Current time:0:00Total duration:7:57
0 energy points
Video transcript
Let's look at the hydration of alkynes. So we're going to start with a terminal alkyne over here. You can see there's a hydrogen on one side of our alkyne. And on the other side of our alkyne, let's say it's an alkyl group, bonded to this carbon on the right. So we're going to add water, sulfuric acid, and mercury(II) sulfate to our alkyne. We're going to hydrate it. Now we've seen the hydration reaction before. We did a hydration reaction with an alkene, and we added an H plus and an OH minus across our double bond. And we're going to do the same thing here. We're going to add an H plus and an OH minus, this time across our triple bond. And the OH minus always adds to the most-substituted carbon. So we saw that in our earlier video. So the regiochemistry shows Markovnikov's Rule. I'll just write Mark's rule here. We've spent a lot of time talking about Markovnikov's Rule in earlier videos. Which means that this OH over here, on the right, is going to add to the most-substituted carbon, which happens to be the carbon on the right, since that's going to be bonded to an aklyl group, over here on the right. The H plus, add it over here to this carbon on the left. So now there are two hydrogens on the left. So we added H plus and OH minus. We added water, essentially, across our triple bond. And we get this molecule as our intermediate. This molecule has a specific name. It is called an enol. The E comes from the fact that there's a double bond present in this molecule, and the ol comes from the fact that there's an alcohol. This OH here. So this enol is the intermediate of this reaction. However, the enol is not the most stable form of this molecule. So it's actually going to rearrange. And we're going to get a ketone for our product. And I'll show you the mechanism for this arrangement in a few minutes here. So we get a methylketone from a terminal alkyne. So this reaction is best used when you're looking to make a methylketone and you're starting with the triple bond here. So let's look at a reaction. Let's start with a terminal alkyne. I'm going to put a methyl group on this side of the carbon, like that. And I'm going to react this terminal alkyne with water and sulfuric acid and with mercury(II) sulfate, like that. If we think about what's going to happen, we're adding H plus and OH minus across our triple bond. We're going to add the OH minus to the most-substituted carbon. So that would be the carbon on the right. So we have carbon double-bonded to another carbon. We're going to add the OH to the carbon on the right. We're going to add the H plus to the carbon on the left, so there's water added across your triple bond. And then we have still another hydrogen on this carbon on the left here, and then the methyl group is still over here on the right. So that's our enol intermediate. Let me just draw this molecule again. So this molecule is equivalent to this molecule right here. So I'm going to draw it in this form, from now on. So this is my enol form. And my enol's going to rearrange. This is not the most stable form of this molecule. So let's see the mechanism of how our enol rearranges. Let's go ahead and redraw our enol down here. Let's get a little room, like that. So here's my enol. Put in my double bond, put in my lone pairs of electrons, like that. So. This enol is present in an acidic mixture. H2O and H2SO4 are going to give me hydronium ions in solution, H3O plus. So there's going to be H3O plus floating around here in solution, so we have our hydronium looking like that. This, of course, is capable of donating a proton. And the pi electrons are going to actually move out here and pick up a proton, like that. And then these electrons are going to kick off onto your oxygen. So this is an acid-base reaction as the first step. So therefore, we go draw our equilibrium arrows here, like that. And let's see what happens. Well, we're going to have-- this is our skeleton. The OH didn't really do anything yet. And we're going to add a proton onto the carbon on the left side of your double bond, like that. So let's go ahead and show what electrons did the moving. The electrons in your pi bond, whichever one of these bonds here is your pi bond. Those electrons are going to move and form a bond with this proton, like this. So we ended up taking away a bond from this carbon, right here. So we had a double bond that took a bond away from that carbon. That's going to make that carbon positively charged. So there's a plus one formal charge on that carbon. And we can draw a resonanced structure for this intermediate. So let's go ahead and draw our resonance bracket and our resonance arrows here. And what can we do with our lone pairs of electrons, to share that positive charge? We can take one of these lone pairs of electrons, and we can move them in here to form a double bond between our carbon and our oxygen. So let's go ahead and do that. A double bond forms between our carbon and our oxygen, like that. Now there's only one lone pair of electrons on this top oxygen. And if we think about what happened to our formal charge, it actually moves out here to this oxygen. So this oxygen now is a plus one formal charge. This is a resonance structure. Let me go ahead and keep this hydrogen in here. Like that. So that's our resonance structure for this molecule on the left. Now, oxygen does not like having a plus one formal charge either. So is there any way for this oxygen to get rid of its plus one formal charge? And of course there is. There's water floating around. So water is going to act as a base. So a lone pair of electrons on water is going to take this proton, just that proton, leaving these two electrons behind on my oxygen, like that. So we have another acid-base reaction. Let's go ahead and show this reaction being at equilibrium. So I draw my equilibrium arrows like that. And so now I have my oxygen. It had one lone pair of electrons around it. Now it has two lone pairs of electrons around it. So now we're done. We formed our ketone version. So this is acetone, our simplest ketone. So we have the keto form of the product. This is the keto form of the product in the form of a ketone. And then we have our enol over here. So the ketone and the enol form are said to be tautomers of each other. And they're in equilibrium with each other, and they can go back and forth. And in this case, we're using acid to catalyze this transition. So this is called acid-catalyzed tautomerization. Like that. And we're going back and forth between the keto and the enol form. So you'll also hear this called keto-enol tautomerization a lot, like that. So the keto and the enol form are in equilibrium with each other. Now the equilibrium is actually going to favor the keto form. So this is actually the more stable form. This is more stable, so that carbon double-bonded to an oxygen, for the keto form, is more stable than the carbon double-bonded to another carbon, in the enol form. And essentially you're changing one proton and one pi bond in this equilibrium. And this is the acid-catalyzed form of this mechanism. So in the next video, we'll see a very similar reaction where we do a base-catalyzed tautomerization.