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

Question

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

Three chemical structures labeled (i), (ii), and (iii) for aromaticity analysis.

(b)

(c)

Three chemical structures labeled (a), (b), and (c) for aromaticity analysis.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem identify each of the structures shown below as non aromatic, anti aromatic or aromatic provide a brief explanation for your answer, we have three different structures but before we look at each of them, we will just define aromatic city in general. What kind of two things we want to look out for whenever we decide whether something is aromatic or not. So if something is aromatic requirement, number one, which is the primary requirement is to have a planer cyclic I system, planer cyclic Hi system. Okay. If a system is not playing are cyclic essentially that implies we can stop considering our metis city and even anti aromatic compounds, there must obey this requirement. But we will see that soon. And number two. The second requirement for aromatic compounds is to follow the Hucles rule essentially we want to have four and plus two pi electrons. Right, so we want to have these two requirements satisfied for an aromatic compound. Anti aromatic, we can say that once again, it must be plainer cyclic by system. But in addition to that, instead of obeying the Hucles rule, we will now have four and pi electrons in each case and is an integer in the jar and here and is an integer. Okay, so let's look at structure. Number one is a planner. We will essentially consider two aspects, is it planner. And how many electrons does it have whenever we are considering those high electrons. First of all, the question is is a planner. So if we look at the these double bonded carbon atoms there sp two hybridized because they are double bonded carbon atoms and we may recall that the double bonded carbon atoms are sp two hybridized, their tribunal planner and geometry based on the V S E P R theory. Right. So we notice that we have carbon atoms that are sp two hybridized. This carbon atom over here. Right, It's not within the ring system. So we're not worried about this sp three hybridized carbon. So you can neglect it. It's fine. We are only interested in those carbon atoms that are within the ring. And now thinking about this carbon atom because the positive charges de localized over during. You can easily notice that there might be a shift of these pi electrons over here to form a double bond at this position and form a positive charge over here. Then that means this carbon atom is also sp two hybridized because it can contain a double bond. So that means we can also indicate that. Now all of the carbon atoms are sp two hybridized. So requirement number one is satisfied right now, requirement number two. We want to calculate the number of pi electrons. So how many do we have? We have? 1234 double bonds and each double bond has two electrons. So we can say that there are eight pi electrons. Now would that satisfy four AM plus two or four? An equation. You may notice that if you equate eight equals for an and you get an equals two as your integer, meaning it satisfies the second equation, not the first one. So not the Hucles rule, meaning in this case if we get four n, this must be anti aromatic as we said. So just to recap we have a planer cyclic pi system. And for and for and pi electrons because we have a total of four double bonds corresponding to eight electrons. So the integer that allows us to solve this equation is too because the integer works for the second equation. We can conclude it as anti aromatic. Okay, well done. Now let's look at number two. Is it plainer? So once again we have our double bonded carbon atoms there sp two hybridized, indicating that so far we have our planer planer structure. And what about this carbon atom? Once again, this lone pair can be de localized to the presence of adjacent pi bonds. So we can essentially say that it's possible to de localize that lone pair into the ring or into our system. And then we would break the pi bond forming our lone pair over here, meaning it's also sp two hybridized because it participates in resonance. Right, It's part of the structure. So number one, the requirement is satisfied. It's our cyclic pi system. How many electrons do we have? So 1234 double bonds. that's eight electrons plus two more from here because they are part of the system. As you can see these electrons can undergo resonance. Right. They are de localized. They do not belong to that carbon only. So we have eight plus two. That's a total of 10 electrons. And that corresponds to four N plus two. Because if you solve that equation, eight equals for N and N equals two. Right. So it corresponds to the first equation meaning is aromatic. Okay, well done. And now let's look at the final one. The question is, is it plain or let's label all of our double bonded carbon atoms Because we know they are sp two hybridized and they are tribunal planner in nature. Let's not worry about this metal group. Once again, it's not within the ring, it's bonded to the carbon that is part of the ring. And we notice that we nearly have all the carbon atoms except from these two. But again, these electrons are de localized into the ring. Right. We may notice that we can easily draw a residence form here and same thing here, Right. We have adjacent pi bonds. So we can do it multiple ways. We can essentially de localize the lone pair into the ring. And we noticed that because there's that formation of double bonds, these carbon atoms are also sp two hybridized. So they are part of that planer cyclic pi systems. So requirement. Number one is satisfied. Right? And therefore requirement # two, we went to calculate how many pi electrons we have total. So if we count our double bonds. 1234567. Right? So that's seven times two. That's 14, 14. And we also said that these two lone pairs are also part of the the localized system or based lower pi system. So if we add 14 and four more from two lone pairs, we can say 14 from seven double bonds, actually 14 from seven pi bonds. That would be more correct. Right? Plus two plus two. That's plus four. We have 18 total. And now let's check which equation that would correspond to definitely not four. And right, because if we divide 18 by four we get 4.5 and and must be an integer. So we can just check whether it's equal to four AM plus two. And indeed if we subtract two we get 16 equals for an and therefore an equals four. This corresponds to an integer meaning it corresponds to the Hucles rule where it says an aromatic compound must have four AM players two pi electrons. So we can conclude that this structure is aromatic. So these are our answers. The first structure is anti aromatic, the second one and the third one, they are aromatic. Now the correct answer choice to this multiple choice question is d. However, if you have any questions, don't hesitate to leave your comment down below in the comments section and thank you for watching.

Explain why each compound or ion should be aromatic, antiaromatic, or nonaromatic. (a)
(b)
(c)

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Explain why each compound or ion should be aromatic, antiaromatic, or nonaromatic. (a) (b) (c)

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Explain why each compound or ion should be aromatic, antiaromatic, or nonaromatic. (a)
(b)
(c)