Benzene can also undergo Nucleophilic Aromatic Substitution via an Elimination-Addition pathway to make aniline. This mechanism requires the formation of a highly unstable aryne (C6H4) intermediate.
Concept: General Mechanism7m
At this point, we’ve learned about two different addition elimination mechanisms for benzene. One is electrophilic and one is nucleophilic. But it turns out that there's another form of nucleophilic substitution out there. But this one is an elimination then addition pathway. It’s weird. This is called the benzyne pathway.
It turns out that benzene can progress through an elimination addition pathway. But it's going to make a very unstable intermediate. That unstable intermediate is called an aryne or benzyne because it literally has a triple bond within the ring. Imagine how unstable it is to have a triple bond. Remember, triple bonds want to have a bond angle of what? 180 degrees. It's going to be forced into 120-degree bond angle on that cyclohexane. It’s going to suck. But it can happen and this is actually a way that we make aniline.
Let me show you the mechanisms that we’ve learned so far and how benzyne is similar and how it's different. It's mostly just different. I’m going to go through this really fast. We know that for EAS, I can add first and then it can eliminate. We get our addition and elimination through an electrophile. We call it EAS.
We know that for SNAr, I can also add first but the way we add is actually through a nucleophilic attack. Then we eliminate using the same anion. We also get addition elimination but this one is nucleophilic, so SNAr. The aryne pathway or the benzyne pathway is much, much different because what happens in the benzyne pathway is that we have to actually eliminate first.
How do you eliminate from a benzene? What you literally have to do in E2, you literally have to do a beta elimination on this benzene. You would grab one of the H’s with a strong base. You can just use a strong base and you would literally do the three arrows for an E2 attack. I would grab the H, make a triple bond and kick out the X. What this is going to make is a very unstable benzyne, again also called aryne because arene is a benzene ring, so a benzyne intermediate. Extremely unstable.
Now we have to add. The way we add is with the conjugate acid. The conjugate acid now can be used to add to the benzyne. The mechanism is a little weird but just bear with me. What we're going to do is we're going to take our conjugate acid and we’re going to attack one side of the triple bond and kick those electrons down to form an anion. What this is going to form is another intermediate that looks like this. I’ve got my benzene is back, now with a negative charge. Then I’ve got BH where now B has a positive change because that's one too many bonds.
To finish this up, this negative can do basically a proton exchange. It can finally grab the H and we can get our substitution. That substitution, it took a long time. It's kind of weird path. We’ve got a base to switch out for a halogen.
What I wanted to show you guys is how we can use this to make an aniline. How could we make an aniline using this mechanism? The base that we use, this is a very typical reaction. It is NH2-, a very strong base. We’re going to do an elimination first. We're going to take an H. We’re going to a beta elimination. This is going to give me my benzyne. There's my benzyne.
What happens next? Notice that now I actually have my base is NH3 and I still have a long pair. It’s still somewhat nucleophilic. I can use the conjugate to attack or add to one side and kick electrons down to that one, to that side of the double bond. What I'm now going to get is a molecule that looks like this. NH2, H+ and a negative charge. Then we do the proton exchange and we're finally going to get what? What you just finally get at the end of this is aniline. This is actually a way to make aniline. You can make aniline using an NH2 base on basically an aryl halide. We started off with an aryl halide plus NH2- can give us aniline.
Crazy, huh? Awesome. Let's go ahead and flip the page.
What happens in we have an asymmetrical benzyne and has substituents? How do we know which side is going to get attacked?
When it comes to the benzyne pathway, there is one more thing we need to learn about because so far all showed you was benzene by itself but what happens if that benzyne is asymmetrical and has substituents on it, how do we know which side of the benzyne is going to get attacked? Well it turns out that this is what we call benzyne regiospecificity and there's actually a trend for it so let's learn it. So here's the issue let's say that we have the following benzyne intermediate, and I'm at the point where my N H 3 is going to attack. Well notice that now I've added an O H group so if my N H 3 attacks from here I would get a lone pair forming here and eventually I'm going to get the N H 2 close to the O H or if my N H 3 attacks here I'm going to get the lone pair performing here and eventually my N H 3 will be meta to the O H, so which one is right? Do I just get a combination of products? Actually no, the identity of where of the location of where it adds is based on the type of group that is next to the benzyne. So guys, I'm going explain the theory but thankfully memorizing is really easy because this follows the same trends that we've seen all throughout benzene chemistry which is that a donating group is going to favor the ortho position and a withdrawing group is going to favor the meta position so we remember how withdrawing groups are always meta directors and donating groups are ortho, para directors, it's the same thing here except we're not going to worry about para because para doesn't even have a benzyne on it so I'm just worried about ortho and meta.
So here's the theory behind it. Notice that what I have here for O H is actually a donating group. So if I have a donating group that would put me into the bottom category here. When I have to form that negative charge I have two different options, I could either form the negative charge here close to the donating group or I could form the negative intermediate lower away from it. Which one do you think is going to be more stable, having it closer to the donating or further? Further guys, so what's going to happen is that the N H 3 is going to selectively pick the ortho position just so that it can put the negative charge further away from the donating group. So that's why when you have an electron donating group, ortho substitution is going to predominate. Now the opposite is true of withdrawing groups, so if I have a strong withdrawing group then we want the negative charge to be as close to it as possible so then that's going to favor meta substitution. So you guys can totally know the theory but you can also just memorize it and that'll be fine.
So let's go ahead and do a practice problem. Now I know this is going to be kind of multistep synthesis, a lot going on, few other reagents. Try to do your best, notice that here you do have to worry about regio-chemistry so try to pick the right answer and then I'll go ahead and I'll give you the right answer.
Example: Activating and deactivating groups4m
Alright guys, so first let's just take care of the benzyne pathway part and then we'll worry about the other reagents. So it says to show the full mechanism so we will. So what we're going to do first is a beta elimination, we have to do it with that hydrogen because the other side doesn't even have a hydrogen so I'll do this, this and this. What that's going to give me is a molecule that now looks like this, triple bond, N O 2, double bond, double bond. So now notice that we have a withdrawing group next to the benzyne meaning that when my N H 3 comes in to attack, I'm going to attack at the meta position. So good job, I'm pretty sure most of you got that right. So now I'm going to get, this is a withdrawing group, so now I'm going to get the intermediate that looks like this.
So I've got N O 2, benzene, benzene, benzene, negative N H 2, H positive so now I'm going to get a proton transfer and I'm going to end this stage with m-nitroaniline. So that's my product here it's meta nitroaniline. Perfect, so now I'm going to react this benzyne derivative with N A N O 2 which, what do you guys think that's going to do to my benzene? N A N O 2 is part of a diazo pathway reaction. So if you haven't gotten here yet or if you don't remember it you can look these reactions up in the diazo section, you can search diazonium reactions in the clutch curriculum, in the clutch search bar and they'll take you there but if you haven't gotten here yet then it's because you just you don't need to know this yet so then just don't worry about this particular part. So what that's going to do is it's only going to react with the N H 2 because the nitro doesn't react with nitric acid that would give me the nitrile remains untouched but that would give me N triple bond N, a diazo compound and that is going to react with my replacement reaction to give me my final compound which would be nitro on one side and a cyano group on the other.
Excellent guys, so the most important part here was the benzyne pathway. That's what I'm concerned about. If you happen to not remember these reagents it's not a huge deal, we'll practice them more in that particular section but you guys did really well with benzyne. Now you guys know three pathways, you guys know EAS, SNAR and benzyne and that's huge, that's pretty much all the mechanisms that benzene undergoes. So awesome lesson, let's move on to the next topic.
Problem: Which is NOT a possible product of the reaction?2m
Problem: Which is NOT a possible product of the reaction?2m
Propose a mechanism that shows why p-chlorotoluene reacts with sodium hydroxide at 350 °C to give a mixture of p-cresol and m-cresol.
Compare the mechanisms of the following reactions:
What is the major product of the following reaction?
Would the following aryl bromide be expected to react with K + NH2- through the benzyne pathway?
Predict the product of the following reaction:
Predict the product of the following reaction:
Which one of the following chloroazulenes would most readily undergo nucleophilic aromatic addition via the elimination addition pathway? Why?
Give step by step mechanism for the following transformation:
What kind of reaction is this? Draw the full arrow-pushing mechanism.
What intermediate is thought to occur in the elimination-addition nucleophilic aromatic substitution mechanism?
a. radical anion
b. radical cation
e. positively charged sigma complex
What is the structure of the intermediate for this reaction?
Which intermediate is involved in this reaction?
Complete the following reaction with the correct structures of starting materials/products. Don’t forget to specify the stereochemistry.
Which reaction follows the benzyne mechanism?