Nucleophilic Aromatic Substitution - Online Tutor, Practice Problems & Exam Prep

Benzene can undergo nucleophilic aromatic substitution (SNAR) in the presence of strong nucleophiles, leading to an anionic sigma complex. This two-step reaction involves an initial nucleophilic attack followed by elimination of a leaving group. The stability of the anionic intermediate can be enhanced by electron-withdrawing groups in the ortho and para positions, forming a Meisenheimer complex. This allows reactions to proceed at lower temperatures, exemplified by the Dow process for phenol synthesis. Understanding these mechanisms is crucial for mastering aromatic substitution reactions.

General Mechanism

By now, we are pros at Electrophilic Aromatic Substitution (EAS). We understand electrophilic aromatic substitution very well. But it turns out that's not the only substitution 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 called SNAr. The addition step is initiated by the presence of a strong electrophile; remember that we're always trying to make that strong electrophile, 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 to actually attack such a negative entity. But then, 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 steps. There are similarities, but then there are also differences. Again, this is called a lot of different things. It's known as nucleophilic aromatic substitution, and it's not abbreviated as NAS. Don't call it that. That's something different. It's called SNAr. The SNAr mechanism is nucleophilic aromatic substitution. Also, in some texts, 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 let's 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, and that positive charge is distributed throughout the entire thing. We could just draw that as a dotted line with a 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, reform the double bond, and get a substitution product that was 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, and it's going to attack the site where there is a strong or a good leaving group. Unlike SN2 where I would have just immediately kicked out my leaving group and it would have been a backside attack and that's done. This is not a backside attack because there is no backside and 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 cationic 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 with 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 leaving group, and I'm going to get in my elimination step, I get my nucleophile substituting where the leaving group was. As you can imagine, this anionic intermediate is extremely unstable because benzene already has so many electrons. Now I'm putting a full negative charge in there, that's going to be difficult. Typically, we're going to need lots of heat and lots of pressure to make these work. An 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 man, they had to work for it. The way they did it was they got chlorobenzene and they reacted it with NaOH which we know is a strong nucleophile at 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 anionic intermediate would look like this. I have chlorobenzene, double bond, double bond, negative charge. That's going to make a bond and break a bond. So I'm going to get something that 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. 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 r

The Meisenheimer Complex

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 that anionic intermediate that are going to make it require less heat and less pressure. The rule that we use for that is WAP. It's going to be withdrawing groups or heteroatoms in the ortho and para positions will stabilize the intermediate. We're basically looking for things. Remember that the anionic intermediate goes through 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 groups. 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 WAP. A Meisenheimer complex is like an ultimate WAP where you have a molecule that has either heteroatoms or withdrawing groups in the ortho and para positions. I'm going to just say that those are synonyms of each other. A Meisenheimer complex is just any benzene that is in a WAP 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. CLOCH₃. Notice what's going to happen is that nitro looks like this. N, so 1 O, negative. 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. Sorry. So much to draw. See how that resonance structure exists, and 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 6 resonance structures. We're not going to draw all of them. That's for sure. But I'm trying to show you guys how a Meisenheimer complex works. Now 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 Cl. 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 was for a nucleophilic molecule that had occurred. Awesome. 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, let me use a different color. If I 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, so 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 2 reactions. It 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.

Which ipso-substitution is more favored

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 WAPs. The one with more WAPs 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 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. And I have a withdrawing group on meta, which again isn't going to help me, so I'm not going to write it down. Overall, 2 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 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.

Drawing resonance structures, what you'd find is that you'd get something like this where oops, I should have a sulfur. You've got chlorine. You've got OEt. You've got negative. That negative charge is going to resonate to where? Well, it's going to resonate. I'm just going to put here 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, this is going to push electrons into my negative. That's terrible.

On this one, double bond, double bond, double bond. Well, actually, let me skip the double bonds actually. What I get for a reaction is this attacks here. I move the electrons down here, so I wind up getting a negative charge on the N. I have chlorine. I have OMe, methoxy. I've got my dotted line representing my anionic sigma complex and my negative charges, oops, I forgot there's a nitro group over here. My negative charges, 2 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 is 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? Cool, guys. Let's move on to the next question.

NAS in the addition-elimination pathway

Let's look at what this question is asking. It's saying which of the following compounds will most readily undergo nucleophilic aromatic substitution addition elimination pathway, which means SNAR. I know it's worded a little bit weirdly. We have to recognize that N, nucleophilic aromatic substitution specifically with addition elimination, is an SNAR. If it's an SNAR, that means what we're looking for is a WAP. It's that easy. We're just looking for the WAPIEST of all. Let's look. We've got A, B, C, D. Which one wins? Let's just take a look. A doesn't have any WAPs. I'm just going to give it an X. B has a heteroatom and a heteroatom. That's good. One heteroatom and one heteroatom in the ortho position. I'm only counting the ones that are ortho-pair. Has a heteroatom in the ortho, a heteroatom in the para, and a withdrawing group in the ortho. That one's looking really good. And then, C has a withdrawing group here in the ortho and then nothing else. This doesn't help me, and this doesn't help me. The best is D. This is essentially a Meisenheimer complex because, in this case, I've got withdrawing groups or heteroatoms in every location, both my orthos and a para. 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. Hope that made sense. Let's move on to the next topic.

Provide the structure of the product formed from the reaction of 1-bromo-2,4,6- trinitrobenzene with one equivalent of sodium methoxide. 

• A.
• B.
• C.
• D.

Provide the major organic product for the following reaction.

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• B.
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• D.

Provide the major organic product for the following reaction. 

• A.
• B.
• C.
• D.

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

• A.
• B.
• C.
• D.

Here’s what students ask on this topic:

What is nucleophilic aromatic substitution (SNAR) and how does it differ from electrophilic aromatic substitution (EAS)?

Nucleophilic aromatic substitution (SNAR) is a reaction where a strong nucleophile attacks a benzene ring, leading to the formation of an anionic sigma complex. This is followed by the elimination of a leaving group. Unlike electrophilic aromatic substitution (EAS), which involves an electrophile attacking the benzene ring to form a cationic sigma complex, SNAR involves a nucleophile and results in an anionic intermediate. EAS is a one-step process, while SNAR is a two-step process involving distinct addition and elimination steps.

What is a Meisenheimer complex in the context of SNAR reactions?

A Meisenheimer complex is an intermediate in SNAR reactions where the anionic sigma complex is stabilized by electron-withdrawing groups or heteroatoms in the ortho and para positions relative to the nucleophile. These groups help to delocalize the negative charge, making the intermediate more stable and allowing the reaction to proceed at lower temperatures. An example is trinitrobenzene, where nitro groups stabilize the negative charge, enabling the reaction to occur at room temperature.

What are the conditions required for the Dow process in SNAR reactions?

The Dow process is an early example of an SNAR reaction used to synthesize phenol from chlorobenzene. It requires harsh conditions due to the instability of the anionic intermediate. Specifically, the reaction needs a strong nucleophile like NaOH, a temperature of around 350°C, and high pressure. These conditions help to overcome the energy barrier associated with forming the anionic sigma complex.

How do electron-withdrawing groups affect the stability of the anionic intermediate in SNAR reactions?

Electron-withdrawing groups (EWGs) in the ortho and para positions relative to the nucleophile stabilize the anionic intermediate in SNAR reactions. These groups, such as nitro (NO₂), pull electron density away from the benzene ring, helping to delocalize the negative charge. This stabilization reduces the energy barrier for the reaction, allowing it to proceed under milder conditions. The presence of EWGs can significantly lower the required temperature and pressure for the reaction.

What is the significance of resonance structures in understanding SNAR mechanisms?

Resonance structures are crucial for understanding SNAR mechanisms because they illustrate how the negative charge in the anionic sigma complex is delocalized across the benzene ring and any attached electron-withdrawing groups. By drawing resonance structures, one can visualize the distribution of the negative charge and identify the most stable configurations. This helps in predicting the reactivity and stability of intermediates, as well as the overall feasibility of the reaction under given conditions.