Hydrogenation of an alkene using hydrogen gas and a platinum (or palladium or nickel) catalyst. The mechanism of syn addition of the hydrogens. Created by Jay.
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- We count Hydrogen's electron when counting Carbon's oxidation number? I don't understand.(14 votes)
- Carbon is more electronegative than hydrogen. So, even though they share electrons in a covalent bond, the electrons spend more time near the carbon than near the hydrogen. Suppose you have a double carbon-carbon bond. The electrons are shared exactly 50-50 (since both atoms are carbon, so have the same electronegativity). If you hydrogenate this double bond (so now you have a C-C single bond, and each C gained a bond to a hydrogen), then, compared to the C=C bond where it had a 50% share in the electrons, each carbon has gained a bit of electron density since it has a bit more than a 50% share of the electrons in the new C-H bond. For oxidation/reduction in organic chemistry, we're just worried about whether electron density was gained or lost, not full transfer of electrons. So for carbon: if it gains a bond to a less electronegative group than itself (or loses a bond to a more electronegative one), those electrons count towards its oxidation state. If it gains a bond to a more electronegative group, then they don't count.(25 votes)
- Why does the alpha pinene look like it has 5 carbons in the ring?(12 votes)
- The alpha pinene molecule which is drawn at8:56doesn't show the structure very well, but if you have a look to the one appearing at10:37, it is absolutely intelligible: a 5-carbons ring and a bridge wearing two methyl groups.(7 votes)
- I'm confused.... at6:30where jay says that Carbon has been reduced as it gained an electron from Hydrogen, I thought C-H interactions were non-polar so they share electrons as oppose to carbon gaining one.... :((3 votes)
- In reality it does not mean that C-H is non-polar. Because of the small difference in electronegativities, the C−H bond is generally regarded as being non-polar. But theres still a difference in electronegativities. But it doesnt really matter in this case because concerning the oxidation state you simply look at the atom which is more electronegative, which gain more electron.(8 votes)
- At9:32, are we supposed to just know that charcoal (C) is needed with our metal (Pd), or is there some rule being followed?(3 votes)
- This is just something to know. When the Pd is spread over carbon, it greatly increases the available surface area, and thus the efficiency, of the catalyst. Thus, practically, a palladium hydrogenation is usually palladium on carbon (written as Pd/C).(7 votes)
- Does adsorb mean anything (0:34) or is it just a different pronunciation of absorb?(2 votes)
- They are different words.
Adsorb means to hold molecules of a gas as a thin film on the outside surface of a solid.
Absorb means to suck up or soak up. For example, a sponge absorbs water.(7 votes)
- Hello I am confused on why in around minute 8 the products are enantiomers . I thought for non imposable mirror images that the up- and down and front-and back directions stay the same for enantiomers. But for this video the hydrogens go from the front to the back . Can someone explain why this is ? thanks !(4 votes)
- It all depends where you put your mirror. If you pull the left hand structure towards you so that your computer screen is the mirror, then the mirror image behind the screen will be identical with the right hand structure. Note that when you do this, wedges turn to dashed lines and dashed lines turn into wedges.
Alternatively, you could rotate the right hand structure 180° about an axis through the top and bottom carbon atoms. Then the right hand structure is a mirror image of the left hand structure if you place the mirror perpendicular to the screen and do a right/left reflection. Note that when you do this, wedges remain wedges and dashed lines remain dashed lines.(3 votes)
- What is exactly syn addition of hydrogens(4 votes)
- If you're doing an addition reaction like hydrogenation, then you're adding something to a C=C double bond like an alkene (or triple bond too with an alkyne). Alkene carbons are sp2 hybridized and have trigonal planar geometry meaning they form a flat plane where the atoms exist. When the hydrogens are added to turn the alkene into an alkane, they can both add from the same side of the alkene plane (top or bottom). Adding from the same side is referred to as syn addition.
We can also have additions reactions where one group adds from one side of the plane, and the other groups adds to the other side of the plane. This is referred to as anti addition.
Hope that helps.(2 votes)
- Can you tell me why metals will be flat ?? why not some other shape(3 votes)
- The metal will be flat because the process literally involves performing the reaction on the surface of the metal catalyst. The diatomic hydrogen (H2) interact with the surface of the catalyst and become adsorbed - resulting in individual hydrogen atoms at the surface of the metal catalyst. These individual hydrogens, which now stick out of the surface, are ready for addition.(3 votes)
- Hello! I was wondering if the two molecules at8:32were actually enantiomers because if you rotate the molecule on the left downward a full 180 degrees, you achieve the molecule on the right. Maybe I am not understanding the concept completely...(3 votes)
- The top group is an ethyl, the bottom group is a methyl. So if you flipped them 180 degrees you wouldn't end up with the same looking molecule. If they were both the same they would be the same molecule.(3 votes)
To hydrogenate an alkene, you need hydrogen gas and a metal catalyst, something like platinum or palladium or nickel. And there are many others, but these are the ones most commonly used. So what happens is those two hydrogens from the hydrogen gas are added across their double bond. And they're added on the same side of where the double bond used to be. So it's a syn addition. Let's take a look at why this is a syn addition of hydrogens. So we have our metal catalyst over here. So let's go ahead and draw our flat metal catalyst. And these metals are chosen, because they adsorb hydrogen really well, which means that if you bubble hydrogen gas through, the hydrogen is going to be adsorbed to the surface of that metal catalyst, like that. And then your alkene comes along. And your alkene is also flat, right? The portion of the molecule that contains the double bond, right? So these two carbons, this carbon and this carbon, these sp2 hybridized, which means that the stereo chemistry around those two carbons is going to be flat. So this portion of the molecule is flat. So you have one thing that's flat approaching something else that's flat. So the only way those hydrogens can add are to add them on to the same side, right? So if this carbon and this carbon, if you add this hydrogen to the carbon on the left and add this hydrogen to the carbon on the right, and then you go ahead and you draw the rest of the bonds, right? This would now be a wedge and then a dash, and then this would be a wedge and then a dash. You can see those two hydrogens have added on to the same side. So these two hydrogens or these two hydrogens for our syn addition. Notice we're also changing from sp2 hybridization to sp3 hybridization over here on the right. So we have to think about stereochemistry for this reaction for your products as well. So let's take a look at an actual reaction here. And let's see if we can follow along. So if this was my reaction, I want to hydrogenate this alkene. So I would add some hydrogen gas and I could choose whichever metal catalyst I wanted to. I would add two hydrogens on the same side, right? So I could add two hydrogens on the same side. So just like I did up there. So we would get now everything changes from sp2 hybridization to sp3. So we have wedges and dashes to worry about. And usually you wouldn't see it drawn like this. That's too much work, quite frankly. It would be much easier just to say, well, all I have to do is take away the double bond, and there's my product. So for some of these reactions, they're very, very, very simple. Just take away the double bond and you'll end up with your alkene-like product. Let's take a look at oxidation states for this reaction. So I'm going to redraw this reaction. And this time I'm going to draw in my atoms. And I'm also going to draw in my electrons here in a second. So I'm just drawing out all the atoms here. So I have all these methyl groups to worry about. And then I have electrons in these bonds, right? Each one of these bonds consists of two electrons. So I'm going to go ahead and put in all of my electrons here, like that. Now, let's assign oxidation states to those two carbons that formed our double bond, right? So let's look at oxidation state for the top carbon. Remember, when you're doing oxidation states, you're worried about electronegativity. So oxidation states are all about electronegativity. So go back and watch the earlier video on oxidation states. So we have here comparing the electronegativities of carbon and carbon. Well, obviously they're the exact same. So in the struggle for these electrons, it gets divided, right? Each one of these carbons is going to get one of these electrons. So that's the case for all of these right here. So that carbon has four electrons around it. It's the exact same thing for this carbon, right? This carbon has four electrons around it. To assign an oxidation state, we take the number of valence electrons that atom usually has, which carbon normally has four, of course. From that we subtract the number of electrons we just drew around it for our dot structure. So that would be four. Each one of those carbons has four. So each of these carbons has an oxidation state of zero. Let's look at the product, and let's see if we can assign some oxidation states for the product. So our product over here on the right, we had a carbon and we had some methyl groups bonded to that carbon. We added on a hydrogen. And so each one of these carbons got a hydrogen added onto it. And let's go ahead and fill in our electrons in these bonds. So once again, each bond consists of two electrons, like that. And now we have a single bond between our carbons. And let's assign some oxidation states. So once again we know that the two carbons have the same electronegativity, right? So the tug of war for these two electrons right here, it's a tie. So it's a tie, it's a tie. What about carbon versus hydrogen? Carbon is actually more electronegative than hydrogen. So in the war over the two electrons in the carbon-hydrogen bond, carbon wins, because it's a little bit more electronegative. So we're going to assign this extra electron here to carbon. And then again, carbon versus carbon. So that carbon gets that electron as well. Same thing down here, right? So it's a tie, it's a tie. Carbon beats hydrogen. And over here, it's a tie. So in the dot structure on the right, the oxidation states that the normal number of valence electrons would be four. From that we subtract the number of electrons in our picture here, which would be five electrons. Each one of these carbons has five electrons around it. So it gained electron. And it's a 4 minus 5 will give us a negative 1. So the oxidation states of these two carbons is negative 1. And we can look at our original oxidation states of being zero, went from zero to negative 1. That's a decrease in the oxidation state, right? A decrease in the oxidation state means a reduction. So this is a reduction reaction. So the alkene is reduced by the addition of these two hydrogens. And you'll see other definitions for oxidation states. You'll see a gain in hydrogens is reduction. That's another definition that's often found in organic chemistry textbooks. And while that's true, to me it makes more sense to go ahead and assign your oxidation states and watch the oxidation states change as you add those hydrogens, as your molecule gains hydrogens. So this is a reduction. Let's look at the stereochemistry of the hydrogenation reaction. So let's do an example involving stereochemistry. So let's say your alkene-- let's do that ring again, it wasn't a very good one-- so let's say your alkene looked something like this. And you're going to react that with hydrogen and with platinum. All right, well, your first thought might be, OK, this is simple. All I have to do is take away that double bond and I'm done. Well sometimes that's true. But in this case, we actually formed two new chirality centers, right? So this top carbon here is a chirality center, and this bottom carbon here is also a chirality center. So sometimes it's not quite that simple. We need to think about the syn addition of those hydrogens when you think about the possible products that would result. So we're going to get two products here. Let's look at the one on the left. Well, one possibility is I can add those two hydrogens on the same side as a wedge, right? So I have one hydrogen as a wedge, the other hydrogen as a wedge. That's our syn addition. And that means that this top carbon here, this ethyl group, must be going away from me. And down here at the bottom carbon, the methyl group must be going away from me. So that's one possible product. The other possibility, instead of having my two hydrogens as wedges, I can have my two hydrogens add as dashes. So there's a hydrogen and then here's a dash, and there's a hydrogen. So at this top carbon here, now my ethyl group is coming out at me. And at this bottom carbon now my methyl group is coming out at me, like this. So I have two possibilities. And if I look at these two products, I can see that they are enantiomers, they are mirror images of each other. So these two would be my enantiomers, and these would be the products of my reaction. So be very careful when thinking about syn additions here. Let's do one more example of a hydrogenation reaction. Let's do a bridged bicyclic compound. So let's look at a famous bridged bicyclic compound. Let's see if we can draw it here. And then I'm going to draw that back carbon a little bit off like that. And my double bond is going to go right here. And then this is going to be a methyl group. And then up here there are going to be two methyl groups, like that. So this is alpha-pinene, found in turpentine. And you can see there's an alkene on this. So if I took this alpha-pinene molecule and I wanted to hydrogenate it, I could use palladium and charcoal, palladium and carbon. And if I think about what happens in this mechanism, I know that my metal catalyst there, my palladium, is going to be flat, like that. And so, when it has those hydrogens, when the palladium adsorbs those hydrogens, it's going to add those two hydrogens to my double bond, think about this guy over here, think about the alpha-pinene as molecules like a spaceship, right? And the spaceship is approaching the docking station. So the spaceship is slowly going down. The spaceship is going to approach the docking station. And there's only one way the spaceship can approach the docking station. And that is the way in which we have drawn it right here. It could not flip upside down and approach it from the top, because of the steric hindrance of these methyl groups. Right? So this is the way that it approaches. In this part of the molecule, your alkene, is the flat part, right? So it's easiest for the molecule to approach in this way. The spaceship analogy always helps my students. So there's only one product for this reaction. And let's see if we can draw it here. And let's see what it would look like. It would look something like this. So we have our two methyl groups right here. So the hydrogens are going to add from below, right? So this hydrogen, let's say it adds right here. That's going to push this methyl group up. So that methyl group gets pushed up when that hydrogen adds right down here. And then this other hydrogen is going to add to the opposite side. And so we can show the addition of that hydrogen. So there's my syn addition of these two hydrogens. And there was something else in that carbon. It was another hydrogen. So another hydrogen got pushed up right here as well. So that is your only product, the only product of this reaction. The hydrogenation reaction is very sensitive to steric conditions.