Practice: Predict the product of the following reaction.
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|Addition to Concave vs. Convex Rings|
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.
Concept #1: General properties of strong oxidative cleavage.
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 KMnO4. Now that's really important to know because KMnO4 can do several things depending on the temperature.
Now if you guys recall KMnO4 in a cold environment is actually going to do a one, two syn diols reaction. It's going to add vicinal syn diols to the double bond, but if I jack up the heat, it's not adding alcohols anymore. It's actually just going to slice right through. You can think of that spaceship I talked about for potassium permanganate, just think that if you rev up the heat, you're just going to crash right through the double bond and split it in half. That's exactly what strong oxidative cleavage does.
So what we're going to do is we're going to take this example. Notice that I have this seven-carbon chain. I'm going to split it in two areas and we're going to kind of inspect each part of it and see what happened.
Now it's important to realize that you're going to have three different types of products when you react with strong oxidative cleavage. You're going to get ketones. That's specifically – and you're going to see exactly why you get each one. Ketones, carboxylic acids, and carbon dioxides. So when would you get each one?
Well, you get ketones for what I call internal double bonds, for internal alkenes. You get carboxylic acid for terminal alkenes. And you get carbon dioxide for one-carbon fragments. If you're just fragging off one carbon, that is going to become a carbon dioxide, fully oxidized and that's kind of the theme here.
I know we haven't really gone over rigorous oxidation introduction chapter yet, but oxidation simply can be simplified as adding oxygens to things. Kind of the prevailing theme here is we're going to add as many oxygens as possible to these cuts.
As you can see, if I go ahead and I slice down those double bonds, cleavage means that they're going to be completely sliced off and I'm going to get completely different chunks of carbons. Well, I would wind up getting a one-carbon chunk here. So keep that in mind. I would get a four-carbon fragment here. And then I would get a two-carbon fragment here.
The question is what does each of these become. Well, this is an example of an internal double bond. Notice that it's surrounded by R groups on both sides. Well, that's not going to change. When it's internal, it becomes a ketone because those R groups aren't going anywhere. As a general rule, we're not breaking off R groups. We're just breaking the double bond and that's it, but R groups are in place.
Now over here, 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. The biggest thing here is that it has at least one H coming off, it's not surrounded by R groups on both sides. It has an H.
So the fact that it has an H means that that H can be oxidized to OH. So that's when we get our carboxylic acids because we had one hydrogen that could be broken off into two different carboxylic acids. It had one hydrogen each.
So basically we saw what happens when you have zero hydrogens, you get a ketone. So let's also write that down. Internal is zero hydrogens. Terminal is when you have one hydrogen. So guess what the last one is, when you have two H's, so when you have two hydrogens that means you're a one-carbon fragment. That's all you have left. And that's when you get fully oxidized to carbon dioxide.
Carbon dioxide is considered the most oxidized form of carbon because all it is is carbon and oxygen. In fact, it's inorganic carbon because it doesn't have hydrogens anymore. So it just goes into the atmosphere. That's called inorganic carbon.
So that's basically the way 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 branched 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 and you cut it in one place, you stretch it out, it's going to become a straight chain. It's not going to break into different pieces. It's just going to be one piece that's a different shape. 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'd get two 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 was a rubber band or some kind of household item and I took it and I cut it in two different places there, 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 hydrogens are coming off the ends of the double bonds to determine the products and also the shape. Tell me how many different fragments you would get. Awesome. 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 :)
Practice: Predict the product of the following reaction.
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