Organic Chemistry

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10. Addition Reactions

Oxidative Cleavage

When you react a double bond with potassium permanganate (KMnO4), your first thought may be to make syn-diols, as seen in previous videos. However, looking at the temperature is extremely important. In heat, KMnO4 will actually cleave the double bond, rather than add diols. Let’s check it out. 



General properties of strong oxidative cleavage.

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in this video, we're going to discuss a really important reaction that happens to double bonds. And that's called strong oxidative cleavage. So strong, oxidative cleavage is the reaction of double bonds with hot can, you know, four. Okay, now, that's really important to know, because Canada four can do several things, depending on the temperature. Okay, Now, if you guys recall camera four in a cold environment is actually going to do a one to sin Dial's reaction, it's gonna add visceral sin Dial's so double bond. But if I jack up the heat, it's not adding alcohol's anymore. It's actually just gonna slice right through. You can think of that spaceship I talked about for potassium permanganate. Just think that, you know, if you rip up the heat, just gonna crash right through the double bond and split it in half, okay? And that's exactly what strong sedative cleavage does. So what we're gonna do is we're gonna take this example, notice that I have this seven carbon chain. Okay, I'm gonna split it in two areas, and we're gonna kind of inspect each part of it and see what happened. Okay, Now it's important to realize that you're gonna get three different types of products. When you react with strong city of cleavage, you're gonna get key tones. Okay, that's specifically, and you're gonna see exactly why you get each one key tones, carb, oxalic acids and carbon dioxide. So when would you get each one? Well, you get key tones for what I call internal double bonds. Okay. For internal Al Keene's, you had carb oxalic acid for terminal, um, for terminal Calkins. Okay. And you get carbon dioxide for one carbon fragments. Okay, so if you're just, if you're just fragging off one carbon, that is gonna be home, a carbon dioxide fully oxidized. And that's kind of the theme here. I know that we haven't really gone over rigorous oxidation and reduction chapter yet, but oxidation simply can be can be simplified as adding oxygen's. The things and kind of the prevailing theme here is that we're just gonna add as many oxygen's is possible to these to these cuts. Okay, So as you can see, if I go ahead and a sliced down those double bonds, cleavage means that they're gonna be completely sliced off, and I'm gonna get completely different chunks of carbons. Well, I would wind up getting ah, one carbon chunk here. Okay, So keep that in mind, I would get a four carbon fragment here, and then I would get a two carbon fragment here. Okay? The question is, what does each of these become? Well, this is an example of an internal internal double bond notice that it's surrounded by our groups on both sides. Well, that's not gonna change when its internal it becomes a key tone because those are groups aren't going anywhere. Okay. And as a general rule, we're not breaking off our groups. Okay? We're just breaking the double bond, and that's it. Our groups are in place. Okay, now, over here. Um, we have examples of terminal. So let me show you what terminal is Terminal would be like this and like this. Notice how they both are at the end of the chain, and they have an h on them. Okay, So the biggest thing here is that it has at least one age coming off. It's not surrounded by our groups on both sides. It has an h. Okay, So the fact that it has an H means that that age can be oxidized to Ohh. Okay, so that's when we get our carb oxalic acids. Okay, Because we had one hydrogen that could be broken off into two different carve oxalic acids. Okay, if you had one hydrogen each. Okay, So basically, we saw what happens when you have zero hydrogen, you get a key tone. So let's also write that down. Internal is zero hydrogen terminal is when you have one hydrogen. So guess what the last one is when you have to. H is right. So when you have two hydrogen, that means that your your one carbon fragment, that's all you have left. And that's when you get fully oxidized to carbon dioxide. Okay, so that's carbon dioxide is considered the most oxidized form of carbon because all it is is carbon and oxygen. Okay, In fact, it's inorganic carbon because doesn't have hydrogen is anymore. So it just goes into the atmosphere that's called inorganic carbon. Okay, so that's basically what we do it now. Also, you have to think geometrically if I started off with a ring instead of a straight chain or whatever, a branch chain and I cut it in one place. What would I expect that ring to become if I just cut it in one place? Well, guys, just think of it like anything around the house. Let's say you have a rubber band, right? And you cut it in one place, you stretch it out, it's gonna become a straight chain, okay? It's not gonna break into different pieces. It's just gonna be one piece. That's a different shape. Okay, so think of it like that. Now, if you took a rubber band and you sliced it right through the middle, both sides now you would get to pieces. But that's if you have two cuts. So that's what I want you guys to be thinking about. Does this make sense? If this is a rubber band or some kind of household item and I took it and I cut it in two different places there, um would it actually make three fragments in this case? Yes, definitely. So that being said, let's move down to this molecule. I want you guys to analyze how many different hydrogen zehr coming off of the end of his old want to determine the products and also the shape Tell me how many different fragments you would get. Okay, So awesome. So go ahead and do that, and then I'll show you the answer.

Note: I added an extra carbon when rotating the structure to a bondline conformation in the video below:)


Predict the product of the following reaction.