These reactions add hydrogen which reduces π-bonds to alkanes.
Concept: General properties of catalytic hydrogenation.6m
All right, guys, so now we're going to talk about an addition mechanism that some of you guys might already be familiar with. But regardless, we're going to talk about it in this section because it has to do with addition and that's called catalytic hydrogenation and also Wilkinson's catalyst. So let's go ahead and check it out.
So basically, catalytic hydrogenation is a way of adding hydrogen to double bonds. So remember that I said that any addition reaction is a reaction that you take one pi bond and you make two sigmas. Well, actually one of the easiest ways that you can do that is just to say let's take that one pi bond and make two hydrogen bonds out of it. That would still be an addition reaction because I'm taking one pi and I'm making two sigma bonds. And that's what happens with both of these reagents.
So how does this work in general? Basically, we going to have a double bond or a triple bond and whenever you have one of those, we're just going to add as many H's to it as possible. So what that means is that the product of these reactions is simply going to be alkanes. So we haven't learned a whole lot of reactions that make alkanes. This is one of them because you can take double bonds and triple bonds, add hydrogens to them until they're fully saturated and become alkanes.
So the stereochemistry for this reaction, both of these is going to be syn addition and what that means is that your end products would be cis products. What that means is that your hydrogens are going to add from the same side of the double bond at all times.
And why is that? Because it turns out that our intermediate or the thing that's going to catalyze this is going to be a metal catalyst. That metal catalyst is going to come in and it's going to coordinate the H's to be on one side of that double bond. So that's why you're always going to get syn addition.
So now for these two things, I'm actually going to kind of ignore them. There's no carbocations, so I'm not going to expect rearrangements. And then I'm adding two of the same thing, I'm always adding two H's, so I don't really have a choice between Mark and anti-mark. In fact, that's not really going to apply since I'm adding two of the same thing.
So let's go ahead and see the general reaction. The general reaction basically looks like this. I have two different mechanisms and two different reagents that do the same thing. This combination of reagents here is called catalytic hydrogenation. How do you know that it's catalytic hydrogenation? Because you got H2 – sorry, a little crash accident there. Because we've got H2 with a metal catalyst such as nickel, platinum or palladium. Whenever you see H2 with one of those three, that means it's going to be a catalytic hydrogenation.
But there's also another reagent that does a very similar reaction and that's if I have H2 over what's called rhodium. This is the Wilkinson's catalyst. I'm just going to write that here. Wilkinson's catalyst. The Wilkinson's catalyst is really going to do the same exact thing, just using a different metal catalyst. It's going to have a rhodium and then three triphenylphosphines and a chlorine. You don't need to be able to draw this.
What's important is that you just recognize that this is going to do the same thing. It's going to add two H's for every double bond that it encounters. But it's not going to do anything else to anything else.
So that means that my general reaction, what I would expect is that I get the same carbon backbone as I had in my beginning. I would get a single bond there. Even my alcohol doesn't change. Nothing changes about that. The only difference is that now instead of having that double bond on this side of the molecule over here, now I'm just going to have two H's. So in fact, I would just draw it like that. I don't need to draw the H's because the H's are implied.
Then over here what I would expect is that I would just have a single bond. Why is that? Because basically there were two pi bonds here. What that means is that I add two hydrogens for every pi bond, so I would have added hydrogen here, hydrogen here and then I would have also added hydrogen here, hydrogen here and that's what I get at the end. What I get is two hydrogens here and – I'm sorry. One of each color. One there and one there. And then I would get a hydrogen there. And a hydrogen there. Those are the four hydrogens that were added through my two hydrogenation or Wilkinson's catalyst reactions.
Maybe you're wondering, “Hey, but Johnny, doesn't this carbon need to have three H's?” Yeah, it does. It has that last H that came from the original compound. Remember there was always one H at the end to begin with, that's that H right there. But now what I did was I added two blue H's from the first equivalent and then I added two red H's from the second equivalent basically giving me an alkane as my end product. Does that make sense? Cool.
So what I want you guys to do now is I want to just provide the products for the following reaction. That means you're going to have to recognize what it is and then think about the stereochemistry if it's required and try to draw the end product.
Now notice that instead of using H2, I'm using D2, but that's really the same thing because remember that deuterium is just heavy hydrogen. So go ahead and try to draw the end product for this and then I'm going to go ahead and explain it to you. So go for it.
Problem: Provide the product of the following reaction.2m
Predict the organic product(s) for the following reaction. Be sure to indicate stereochemistry when appropriate. If stereoisomers are produced draw one and state the relationship between the otherstereoisomer formed. (enantiomers, diastereomers, etc.).
Complete the following reaction by drawing the structure of the principal major product. Indicate relative stereochemistry where necessary. If there is no reaction, write NR.
Circle the major product of the following reaction.
Predict the product(s) or reagents of the following reactions.