Basicity of Aromatic Heterocycles - Online Tutor, Practice Problems & Exam Prep

Aromatic compounds, particularly heterocycles like imidazole, exhibit remarkable stability and typically resist reactions. When reacting with strong acids, only the lone pairs not involved in maintaining aromaticity can participate in acid-base reactions. For imidazole, the sp² hybridized lone pair on the red nitrogen is basic and can react with the acid, while the blue lone pair is essential for aromaticity and remains unreactive. This selective reactivity ensures the final product retains aromatic characteristics, adhering to Hückel's rule of 6 π electrons.

Heterocycles - Which lone pairs react?

So far, we haven't learned any reactions that aromatic compounds undergo because they're so crazy stable. They don't like to react with anything really. However, you might be asked to react an aromatic heterocycle with a strong acid. In that case, this is going to be an acid-base reaction and nothing more. Nothing actually happens to the aromatic ring. We're just doing an acid-base reaction with lone pairs on the outside of the ring. Let's see how this works. Heterocycles, as you have seen previously, often have multiple lone pairs that are available to react with acids. The question that we have to ask ourselves, and the question you're going to be asking yourself on the exam, is which lone pair do I react with? Because it can be very confusing. These lone pairs don't just say me, me, me. You have to think about it. You have to conclude which lone pair is going to want to react with the acid.

Let's take this example. Here I have a molecule called imidazole. It has 2 nitrogens with lone pairs. It's got this nitrogen with the red lone pair and this nitrogen with the blue lone pair. I'm reacting this with a strong halohydric acid, HX. This could be HCl. It could be whatever. We know that this lone pair is going to be attracted to which atom? That one of the lone pairs would be attracted to which atom on the HX. You've got this incredibly strong dipole. You've got a partial negative and a partial positive. Now these lone pairs are basic, so they're going to be tempted to attack the H and do basically a proton grab. They're going to be attracted to the proton. The question is, will the H attach to the red end or will it attach to the blue end? Or will it attach to both? How do we solve this question? Okay?

The Sp2 Rule

The rule that we're going to use for this is that acids can only react with the lone pairs that are not necessary for aromaticity. Like we just said, if a lone pair is making a molecule stable, why would it make sense for that lone pair to react with an H and then make the molecule non-aromatic? That molecule wants to stay aromatic, so the lone pair that is being donated to the ring is not available at all to be reacted with the HX. That leaves what's left over. What can we actually react with? Sp₂ hybridized lone pairs are basic because if you remember, the sp₃ lone pairs are the ones that are able to donate. Sp₂s were never allowed to donate. So that means if you have an sp₂ hybridized lone pair, that is a basic lone pair because that one is not required to maintain aromaticity on the ring. Meaning that if we were to draw a final product here, would you draw the H attaching to the red nitrogen or to the blue nitrogen? The answer is that to maintain aromaticity, the final answer to this question would simply be this. With a hydrogen here and a plus charge and probably an X negative hanging by pretty close. This looks like it's a reaction; it looks like maybe you're supposed to add an X somewhere. Guys, it's really a lot less complicated than that. It's literally just an acid-base reaction, but you have to pick the right heteroatom. That's the only challenge. If you pick the blue lone pair, you would have been wrong because that would have made a non-aromatic compound. Now notice, let's just double-check. Is this product still aromatic? What do you think? Is this molecule still aromatic? What do we have? Is it a ring? Yes. Is it planar? Yes. Is it fully conjugated? It's only fully conjugated if this lone pair donates. If that lone pair donates, does it have a Huckel's rule number of electrons? It has 2. It has 4. That positive charge doesn't count towards anything. It doesn't add electrons and now it's got 6. This molecule is still aromatic. That means I drew it right. That means I drew this reaction correctly. Now I'm going to give you guys a practice problem. It's a little bit like the next step of this question. Hint: there's only one correct answer. Go ahead and do that and then we'll answer the question.

Clarification Video

Hey, guys. I'm back with a quick correction video. A few of you reached out to me saying that you were confused by a few of the comments that I made in the prior two videos. So here's what happened. In the first video, when I was talking about the reactivity of HX as a strong acid, I mentioned that the lone pairs on imidazole are basic and that they could react with HX. I made it sound like both the red lone pair and the blue lone pair are basic. But then in the second video when I was explaining the rules for basicity, I went on to say that only the red one is basic and the blue one is not basic and is not going to react. So, it might have been a little confusing. Maybe you didn't even catch on to it, but a few students were like, "hey, that's a bait and switch. You told me that they were both basic and now you're telling me that only the red one is basic." So let me explain. Actually, both statements were kind of true, just if you look at it in different ways. The truth of the matter is that those two lone pairs are the most basic part of the entire molecule. Okay? Carbons are not basic. Nitrogen atoms themselves are not basic. It's the lone pairs that are basic, okay? So the truth of the matter is that both of those lone pairs are more basic than anything else and they both have the best shot of reacting with the strong acid. Okay? I wouldn't even consider the carbons to react with the strong acid. However, it turns out that after we learn the rule of basicity, one of the lone pairs is significantly more basic than the other. And the one that is way more basic is the red one because the red one is not necessary for aromaticity whereas the blue one is necessary. So if you have to pick one lone pair, you're going to pick the red one and not the blue one. Okay? Does that make more sense? So I'm just trying to show you how both statements were kind of true and now you have a better idea of how those two statements could make sense at the same time. Could they how could they both be true, okay? So that's it for this video. Let's move on to the next.

Draw the acid/base reaction product

Let's just start off by drawing all the lone pairs. What we notice is that oxygen, since it only has room for 1 lone pair here, that means it must have a positive charge because oxygen is only neutral with 2 lone pairs. If that threw you off, I'm sorry but you just have to write that in yourself on this one. So then what else? This nitrogen has a blue lone pair and this nitrogen has a green lone pair. Basically, it's red versus blue versus green. Only one of them is going to be the winner. Who is it going to be? You could use the rules that I gave you above which is going to at least narrow it down to 2. Let's just talk about that for a second. We said that you're only going to use electrons that are not contributing to the ring. Well, right now, is this ring or is this molecule, this structure, is it aromatic? Is it cyclic? Yes. Is it fully conjugated? If this nitrogen donates, it is. Is it planar? Yes. If this nitrogen donates the green lone pair, would it be a Huckel's rule number molecule? We would have 4 and then 6 with the addition of the green lone pair. What about the red lone pair? Would that one count? No, because it's sp^2. How about the blue lone pair? Would that one count? Again, sp^2, we never count those sp^2 lone pairs because they can't contribute no matter what. This is aromatic right now. You want to make sure that the product of this thing is also aromatic. Is there a lone pair that I can definitely cross out that we're not using? I heard over half of you guys say green. I'm just going delusional at this point but I'm pretty sure I heard that. Green is not available. I'm going to cross that one out. That means we have a choice. It's either red or it's blue. That's what we have to decide between. Are we going to use the oxygen to take an H or are we going to use the nitrogen to get an H? The answer to this one is actually pretty simple. It just has to do with stuff we learned in the acid and base chapter a very long time ago. But it could simply be answered as which of the atoms is more basic? Which of the lone pairs is more basic? Is a nitrogen lone pair more basic or is an oxygen lone pair more basic? What do you think? You got it. You always hear nitrogen is more basic than oxygen. Obviously, if we have a choice between 2 sp^2 lone pairs, you're going to pick the more basic one. The one that's more likely to react with a hydrogen. That being said, that means that my final molecule would look like this and hanging around and this still has a positive charge. It wouldn't look like the H being on the oxygen. So of course the mechanism would have been this: giving the H to the nitrogen meaning that that nitrogen was more basic than the oxygen, so it's obviously going to be the one that attacks the hydrogen. On top of that, the big mistake you really can't make is that you can't say that the top nitrogen would have taken it because that means that you don't have any understanding of aromaticity, and you're just lost. But hopefully not at this point. Okay? So let's just go to the next topic.

Here’s what students ask on this topic:

What determines the basicity of lone pairs in aromatic heterocycles?

The basicity of lone pairs in aromatic heterocycles is determined by their involvement in maintaining the aromaticity of the molecule. Lone pairs that are not required for aromaticity are more basic and can participate in acid-base reactions. For example, in imidazole, the sp² hybridized lone pair on the nitrogen not involved in aromaticity is basic and can react with acids. In contrast, lone pairs that contribute to the aromatic π-electron system are less basic because reacting with them would disrupt the aromatic stability.

Why is the sp² hybridized lone pair on imidazole more basic than the sp³ lone pair?

The sp² hybridized lone pair on imidazole is more basic than the sp³ lone pair because it is not involved in maintaining the aromaticity of the ring. The sp³ lone pair contributes to the aromatic π-electron system, making it less available for protonation. Reacting with the sp³ lone pair would disrupt the aromatic stability, whereas the sp² lone pair can react without affecting the aromaticity, making it more basic.

How does the basicity of lone pairs affect the reactivity of aromatic heterocycles with strong acids?

The basicity of lone pairs affects the reactivity of aromatic heterocycles with strong acids by determining which lone pairs can participate in acid-base reactions. Only the lone pairs that are not essential for maintaining aromaticity can react with strong acids. For instance, in imidazole, the sp² hybridized lone pair on the nitrogen not involved in aromaticity is basic and can react with acids, while the lone pair involved in aromaticity remains unreactive. This selective reactivity ensures the aromatic heterocycle retains its stability and aromatic characteristics.

What is the role of Hückel's rule in determining the aromaticity of heterocycles?

Hückel's rule plays a crucial role in determining the aromaticity of heterocycles. According to Hückel's rule, a molecule is aromatic if it has a planar ring structure with a continuous overlap of p-orbitals and contains (4n + 2) π-electrons, where n is a non-negative integer. For example, imidazole has 6 π-electrons (n = 1), making it aromatic. The rule helps identify which lone pairs are involved in the aromatic π-electron system and which are available for reactions, thus influencing the basicity and reactivity of the heterocycle.

Why is it important to maintain aromaticity in reactions involving aromatic heterocycles?

Maintaining aromaticity in reactions involving aromatic heterocycles is important because aromatic compounds are highly stable due to their delocalized π-electron system. Disrupting this aromaticity would result in a loss of stability. For example, in imidazole, only the lone pairs not involved in maintaining aromaticity can react with acids. This selective reactivity ensures that the final product retains its aromatic characteristics, adhering to Hückel's rule of 6 π-electrons, and thus remains stable.