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Electrophilic Aromatic Substitution
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0:00
Hi my name is David Van Vranken. I'm here UC Irvine in the Physical Sciences lecture hall and I came here to teach organic chemistry to 448 amazing students, but the coronavirus has changed all of that. Our lecture halls are empty. Our labs are empty. Our buildings are empty. Our campus is empty. We've gone remote on the UCI campus and we're finding ways to make it work - to deliver instruction any way that we can. Your job during this critical period is
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to stay healthy, and stay engaged with the material, and be interactive with the faculty, with the TAs and with the other students. I look forward to working with you this quarter. All right let's go ahead and get started with the content for Chemistry 51 C we're starting with chapter 18 called electrophilic aromatic substitution and I want to try to remind you of what you already know throughout this first two quarters what you've learned is substitution of
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electronegative leaving groups you've been trained to look for and identify your bonds two halogens like chloride bromide or iodide or bonds to two leaving groups liked tosylate. You've been trained to think about this process where a nucleophile can either attack the carbon atom and make a bond or have the leaving group leave first but ultimately you've been taught to think about this idea of substituting electronegative leaving groups so that's
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the way to form bonds, like this bond that I'm darkening, in organic chemistry but you have to get over that. Let me just elaborate on this by saying that while you learned that you do nucleophilic substitution of electronegative leaving groups bonds - to oxygen, bonds to halogen - you never ever do nucleophilic substitution to bonds that are directly attached (this is supposed to be a double bond here) directly attached to sp2 hybridized carbon atoms like carbon-carbon double bonds or even more
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especially onto benzene rings so you would never do SN2 substitution even with the best leaving group here with an iodide you would never do an SN2 substitution on these types of substrates so I'll just write never SN2. So we never do that. So this is what you should have should have picked up over chemistry 51a and 51b and now we're going to change all of that in a way it's a shame that we spend so much effort trying to teach you to learn how
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to substitute electronegative leaving groups through SN2 reactions or even SN1 reactions because now let me just say or SN2 or SN1 here. You don't substitute those types... We're going to teach you different ways to make bonds to sp2 hybridized carbon atoms so starting in chapter 18 let's just forget about SN2 we're going to teach you about a completely different type of substitution where there's no electronegative leaving groups, so in
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this chapter we're going to focus on the substitution of hydrogen atoms that are attached to benzene rings and the biggest challenge for you that's different from chemistry in [Chem] 51a and [Chem] 51b is we don't usually draw the atoms that get substituted. So let's go ahead and take a look at this process. So here's a substitution reaction I'm using E as a symbol for some sort of a powerful electrophile, and I'm not going to tell you what that electrophile is just yet,
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you have to wait just a minute for me to elaborate on that, but here, a substitution has occurred and what makes it tricky is the thing that got substituted is the atom that we don't draw - it's the atom that we assume that you know is there and it's an H atom. We typically don't draw H atoms and somewhere along the line you should have become so good at identifying how many H's are attached to every carbon that you should know that
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there was one H atom there so if I ask you what happened here, if all of you say you formed a bond between carbon and electrophile then you're missing the point that you BORKE the bond between the carbon and the H. That's just as important as forming the bond between the carbon and the symbolic electrophile. So that's what we're going to do this quarter - this chapter, chapter 18 - is we're going to learn about this completely different type of substitution reaction where
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we're not substituting electronegative leaving groups. We're substituting hydrogen atoms, more specifically protons on benzene rings. Okay what I've arranged for you here is a list of five electrophilic aromatic substitution reactions. That's what this chapter is about so we're taking the electrophilic reagents and we are substituting H atoms on benzene rings. I've done you the courtesy of drawing those H atoms here but as we go through the chapter I'm
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going to stop drawing those Hs every time. I expect you to know that there's six Hs on every one of these benzene rings now this chapter is really about five reaction recipes. We're going to teach you five recipes for substituting age with halogen, H with a nitrogen atom, H with a carbonyl group, H with a sulphonic acid, or H with an alkyl group; five recipes - that's what this whole chapter is about. It's going to seem like an insurmountable amount of work to
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you because the challenge here is all about not really memorizing these recipes but about all the exceptions. Let's go and start off with those recipes so that we can see what's going on in the reaction. Okay so here's our recipes. The first recipe for forming a bond between a benzene ring and a bromine for substituting in each atom is going to be the addition of the halogen. So this is a reaction that will also work with
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chlorine so I'll just write or Cl - but this isn't enough - you should have learned sometime way back in a previous chapter like chapter 10 on addition to alkenes that benzene rings don't react like alkenes [benzene rings] are UNreactive you can't just take Br2 and add it to a simple benzene ring and get any sort of a bond forming reaction to occur you need a super-secret reagent here and that reagent is iron tribromide or more correctly ferric bromide so if you're
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doing Cl2 addition you usually use FeCl3. This is a secret catalyst without this catalytic FeBr3 there's no way that this reaction will work and you simply have to memorize that additional reagent. OK, that's your first recipe and later will tell you some really important reactions that you can do with the bromine you can't do SN2 on this we'll teach you how to convert that into a SUPER POWERFUL nucleophile. OK what about carbon nitrogen bonds when you look at drugs, a large number of the
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pharmaceuticals that are sold in the United States and around the world have bonds between benzene rings and nitrogen atoms ,and so it's useful to know how to put those carbon-nitrogen bonds on benzene rings. So the secret recipe that we use there is to use nitric acid, HNO3. Nitric acid is a super dangerous acid we have concentrated nitric acid in our laboratory but it's mainly dangerous not just because it's a strong acid but also because it's an OXIDIZING acid but
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it's not the strongest acid that we have in the lab and it is not strong enough to nitrate a benzene ring if you want this reaction to work you have to remember to add an acid catalyst that is more powerful, more acidic, than nitric acid and the acid that's more powerful and more acidic than nitric acid is sulfuric acid. So you have to memorize this two-part recipe for substitution of H atoms with nitro groups and if I draw the structure of that nitro group -
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maybe you're not used to drawing out nitro groups - we'll get into this in more detail shortly. Here's the lowest representation for that nitro group. There's more than one resonance structure possible. Okay what about forming carbon-carbon bonds to a benzene rings. Here's the best reaction for forming a carbon-carbon bond to a benzene ring. It involves substitution of a functional group called an acid chloride, an R can be alkyl (R can''t
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be a heteroatom) but could be methyl, ethyl, isopropyl, another benzene ring, so when you have this type of functional group called an acyl chloride or an acid chloride, that can be used to form bonds between benzene rings and carbonyl groups but again we need a special secret acid catalyst as part of this recipe and you simply have to memorize this, and that acid catalyst that's a Lewis acid, not a protic acid, is aluminum trichloride (AlCl3). That's a super powerful
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Lewis acid it's going to grab onto that chlorine and help yank it away. At some point in the mechanism and we'll talk about the mechanisms for these substitution reactions. Okay next is the addition of a sulphonic acid group. I think we've all we've already learned about tosylate leaving groups. The way that you make toluene sulphonic acid - make tosylate leaving groups - is through this sulphonation reaction. The secret recipe here is to use sulfur
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trioxide (SO3) so if you burn coal that has high sulfur content the sulfur ends up getting oxidized and burning; and when that comes into contact with water it creates sulfuric acid. This is a powerful electrophile but it's not powerful enough. Again, we need a super powerful acid catalyst and the acid catalysts that we use is sulfuric acid. In fact you can buy bottles of sulfuric acid that have sulfur trioxide saturating them, so in a way you can get
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a pre-made mixture of that; but sulfur trioxide helps substitute the H- carbon bond with a sulfur-carbon bond or here our last recipe is a little bit redundant. I showed you in this third recipe how to make up done between a benzene ring and a carbon atom; but here I'm telling you how to put an alkyl group directly attached on the carbon atom - t-butyl, isopropyl, or groups like that - and the recipe looks a little bit similar to the acid chloride. Alkyl chlorides or alkyl bromides or
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even alkyl iodides can be used to do this and we use (seemingly) the same recipe that we did for that acyl chloride over in this third recipe but the problem is, with this reaction, is that there are lots of exceptions to this, and the most important exception that makes it not used very often, is that you would have to use the benzene in excess you have to use this as the solvent for the reaction in order to get a little bit of the product and so you're really
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throwing away most of the benzene that you put into the reaction that doesn't end up reacting, so that [excess] is absolutely required. Often times, I indicate excess in organic chemistry by writing XS. You have to remember to do that if you want to convince me that you you have an efficient process. Okay this is what chapter 18 is about. It's about these five reactions. You need to get out of here and you need to go memorize these reaction recipes, so that you have
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instant recall. That's what this chapter is about. Now let me be straight with you these are not all equally useful so it turns out that the first three reactions on this list are going to be very useful - I'll just write most useful - and these are MONEY reactions for making bonds to carbon nitrogen or to bromine iodine but the chlorine so again this would be Cl2 plus FeCl3 over here, the ones on the bottom it's just not that common to make carbon-sulfur bonds attached to benzene
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rings the reaction works it's fine it's just not that common there's a few drugs that have on there like Viagra, or sildenafil , and some others but it because it's not a very common functional group. I would say that this is a less useful reaction you're going to see less throughout the quarter then this last version of forming a a carbon bonded to a benzene ring is less useful and has more exceptions than this example up above where you use an acyl
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chloride. Usually if you want to have some alkyl group or isopropyl group you can put that on there through a multi-step sequence starting with this reaction so that it'll turn out to be more efficient. So really these top three reactions are the ones that you're going to find yourself using most throughout the rest of the quarter but you need to learn all of these recipes for this chapter, and you need to get out of here and go practice those recipes, and there's
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plenty of problems in the textbook and in Sapling [homeowork system] to help you learn those. We talked about the mechanism for these processes but they all kind of have a very a similar mechanism, so... let's go ahead and say what's going on here. Okay they all involve a two-step mechanism and I'm going to be generic about this mechanism I'm going to symbolize some sort of a POWERFUL electrophile that we're generating in this reaction mixture; all those recipes are designed to generate
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super-powerful electrophiles - more powerful than any electrophile that you seen in [Chem] 51A or [Chem] 51B and we need those powerful electricals because the double bonds in benzene rings are not like regular double bonds; they're not very reactive and so we have to have super-powerful electricals. Look at this I'm drawing the H for you because I think you're going to forget that it's there if I don't draw it at this point. So let me go ahead and symbolize the
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super-powerful electrophiles that were generating with those reaction recipes by writing E-plus (E+), and whether you're forming a carbon- sulfur bond or a carbon-nitrogen bond the only thing that we're changing is E+ in these reactions. The first step is the slowest step of the mechanism. If a reaction takes two days then it takes two days because you're just waiting for that double bond of the benzene ring to attack E pluses (E+s) that are
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floating around in the reaction. So step 1, the slow step, is to get these super stable pi bonds in the benzene ring to attack an electrophile, so when we do this when we take electrons away from one of these carbons over here - I'm going to leave a carbocation over here - we are going to lose aromaticity and we're going to generate a very special carbocation called an arenium ion on that it's got its own name. So according to my arrow pushing I'm going to leave... since
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I'm taking the electrons away from this top carbon I'm going to leave a carbocation there. Now notice how I drew that H atom here? The reason I draw that H atom here is because I'm going to need it in this step and if I don't draw it here, you'll find that you're going to forget to what to do with this step if you haven't drawn that H there. That's why it's so critical to to draw the H at this position because you need that for this next step. In all of these reactions
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there is some sort of a conjugate acid that can pull a proton off. It can be a solvent molecule; it can be an ether molecule; it could be an oxygen from the glassware; and we don't care what the identity of that conjugate base is that's left over from our acidic conditions. So I'm going to symbolize that with the letter A could be a sulfonate anion; it could be a FeBr4-, we don't care. It's not worth your time worrying about that. The important
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point is that this angry arenium ion. Let me go ahead and write that out - arenium ion - so we can get this this language down. The point is, that we've... that anything can pull this proton off. It really wants to be an aromatic system again and the only way to regain aromaticity is to pull off this proton and give the electrons back to this carbocation - this carbenium ion ( the technical term for a carbon that only has three bonds and a positive
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charge). So you have to pull this proton off to regenerate aromaticity and that's easy. Wow. We're not worried about what the A- is because anything can do that but we have to symbolize it with something so when we do that with regenerated aromaticity and I now have replaced an H atom on a benzene ring with some sort of a powerful electrophile. All five of those reaction recipes that we showed you go through that reaction mechanism So the the reaction mechanism here. Okay,
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you have to get used to drawing resonance structures. Let's go ahead and start off by going into a little more detail on this areinium ion so let me go ahead and redraw that arenium ion here so we can think about the fact that there's another alternative reaction that could have occurred, but never occurs, so here's one possible resonance structure for that arenium ion and something that you never see happen right if you have A- floating
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around it ALWAYS pulls off the proton what you never see happening and I'm just gonna - I hate to even draw this - is you never see that A- (I'll just write NO)... you never see that A- attack that carbocation attack carbon; instead it pulls off the proton as I showed you up above here. If you attack here you've lost aromaticity and you'll never get it back, so the the favorable reaction pathway is to deprotonate and regenerate aromaticity. There's two other key
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resonance structures we can draw for this simple substituted arenium ion where we added an E. Let's go and practice drawing those. Remember with a resonance arrow, it's a double-headed arrow, not an equilibrium arrow, but an arrow two heads on the line. The other resonance structures that we can draw, we can shift this double bond over here and leave the positive charge over at the seven o'clock position on that ring. So let's go ahead and practice with that. Here we go.
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You know, since I didn't have one... I don't like to have more than one set of arrow pushing here, since this isn't a real arrow I feel to draw a curved arrow to represent this process. Never have two different TYPES of processes depicted with curved arrows only one [process]; but one of those is a dashed line which indicates it's not really happening, and so this next arenium ion resonance structure we have would look like this, where I now have the
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positive charge over here at this lower left position, and there's one more resonance structure that we can draw and we use... again, we use a resonance arrow to depict that; and that is if we shift this double bond over an attack, in a sense we attack that carbenium ion, and when we draw that last arenium ion resonance structure it has a positive charge or a carbocation on the other side of this atom that got substituted. Now all of these resonance structures are pretty similar, and not
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any of them is more stable, these are just three different depictions - three different depictions of one single arenium ion. They're just different depictions that are not different compounds. There's no rate or speed at which all of this occurs; they're just all different ways to depict the same arenium ion. So again when you get to this mechanism (because you spent so much time doing SN1 reactions) every fiber of your being is going to make you want to
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attack that carbocation. But don't do it; don't attack that arenium ion. You deprotonate. That's what leads to the substitution product and so fight that temptation.
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Right. That's what we're going to finish up with here in lecture one. When we come back on Wednesday we're going to be talking about the detailed mechanisms as we go through each of those five different recipes for electrophilic aromatic substitution. So I look forward to seeing you on Wednesday and and in discussion section this week and at office hours.
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