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Studying for a test? Prepare with these 7 lessons on Aromatic compounds.
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Here's the general reaction for the nitration of benzene. So we start off with benzene, and to it, you add concentrated nitric and concentrated sulfuric acids. And that puts a nitro group onto your benzene ring, in place of this proton. Let's look at the mechanism for the nitration of benzene. So we start over here with the dot structure for nitric acid. And here's the dot structure for sulfuric acid. And sulfuric acid is actually a stronger acid than nitric acid. So the first step is sulfuric acid is going to function as a Bronsted-Lowry acid and donate a proton. So I'm going to say, it's this proton right here. And nitric acid is actually going to function as a base and accept that proton. So we can go ahead and show this lone pair of electrons picking up this proton and these two electrons in here remaining behind on that oxygen. So let's go ahead and show that acid-base reaction. So we would have our compound over here. So let's go ahead and draw in these atoms. So this oxygen is still going to have a negative 1 formal charge on it. And then over here on the right, this oxygen already had one hydrogen bonded to it. It just picked up another proton. So it still has one lone pair of electrons, which gives that a plus 1 formal charge. So let me go ahead and highlight these electrons here. So I'm saying that this lone pair of electrons right here picked up a proton like that, which gives that oxygen a plus 1 formal charge. And notice that forms water as a leaving group. And so if these electrons were to move in here, that would kick these electrons in here off onto the oxygen. So let's go ahead and draw the result of that. So we would have water that left. So let's go ahead and show H2O over here on the right. So we'll go ahead and draw in those electrons like that. So I'm saying that these electrons in here in magenta are the ones that came off onto the water molecule, so water as our leaving group. And if water leaves, that leaves what's called the nitronium ion behind. So let me go ahead and draw the nitronium ion here, so it looks like this. Now, let me go ahead and highlight those electrons. So I'm saying that these electrons in here moved in to form a pi bond. So I'm saying it could be represented by those electrons down here on my structure. Now, for formal charges, this nitrogen has a plus 1 formal charge. And that nitrogen is actually SP hybridized, which makes the nitronium ion linear. And the nitronium ion is positively charged. This is going to be the electrophile in our mechanism. So this is our electrophile, our positively-charged ion. So up here, if we think about it, we would also create the conjugate base to sulfuric acid, so HSO4 minus would be over here. All right, so let's go ahead and show what happens now that we formed our electrophile. So the electrophile is going to add to the benzene ring. So we're going to have a nucleophile-electrophile reaction. So let's go ahead and draw our benzene ring over here. And I draw in my hydrogen on my benzene ring. We've now formed our nitronium ion. So the point of the catalyst was to produce our electrophile here. So we have a positively charged nitronium ion like that. And so now we have a nucleophile-electrophile situation where, once again, the pi electrons in our benzene ring are going to function as a nucleophile and attack our electrophile. So those electrons are going to attack that positively-charged nitrogen, which will kick these electrons in here off onto that oxygen. So let's go ahead and draw the results of our nucleophilic attack. So we have our ring right here. We have a hydrogen. And once again, just for convention sake, I'm going to show our electrophile adding to that top carbon double bond there, of what used to be the double bond. And so now we have a nitrogen there. We have a nitrogen double bonded to our top oxygen. And then over here, we would have an oxygen with three lone pairs of electrons, giving that a negative 1 formal charge. And the nitrogen, of course, is still going to have a plus 1 formal charge like that. All right, let me go ahead and highlight those electrons. So once again, these pi electrons are going to be attracted to the positive charge, nucleophile-electrophile. And those pi electrons are going to form this bond right here to our nitro group. Well, once again, as we've seen several times before, we took away a bond from this carbon. So that's where our plus 1 the formal charge is going to go like that. And so we can draw some resonance structures. So let's go ahead and show a resonance structure for this. We could move these pi electrons in over here. So let's go ahead and draw that. So we had a hydrogen up here. And you could just show a nitro group as NO2. So I'm just going to go ahead and do that to save some time. These pi electrons over here are still there. And I'm saying that those pi electrons moved over to here. So let me go ahead and highlight those. So these pi electrons in blue move over to here, took a bond away from that carbon. So now we can put a plus 1 formal charge at that carbon like that. We can draw yet another resonance structure. So I could show these electrons over here moving to here. So let me go ahead and draw that. So we have our ring. We have our nitro group already on our ring. We have some pi electrons right here. And we have some more pi electrons moving from here to here, which, of course, takes a bond away from this top carbon. So that's where our positive 1 formal charge is now. So now we have our three resonance structures. And remember, once again, that the sigma complex is a hybrid of these three. And we're now ready for our last step, so deprotonation of our sigma complex. So if we go back up to here, we think, what could function as a base? Well, the water molecule here could function as a base. So a lone pair of electrons on our water molecule are going to take that proton, which would cause these electrons to move in here to reform your aromatic ring. So let's go ahead and show that. So we're going to reform our benzene ring here. And we took off the proton. So deprotonation of the sigma complex yields our product with a nitro group substituted in. So let me go ahead and highlight those electrons again. So this time I'll use green. So these electrons right in here, when that sigma complex is deprotonated, those electrons are going to move in here to restore the aromatic ring, and we have created our product. We have added in our nitro group. So that's the mechanism for nitration. Now, once you form a benzene ring with a nitro group on it, sometimes when you're doing synthesis, it's helpful to turn the nitro group into an amine. So let's go ahead and real quickly look at another useful reaction here. So once you form your nitro group like that, you can turn that nitro group into an amine a couple of different ways. So one of the classic ways to reduce a nitro group would be, in the first step, to use either iron or tin and a source of protons, so something like HCL will work well. And since you're doing this in an acidic environment, in the second step, you would need to neutralize it with something like sodium hydroxide. And those steps will reduce the nitro group to an amine. So let me go ahead and show the product here. So instead of an NO2 on the ring, now you have an NH2. So of course, this would be the aniline molecule. And there are other ways to reduce a nitro group to an amine. This is just one of the classic ways to do it. So it could be useful for synthesis problems, which we will study later in this tutorial.