Current time:0:00Total duration:7:20
0 energy points

Halogenation and ozonolysis of alkynes

Video transcript
Let's look at two more reactions of alkynes, and we'll start with the halogenation of alkynes. So I start with my alkyne over here, and I add to my alkyne one more equivalent of a halogen, so x2. My solvent is going to be carbon tetrachloride. And I'm going to add those two halogen atoms across my triple bond in an anti addition, so those two halogens end up on opposite sides from each other. So this is an anti addition of my halogens, like that. Now, I could add two molar equivalents of my halogen. And if that happens, each of these carbons is going to end up with two bonds to halogens, like that, for my product. Now, the mechanism of the halogenation of alkynes is not completely understood, so because of that, we're just going to move on to one practice example instead of showing the mechanism. So let's look at an alkyne. So we'll go ahead and draw an alkyne over here. So here I have my carbon triple bonded to another carbon. I'm going to put a methyl group on one side here, and let's write a CH3 in here. And I'll put an ethyl group on this side, so CH2CH3. So to that alkyne, I'm going to add bromine. And I'm going to use carbon tetrachloride as my solvent, and I'm going to say one more equivalent of my bromine is added. So when you do your stoichiometry, just one more equivalent like that. So I'm going to add my two bromines on anti to each other, right? So let's go ahead and show my triple bond became a double bond. And my two bromines are going to add on anti to each other. So they're going to add on opposite sides of my double bond, like that. And I still have a methyl group over here on the connection to the carbon on the left. So I'll go ahead and put in my methyl group, like that. And the carbon on the right still has an ethyl group attached to it, so CH2CH3. So that would be the result of the halogenation of this alkyne. Let's take a look at one more reaction of alkynes. Let's look at ozonolysis, the ozonolysis of alkynes. Let me go ahead and write ozonolysis right here. And we've seen this reaction before, similar reaction, when we did this with alkenes. So this time we're going to do it with alkynes. Let's take a look at an alkyne, so there's an alkyne. I'm going to say it's an internal alkyne, meaning the triple bond is found in the interior of the molecule. It's not on the end of the molecule. So let's look at ozonolysis of internal alkynes first. And when you're doing ozonolysis, you're adding ozone to the molecule in the first step. So we went in a very, very detailed mechanism for the ozonolysis of alkenes, and you go back and watch that video. For this video we're not going to go through any kind of a mechanism. We're just going to go for the products. So we add ozone in the first step, and in the second step, we're going to add water. And what this does, is this cleaves your triple bond and gives you carboxylic acids as your products, so two of them. So let's go ahead and draw those two carboxylic acids, and then we'll try to point out where everything comes from. So here's one of the carboxylic acids, and then here is going to be the carboxylic acid that's going to result on the right side. So I'll make this R prime over here. So let's go ahead and point out which carbons are what here. So let's show that this carbon over here is bonded to an R group. So that's this carbon bonded to an R group. And then over here on the right, this carbon is the one bonded to an R prime, right? This carbon is the one bonded to an R prime. So you cleave your triple bond, right? You break your triple bond. And you're going to create two separate molecules from this, so you get carboxylic acids from this reaction. Let's look at ozonolysis of terminal alkynes now. Instead of the triple bond being in the interior of the molecule, now the triple bond is on the end of the molecule. So that makes this a hydrogen, right here. So let's add, once again, ozone in my first step and water in my second step. And on the left side, the left side is going to give us the same product as before, right? So let's go ahead and identify that in blue, this carbon and this R group. We saw before that's going to give us a carboxylic acid, like that. So let's go ahead and draw that again. So it's the same. That portion of the molecule gives us the same product as before. So we get a carboxylic acid. And once again, the carbons-- this is the carbon in blue, and that R group in blue-- those are the same ones on the left side of your reaction. Now, on the right side of my reaction-- just go ahead and put that hydrogen in there. On the right side of the reaction, the terminal alkyne portion, you're actually going to get carbon dioxides. So now you have only one carbon to think about. This is the only carbon you have right here. So you're going to get CO2 out of it. So let's go ahead and draw CO2 as your product, like that. Remember, it's a linear molecule. And so this reaction was used decades ago for structure determination, right? If a terminal alkyne was present and you reacted it with ozone, then you would get some carbon dioxide. And you could also analyze the molecule by the carboxylic acids that you get. For example, you could see how many carbons are in your R group. And then that would give you an idea about the structure and all those kinds of things. So this used to be a very important reaction in organic chemistry. Now, with all spectroscopy stuff that organic chemistry has, this reaction isn't really used as much anymore. So let's look at one example of an ozonolysis reaction. So let's look at this one right here, like that. So we're going to take that terminal alkyne. We're going to add ozone to it in the first step, and then we're going to add water to it in the second step. And let's see how many carbons we're dealing with here. Sometimes that's what confuses students, right? So there's one carbon, two carbons, three carbons. And on this side is our fourth carbon. And that carbon on the far right is bonded to a hydrogen, making this a terminal alkyne. So I have four carbons to worry about. And when this undergoes alkyne cleavage, it's going to cleave the molecule here. It's going to break that triple bond. And so you're going to get two products. You're going to get one product with three carbons, right, over here on the left. And then one product with one carbon on the right. So the product with three carbons on the left is going to be a carboxylic acid. So all you have to do is draw a three carbon carboxylic acid. So let's go ahead and do that. So here's our three carbon carboxylic acid. That's just one of our products. And the terminal alkynes will give us CO2, which takes care of the other carbon on the right over here. So we're going to get CO2 as the other product of this reaction. So that sums up all of our reactions of alkynes. In the next video, we're going to take a look at some synthesis problems, using all the stuff that we've learned in the first semester of organic chemistry, especially including some of the alkyne reactions.