Concept: EAS Halogenation7m
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: EAS Nitration7m
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: Reduction of Nitro Groups4m
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.
Concept: Friedel-Crafts Alkylation7m
Now we're going to explore the mechanism for Friedel-Crafts Alkylation. So Friedel-Crafts Alkylation is going to be the reaction of a alkyl halide with a strong Lewis Acid, coupled together they are going to make a strong electrophile but that active electrophile for this molecule is actually a true carbocation. Remember that carbocations are you know definitely electron loving because they have an empty P orbital but they're also very tricky because what happens when we have carbocation intermediates? We have to watch out for rearrangements. So that's going to be kind of the annoying part of this reaction that we can what we will observe carbocation rearrangements when it's favorable for the reaction. So that is going to complicate or that could complicate our products quite a bit. Now we're not going to focus on rearrangement in this mechanism because what I'm going to do is I'm going to show you the EAS part but later on when you're drawing products for alkylation, it's going to be important that you always think about those rearrangements before drawing your final product. So as you guys can see we have an alkyl halide and a strong Lewis acid. We say here aluminum trihalide but usually that's going to be a chlorine. So let's look at this following alkyl halide and how it's going to react.
The very first thing is going to be that we want to generate carbocation, right? So we're going to take the electrons from the carbon chlorine bond and give them, donate them to that empty orbital on the aluminum. Now this mechanism is similar to when I told you guys that you could just pick up your electrons and then give them away to the aluminum, it's the same thing I'm just taking them and giving them directly to the aluminum. Now what that's going to give me is it's going to give me a benzene ring, with I'm going to get ALCL4 minus but I'm also going to get a carbocation so R plus. So I'm going to get R plus and then I can do the rest of my mechanism so I'm going to get that that the double bond attacks the R plus and I'm going to get my sigma complex. Let's draw it. So I've got my double bond, double bond, H, R, oops I forgot to draw the positive charge so let's just do that. Great so now I've got my resonant structure that's got a positive here, you're going to draw one more and there you have it that is our sigma complex. Now what do you think is going to be left over to eliminate in the second step of my reaction?
You got it, the Lewis acid catalyst that's negatively charged. So I could draw this as CLALCL3 negative and what we're going to do we're going to take the electrons from that bond and use them to do my beta elimination. What that's going to give me as a product is now I'm going to have an alkylated benzene with what? Well with my ALCL3 catalyst, notice that I regenerated the same catalyst as I had in the beginning so it truly is a catalyst and I'm also going to have HCL so an acid being generated as a byproduct. So really straightforward mechanism now I do want to show you guys one thing though. It turns out that when you have a primary alkyl halides, primary alkyl halides, the mechanism does get a little more tricky because primary carbocations are unstable and what did we just say about this? Let me just put a sad face.
Primary carbocations are unstable. What did we just say about this mechanism? It's carbocation intermediated. So how does this reaction happen if primary carbocations are so high energy they're so difficult to create? Well in that case what we're going to do is we're going to make the mechanism very similar but we're going to have to do the carbocation shift actually in the actual mechanism. So let me just show you really quick what happens if this was my carbocation or this was my alkyl halide and I'm reacting that with ALCL3 and I've got my empty P orbital at the top. Well in this case it would actually be a mistake your professor would be unhappy if you just grab these electrons and put them into there and that's it because what you're going to get is a product, don't draw this, what you would get as a product is you get this and that as we know that's a primary carbocation and that's not very stable so most professors don't like to see that on your page so how do we fix this? Well it's pretty easy to fix guys all you have to do is if you have the situation you have to draw the carbocation shift in the same mechanism so would this carbocation want to shift if it was formed non-primary? Yes you would shift to the secondary so then we would just draw the shift in the same mechanism so I would draw that, hey, this is going to give its electrons away but now this bond is going to give its electrons away so that eventually in the same step we get our rearranged carbocation plus our ALCL4 negative and then at this point this is the active electrophile that my benzene would react with not the primary. So if you have a primary carbocation then draw shift within mechanism. Not so bad just a little contingency there because as you're going to see as a general pattern we're going to avoid primary carbocations at all costs in this course because they're so high energy they really don't, they're really difficult to generate them in a lab. Awesome guys that's the end of that mechanism. Let's move on to the next one.
Concept: Friedel-Crafts Acylation6m
Let's take a look at the exact mechanism of Friedel-Crafts Acylation. So Friedel-Crafts Acylation is going to involve an acyl halide typically acid chloride complexing with a Lewis acid catalyst to produce an electrophile but in this case my active electrophile is not going to be a carbocation like alkylation, it's going to be an acylium ion. Now what does an acylium ion look like? It looks like this, you've got a carbon with a double bond to O and an R group with a positive charge. So that is an acylium ion. Can you think of why that would be a good electrophile? Full positive charge. Now it is resonant stabilised so there's two different ways that it can be drawn. That's one way, another way would be to take these electrons and move them into that carbon to become a triple bond and the other way to represent it would be to now move the positive up to the O. So just so you guys know these are both acylium ions obviously the hybrid is going to be a blend of both of these it's going to look like you know some mixture of those contributing structures but the one that's easiest for us to use that helps us to visualise the mechanism the best is this first one because the first one shows the positive on the carbon that's we're going to be attacking. So when I draw the acylium ion I'm going to draw the first resonant structure. Awesome, so notice that what are we trying to get at the end? We're trying to make ketones.
So we're, our end product will be a ketone on the benzene ring. Let's go ahead and take a look at this mechanism. Okay guys, so really there's nothing tricky about this mechanism. What we're going to do is we are going to donate our chlorine to the Lewis acid catalyst and what that's going to give me is an acylium ion. I'm going to get carbon, double bond O, single bond R positive plus my ALCL4 negative. So as you can imagine this is a wonderful electrophile for a benzene to react with and it's going to grab the carbon. Now something to keep in mind guys is that this is not a carbocation intermediate reaction so that means we don't have to worry about shifts so there's no rearrangements. You might say well why, that's not how I spell it let's try it again, arrangement. Guys, the reason that it's not going to rearrange is because this is a resonant stabilised electrophile so it doesn't want to break resonance by moving to the R group it just wants to stay where it is because where it is it can resonate so you don't have to worry about that so you don't worry about any of the complications of rearrangement which is great and we're going to discuss more later why that's so important. So let's go ahead and draw this full mechanism, that's it I'm just going to make a bond. Do I have to break a bond? No I have a full cation there, empty orbital, so I don't have to break a bond I can accept those electrons.
Now I'm going to draw my sigma complex which would look like this, now with C, double bond O, R positive and you guys know what to do. Move it over, draw your double bond, draw my cation here, move it over again and I get my cation up here. Now actually I mean this is right but always remember you need to draw that hydrogen. The hydrogen is important because it's part of the elimination step so let's just draw the hydrogen facing up. What do you think we're going to use, by the way I didn't draw the hydrogens in the other ones it's not necessary you can just draw the hydrogen in the last one if you want because hydrogens can always be implied in a bond-like structure. So what's going to happen in this last step? Yeah guys, you just use your conjugate base, your ALCL4, A L C L oops that's not what I meant let's erase, CLALCL3 negative and as always I'm just going to grab the electrons from that bond and do my beta elimination. What that's going to give me is now a ketone plus I'm going to get my Lewis acid catalyst that's regenerated plus I'm going to get my acid as a byproduct. Once again guys no rearrangements are possible here so there's less to think about, you're always just going to get your ketone on the benzene ring and that's it. Alright, so that's it for this mechanism. Let's move on to the next one.
Concept: Any Carbocation8m