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Video transcript

In the last video I promised you that I would show a concrete example of electrophilic aromatic substitution. So let's do that right here. So let's say we have some benzene. And it's a solution with some molecular bromine. So I'll draw the molecular bromine like this. So it's one bromine right there. It has one, two, three, four, five, six, seven, valence electrons. And it's bonded to another bromine that has one, two, three, four, five, six, seven electrons. These two guys in the middle are bonded to each other. They can kind of act as an electron pair or make bromines feel like they have eight electrons. This guy thinks he has a magenta one. That guy thinks he has the blue one. And let's say we have some iron bromide in our mix as well. And what we're going to see is that this is going to catalyze the reaction. So the iron bromide. We have some iron and it's bonded to three bromines just like that. So we haven't seen iron bromide much. But the way I think about it is, bromine molecules are much more electronegative than the iron. So even though these look like fair covalent bonds, if we had thought about oxidation, these guys are going to be hogging the electrons. This actually would have an oxidation number of three. And in reality they are hogging the electrons. They are more electronegative. So the way I think about is that this iron will have a slightly positive charge. Because the electrons are being hogged away from it. So it wouldn't mind to gain an electron. Or it might want to accept an electron. It might want to act as a Lewis acid. Remember a Lewis acid will accept an electron. So this might want to act as a Lewis acid. So who can it nab electrons from? Well maybe it can nab an electron from this bromine right here. And I'm not saying that this always going to occur, but under the right circumstances if they bump into each other with the right energies it can happen. So this electron-- let me do it in a different color that I haven't used yet, this green color-- let's say this electron right here gets nabbed by that iron. Then what do we have? Well then we have a situation, we have this bromine-- the blue one-- with one, two, three, four, five, six, seven valence electrons. We have the magenta bromine with one, two, three, four, five. Now it only has the sixth valence electron right here. The seventh got nabbed by the iron. So the iron has the seventh valence electron and then you have the rest of the molecule. So then you have your iron and it's attached, of course, to the three bromines. Just like that. And then our bonds, these guys were bonded. They still are bonded. And now these guys are bonded. These were in a pair. This electron jumped over to the iron. And now we have another bond. But because this bromine lost an electron-- it was neutral, it lost an electron-- it now has a positive charge. And the iron, now that it gained this electron, now has a negative charge. So let's think about what's going to happen now. Now we're going to bring the benzene into the mix. So let me redraw the benzene. And we have this double bond, that double bond, and then just to make things clear, let me draw this double bond with the two electrons on either end. So we have the orange electron, you have your green electron right over there, and I'll draw the double bond as being green. Now let's think about this molecule right here. We have a bromine with a positive charge. Bromines are really, really, really, electronegative. You might see them with a negative charge, but the positive charge, it really wants to grab an electron. And in the right circumstances, you can imagine where it really wants to grab that electron right there. So maybe, if there was just some way it could pull this electron. But the only way it could pull this electron is maybe if this-- because if it just took that electron than this bromine would have a positive charge, which isn't cool. So this bromine maybe would want to pull an electron. If this guy gets an electron, then this guy can get an electron. So you can imagine this thing as a whole really, really wants to grab an electron. It might be very good at doing it. So this is our strong electrophile. So what actually will happen in the bromination of this benzene ring-- let me draw some hydrogens here just to make things clear. We already have hydrogens on all of these carbons. Sometimes it's important to visualize this when we're doing electrophilic aromatic substitutions. So we already have a hydrogen on all of these molecules. So maybe this is so electrophilic it can actually break the aromatic ring, nab this electron right there. So maybe, this electron right there goes to the bromine. Maybe I should even do it this way. Just to make it clear. Kind of replaces that one. Although, obviously the electrons are a bit fungible. So maybe it goes over there. And then when it goes to this-- let me make it clear it's going to the blue bromine. So the electron right here goes to the blue bromine. If the blue bromine gets an electron then it can let go of this blue electron. So this blue electron can then go to this bromine right over here. And then what is our situation look like? Well if we have that, then let me draw our benzene ring first. We have our benzene ring first. So let me draw the benzene ring. This double bond, that double bond. Let me draw all of the hydrogens. I want to do them in purple. So we have one hydrogen, two hydrogens, three hydrogens, four hydrogens, five hydrogens, and six hydrogens. This orange electron is still with this carbon right here. But that electron got nabbed by this bromine. So that electron got nabbed by this bromine right over here. I've kind of flipped it around and now it has its other six valence electrons. One, two, three, four, five, six. The electron got taken away from this carbon. So now that carbon will have a positive charge. But we saw in the last video, it's actually resonance stabilized. That electron could jump there. That electron can jump there. So it's not as stable as a nice aromatic benzene ring like this. But it's not a ridiculous carbocation. It's stable enough for it to exist for some small amount of time while we kind of hit our transition state. And then this molecule over here, what's it going to look like? Well this bromine had a positive charge. Now it gained an electron. Let me draw it. So you have your bromine. It gained an electron. So now it is neutral again. So let's see, it had the one, two, three, four, five, six. Now it gained this blue electron. So now it's seven valence electrons. Back to being neutral. It's bonded to the iron bromide. Let me draw the bromines. One, two, three. And so we've given this blue bromine to the benzene ring. But it's not happy here. It doesn't want to break it's aromaticity. It wants an electron back. So how can it gain an electron back? Well this thing-- actually let me make it very clear. Let me make it clear. This thing had a negative charge. So you can imagine-- and we had this electron right here. So maybe this thing right over here can now act as an actual base. It can nab a proton off of the benzene ring. Just like we saw in the last video. This is now the base. This whole complex, to some degree, acted as a strong electrophile. Now that we got rid of one of bromines, this thing might want to grab a proton now, since it is positive and will act as our base. And will nab one of these protons. If it nabs the proton, than the leftover electron is still there. That electron is still there. Let me do that in a different color. I'll do it in red. That electron is still there. And then that electron can be given to that original carbocation and we'll have a nice aromatic ring again. So how would that look? So you could imagine a situation where this electron, this green electron right here, gets given to the hydrogen nucleus. If it's given to that hydrogen nucleus, then this red electron right here can then be given to the carbocation And then what are we left with? Well, we have our-- let me draw what we started with-- so we have our ring. We have this double bond and that double bond right there. Now let me draw all of our hydrogens. We have this hydrogen, that hydrogen, this one over here, this one over here, and we have this one over here. Now this hydrogen just now got nabbed. So this hydrogen over here got nabbed, it got given-- actually the hydrogen nucleus got given this green electron. It got given this green electron, which is paired with this magenta one right over here. So it is now bonded to the bromine. We now have hydrogen bromide. And this had one, two, three, four, five, six other valence electrons. I'll keep the colors consistent. And this is now bonded. Now this electron went away from the iron. So the iron will now lose its negative charge. The iron bromide. So now it is back in its original form. So we have our iron bonded to three bromines. We are bonded to three bromines just like that. It has lost its negative charge. And this electron right here has now gone to the carbocation So that electron right there has now gone to the carbocation. It is right there. So this bond, you can imagine it now being this double bond. So it was magenta, so I'll draw it in magenta just there. And we can't forget the whole point of this whole video was the bromination. So we now have this bromo group there. So we have this orange electron is right over here. And it is bonded to that bromine. Just like that. And we have brominated the benzene ring using-- and just let's be clear-- we had a negative charge and a positive charge. Now it's canceled out. The thing with the negative charge gave an electron to the positive charge. It is now neutral. That is now neutral. Now let's be clear on what happened here. Just to map it to what we saw before. We had a benzene ring. Right here we have no strong electrophile, or strong base yet. It had to become a strong electrophile. When this thing gave an electron to the iron bromine, it became this larger molecule. Now this whole thing can act as a pretty strong electrophile. This grabbed an electron. It broke the aromaticity of the benzene ring. But just long enough for this bromine to form. But once this bromine, it gained an electron and then bonded to the benzene. And then gave up an electron to this other bromine that really wanted to get an electron because it had a positive charge. And then once it got it, now this whole thing acted as the base. So we kind of have the same molecule changing up a little bit, acting as an electrophile or acting as a base. And then once it acts as a base, this bromine, this magenta bromine, nabs the proton, allows this electron to go back to that carbocation and then we're left with the iron bromide again. So this thing really didn't change through the whole reaction. That's why we can call it a catalyst. It wasn't one of the reagents, one of the reactants in the reaction. Hopefully you found that vaguely interesting.