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Sn2 reactions. Created by Sal Khan.
Video transcript
We're now going to explore one of the most fundamental reactions in organic chemistry. And first, let's look at what are the things that are going to react. So over here, I have a hydroxide anion. It has a negative 1 charge. And that negative 1 charge comes from the fact that oxygen is neutral if has six valence electrons. But this has one, two, three, four, five, six, plus half of this bond with hydrogen. So you could think of that as another electron. Or it's got half of these two, so that's another electron, which gives it a negative 1 charge. Now, this right over here, it has one carbon-- I guess that you could say-- in its backbone. It's a methane. And it has a group right over here consisting of bromine. So this is bromomethane. So let's think a little bit about what might happen under the right conditions. So this hydroxide anion-- we've already gone over the fact-- that you could kind of view it as having an extra electron, which gives us this negative charge. Well, if you have a negative charged anion-- like this-- it will be attracted to positive things. And one example of a positive thing is a nucleus. And so in this reaction, we are going to refer to this hydroxide anion as the nucleophile. Phile literally refers to things that something likes. This thing, it's a nucleus-lover, I guess you could say. And so it might want to share maybe a pair of electrons with a nucleus. Now, over here, I have this bromo group in the bromomethane. And bromine is quite electronegative. It is more electronegative than carbon. So even in this bond, we have two electrons in this bond right over here. Bromine is going to hog the electrons a little bit more than carbon. And also, electronegativity says that, if this bond were to break, it's more likely that this pair-- the pair that's in this bond-- is going to go to the bromine than it's going to go to the carbon. So the way that this reaction works-- and it's really kind of happening in one step. And this is key to the name of this reaction, as we'll see in a second-- is that this nucleophile right over here might say, hey, I got this extra charge. Let me share some of that charge with a nucleus-- in this case, the carbon. Let me draw it like this. So it would share this pair, which would form a bond. Carbon would have half of the pair. Oxygen would have half of the pair. So it would share this pair with carbon. Now, if carbon is now sharing half of this, it's essentially kind of getting an electron. It doesn't need this electron, which bromine is already hogging a little bit. And so bromine would take this entire pair back. Whenever we draw full arrows like this-- just normal arrows that you're used to seeing-- it's referring to the action of pairs. So this whole pair is going to bromine. This pair right over here is now going to be shared with the carbon. It's going to form a bond with the carbon. So when we do both of these steps, after both of these steps, it would look like this. Let me draw my bromine first, which is now going to become a bromide anion. So bromine one, two, three, four, five, six. One, two, three, four, five, six. And now, it gets both of these electrons from this bond. So it has both of those electrons right over there. Now, this will have a negative 1 charge. How do we know that? Well, bromine is neutral, but it has seven valence electrons. And this thing now has eight. So this now has a negative charge. Now, let's think about what happens to all of this business. So let me draw my carbon first. And let me draw this hydrogen that's popping out. So this hydrogen that's popping out, let me draw it like that. So that's the bond. I'm trying to show that's popping out of the page. So that's that hydrogen. I just made them different colors so we can keep track of the different hydrogens. Then the hydrogen that's going into the page, that's that one right over there. And you have this hydrogen that's going up. So that hydrogen right over there. And now, the hydroxide anion has bonded to this carbon. Now it'll be a hydroxyl group. So this thing has now bonded. So let me draw the bond and the color of these two electrons. It has now bonded. So those two electrons now make up this blur, this bond right over here. And now, we have our oxygen but with those two, these two right over here, these two right over here, and then, the bond to the hydrogen. Now, what just happened here? The oxygen is now neutral. Why is it neutral? Well, it has one, two, three, four. And then, it has half of this bond. So you could view that as an electron or the equivalent of electron charge. And then, it has half of this bond. So it has the electron charge equivalent of six valence electrons-- which makes oxygen neutral-- and now bromine is negative. So one way to think about it is that the negative charge has been transferred to the bromine in this reaction. Now, in this reaction, bromine has left. And that's why we call this the leaving group of the reaction. This thing right over here is the thing that is going to leave. That is the leaving group. Now, the other thing that you might find interesting-- it looks like kind of this hydroxide anion. Some people would even say attacked this bromomethane, the bromomethane left. The bromine left-- not the bromomethane left-- that attacked the bromomethane, the bromine left. So in some ways, you would say that there's really two reactants over here. You have the hydroxide anion, and you have the bromide anion. And this carbon right over here with the hydrogens, this was kind of the thing that to some degree facilitated the reaction. This thing bumped into it and attached. And then, this thing the left. And so in this context, we would call the carbon-- with the three hydrogens-- we would call this the substrate. Now, what should we call this reaction? Well, one way to think about it is we substituted this bromo group with a hydroxyl group. So it might make sense to call this a substitution reaction. Substitution has occurred. Now, what did we substitute with? Well, we substituted with a nucleophile. And how many reactants were involved in the rate-determining step? And I know that might not make a lot of sense to you. But this had, essentially, one determining step. This had to want to bump in the right way. And this would want to have to leave right at the same time, or roughly the same time, in order for this reaction to occur. So this reactant and this reactant were involved in figuring out, well, how fast is this reaction going to happen? And we're going to see future reactions where you don't have both reactants involved in the rate-determining step. We'll see only one reactant involved in the rate-determining step. But I'll just write this right over here, 2 reactants in rate-determining step. There's a garbage truck outside. OK. I think the garbage truck has left now. Safe to resume. So as we were saying, there are two reactants involved in the rate-determining step. And so the shorthand for this reaction is to call it an Sn2 reaction. It's substitution with a nucleophile where both reactants are involved in the rate-determining step.