SNAr Mechanism

Concept: Concept: General Mechanism

7m
Video Transcript

By now, we are pros at EAS. We understand electrophilic aromatic substitution very well. But it turns out that that's not the only substitution mechanism that benzene can undergo. It turns out that in the presence of strong nucleophiles, benzene can actually be made to do a nucleophilic aromatic substitution also call SNAr.
Unlike EAS where the addition step is initiated by the presence of a strong electrophile. Remember that we’re always trying to make that strong e plus and then the benzene attacks. It turns out that you can also start another addition elimination reaction with benzene by a strong nucleophile attacking the benzene. You might imagine, this must be a very strong nucleophile. It actually attacks such a negative entity. But the thing is you also need a good leaving group on your benzenes.
You’re going to need something that you can kick out when the electrons get there. In some ways, this reminds us a little bit of SN2 because remember that SN2 was a backside attack. It was nucleophilic substitution. This is also nucleophilic substitution. The only thing is that it's not concerted. It's a two-step reaction. It’s distinct addition and elimination step.
There are similarities but then there are also differences. Again, this is called a lot of different things. It's known as instead EAS, it’s known as nucleophilic aromatic substitution. It's not abbreviates NAS. Don’t call it that. That’s something different. It’s called SNAr. The SNAr mechanism is nucleophilic aromatic substitution. Also in some text, it’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.
Let's go ahead and remind ourselves of EAS. We don’t really need to but we’ll quickly run through it so you guys can see how SNAr is similar and different at the same time. Remember that your first step is always the slow step to create the sigma-complex. This is a cationic sigma-complex because you get a positive charge. That positive charge is distributed throughout the entire thing. We could just draw that as a dotted line with the positive in the middle.
Remember that after the resonance structures, etc. you get an elimination step and that’s a beta elimination where you grab an H, you reform the double bond and you’d get a substitution product that we started off by an electrophile, so EAS.
With SNAr, the reaction is really much, much different. What we get is that the reaction starts off with a nucleophile attacking the benzene. It’s going to attack the site where there is a strong or a good leaving group. Unlike SN2, 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 backside attack because there is no backside. It’s two steps, meaning that what we're going to do is we're going to break a bond on the benzene and make an anionic intermediate. We're going to get instead of a catatonic sigma complex, we’re going to get a negative charge that’s distributed throughout the same five atoms. This is what we call the anionic sigma complex. It’s similar but it's a negative charge instead of a positive charge.
Then what happens? Eventually the negative charge is going to reform a double bond. Eventually the negative charge is going to reform a double bond, kick out my X and I'm going to get in my elimination step, I get my nucleophile substituting where the X was. As you can imagine, this anionic intermediate is extremely unstable because benzene already has so many electrons, and putting a negative charge in there, that’s going to suck.
Typically, we're going to need lots of heat and lots of pressure to make these work. In early example of this was a reaction called the Dow Process that was actually started by the company Dow. It was an early method to make phenol. But they had to work for it. The way they worked it was they got chlorobenzene and they reacted NaOH, which we know is a strong nucleophile with 350 degrees Celsius and high-pressure. We need all that so that we can actually make the intermediate. Remember the intermediate would look like this. You attack and then you break off into a lone pair on the top.
What I want to do is draw the resonance structures. Let's do that. The resonance structures for this would look for this anionic intermediate would look like this. I have Cl, OH, double bond, double bond, negative charge. That's going to make a bond and break a bond. I’m going to get something that looks like this. That is going to again make a bond and break a bond. Then I'm going to get something like this.
What happens at the end? At the end, remember I said there’s an elimination step. The elimination is that it's self-eliminating. The lone pair reforms a double bond and kicks out the chlorine. That is going to give us basically phenol. In this case because it happened in a basic environment, deprotonation does take place. Usually you’re going to need an extra equivalent of acid to turn the phenoxide into phenol.
This is an early method of making phenols that used the SNAr mechanism. But it wasn't very efficient. This is not how modern-day phenols are made. Made much more simply in other ways. Now that we understand the general mechanism of SNAr, let's explore a little bit more on the next page.

Concept: Concept: The Meisenheimer Complex

7m
Video Transcript

Now that we understand a little bit about the SNAr mechanism, let's talk about something called the Meisenheimer Complex.
As mentioned earlier, the Dow Process was a typical SNAr reaction but it required tons of heat and pressure to proceed forward. This is due to the instability of the anionic intermediate. But it turns out that scientists figured out that there are ways to naturally stabilize the antibiotic intermediate that are going to make it require less heat and less pressure. The rule that we use for that is WHOP. It's going to be withdrawing groups or heteroatoms in the ortho para positions will stabilize the intermediate. We're basically looking for things. Remember that that anionic intermediate goes to the ortho and para positions relative to the nucleophile.
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 ortho and para positions to stabilize that negative charge. A classical trinitrobenzene. Think about it. Is nitro a good withdrawing group? It’s the best. It's one of the best electron-withdrawing group. If you use a trinitrobenzene Meisenheimer complex, the reaction can actually proceed forward at room temperature.
What is a Meisenheimer complex? A Meisenheimer complex is just what I'm saying, a WHOP. A Meisenheimer complex is like an ultimate WHOP where you have a molecule that has either heteroatoms or withdrawing groups in the ortho and para position. I'm going to just say that that's literally those are synonyms of each other. A Meisenheimer complex is just any benzene that is in a WHOP formation.
Let's take a look at this. Here I have, once again, a strong nucleophile, OCH3- and I have a leaving group. But notice that on my ortho and para positions, I have all withdrawing groups.
Normally for the Dow Process, I would have required 350 degrees Celsius to proceed forward. But it turns out these withdrawing groups are so stable, are so strong that I'm actually going to be able to proceed forward at room temperature. 35 is a little bit warm for room but it could be a hot room. Let's look at this mechanism. Basically your negative is going to attack the leaving group but you're going to make an anionic intermediate. You're going to make a negative charge that's now stabilized by my withdrawing group and we can draw resonance structures for this. We would have resonance structures, tons of resonance structures.
Let me just show you a few of them. We're not going to draw all of them because there's a lot – Cl, OCH3. Notice what's going to happen is that nitro looks like this. Not only will the negative charge be able to resonate through the ring, it can even resonate with the nitro group. We can get something that looks like this. We can get resonance structures that form within the nitro groups giving us something like this. By the way, that was supposed to have a plus.
See how that resonance structure exists. We can also draw resonance structures of the negative charge moving to the next nitro. Then that would be another resonance structure. Altogether there's going to be like six resonance structures. We're not going to draw all of them, that's for sure. But I'm just trying to show you guys how a Meisenheimer complex works. Anywhere that this negative charge goes, it's stabilized by withdrawing groups.
Eventually what winds up happening is that the negative charge is going to reform a double bond and it's going to kick out the CO. What you're going to wind up getting is an SNAr product because you've got a substitution that occurred but it was for a nucleophilic reason. It wasn't for an electrophilic molecule. It’s
for a nucleophilic molecule that had occurred.
That's what a Meisenheimer Complex is. It doesn't just have to look like this. It could be any combination of withdrawing groups and heteroatoms. That means if I just put a nitrogen inside the ring here, that would qualify as a heteroatom because now I have a non-carbon atom inside the ring and non-carbon atoms are more electronegative. They're also good at stabilizing negative charges. It's not just withdrawing groups. It’s also heteroatoms that will help.
We're going to look at the following two reactions that says use resonance structures to determine which of the following ipso-substitutions is more favored. Remember that ipso-substitution is just another name for SNAr. Go ahead and look at both of these. Try to draw resonance structures and then figure out which one is going to be the more favored reaction.

Concept: Example 1: Which ipso-substitution is more favored

4m
Video Transcript

First of all, I don't really need to draw resonance structures to know the answer to this question because all I'm looking for with an SNAr mechanism is I'm looking for WHOPs. The one with more WHOPs is going to win.
Let's analyze the first one. The first one, I've got a heteroatom on the meta. That's not really going to help me. That doesn't really help. Then I've got a donating group on the ortho. I have a heteroatom on meta, donating group on ortho. EDG on ortho. Does that help me? No. That's terrible. This one isn't very favored.
Let's go to the next one. The next one I have a heteroatom on ortho. I have a heteroatom on meta but that's not really going to help me, so I’m not going to write it down. I have a withdrawing group on meta which again isn't going to help me. I'm not going to write it down. Overall, two of these groups don't really help. That nitrogen on the meta and the withdrawing group on the meta don't really help because notice that, I’ll draw the resonance structure in a second. The resonance structure isn't going to hit those carbons. But I do have this one atom here that's going to make it more favored 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 and one good location.
Drawing the resonance structures, what we'd find is that you get something like this where you’ve got chlorine, you've got OEt. You’ve got negative and that negative charge is going to resonate to where? It's going to resonate. I'm just going to put the double bond here. It's going to resonate to there. It’s going to resonate to here and it's going to resonate to here.
I've actually got a bad group in place because this is going to push electrons into my negative. That's terrible. What I get for the reaction is this attacks here. I move the electrons down here. I wind up getting a negative charge on that end. I have chlorine. I have OMe, methoxy. I've got my dotted line representing my anionic sigma complex and my negative charges. I forgot there's a nitro group over here. My negative charge is two of them aren't stabilized but one of them is. This one is stabilized because it’s on a heteroatom. This one is going to be more favored. This one’s going to be less favored.
In terms of favorability, you could actually relate that to temperature. Meaning that this reaction could occur possibly at a lower temperature than 350. The one above may need more than 350 because it's got that donating group that's destabilizing it.
Makes sense? Let's move on to the next question.

Concept: Example 2: NAS in the addition-elimination pathway

3m
Video Transcript

So guys, let's look at what this question is asking. It's saying which of the following compounds will most readily undergo nucleophilic aromatic substitution in the addition-elimination pathway. It's important for us guys to recognize exactly what's being asked, it says nucleophilic aromatic substitution in the addition-elimination that specifically means S N A R. So I know it's worded a little bit weird but we have to recognize that N, nucleophilic 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 whop. It's that easy. So we're just looking for the whoppiest of all, so let's look.

We've got A, B, C, D, which one wins? So let's just take a look. So A doesn't have any whops so I'm just going to give it an X. B has a heteroatom and a heteroatom so that's good. So one heteroatom and one heteroatom in the ortho position, I'm only counting the ones that are ortho or para. C has a heteroatom in the ortho, a heteroatom in the para and a withdrawing group in the ortho. So that one's looking really good and then D has a withdrawing group here in the ortho and then nothing else. This doesn't help me and this doesn't help me. So the best is going to be C. This is essentially a meisenheimer complex because I've got in this case I've got withdrawing groups or heteroatoms in every location, both my ortho's and a para so I would expect this to run at a very normal temperature, I wouldn't need ridiculous pressure and heat to do this. It would probably run at a very reasonable temperature. Cool, awesome guys so hope that made sense. Let's move on to the next topic.

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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Problem: Provide the major organic product for the following reaction.

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Problem: Provide the major organic product for the following reaction. 

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Problem: Which of the following compounds is most likely to undergo nucleophilic aromatic substitution via the addition-elimination pathway? 

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Problem: Which of the following compounds is most likely to undergo nucleophilic aromatic substitution via the addition-elimination pathway? 

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SNAr Mechanism Additional Practice Problems

Rank the following compounds in descending order of reactivity toward hydroxide ion with the most reactive first and the least reactive last.

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 Predict the product(s) for the following reaction. When appropriate, Iabel major and minor.

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Which of the following is not true about Meissenheimer complexes?

a. They are resonance-stabilized anions

b. They are formed upon addition of a nucleophile to aryl halides

c. They are aromatic

d. They are intermediates in nucleophilic aromatic substitution reaction which take place by an addition-elimination mechanism 

e. All of them are true

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Predict the major product for the following reaction paying attention to the regio- and stereochemistry.

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Predict the major product for the following reaction paying attention to the regio- and stereochemistry.

 

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Place the following molecules in order of their reactivity with CH3S– Na+ via an SnAr mechanism. Label the most reactive as #1 and the least reactive as #4. Work carefully. There will be no partial credit or regrades on this problem.

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Draw the organic product(s) for the following reaction. Indicate stereochemistry where appropriate. Assume an aqueous workup, when necessary. A reasonable answer may be “ No Reaction.”

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Draw a reasonable mechanism for the reaction below. 

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Predict the principal organic product of the following reaction.

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Predict the principal organic product of the following reaction.

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Draw the structural formula of the major organic product in the box below.

 

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Predict the major product for each of the following reactions paying attention to the regio- and stereochemistry. If there is no reaction, write just “No Reaction.”

 

 

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