic, antiaromatic, or nonaromatic.
(d)
(e)
(f) the [20]annulene dication

Key Concepts

Aromaticity

Antiaromaticity

Nonaromaticity

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