EAS: Generating Electrophiles

Concept: Concept: EAS Halogenation

7m
Video Transcript

Let's dive into the exact mechanism of EAS halogenation. EAS bromination and chlorination are both going to have to complex with a Lewis acid catalyst before any reaction can take place. Remember that in our general reaction, you need that Lewis acid catalyst in order to go anywhere. We're going to be using that to start off this reaction.
In our very first step for our mechanism, before the benzene can get involved at all, we need to complex our diatomic halogen with the Lewis acid catalyst. That happens through the bromine sharing some of its electrons with the Lewis acids. It's missing electrons. It's a great electron pair acceptor.
Now what this is going to do is it's going to from a complex that's very electrophilic. Let's see what it's going to look like. It's going to be a bromine attached to a bromine, attached to an iron, which is attached to three bromines. I'm just going to put Br3. Is that fine? You guys know that that stands for all three bromines. Awesome.
Now are there any formal charges included in this molecule? Yeah, well, now the iron was neutral before. It has an extra bond. That's going to get a negative charge. Bromine, as we know, likes to have seven valence electrons, right now it only has six because it has two lone pairs. Two bonds, two lone pairs, that's going to get a positive charge. It's missing a valence electron. This is our active electrophile. This is our active electrophile. This is the one that reacts with benzene.
What's going to happen in this mechanism is that my benzene is the nucleophile. Now it's going to attack the electrophile. What might seem a little bit weird is that you would think that would go straight for the positive charge because usually negatives attack positives. But actually, let's just bring down the benzene here. What's going to happen is that the benzene is not going to attack the positive, it's going to attack the bromine next to the positive. Why? Because if it can attack that one and remove it, then this bromine can donate its electrons to the one that's missing some, which is the actual positively charged bromine.
Now, going down, what this is going to cause is an interruption of aromaticity. This is going to be our intermediate. So now what we're going to have is one double bond here, one double bond here. We had an H before, but now we've also got a bromine. Here we also have an H, but it's missing it's fourth bond, so that's going to be where our carbocation goes and this specific carbocation is called what? This is our arenium ion or sigma complex. This is going to be our sigma complex.
As we know that sigma complex has resonance structure, so let's just draw those really quick. We know that this double bond can resonate to three different positions. It's going to move over. HBr. And what we're going to end up with is three resonance structures that are stabilizing that intermediate. As you guys recall, this is the slow step of the reaction, making the intermediate.
All right. Now we want to do the elimination step. That was addition. Let's do the elimination step and end this reaction. What do you guys think is going to be the nucleophile that reacts with my arenium cation? Good, it's going to be the FeBr4 that's negatively charged. We've still got this FeBr3 that has an extra Br on it. And the whole thing is negatively charged. So how is that going to react? Well, we can use the electrons from the bond, from the extra bond to eliminate the hydrogen.
Now, to me personally, the mechanism that makes the most sense and you're going to see this a lot in this chapter is what I think makes sense is the bromine grabs its electrons. Says, I'm taking them back. And then it gives its electrons to the H. Actually, don't draw this. Use you eraser really quick. Then it gives its electrons to the H. To me, if I were writing your textbook, that's the way I would have drawn it because it makes the most sense that it takes it's electrons and then it gives them to the H.
But the way that textbooks usually write this mechanism is all in one shot. They'll write that the electrons just go straight for the H. But it's the same thing. This notation of showing that the electrons go from that bond to the H, literally just means that the bromine is taking it's electrons and giving them out to that H and now making a bond with the H.
Now we know that hydrogen can't make two bonds, so if we make that bond, we would then break this bond and reform the aromatic compound.
That was just a note to say guys that if you ever see me in the next few mechanisms drawing straight from a bond, that means that you can just think of it as that two arrow mechanism instead.
So what we're going to get now is we're going to get, since I reacted with this resonance structure, I'll draw it like this, bromine here. Plus we're going to get what else? We're going to get FeBr3. Notice that this is why it's called a Lewis acid catalyst because we regenerated it at the end. And we're going to get HBr. Excellent guys.
If you're wondering if the resonance structure matters, no it doesn't, guys. You could have drawn that resonance structure however because it's constantly – it's a resonance structure, you can draw it however you want. You could just draw it as a circle if you want. But that is our product. And that's it. So let's go ahead and move on to the next mechanism. 

Concept: Concept: EAS Nitration

7m
Video Transcript

Now let's take a deeper look into the mechanism of EAS nitration. So as mentioned earlier, nitration always has to proceed through the creation of a strong nitronium ion electrophile. A nitronium ion electrophile looks like this. It's going to be an N with a double bond O at the top, double bond O at the bottom and a positive charge.
Why do you think that's going to be a strong electrophile? Guys, it has a full positive charge. That's one of the strongest electrophiles possible. That's the perfect type of molecule that benzene wants to react with.
Remember that we said there's two different common ways to generate this nitronium. You could use concentrated nitric acid by itself. By the way, heat never hurts. Heat is going to help this reaction regardless of what reagents you're using. Or you could also just use nitric acid and sulfuric acid together. I included both of these in our mechanism just so that you guys can see how they're really the same exact thing.
In this mechanism, we have one equivalent of acid reacting with another. Regardless of which acid it is, they really do the same thing. One is going to be a proton donor and one is going to be the base. With the sulfuric acid and the nitric acid that makes sense. Sulfuric acid is a much stronger acid than nitric acid. It makes sense that sulfuric acid is going to be the proton donor.
Now for nitric acid, we would just imagine that at equilibrium some of this nitric acid is going to be donating protons to the other.
Let's go ahead and see exactly which part of the nitric acid would be basic. It's going to be this oxygen here with the lone pairs. The reason being that you're going to see later on it's going to become a great leaving group. We would have that if you're proceeding through sulfuric acid, that this oxygen would grab one of the hydrogens on sulfuric acid. But you could also do the same thing guys for nitric acid up here. It's going to have the same net result.
What we're going to get is we're going to wind up getting, here's our benzene ring. I'm just bringing it down. And now we're going to get nitric acid that looks like this. It's protonated. It's now going to have two hydrogens on that oxygen. That has just created a water leaving group.
So in order to generate the nitronium ion, all I have to do is eliminate with the O-. What I can do is I can do an elimination reaction. Bring down – make a new pi bond and kick out the water leaving group. What this is now going to give me is a nitronium ion plus water. Cool?
Now I can go ahead and I can do the rest of my mechanism. At this point, benzene, had to bring it down a few times now. I'm drawing too much. Benzene is going to attack my nitronium ion. What is that going to look like? It's going to do this. By the way, positive. It's going to attack the nitro or the N. And then it's going to kick out one of these pi bonds and make them into a lone pair because you have to break a bond.
This is going to lead us to our sigma complex. So let's draw our sigma complex. I know it's a little annoying, but you guys should get practice with it. So now this is NO2. We've got double bond here, double bond here, positive charge, and now we've just got to draw the whole complex. I'm going to move it over. I'm going to move it to this location. And now this is my last resonance structure. All right, so we're done with the sigma complex. There we go. That's our full resonance structure.
Now, what are we going to use as the conjugate to eliminate this hydrogen? Remember we have to do basically what's a beta-elimination on this hydrogen. Beta to the carbocation. Even though you totally could use the conjugate of sulfuric acid, so you could use the negatively charged OSOH4. I think I said that right. Whatever, I could write it down correctly. But actually, we're not going to use that because typically most textbooks and most professors are actually going to use the water that left in the nitro group. It really doesn't matter guys. You could use the water. If you want, you could use the conjugate base of the acid that you used. I don't really care, but just typically it's the water that's used in these reactions. It doesn't matter because at the end of the day this would make HNO3, I'm sorry HO3+, which is just basically aqueous acid. So it doesn't matter.
I'm going to go ahead, come down and eliminate my H and put the double bond there and we get our final product. What our final product is going to be is it's going to be our nitric acid. I'm sorry, it's going to be our nitro group, nitrobenzene plus we're going to get H3O+ and then you would get I guess the conjugate of your sulfuric acid if you had used that which is OSO3H. Awesome guys.
That's it for the nitration mechanism. So let's move on to the next reaction. 

Concept: Concept: Reduction of Nitro Groups

4m
Video Transcript

Guys, it's worth noting that nitro groups or nitrobenzene is often used as a precursor to get to aniline. So remember that aniline is an amino group on a benzene ring. That's called an aniline molecule. Nitro groups can be easily reduced to aniline. As you can see a reduction reaction would remove oxygens and add hydrogens and make aniline.
Even though we're going to discuss this more in your amines chapter, I do want to go through it right now and just kind of clue you guys in to some of the most important reducing agents that can make this conversion happen.
Now, the way we always want to start with and probably want to be our default whenever we think reduction is lithium aluminum hydride. That's just because this is the most common reducing agent of all organic chemistry. It's also one of the strongest. Lithium aluminum hydride will absolutely get the job done and it will absolutely turn a nitro group into aniline.
But there are a few other types of reagents that can do the same thing that you also might see. Do you guys recognize H2 and a palladium catalyst? This also goes for nickel or a platinum catalyst. These would be the reagents used in catalytic hydrogenation. I'm just going to put here, these are the reagents for catalytic hydrogenation and that will definitely reduce your nitro group to an aniline.
Now one that's actually really special, kind of important here is Tin, two chloride, and water, or what's also known as Stannous chloride. Benzene is just going to have to get written on because I don't have that much room. Stannous chloride. Now this one is particularly special here because we're going to talk a little bit more about this later. This is actually your only chemoselective reducing agent.
What does that mean? What it means is that, by saying that it's chemoselective, what I'm saying is that it has a tendency to only reduce nitro groups and nothing else. It's kind of talented at doing that. It really doesn't like to reduce many other types of groups so that's going to be important when we have other groups that are vulnerable to reduction, Stannous chloride is a great choice because it really just hones in on the nitro groups and turns them into aniline.
Finally, really common reducing agents are either iron or zinc in the presence of HCL. You'll see this all the time. These reagents turn into strong reducing agents that will reduce a nitro group into anilines.
Really, the exact reducing agent that you're going to wind up using the most is going to probably be up to your professor more than anything else, but bear in mind that all these reagents could be used in some way or another to reduce a nitro group to an aniline. My personal favorite is going to be the tin, two chloride, the Stannous chloride. That's the one I'm going to use the most often in this course because I know that it's chemoselective specifically for the nitro group, so it has very high yields of aniline when we use it.
Let's move on to the next topic.