Nucleophilic Substitution - Online Tutor, Practice Problems & Exam Prep

Remembering general patterns of reactions.

Hey guys. So I just want to kick off this topic by describing substitution in the most general terms possible and also by relating it to reactions that we've already learned previously in Organic Chemistry 1. So let's get started. All right. So previously, when we were talking about acids and bases, we talked about how electrons would always move in a very predictable fashion. They would always move from one thing to the other. Okay? And what we said was that basically electrons would always travel from regions of high density to regions of low density. Okay? I've been saying that pretty much all semester, but it holds true again for substitution. Okay? So what that means is that nucleophiles or things with negative charges are going to be attacking electrophiles. All right? So there's actually a lot of different ways that this can happen. Like I've said earlier, a lot of organic chemistry can be broken down just into nucleophiles and electrophiles, but the exact way that they react together is going to be what we actually name as a reaction. Okay?

Nucleophiles and Electrophiles can react in Bronsted-Lowry Reactions.

Let's talk about one of the simplest ways they could react, which would be to react in an acid-base with a Brønsted-Lowry reaction. So when a nucleophile and electrophile react together to exchange a proton, we call that a Brønsted-Lowry reaction. This is an example that we used earlier for acids and bases. You could see here I have a negative charge and I have a neutral substance. If I were to figure out which one is the nucleophile, what would you say that is? Well, the nucleophile was always the thing that was good at donating electrons, so let's say that would be this thing right here. That species is my nucleophile, which means that invariably the other thing needs to be my electrophile. Even if it's difficult for me to see how it's electrophilic, but it must be because the other thing is a better nucleophile. Let's go ahead and say this is my electrophile here.

Now in this reaction, I would have to figure out, okay, now I know my nucleophile and my electrophile, where does that arrow start from? Remember that with mechanisms, we always start from the nucleophile, so I would know that I need to draw an arrow starting from this negative charge. Now the question is where does it go to? To figure out which atom it's going to go to because there's no positive charge directly drawn. If there were a positive charge, then I would just go ahead and attack that. But the electrophile doesn't have a positive, so I'm going to have to use dipoles to figure out what's the most positive atom in this molecule. So I would say I've got a few different bonds. I've got a carbon-sulfur bond. I've also got a hydrogen-sulfur bond. Which direction would those dipoles go? They would both go towards the sulfur. So what that means is that eventually, my sulfur would have a partial negative and both of these would have partial positives. Now notice that I have a positive on a hydrogen. That means that it's the same thing as me saying I have an acidic hydrogen. Why is that acidic? Because it's going to be easily donated because it already has a partial positive charge, so it's looking for something negative that can attack it. So to finish off this arrow, since I have an acidic hydrogen, that's going to be my electrophile right away. I'm going to go ahead and attack the H. You guys could have predicted that's what's going to happen because I just told you we're doing a Brønsted-Lowry reaction.

So I go ahead and grab that H. What's the next question that I ask myself? Well, the next question is always are we done with the mechanism or do we need to keep drawing? Do you think that we're done with the mechanism just the way it's drawn? No, we're actually not. And the reason is that remember that arrow I just drew represents the sharing of two new electrons. So this is two electrons that are now going to be attached to that H. How many electrons does the H want to have? In total, it only wants to have two electrons. So now it would have four electrons if I donated this new lone pair. So that means is that we're going to follow that predictable rule, which is that if I make a bond, I have to break a bond to preserve the octet of the hydrogen. So obviously, the only thing that I can break is this bond to the sulfur, So I would go ahead and dump the electrons onto the sulfur and since this is a Brønsted-Lowry reaction, I would draw equilibrium arrows, and what I would wind up getting is that now I have my O that now has a new single bond, and that new single bond is to an H because it pulled that H off of the acid. Then I also have to draw my conjugate base. Remember that? And my conjugate base would just be the thing that now it doesn't have a hydrogen anymore, so I would draw my ring structure and then I would draw an S and then I would ask myself how many electrons did the S have before? It had eight. It had let me just draw them in. It had a lone pair here and a lone pair there. So this S would still have those blue electrons from before. But now it's going to have one extra lone paired that came from the breaking bond. So I'm going to go ahead and add that lone pair here and then I would use the formal charge rules to figure out what kind of charges these should have. So the oxygen should be neutral because the oxygen wants to have six electrons, and right now it does have six. But the sulfur should have a negative charge because the sulfur wants to have six as well. It's actually in the same column as oxygen, but it has seven electrons. Notice that this is a very predictable Brønsted-Lowry reaction because what I'm doing is I'm reacting a nucleophile-electrophile, and what I'm getting is an exchange of a hydrogen and the exchange of a lone pair. That's easy. This is what we've already done in acids and bases. Is that cool?

Nucleophiles and Electrophiles can react in Lewis Acid-Base Reactions.

Now what I want to show you guys is how this relates to substitution. Well, remember that we also had the Lewis acid and base definition. That didn't have to do with protons. And what that means is that sometimes you're going to have electrophiles and nucleophiles that want to react together, but there's no acidic hydrogens that they can react with. Does that mean that you give up? No. Okay. You still react and this is an example that I also used when we were talking about acids and bases. We said like in this compound, which one would be the Lewis acid and which one would be the Lewis base? Do you remember? Well, remember that Lewis acid, so I'm just going to write here, is actually the definition of a Lewis acid is that it's an electrophile. Those two words are actually synonymous with each other. Lewis acid means it's a good electron acceptor. Okay? Remember that a Lewis base is synonymous with nucleophile. It means that it's good at giving away electrons. Okay? So in this case, which one would be good at giving away electrons? The double bond. I've mentioned this several times during the span of this course, but I keep saying double bonds are really good sources of electrons. Not electrophiles. Of electrons because they have these 2 free electrons in the pi bond. Okay? So I know that I'm going to start from there. Okay? On top of that, is boron a good electrophile? Actually, yes. Remember that boron and aluminum were 2 special atoms that I keep pointing out that happened to have an incomplete octet or basically they don't have 8 electrons, they only have 6 and they have an empty p orbital. Now if this is the first time that you are hearing me say that, that's okay. That just means that maybe you didn't get to watch the old reviews. Okay? But from now on, this is going to be a very important fact for the rest of Orgo. You need to remember that aluminum and boron are very good at accepting electrons because they just have this empty p orbital that's just waiting to have some electrons in it. So I'm going to go ahead and draw the rest of my mechanism. My electrons would go straight into that orbital. Okay?

So my end products here, when it's a Lewis acid Lewis base, we actually don't use the equilibrium arrows. We use just a forward arrow. The reason is because what we're going to get is a new covalent bond without the exchange of hydrogens. When you have an exchange of hydrogens, you use an equilibrium sign because the hydrogen could go from one place to another and then it could go back. But with Lewis acid, Lewis base, there's no exchange. Okay? So as you can see from my description, I didn't read it, but that's because I wanted to show you guys. When a nucleophile and electrophile react with an empty orbital, that's called a Lewis acid Lewis base. So this is what we were used to doing in the acid-base chapter. When we had an empty orbital, we would just draw this and what I would get now is a square again. Okay? A cyclobutane. Now you guys know how to name that. And I would get now BH3. Okay? Remember that basically every arrow always has to turn into a sigma bond. Alright? Or a single bond. So now I have that and I just have to figure the formal charges. Are there any formal charges here that I have to worry about? And yes, there are because let's look at the double bond for a second. Anytime you break a double bond, what that means is that you are going to be removing electrons for 2 atoms, not just 1. So this top carbon would have had a hydrogen. This bottom carbon would have also had a hydrogen. Why? Because remember that carbon wants to have 4 bonds, so obviously according to bond line, they need 1 hydrogen each. After the reaction, does that change? Absolutely not. They still have 1 hydrogen each there and there. The only difference is that now one of the carbons is happy. Its octet is filled because it has 4 bonds. The other one is not happy because it only has 3. Okay? So what are the formal charges that I'm going to have to put here? There's going to be a positive charge here because that carbon is missing electrons. There's going to be a negative charge here because boron wants to have 3 bonds and now has 4. Easy. Now we're going to leave this right here. Later on in future chapters, we're actually going to continue. Okay? I'm just going to put a question mark because we don't know what that is yet. Okay? But in the addition chapter, once we get there Oops. Addition. What we're going to find is that this is the precursor to a very important reaction. Okay? So but we're not there yet. But I just want to show you guys that this is another example of an acid base, but this is the Lewis definition. Okay? So once again, you're like okay Johnny, I get it. What does this have to do with substitution? Okay? Finally, let's get to it.

Nucleophiles and Electrophiles can react in Substitution Reactions.

So a substitution reaction is simply a reaction that takes place when you have a nucleophile and an electrophile, like before. The only difference is that it does not have an empty orbital. If it does not have an empty orbital, that presents a problem. What that problem is is that before, when I made the bond for this Lewis acid-Lewis base, did I have to break a bond? No, I didn't because if you have an empty orbital, that means you can accept electrons and you don't need to break a bond after because you're not violating any octets. That boron was ready to accept some electrons. It didn't need to break anything. But now for a substitution reaction, if you don't have an empty orbital, what that means is that if you attack an atom, that atom is going to have too many bonds because there were no empty orbitals. So what that means is that in order to make a bond, I'm going to have to break a bond. And since you have to break a bond, what that means you always have to break a bond in substitution. What that means is that we're going to get a new type of compound called the leaving group. The leaving group is going to simply be the thing that always leaves after the reaction is over. Now we have dealt with leaving groups before. The name that we used to use for the leaving group was just the conjugate base. The conjugate base, if you think about it, it's always the thing that leaves, but that's when you have an acid-base reaction. In these types of reactions, it's not going to be a Brønsted-Lowry acid because we're not going to be exchanging hydrogens but we still are going to have the general principle of something that needs to leave at the end. All right? So I know I've been doing a lot of talking. Let me show you a substitution reaction in action. That rhymed. Wow.

Predict the product

So let's go for it. Nucleophile, electrophile. Which one's the nucleophile? Obviously, my O negative. So let's do that. Nu negative. Which one's my electrophile? Well, it must obviously be the other thing, but which atom is the electrophile? That's the hard part. So what do you guys think? We know that the arrow is going to start where? It has to start at the O. But then where is it going to go? Is it going to go to a carbon, a iodine, a bond? What do you guys think?

In order to listen, if there was a positive charge present on this molecule, that would be easy. But there's no positive charge, so that means my only choice is to use to draw a dipole. So what I have to do here is I have to draw a dipole and the dipole is going to pull in which direction? Well, all these carbon-hydrogen bonds I don't worry about. They're all the same. Okay? They all have very they're all covalent, but I do have a carbon-iodine and that dipole is going to pull towards the iodine. So what that means is that I'm going to have a partial negative here, a partial positive there. Cool so far?

All right. So that means that where is my arrow going to actually attack? It's going to attack the positive because this is a negatively charged nucleophile, which that's kind of the definition of a nucleophile. So now I'm going to go ahead and attack this carbon. Okay? That's already the first common mistake the students make. They don't draw the dipoles and they end up attacking the iodine instead. That doesn't make sense because the iodine has a negative charge already. Why would a negative want to attack a negative?

So now we come to the part with substitution. Okay? Does that carbon have an empty orbital? No. Actually, this carbon already has 4 bonds. Where are those 4? Well, it has 2 to the ring, has 1 to the iodine and it also has one H that we never drew because it was implied. So now I've got a new bond being made to that carbon. How many bonds will that carbon have after that arrow is formed? 5. Does carbon like to be what's called pentavalent or 5 bonds? Absolutely not. That carbon hates its life right now. So what are we going to do to fix this situation because it sucks?

In order to make this bond, I'm going to have to break a bond. In order to break a bond, that means I have to kick out a leaving group. So in this case, I've got 5 bonds. I can choose to break any one of them, but the one that's easiest to break is the one that already has a dipole because the one that already has a dipole is the one that's already pulling electrons away from it, so the bond that I break is the one to the iodine. Afterwards, what I'm going to get is a forward direction arrow not equilibrium. Why? Because this is not the exchange of a proton. It's literally just a nucleophile reacting with an electrophile.

This is like the Lewis acid Lewis base reaction that we did on top. Now all we have to do is we have to draw our products. Well, what would they look like? I would have this ring still, so let's go ahead and draw that ring. I drew it a little bit different. Still a 5-membered ring, though. And what we would notice is that what's attached to that ring? Well, it still has that H, so that H is still there. It doesn't have the I anymore. So let's not draw the I yet. But what it does have is now it's going to have a single bond and that single bond is now going to be attached to my nucleophile, which is O and then an ethyl group. So there we go.

We have a completely new product. In fact, if you were to look at this functional group, this is an ether. Did I start off with an ether? No. I started off with an alkyl halide and now I have an ether. So you can see already we haven't even gotten to the mechanisms, the full mechanisms yet and you can already see how this would be useful. I can make a completely new functional group out of this.

All right? So I've got that thing connected. Is there anything else that I'm missing? The leaving group. Something left to make this reaction possible and that was my I negative. Why is it an I negative? Because now it's going to have one extra lone pair that it didn't have before. Congrats. You guys just drew your first substitution reaction because you were able to use I know I coached you a lot through it, but I'm just letting you guys know we used all the same principles that we used from acid base and now we applied it to substitution and it works. Okay? And the reason is because all the substitution reaction is is it's a Lewis acid Lewis base, electrophile and nucleophile, but when you don't have an empty p orbital. So that means you have to break, make up on and then break up on. And the one that you break is simply the I.

By the way, how could you tell this with substitution? Because notice that everything traded places. My iodine used to be on the ring, now it's in solution. Okay? My negative used to be on the o, and now it's on the I. Okay? So what I'm trying to say here is that everything switched places. The ring used to have an iodine, now the ring has an o. Okay? The O used to have a negative and now the I has a negative. So like everything's perfectly switching places. Okay? And that's the definition of a substitution. Substitution means you're trading things. All right? So a few more facts and then we'll be done with this video really quick. In a typical acid-base reaction, remember that we would use the stability of the conjugate base to determine if it was favored. Okay? What we would say is we would compare pKa's and we'd say which one's stronger, which one's weaker, all that stuff. Well, it turns out that because the conjugate gets a new name in these reactions, we call it a leaving group. We're just going to use the same principle to figure out the reaction rate. We're going to say that the strength or the stability of the leaving group is going to tell us if this is supposed to be a fast reaction or a slow reaction. If it's going to be favored to happen quickly or not favored to happen quickly. All right?