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Ch. 9 - Alkenes and AlkynesWorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins
Ch. 26 - Transition Metals
Alkene Stability
Zaitsev Rule
Double Elimination
Hydrogenation of Alkynes
Dehydration Reaction
POCl3 Dehydration
Alkynide Synthesis

Hydrogenation reactions are a subtype of reduction reactions that add hydrogens to double and triple bonds. These differ in the types of products made and the stereochemistry of the mechanism.  

Concept #1: The definition of hydrogenation.


Now we're going to discuss a few reactions that add hydrogens to triple bonds. And these are going to be called, in general, hydrogenation reactions. The major difference between them is going to be how many hydrogens they add and in what stereochemistry. So let's go ahead and check them out.
Basically, there's three major types of hydrogenation that we look at and these are going to be reactions that turn triple bonds into either double bond or single bonds. So basically, we're taking one type of hydrocarbon and making it a more saturated version of that hydrocarbon. Remember that the definition of saturation or degree of unsaturation has to do with how many hydrogens are on that molecule. So in all of these reactions what we're doing is we're increasing the saturation of the molecule meaning that we're adding hydrogens to it. 

Concept #2: Using Catalytic hydrogenation or Wilkinson’s Catalyst to turn alkynes to alkanes. 


So let's look at the first and probably simplest form of hydrogenation and that would be what we call full saturation. What that means is that I'm taking a double bond or a triple bond and I'm fully saturating it with hydrogens.
We can do this by two different reagents. We can do this with either catalytic hydrogenation which would be these reagents right here in the first arrow. That would be basically H2 over a platinum, palladium or nickel catalyst. Or there's another reaction we can use as well called Wilkinson's catalyst. And Wilkinson's catalyst is used really for the same purpose, to take double bonds and triple bonds and fully saturate them to alkanes.
As you can see the reagents for Wilkinson's catalysts are a little bit confusing. It's got the H2 because we're adding hydrogen of course, but the catalyst is way different. In this case, we're going to use rhodium with these three triphenylphosphines and then a chlorine. It looks messy. I'm not going to draw the whole thing for you as long as you can recognize that that's Wilkinson's, that's fine.
Now the only thing about full saturation is that this is a little bit of a misnomer because remember that rings also count towards unsaturation. A ring is something that it's missing two hydrogens because two ends are attached to each other. And actually, these reactions do nothing to rings. So what's going to happen is that all the double bonds and triple bonds would be gone, but the rings will stay intact.
One more thing before I draw these products, notice that the stereochemistry says syn addition. Syn addition is another way to say that we're going to get cis products. What that means is that I would expect the hydrogens that we're adding to add from the same side of the double bond or the same side of the triple bond.
So let's go ahead and draw our products. Basically, I would get the same exact sigma framework. When I say sigma framework, I'm just saying all the sigma bonds are the same. But I would get a single bond over here so that triple bond just turned into an alkane and I would get that this alcohol is still here. And then I would finally get that that double bond is also gone, meaning that there's no pi bonds left. What this is really good at is getting rid of pi bonds. That's pretty much it.
Here, I didn't include stereochemistry because there are no chiral centers, so I don't really have to. But if you're wondering where do these H's come from, especially maybe on the double bond, they could have both come from the same side. So that means that – let's say that one of the H's came from here and one of the H's came from here. Notice that they came from the same side of the double bond then that means that my alcohol would be forced to go down. Alternatively, I could have also had the H's add from the bottom, which means that the OH would have gone up as well.
Same thing with the triple bond, except that the triple bond I added four hydrogens, I didn't just add two. So I basically added let's say, two from the front, H, H, and I also added two from the back, so let's say, H, H. So what I wound up getting was instead of – and then also – so now that – this carbon here makes sense. That's CH2. But this one looks like it doesn't make sense because it's also CH2. But remember there was already one H to begin with on that triple bond. So that last H is still there and it's right next to my head. Does that make sense? Basically, I added the two blue hydrogens. Then I added two green hydrogens and then this black H was just always there.
So basically, we can think that full saturation or one of these two catalytic or Wilkinson's is going to completely get rid of all pi bonds, all double bonds, and all triple bonds. Easy. Cool. 

Catalytic Hydrogenation or Wilkinson’s Catalyst: Alkane products

Stereochemistry: Syn Addition

Concept #3: Using Dissolving Metal Reduction or Lindlar’s Catalyst to turn alkynes to alkenes.


Let's move on to the next reaction. And actually, these next two are going to be both considered partial saturation. So what that means is that for these next two I'm not saturating all the way to an alkane. In fact what I'm doing is I'm taking a triple bond and I'm only reacting it to a double bond. So I am adding two hydrogens, but I'm stopping there.
And the first of these – really the biggest difference between them other than the reagents which are totally different, are the stereochemistry. Notice that my first one is anti addition, notice that my second one is syn addition. That's going to have a difference in what my products look like.
So let's look at this first one. The first one's called dissolving metal reduction. Dissolving metal reduction is just another name for these reagents. What it is is that we have some kind of metal, lithium or sodium, your professor can use either one, and some kind of liquid amine source. Many times that amine is going to be NH3, but there's a lot of other things it can be. It could be NH2CH3 or it could be just any kind of usually primary or secondary amine. All right? Cool.
So we've got that part down. What does it do? Well, we learned – I just taught you guys that partial saturation means that it's only working on triple bonds. So notice that I have the same or very similar molecule here and what I'm wondering is, how should I draw this. I have the same sigma framework, but for example, should I draw the double bond or should I leave it there. I should leave it there because it turns out that since this is partial saturation it has no effect on alkenes. No effect on alkenes. It's only going to work on triple bonds.
So now I've got this triple bond. Is it going to work on this one? Absolutely. Well, what does anti addition mean? Anti addition, if you remember that just means I have trans products. So what that means is that I would expect that my hydrogens are going to add from different sides of this double bond. What that means is that I should actually draw the double bond in a trans position. Why? Because the two hydrogens that I added must have added on different sides. I have different sides of the double bond, which means that they were anti addition. Is that making sense?
So what that means is that dissolving metal reduction is important because this is a way that we can turn triple bonds into trans-double bonds. That's going to be your product. Trans-double bonds. Pretty cool, right? So that's something you have to keep in mind because for synthesis later when we start trying to make molecules, that's going to be really the only way that we know how to turn a triple bond and make it trans. Cool?
So now let's go to this next one which I'm sure you guys can kind of already anticipate what's going to happen. Notice that this one's called Lindlar's catalyst. Lindlar's catalyst has a lot of different reagents that you could use. Unfortunately, different textbooks, different professors have their own way of writing it.
Some professors are going to be really easy. They're going to be like, you know what? F all of these reagents. We're just going to write Lindlar's. They're going to put H2 and Lindlar's and you're going to know that that's Lindlar's. But some professors are going to use the actual reagents that are in the textbook, so some textbooks have this as the P-2 catalyst, so the P-2 catalyst is one form of Lindlar's. Another form is actually, with lead acetate and quinoline. These are reagents that you just have to recognize that they are Lindlar's. You don't need to know exactly what they look like or draw them or even know the mechanism, but what you do need to know is recognize that these are Lindlar's.
These are also, just so you guys know, another term that you may hear is that these are poisoned catalysts. Now what could – oops, catalysts. Wow. Okay. Sorry, guys. Spelling isn't my forte. So poisoned catalysts. What does poisoned mean? What it means is that these catalysts have been created in such a way that they're not going to reduce all the way or they're not going to hydrogenate all the way, instead, they're going to stop at a given point. That's why they're called poisoned.
So these are three different ways that you could see this reagent just be aware of them. Personally, I hope that your professor just writes Lindlar's, but you may see these other two reagents come up.
So now let's finally get into the products. What should I draw for my product? Well, what I should draw is that I should still keep that double bond because whenever you have partial saturation that means you're going to have no effect on double bonds. That means the only difference – the only thing that I'm really touching here is the triple bond. So how should I draw it? Well, in this case, I have syn addition and what syn addition means is that I'm going to get cis products. So what I would expect is that I have to draw this double bond now in a way that has my R group cis to whatever that double bond is.
So I'm going to get out of the way here and draw this for you. I would actually want to draw my stick down this time. Why? Because if I create like a little fence parallel to this double bond, what I see is that now both of my groups are going to be on the same side of that fence, so this creates cis double bonds.
So I know that was a lot of reagents. I'm just going to stay here for a little bit so that you guys can write that down. But anyway, I know that seemed like a lot of reagents, but really it's not that bad, especially when you're not having to consider any mechanisms at this point yet. I just want you to know the reagents, recognize what they do and be able to draw them. Be able to draw the products. And keep in mind the stereochemistry because the stereochemistry really is the most important part of these reactions.
So let's move on to the next topic. 

Dissolving Metal Reduction: Alkene products

Stereochemistry: Anti Addition

Lindlar’s Catalyst: Alkene products

Stereochemistry: Syn Addition

Practice: Predict the product of the multistep synthesis