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Steric hindrance and how it affects whether a molecule will participate in an Sn2 reaction. Created by Sal Khan.
Video transcript
We've got four scenarios right over here. In each scenario, we have a hydroxide anion. And then, we have another molecule. And right over here, this molecule, this is a bromomethane. So it's a bromo group attached to a carbon. And this carbon isn't attached to any other carbons. It's only attached to other hydrogens. So carbon like this we call a primer-- actually, not a primer. We call this a methylcarbon because it is attached to no other carbons. Now, let's think about this right over here. What's the difference between this scenario and that one? We still have our hydroxide anion. We still have the bromo group. But now this carbon is attached to another carbon. So this right over here is a primary carbon. It's attached to one other carbon. And I think you see a pattern here. What about this one? Well, this carbon right over here is a secondary carbon. It's attached to 2 other carbons. And then, finally, this carbon right over here is a tertiary carbon. It's attached to 3 other carbons. And as we'll see in a few seconds, the fact that it has these other carbons and that these other carbons kind of might block some or all of the reaction, it will change what actually goes on. So let's go, first, to the first scenario right over here. We have a hydroxide anion. We've already seen that a hydroxide anion is a decent nucleophile. So this right over here is a nucleophile. And bromine is a decent leaving group. And so assuming that it's in a appropriate solution-- we'll cover that in future videos-- you could imagine that one of these, if this hydroxide kind of bumps into this bromomethane in just the right way, that this pair right over here could form a bond with this carbon. And just as that is happening, essentially the carbon is gaining half of a bond, which is kind of the equivalent of a electron charge. So then, this bromine could take away this entire pair. Instead of just kind of sharing it and having the equivalent of 1 electron charge from this pair, it has both of them. And then, it'll become the bromide anion. And we've already seen that before. That is an Sn2 reaction. It's a nucleophilic substitution where both reactants-- so the nucleophile and the leaving group-- are involved in the rate-determining step. And this will actually be quite fast. Right over here. Now, let's look over here. The only difference-- we already said-- is the existence of this CH3. Now, even though we've drawn it, written it down, CH3, when it's written down, doesn't look a lot bigger than an H. Remember, CH3-- I mean, this is roughly the size of this thing right over here. So if we were to actually draw it out with the hydrogen bonds, this is quite large. And remember-- in chemistry and in organic chemistry-- we always draw these neat and clean reactions, but that's not the way it happens in real life. These things are all bumping in a solution and bumping from every direction. And depending on how they bump and their relative energies, the bonds, the reactions may or may not happen. And so in this scenario, this fairly big part of this molecule might block the approach of a hydroxide anion coming from that direction. The good thing is-- well, I guess, if you view Sn2 reactions as good-- is that there's still other directions that it can come in from. So you could still see the Sn2 reaction, but it's going to be a bit slower. So maybe here, we could write a bit slower, but still fast. Now, let's think about this scenario. And once again, I think you can see where it's going. The only difference now is it's the secondary carbon. We have these two groups here. And remember, these are much larger than the way that I've drawn it. You have a carbon attached to 3 hydrogens. So now we're blocking from two directions. So once again, you can still have your Sn2 reaction, but it's going to happen a lot slower. So now, you're going to have a slow Sn2 reaction. And now, finally, what's happening here? Well, here, this entire side is blocked. You have a CH3, CH3, CH3. This hydroxide anion isn't going to be able to get through. It is being hindered. So here we have no Sn2. Now, you might be saying, wait, why didn't it come from the other end? Well, remember Sn2 reactions-- the nucleophile is attacking from one side, and the leaving group is leaving from the other. If the nucleophile were coming in from here, the leaving group would have trouble leaving from there. It wouldn't work out as well. Sn2, nucleophile from one side, leaving group from the other. Now, the only thing that could reasonably happen here-- and it is reasonable. And actually, we've even seen it when we looked at Sn1 reactions-- is maybe the leaving group leaves first. And actually, the tertiary carbon is bad for the Sn2, but it's good for having a stable carbocation. If this bromine takes that pair and becomes bromide, then this becomes a carbocation, which will be reasonably stable because that positive charge can kind of be spread amongst its brothers. And then, at that point, it's not crazy for a nucleophile. And you could even have a weaker nucleophile than the hydroxide anion to come in and actually form a bond. Now, this general idea that we're talking about in this video-- this idea that a reaction happening or not happening because of three dimensions of something hindering the reaction-- this is called steric hindrance. Steric refers to what's happening in three dimensions-- the size and shape of molecules. Hindrance is that that three-dimensional size and shape is hindering the reaction. So the steric hindrance is greatest right over here, and it's the least right over here for the Sn2 reaction.