Aromatic Heterocycles

There's only one more type of molecule that we need to be able to predict air Metis City for, and that's hetero cycles. So what is a hetero cycle? Well, that's just gonna be any ring that contains at least one hetero Adam within it. Now, recall what a hetero atom is. That's just gonna be any non carbon atom. That could be something like nitrogen or oxygen. But also phosphorus or sulfur thes air. All very common Adams we found within rings. And when you find that it's called a hetero cycle, Okay, Now, one of the best examples of a hetero cycle that I could think of is pure Dean Perdon was a base that we commonly used in organic chemistry one. And if you guys remember it had this basic lone pair that could be used for reactions, especially for acid based type reactions. Well, later on in this section, we're actually gonna discuss why that lone pair is basic. But for right now, we have to understand that hetero cycles are going to present one extra complication to figuring out their Metis city, which is that typically a hetero atom is gonna have one or more lone pairs on it, and the question is gonna be Do I count that lone pair towards the pie conjugated system? So, for period Dean, would I go ahead and count this lone pair towards the total sum of electrons to determine Hucles rule? Or would I ignore it? Well, it turns out that it's not a clean and simple rule. There's actually a few steps that have to go through to figure that out. It's not just that they either donate or that they don't donate. There's situations in the middle, so let's look at what the rules are. Okay? Hetero atoms can choose to donate upto one lone pair each. That means, for example, oxygen has two lone pairs, but on Lee, one of them is able to be donated into the ring. Now, why would oxygen want to donate one of its lone pairs to the ring? Let's take a look. One the oxygen or whatever. Hetero atom already has to be SP three hybridized to do this, so that means that if it was SP two or SP hybridized thin, that lone pair is definitely not getting donated. It's on Lee getting donated. If the Adam was SP three to begin with. All right. But that's not the only thing. We have a second criteria. So one the hetero atom needs to be sp three. But to you're on Lee gonna donate if it helps to create Ara Metis ity, meaning that you're not going to donate a lone pair if it goes against Hucles rule. And if you wind up getting a number of pie electrons that makes it anti aromatic or non aromatic, you would Onley donate if it makes it aromatic. Okay, so we could just go back to this example of Paradyne. I've already given you some clues. Why don't you guys try to solve for question A. Whether you think that pure dean is an aromatic compound or not, but also predict whether you think this lone pair will donate to the ring or won't donate? Basically, should this loan pair count towards Hucles rule or should you just ignore it? Go ahead and try to use those two rules and then I'll explain the logic behind it. It's all your turn now

So first of all, do you think parodying is aromatic? I gave that one away already, so yeah, it's aromatic. But how so? Why is it aromatic? Is that lone pair gonna donate or not? Actually, for two reasons, it's not going to donate its lone pair. First of all, let's look at the hybridization of this nitrogen. What is that? Hybridization, Guys, that hybridization is s p two, I told you, is explicitly that you're never going to donate a lone pair unless it's s p three. So this one cannot donating, cannot donate, so I'm not even going to consider it. Okay, Second of all, even if it waas sp three and it donated how maney pie electrons would you then have we already have 246? If I were to donate these electrons to the ring, I would get eight. So for two reasons, that lone pair is just gonna sit there and it's gonna be highly accessible. It's not gonna be involved with the ring at all. So the answer was, this would have zero lone pairs donated. Cool. Okay, So go ahead. Try to do the second problem. Try to use the same logic and predict what the narrative city would be

Alright, guys. So when you put for electricity, this is aromatic. Good job. Okay, so this is an aromatic molecule. How so? Well, notice that the nitrogen has one lone pair and we have to ask ourselves, would that lone pair donate or not? So, first of all, I have to look at the hybridization. What is the hybridization of that nitrogen? It's s p three. So that means it's a candidate to be donated. Okay, I'm not saying that it absolutely will be, but it's a candidate. Okay, to if I donate those electrons, will it become Ah four n plus two. Number of by elections already have +24 Yes, it will. If I donate those electrons, I'm going to get six. So I would get four end plus two. So it's gonna be aromatic. So the answer here is that aromatic and one lone pair e should keep it the same way. One lone pair Will Dhoni. Awesome! Okay, Does that make sense? Guys, we're just going straight off of the rules. Go ahead and try to apply them. Thio problem. See

and the answer for C is non aromatic. Very good, non aromatic. Now why is that okay? Because let's just look at the hetero atoms we've got to hetero atoms. This time we know that they could each donate one lone pair if the conditions are right. So let's look at the hybridization is first. So let's just draw these completely to lone pairs to lone pairs. My question is, will one from each donate So first of all, the hybridization of both of these happen to be SP three, so both of them can basically qualify to donate a lone pair. Now the question is, will it be beneficial for them both to donate a lone pair? And the answer is no. Because if you get this one donating one lone pair, this one donating one lone pair how many electrons do you get? You get eight pi electrons in total because you're gonna have to from the door to and to from door once that's four, and you're gonna get an extra four from the lone pairs. Okay, so that means that it's going to choose to not donate its electrons, because that way you can stay non aromatic instead of becoming anti aromatic. Okay, you might be wondering, but, Johnny, why doesn't it just have one of the lone pairs donate, and then the other one stays the way it iss? Well, guys, remember that if this lone pair doesn't donate, if both of them don't donate, it's not fully conjugated. So if all I did was donate the top one and I kept the bottom one the way it is, then this is not a fully conjugated molecules because that lone pair isn't participating in conjugation. In order to participate in conjugation, the lone pair has to flip into the ring. So anyway, it wouldn't be beneficial toe Onley donate one lone pair. Since it's not gonna be aromatic anyway, it's not fully conjugated, so it's kind of an all or nothing. Either they both donate or they both don't donates. This would be no or zero lone pairs donate. Okay, side note this molecules called 14 dioxin. And if you look at the Wikipedia page for it, um, it used to say that it was an anti aromatic molecule and I was like, That's wrong. So I went to some primary literature. I looked up like this scientific book that actually analyze the bond lengths and through the bond lengths they were able to determine that this is a non aromatic molecule. So I actually edited the page in. Now If you go toe 14 dioxin, it says that it's a non aromatic molecules. So I'm so nerdy that I'm actually writing Wikipedia articles about molecules. All right, so just kind of a side note for you guys to know Wikipedia isn't always right. It's constantly being updated. Thankfully, you got smart people on there, that air checking things. But anyway, so that's my good deed for the day I was. I saved Wikipedia from one small, tiny little error. All right, so let's go ahead and go to molecule D and see if you guys can predict the right air electricity. There

so molecule D is non aromatic. Great job. I know a lot of you got that? Why? Because, guys, it is not fully conjugated. Notice that this carbon has to ages. So not every atom can participate in resonance. If it's not fully conjugated, could this ever be aromatic? No. Okay, If it's not gonna be aromatic, then why would these lone pairs ever donate? I told you the only reason it donates is to help it to become aromatic. But this was a lost cause from the beginning, because not even fully conjugated. It's like me taking a straight chain and asking you if the lone pairs gonna donate, it's not because it doesn't fulfill the four the four tests of their authenticity. So anyway, this is non aromatic, and once again, zero lone pairs. Well, Donnie, perfect. Next question. All yours.

and the answer for e iwas non aromatic. Okay, so why is it non aromatic? Maybe you're getting the hang of this by now. Because guys, let's analyze the hetero Adam. Okay, We said that the hetero atom will only donate if it fails to criteria. One is it s p three hybridized. Yes, it iss So then we go to the second one to Would donating one lone pair help it to become aromatic? No, that would make it anti aromatic, right, That would make it have ate pie electrons, which we don't want. So then the answer is gonna be that these lone pairs, we're going to remain outside of the ring. They're going to remain horizontally position to the ring and they're not gonna participate. But here's the problem. You might be saying, Johnny, why isn't this an aromatic molecule? Since I have 246 electrons and I'm not counting these, right, But that's what I was trying to say earlier. If you don't count the electrons on the O, then this is not fully conjugated, it's not fully conjugated. That's a G fully conjugated. So if it's not fully conjugated, then there's no way that it could be aromatic. Get it? Conjugation happens when you have a lone pair flip into the ring so that it participates in conjugation. Okay, so the answer is that once again, this would be, you know, this would be, ah, four n number if you did. Don't eat. So zero lone pairs, Tony. Excellent. So let's move on to the next question.

This was a tricky one. The answer is actually aromatic. Okay, so you might be scratching your brains how that happened. It's not that difficult. So first of all, we have to figure out What does this molecule even look like? What kind of lone pairs doesn't have? Well, nitrogen has one lone pair. Does Boron have a lone pair? No. Born has an empty P orbital. Okay, Boron aluminum er special for always having that empty p orbital. That's kind of like a cat ion, right? It doesn't have a charge, but it's an empty orbital. Okay, so can an empty orbital participate in resonance? Sure. I could totally put my electrons into it. Okay. And that wouldn't be a problem. Meaning that this molecule is fully conjugated if the nitrogen will donate. It's alone. Pair. Let's see if it will. So one. We're looking at the nitrogen. What kind of hybridization does the nitrogen have? S p three to Will it help to create a foreign plus two number if my loan pair donates? Well, let's count it up. I've got to I've got four This orbital here that is just sitting here. It does participate in conjugation but it doesn't add any electrons. So that orbital with the orbital I still just have four from the two double bonds. Now, if I flip this lone pair into the ring, that because my sixth electron, my 5th and 6th pie electrons. So the answer is that, yes, I would get foreign plus two. So one lone pair will. Tony. Interesting, right? So that time I tricked you a little bit with Boron because you probably weren't thinking that boron could be part of conjugation, but it has an empty orbital's that's the same assaying a crumble cat. I am basically empty orbital that you could stuff electrons into All right, so let's move on to problem G.

so guys, you might have not realized how kind of tricky this problem is. But remember that whenever you're dealing with eight members rings, would you have to think about. We're just actually any large annual lean. You have to start thinking about play narrative, right? And we said that an eight member annually in an eight annual inner cycle, Octa tetra in likes to fold like a taco. So if this folds like a taco than that nitrogen, donating electrons wouldn't seem to help much because it's not gonna be aromatic anyway. But then remember, we also talked about another rule that said, But if the taco could get enough electrons to be aromatic, it will flatten out again. Kind of like a tortilla. Alright, so let's see what goes on here. We've got this molecule that's got to 46 eight pi electrons. Yeah, that triangle could count. There's awfully conjugated, but then you've still got this lone pair on the nitrogen and I'm wondering what's gonna happen with that loan. Perisic and donate. Is it not going to donate? Well if it donates. First of all, let's just look at the rules is it s p three hybridize. Yes. Sorry. So is it s p three hybridized? Yes, to, um if it donated into the ring, would it give you the right number of electrons? The Hucles rule number? Yes, it would. It would give us 10 pie electrons, which would be a four end plus two number. So what did I say? Happens to an eight member during when it has the right number of electrons. It flattens out like a tortilla. So it turns out that this actually would be aromatic because of the fact that those electrons can cause it to take the right confirmation. And now all all those orbital's will be able to communicate with each other and congregate with each other. Okay, so these last ones were getting tricky, guys, So just kind of do your best, and then I'll explain them along the way. Here's another tricky one. Try to do your best with H, and then I'll explain it

So what was the answer here, Guys, this nitrogen was definitely a little weird compared to the other hetero out of if you looked at now, if you just looked at the positive charge and didn't think about lone pairs, you could probably confuse yourself on this question. But the fact that this nitrogen with two hydrogen on it has no lone pairs should be a giveaway that this thing cannot participate in residence. Why? Because I said that Ah, hetero atom. Can Onley participate in residence if it donates one lone pair, This one has no lone pairs to donate. So the answer is that this is gonna be non aromatic. So let me just walk you through this. First of all, we always go through this thing of saying What's the hybridization with the lone pair donate. But I can't do any of that because I don't even have a lone pair. So let's look at the rest of the molecule. Will the rest of the molecule You're right, it has six pile electron. So you're thinking maybe aromatic. But it's not fully conjugated once again, not fully conjugated because this nitrogen doesn't have any empty orbital's. It has literally four Sigma bonds, and in order to participate in residence, we have to break a bond toe a carbon or toe a hydrogen that doesn't make sense. So this is non aromatic because it cannot participate in resonance. With every atom, the nitrogen would be excluded. Alright, super tricky. So let's move on to the last question and then we'll move on to another section.

Alright, guys, in the final question, I'm going to go ahead and take myself out of the frames so that we can have plenty of space to draw. But as you see, there's actually three hetero atoms to consider here. And we're gonna have to Do you know, all of the different criteria with, you know, all three of these. So first of all, they each have a lone pair, so it struggles in. I'm just gonna erase that. Each of them have a lone pair. We have to take all of their hybridization and all that stuff. So the hybridization of the first one that I drew was SP two. Then I have SP three and then I have SP two. Okay, so right off the bat. How many of these lone pairs are available to donate? The answer is on Lee one on Lee. One of these lone pairs is available. The others are not available because there s p two hybridized, So that means I don't even have to think about those. So that means the first question was, you know, I've got one s p three. I'm gonna put here one sp three. So now my second question is about pie electrons about Would it make it aromatic if I added those pie electrons in? So let's start counting up rpai electrons. So I know that for sure. I've got to pi electrons with that double bond for with that double bond. Now the pie electrons on the other nitrogen is Do they count? Should I go ahead and say six and eight? Absolutely not. Guys, remember that we said Thies nitrogen is cannot contribute. So I should not count those lone pairs. So so far all I have is for so what if I add these lone pairs? That's exactly right. They will help toe contribute Tau Hucles rule number of pie electrons meaning that they will donate and this will be aromatic. Okay, so we would put here to I know there's not a lot of space. I'll just add it over here. One lone pair donates crazy, right? So hopefully this has taught you guys how to navigate molecules with multiple hetero atoms. Okay, it's not that hard. If you use my system barely any thinking involved, you just have to be consistent about how you apply the rules. Okay, so that's pretty much. I threw the hardest problems I could at you, so he should be probably harder than anything you'll experience. Let's go ahead and move on to the next topic.