Time to start learning about some of the most important reactions in all of organic chemistry. In fact, you’re never allowed to forget this one!
1
concept
Drawing the SN2 Mechanism
8m
Play a video:
Was this helpful?
Alright, guys, Now we're gonna jump into one of the most important mechanisms in all of organic chemistry, and it's a mechanism that you're never allowed to forget. So when I teach you in the next twenty minutes, like, it's gonna stick because you're gonna need it for your obviously for or go for your graduate exams. And even in graduate school, if you're planning going to graduate school for anything pre health, you're still gonna need to know this reaction. And that's called the S and two mechanism. So let's dive right into it. All right? So what if I were Just give it a tag line and just say in one sentence, What s and two mechanism is Okay. What it is, is that a negatively charged nuclear file? Hopefully, all of that is words that you should be comfortable with. Negatively charged nuclear file reacts with an accessible leaving group. Okay, now, leaving group, you should know what it is accessible. Maybe a little confused, but will define Define it. Okay. To produce substitution. You know what that is in one step. All right, so let's go ahead. I just wanted get right into it. Let's just draw this mechanism out. All right, So I have this nuclear file that I'm just generally putting his n you negative. There's a lot of different nuclear files out there. It doesn't really matter the identity right now. Okay. Not to figure out I'm reacting with an alcohol. Hey, like, what did I say? Alcoholics. We're good at leaving. Okay, So what that means is, I have to figure out what's the Electra Filic part of this molecule? Because this is my Electra file. And what is it gonna look like after it reacts? Okay, So how do I find out which parts Electra, Philip, Does it have a positive charge already on it? No. So I'm gonna have to draw the dye poll. What does the dipole look like? Well, remember that halogen is pretty much always pull away from whatever they're attached to. So I would have My only major die poll is pulling away from the carbon. So what I would have is a negative here. A partial positive there. Where is my nuclear file? Gonna want to attack? It's gonna want attack the carbon. Okay, So the Electra Philip part is not the X. It's the carbon. Okay, so I know I'm going to start off my arrow from my nuclear file, and I know I'm going to attack that carbon. But now, actually, we have a choice, okay? Because what we have is a distinct set of sides. Okay, so let's think about it this way. This is my carbon. And in the past, I haven't really worried about exactly how I draw my arrows because I haven't been very picky. Okay, But if you think about it, there's actually two different sides. This carbon. Let's say that the X side, the one that the halogen is called the front side. Okay, so the X has three lone pairs, one to three. Okay, so that would be what I would consider the front side. So I'm just gonna right here. That's the front, okay. And the back side would be everything that's on the other side over here and the backside. What it's gonna have is just like a hydrogen, and then some outfield groups. This would be a metal group in an ethnic group. Okay, which of these two sides? Front or back? Do you think is gonna be the easiest for my nuclear file to approach. Okay? And let's think about it this way. We know that it wants to hit the carbon, so no matter what, it's going for the carbon. But what? All I'm asking is is it gonna try to go from the front side of the backside? It turns out that the front side is really bad option. Why? Because the nuclear file, remember, it already has extra electrons. Okay, It's got extra election was trying to get rid of. In order to go through the front side, it would need to pass through a bunch of electron clouds from the halogen. Okay. Do you think that's gonna be very easy to Dio? It's actually gonna be almost impossible. It doesn't happen. Okay, Those electrons are going to repel each other like crazy, so front side attack is actually impossible. It's never gonna happen. Okay, so what that means is that this is gonna lead us to one of the most inappropriate phrases in all of science, and that is backside attack. All right, so as messed up as that sounds, alcohol highlights are totally down with it. All right? Backside attack is something that they're all about, and We're gonna be doing this every day, all right? For the rest of organic chemistry. So I hope you guys were cool with that. You got to get used to it pretty quick. So backside attack is the way to go. Because it's the way that's less basically, less hindered. It's gonna be ah, lot easier for those electrons to pass through the backside. Whether are not as many electrons as the front. Okay, so now what we need to do is we have to draw the transition state of what this is gonna look like. Okay, because we let me just ask you, um, I done with this mechanism. Do we Do we need to draw any more arrows, or, um, I done? No, we should draw some more arrows. Why? Because remember that this is gonna be a nuclear phone Electra file that does not have an empty orbital notice that there's no empty orbital here. This carbon already has four bonds. Okay, so this carbon already has four bonds. If I make a new one, that's five. So if I'm making this bond, I'm gonna have to break upon. And you guys already know what I'm gonna break. I'm gonna break the the alcohol. I mean, the hey, light off. I'm gonna break the halogen off. All right? So that means that I'm making a bond and I'm breaking the bond at the same time This is gonna lead you lead to something called a transition state. Okay, a transition state is just, ah, high energy. Um, phase of the reaction that is very, very short lived. Okay, What it means is that it never even really happens, Or what I'm trying to say is it cannot be isolated. Okay? It's a very high, high energy thing that it must happen because we know that it must go from one state to another. But if I tried to just isolated in the test tube, I would never be able to isolate transition states. Okay, so let's go ahead and draw what it would look like. It would be a carbon, okay. And it would be attached to three things for sure that air just signal single bonds would be attached to a metal group. That's the one in the front. Ah, hydrogen in the back on. Then in Ethel Group, right. And I'll just put the ethnic group facing down because I'm gonna need all this space I can catch. All right, now, this is the interesting part. We just said that I'm making a bond and I'm breaking a bond, and it's all happening in one step. Okay, So that means that I'm gonna have to draw a partial bonds. So that means that my nuclear file is partially making a bond to that carbon. And my hey, lied is partially breaking a bond to that carbon. Okay, on top of that. Now, this carbon has too many bonds. Okay? It has five instead of four. So I'm gonna have to put partial negatives on the's. Adams. Okay. What that means is that remember that Carbon wants to have four. Bonds now has five. So in this transition state, it's extremely unstable. Carbon does not like to have this many bonds. I have to indicate that has one too many by putting negatives that are distributed. Okay, so there we go. That's our transition state. If you ever see this little like double dagger, that means transition state. Okay, Like I said, this is something that it must exist for, like, a nanosecond. But it's not something you could isolate. Okay, after this reaction is done happening after it's all completed, all happens at one time. Now I have to figure out what my products are gonna look like. Well, what I'm gonna have now is that I'm gonna have my nuclear file. Okay? But now my nuclear follow is going to be attached to have a single bond. I'll draw it in blue because that indicates the arrow that was just made to that red carbon. And what is that Red carbon gonna be attached to? The same three things it was attached to before. So it's gonna have that metal group. It's gonna have that hydrogen. And now the Ethel Group, because the nuclear fall came from the back. My ethel group is getting pushed towards the front. Okay, so that's gonna be important. Let's just hold onto that thought. Okay? On top of that, is there anything else that we need to draw are leaving groups or a leaving group is going to get xnegative. Okay, so first of all, how can I tell that a substitution reaction just took place? Well, I can tell because things substituted before I had a carbon with an ex. Now I have a carbon with a nuclear file. Before I had a nuclear file with a negative. Now I have an ex with a negative. See how everything perfectly swapped. So this is definitely substitution, okay?
Summary: A negatively charged nucleophile reacts with an accessible leaving group to produce substitution in one-step.
2
concept
Understanding the properties of SN2.
11m
Play a video:
Was this helpful?
now what I want to do. So that's basically that. Okay, Now what I wanna do is you guys know the mechanism. Now, this is gonna be the mechanism that we use every time. It's very important that you guys understand everything about this mechanism, but on top of that, there's a lot of facts that we need to memorize about it, too, because your professor is gonna want you to know a lot of conceptual stuff about this is well, okay. And in fact, you're really not gonna able to understand it unless you know all the concepts behind it. So now what I wanna do, it's start breaking all these different concepts down. Let's start off with nuclear file. Okay? You could already guess what this is because I already talked about it. But do you think we need a strong nuclear file or a week nuclear file to start off this reaction? And the answer is that I'm going to say whenever something's negatively charged, that is strong, Okay, because that means is that means it is actively trying to get rid of electrons. So in the description, I told you guys, it's always a negatively charged nuclear file. Okay, so that means it's strong. All right, so that means a week nuclear file one that is neutral or doesn't have a negative charge would not be a good candidate for s and two Good. So far. Cool. So then the leaving group is that gonna be better if it's un substituted or if it's highly substituted? Now, just you guys know Substituted has to do with our groups. Okay, That's what substituted means un substituted. Means that you have less. Are groups coming off the alcohol? Hey, lite believing group. Okay, highly substituted means that you have a lot of our groups. Okay, So which one do you think is gonna be better? Remember that the mechanism is backside attack. So which one do you think is gonna be better for backside attack? Having not a lot of stuff on the back side, like just hydrogen is that are really small or having these big, bulky groups. Ah, bunch of them that are gonna take up a lot of space when I word it That way, it sounds like the best option would be unsubs itude ID. And that's exactly right. Okay. For backside attacks be favored. You need to have a lot of room on the back side. All right, Otherwise, it's just gonna be too cramped, and you're not gonna be ableto get it in. All right, so now let's go on to the next one. So let's those air really big points strong and un substituted. Okay, next reaction. Coordinate. If I were to draw an energy diagram of this, would the highest point Would it be a transition state, or would it be an intermediate? Is that which one is the one that I'm passing through to get to my product? And the answer is I told you guys, Transition State, Okay, we're gonna use a transition state because this is all happening at one time. So that means I have one molecule that is kind of in between both sides. OK, and that helps us flow into the next question. Is this a concerted mechanism or is it a two step mechanism now, just you guys know concerted just means one step. Okay? Concerted is actually a word that we can use in English. It just means everything happens at once. Okay, So it's not just a chemical chemistry word. It's actually, just like a normal word. So concerted. One step. Absolutely. I mean, I already told you guys, it's one step, but you could have just looked at my little sentence. But also, that makes sense because there's no distinct first step and second step right? Is there, like, step one? Step two? No, I told you, you're making upon and you're breaking a bond at the same exact time. All right, so that means it's all happening at the same time. Cool. So now we're gonna get into is rate questions. Okay, So all this, this stuff down here has to do with rate and rates. We haven't really talked about too much, so I'm gonna have to explain this. All right, So is the rate gonna be uni molecular or by molecular? A really easy way to remember. This is that s And to the two in s and two stands for by molecular. In fact, what s and two stands for is substitution. Then it says nuclear filic. Why? Because the nuclear follows starting it and then by molecular. Okay, that's what it actually means. That's what s and two means. Okay, so I know that the rate is by molecular. But what the hell does that I have actually mean? Okay, well, what it has to do with is it's basically saying the rate. Remember that rates are always based on. You can trace them back to rate constants. Okay? And you can say, if I increase the concentration of a certain re agent, is that going to affect the rate at which I make my products? Okay. And if it's by molecular, what that means is that there's gonna be not just one species that the rate depends on, but it's gonna be two species, okay? And the best way that I can illustrate this is I always think of an arrow and a target. Okay? And I think that the arrow is the nuclear file, Okay? Because it's the one that's kind of starting this all off. And the target is the electric file or what we call the leaving group. Okay, cool. So I've got my nuclear file. I've got my Electra file. Okay. Also, that just you guys know the leaving group is also my alcohol. Hey, lied. Right? Because I said the alcohol. Hey, loads of the most common type of leaving group. Okay, Doesn't have to be, but in most cases, right. So now here's the question. If I double the amount of arrows that I'm shooting, so instead I've got a bone marrow. And now, instead of shooting one arrow, I've got another guy next to me, and he's shooting another one. Okay. Is that gonna increase the chances that I hit the target or they hit a bull's eye? Hell, yeah. It's gonna increase the chances, Okay? Because now, even if I miss, maybe the other guy hits it. Okay, Now think about it this way. Molecules are not smart. In fact, they're super stupid. In fact, they're the most stupid ever. They don't have brains. I know that's going to come as a shock to you. All they are. All these reactions are dictated by random motion. Okay, so that means I have a test tube and I have a bunch of arrows and I have a bunch of targets. Guess what's happening. They're not aiming for the target. They're randomly colliding. They're hitting the wall, they're hitting each other. Then they hit the front side and they bounce off. Why would they bounce off if they hit the front side because, remember, there's electrons there and the electrons repel so they keep bouncing, bouncing. Eventually, something hits the backside. Boom. That's my reaction. And I go and I make product. All right. If I double the amount of arrows I'm shooting with, I'm doubling the chances that I'm randomly gonna make a product cool so far. All right, Next question. If I were to double the amount of my leaving group of my targets, okay, would that double the amount of chances of or would that increase the chances of getting a bull's eye? What do you think? I could tell you guys were really thinking about this one. Yes, it would. Okay, because let's say I'm only shooting with one arrow, but now there's two targets. Hey, I could be a really bad shot, remember? I'm dumb. Okay, so I missed the first one, but actually hit the second one because, you know, I was really bad, all right? And that's the way it also works. So if I double the concentration of my leaving group, I'm going to double the chances that I get a collision that leads to a backside attack. Alright. How about if I were to double both. So now I have twice as many arrows, and I have twice as many targets. What would that dio? That would wind up quadrupling the rate of my reaction? Because now what would happen is that I have four times the chances of hitting a bullseye. Is that making sense? And what I'm trying to say is hitting the bull's eye is the equivalent of backside attack. It means that you get a product. All right, So what that means is that the rate of my reaction is dependent on both the nuclear file and the leaving group of my Electra followed. By the way, this is the same as Elektra file. It's got a lot of names, but it's the same thing. Okay, so what that means is that this is by molecular, and that means that my rate is going to be equal to the concentration of both my re agents. Not just one, that's what by molecular means. All right. Are you guys cool with that? So we just have to memorize that for now. We're not gonna do a whole lot of calculations, but you do need to be able to answer questions about if I increase the rate. The regent this have regent, would that increase? Would that double with that triple? Whatever. Okay, cool. So then we get until one last type of question, which is stereo chemistry. Now, remember that we had a whole chapter dedicated to Cairo Ality. Alright. The Chire ality here isn't going to get back confusing. But if this is a Cairo center, by the way, is this a Cairo center? The carbon that's red. Yeah, it is. That's actually a Carl center right there. That's got the green carbon Now. I just drew a green has four different groups on it. Okay, so if you have a Cairo center at the beginning, okay, and you do a backside attack, notice that at the end two of my group swapped places. Okay, notice that my H and my methyl group are still in the same place, so they're good. But now my Ethel group is on the right side and my high priority group, my nuclear power, what used to be an ex is on the left side. So what that means is that these switched places Okay, Any time that you switch the orientations of the positions of two groups. Guess what you do. You flip the configuration. So what that means is that originally, if this wasn't our if the first one was in our by the way, I'm not going to calculate Rs here. Okay? Cool with that. I'm just gonna say that. Imagine that. Whatever. Okay, I will calculate it cause it's super easy. It would just be that this is one this is to this is three. So it would be an are. Okay, So if this first one is our afterwards, it's gonna turn into an s. Okay? And if you want to prove that, you could just say Okay, this is one. This is two. This is three. Now it's gonna go this way. Okay, So if you start off with a are you gonna get an s at the end and guess what that's called? It's a huge, huge thing. Very important. That is called inversion of configuration. Okay, so now any time you hear the word backside attack, all right? You're gonna have to get used to the fact that backside attack always means this one thing with Chire ality. It always means inversion of configuration. Okay, now, what would retention? What would these other words mean? Retention just means that you get the same thing, our turns into our afterwards. That's not this reaction. So don't worry about it. Okay? What does receive McMeen? Remember that regime? It means it's a combination or a perfect mixture of both an anti MERS. So regime, it would be that I get 50% of our 50% of s. Is that what's happening here? No, I'm getting 100% of the other and anti. Um er why? Because I'm flipping the orientations of two groups. Is that cool? All right, so one last thing, And then we'll just do a really quick practice problem. The nickname of this reaction is gonna be I've said it 10 times. Backside attack. Okay, Whenever you hear backside attack, you're always thinking s and two. And that just has to come very naturally to you at this point. Okay? And as the semester rolls on, all right, so I want to do a practice problem. I will give you guys a chance to pause the video or Thio answer, and then we'll go into the next video I want you to rank the following alcohol highlights in terms of their reactivity. Okay, So figure out which of these would be the most reactive, which would be the least reactive towards an s and two reaction notice that all of the leaving groups of the same in terms of the the actual strength of the leaving group. So there must be something else that's gonna make it good or bad at reacting. Okay, So go ahead and try to figure that out, and then I'll answer for you Go.
Properties of SN2 reactions:
Nucleophile = Strong
Leaving Group = Unsubstituted
Reaction coordinate = Transition State
Reaction = Concerted
Rate = Bimolecular
Rate = k[Nu][RX]
Stereochemistry = Inversion
Nickname = Back-side attack!
Rank the following alkyl halides in order of reactivity toward SN2 reaction.
3
example
Ranking reactivity toward SN2
4m
Play a video:
Was this helpful?
So would all these have the same exact rate towards an S and two reaction? Or would some of them be faster? Slower? And the way to figure that out is really just Look at the degree of the Al Kyohei lied. So if you'll notice all the way to the left I have a flooring attached to a carbon that's attached to zero other carbons. No other carbons. This is what we would call a zero degree alcohol. Hey, lied or that's also called a metal. Okay, um ethyl alcohol. Hey, light. Alright, then all the way in the opposite end of the spectrum, we've got a carbon attached row flooring. That's three other carbons coming off that. That's actually what we call a tertiary alcohol. Hey, light. All right, So would they all have the same rate? Absolutely not, because one of these is going to be really good at a backside attack, and then some of them are gonna be really bad, OK? And the answer is that my method aqeel highlight is the best, because it's the one that has the most free backside kids. The one that is gonna be easiest to attack as I start putting in our groups. So as I start going from primary to secondary to tertiary, my backside is going to get more and more clogged. So, as you can see, for example, my backside is the same. But now I've got this big group over here. So now if I come in from the right side or from the bottom side, I'm gonna bounce off. Okay? The only way I could hit the back side now is from the top. Okay, so that means that now the rate of destruction is gonna be slower, because the chances of me hitting that spot are gonna be less. Now, when I get to secondary, it gets even worse. Because now I've got this group in the top, this group in the bottom. Really. The only way that I can hit the backside is if I thread the needle perfectly. Okay, so now for a secondary, this is gonna be even harder. It's gonna be way, way less likely that I hit the back side so this won't go even slower then finally, if you look at tertiary, tertiary is just getting freaking impossible. Like I've got the top blocked off. I've got the bottom blocked off, backed off, and then I've even got the back completely, like, backed up. Okay, so basically, this one is impossible to do a backside attack on. And in fact, tertiary is don't even happen. Okay? A tertiary alcohol haloed will never do a backside attack because it's too difficult for the nuclear file to access the backside. All right, so I just wanna make one more point. So we know what the answer is that my method is the best. Okay, now, the other question is, why would this affect the rate? Okay, I told you is that the rate of this of these reactions is K. Okay, the rate constant times the concentration of my nuclear file times the concentration of my Rx or my leaving. Okay, so, um, I changing the concentrations by changing the degrees? No. Okay. Like the nuclear follow. Staying the same. So it's not changing the rate constant staying the same. Okay, the only thing that's changing is that I'm increasing the strength of my leaving group as I go towards the left, because, yeah, the f might be the same Adam in either case, but it's gonna be easier for me to hit that back side for the method. Okay? And remember that the backside is what really determines if destruction happens or not. Okay, so what that means is that effectively, by using a method or primary, I am kind of jacking up the amount of alcohol that I have because I'm jacking up the amount of Okay. Hey, like, that's actually possible to react. Okay, that's gonna be that's gonna allow me toe hit more backsides and get more reaction. Okay, so that's what I'm trying to show you That the rate actually would go up as my alcohol. It becomes more like a metal. All right, Cool. So I hope that made sense. Let's move on to the next video.
4
Problem
Predict the product of the following reaction.
A
B
C
D
The product must contain inversion of configuration if the original leaving group is located on a chiral center.