Nucleophilic aromatic substitution II

In the last video, we looked at nucleophilic aromatic substitution with an addition-elimination reaction. In this video, we're going to look at an elimination-addition reaction, also called the benzine mechanism. And we start with bromobenzene here. And to bromobenzene, we add some sodium amide, which is a strong base, and some liquid ammonia. And you can see that we have substituted an amino group for our halogen on our ring to form aniline as our product. Let's go ahead and look at the mechanism here. And so we're going to start with the elimination parts. Let me go ahead and write down here elimination. And the sodium amide functions as a base. So we go ahead and draw the amide anion here, with a negative 1 formal charge on this nitrogen. And so it's going to take this aromatic proton right here, which leaves these electrons behind on this carbon. So let's go ahead and show the result of our acid-base reaction. So we have our ring. We still have our leaving group for the moment. And now we have a negative 1 formal charge on this carbon. So we form a carbanion. Let me go ahead and highlight these electrons. So these electrons in here are now on this carbon, forming a carbanion that's ortho to our leaving group-- so an ortho carbanion. So when these electrons in magenta move into here, that allows these electrons to kick off onto our halogen, so it leaves as an anion. And we create what's called the benzine molecule-- so a triple bond in our molecule like that. And so these electrons in magenta have moved in here to form a triple bond. This is a little bit different from usual triple bonds. This happens to be an unstable intermediate. So this benzine molecule turns out to be very reactive. And so in the next step-- we'll call this the addition portion of our mechanism-- our benzine molecule is right here. And it's going to react with amide, the amide anion. But this time, the anion is going to function as a nucleophile. And so when I go ahead and show my amide anion like that, I'm going to show it functioning as nucleophile attacking our triple bonds--so attacking one of the carbons. So I'm going to say on that side, which would push these electrons back onto this carbon. And so we can go ahead and show that. So we have our ring. Right now, we have our NH2 attached to our ring. And we have these electrons move out onto here, onto this carbon. So once again, the electrons in magenta have moved out here to form a carbanion-- again, an ortho carbanion. Our last step would be to protonate our ring. And so we have ammonia. Ammonia comes along. That's going to function as an acid and donate a proton. So these electrons in magenta are going to pick up a proton here, leaving these electrons behind. And we protonate our ring and we're done with our mechanism. We have formed aniline as our product, like that. All right. So that's the mechanism for elimination-addition. Let's see if we can go ahead and do a practice problem here. And here we have a disubstituted ring. And our two substituents are para to each other. So we're going to add our strong base-- so sodium amide, so negative 1 formal charge here. First step is elimination. So the first up is an acid-base reaction, where the amide takes a proton that's next to our halogen. And so we have two choices here. We could take this proton that's next to our halogen or we could take this proton that's next to our halogen. In this case, it doesn't matter, because they are equivalents, pretty much the same thing, if you think about the symmetry of the molecule. And so this is going to take this proton, to make our lives easier here. And then these electrons would remain behind on this carbon. So let's go ahead and show our acid-base reaction here. So we have our ring. We have our methyl group here on our ring. We have our halogen here on our ring. And now we have a negative 1 formal charge on this carbon. So we have electrons on that carbon, negative 1 formal charge. These are going to move into here to form a triple bond, and these electrons kick off onto our leaving group. So the chloride anion leaves and we form benzine. So we form our benzine intermediate. And so I can go ahead and show that right here. So there's benzine. There's still a methyl group attached to our ring. Now once again, I chose to show this proton participating in the acid-base reaction. It could have been in this one, but it's just helpful the way I drew my benzene ring. I happen to show my pi bond right here, and so it's easier to show it there. But you only get one benzine intermediate here. And so here is our benzine intermediate, which next is going to react with the amide anion, which is now going to function as a nucleophile. So I have a negative 1 formal charge on my nitrogen. And it's going to function as a nucleophile. And it could attack either side of that triple bond. I'm going to go ahead and show it attacking this side of our triple bond. And so these electrons would come off onto there. So let's go ahead and show the results of that. And so I have my ring. And I'm going to add on my NH2 to this top carbon this time. And my electrons went off over here onto to this carbon. So I still have a methyl group, like that. Now, I could have showed the amide anion attacking the other side of the triple bond. That's possible, too. So I could have thought about my nucleophile attacking over here, which would push these electrons off onto that carbon. So let's go ahead and show that possibility as well. So once again, we have our ring. So we have these electrons on our ring. We have our methyl group. This time, the NH2 added onto the right side of where our triple bond used to be. And these electrons ended up over here. And so of course, the final step, which I won't show the mechanism for, we just need to protonate that anion. And we end up with our two final products. And so let me go ahead and draw those. So we have our two groups, and in our top example, they are para to each other. And then our other products, our two groups on our ring, are meta to each other. And so you get a mixture of products here. So thinking about your benzine mechanism. So let's look at one more type of elimination of a reaction using the benzine mechanism. So another famous one. And you can see this time, instead of using sodium amide, we're using sodium hydroxide. But we know that this is a strong base, and so that will work. You have to heat it up a lot and add some water, some proton sources. And you can see that, eventually, you're going to get an OH group substituting in for your halogen. So this is a very famous reaction. Let's see if we can do this problem down here, where we have a methyl group on our ring and we have our chlorine like that. So this is a little different from the one that we just did. So these two are meta to each other, and those other ones were para. So once again, we're going to think about the hydroxide anion functioning as a strong base. And it's going to take a proton off of our ring. Remember, it's going to take a proton off next to our leaving group. And so it could take a proton off from here, or it could take a proton off from here. And those are two different positions this time. So we're going to get different benzine intermediates. So we need to think about two different benzines. So that's different from the previous example. We only had to think about one enzyme. So first I'm going to think about the proton on the right. So if this proton leaves, you can think about these electrons in here moving in to form your triple bond. So let's go ahead and draw that one first. So the elimination of your halogen would give you your benzine intermediate and your benzine intermediate would look like this, with your methyl group being right here. And so that's one possible benzine intermediate. Of course, we could have taken off this proton-- so sodium hydroxide, could've taken off that proton. And your halogen could've left in the elimination step to form another benzine. And this time, it would be between this hydrogen and the chlorine. So it would be over here. So benzine would have to be over on that side. And so we could go ahead and draw our benzine like that. Now, notice I'm showing a different resonance form of the pi electrons in the ring. But we know that that's OK. I just happened to start out with my pi electrons in my benzene ring like this. But I could have just as easily have started out the opposite way. So what I drew here is not incorrect. It's just a different way of showing it. And it's necessary for us to show the benzine intermediate this way. And so now we have two benzine intermediates. And in the next step, hydroxide would function as a nucleophile, and it could attack either side. So let's go ahead and think about the hydroxide anion functioning as a nucleophile. And let's go ahead and draw the possible products. I could think about hydroxide attacking this side of the triple bond, in which case our final product would look like this. We would have our ring. We would have our methyl group here. And then we would have our OH group para to that methyl group. So that's one of our possible products. So let me go ahead and use a different color so we can see that better. So a nucleophilic attack here would give us this product, nucleophilic attack here would give us a different product. So let's go ahead and draw that. We would show our ring, and we would show our pi electrons. We'd show our OH adding here. And we would still have this methyl group. And so you can see these are different products. So we have two different products so far. And once again, we could go over to this molecule, and we could show the same thing. We show hydroxide attacking the left side of our triple bond. So let's go ahead and draw that product. So if hydroxide attacked the left side, it would add on to the left side of where our triple bond used to be. And now we would have this product. So that's another possible product. And now, of course, we could have attacked on the right side of our triple bond for this one. So let's go ahead and draw the fourth possibility here. And so now we have our benzene ring. We have our OH group added onto here. And then we have a methyl group right here. And so if you look at it, you can see that these two are actually the same molecule, because these benzene rings-- I've just drawn the pi electrons in different places, but we know that those are just resonance structures of each other. And so we get a total of three products for this reaction. So this would just be one molecule, and then this would be two, and then this would be three. So this would be my approach to answering this question on an exam. You don't have time to draw out the full mechanism, so think about eliminating your halogen to form the benzine intermediate. Think about how many benzine intermediates are possible. And then, finally, think about adding the nucleophile to either side of your triple bond, and that will give you your possible products for your answer.