In the presence of strong nucleophiles, benzene can do a Nucleophilic Aromatic Substitution, also called SNAR. Unlike EAS, where addition is initiated by the presence of a strong electrophile, addition-elimination can also be initiated by a strong nucleophile in the presence of a good aryl leaving group.
1
concept
General Mechanism
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guys by now. We are pros at E A s. We understand Electra Filic. Aromatic substitution. Very well. But it turns out that that's not the Onley substitution mechanism that bending can undergo. It turns out that in the presence of strong nuclear files bending can actually made to do a nuclear filic Aromatic substitution also called S n a R. So, unlike e. A s, where the addition step is initiated by the presence of a strong electrified, remember that we're always trying to make that strong e plus, and then the benzene attacks. Well, it turns out that you can also start another addition elimination reaction with benzene by a strong nuclear file attacking the benzene. Okay, you might imagine this must be a very strong nuclear file. It's actually attacks such a negative entity. But the thing is, you also need a good leaving group on your bending. So you're gonna need something that you can kick out when the electrons get there. Okay, In some ways, this reminds us a little bit of s and two, because remember, the S and two was a backside attack. It was nuclear. Felix, Substitution. This is also nuclear. Felix substitution. The only thing is that it's not concerted. It's a two step reaction. It's distinct addition and eliminate nation steps. So there's there are similarities. But then there are also differences. Okay, so again, this is called a lot of different things. It's known as instead of E. A s. It's known as nuclear filic aromatic substitution. And it's not abbreviated nes, so don't call it that. That's something different. It's called S N a R. Okay, so the s n a r mechanism is nuclear filic aromatic substitution. Also in some text that's even called the ipso substitution. That just refers to the fact that you have two groups sharing a carbon for a little bit in the intermediate. Okay, so let's go ahead and remind ourselves off E A s. We don't really need to, but quickly run through it so I can see how S N a. R is similar and different at the same time. So remember that your first step is always the slow Step two, create the Sigma complex. Now, this is a catatonic sigma complex because you get ah positive charge on that positive charges distributed throughout the entire thing. So we could just draw that as a dotted line with a positive in the middle. Okay, remember that after the resident structures, etcetera, you get an elimination step, and that's a beta elimination where you grab in age, you read from the double bond and you get a substitution product that was started off by an electric file. So e a s. Well, with S n a r, the reaction is really much, much different. So what we get is that the reaction starts off with a nuclear file attacking the benzene, and it's going to attack the site where there is a strong or a good leaving group. Okay, now, unlike s and to where I immediately would have just kicked out my leaving group and it would have been backside attack. And that's done. This is not a backside attack because there is no backside. And it's two steps. Meaning that we're gonna do is we're gonna break a bond on the benzene and making an Ionic intermediate. So we're going to get instead of a catatonic sigma complex, we're gonna get a negative charge that's distributed throughout the same five atoms. Okay, so this is what we call the an ionic signal complex. So it's similar, but it's a negative charge instead of a positive charge. Okay, So, guys, then what happens? Well, eventually the negative charge is going to reform a double bond. Okay, so eventually, in a negative charge is going to reform a double bond kick out my ex, and I'm going to get in my elimination step. I get my nuclear file substituting where the ex waas. Okay, So, guys, as you can imagine, this an Ionic Intermediate is extremely unstable cause Benson already has so many electrons non putting a full negative charge in there. That's gonna suck. So typically, we're gonna need lots of heat and lots of pressure to make these work. Okay. And in early example of this was a reaction called the Dow process that was actually started by the company dow. And it was an early method to make final, but man, they had toe work for it. So the way they work did it was they got chloral benzene and they reacted Noh, which we know is a strong nuclear file, right, with 350 degrees Celsius and high pressure. So we need all that. So that we can actually make the intermediate. And remember, the intermediate would look like this. You attack and then you break off into a lone pair on the top. Now, what I wanna do is draw the resident structures. Let's do that. So the resident structures for this would look for this? An Ionic Intermediate would look like this. I have C l o h double bond, double bind, negative charge. That's gonna make a bond and break a bond. So I'm gonna get something. Looks like this. And that is going to again make a bond and break a bond, and then I'm going to get something like this. Okay, So what happens at the end at the end, remember? I said there's an elimination step. The elimination is that itself. Eliminating the lone pair reforms a double bond and kicks out the chlorine. Okay, that is going to give us on, basically final now, in this case because it happened in a basic environment, Deep Throat nation does take place. Okay, So usually you're gonna need an extra coolant equivalent of acid to turn the fan oxide. Yeah, into final. Okay, So this was an early method of making females that used the S N a R mechanism. But it wasn't very efficient. This is not how modern day females are made made much more simply in other ways. Okay, so now that we understand the general mechanism of S N A r, let's explore a little bit more on the next page.
The Dow Process (mentioned above), a typical SNAr reaction, requires tons of heat and pressure to proceed forward. This is due to the instability of the anionic sigma-complex. However, Withdrawing groups or Heteroatoms in the Ortho or Para positions (WHOP) will stabilize the intermediate.
2
concept
The Meisenheimer Complex
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Now that we understand a little bit about the snr mechanism, let's talk about something called the Mizzen Heimer complex. So, as mentioned earlier, the Dow process was a typical snr reaction, but it required tons of heat and pressure to proceed forward. Okay, this is due to the instability of the antibiotic Intermediate. But it turns out that scientists figured out that there are ways to naturally stabilize the antibiotic intermediate that are gonna require that are gonna make it require less heat unless pressure and the rule that we use for that is what? So it's gonna be withdrawing groups or hetero atoms in the Ortho pair of positions will stabilize the intermediate. We're basically looking for things. Remember that that an Ionic Intermediate goes through the Ortho and pair of positions relative to the to the nuclear file. Okay. And that's exactly what we're trying to do here. What we're trying to do is we're trying to use atoms in those or thrown pair of positions to stabilize that negative charge. Okay, so a classical trying nitrobenzene, So think about it. Is nitro a good withdrawing group? It's the best. It's one of the best electron withdrawing groups. So if you use a try nitrobenzene Mayes and heimer complex. Um, the reaction can actually proceed forward at room temperature. So what is amazing? Heimer complex. Well, ah. Mazing heimer complex is just what I'm saying. Ah, whop amazing heimer complexes like an ultimate whop where you have a molecule that has either hetero atoms or withdrawing groups in the Ortho and para position. So I'm gonna just say that that's literally those are synonyms of each other. Amazing time of complex. It's just any benzine that is in a whop formation. Okay, so let's take a look at this. So here I have, once again a strong nuclear file. Oh, ch three negative. And I have a leaving group, but notice that on my Ortho and parent positions, I have all withdrawing groups. Okay. Normally, for the Dow process, I would have required 350 degrees Celsius to proceed forward. But it turns out thes withdrawing groups are so stable are so strong that I'm actually gonna able to proceed forward at room temperature. Okay. 35 is a little bit warm for room, but it could be a hot room. Okay, so let's look at this mechanism, Basically, your negative is going to attack the leaving group, but you're gonna make an an ionic Intermediate. Okay, so you're gonna make a negative charge that's now stabilized by my withdrawing group. And we can draw resident structures for this, right? So we would have resident structures, tons of resident structures. Um, let me just show you a few of them were not gonna draw all of them because there's a lot C l Oh, ch three Now notice. What's gonna happen, Guys, is that nitro? Looks like this, and it's on 10 negative. So not only will the negative charge be able to resonate through the ring, it can even resonate with the Nitro group. Right? We can get something. Looks like this so we can get resonant structures that form within the nitro groups giving us something like this, by the way, That was supposed to have a plus. Sorry. So much to draw. Okay, Okay. So see how that resident structure exists, and we can also draw resident structures of the negative charge moving to the next nitro. So then that would be another resident structure. Okay, So altogether, there's going to be like six resident structures. We're not gonna draw all of them, that's for sure. But I'm just trying to show you guys how amazing heimer complex works. So now, anywhere that this negative charge goes, it's stabilized by withdrawing groups. Okay. Oops. Oh, ch three. So, guys, eventually what winds up happening is that the negative charge is going to reform a double bond. Um, it's gonna reform Taliban, and it's gonna kick out the seal, right? And what you're gonna wind up getting is an S N a r product, right? Because you've got, ah, substitution that occurred. But it was for a nuclear filic reason. Okay, It wasn't for an Electra Filic. Um, you know, molecule was for a nuclear molecule that it occurred. Cool, Right. Awesome. So that's what amazing heimer complexes, guys, it doesn't just have to look like this. It could be any combination of withdrawing groups and hetero atoms. So that means, like, if I just put a nitrogen inside the ring here, let me use a different color. If I put wow, if I put a nitrogen inside the ring here, that would qualify as a hetero atom. Okay, because now I have a non carbon atom inside the ring and non carbon atoms arm or election negative. So they're also good at stabilizing negative charges. Okay, so it's not just withdrawing groups. It's also hetero atoms that will help. So we're gonna look at the following two reactions and says use resident structures to determine which of the following ipso substitution is more favored. Remember that ipso substitution is just another name for S n a R O K. So go ahead and look at both of these. Try to draw resident structures on, then figure out which one is going to be the more favored reaction.
3
example
Which ipso-substitution is more favored
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Alright, guys. So, first of all, I don't really need to draw resident structures to know the answer to this question. Because all I'm looking for within S n a r mechanism is I'm looking for wops, right? I'm looking for wops. So the one with more wops is gonna win. Let's analyze the first one. The first one. I've got a hetero atom on the meta, so that's not really gonna help me. Okay, that doesn't really help. And then I got a donating group on the Ortho, so I have head arado on meta donating group on Ortho E d g on Ortho. Does that help me? No, that's terrible. So soon isn't very favored. Let's go to the next one. The next one. I have a hetero atom on Ortho. I have a hetero atom on met up, but that's not really gonna help me, so I'm not gonna write it down. And I have a withdrawing group on meta which again isn't gonna help me, so I'm not gonna write it down. Okay? So, overall, two of these groups don't really help the nitrogen on the meta and the withdrawing group on the meta Don't really help Because notice that, well, I'll draw the resident structure in a second. The resin structure isn't gonna hit those carbons. But I do have this one Adam here that's going to make it more favorite than the other because the other didn't have anything going for it. In fact, it had a bad group. This one at least, has one good group in one good location drawing resonance structures. What you find is that you get something like this where Oops. I should have a sulfur. You got chlorine. You've got O E T. You've got negative. And that negative charge is gonna resonate toe where? Well, it's gonna resonate. So I'm just gonna put here the double bond here. It's gonna resonate to there. It's gonna resonate to here, and it's gonna resonate to here. Okay, so I've actually got a bad group in place because this is gonna push electrons into my negative. That's terrible. Okay, now, on this one, double bond, double one double bond. Well, actually, let me skip. It's all want, actually, what I get for reaction is this attacks here. I moved the electrons down here, so I'm getting a negative charge on that end. Have chlorine. I have O M e myth ox, meth, Oxy. I've got my dotted line representing my an Ionic Sigma complex and my negative charges. Oops, I forgot there's a natural group over here. My negative charges. Two of them aren't stabilized, but one of them is. This one is stabilized because it's on a hetero atom. So this one is gonna be more favored, and this one's gonna be less favorite. Okay. Now, in terms of favorability, you could actually relate that the temperature Meaning that this reaction could occur, possibly at a lower temperature than 3 50. The one above may need more than 3 50 because it's got that donating group that's destabilizing. It makes sense. Cool, guys, let's move on to the next question.
4
example
NAS in the addition-elimination pathway
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So guys, let's look at what this question is asking its saying Which of the following compounds will most readily undergo nuclear filic Aromatic substitution in the addition elimination pathway? It's important for us guys to recognize exactly what's being asked. It says Nuclear Filic. Aromatic substitution in the addition elimination that specifically means S N a R. Okay, so I know it's word a little bit weird. We have to recognize that en nuclear aromatic substitution, specifically with addition elimination is an s n a r. And if it's an S n a R, that means what we're looking for is a what? Okay, it's that easy. So we're just looking for the wa p ist of all. So let's look, we've got a be c d wishing winds. So let's just take a look. So a it doesn't have any wops, so I'm just gonna give it an X. Okay. B has ah hetero atom and a hetero atoms. That's good. So one hetero atom and one hetero atoms in the Ortho positions. I'm only counting the ones that are Ortho apparel came See has a hetero item in the Ortho ah hetero atom in the Ortho in the para and withdrawing group in the or so Someone's looking really good and indeed has a withdrawing group here in the Ortho, and then nothing else. This doesn't help me, and this doesn't help me. Okay, so the best is gonna be See, this is essentially amazing heimer complex, Right? Because I've got in this case, I've got withdrawing groups or hetero atoms in every location. Both my or or those in a pere. So I would expect this to run at a very normal temperature. I wouldn't need ridiculous pressure heat to do this. It would probably run at a very reasonable temperature. Cool. Awesome, guys. So hope that makes sense. Let's move on to the next topic.
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Problem
Provide the structure of the product formed from the reaction of 1-bromo-2,4,6- trinitrobenzene with one equivalent of sodium methoxide.
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B
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D
Sorry for any confusion, guys. Even though Jules uses the abbrevation "NAS", just know that SNAR will be the correct, more direct term to use moving forward:).
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Problem
Provide the major organic product for the following reaction.
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B
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D
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Problem
Provide the major organic product for the following reaction.
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B
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D
8
Problem
Which of the following compounds is most likely to undergo nucleophilic aromatic substitution via the addition-elimination pathway?