Nucleophilic Substitution

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 orgo one. So let's get started.
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. What we said was that basically electrons would always travel from regions of high density to regions of low density. I've been saying that pretty much all semester, but it holds true again for substitution.
What that means is that nucleophiles or things with negative charges are going to be attacking electrophiles. There's actually a lot of different ways that this can happen. Like I said earlier, a lot of organic chemistry just 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. 

So let's talk about one of the simplest ways that they could react which would be to react in an acid/base with a Bronsted-Lowry reaction. So when a nucleophile and electrophile react together to exchange a proton, we call that a Bronsted-Lowry reaction.
So this is an example that we used earlier from 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 I'd 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. So let's go ahead and say this is my electrophile here.
Now in this reaction I'd have to figure out now I know my nucleophile and my electrophile, where does the 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 was 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 on 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. 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 that's the same thing as me saying that I have an acidic hydrogen. Why is that acidic? Because it's going to be easily donated because of the fact that it already has a partial positive charge, so it's looking for something negative that can attack it.
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. And you guys could have predicted that's what's going to happen because I just told you we're doing a Bronsted-Lowry reaction, so this is kind of a recap of the acid/base chapter. So I go ahead and I grab that H.
Now, what's the next question that I ask myself? Do you guys remember? Well, the next question is always are we done with the mechanism or do we need to keep drawing. Do you think we're doing with the mechanism just the way it's drawn? No, we're actually not. The reason is because remember that that arrow that 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 what 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 in order 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 on to the sulfur and since this is Bronsted-Lowry I would draw equilibrium arrows and what I would wind up getting is 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 8. 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 pair 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 oxygen wants to have 6 electrons and right now it does have 6. But the sulfur should have a negative charge because the sulfur wants to have 6 as well. It's actually in the same column as oxygen, but it has 7 electrons.
So notice that this is a very predictable Bronsted-Lowry reaction because what I'm doing is I'm reacting a nucleophile and 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 base. This is really just a recap. Is that cool?

So now what I want to show you guys is how does this relate to substitution. Well, remember that we also had the Lewis acid and base definition and that didn't have to do with protons. 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. You still react.
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, which one would be the Lewis base. Do you remember?
Well, remember that Lewis acid, so I'm just going to write here, 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. Remember that a Lewis base is synonymous with nucleophile. It means that it's good at giving away electrons.
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 electrophiles, of electrons, not electrophiles, of electrons because they have these two free electrons in the pi bond. So I know that I'm going to start from there.
On top of that, is boron a good electrophile? Actually, yeah. Remember that boron and aluminum were two special atoms that I keep pointing out that happen 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're hearing me say that, that's okay. That just means that maybe you didn't get to watch the old reviews. 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. 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 can go from one place to another and then it could go back. But with Lewis acid/Lewis base there's no exchange
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. That's cyclobutane. Now you guys know how to name that. And I would get now BH3. Remember that basically ever arrow always has to turn into a sigma bond or a single bond.
Now I have that and I just have to figure out the formal charges. Are there any formal charges here that I have to worry about. 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 two atoms, not just one. So this top carbon would have had a hydrogen. This bottom carbon would have also had a hydrogen. Why? Because that carbon wants to have four bonds, so obviously according to bondline they need one hydrogen each.
After the reaction, does that change? Absolutely not, they still have one hydrogen each, there and there. The only difference is that now one of the carbons is happy. Its octet is filled because it has four bonds. The other one is not happy because it only has three.
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 three; boron now has four. Easy.
Now we're going to leave this right here. Later on, in future chapters, we're actually going to continue. I'm just going to put a question mark because we don't know what that is yet. 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. 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. So once again you're like, “Okay, Johnny, I get it. What does this have to do with substitution.” Finally, let's get to it. 

Alright so a substitution reaction is simply a reaction that takes place when you have a nucleophile not purple, a nucleophile and electrophile like before, Ok? Same thing the only difference is that it does not have an empty orbital, OK? If it does not have an empty orbital like that presents a problem to us, OK? What that problem is that before when I made that 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 it because you're not by leaving any octets, OK? That Boron was ready to accept some electrons it didn't need to break anything but now for 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 that in order to make a bond I'm going to have to break a bond, OK? 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 and the leaving group is going to simply be the thing that always leaves after the reaction is over, OK? Now we have dealt with leaving groups before the name that we used to use for the leaving group was just the conjugate base, OK? 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 Bronsted acid.... Bronsted Lowry acid base because we're not going to be exchanging hydrogens, OK? But we still are going to the general principle of something that needs to leave at the end, alright? So I know I've been doing a lot of talking let me show you a substitution reaction in action, OK? That rhymed wow.