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So in a hydration reaction, water is added across a double bond. And the OH adds in a Markovnikov way. So according to Markovnikov's rule. So let's go ahead and write that down here. So you have to think about Markovnikov when you're doing this reaction. And this is an acid catalyzed reaction. So technically, this reactions is at equilibrium, and we will cover that at the very end of the video here. So let's look at the mechanism for the acid catalyzed addition of water across a double bond. So here I have my alkene, and I have water present with sulfuric acid. So sulfuric acid, being a strong acid, will donate a proton in solution. And let's say the water molecule picks up that proton. So H2O would go to H3O+. 3 So I'm just jumping ahead to the H3O+ ion called the hydronium ion. So here's my H3O+ ion. So put my one pair of electrons on there, and it's positively charged. So what's going to happen is the pi electrons, the electrons in this pi bond here are going to function as a base. They're going to abstract a proton. They're going to accept a proton. They're going to take, let's say, this proton right here, which would cause these two electrons in this bond to kick off onto your oxygen. So acid base equilibrium. So I'll go ahead and make this an equilibrium arrow here. And what are we going to get if that happens? Well, let's say that the proton added to the carbon on the right-- so the proton added to the carbon on the right here. So I'm saying that the blue electrons on the left are going to be these electrons right here, like that. And that would mean that I took a bond away from the carbon on the left. This carbon over here on the left, this carbon right here, used to have four bonds to it. Now it has only three bonds to it. So it ends up with a plus 1 formal charge. So we have a carbocation in our mechanism. So what's left? In this acid base reaction, we took a proton away from H3O+, which leaves us H2O. So here we have H2O over here, so I'll go ahead and put lone pairs of electrons in on our water molecule. And we know that water can act as a nucleophile here. So this lone pair of electrons is going to be attracted to something that's positively charged. So nucleophilic attack on our carbocation. And this is technically at equilibrium as well, depending on the concentrations of your reactants. So let's go ahead and show that water molecule adding on to the carbon on the left. So the carbon on the right already had a hydrogen or proton added onto it, and the carbon on the left is going to have an oxygen now bonded to that carbon. Two hydrogens bonded to that oxygen, and there was a lone pair of electrons on that oxygen that did not participate in any kind of bonding. This gives this oxygen right here a plus 1 formal charge. So our oxygen is now positively charged, like that. And we're almost to our product. So we're almost there. We need one more acid base reaction to get rid of that proton on our oxygen. So water can function as a base this time. So water comes along, and this time it's going to act as a Bronsted Lowry base and accept a proton. So let's get those electrons in there. So this lone pair of electrons, let's say it takes that proton, leaving these electrons behind on my oxygen. Once again, I'll draw my equilibrium arrows here, acid base reaction. And I'm going to end up with an OH on the carbon on the left, and the carbon on the right there is a hydrogen, like that. So I added water. I ended up adding water across my double bond. And to be complete, this would regenerate my hydronium ion. I'd get H2O plus H+ would give me H3O+. So hydronium is regenerated. And so there you go. So remember, a carbocation is present, so you have to think about Markovnikov addition. And since a carbocation is present, you have to think about possible rearrangements. So Markovnikov and rearrangements. Let's take a look at an example where you have a rearrangement here. So let's look at a reaction. So let's look at this as our starting alkene. And let's go ahead and think about the mechanisms. So we know H3O+ is going to be present. So H3O+ right here. So we're adding our alkene to a solution of water and sulfuric acid. And our first step in the mechanism, the pi electrons are going to function as a base and take a proton from our hydronium ion, leaving these electrons in here letting them kick back off onto the oxygen. So let's see what we would have from that acid base reaction. And I realize I didn't draw an equilibrium arrow here. I'm more concerned with getting the right product. So this is our carbon skeleton. And which side do we add the hydrogen? Which side of the double bond-- do we add the proton to the left side, or do we add the proton to the right side? Well, we want to form the most stable carbocation we possibly can. And if we add the proton to the left side of our double bond, we end up with a secondary carbocation. This carbocation right here is secondary, because this carbon that has the positive charge is bonded to two other carbons. So this is a secondary carbocation. If we had added the proton on to the other side of the double bond, would have a primary carbocation. So secondary carbocation is more stable. But can we form something that's even more stable than a secondary carbocation? Of course we can. We can form a tertiary carbocation if we think about the possibility of a hydride shift. So right here there is a hydrogen attached to that carbon. And if the proton and these two electrons are going to move over here and form a new bond with our positively charged carbon, so we get a hydride shift at this point. So let's draw what would result from that hydride shift. We moved a hydrogen over here. That took a bond away from this carbon. So that is the carbon that's going to end up with the positive charge now. We added a bond to what used to be our secondary carbocation carbon. And so that formal charge goes away. The formal charge moves to this carbon right here, which is now a tertiary carbocation. If you look at the carbons connected to that carbon, this is a tertiary carbocation. So we know tertiary carbocations are more stable than secondary. So now we're at the step of the mechanism where a water molecule is going to come along. So we have a water molecule, which is going to function as a nucleophile and attack our positively charged carbon, like that. So let's go ahead and draw what the result of that nucleophilic attack would look like. So we have our carbon skeleton, and we have an oxygen atom now bonded to that carbon. So two hydrogens here, and once again, one lone pair of electrons now participate in that reaction, giving this oxygen a plus 1 formal charge. And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons off of our positively charged oxygen and gives us our major product with the OH adding on to this carbon right here. So this is a major product. This is our major product. And we would get some of the alcohol that forms from the secondary carbocation. So a minor product, that's what we would get if this oxygen had attacked our secondary carbocation. And you will get some of that. OK. But if your test asks for the major product, you should show the product of this rearrangement. Now, we're lucky in this instance because our product here, this carbon right here, ends up not being a chirality center. Because I have two methyl groups attached to that carbon, so I don't have to worry about my stereochemistry here. Let's do a reaction where we do have to worry about stereochemistry. OK. So let's look at this reaction right here. So we take this as our starting alkene. So I'll put the double bond right there. And let's make that a little bit more clear here. So the double bond is between these two carbons right here. And once again, we're going to add water and sulfuric acid. So H2SO4. And when we think about the mechanism, we know that we're going to add a proton to one side of the double bond and the other side of the double bond is going to end up being our carbocation. So the first thing to think about is OK, which one of these two sides is going to get the proton. We want to form the most stable carbocation we can. So the proton's going to add on to the carbon on the left. So if I can go ahead and show the intermediate here. So for an intermediate-- don't need that arrow. We'll just go ahead and show the proton adding on to the carbon on the left. So we get an H here. And then the carbon on the right of the double bond now ends up being our carbocation. So this is now positively charged, like that. And remember, when we have a carbocation, this carbon is bonded to three other atoms. You have to think about what that looks like. So remember, a carbocation-- when something is bonded to three other carbons, you get this situation where everything is in the same plane. Sp2 hybridized carbon exhibits trigonal planar geometry. Also with your unhybridized p orbital, like that. So this is your sp2 hybridized carbocation situation here. So when your water molecule comes along and acts as a nucleophile, your water molecule could end up attacking from the top here. Or it could end up attacking from below here. So that's where the stereochemistry comes in. So let's go ahead and take our carbocation and let's see if we can draw the products that would result from our nucleophilic attack of water. And then we'll just go ahead and think about the proton going away in our heads for the mechanism. So when you have enough practice, you can do steps of the mechanism in your head. So let's see. What would we get for our two possible products? Well, the OH could've added this way, which would push that methyl group there away from us. So the methyl group would be going away from us. Or the OH could've added from the opposite side. The OH could've been the one back here, and that would've pushed the methyl group out like this. OK. And we know that these are chirality centers. We know that this is our chirality center on these guys. So we get enantiomers here. 50% racemic mixture for our products. And we see that the OH adds on in a Markovnikov fashion. It adds on to the side that's the most stable carbocation here. So one more thing about this reaction. So let's just do one that doesn't have any stereochemistry to worry about. And we'll try to make a different point about this reaction here. So this is our reaction. So we're going to add water to this. And we'll put sulfuric acid up here. And we'll make our arrow a little bit different to illustrate the point here. So we could go-- let's go ahead and draw an equilibrium arrow here. So let's say this whole reaction is at equilibrium. So let me get this equilibrium arrow in here. And our product. So if we don't have to worry about stereochemistry, we think OK, really all I have to do is think about which side of the double bond do I put my OH. And again, It's Markovnikov addition. The more substituted carbon is the one that's going to get your OH. So the more substituted carbon would be the one on the left here. So if you were a product, you would say OK, I know all I have to do is really just go ahead and put my OH in there on the more substituted carbon and I'm done. I don't have to worry about stereochemistry for this reaction. I don't have to worry about rearrangement, since it's the tertiary carbocation. So that takes care of it. Now, this reaction is technically at equilibrium. And you could think about water as being one of your reactants. So if water is one of your reactants and you think about general chemistry, Le Chatelier's principle, how do you shift in equilibrium? If you want to make more of your product, if you wanted to make more of this, your product or your alcohol, one way to do it would be to increase the concentration of water. So if you increase the concentration of water, the equilibrium will shift to decrease the stress that was put on the system. So you're going to get a shift to the right, and you're going to form more and more of your product here. But remember, if you have an alcohol for a product, and if you react this alcohol with sulfuric acid, that's an E1 elimination reaction that we saw in earlier videos. So acid catalyzed dehydration, the addition of concentrated sulfuric acid to your alcohol can actually form your alkene. So that's a reaction that we saw earlier, an E1 elimination acid catalyzed dehydration. Which your major product would be your most substituted alkene here. So you could go back the other way. You could go back to the left. Let's say you decreased the concentration of water to shift the equilibrium to the left, and you'd actually form your alkene here. So the way to control your equilibrium is if you want to go to the right, you just dilute your sulfuric acid. You add more water to it, which would increase this concentration. If you want to shift the equilibrium to the left, you decrease your concentration of water, which means using concentrated sulfuric acid. So I could just write concentrated sulfuric acid here. And that would shift your equilibrium to the left and make more of your alkene. So it all depends on what you're trying to make. And so you have to remember all that general chemistry, shifting equilibrium stuff.