Basicity of Aromatic Heterocycles

So far, we haven't learned any reactions that aromatic compounds undergo because there's so crazy, stable they don't like to react with anything. Really, however, you might be asked to react an aromatic, hetero cycle 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. So let's see how this works. Hetero cycles, as you have seen previously, often have multiple lone pairs that are available to active acids. Okay, the question that we have to ask ourselves and the question you're gonna be asking yourself from 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 tow conclude which long pair is gonna want direct with the acid. Let's take this example here I have a molecule called amid YSL amid is all right, and it has to nitrogen is with lone pairs. It's got this nitrogen with the red lone pair, and this nitrogen with the blue lone pair. Okay, I'm reacting. This with a strong halo. Heidrick Acid. Okay, um h X, This could be hcl. It could be whatever. Okay, so we know that this lone pair is gonna be attracted toe which Adam That one of the lone pairs will be attracted to which Adam on the H X. Well, you've got this incredibly strong die poll, right? So you've got a partial negative and a partial positive. Now, these lone pairs are basic, so they're gonna be tempted to attack the h and do basically a proton grab right there gonna be attracted to the proton. So the question is, will the h attached to the red in or will it attached to the blue end, or will it attached to both? How do we solve this question? Okay.

So the rule that we're gonna use for this is that acids can Onley react with lone pairs That air not necessary for air Metis ity, 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 that lone pair that is being donated to the ring is not available at all to be reacted with the H X. So that leaves what's left over right? What can we actually react with? Well, s p to hybridize. Lone pairs are basic because if you remember the SP, three lone pairs of the ones that are able to donate SP two s were never allowed to donate. So that means if you have an SP to hybridize lone pair, that is a basic lone pair because that one is not required to maintain air Metis city on the ring, meaning that if we were to draw final product here, would you draw the h attaching to the red nitrogen or to the blue nitrogen? And the answer is that to maintain Air Metis ity. The final answer to this question would simply be this with ah hydrogen here and a plus charge and probably an xnegative pain by pretty close. Okay, so this looks like it's a reaction. It looks like maybe you're supposed to add an ex somewhere. Guys, it's really a lot less complicated than that. It's literally just a acid base reaction. But you have to pick the right head. Arata. 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? Well, what we have Is it a ring? Yes. Is it plainer? Yes. Is it fully conjugated? Well, it's on Lee fully conjugated If this lone pair donates, if that lone pair donate, doesn't have Ah Hucles rule number of electrons has to has four. That positive charge doesn't count towards anything. It doesn't add electrons. And now it's got six. So this molecule is still aromatic. That means I drew it right. That means I drew this reaction correctly. Okay, so Now I'm going to give you guys a practice problem. It's a little bit. It's like the next step off this question where there's a little bit more to think about. But I believe in you. Okay, think about everything you know about acids and bases. To try to predict with the exact product of this reaction is hint. There's only one correct answer. So go ahead and do that and then I'll answer the question.

Hey, guys, I'm back with a quick correction video. Ah, few of you reached out to me saying that you were confused by a few of the comments that I made in the prior to videos. So this is what happened in the first video when I was talking about the reactivity of H X as a strong acid. I mentioned that the lone pairs on amid YSL our basic and that they could react with HX. I made it sound like both the red lone pair and the blue lone pair, our basic. But then in the second video, when I was explaining the rules for basis ity, I went on to say that on Lee, the red one is basic and the blue one is not basic and it's not going to react. So it was a little bit 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 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 are not themselves are not basic. It's the lone pairs that air 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 learned the rule of basis city, 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 air metis ity, whereas the blue one is necessary. So if you have to pick one lone pair, you're gonna pick the red one and not the blue one. Okay. Does that make more sense? So I'm just trying to show you how, um both statements were kind of true. And now you have a better idea off how those two statements could make sense at the same time. Could how could they both feature? Okay, so that's it for this video. Let's move onto the next one.

So let's just start off by drawing all the lone pairs. And what we notice is that oxygen. Since it on, Lee has room for one lone pair here. That means it must have a positive charge. Right, Because oxygen is on Lee neutral with two lone pairs. So if that threw you off, I'm sorry, but you just have to write that in yourself on this one. Okay, so then what else? This nitrogen has a blue lone pair, and this nitrogen has a green lawn care. Okay, so basically, it's red versus blue versus Green on Lee. One of them is gonna be the winner. Who is it gonna be? Okay, Well, I mean, you could use the rules that I gave you above, which is gonna at least narrow it down to two. So let's just talk about that for a second. We said that you're on Lee going to use electrons that air not contributing to the ring. Well, right now is this ring or is this molecule this structure? Is it aromatic? Well, is it cyclic? Yes. Is it fully conjugated if this nitrogen donates it? ISS okay. Is it plainer? Yes. If this nitrogen donates the green lone pair. Would it be Ah Hucles Rule number molecule. So we would have to four and then six with the addition of the green lone pair. What about the red lone pair with that one count? No, because it's SP two. How about the blue lone pair with that one count again? SP two, We never count those sp two lone pairs because they can't contribute no matter what. This is aromatic right now. So you want to make sure that the product of this thing is also aromatic. So 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. Okay, I'm just going delusional at this point, but I'm pretty sure I heard that. So green is not available. I'm going across that one out. That means we have a choice. It's either read or it's blue. That's the one that we that's what have to decide between. Are we going to use the oxygen to taken age? Or we're gonna use the nitrogen to get an age? Okay. And the answer this one is actually pretty simple. Um, 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 is which of the atoms is more basic, which the one pair is more basic. Is a nitrogen loan Paramore 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. So obviously, if you have a choice between two SP two lone pairs, you're gonna pick the more basic one, the more that's the one that's more likely to react with the hydrogen. Okay, so that being said, that means that my final molecule would look like this and b are hanging around, and this still has a positive charge, and it wouldn't look like the h being on the oxygen. So, of course, the mechanism would have been this okay, 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. Okay, 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 their mentis ity, and you're just lost. But hopefully not at this point. Okay, so let's just go to the next topic.