Aromatic Heterocycles - Online Tutor, Practice Problems & Exam Prep

Heterocycles are ring compounds containing at least one heteroatom, such as nitrogen or oxygen. Aromaticity in heterocycles, like pyridine, involves evaluating lone pairs on heteroatoms. A heteroatom can donate one lone pair if it is sp³ hybridized and if this donation supports aromaticity according to Hückel's rule. Understanding these criteria is essential for predicting the aromatic nature of heterocycles and their reactivity in organic chemistry.

Heteroatoms

There's only one more type of molecule that we need to be able to predict aromaticity for, and that's heterocycles. What is a heterocycle? Well, that's just going to be any ring that contains at least one heteroatom within it. Now recall what a heteroatom is. That's just going to be any non-carbon atom. That could be something like nitrogen or oxygen, but also phosphorus or sulfur. These are all very common atoms to be found within rings, and when you find that, it's called a heterocycle. One of the best examples of a heterocycle that I could think of is pyridine. Pyridine was a base that we commonly used in Organic Chemistry 1. If you guys remember, it had this basic lone pair that could be used for reactions, especially for acid-base type reactions. Later on in this section, we're actually going to discuss why that lone pair is basic. But for right now, we have to understand that heterocycles are going to present one extra complication to figuring out aromaticity, which is that typically a heteroatom is going to have one or more lone pairs on it. The question is going to be, do I count that lone pair towards the pi conjugated system? For pyridine, would I go ahead and count this lone pair towards the total sum of electrons to determine Huckel's 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 you have to go through to figure that out. It's not just that they either donate or that they don't donate. There are situations in the middle. Let's look at what the rules are. Heteroatoms can choose to donate up to one lone pair each. That means, for example, oxygen has two lone pairs, but only 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 heteroatom already has to be sp3 hybridized to do this. That means that if it was sp2 or sp hybridized, then that lone pair is definitely not getting donated. It's only getting donated if the atom was sp3 to begin with. Alright? But that's not the only thing. We have a second criterion. So one, the heteroatom needs to be sp3, but two, you're only going to donate if it helps to create aromaticity. Meaning that you're not going to donate a lone pair if it goes against Huckel's rule and if you end up getting a number of pi electrons that makes it anti-aromatic or non-aromatic. You would only donate if it makes it aromatic. We could just go back to this example of pyridine. I've already given you some clues. Why don't you guys try to solve the question whether you think that pyridine 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 lone pair count towards Huckel's 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.

Determine heterocycle aromaticity

First of all, do you think pyridine is aromatic? I gave that one away already. Yeah, it's aromatic. But how so? Why is it aromatic? Is that lone pair going to 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 sp². I told you explicitly that you're never going to donate a lone pair unless it's sp³. This one cannot donate, so I'm not even going to consider it. Second of all, even if it was sp³ and it donated, how many pi electrons would you then have? We already have 2, 4, 6. If I were to donate these electrons to the ring, I would get 8. For two reasons, that lone pair is just going to sit there and it's going to be highly accessible. It's not going to be involved with the ring at all. The answer was this would have 0 lone pairs donated. Cool. Go ahead. Try to do the second problem. Try to use the same logic and predict what the aromaticity would be.

Determine heterocycle aromaticity

All right, guys. What did you put for aromaticity? This is aromatic. Good job. 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? First of all, I have to look at the hybridization. What is the hybridization of that nitrogen? It's sp³. That means it's a candidate to be donated. I'm not saying that it absolutely will be but it's a candidate. Second, if I donate those electrons, will it become a 4n+2 number of pi electrons? I already have 2, 4. Yes, it will. If I donate those electrons, I'm going to get 6. I would get 4n+2, so it's going to be aromatic. The answer here is that it's aromatic and one lone pair, I should keep it the same way, one lone pair will donate. Awesome. Does that make sense, guys? We're just going straight off of the rules. Go ahead and try to apply them to problem c.

Determine heterocycle aromaticity

And the answer for c is non-aromatic. Very good. Non-aromatic. Now, why is that? Okay. Because let's just look at the heteroatoms. We've got 2 heteroatoms this time. We know that they could each donate one lone pair if the conditions are right. Let's look at the hybridizations first. Let's just draw these completely. 2 lone pairs, 2 lone pairs. My question is will 1 from each donate? First of all, the hybridizations of both of these happen to be sp₃. 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? The answer is no because if you get this one donating 1 lone pair, this one donating 1 lone pair, how many electrons do you get? You get 8 pi electrons in total because you're going to have 2 from the dual 2 and 2 from the dual one so that's 4 and you're going to get an extra 4 from the lone pairs. That means that it's going to choose to not donate its electrons because that way, it can stay non-aromatic instead of becoming anti-aromatic. 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 is? Remember that if this lone pair doesn't donate, if both of them don't donate, it's not fully conjugated. 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 molecule 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, it wouldn't be beneficial to only donate 1 lone pair since it's not going to be aromatic anyway. It's not fully conjugated. It's kind of an all or nothing. Either they both donate or they both don't donate. This would be no or 0 lone pairs donate. Side note, this molecule is called 1,4-dioxin. If you look at the Wikipedia page for it, 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 analyzed the bond lengths and through the bond lengths, they were able to determine that this is a non-aromatic molecule. I actually edited the page and now, if you go to 1,4-Dioxin, it says that it's a non-aromatic molecule. 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 are checking things but anyway, so that's my good deed for the day. 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 aromaticity there.

Determine heterocycle aromaticity

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 2 Hs. Not every atom can participate in resonance. If it's not fully conjugated, could this ever be aromatic? No. If it's not going to be aromatic, then why would these lone pairs ever donate? I told you the only reason it donates is to help it become aromatic. It's not because it doesn't fulfill the four test tubes. It's not. It's not because it doesn't fulfill the four tests of aromaticity. Anyway, this is non-aromatic and once again, 0 lone pairs will donate. Perfect. Next question. All yours.

Determine heterocycle aromaticity

And the answer for e was 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 heteroatom. We said that the heteroatom will only donate if it fulfills two criteria. First, is it sp3 hybridized? Yes, it is. Let me go to the second one. Second, would donating one lone pair help it to become aromatic? No. That would make it anti-aromatic, right? That would make it have 8 pi electrons which we don't want. So then the answer is going to be that these lone pairs are going to remain outside of the ring. They're going to remain horizontally positioned to the ring and they're not going to participate. But here's the problem. You might be saying, "Johnny, why isn't this an aromatic molecule since I have 2, 4, 6 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. Fully conjugated. 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 flipped into the ring so that it participates in conjugation. The answer is that once again, this would be a 4n number if you did donate. So 0 lone pairs donate. Excellent. Let's move on to the next question.

Determine heterocycle aromaticity

This was a tricky one. The answer is actually aromatic. You might be scratching your brains how that happened. It's not that difficult. First of all, we have to figure out what does this molecule even look like. What kind of lone pairs does it have? Well, nitrogen has one lone pair. Does boron have a lone pair? No. Boron has an empty p orbital. Boron and aluminum are special for always having that empty p orbital. That's kind of like a cation, right? It doesn't have a charge, but it's an empty orbital. Can an empty orbital participate in resonance? Sure. I could totally put my electrons into it, and that wouldn't be a problem. Meaning that this molecule is fully conjugated if the nitrogen will donate its lone pair. Let's see if it will.

• 1, we're looking at the nitrogen. What kind of hybridization does the nitrogen have? sp3.
• 2, will it help to create a 4+2 number if my lone pair donates? Well, let's count it up. I've got 2. I've got 4. This orbital here that is just sitting here, it does participate in conjugation but it doesn't add any electrons. With that orbital, I still just have 4 from the 2 double bonds. Now if I flip this lone pair into the ring, that becomes my 6th electron, my 5th and 6th pi electrons.

The answer is that yes, I would get 4n+2, so one lone pair will donate. Interesting. That's how 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. So that's the same as saying a carbocation basically. Empty orbital that you can stuff electrons into.

Alright. So let's move on to problem g.

Determine heterocycle aromaticity

So guys, you might not have realized how tricky this problem is. But remember that whenever you're dealing with 8-membered rings, what do you have to think about? Or just actually any large annulene, you have to start thinking about planarity. We said that an 8-membered annulene, an 8 annulene or cyclooctatetraene likes to fold like a taco. If this folds like a taco, then the nitrogen donating electrons wouldn't seem to help much because it's not going to be aromatic anyway. But then remember, we also talked about another rule that said, but if the taco can get enough electrons to be aromatic, it will flatten out again kind of like a tortilla. Let's see what goes on here. We've got this molecule that's got 2,4,6,8 pi electrons. Yeah. That triangle could count. They're all fully conjugated. But then you've still got this lone pair on the nitrogen. I'm wondering what's going to happen with that lone pair? Is it going to donate? Is it not going to donate? Well, if it donates, first of all, let's just look at the rules. Is it sp³ hybridized? Yes. 2, if it donated into the ring, would it give you the right number of electrons, the Huckel's rule number? Yes, it would. It would give us 10 pi electrons which would be a 4n+2 number. What did I say happens to an 8-member ring when it has the right number of electrons? It flattens out like a tortilla. It turns out that this actually would be aromatic because of the fact that those electrons can cause it to take the right conformation, and now all of those orbitals will be able to communicate with each other and conjugate with each other. These last ones are getting tricky, guys. Just 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.

Determine heterocycle aromaticity

So what was the answer here, guys? This nitrogen was definitely a little weird compared to the other heteroatoms 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 2 hydrogens on it has no lone pairs should be a giveaway that this thing cannot participate in resonance. Why? Because I said that a heteroatom can only participate in resonance if it donates 1 lone pair. This one has no lone pairs to donate. The answer is that this is going to be non-aromatic. Let me just walk you through this. First of all, we always go through this thing of saying what's the hybridization, would the lone pair donate. But I can't do any of that because I don't even have a lone pair. Let's look at the rest of the molecule. The rest of the molecule, you're right. It has 6 pi electrons, 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 orbitals. It has literally 4 sigma bonds and in order to participate in resonance, we'd have to break a bond to a carbon or to a hydrogen. That doesn't make sense. This is non-aromatic because it cannot participate in resonance with every atom. The nitrogen will be excluded. All right. Super tricky. Let's move on to the last question and then we'll move on to another section.

Determine heterocycle aromaticity

Right, guys. And the final question, I'm going to go ahead and take myself out of the frame so that we can have plenty of space to draw. But as you see, there are actually 3 heteroatoms to consider here and we're going to have to do all of the different criteria with all three of these. First of all, they each have a lone pair. Let's draw those in. I'm just going to erase that. Each of them has a lone pair. We have to take all of their hybridizations and all that stuff. The hybridization of the first one that I drew was sp₂. Then I have sp₃ and then I have sp₂. Right off the bat, how many of these lone pairs are available to donate? The answer is only 1. Only one of these lone pairs is available. The others are not available because they're sp₂ hybridized. That means I don't even have to think about those. That means the first question was I've got 1 sp₃. I'm going to put here 1 sp₃. Now my second question is about pi electrons. Would it make it aromatic if I added those pi electrons in? Let's start counting up our pi electrons. I know that for sure I've got 2 pi electrons with that double bond, 4 with that double bond. Now the pi electrons on the other nitrogens, do they count? Should I go ahead and say 6<8>? Absolutely not guys. Remember that we said these nitrogens cannot contribute, so I should not count those lone pairs. So far all I have is 4. What if I add these lone pairs? That's exactly right. They will help to contribute to Huckel's rule number of pi electrons, meaning that they will donate and this will be aromatic. We would put here 2. I know there's not a lot of space. I'll just add it in over here. One lone pair donates. Crazy, right? Hopefully, this has taught you guys how to navigate molecules with multiple heteroatoms, 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 these should be probably harder than anything you'll experience. Let's go ahead and move on to the next topic.