Concept: Concept: EAS Reactions Overview12m
In this video, I'm going to introduce all of the different reactions that can occur to benzene through an EAS mechanism.
As previously discussed, benzene is going to require very strong electrophiles in order to react through EAS. Some of these agents are going to require catalysts in order to become strong enough for the benzene to react with them.
Let's just take the first one into account, EAS halogenation. Halogenation would be a reaction with a double bond and a diatomic halogen, so something like X2. But actually, X2 is not a reactive enough electrophile to react with a benzene. So guess what? If you just put X2 in a mixture of benzene, you're not going anywhere. You're not going to make a substituted benzene.
So how do we get the X2 to react? We need a catalyst, guys. This is going to be a Lewis acid catalyst, FeX3. This is going to work for bromine. I'm going to put here where X is equal to specifically bromine or chlorine. You always want to make sure that when you're performing one of these EAS halogenations, that you're using the same hydrogen in your catalyst as you are in the actual electrophile. That's going to become important later when we look into the mechanism.
As you can see, the product is now going to be a mono-halogenated benzene, which is great. That's what we wanted. It's important to distinguish this mechanism from another reaction we learned in the past, which is called halogenation. It's important that when you discuss this mechanism, you call it EAS halogenation because specifically, this is the one that requires a Lewis acid catalyst.
All right. Let's move on to the next one. It turns out that iodination is a little bit tricky. Iodine does not – I was going to use a different color. Let's use red. Iodine does not react with the Lewis acid catalyst of iron and three halogens to perform this EAS reaction. So we're going to have to use nitric acid instead, HNO3. That's one little peculiar thing that you have to memorize that iodination, in order to put an iodine on a benzene, we're going to need to use nitric acid, not our iron Lewis acid catalyst.
Let's move on. The next one we want to discuss is nitration. Now a nitro group is not a group that we've discussed a whole lot at this point of the course. I do want to just help you guys understand what a nitro group is by drawing it. It would be something like this, we have a nitrogen with a double bond O and a single bond O and there's going to be formal charges. The single bonded O has a negative charge. The nitrogen, since it has four bonds has a positive, but notice that this has no net charge because they cancel out. That's why when you draw NO2, you draw it neutral without charges even though those charges do exist.
Now there are two different ways to add a nitro group to a benzene. But they're really the same thing. It's just that different websites, different textbooks, different professors like to use different combinations. The most common would be using nitric acid, HNO3, and sulfuric acid. A combination of sulfuric acid and nitric acid is going to react together to make the active electrophile as we'll see later when we go into the mechanism.
But just keep in mind that you can also generate this electrophile just through using concentrated HNO3. Where really in the first example you have one acid reacting with another, when you use concentrated HNO3, it just reacts by itself. One HNO3 molecule reacts with another one and makes that active electrophile. Regardless of how you draw it, it's really going to produce the same product. That will be EAS nitration.
Also, one thing to keep in mind is that nitric acid can also be written differently. It can be written as HONO2. This is just a little technicality that actually could make you guys stumble on an exam because you might not be recognizing it as HONO2, but it's the same compound as nitric acid, HNO3 just written different. It's basically written on the connectivity in a condensed structure instead of just based on the molecular formula.
Let's go onto sulfonation, which is actually going to be kind of the most special reaction we're going to talk about on this page because sulfonation is the only one with these two arrows. Let's discuss what that is.
Well, first of all, the product of sulfonation is going to make what we call a sulfonic acid group. That's what we have at the top. By the way guys, I forgot to write what the product of nitration would be. It would be a nitro group. We've got nitro and we've got sulfonic acid. These are literally functional groups that we're just learning in this course.
Sulfonic acid. The way that we do a sulfonation is to use sulfuric acid, so H2SO4 and SO3, sulfur trioxide. Now it turns out that when you heat up H2SO4, it's going to actually make the sulfur trioxide, the SO3, so sometimes it's just written as fuming H2SO4. However you see it, that's fine, as long as you have fuming H2SO4 or H2SO4 and SO3, it's the same thing. It just means that it's actually fuming. It's evolving an SO3 gas that's going to react. Now that's going to make your sulfonic acid.
Now, when you want to reverse it, if you want to go in the reverse reaction, that would actually be called a desulfonation, or EAS desulfonation. To desulfonate it's actually even easier. You're just going to use a dilute form of the acid, so you might see it written as dilute H2SO4. Or because of the fact really it's usually heat that winds up taking the sulfonic acid group off, you might even see it written as just steam because steam can be used to basically take off the sulfonic acid group and reverse it back to benzene.
Again, I'm writing it in multiple ways so that you're not going to get stumped on an exam or on an online homework. There's different ways to represent these reactions, but they're all really the same thing, steam, dilute H2SO4, that would take it to the desulfonation back to the original benzene. Awesome.
Now let's work on the Friedel-Crafts reactions. The Friedel-Crafts reactions are ways to make R groups or put R groups onto a benzene. Now guys, this is going to be very important because any time that you can add an R group to a smaller molecule, that's really important for organic synthesis. Aromatic compounds are extremely important for all kinds of pharmaceuticals. If you can add R groups and make them bigger, that's awesome. You can make a million bucks. Maybe making the next Viagra.
Anyway, let's go ahead and talk about alkylation and acylation. Alkylation would be the reaction of an alkyl halide so that would be RX with a Lewis acid catalyst. That Lewis acid catalyst for Friedel-Crafts reactions is going to be an aluminum catalyst, so it's going to be AlX3. Once again, X is only really equal to only bromine and chlorine in these cases. And you want to use the same X in both of them as you're going to see in the catalyst, it might actually matter. So we want to go ahead and use the same X in both, but typically guys, the one we're using is chlorine. For the rest of the reactions that I work with you guys, I'm always going to be working with AlCl3. It's the most common version.
Now what we see is that the X actually doesn't end up on the benzene. What ends up on the benzene is the R group. Just keep in mind that the R group is what stays, whatever we had as an R group. And then we're going to get some kind of acid byproduct.
Then we use, we have Friedel-Crafts acylation. Friedel-Crafts acylation would use an acid chloride. Now an acid chloride, if you're not familiar with it is a functional group that we're going to use a lot in organic chemistry two. And we're going to react it over that same Lewis Acid catalyst. In this case, AlCl3 because we're using a chlorine. And that's going to give us actually what we call an acyl group. Alkylation gives us an alkyl group, literally just a R group. Whereas acylation gives us a acyl group. So a carbonyl group. A ketone. Usually a ketone.
Those are the two Friedel-Crafts reactions. They have a lot of similarity between them. As you can see they use the same reagents or the same Lewis acid catalyst, but there are some differences that we'll explore later.
Finally, we have any carbocation So what's the big deal with this. Well, carbocations by definition are super strong electrophiles because they have an empty p orbital. They literally are saying feed me electrons. Any reaction that you've ever learned that involves carbocations could theoretically work with benzene. Now, we're not going to have to go back and remember every single carbocation intermediated reaction. But there are two different combinations that I want you to focus on.
That would be one, these are very common. One combination would be HF and a double bond. When you have HF and a double bond, like for example cyclobutene, that could generate a carbocation because you're going to get an addition reaction. You're going to wind up getting a carbocation on the square on cyclobutane and then that winds up getting attacked by the benzene.
Another combination that can produce carbocations which be actually BF3, so a strong Lewis acid with alcohol. We're going to see later how BF3 and alcohol can also produce carbocation-like structures that will react with benzene and produce an R group.
The whole point here is at the end that we're adding some kind of R group to our benzene ring. Some professors and texts might even refer to these as Friedel-Crafts reactions, but I prefer to discuss more their mechanism and highlight the fact that they are carbocation intermediated reactions, so you can really instead of trying to remember how that's a Friedel-Crafts, think of it more just as a carbocation reaction that will basically tempt the benzene to react with it.
Hope that makes sense so far. Let's move on to the exact mechanisms of EAS reactions.
If electrophilic addition to benzene is overall an endergonic reaction, how can electrophilic addition to an alkene be overall an exergonic reaction?
Draw the arenium cation intermediate that results from addition of an electrophile E+ to benzene
What is the best method for carrying out the following reaction?
Hexadeuterated benzene (C6D6) is a very useful solvent for 1H NMR spectroscopy. Suggest a method (reagents) for the preparation of C6D6 starting from benzene (C6H6) and propose a representative mechanism.
The electrophilic bromination or chlorination of benzene requires, in addition to the halogen:
A) a hydroxide ion.
B) a Lewis base.
C) a Lewis acid.
E) ultraviolet light.