Mm-hm. Right. >> All right. Good afternoon. The group as a whole is off to a pretty good start. Obviously, a lot of people did very well on this exam scores. >> At the high end. >> There are a few people at the low and middle, little bit God, at this point. But on the whole, pretty normal performance for the first exam. And hey, if it makes you feel any better, I didn't get a 100 on this exam either. >> There's a mistake on the answer. >> If you haven't already picked up a copy of the answer key, please feel free to do so before you leave. If you have picked up a copy, the answer turned out page for the answer to question number 11, declined. A. Well, there's an obvious mistake that needs correcting and since I wasn't typing, I can't call it a title. So I'll call it a script that I wrote. But main thing you should do is compare your answers to what's on the answer key, C major mistakes. If you have any questions about how anything was greater, please don't hesitate to bring that to my attention. In fact, before we go any further, I should probably call your attention to the regrading policy that appears in your syllabus, which we kind of glossed over on the first day of class. But now that you have an exam back, maybe it's time to say something about how this works. Now it's not quite as applicable to this class as it is normally because the TAs didn't grade your exams? >> I did. >> But nonetheless, if you believe that an error has been made and grading your examination, you may submit the examination to me for regrading the diesel. Simply circle the numbers of the questions that you would like to have the reconsider and return the entire examination to me at the next class meeting or some other future tab, I will reconsider the grading of the circle questions, make any necessary adjustments to your grade, and return the examination to you in class at some future plot. It is strongly recommended that you consult the answer key before you submit your exam for grading. So my advice, compare what you have and what's on the answer key, that you think you deserve more points than you got for a question or two or three. >> Just circle the numbers of those questions that you can turn it back into me for reconsideration beginning tomorrow. >> And feel free to read over the entire regrade policy whenever you get a chance. That's on the last page of your syllabus. On the whole, I'm reasonably pleased with the group from here and to see how things go for the next few weeks. Alright, for today, there's a few topics in chapter five on stereo chemistry that I want to finish up talking about. But where we're going mostly from there for the rest of today's into chapter six, which is the first chapter that looks seriously at chemical reactions involving organic compounds, will make your day right off the bat. We will pick up more or less where we had left off. Section 5.12 talks about, among other things, had a name compounds that have more than one stereo center, such as 2R, 3R, T3. And then we move into section 5.13, which is the Fischer projection. >> For this one, I'm going to tell you flat out ignore efficient projection formula, but now I have to tell you what they are. >> So you know, what do they look. So let me just show you what one. And here's a molecule we were looking at last time. This is 2-3 di Roma butane, where the red balls represent bromines, on the white balls represent hydrogens, and everything that's black represents a carbon f. Now this is a 3D model of it. >> Here's one way of drawing the structure, and I hope it's clear that my picture and the model represent the same thing, but everybody agree with that. >> Okay, now that's one way of representing the compound. >> And it's the way I think most people would try to represent the compound using a two-dimensional piece of paper to represent the three-dimensional molecule. But here's an alternative. If I take this model and turn it like this, and let me just position this properly. >> Now the idea here is that >> This bond where my fingers are right now is lying more or less in the plane of the page, which means on the right side, red ball, red ball out toward you, on the left side, white ball, white ball out toward you. And those two methyl groups are now away from you. In other words, here's a alternative three-dimensional representation for the same thing. >> So again, when everybody can see that this drawing and the model represent the same thing. >> Okay? >> Now here's the point. Neither one of these is a Fischer Projection. You got anything that has a wedges and dotted lines in it is not a Fischer Projection. Here's the Fischer projection for what I just drew. Now the point is what a Fischer projection is, is a drawing that looks something like this, where each little crossbar here represents not just a carbon atom, but specifically an sp3 hybridized carbon atom. And even more specifically, the horizontal bonds are interpreted as being out toward you and the vertical bonds are interpreted as being away from you. >> So this is a Fischer Projection, and there are two reasons why I'm going to say do not use Fischer projections. >> One is nobody does any place you ever cedar Shane's was in textbooks. >> You're going to take a walk down the hall to the next building where all the organic graduate students do their research and thumb through the pages of their laboratory notebooks, I guarantee you, you would not find any Fischer projections. >> They have their purpose. >> They certainly worked well for email Fisher a 100 years ago. >> And he was developing these things to talk about carbohydrate chemistry. But even carbohydrate chemists don't really use these things anymore. They draw things out using chair conformations are wedges and dotted lines like we do here. And the other thing I don't like about Fischer projections is they tend to be a bit misleading. Let me show you an example what I mean. Here's a somewhat simpler chiral molecule along the lines of what we were looking at last time. >> We have a central sp3 hybridized carbon atom. >> We have green ball, red ball, white ball, gray thing attached to it. Alright, let's turn this suite of matter which way? Let's turn it something like this. And now let's draw a picture of what that looks like and Fischer projection for and again, the understanding is horizontal bonds are the ones out toward you. So green and white, like this vertical bonds away from you read >> Right now, we haven't said specifically what red, gray, white, green stand for. >> So we'll just make up something and we'll say, we'll call this R. Now the point is that this represents the R form. But whatever molecule this is just turning this different directions or turning it over doesn't change the fact that it's still aren't. The only way to do that is to switch to groups at the stereo center. Like we described before, it's sort of like saying this is my right hand and I can turn it this way or that way or flip it upside down. It's still my writing it. >> But if you rotate a Fischer projection 90 degrees in any direction, so I'll rotate this 90 degrees clockwise. >> We'll call that red. >> They're white, they're green. Well, the point is, whatever your priorities were over here, that leads you to the conclusion that this is our pictures s, Let's not especially real-world. >> You can't rotate a molecule 90 degrees and haven't changed configuration. So more of the story is, I think Fischer Projections cause more confusion than they solved. And I will strongly recommend that you not, if you want to represent a molecule in three dimensions on a two-dimensional surface, grow out wedges in dotted lines like we've been doing. So I think in the long run typically a lot more clear. So you can pretty much get Section 5.13. Now that you know what Fisher prediction formulas on. >> All right, section 5.14, stereo isomorphism of cyclic compounds. >> Let me just say two things about this. First of all, it doesn't make any difference whether a stereo center is part of a ring or not. It still behaves like a stereo set. But a lot of people, for whatever reason, freak out when they realize the stereo center they're looking at is part of a rate. So here's my advice on how to deal with that situation. Well, if you have a molecule like this one, that is a fine numbered ring in stereo center here. >> Stereo center here turns out to still be amazing. >> Compound, one of those two stereo centers, as are the other ones, this. >> But people get a little freaked out by the fact that it's part of a writing, especially withdrawn it bond line notation, when it's drawn out more detail might be easier to see. My advice. >> If you find yourself not to have an anxiety attack because the stereo center as part of a ring, OK, now support a ring anymore. >> They've got a carbon atom. That makes it easier to visualize fine. But if you're asked to draw this molecule, don't forget to put carbon back in before you're done. >> Yes or no, you can't either side. >> Let's draw a picture. >> Let's take this exact same molecule, just showing it and let me just rebuild a little cycle gang here. >> Okay. >> Now again, would you agree that a model and the picture represent the same thing? >> Everybody. Okay, now here are the two stereo centers. >> Let's assign a configuration at this. >> One highest priority group attached to that stereo center is, well, let me do this in two different colors. >> We'll label this green carbon atom is this stereo center here. And bromine gets the highest priority. Okay, well, it's the lowest priority. >> That stereo setup, the hydrogen four. >> Now the other two things attach there are this turbine and this carbon. >> Carbon versus carbon. >> You have a tie, but which one of those two is going to have the higher priority, but one that has the bromine attached to it. >> So this is number two, this is number three. >> And as this is drawn, is this stereo center R or S S. >> So the point is assigning our arrest you a stereo center that's part of a ring is no different from assigning artist to stereo center. >> That's not Puerto Rican carbon now, because the tie is broken by the time you get to this point. >> And that's, that's an important point about assigning ART s In general, the tie is broken at the first of difference. >> Once you break the tie, you don't have to worry about what's beyond that. >> Now, just to finish this off, let's do the other stereo center, which I will label in red. Highest priority group attached to that stereo center is bromine. >> Again, lowest hydrogen carbon versus carbon tie. >> Which one wins? But one with a bromine. >> Item number two. >> Number three, read stereo center is what? R or F? R, which you kind of knew because a moment ago I said it was a stereo, but instead it was a meso compound and by definition a mezzo combative and asked to stereo centers. >> One of them must be o and the other one must be yes, because they meso compound has a symmetry when there's a plane of symmetry in that contract. >> And yes, planes of symmetry can cut through atoms, which this one clearly does make sense. Now, one of the other things they will point out, especially for cyclohexane derivatives Let's just draw something here. Suppose we have something like this. Now it turns out and not be obvious from how it's drawn. >> But it turns out, just like this guy has to stereo centers, this guy has to stereo centers. >> However, visualizing that and assigning ours and S is to stereo centers. >> When the compound is drawn in something like a chair. >> Former boat form can be very, very difficult. So my advice here, It's hard to visualize what's our, what's s. When you're looking at a picture like this, redraw it, redraw it in the flat form, sort of like we drew it in the previous case. And by analogy to the previous example, I hope it'll be clear that this one is R and this one of this, but it's probably a lot easier to visualize when you draw it this way. But it isn't, you draw it something like this concept makes sense, also makes it a lot easier to see the plane of symmetry in this combat. This compound is a mezzo Kanban, but it's very hard to see a plane of symmetry when it's drawn like this. When you draw it some other way, then the plane of symmetry becomes a little bit more obvious. So the point is, don't hesitate to use whatever tools you have at your disposal to make the molecules easier to visualize. And that's pretty much all I had to say about what's in Section 5.14 and your textbook. >> But there are other examples of the sorts of things. >> On pages 21819 of your book, section 515, the title of which is the somewhat convoluted English relating configurations through reactions in which no bonds to that chirality center or broken. Alright, what this really does is open the door to a discussion of how stereochemistry relates to chemical reactions in general, which we will certainly talk about in more detail throughout the course of the lecture today. >> But let me just emphasize a few general principles that apply in general to use stereo chemistry and chemical reactions. >> Now, just to speak to what it is, the title of Section 5.15 is trying to say. The basic principle they're trying to emphasize and Section 5.5 td is if we do not break or form any bonds stereo center in the course of a chemical reaction. >> And the stereo center will retain whatever configuration at hat. In other words, suppose just for argument's sake, that, alright, say that this are stereo center right here. >> And whatever chemical reaction we do to this molecule takes place over here somewhere. >> But it doesn't involve forming or breaking any bonds to this stereo center right here. Well then what that means is this stereo center will still be r By the time the reaction is done. >> And likewise, if you don't break or form any bonds at this stereo center, then that'll still be s By the time the reaction is done. >> So the point is in general, whatever configuration R or S your stereo center has, it will retain that configuration if you don't break or form any bonds to that stereo center during the course of the game. >> That much Makes sense. >> Yeah, what did the reaction adds? Who's able to carve out? >> And that's right below, like right here saying, okay, well, we're going to talk about reactions like that as the semester unfolds. But right now I'm just illustrating a couple of general principles, but you raise an interesting point. >> So that brings up some other principles. >> Now for starters, is this molecule optically active or optically in the molecule as a whole? Optically active or optical, You know, I thought you said optically active. >> That's why I say that. >> Okay. I'm sorry, I didn't hear the end. Yeah, the point is, it has a plane of symmetry. >> So the point is anytime it has a plane of symmetry, it must be optically good. >> Fundamental principle that applies here is to say that optically inactive reactants yield optically inactive products. Now just to use the reaction you suggested as one example of that sort of thing. >> Alright, suppose you're going to do some reaction at this carbon atom right here that puts on a chlorine and does whatever it does to the chirality of the rest of the molecule. >> But the thing is, since this thing has this plane of symmetry, if you can attach a chlorine over here, you're going to catch a chlorine over here. And because of the symmetry of this molecule, there is absolutely nothing to favor attack over here versus attack over here. So the point is, whatever happens on this side will happen with equal probability on this side and the products will still be optically Again, we'll see other more specific examples of that a little bit later on. But are the reactants are optically active, all bets are off. >> That is to say that there are specific outcomes for each reaction. >> But there is no corresponding general principle that applies to situations where the reactants are optically active. So while we can state with certainty that optically inactive reactance will always yield optically inactive products. >> Optically active reactance might give you, optically active products might not. >> Depends on what kind of reaction is taking place. >> And more specifically, it depends on what the mechanism of that reaction. >> Before today is over, I'm hoping to show you some examples of each of those kinds of circumstances. >> We're now due to general principles. >> Make sense? >> Okay? Well, I have a lot more to say about these things as time goes on. All right, what else do we have here? In Chapter five? We go Section 5.16, separation or resolution of IV antibiotics. We talked before about the thalidomide situation politically. >> It was a drug marketed to pregnant women in the 19 fifties. And the point is, when you, Nancy Marie, had the beneficial effects for which the drug is really prescribe. The other, an antibody produced severe birth defects. >> So ideally, we would like to not market the drug to the public as a receiving mixture, but take the bad stuff out to the shed and shoot it, put the good stuff out there where it can do good things. >> But how do we do that? >> Because as we said before, enantiomers have pretty much identical chemical and physical properties. >> There's only two exceptions. Well, it turns out that there is a way of doing this and it takes advantage of one thing. While enantiomers may have virtually identical chemical and physical properties, melting point, boiling point for water solubility, et cetera. >> It do compounds are diet staring mirrors of each other. >> They will have very different chemical and physical properties, different melting points, different boiling points, different solubilities in various solvents, etc. >> So let's just suppose that you had isolated some compound from some plant source. >> Which is why I'm putting it in here in green. >> And it occurs in nature as a receiving mixture. >> 50% are 50% S turns out just to pick something, they are form cure Parkinson's disease. >> So you'd like to isolate the form, get it out to the public, and not only cure a lot of Parkinson's patients, but probably win the Nobel Prize for medicine. Problem is the S form causes Parkinson's disease. So you gotta get rid of the S before you get the r out. And how do we do it? >> Well, what we do is because we're really good at Organic Chemistry, we know about some other compound represented in red here that occurs in nature as a single stereo isomer. >> And by the way, this is not unusual, as we said before. For example, in the case of the amino acids, nature mostly makes the S form of most amino acids. And there are many other compounds that occur in nature as either single stereo isomers or predominantly one stereo isomer. So we've isolated this compound from some other source. >> And furthermore, being skilled organic chemists, we know when green thing reacts with red thing, they attach to each other. >> But we don't change the configuration at any of the stereo centers. >> So when this reaction takes place, when green our reacts with red s, The product as we configuration or S. >> But when green S reacts with red s, The product is the configuration SLS. The point is the original mixture of the two stereo isomers, but they were CMYK mixture, which is a mixture of two enantiomers by definition. >> But what's the relationship between RF and SS, RBI stereotypes? >> And the point is, since dice streamers have different chemical and physical properties, we should be able to find some separation technique that you've been learning about in the lab, like recrystallization or distillation or scratching or something to be able to separate those two. >> So the point is separating a mixture of dietary emirs was a lot easier than separating a mixture. >> So the point is we can now get these things into separate beakers. And then fortunately, because we are stilled organic chemists, we also know about some other compound which we'll call blue X. And blue X's mission in life is to react with that red thing wherever can find it and separate it from wherever else statute. So we take our RS and react it with x. >> You get pure are sitting in the beaker. >> The byproduct is SAX, which we don't care about because what we care about is getting this stuff out on the market and hearing Parkinson's disease. So the moral of the story is with minor variations. >> Essentially the way you separate a mixture of enantiomers is to convert it to a mixture of dietary mirrors, which have different chemical and physical properties. >> Separate that and then somehow reverse your original reaction to get your single bathroom with the logic make sense. Now there are again, variations on the theme that can be used. If you want more details, read Section 5.16. In fact, Report section 5.16 EI. Anyway, the first separation of a pair of enantiomers was the two different antibiotic forms of tartaric acid separated by a fellow you may have heard of, Louis Pastor. Pastor did it. He crystallize them out. So he had solid forms, the two. And then he got out his magnifying glass. And notice that the two different kinds of crystals were left-handed and right-handed. When he got out tweezers and separated the left handed crystals from the right handed crystals. It was the time it was. >> All right. >> All right. >> A couple of other little details to finish up. >> Chapter five, section 5.18, paragraph long. >> And what that one paragraph fundamentally says is other atoms, then carbon can also be stereo centers. They show examples here where silicon, germanium, nitrogen surfer in various incarnations can be stereo centers. Now let me just point out one thing about nitrogen. >> Okay? >> They show what's referred to here as a quaternary ammonium ion. >> And by that we mean nitrogen with four carbon atoms attached to it, R1, R2, R3, and R4 for different things like methyl ethyl purple and Buto or something like that. And then here's the x minus over here. Because if you have four things attached to nitrogen, that nitrogen has a net positive charge. >> That said, a nitrogen atom like this with a lone pair like we'd normally expect to see. >> For a nitrogen atom is not a stereo set. Even though in principle it has four different things attached, R, R prime, R double prime, and the lone pair. And there are other atoms with lone pairs that can be stereo centers, like the sulfur with alone bear that's shown over here. >> But not nitrogen. >> Because what nitrogen tends to do through a process called quantum mechanical tunneling to undergo inversion to its other anti-American form. >> But the problem is these two are in equilibrium with each other. You want to think of it as the lone pair sort of burrowing through the nitrogen atom, coming out the other side and flipping everything around. >> That's fine. >> But the point is the mean flip flops back and forth between these two. You cannot isolated it either form, but get rid of that lone pair by putting a fourth alkyl group on that nitrogen atom. >> And then yes, it can be a stereo set. >> So a nitrogen atom to be a stereo center must have a positive charge. >> But the main point is it doesn't have to be a carbon atom to be a stereo center. There are other sp3 hybridized atoms that will do the job just to go. >> That's it since finally Section 5.18. >> Now, when we were first introducing all of the important concepts in this chapter, we said that for the most part, chiral molecules have stereo centers. Molecules that are chiral don't have stereo centers. And in general that's true, but we have seen exceptions. Meso compound is a compound that has stereo centers, but is nonetheless a chiral because it has a plane of symmetry. And likewise, it's not necessarily the case that a chiral molecule has to have a stereo set. >> Just to show you one example, you will see others in your textbook, just for the heck of it, naming this how many carbons in the main chain or a double bond to the last part of the name is you can die for four carbons, di, for two double bonds. >> Where are the double bonds? Obviously, we're going to call this carbon number one. First-level Bob begins the number one. Second-level. Bad begins at number two. And the rest of it should be pretty obvious. >> This is one bromo chloro 12V die. >> How many stereo centers are there in that molecule? Not because by definition, a stereo center must be an sp3 hybridized atom. >> There's a grand total of one sp3 hybridized atom and molecule. And it has three hydrogens on it as they have four different things on it to be a stereos. And yet I submit to, if you build a model of this compound and you build a model of its mirror image, those two will not be superimposable. And here's what. What's the hybridization of the central carbon atom between the two double bonds, S, P. Now let's think about what that means. >> It means that to form these bonds and the double bonds, you have an SP orbital going in this direction, and an SP orbital going in this direction overlapping with whatever else. You have a carbon tax, but it also means you have two P orbitals leftover, not used in the hybridization process. And those two p orbitals must be at 90 degree angles to each other. One of them is up and down, the other one is out toward you and back into the plane of the page away from $0 to borrow a couple of erasers for me. Now obsolete chalk tray. Those two double bonds, or I should say those two double dot pi bonds, whatever, are arranged at 90 degree angles to each other by January through something like this. This should give you an idea what the actual geometry of those things looks like. And what that means is that unlike what you would normally expect for something that's normally SP2 hybridized all the way and completely flat. If this metal and this hydrogen are in the plane of the page. And what that means is, for example, the bromine will actually be out toward you and the chlorine will actually be a waveform or the other way around. And the point is, it will take a whole lot of staring at this to realize that those two are mirror images, but not the same thing. But again, don't take my word for it. Go home and build models and convince yourself that that's good. So the point is, these two are enantiomers of each other, despite the fact that neither one of them has a stereo set. And there are other examples. For example, these two so-called bio molecules on page 224 will also fall into the same category. So in general, chiral molecules have stereo centers. A chiral molecules don't, but there are exceptions to that statement. Mezzo compounds have stereo centers that are eight chiral because of their plane of symmetry. Molecules like this are chiral despite the fact that they had mostly reserves. So there are rules, but there are also exceptions. Make sense? >> Okay? >> That is officially the end of chapter five. You haven't yet started the problems at the end of chapter five, please feel free to work on those. And again, when you're working these out at home, don't hesitate. Could use models as necessary so you can see what's going on. >> Alright, so for the rest of today, for about ten minutes or so that we had before. >> We're going to take our break. An introduction to what's going on in Chapter six. Misleadingly simple title of which is ionic reactions. Really, the subtitle tells you more about what's going on in this chapter. Nucleophilic substitution and elimination reactions about what he likes. Yeah, that's about half a dozen interesting terms right there. So let me just spend the next little bit of time before we take our break explaining what those things mean. First of all, by definition, an alkyl halide is simply a compound like this one, in which there's a halogen atom attached to an sp3 hybridized carbon alkyl should sound like alkanes, alkenes and sp3 hybridized carbon. So what's the name of this Kanban? Yeah, two bromo propane be the IUPAC name a way think of another way to name this campaign. Yeah, you could also call it isopropyl bromide. Civil alcohol, I like this one, are commonly named by just naming the alkyl group and then whatever the halogen is with the IDE suffix. And just one more point, because this will turn out to be important to this chapter, is this eight primary, secondary or tertiary alkyl halon secondary. Remember, we can classify alkyl halides as primary, secondary, or tertiary, depending on what kind of part of an atom as the halogen atom attached to it. So here's your halogen atom, here's the carbon attached to it. How many carbon atoms are attached to that carbon? The answer is 12. So secondary output. >> Now >> But the all this compound to react with something like potassium cyanide. Here's what we get. The inorganic byproduct that we don't much care about is potassium bromide. Here's the organic product. What functional group is represented by this molecule? This is a nitrile. But more importantly, notice what's happened here. E cn piece, which makes the nitrile functional group came from the cyanide. And what happened in this reaction is the sign I took the place of the bromine was originally attached to that secondary carbon five. That's what the title of chapter six means when he talks about a substitution reaction on an alcohol a lot, when something takes the place of something else, That's referred to as a substitution reaction. And the big chunk of what we're going to be talking about today is the different kinds of substitution reactions that alpha e lights can build. However, before it's over, we're also going to discuss the other possibility. Let me know about sodium hydroxide strong base. Now, in principle, if you throw in sodium hydroxide with the same secondary alcohol library used in the first example, the same kind of reaction to take place. And if it does, what compounds should be formed? Those we do a substitution reaction here. What do you think it's going to take the place of the Baroque, the, OH, and if it does, what functional group will be born? >> Alcohol? >> Specifically, since this is isopropyl bromide, do you think will be formed? Isopropyl alcohol. But there's a reason I'm not writing that, that that might happen to a small extent. Maybe about 10% of the product might be isopropyl alcohol, but the vast majority actually turns out to be this. >> And of course, this Kanban is an alkene. >> That is the reason I asked you. I originally asked you what you knew about sodium hydroxide and directly responded with strong base. What do strong bases do? Dig back to either chapter three in your textbook or your general chemistry class? Yes, Sagan. Ok. And a fully ionized. So if you dissolve solid sodium hydroxide water to the law of one mole of water, you get one mole of sodium ions and hydroxide ions. But more specifically, in terms of chemical reactions with other things, what these strong bases do. Yeah, yeah, the point is by definition, what HBase does is remove a proton, right? Lowry Bronsted definition of acids and bases acids donate protons basis, accept protons. Strong bases tend to yank a proton off of something else. That's pretty much what's happening here. Hydroxide is grabbing off one of the hydrogens from this end, CH3, which is why in the product that winds up as a CH2. And we also get rid of the bromine as b r minus. And so that all the carbon atoms can add. The number of bonds is supposed to have. >> We put a double bond between those two carbon atoms. >> We're starting with the compat is formula is C3H8. Seven, we're winding up with a cop out is formula is C3H6O. In the process we've gotten rid of a hydrogen and the bromine. We've eliminated a HBR. That's why reactions like these, I'll refer to as elimination reactions of alkyl highlights. Again, we're back to the title of chapter six. That's what an elimination reaction is. It is typically a reaction that involves a strong base and involves the formation of typically an outcome. Are we clear on what substitution and elimination meaning in the context of what we're looking at here. Okay, because that's clear then really, most of the rest of chapter six is discussing the details. Now let me just say one more thing about this, just to introduce some terminology and we'll go ahead and take our break. Typically, what happens in a substitution reaction involving alcohol is the alcohol itself, which we can generally summarized by r x, where r is the alcohol piece. So in this example, r would be the isopropyl group and X is the halogen bromine. In this case Reacts with something else. And something else generally gets the symbol N U, which is short for the nuclear file that we've seen nuclear files before. All neatly files are, are Lewis bases, things that can donate a pair of electrons. Many of them have negative charges. But that's not an absolute requirement for nuclear file. The only thing it really needs to have is a lone pair that it can use to form a bond with the r. >> In our next, the alkyl halide that participate in this kind of reaction is referred to generically as the substrate. >> So in this specific example up here, isopropyl bromide is the substrate. And essentially what happens during a so-called nucleophilic substitution reaction is the nuclear file takes the place of x. And since x departs from the substrate and winds up floating around in the reaction mixture as a free. I'm anxious, commonly referred to as the leaving group because that's what it does. It leaves the substrate. So the point is, this first reaction we showed you was a specific example of a nucleophilic substitution reaction. Isopropyl bromide is the substrate. The cyanide ion is the nuclear file. Potassium is pretty much just a spectator ion here, just floats around in the reaction mixture. And the leaving group is the bromide ion, because it leaves the substrate as the cyanide attaches. With the terminology make sense? Okay, I just spent the better part of ten minutes explaining what the title of chapter six means. But it's worth it if you understand, because you have to have those basic principles down before we can make sense out of it. Let's take a short break. When we come back, we're going to dig more into nucleophilic substitution reactions. If you haven't already picked up your exam and the answer, please feel free to do so that even if we have one more hand out to pick up, so you can pick that up during the break or after class. >> Thank you. Before you. Do you never hear about here they are. I don't think you can be here now. Where do they go? Gate to get really pretty Berg. Okay. Next I'm going to be like, yeah, I forgot. Oh, that's broken. Is that what you're required by law that says they didn't get it. But to just get back to negative, i yeah, I'm just kidding. That was the feeling that I was really pretty good. We're going to actually go out there. And I thought, oh, maybe, what do you think that, okay, as we get back into this, let me just call your attention to the first few sections of chapter six in your book, section 6.1. >> Perhaps the batch organic halides in general shows examples of fluoride, chloride, bromide, iodide, that here. But one thing they point out in this section that I think is worth calling your attention to. And we kinda did so and we do all this stuff add up here for the purposes of the kinds of substitution reactions we're going to be talking about. Chapter six. It's important for the carbon atom that has the halogen attached to it to be sp3 hybridized. Your textbook does show homepage T3, T2. But there are some compounds in which x, the halogen atom attached to a carbon atom with a different hybridization, it's attached to a double bond. It's called a vine electric belive that it's attached to an aromatic ring. It's called a fan or arrow. But the point is the kinds of substitution reactions and for that matter, the kinds of elimination reactions that we're talking about in this chapter don't work especially well for compounds like this. So we will restrict our discussion to compounds like this one in which the halogen or the leaving group is attached to an sp3 hybridized carbon out. Section 6.2 is your introduction to nucleophilic substitution reactions. And much of the terminology that we're just introducing you to. Section 6.3 talks about different kinds of nuclear thoughts. Now, as it says here, a file as a reagent but seeks a positive center. That's what the word nucleo phile means. Nucleus, loving nucleus has a positive charge. So ideally, something with a negative charge, or at least something with a lone pair of electrons, should be looking to form a bond to something that has a positive charge. And it's worth pointing out that this actually takes us back to some of the concepts we talked about way back in chapter one. >> Because when you look at a molecule like this >> What's the most electronegative atom in this molecule, the bromine shirt. So if the bromine has a partial negative charge, which Adam has a partial positive charge. >> Apartment, specifically this carbon. >> It. >> As we suggested before, one big thing that makes 90% of organic reactions work is the attraction between positive charges and negative charges. So when it comes time for the negative sign itis, aesthetically the negative carbon of the negative cyanide to react with something. Hopefully it won't be too surprising to find that it's the partially positive carbon atom of the substrate. >> What else? >> Section 6.4 talks about leaving groups. Now, as it says here, a good leaving group is a substituent that can leave as a relatively stable, weakly basic molecule, or I are. Again, for the most part, we're going to be talking about alkyl halides. And in general it's the halogen atom but does the leaving because chloride ions, bromide ions, iodide ions are reasonably stable, very weak bases, therefore good leaving groups. We will see examples of other leading groups, but for the time being, we're going to focus mostly on Kayla's already brings us to Section 6.5, kinetics of a nucleophilic substitution reaction an S into reaction rate. Let's back up. That's more terminology. And we do bring together a few ideas. >> So let's do that. >> Let me redraw the basics of the reaction again. Now one thing that I hope is clear when you look at this is that in the course of a typical nucleophilic substitution reaction, represented in generic form up here. There's a total of two bond forming and bond breaking processes that happen in the course of this reaction. The bond between substrate and leaving group breaks. The bond between nucleate file and substrate forms. Now one thing we're going to be talking about throughout chapter six is the mechanisms by which nucleophilic substitution reactions take place. And one way of thinking about what a reaction mechanism is, is to describe the order in which bonds form and break. But since we only have two bond forming and bond breaking processes happening in this reaction. There are only three possibilities for the order in which these things formed break. That is, there's only three possible mechanisms by which this reaction can occur. >> Possibility a is to first break the bond between substrate and leaving group, then form the bond between substrate and nuclear possibility the other way around, formed a bond between substrate and nuclear file, then break the bond between substrate. >> And the third possibility is that the two processes occur simultaneously. We form the bond between substrate and nucleate file at the same time as we break the bond between substrate and leave it. Now knowing what we know about organic compounds, which of those three possibilities is really not possible for organic compounds? >> My yeah, because usually it will be a carbon attached to its new. First of all, when you bond exactly. >> If we go back to the picture we had up here a few moments ago, look at isopropyl bromide as our example. If we follow Option B and attach the cyanide and a carbon without anything else happening at what's five bonds on target? Never, never, never. Some of you tried it on the exam, but never? Never. So option B is really not an option for organic compounds, and you cross that out a little bit more vigorously. But the other two options are very possible, and they are known by the terminology S and one for option a and S n2 for option B. And let me show you examples of both. In each case, the S and the N means the same thing. Substitution by nucleate file. What the one and the two refer to are the kinetics of these reactions. The definition of an S and one reaction is a substitution by a nuclear file that follows first-order reaction kinetics. And they give you an example of an SN one reaction. >> And I will talk more about what the kinetics means if we take two CO, two methyl propane or tertiary doodle chloride if you prefer and allow it to react with water. >> The organic product that forms is tert-butyl alcohol. Two methyls who've broken all the byproducts here are H plus Cl minus essentially hydrochloric acid. But here's what we mean by the kinetics. Now, hopefully from your general chemistry class, you remember what kinetics are. It has to do with the rates at which chemical reactions take place. And you should be aware of the fact that any chemical reaction has a rate equation, which includes a rate constant k multiplied by concentration terms for the reactants in the reaction. Now there are ways of doing experiments in the laboratory to determine what the rate equations for reactions look like. If you took your general chemistry classes here at the University of Delaware, you probably did somebody's things in a laboratory method called the method of initial rates for point is we do a method of initial rates study on this reaction. What you find is that its rate equation looks like this. The rate of the reaction is equal to the rate constant k times the concentration of the substrate, C4H9 Cl. But water, even though it's one of the reactants, the reaction does not enter into the rate equation. The order of a reaction is simply the sum of the exponents on the concentration terms in the rate equation. And in this case, there's only one concentration term. And we didn't bother writing and its exponent because its exponent is one. So in this case, the sum of the exponents is one. That's what we mean by first order reaction kinetics terminology makes sense. Okay, back in that reaction a few moments. But it may also give you an example of an S into reaction. And as you've probably already guessed, what this means is substitution by an equally, if I'll following second-order reaction kinetics. And here's an example. Now actually, I'm going to ask you to help me finish that. Here's our substrate over here, simply ethyl bromide or bromo ethane reacting with sodium acetate. Now, how would you draw the structure of the organic product of this reaction? Bearing in mind what we said before about positive and negative charges attract each other. >> Yeah. >> Okay. If you want, sodium bromide is the byproduct, so you can get rid of the sodium and get rid of the Roman. >> And then, and then you would not find exactly. >> Now it turns out it's not quite that way, but conceptually it, it helps you to think about it that way. That's perfectly fine. Like we said before, Bromine is the most electronegative atom in this molecule, which means the carbon atom attached to it has a partial positive charge. I hope it's obvious that the most negative Adam and the acetate ion is the oxygen with a full negative charge on it. So what's going to happen here is we redraw our substrate except without believing root, because the leaving group leaves. In place of the leaving group, we attach the negative oxygen atom through the carbon atom that add the positive charge, well at the bromine there. And then we just connect up everything else. You're right. Sodium bromide is the byproduct, but this is the major organic product. And just for the heck of it, what functional group is this molecule represent? Yeah, this is an ester. Now one thing that I hope is starting to become obvious here. There's a broad spectrum of different kinds of functional groups that can be made by these reactions. We've already shown how you can make. Wow, just page back here for a moment. We've shown how you can make a nitrile this way. It is SN, one reaction we showed you a few moments ago. >> And a functional group is represented by the product and alcohol. And in this reaction, we just made an ester. There are plenty of other possibilities. One of the things that hopefully you will get out of Chapter six is an understanding of how it's possible to create all kinds of different functional groups from alkyl halides, depending on what particular nuclei you happen to use it. But back to the meaning of S into when I see this reaction takes place via second order kinetics. What I mean is if you go into the laboratory and conduct a method of initial rates study on this reaction. The rate equation looks like this. It turns out that the rate of this reaction is dependent on the concentrations of both reactants. And each of these concentration terms is raised to the first power concentration of ethyl bromide, to the first power concentration of acetate to the first power. The point is the order of the overall reaction is the sum of the exponents in the rate equation, one plus one is two. >> So second-order reaction >> Still making sense, at least in terms of what Modernists into me. All right, now the obvious question that results from all of this is, why should this reaction of a second order kinetics when the one we were looking at a few moments ago obeys first-order kinetics. Well, the simple answer to that question is they go by way of two different mechanisms that we've kind of already told you what those mechanisms are. >> But let's see how this plays out for these specific reactions. >> Given that this is an SN one reaction down here, and given what we told you before about what that means, What's the first step in the mechanism of this reaction going to look like the first step in and SN, one reaction mechanism. And break the bond between substrate and leaving group. And in this case, the leading group is the chloride ions. So the first thing we're going to do is get rid of the chloride I. And again, we're going to use these cute little curved arrows to indicate how electrons move. But the point is we break that bond and move the electrons towards the chloride. >> I look towards the chlorine atom to make the chloride ion out of it. >> What we leave behind looks like this. We've seen this kind of thing before. What's it called? When there's a full positive charge on the carbon atom. >> Carbo. >> Can I be more specific? >> Is this eight primary, secondary, or tertiary carbo, tertiary because the tertiary carbon atom as the positive charge. So this is a tertiary carbon cat eye. And when shouldn't come as a real big surprise because as the original substrate, a primary, secondary, or tertiary alkyl highlight tertiary, Of course. So step one, get rid of the leaving group. >> Let's step to yeah. >> Okay. Specifically between what and what I tried. Okay, and let me just back up a few moments to general chemistry. Alright, we have water here. Now, to some small extent, water molecules do dissociate into hydronium ions and hydroxide ions. But assuming this is pure neutral water, What's the concentration of hydroxide ions? >> Remember, what's the pH of neutral water? >> Seven. So what's the p o h of neutral water? Also 70 because pH and pOH, I have to add up to 40. >> But what does that say about the actual concentrations of either the hydronium ion or the hydroxide ion in neutral pure water. >> Yeah. >> Ten to the negative seven. That a large number or a small number? That's a very small number. The only problem with attaching a hydroxide ion is you don't have very many hydroxide ions to begin with. What you have a lot more of is intact water molecules. >> The next step is indeed to attach the nucleoside. >> But what the nuclear file is, is a molecule of water. It doesn't have to have a full negative charge because it has a partial negative charge, which it does because the oxygen is the most electronegative atom, and that's all it really needs to be able to attach itself to the fully positive carbon atom. And the arrow shows that we share a lone pair of electrons to form a new bond to that positive Portman. And let's stop and draw a picture of what that looks like. And we've seen things that look like this. What's it called when there's three bonds and a positive charge on the oxygen. >> Remember? >> Yeah, and OK, sodium ion. Now those are fundamentally the two main steps of any SN one reaction. >> Step one, lose the leading group, generate a carpet canine step to attach the nucleotide. >> But in this particular case, we're not done yet because we don't quite have our final product. What's the last step? Ok? In other words, to arrive at, this is our final product. We have to break an OH, bond. So we break that. By now you're probably correct in the sense that the proton doesn't simply fall off. Something comes along and removed. And if you wanted to indicate that by having another water molecule come along and share one of its pairs of electrons and take that off and generate a hydronium ion. >> And the process that's fine. >> But as far as I'm concerned, since really all we're trying to do with a mechanism to show where the final product comes from. Just showing this arrow right here would be fine. In the last step, we lose a proton. H plus means we get rid of an H plus ion, which again, doesn't just disappear into thin air. >> Something comes along and pulls it off. >> But the point is, when we're done, we have our neutral alcohol final product makes sense with the clear what each of those pictures is trying to show. >> Is it clear what each of those little curly red arrows is trying to indicate is clearly on a question. >> All right. >> Now let's think about something else for a moment. >> This particular mechanism is a three-step process, and I will label the three steps, step one, step two, step three. Which of those strikes you is most likely to be the slowest overall step in the mechanism. >> Why like, like, like you're on the right track. >> It is indeed the case that step one is the slowest overall step, the others relatively fast. And there's a number of different ways of trying to explain why. One is you're gonna correct. >> That's kind of the logical one. >> Step one is done. You have a positive ion and a negative ion. A positive ion and a negative ion anywhere near each other, what do they tend to do? >> Attract? >> Yeah, if anything, these two should be getting together and forming a bond, not breaking a bond. That's kind of the opposite of what you would expect. Plus it could be argued that you have this perfectly stable, reasonable organic molecule just sitting here and all of a sudden the bond breaks. Why does that happen? >> We'll talk more later about why it happens, but you could argue that that's just not something you would normally expect to covalently bonded molecule to do. >> And here's another way of thinking about it. Obviously in step one, we're breaking a bond. Obviously in step two, we're forming a bond in general, is bond breaking exothermic or endothermic? Somebody said something or you were vocal for the first two weeks and I give you a one exam back and really shuts up 50-50 chance exothermic or endothermic? >> Sorry. Tale. >> Yeah, that's what he said. Now in general, bond breaking is endothermic. If you want to break a bond, you'd have to put energy in. In general, if you want to form a bond, especially if you're doing something logical, like bringing together a negative atom and a positive atom that tends to be an exothermic process. Now, which of those tends to happen faster? >> Exothermic or endothermic processes, exoplanets. >> Yeah, so that's another reason why step one is probably the slow step. It costs energy to break a bond. You get energy out what a bond force in general, exothermic process is like, step to happen faster than endothermic processes like step one. So for any of a number of reasons, you can pick whichever of those explanations you like. Hopefully it's reasonable to think, but step one is the slow step and the overall brightness. Now again, from your discussion of kinetics in general chemistry, what's another word for the slowest overall step in a reaction mechanism. >> Yeah, the rate-determining step >> There are many analogies for this kind of situation. Let me just show you the one that appears in your textbook. >> Find it. >> There we go. Page 247. Let me zoom in on this because here's one for those of you who might be fans of days of our lives, like send through an hourglass. Only trouble is, this particular hourglass has several chambers. The one up top relatively skinny, the other two down here somewhat larger. What determines how fast the sand gets from the top to the bottom? The answer is how fast it makes it through the narrow neck. What happens after that is completely irrelevant. So the slow step is the rate-determining step. And what happens after that makes no difference. That's, that makes sense. >> All right, now let's apply that to the reaction we were looking at. >> If this step is the slow step and therefore the rate determining step than anything that happens after that step, has no impact whatsoever on the rate of the reaction. >> Where does the water get involved in step two? >> That comes after the rate-determining step. The reason you don't see a term in the rate equation for water. And in general, the reason you don't see a term for the nuclear file and the rate equation for any asset. One reaction is a nuclear file, doesn't get involved until after the rate-determining step is already over make sense? So the point is, if this mechanism is what's going on at the first step is the slow step, which for us, and one reaction usually this, that it makes sense that the rate equation should look the way it does. Now let's go back to our S into reaction and remember what we said before was the basic deal about what's going on mechanistically furnace into reaction. That is, things are happening simultaneously. In other words, once again, we have to bond forming and bond breaking process is involved. Here we break the carbon bromine bond. We form the bond between the oxygen and the carbon. But those two things happen at the same time, which is to say that an S into reaction is pretty much a one-step process. Now for a one step process, you can't draw out step one, step two, step three, because it's a one-step process, the best we can do is show you a picture what the transition state looks like. So here's my best shot at showing you what the transition state looks like for this particular reaction. Now, let me just take a moment and find the corresponding pictures in your textbook. >> I'm sure there are soon. >> Now here's something similar on page 239 for a different reaction, but you can see what the transition state looks like right here in the middle of the page. Now, just for purposes of trying to be clear about what the artwork is trying to show. >> Oops, that's not the bond that's breaking the bond, coordinate bond that's breaking is over, as you said a few moments ago, that S into reaction. >> We formed a bond between substrate and nuclear file at the same time as we break the bond between substrate leaving group. So if the transition state that we've drawn here, these two horizontal dotted lines represent bonds that are in the process of forming and breaking. We are forming this BOD at the same time as we're breaking, despite not to be confused with this dotted line right here, which indicates that carbon-hydrogen bond is away from you. The wage, of course, means that carbon-hydrogen bond is out toward you. So we throw things together, form this bond, break this bond splat product. >> Done one step. >> And of course that one step involves getting the substrate and the file to bump into each other and actually start this process going. Which is why the rate equation for this reaction depends on the concentration of both, because you have to have both colliding with each other in order for this to work. And so are the reason we see two different rate equations. And then we have two different kinds of mechanism. That's, it makes sense. Any questions about any of my lousy pictures is trying to show. Again, if you want better pictures of all this, check out the examples in your textbook. >> But this gives you the basic idea. >> Alright, so everybody reasonably comfortable at this point, although we mean by SNP1 and SNP2 reactions. >> Alright? Now the point is one thing you'd like to be able to do is take a look at a reaction. >> Without going into the laboratory and conducting method of initial rates studies, which is the only foolproof way to do it, be able to make some sort of a prediction as to whether a particular substrate nucleate file combination is going to do S one or S into. And there are a number of factors that come into play here, will be discussing these in more detail as we go, but let me just call your attention to a couple of them right off the bat that can help you figure out what's going on. Certainly one thing you should look at is the structure of the substrate. And especially what you should do is decide whether your alcohol, hey, why is a primary, secondary, or tertiary alkyl halo that turns out to make a big difference. It turns out that in general for SN one reactions, tertiary substrates do these reactions faster than secondary substrates due them, which in turn do these reactions a lot faster than primary substrates through, which is part of the reason why when I showed you innocent one reaction a few moments ago, I chose a tertiary alkyl. Hela is my substrate. Tertiary alcohol i's are the best that we'll be able to do. An s And one final branch. Now there's a reason why this is true, and it goes back to the concept of stability. Let me just see if they show you this in your textbook, and they do a lot of page 249. >> Let me zoom in on this because this is pretty important. >> Section 6.11 b, thoughts about the relative stabilities of parvo catalogs. And suffice it to say, this, tertiary carbo cat ions are in general more stable than secondary ions, which are in general more stable than primary garb. Again, ions which are generally more stable than the metal guard. But now you have to remember that what we say more stable then we just mean lower in potential energy. Carb a cat ions are not stable. You can't put one in a bottle and come back in a few minutes or even a few milliseconds and expect it to still be there. It's a reactive intermediate. Here's what we mean by this. If you look at some of the potential energy diagrams in your textbook, and for that matter, I'll just show you page 248 when I say potential energy diagrams of looking at things that look something like this. Well, the point is, if you start with your substrate and then lose your leaving group to generate a carpet. Canine. Canine is going to be higher in potential energy. So it's going to be up here somewhere. But the point is, when we say that a tertiary carbocation is more stable than secondary, is more stable than primary. We mean that a tertiary carbo cad ion might be about here for its potential energy as opposed to a secondary, which might be that there, as opposed to a primary which might be up here some place. In other words, relatively speaking, it's easier to form a tertiary carbon paradigm than it is to form a secondary or primary simply because its potential energy is lower. The energy of activation, the energy hill that you have to climb to get there is somewhat lower. Bottom line is if your substrate is tertiary, chances a reasonably good that you can form a reasonably stable Carver CAD I am, and you lose the leaving group. >> So that's why tertiary substrates tend to do AES and one 3x or S into reaction is the order of reactivity is exactly the opposite. >> Metal substrates react very rapidly by means of S into reactions. So do primary substrates. Secondaries Do them but somewhat more slowly. And it's very hard to get a tertiary substrate to do an S into reaction. But the reason why has nothing to do with carve a cat ions because if you look up here at the US into the reaction mechanism, there are no carb account, I think turns out to be true for complete other reason. And you see if I can find a good picture your textbook that might explain the reason why I was looking for a good orbital diagram. Maybe this'll do. >> Looking at the bottom of page 238 for a moment, let's specifically this picture. >> Now, what this is trying to show is, here's the incoming nuclear file approaching the carbon atom, where the leaving group is leaving group over here someplace. And the point is, as we form a bond, these orbitals overlap. As we break this bond, these orbitals cease to overlap. But the point is, for these two things to happen simultaneously, the geometry of all of this must be pretty much as shown. And that was pretty much what I was trying to show, what I drew the picture, the bonds forming and breaking simultaneously in the transition state. Specifically the angle between the bond that's forming, the bond that's breaking in the S N transition state is a 180 degrees. More specifically, what that means is the incoming nucleate file must approach from the opposite side for which the leaving group leaves. Now let's think about what that means for a moment. Here's your nucleate file rear. Here's your substrate with your leaving group on the opposite side getting ready to leave. But remember, this is an sp3 hybridized carbon atom. Tetrahedral geometry, 109 degree bond angles, looks something like this. >> Now what the nuclear file is trying to do is get in here to the carbon out of the substrate. >> For the mock, the question is, how easy is that going to be? Well, if all we have stick an ad here from the direction that I was approaching these little dinky hydrogen atoms. >> Let me pretty easy. >> This hydrogen atoms are very small. And even when we make one of those into a methyl group, in fact, that's pretty much what we have up here. This is now F0 x. Still not a big deal for that squeeze in between those two will Nikki hydrogen atoms there. But if two of those things are methyl groups, secondary substrate starting to get a little crowded back there. If all three of them are methyl groups. Tertiary substrate is a fairly dense thicket of hydrogen atoms through which the nucleate file has to wait before it can make its way into the carbon atom or the leaving group is in fact, let me just draw this out in full detail for a moment. When I say full detail, I'm going to pencil in all the carbon-hydrogen bonds and don't take my word for any of this. Go home and build a model and convince yourself that these things are true. But the point is, if you're some nuclei if operand to come in from this direction, chances are pretty good you're going to start bumping into somebody's hydrogen atoms back here before you ever make it to that carbon out. We were talking before about different kinds of molecular strain. What's the strained involved when atoms start bumping into each other, scare x-ray, this phenomenon of the nuclear file having trouble making it into the reactive carbon on a tertiary substrate because it starts pumping into the hydrogen atoms that are in the way, is called steric hindrance to approach by the nuclear fire. And that's why tertiary substrates don't do us into reactions very well, because there's a lot of steric hindrance to approach by the nuclear bomb. By contrast, for methyl or primary substrates, there's relatively little steric hindrance to approach by the birthday. So the moral of the story is, if the substrate is tertiary, chances are pretty good. It's going to do. And SN one reaction If the substrate is methyl or primary, chances are pretty good it's going to do an SIMD react. If the substrate is secondary, it's kind of on the borderline, might be less than one. Make us into there are other factors involved that we'll say more about as time goes on for now. Is everybody reasonably comfortable with this much? That's part of the reason why when I wanted to show you an S into example over ER, I chose a primary substrate because primary substrates us into Reactions. The last thing I'll talk about for today is how stereochemistry influences an S. And one. Or in other words, same question as before. How can you tell whether the reaction is going to be S one or S into? Well, another factor that comes into play when there's a stereo set are involved is what's the stereochemistry of the product compared to the stereochemistry of the starting material. >> For example, Let's look at this compound, primary, secondary, or tertiary here. >> So most likely what S one or S into, based on what we just said, that sin one now we know what's going to happen when it reacts with methanol. Same basic thing has happened in the previous case. We get rid of a leaving group. We attach the oxygen and we wind up losing a proton from the oxygen. >> I'll let you draw out the details of the reaction mechanism again, notice that as I draw this product, and by the way, what functional group does this represent? This is an ether, so yet another functional group that can be made by these kinds of reactions. >> But you'll notice I haven't draw any wedges are dotted lines over here because I'm being deliberately vague about the stereochemistry. >> But I want you to do is think about what happens during an S and one reaction and tell me, well, first of all, is the study material R or S as drawn our product would you expect the product to be are S mixture the two. >> What do you think? >> Yep, you're half right? So half of its are the other hat is f. What do we call a 50-50 mixture of R and S, a receipt mixture. And if you think about what's going on mechanistically and an SN, one reaction, this shouldn't be a big surprise. First thing that happens, we lose the leaving group and generate a carbo cation, tertiary carbon cat ion, of course. But here's the thing. What's the hybridization of the positively charged carbon atom in the tertiary garb, Again, I as P2, which suggests what geometry? Trigonal, planar or in other words, here's another way we can draw this. By definition, if that carbo cation is SP2, hybridized carbon atom, I should say that means it's a leftover P orbital with a lobes sticking above the plane and a lobes sticking below the plane occupied by the three alkyl groups. Now from here, we're going to bring in our nuclear file, which is the methanol and ethanol course is going to be for Nevada that carbon atom. But it has two options. >> We can either approach from on top or can approach from underneath. >> And if you work out the details, you will find that approaching from on top gives, you are approaching from underneath gives us. But the point is there is absolutely nothing to favor either of those pathways over the other. Which means 50% of the time you get are 50% of the time you get S net result, we get our CMYK mixture. And the fancy word that is used to describe the stereochemical outcome of an SN. One reaction It's to say that proceeds with randomization, which simply means that if you start with a chiral substrate, you get either CMYK product. One thing we said earlier today is that optically active starting materials may or may not give you optically active products. >> Starting material optically and at least be optically active, pure art about the product optically enact concept makes sense. >> Alright, that fluoresce in one. We are down to about the last three minutes. I'm just going to introduce the f n2 problem. In fact, now actually I take that back and I think that we will see more about this tomorrow. But let me just call your attention to one thing. You haven't already picked one up. We have a handout up here that looks like this. And a large part of what we've had to say so far that S N1 and N2 reactions is summarized in this table that appears at the bottom of that handout will do more with all of this. >> That would be the outlook. Why don't I just jot it down? >> This is to watch the video. >> Yeah. Well, actually we're really close to me is that I put F. Ok. What do I do?
Lecture from Jun 21, 2010
From Dana Chatellier March 03, 2020
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