Understanding Alkene Reactions and Mechanisms

Hello organic chemistry students! In this video we're going to cover the reactions of alkenes. Now we already started talking about this concept in the previous video in which carbocations were introduced. Let's go ahead and review that ever so slightly. In the last video we talked about how a double bond, which is a nucleophile, can attack a hydrogen ion and if this hydrogen goes onto this carbon here, C3, you form this carbocation right here because now there's one hydrogen. Let's go ahead and show that hydrogen. So here there's one hydrogen I'll show that blue hydrogen. Show that blue hydrogen right there. When the reaction is done, there is the white hydrogen that added onto that carbon. Therefore, the other carbon becomes the carbocation. In the opposite case, the proton adds on to C2, as we're seeing right here, and we form our secondary carbocation. Now of the two, we know tertiary is more stable than secondary because of hyperconjugation and inductive effects. Yes. Let's go into this in a little bit more detail. I know it seems a little weird saying that double bond seems like it has a choice on if it's going to attack left or right. It really doesn't. What's actually happening here is that this double bond is attracting the hydrogen ion or the electrophile to the pi orbitals of the double bond. And we formed this system right here. Let me go ahead and put a positive charge on that hydrogen there. Let's go ahead and put it inside the orbital. Actually, no, we're not going to put that in. I'm so sorry. Let me take that out right there. The whole system has a partial positive charge right there. So now, what's happening here is that the hydrogen atom's s orbital is interacting with the pi orbital. So there's one electron being shared by this s and p, one electron here. That's too many, right? Oxygen can't, I'm sorry, hydrogen cannot have two bonds. It can only have one. This is where the decision needs to be made. Does the hydrogen and the electrons move to carbon too? to form this species here, or does the hydrogen and this electron move to carbon 3 and form this tertiary? So what's going to dictate that is which carbon is more stable with the positive charge. So if this electron starts to move to this orbital, this carbon is going to start to become partial positive, and it's tertiary. This one, if we lose this electron to this p orbital, this carbon will start to become partial positive, a secondary. And that's how we get our tertiary and secondaries. The tertiary occurs more frequently because it's a greater degree of stabilization by this carbon being a tertiary system. Now, new concept on top of this since we're talking about alkene reactions. Once we've formed these carbocations, we took this alkene, which is a nucleophile, created a carbocation intermediate, which is an electrophile. A nucleophile can now attack this carbocation or this carbocation. And don't forget carbocations are sp2 hybridized with a completely open p orbital ready for attack. So when the nucleophile attacks that carbocation we form this species here. When the nucleophile attacks this carbocation we form this species right here with a squiggly line. We're going to talk about that squiggly line here in a little bit. Don't worry about it right now. Now the first big question I want to ask is in this starting material right here, was there a specific region of the molecule that underwent chemistry? or could any area undergo chemistry? Meaning the carbon-hydrogen bond here, or this carbon-carbon bond, this one or this one? And the answer is no. Only this area of the molecule underwent any type of chemical transformation, and that is termed regioselectivity. The portion of the molecule that can undergo a reaction. Portion that undergoes a reaction. So not that the entire molecule can't do it, it's only part of it can. Now the next part is going to seem a little weird. If we think back to when we were doing Newman projections, we were talking about the left side, the right side, and then chair conformations, the top and the back side. That's facial selectivity. When you have this carbocation right here, let's go ahead and show that intermediate. I'm going to show the p-orbital on it. That p-orbital can be attacked from the top face or it can be attacked from the back face. Either or is 100% possible. We don't have to do one versus the other. Oops, sorry, just trying to move this part which I can't seem to do. Oh, I'm making more of a mess. Let me just take this back to where it was. Okay, nucleophile right there. So while we can attack the top face or the back face of a p orbital, that's implying facial non-selectivity. So if you can attack both faces of a molecule, there is no facial selectivity. But if one face can be attacked more than the other side, that's what we mean by facial selectivity. Attack on one side of the molecule, attack on one side, one face, more than the other. I was trying to move that other one so we can have more room right there. I apologize. And that is facial selectivity. All right. With this basic scheme right here, we've actually, believe it or not, covered the basics of alkene chemistry. Now we have to go through some specific alkene reaction types. Let's go ahead and start that on the next slide. The first one that we're going to talk about is one that is called hydrohalogenation. So hydrohalogenation is going to take an HX and add it across the double bond. So here's this double bond right here. I'm going to take HX and I'm going to go ahead and show you what the products are and we're going to walk through how we get there. So give me one moment to write them all down and these two right here we can abbreviate with that squiggly line. I know that seems weird. We're going to get there. Let's push through the mechanism. The carbon-carbon double bond system right here, it can attack this H from the HX because hydroacids are acidic. So what we are going to see here in red is the double bond attacks the proton, and we cleave that sigma bond, just like we saw in the previous slide. Nothing is different. So we form a tertiary carbocation, and we form a secondary carbocation. Now of the two, which one do we form more of, the tertiary or the secondary? And the answer is the tertiary because the tertiary carbocation has more inductive and more hyperconjugation to stabilize itself. Why do we care about that? That's going to make a huge difference on our end products. If we form more of this intermediates, we have to form more of the product that corresponds to it. Now enter in the nucleophile X minus. X minus will come in, attack this carbocation from the top and back face. Attack this carbocation from the top and back face and we form these three compounds. That seems a little weird. I said that we're attacking the top and back face twice in each intermediate, didn't I? So we should form four compounds. What's different here? If we look at this compound right here and specifically this carbon that was attacked, this came from this carbon right here, how many different groups are on that carbon? We have the halogen, we have a carbon, and we have the isopropyl. the whole group's attached once again. So if we have the halogen, a hydrogen going in the back, a methyl, and the isopropyl, there are four different groups on that carbon. Whenever you have four different groups on a tetrahedral sensor, we must show stereochemistry. If it's coming out of the plane or going into the plane. This one attacked coming out of the plane, this one right here attacked going into the plane. And that's how we come up with these two structures that we abbreviate here. Why abbreviate? We'll get into that in a moment. But over here, we attack the top and back face. Why don't we show two? If we look at this carbon again, we have a halogen, we have an ethyl group right here, a methyl, and a methyl. Are those four totally different groups, or do we have a repeating group? We have a repeating group. We have to have four totally different groups or substituents attached to that carbon. So ignore stereochemistry. So because we have these two methyls right here, we ignore the stereochemistry, and that's why we just show it as a standard line. Now, when this double bond attacks this proton right here, we form more of this major and some of this minor. When the halogen attacks this carbocation from the top or back face, what do you think the ratio of that is? And you're right, it's 50-50. So we form 50% of this, 50% of this. So a way that we can abbreviate this to save us some time is that we draw the squiggly line. This squiggly line right here means that we formed 50-50 of the two different stereochemical outcomes coming out of the plane and going into the plane. So I'll draw the wedge and the dashed line right there. 50-50. Not the percentage. It's just saying that both of them form in an equal amount. Very important. Now, in this reaction, if I call this 1, 2, and 3, I would like you to rank these in terms of major to minor products. How do we do that? Look at the intermediate of the reaction. This intermediate right here is our most major. This is minor because of carbocation stability. So this major one is going to form our major product one since it comes from it and then two and three are going to be formed in equal amounts because they come from the same intermediate and it's just attacked from the top and back face of the carbocation. And believe it or not that is our first type of reaction. We can do this with HCl, HBr. H-I, not H-F though. H-F doesn't participate. It's a horrible, horrible nucleophile. So, that is hydrohalogenation. Now, let me go ahead and... leave you with. I'm going to give you things to think about so you can practice. And then if you come to the discussion section, we can talk about it. Actually, I'm going to take that out and I'm going to put this group right here. And I'm going to write HBR. This is going to be a more challenging problem. We have a tert-butyl group on this ring. So any attack that comes from the top face has to get past the tert-butyl group, doesn't it? So now, could we bias one side versus the other in terms of facial selectivity? Yes, we can. Go ahead and try this problem. If you want to email me your solution to it, I'll be happy to look at it. Or ask me to go over in the discussion section, and I'm happy to show you what the answers are to this problem right here. It's definitely more of a harder type of problem. Now, that is hydrohalogenation. Let's get into the next type of reaction. The next one is called hydration. Oops and that should be in yellow. Not that it really matters, but just trying to keep the same format. So hydration. We're going to keep the same starting material like I said before just so we can keep that consistent. And we're going to react this with with H plus and H2O. Or we could just say the hydronium ion itself because that means the same exact thing. One H plus sitting there with water. Now We have to rank our nucleophiles. The double bond is a nucleophile. Water is a nucleophile because of those lone pairs. Hydrogen is an electrophile. Of the double bond versus oxygen, which one is the stronger nucleophile? And it's going to be the double bond. Because remember, oxygen is the second strongest electronegative atom. It doesn't necessarily want to react and give away electron density. It will only do so if it needs to. So the double bond is our stronger nucleophile here, so it attacks the proton. Wait, that sounds familiar. When we attack that proton, we end up forming this tertiary carbocation and this secondary carbocation. Wait a second, this sounds just like hydrohalogenation. And it is. There's just one small new step in it that we're going to talk about. So now, our water is sitting in solution with lone pairs. Now it has to attack the carbocations. The carbocations are stronger electrophiles. and it's going to draw it into a reaction. Previously, we were trying to get water to attack H+. It doesn't need to be attacked, and water doesn't want to attack. Down below, these carbocations need electrons to stabilize them, so it draws it into the reaction, kicking and screaming a little bit. So now, what do we form? Another intermediary set of compounds. We're not done yet. And why are we not done? Once that water attacked, we formed a positive charge. on that oxygen. So let's do a little squiggly line here and the positive charge. What has to happen? Hydrogen will give electrons back to oxygen, one of those hydrogens, doesn't matter which one, to stabilize that positive charge. And what we end up forming is this tertiary alcohol and a mixture of these two secondary alcohols in an equal amount. Now as a whole, was this really that different? compared to the hydrohalogenation reaction? And the answer is no, it really wasn't. It's the same exact thing minus this last step where we have a proton leave the molecule. That's the only thing that's different there. All right, so now, hydration, hydrohalogenation, pretty much the same type of reactions. What I'd like to do is take this hydration reaction, and we're going to give you another problem to try here. Let's see if I can actually move this up. I've been having issues with it. There we go. I'm going to go ahead and give you a six-membered ring. I'm not going to put a tert butyl group on it. I'm just going to give you this carbon-carbon double bond system. And the reagents are going to be H plus and ethanol. Wait, that looks totally different from up above. We're going to find it's the same thing. The double bond will attack the proton, and we are going to form as intermediates a secondary carbocation. Now if I show the secondary carbocation on this carbon or this one, aren't they the same? They are. There's nothing else on the ring making this these different molecules. So here I just get to show it once and call it good. Now this R group, the two carbon chain, and the OH group has lone pairs which can attack that carbocation. We attack from the top and back face but I'm only going to draw one molecule. I know you might be saying, wait a second, why? If we look at it, this carbon has a hydrogen not being shown, this protonated oxygen and ethyl groups, that's two different groups, and a CH2CH2CH2CH2CH2. Okay, and then down below is a CH2CH2CH2CH2 and, ah, CH2. They're identical. We don't have to show stereochemistry here because they're the same. The hydrogen then leaves, giving electrons back to oxygen, and we have just formed our ether type of molecule right here. Now, if I shrink this down, when we look at the reaction mechanism, Isn't it identical? It is. The only thing that's different here is instead of using water, I'm using an alcohol for the reaction. The important concept here is that the oxygen, the sulfur, the nitrogen, any of those species will do this hydration-like reaction. Must have a hydrogen on it. Why is that so important? This group right here can be sulfur, nitrogen, oxygen, with those lone pairs of electrons. Has to have a lone pair on it too. But why does it have to have a hydrogen ion? If there is no hydrogen ion on this oxygen, once we get to this step, can we lose a hydrogen and stabilize the molecule? No, and the reaction would reverse itself back to the starting materials. So we must absolutely have an atom that has lone pairs of electrons to do a hydration reaction, and we must absolutely have a hydrogen on that atom so we can lose it to stabilize the cationic charge. And that is hydration reactions. The next type of reactions that we are going to cover is diatomic halogenations. Now, it's more than likely that you will watch this and go, what on earth was he just talking about? And that happens a lot. Go back and watch it one more time. It will make more sense the more you practice it. So in a diatomic halogenation reaction, I'm going to go ahead and show our standard double bond system. What we're going to form here, and actually I'm going to erase this for a second. Oops, maybe if I can. Sometimes this program is iffy. Okay, I'm going to switch it up. We're actually going to put in a six-membered ring and just a double bond like that. In the diatomic halogenation HX, I'm sorry, X squared, we're going to use this on chlorine, bromine, or iodine. And what we are going to form here is one halogen going up, one halogen going down. That's it. Nothing else. Now, this is unique because there's no other substituents on this molecule. So, let's go ahead and come back here, and I'm going to put a methyl group on it. Now, because of that methyl group, don't worry, we're going to push through the mechanism on this. The methyl group remains going up, but the X could be going down, and now this X could be coming up. The important thing is that these two halogens have to be trans to one another. This is going to be the first reaction that we see that's not governed by carbocation intermediates. It's governed by a trans mechanism. And let's go ahead and get right into it so we can see this. When we look at x squared, they're both the same halogen, right? Now, if there are mixed halogens, this would be a lot easier because if it was mixed, we look at the most electronegative and we design it partial positive, the other one partial negative. We don't have that here. But if we remember back to Gen Chem, Don't we have temporary dipole moments, those London dispersions? We do. So for a temporary second in time, that carbon or that chlorine or bromine or iodine can be partial negative, and this one can be partial positive. When this happens, does this become an electrophile? It does. The double bond can attack it. But let's not forget, these halogens also have lone pairs, and they are absolutely huge species. So, So, As the double bond attacks that halogen, we cleave the halogen bond. That looks just like hydration and hydrohalogenation. Same thing. But the unique part here is that halogen that's being attacked will also attack the double bond. So the carbocation that we start forming in the double bond is immediately attacked by the lone pairs in the halogen. Because it's so large, it's sitting on top of that carbon-carbon double bond system. So what do we end up getting here? The halogen could have attacked from the back face, at which point it's coming into the plane and now it's positively charged because it used one of its lone pairs, or it could be coming from out of the plane being attacked from the top face of this molecule. So going over this again, the carbon-carbon double bond attacks the halogen, we cleave the halogen bond, the sigma bond, And the halogen that was attacked, one of its lone pairs, attacks the carbon-carbon double bond, specifically the carbon that's starting to become a carbocation. And what we form here is this three-membered ring going towards the same side, either down or up, just like that. Now, we have x minus floating in solution. x minus now needs to come in and attack. either this carbon right here or this carbon right here. They're both secondary carbons, so there's no difference whatsoever. If there was a tertiary versus secondary, there's absolutely a difference, and your textbook illustrates this very nicely. Remember, our video is to give you the good, firm foundational knowledge so when you're reading the textbook, it makes a little bit more sense. If you have more questions on this, always send an email, come to the discussion session. or come to office hours. Please, please, please. Now, when this X minus attacks this carbon, is it going to attack the same side as this X plus, or will it attack the opposite side? Keeping in mind that this bond already has two electrons going down, and two electrons is the maximum occupancy. That means that this X minus has to attack from the opposite side of this group right here. When this happens, we break that carbon halogen bond and what we have just formed is, and let me do it this way from this arrow, that compound right there. If we attack on the opposite side, we will be forming this compound right here, one of each. Remember, keep that in mind. Now, let's do the same thing on the opposite side. Here is our X minus. We can attack this lower carbon. We can attack, oops, or we can attack this top carbon right here. As this happens, we will be breaking the carbon halogen bond in both cases. The lower one, when we attack it, we will be forming this intermediate, or this compound right here, and the top one will form this one right here. So what we notice here is that while we have four different products being formed, they coalesce into the same exact molecules. So that's why I just drew those two at the top. Now, a million dollar question that I know several of you are asking. Why did it not attack the X plus? X plus is positive, right? It wants electrons, so X minus should attack it. Remember, there's still three lone pairs on this X, and it donated one to this bond. That's why it's positive. There's still an abundance of electrons around that halogen. It's not deficient of electrons. We can't add more in there. And that's why we attack the carbon that the halogen is attached to. The next thing. Out of the two intermediates that I've shown in red, this intermediate here, ignoring all the arrows, and this intermediate here, which intermediate do we form more of, do you think? Now, carbon-carbon double bonds have no facial selectivity, and I want to write that down. Carbon-carbon double bonds have no facial selectivity in their own rights. So just by themselves, there's no selectivity. But Isn't there a methyl group right here on the carbon directly attached? It sure is. Now, double bonds have no facial selectivity, but won't this methyl group kind of block one face versus the other? It will. So out of these two intermediates, intermediate one or intermediate two, which one are we going to form more of? And it's going to be this first one right here, where the halogen is attacked from the back side, far away from the methyl group. Now, once we attack this, We can form these two compounds, but we saw that both of these arrows right here give us both products. So how do we know which one is major, which one is minor? We've established which one is the major intermediate. Let's look at the two attacks. This X- is attacking the top side of the molecule. This X- is attacking the top side of the molecule, right next to the methyl group. If this X- is approaching this carbon, isn't that methyl group going to... impede it getting to that carbon? It is. So as a whole, the major attack pathway is going to be via this lower carbon right here that I'm trying to highlight in red. So when we look at that, what is our major product of this reaction? This one right here. It's the most major because it's coming from our most major intermediate. And then the carbon that we're selecting is the major one because it's the one that is sterically less hindered. That's the important concept right there. And that is diatomic halogenation and how it works. Wonderful, wonderful, powerful reaction. Once again, highly recommend go back and watch this one more time, maybe even twice, to get over the concepts of it. We're going to do a variant of this one, just like when we had hydration, we had that variant with the alcohols. Move this up. Let me go ahead and give you this six-membered ring, and this time I'm going to give you a different type of double bond. We clearly have a tertiary versus a secondary carbon. I'm going to throw in Br2, which is going to be diatomic halogenation, but now I'm going to throw in, actually let me do it this way, there it is, HRHOR, where R can be a hydrogen or a carbon chain. The oxygen could also be sulfur or nitrogen. But once again, you have to have a hydrogen on this atom so it can leave. What happens here? In this reaction, we'll get rid of that little Br2 and just draw it out like this. The carbon-carbon double bond is going to attack one of the bromines that temporarily builds up a partial positive charge. That same halogen will attack the carbon-carbon double bond. There's no methyl group, no other group causing steric bias. So this bromine is going down. positively charged and then this bromine over here is coming out of the plane with a positive charge. I'm forgetting something. What about the methyl group? When this Br is attacking from the back side the methyl group is going to get pushed out of the plane in the opposite direction. When the Br2 is coming from the top face to form the second intermediate the methyl group is going to get pushed into the plane because we're forming a tetra hydrahedral center and that's a very important concept in these reaction pathways. Now, oops, positive charge, I'll circle that. We now have Br- sitting in solution or HOR. Which one's the better nucleophile? A halogen with a negative charge or a neutral oxygen? And keep in mind, halogens being so large with their valence shells filled are completely happy. They don't want to do much of anything unless they're dragged into it. Here, the oxygen is going to be more reactive. We can attack this lower pathway here that I'll call 1, or we can attack this top pathway that I will call 2. I'm going to pause for one moment and get a drink. All right, I'm back. I know it was instantaneous. So we have two pathways that we can follow, pathway 1 or pathway 2. Before I get there, out of the two intermediates, which one is more major, if either? And the answer is they're both formed in an equal amount. They're both equal because carbon-carbon double bonds have no facial selectivity, and there's nothing on this molecule to cause a bias. So we form 50% of this one, 50% of this one. Let's look at our reaction now. This oxygen can attack this lower carbon, break open this bond, and will form this product right here. I'm going to write something over the arrow that I'll get to here in one moment. So that's going to happen. Here's the Br. There's the O and the R and that's if pathway one occurs. For pathway two we attack the opposite side. We'll draw it down here. Hopefully I don't get this too scrunched in. The bromine will be on the back face. The O R is on the top face and when the oxygen attacked it pushed the methyl group into the back face. So as it's approaching this carbon right here to break that bond We push the methyl group down so we can attack from the opposite side of this carbon-bromine bond to form this system right here called compound 2. Now, where did the hydrogen go on the oxygen? We know it has to be lost in this reaction, right? And that's what we call a proton transfer, PT. If you're ever just removing a hydrogen ion in the mechanism, you can write PT instead of showing that step. It's totally up to you though. So now, These are our two products from this intermediate right here. The question I have for you is which is major and which is minor? Uh-oh, that sounds horrible. Let's go back and look at this. Before, we looked at major and minor bicarbocatines. We do not have a carbocation here. This bromine is positive, so it's pulling electrons from this bond and this bond. So this is going to start to become partial positive, and this lower one is going to start to become partial positive. Now, if you have a partial positive carbon that's on a tertiary versus a partial positive on a secondary, which one do you think will be more stable? Ah, just like carbocations, right? The tertiary. So we're going to see pathway two being our major path because we formed something that looks more like a tertiary carbocation. And that's why we get major this one, minor this lower one right here for these two products. Let's go ahead and see what happens in the other four compounds. So we're going to follow the same thing. We'll attack this lower one as pathway three. Top one as pathway four. Going from pathway three, we attack from the back side. So the OR is going down. Here's the methyl group and there is the bromine coming out of the plane. That's from three. Now we attack via pathway four. We attack, the methyl group gets pushed up and the OR is down and the BR is coming out of the plane. So if you notice, The bromine doesn't change the side that it's on. It doesn't. It switches the carbon it's located, but if it started down, it stays down. If it started up, it stays up. So now, if we look at these, which is going to be the major pathway? This first one forming compound three, or this second one forming compound four? And it's going to be this lower one, right? Because that is the tertiary carbon, more stable with a partial positive charge, therefore it will generate more partial positive, and we can attack it more easily. So when we look at this, we have now just determined our major products and our minor products. So I should technically write minor in between both of them. Both of them are minor in equal amounts. Both of them are major in equal amounts. But wait, why? Because the intermediates of this reaction right here, aren't they both formed in equal amounts, 50-50? Yes, they are. And because they're formed in equal amounts, that's why we get... this type of product distribution. And this is the type, this is called halohydrin formation. So halohydrin formation is just a variant of diatomic halogenation. Halohydrin formation. But as a whole, if I shrink this down, isn't the mechanism the same here? It is. It's the same exact mechanism in both cases. We have the double bond attack one of the halogens. The halogen being attacked attacks the double bond, and we cleave the sigma bond up above. We do the same thing down here. We form two intermediates that has the halogen coming into the plane or out of the plane, into the plane, out of the plane. Then, up above, the X- attacks this lower carbon more than the top one because of the methyl group. Here we have two secondary carbons, so carbocation-like doesn't matter. If this was tertiary, though, that changes things. And that gets a little bit more complicated. We're not going to dive into that much detail in this class, since I just want to give you the basics that we need so we can talk about biological chemistry. But there is a lot more here that we could talk about. But as a whole, we have the same exact reaction pathway between diatomic halogenation and halohydrin formation. Alright, the next one that we are going to discuss is one of the ones that most students hate. And I hated it as well when I first was learning this information, until I actually went back and I realized, oh, if I just learn the why, this makes sense. So I'm going to start off with our standard carbon-carbon double bond, and I'm going to treat it with H plus and H2O. And what are we going to form right here? This is hydration conditions. I know that. This is not hydroboration. I'm just showing this right here. So here's this. So under these conditions right here, which is hydration conditions, this would be major and this would be minor. That's wonderful. If we treat this under BH3 and hydrogen peroxide OH-, this will be minor. and this will be major. We have to be switching the regioselectivity of the reaction, and that's what we're going to do here. Up until this point, the first step was to have our double bond attack an H plus or our double bond to attack a halogen. We don't have that here. We have BH3. That's reagent one. So let's go ahead and push this up and get into this reaction. So there's that double bond. Here is the boron. Here are the three hydrogens on it, and it has a 100% open p orbital. Boron violates the octet rule. It has an open p orbital. So as it attacks... that open p-orbital, the double bond, one of those carbons starts to become a carbocation. Now, once the double bond attacks the boron, it starts to become negative. It doesn't want to be negative. It violates the octet rule purposely, which means it needs to get rid of one of the hydrogens and the electrons in its bond. So as the double bond attacks the open p-orbital, one of the hydrogens and the electrons attack the double bond. Wait, wait, wait. Doesn't this almost sound like diatomic halogenation? It does. Just instead of one atom doing the attack and being attacked, it's going to be one atom being attacked, the other one attacking. And what we get here is the boron added onto this carbon. It can add from the top or back face right there, or we can get it added like this. Wonderful. Now, Actually, let me take a quick second for one moment, and I'm just going to erase that boron here and here. You'll see why in a moment. I'm going to make the boron right here and put the boron with the squiggly line right there. How did this happen? Did the double bond not attack the open p orbital? It did. And while it's attacking it, the electrons in one of the hydrogen-boron bonds will attack the carbon of the double bond at... the same time. This is what's called a concerted reaction, meaning all of these steps are happening at the same time. Now, if they're happening at the same time, do you think the boron that's being attacked and the hydrogen that's being added onto will be added onto the same side or opposite side? And the answer is it will be the same side. So a concerted process will always give you a syn reaction pathway, or what we call a, oops, ah. And of course I can't erase it. Oh, using the wrong one. There it is. We can use a syn reaction pathway or we form a cis product between the boron and the hydrogen that was added in. Let's go over that one more time. As the double bond attacks the open p orbital in the boron, the electrons in the hydrogen-boron bond attack the opposite carbon of the double bond at the same exact time. So if the boron added from the top face, the hydrogen adds from the top face. If the boron adds from the back face, the hydrogen adds to the back face. And we form these two products right here. Now, you're going to get to know that I hate having black box mechanisms. But I'm going to have you have one mechanism right now that's a black box. We do not need to know the chemistry here. If you want to know this chemistry, ask me in the discussion section or over email, and I will show it to you. happily, but I don't want to put it in a video where it's testable material. So we're not going to go through the mechanism for the oxidation. I'm going to give you the basic trend. When you treat it with H2O to an OH-, the BH2 is pulled off and OH- is put in its place. And we form these two products right here. And if I shrink this down, ah, it's the same products that we see up above, but But still, this is our major, this is our minor. That seems weird. Tertiary carbocations are more stable than secondary carbocations, so this should be our major. Now what I just said is still true. Tertiary carbocations or tertiary carbocation-like systems are going to be more stable than secondaries and primaries and stuff like that. Let's look at this reaction a little bit more. As the double bond attacks the p-orbital, I know you're like, oh dear gosh, not again. As it attacks the p-orbital, this carbon can build up a partial positive, or this carbon can build up a partial positive. Which one's better holding a partial positive charge, the tertiary or the secondary? And the answer is the tertiary, right? So the bond is going to form between the secondary carbon and the boron to create this partial charge here, that which the hydrogen adds in onto that carbon, forming this. as our major intermediate or major product of the first step of the hydroboration step itself. We're still governed by the carbocation-like systems. The reason why we're getting the opposite regioselectivity is because the hydrogen, in the form of a hydride, is attacking that carbon last. Very important distinction, right? Now that is hydroboration. Let's go ahead and open up another page. I know you think this is never going to end, but we are almost there. The next one that we want to talk about is called hydrogenation. A hydrogenation reaction is going to take a carbon-carbon double bond, and we're going to reduce it with hydrogen gas, and we're going to use either palladium or platinum. And what we end up forming here is an alkane. What is the role of the metal? The metal's job... The palladium or the platinum, you can never draw lead. Lead PB does not work. The palladium or the platinum, its whole job is to hold that hydrogen gas in place. a gas and it's moving around like crazy. So if the hydrogen gas is being held, it allows the double bond to attack one of those hydrogens, and then the other electrons and the hydrogen attack the opposite side of the carbon, just like we saw in hydroboration. And that is how we form our alkane from an alkene. And now with that, we are done with reactions for alkenes. There are many other reactions for alkenes. There are. I'm just covering these basic ones right here, and this is basic even for full-year organic chemistry. We don't need the more detailed ones. None of your majors require it, nor any application into biological chemistry. I know this was a long video. I highly recommend they'll go back and watching it again. Practice the problems in your textbook. Read the textbook. Come to office hours and discussion sections, and we can talk about this. I hope each of you are doing well. And I look forward to seeing you in a session sometime this semester.

In the last video we talked about how a double bond, which is a nucleophile, can attack a hydrogen ion and if this hydrogen goes onto this carbon here, C3, you form this carbocation right here because now there's one hydrogen. Let's go ahead and show that hydrogen. So here there's one hydrogen I'll show that blue hydrogen.

Show that blue hydrogen right there. When the reaction is done, there is the white hydrogen that added onto that carbon. Therefore, the other carbon becomes the carbocation.

In the opposite case, the proton adds on to C2, as we're seeing right here, and we form our secondary carbocation. Now of the two, we know tertiary is more stable than secondary because of hyperconjugation and inductive effects. Yes.

Let's go into this in a little bit more detail. I know it seems a little weird saying that double bond seems like it has a choice on if it's going to attack left or right. It really doesn't.

What's actually happening here is that this double bond is attracting the hydrogen ion or the electrophile to the pi orbitals of the double bond. And we formed this system right here. Let me go ahead and put a positive charge on that hydrogen there. Let's go ahead and put it inside the orbital. Actually, no, we're not going to put that in.

I'm so sorry. Let me take that out right there. The whole system has a partial positive charge right there. So now, what's happening here is that the hydrogen atom's s orbital is interacting with the pi orbital. So there's one electron being shared by this s and p, one electron here.

That's too many, right? Oxygen can't, I'm sorry, hydrogen cannot have two bonds. It can only have one.

This is where the decision needs to be made. Does the hydrogen and the electrons move to carbon too? to form this species here, or does the hydrogen and this electron move to carbon 3 and form this tertiary?

So what's going to dictate that is which carbon is more stable with the positive charge. So if this electron starts to move to this orbital, this carbon is going to start to become partial positive, and it's tertiary. This one, if we lose this electron to this p orbital, this carbon will start to become partial positive, a secondary. And that's how we get our tertiary and secondaries.

The tertiary occurs more frequently because it's a greater degree of stabilization by this carbon being a tertiary system. Now, new concept on top of this since we're talking about alkene reactions. Once we've formed these carbocations, we took this alkene, which is a nucleophile, created a carbocation intermediate, which is an electrophile.

A nucleophile can now attack this carbocation or this carbocation. And don't forget carbocations are sp2 hybridized with a completely open p orbital ready for attack. So when the nucleophile attacks that carbocation we form this species here.

When the nucleophile attacks this carbocation we form this species right here with a squiggly line. We're going to talk about that squiggly line here in a little bit. Don't worry about it right now. Now the first big question I want to ask is in this starting material right here, was there a specific region of the molecule that underwent chemistry?

or could any area undergo chemistry? Meaning the carbon-hydrogen bond here, or this carbon-carbon bond, this one or this one? And the answer is no.

Only this area of the molecule underwent any type of chemical transformation, and that is termed regioselectivity. The portion of the molecule that can undergo a reaction. Portion that undergoes a reaction.

So not that the entire molecule can't do it, it's only part of it can. Now the next part is going to seem a little weird. If we think back to when we were doing Newman projections, we were talking about the left side, the right side, and then chair conformations, the top and the back side.

That's facial selectivity. When you have this carbocation right here, let's go ahead and show that intermediate. I'm going to show the p-orbital on it. That p-orbital can be attacked from the top face or it can be attacked from the back face.

Either or is 100% possible. We don't have to do one versus the other. Oops, sorry, just trying to move this part which I can't seem to do. Oh, I'm making more of a mess. Let me just take this back to where it was.

Okay, nucleophile right there. So while we can attack the top face or the back face of a p orbital, that's implying facial non-selectivity. So if you can attack both faces of a molecule, there is no facial selectivity.

But if one face can be attacked more than the other side, that's what we mean by facial selectivity. Attack on one side of the molecule, attack on one side, one face, more than the other. I was trying to move that other one so we can have more room right there. I apologize.

And that is facial selectivity. All right. With this basic scheme right here, we've actually, believe it or not, covered the basics of alkene chemistry.

Now we have to go through some specific alkene reaction types. Let's go ahead and start that on the next slide. The first one that we're going to talk about is one that is called hydrohalogenation.

So hydrohalogenation is going to take an HX and add it across the double bond. So here's this double bond right here. I'm going to take HX and I'm going to go ahead and show you what the products are and we're going to walk through how we get there. So give me one moment to write them all down and these two right here we can abbreviate with that squiggly line.

I know that seems weird. We're going to get there. Let's push through the mechanism.

The carbon-carbon double bond system right here, it can attack this H from the HX because hydroacids are acidic. So what we are going to see here in red is the double bond attacks the proton, and we cleave that sigma bond, just like we saw in the previous slide. Nothing is different.

So we form a tertiary carbocation, and we form a secondary carbocation. Now of the two, which one do we form more of, the tertiary or the secondary? And the answer is the tertiary because the tertiary carbocation has more inductive and more hyperconjugation to stabilize itself.

Why do we care about that? That's going to make a huge difference on our end products. If we form more of this intermediates, we have to form more of the product that corresponds to it. Now enter in the nucleophile X minus. X minus will come in, attack this carbocation from the top and back face.

Attack this carbocation from the top and back face and we form these three compounds. That seems a little weird. I said that we're attacking the top and back face twice in each intermediate, didn't I?

So we should form four compounds. What's different here? If we look at this compound right here and specifically this carbon that was attacked, this came from this carbon right here, how many different groups are on that carbon?

We have the halogen, we have a carbon, and we have the isopropyl. the whole group's attached once again. So if we have the halogen, a hydrogen going in the back, a methyl, and the isopropyl, there are four different groups on that carbon.

Whenever you have four different groups on a tetrahedral sensor, we must show stereochemistry. If it's coming out of the plane or going into the plane. This one attacked coming out of the plane, this one right here attacked going into the plane. And that's how we come up with these two structures that we abbreviate here. Why abbreviate?

We'll get into that in a moment. But over here, we attack the top and back face. Why don't we show two?

If we look at this carbon again, we have a halogen, we have an ethyl group right here, a methyl, and a methyl. Are those four totally different groups, or do we have a repeating group? We have a repeating group. We have to have four totally different groups or substituents attached to that carbon.

So ignore stereochemistry. So because we have these two methyls right here, we ignore the stereochemistry, and that's why we just show it as a standard line. Now, when this double bond attacks this proton right here, we form more of this major and some of this minor. When the halogen attacks this carbocation from the top or back face, what do you think the ratio of that is? And you're right, it's 50-50.

So we form 50% of this, 50% of this. So a way that we can abbreviate this to save us some time is that we draw the squiggly line. This squiggly line right here means that we formed 50-50 of the two different stereochemical outcomes coming out of the plane and going into the plane.

So I'll draw the wedge and the dashed line right there. 50-50. Not the percentage. It's just saying that both of them form in an equal amount. Very important.

Now, in this reaction, if I call this 1, 2, and 3, I would like you to rank these in terms of major to minor products. How do we do that? Look at the intermediate of the reaction.

This intermediate right here is our most major. This is minor because of carbocation stability. So this major one is going to form our major product one since it comes from it and then two and three are going to be formed in equal amounts because they come from the same intermediate and it's just attacked from the top and back face of the carbocation. And believe it or not that is our first type of reaction. We can do this with HCl, HBr.

H-I, not H-F though. H-F doesn't participate. It's a horrible, horrible nucleophile. So, that is hydrohalogenation.

Now, let me go ahead and... leave you with. I'm going to give you things to think about so you can practice.

And then if you come to the discussion section, we can talk about it. Actually, I'm going to take that out and I'm going to put this group right here. And I'm going to write HBR.

This is going to be a more challenging problem. We have a tert-butyl group on this ring. So any attack that comes from the top face has to get past the tert-butyl group, doesn't it? So now, could we bias one side versus the other in terms of facial selectivity?

Yes, we can. Go ahead and try this problem. If you want to email me your solution to it, I'll be happy to look at it. Or ask me to go over in the discussion section, and I'm happy to show you what the answers are to this problem right here.

It's definitely more of a harder type of problem. Now, that is hydrohalogenation. Let's get into the next type of reaction.

The next one is called hydration. Oops and that should be in yellow. Not that it really matters, but just trying to keep the same format.

So hydration. We're going to keep the same starting material like I said before just so we can keep that consistent. And we're going to react this with with H plus and H2O. Or we could just say the hydronium ion itself because that means the same exact thing.

One H plus sitting there with water. Now We have to rank our nucleophiles. The double bond is a nucleophile. Water is a nucleophile because of those lone pairs. Hydrogen is an electrophile.

Of the double bond versus oxygen, which one is the stronger nucleophile? And it's going to be the double bond. Because remember, oxygen is the second strongest electronegative atom.

It doesn't necessarily want to react and give away electron density. It will only do so if it needs to. So the double bond is our stronger nucleophile here, so it attacks the proton. Wait, that sounds familiar. When we attack that proton, we end up forming this tertiary carbocation and this secondary carbocation.

Wait a second, this sounds just like hydrohalogenation. And it is. There's just one small new step in it that we're going to talk about.

So now, our water is sitting in solution with lone pairs. Now it has to attack the carbocations. The carbocations are stronger electrophiles.

and it's going to draw it into a reaction. Previously, we were trying to get water to attack H+. It doesn't need to be attacked, and water doesn't want to attack. Down below, these carbocations need electrons to stabilize them, so it draws it into the reaction, kicking and screaming a little bit.

So now, what do we form? Another intermediary set of compounds. We're not done yet. And why are we not done?

Once that water attacked, we formed a positive charge. on that oxygen. So let's do a little squiggly line here and the positive charge.

What has to happen? Hydrogen will give electrons back to oxygen, one of those hydrogens, doesn't matter which one, to stabilize that positive charge. And what we end up forming is this tertiary alcohol and a mixture of these two secondary alcohols in an equal amount.

Now as a whole, was this really that different? compared to the hydrohalogenation reaction? And the answer is no, it really wasn't. It's the same exact thing minus this last step where we have a proton leave the molecule. That's the only thing that's different there.

All right, so now, hydration, hydrohalogenation, pretty much the same type of reactions. What I'd like to do is take this hydration reaction, and we're going to give you another problem to try here. Let's see if I can actually move this up. I've been having issues with it.

There we go. I'm going to go ahead and give you a six-membered ring. I'm not going to put a tert butyl group on it. I'm just going to give you this carbon-carbon double bond system. And the reagents are going to be H plus and ethanol.

Wait, that looks totally different from up above. We're going to find it's the same thing. The double bond will attack the proton, and we are going to form as intermediates a secondary carbocation.

Now if I show the secondary carbocation on this carbon or this one, aren't they the same? They are. There's nothing else on the ring making this these different molecules.

So here I just get to show it once and call it good. Now this R group, the two carbon chain, and the OH group has lone pairs which can attack that carbocation. We attack from the top and back face but I'm only going to draw one molecule. I know you might be saying, wait a second, why?

If we look at it, this carbon has a hydrogen not being shown, this protonated oxygen and ethyl groups, that's two different groups, and a CH2CH2CH2CH2CH2. Okay, and then down below is a CH2CH2CH2CH2 and, ah, CH2. They're identical.

We don't have to show stereochemistry here because they're the same. The hydrogen then leaves, giving electrons back to oxygen, and we have just formed our ether type of molecule right here. Now, if I shrink this down, when we look at the reaction mechanism, Isn't it identical?

It is. The only thing that's different here is instead of using water, I'm using an alcohol for the reaction. The important concept here is that the oxygen, the sulfur, the nitrogen, any of those species will do this hydration-like reaction.

Must have a hydrogen on it. Why is that so important? This group right here can be sulfur, nitrogen, oxygen, with those lone pairs of electrons.

Has to have a lone pair on it too. But why does it have to have a hydrogen ion? If there is no hydrogen ion on this oxygen, once we get to this step, can we lose a hydrogen and stabilize the molecule?

No, and the reaction would reverse itself back to the starting materials. So we must absolutely have an atom that has lone pairs of electrons to do a hydration reaction, and we must absolutely have a hydrogen on that atom so we can lose it to stabilize the cationic charge. And that is hydration reactions. The next type of reactions that we are going to cover is diatomic halogenations. Now, it's more than likely that you will watch this and go, what on earth was he just talking about?

And that happens a lot. Go back and watch it one more time. It will make more sense the more you practice it.

So in a diatomic halogenation reaction, I'm going to go ahead and show our standard double bond system. What we're going to form here, and actually I'm going to erase this for a second. Oops, maybe if I can.

Sometimes this program is iffy. Okay, I'm going to switch it up. We're actually going to put in a six-membered ring and just a double bond like that. In the diatomic halogenation HX, I'm sorry, X squared, we're going to use this on chlorine, bromine, or iodine.

And what we are going to form here is one halogen going up, one halogen going down. That's it. Nothing else. Now, this is unique because there's no other substituents on this molecule. So, let's go ahead and come back here, and I'm going to put a methyl group on it.

Now, because of that methyl group, don't worry, we're going to push through the mechanism on this. The methyl group remains going up, but the X could be going down, and now this X could be coming up. The important thing is that these two halogens have to be trans to one another.

This is going to be the first reaction that we see that's not governed by carbocation intermediates. It's governed by a trans mechanism. And let's go ahead and get right into it so we can see this.

When we look at x squared, they're both the same halogen, right? Now, if there are mixed halogens, this would be a lot easier because if it was mixed, we look at the most electronegative and we design it partial positive, the other one partial negative. We don't have that here. But if we remember back to Gen Chem, Don't we have temporary dipole moments, those London dispersions?

We do. So for a temporary second in time, that carbon or that chlorine or bromine or iodine can be partial negative, and this one can be partial positive. When this happens, does this become an electrophile? It does. The double bond can attack it.

But let's not forget, these halogens also have lone pairs, and they are absolutely huge species. So, So, As the double bond attacks that halogen, we cleave the halogen bond. That looks just like hydration and hydrohalogenation.

Same thing. But the unique part here is that halogen that's being attacked will also attack the double bond. So the carbocation that we start forming in the double bond is immediately attacked by the lone pairs in the halogen. Because it's so large, it's sitting on top of that carbon-carbon double bond system. So what do we end up getting here?

The halogen could have attacked from the back face, at which point it's coming into the plane and now it's positively charged because it used one of its lone pairs, or it could be coming from out of the plane being attacked from the top face of this molecule. So going over this again, the carbon-carbon double bond attacks the halogen, we cleave the halogen bond, the sigma bond, And the halogen that was attacked, one of its lone pairs, attacks the carbon-carbon double bond, specifically the carbon that's starting to become a carbocation. And what we form here is this three-membered ring going towards the same side, either down or up, just like that.

Now, we have x minus floating in solution. x minus now needs to come in and attack. either this carbon right here or this carbon right here. They're both secondary carbons, so there's no difference whatsoever.

If there was a tertiary versus secondary, there's absolutely a difference, and your textbook illustrates this very nicely. Remember, our video is to give you the good, firm foundational knowledge so when you're reading the textbook, it makes a little bit more sense. If you have more questions on this, always send an email, come to the discussion session.

or come to office hours. Please, please, please. Now, when this X minus attacks this carbon, is it going to attack the same side as this X plus, or will it attack the opposite side? Keeping in mind that this bond already has two electrons going down, and two electrons is the maximum occupancy. That means that this X minus has to attack from the opposite side of this group right here.

When this happens, we break that carbon halogen bond and what we have just formed is, and let me do it this way from this arrow, that compound right there. If we attack on the opposite side, we will be forming this compound right here, one of each. Remember, keep that in mind.

Now, let's do the same thing on the opposite side. Here is our X minus. We can attack this lower carbon.

We can attack, oops, or we can attack this top carbon right here. As this happens, we will be breaking the carbon halogen bond in both cases. The lower one, when we attack it, we will be forming this intermediate, or this compound right here, and the top one will form this one right here. So what we notice here is that while we have four different products being formed, they coalesce into the same exact molecules.

So that's why I just drew those two at the top. Now, a million dollar question that I know several of you are asking. Why did it not attack the X plus? X plus is positive, right? It wants electrons, so X minus should attack it.

Remember, there's still three lone pairs on this X, and it donated one to this bond. That's why it's positive. There's still an abundance of electrons around that halogen. It's not deficient of electrons. We can't add more in there.

And that's why we attack the carbon that the halogen is attached to. The next thing. Out of the two intermediates that I've shown in red, this intermediate here, ignoring all the arrows, and this intermediate here, which intermediate do we form more of, do you think?

Now, carbon-carbon double bonds have no facial selectivity, and I want to write that down. Carbon-carbon double bonds have no facial selectivity in their own rights. So just by themselves, there's no selectivity. But Isn't there a methyl group right here on the carbon directly attached?

It sure is. Now, double bonds have no facial selectivity, but won't this methyl group kind of block one face versus the other? It will. So out of these two intermediates, intermediate one or intermediate two, which one are we going to form more of?

And it's going to be this first one right here, where the halogen is attacked from the back side, far away from the methyl group. Now, once we attack this, We can form these two compounds, but we saw that both of these arrows right here give us both products. So how do we know which one is major, which one is minor? We've established which one is the major intermediate.

Let's look at the two attacks. This X- is attacking the top side of the molecule. This X- is attacking the top side of the molecule, right next to the methyl group.

If this X- is approaching this carbon, isn't that methyl group going to... impede it getting to that carbon? It is.

So as a whole, the major attack pathway is going to be via this lower carbon right here that I'm trying to highlight in red. So when we look at that, what is our major product of this reaction? This one right here.

It's the most major because it's coming from our most major intermediate. And then the carbon that we're selecting is the major one because it's the one that is sterically less hindered. That's the important concept right there.

And that is diatomic halogenation and how it works. Wonderful, wonderful, powerful reaction. Once again, highly recommend go back and watch this one more time, maybe even twice, to get over the concepts of it.

We're going to do a variant of this one, just like when we had hydration, we had that variant with the alcohols. Move this up. Let me go ahead and give you this six-membered ring, and this time I'm going to give you a different type of double bond.

We clearly have a tertiary versus a secondary carbon. I'm going to throw in Br2, which is going to be diatomic halogenation, but now I'm going to throw in, actually let me do it this way, there it is, HRHOR, where R can be a hydrogen or a carbon chain. The oxygen could also be sulfur or nitrogen.

But once again, you have to have a hydrogen on this atom so it can leave. What happens here? In this reaction, we'll get rid of that little Br2 and just draw it out like this. The carbon-carbon double bond is going to attack one of the bromines that temporarily builds up a partial positive charge.

That same halogen will attack the carbon-carbon double bond. There's no methyl group, no other group causing steric bias. So this bromine is going down.

positively charged and then this bromine over here is coming out of the plane with a positive charge. I'm forgetting something. What about the methyl group?

When this Br is attacking from the back side the methyl group is going to get pushed out of the plane in the opposite direction. When the Br2 is coming from the top face to form the second intermediate the methyl group is going to get pushed into the plane because we're forming a tetra hydrahedral center and that's a very important concept in these reaction pathways. Now, oops, positive charge, I'll circle that. We now have Br- sitting in solution or HOR.

Which one's the better nucleophile? A halogen with a negative charge or a neutral oxygen? And keep in mind, halogens being so large with their valence shells filled are completely happy. They don't want to do much of anything unless they're dragged into it. Here, the oxygen is going to be more reactive.

We can attack this lower pathway here that I'll call 1, or we can attack this top pathway that I will call 2. I'm going to pause for one moment and get a drink. All right, I'm back. I know it was instantaneous. So we have two pathways that we can follow, pathway 1 or pathway 2. Before I get there, out of the two intermediates, which one is more major, if either?

And the answer is they're both formed in an equal amount. They're both equal because carbon-carbon double bonds have no facial selectivity, and there's nothing on this molecule to cause a bias. So we form 50% of this one, 50% of this one.

Let's look at our reaction now. This oxygen can attack this lower carbon, break open this bond, and will form this product right here. I'm going to write something over the arrow that I'll get to here in one moment.

So that's going to happen. Here's the Br. There's the O and the R and that's if pathway one occurs.

For pathway two we attack the opposite side. We'll draw it down here. Hopefully I don't get this too scrunched in. The bromine will be on the back face.

The O R is on the top face and when the oxygen attacked it pushed the methyl group into the back face. So as it's approaching this carbon right here to break that bond We push the methyl group down so we can attack from the opposite side of this carbon-bromine bond to form this system right here called compound 2. Now, where did the hydrogen go on the oxygen? We know it has to be lost in this reaction, right?

And that's what we call a proton transfer, PT. If you're ever just removing a hydrogen ion in the mechanism, you can write PT instead of showing that step. It's totally up to you though.

So now, These are our two products from this intermediate right here. The question I have for you is which is major and which is minor? Uh-oh, that sounds horrible.

Let's go back and look at this. Before, we looked at major and minor bicarbocatines. We do not have a carbocation here.

This bromine is positive, so it's pulling electrons from this bond and this bond. So this is going to start to become partial positive, and this lower one is going to start to become partial positive. Now, if you have a partial positive carbon that's on a tertiary versus a partial positive on a secondary, which one do you think will be more stable?

Ah, just like carbocations, right? The tertiary. So we're going to see pathway two being our major path because we formed something that looks more like a tertiary carbocation.

And that's why we get major this one, minor this lower one right here for these two products. Let's go ahead and see what happens in the other four compounds. So we're going to follow the same thing.

We'll attack this lower one as pathway three. Top one as pathway four. Going from pathway three, we attack from the back side. So the OR is going down.

Here's the methyl group and there is the bromine coming out of the plane. That's from three. Now we attack via pathway four. We attack, the methyl group gets pushed up and the OR is down and the BR is coming out of the plane.

So if you notice, The bromine doesn't change the side that it's on. It doesn't. It switches the carbon it's located, but if it started down, it stays down.

If it started up, it stays up. So now, if we look at these, which is going to be the major pathway? This first one forming compound three, or this second one forming compound four?

And it's going to be this lower one, right? Because that is the tertiary carbon, more stable with a partial positive charge, therefore it will generate more partial positive, and we can attack it more easily. So when we look at this, we have now just determined our major products and our minor products. So I should technically write minor in between both of them.

Both of them are minor in equal amounts. Both of them are major in equal amounts. But wait, why?

Because the intermediates of this reaction right here, aren't they both formed in equal amounts, 50-50? Yes, they are. And because they're formed in equal amounts, that's why we get...

this type of product distribution. And this is the type, this is called halohydrin formation. So halohydrin formation is just a variant of diatomic halogenation.

Halohydrin formation. But as a whole, if I shrink this down, isn't the mechanism the same here? It is. It's the same exact mechanism in both cases.

We have the double bond attack one of the halogens. The halogen being attacked attacks the double bond, and we cleave the sigma bond up above. We do the same thing down here. We form two intermediates that has the halogen coming into the plane or out of the plane, into the plane, out of the plane.

Then, up above, the X- attacks this lower carbon more than the top one because of the methyl group. Here we have two secondary carbons, so carbocation-like doesn't matter. If this was tertiary, though, that changes things. And that gets a little bit more complicated. We're not going to dive into that much detail in this class, since I just want to give you the basics that we need so we can talk about biological chemistry.

But there is a lot more here that we could talk about. But as a whole, we have the same exact reaction pathway between diatomic halogenation and halohydrin formation. Alright, the next one that we are going to discuss is one of the ones that most students hate. And I hated it as well when I first was learning this information, until I actually went back and I realized, oh, if I just learn the why, this makes sense.

So I'm going to start off with our standard carbon-carbon double bond, and I'm going to treat it with H plus and H2O. And what are we going to form right here? This is hydration conditions.

I know that. This is not hydroboration. I'm just showing this right here.

So here's this. So under these conditions right here, which is hydration conditions, this would be major and this would be minor. That's wonderful. If we treat this under BH3 and hydrogen peroxide OH-, this will be minor.

and this will be major. We have to be switching the regioselectivity of the reaction, and that's what we're going to do here. Up until this point, the first step was to have our double bond attack an H plus or our double bond to attack a halogen. We don't have that here.

We have BH3. That's reagent one. So let's go ahead and push this up and get into this reaction.

So there's that double bond. Here is the boron. Here are the three hydrogens on it, and it has a 100% open p orbital.

Boron violates the octet rule. It has an open p orbital. So as it attacks... that open p-orbital, the double bond, one of those carbons starts to become a carbocation.

Now, once the double bond attacks the boron, it starts to become negative. It doesn't want to be negative. It violates the octet rule purposely, which means it needs to get rid of one of the hydrogens and the electrons in its bond. So as the double bond attacks the open p-orbital, one of the hydrogens and the electrons attack the double bond.

Wait, wait, wait. Doesn't this almost sound like diatomic halogenation? It does. Just instead of one atom doing the attack and being attacked, it's going to be one atom being attacked, the other one attacking. And what we get here is the boron added onto this carbon.

It can add from the top or back face right there, or we can get it added like this. Wonderful. Now, Actually, let me take a quick second for one moment, and I'm just going to erase that boron here and here.

You'll see why in a moment. I'm going to make the boron right here and put the boron with the squiggly line right there. How did this happen?

Did the double bond not attack the open p orbital? It did. And while it's attacking it, the electrons in one of the hydrogen-boron bonds will attack the carbon of the double bond at...

the same time. This is what's called a concerted reaction, meaning all of these steps are happening at the same time. Now, if they're happening at the same time, do you think the boron that's being attacked and the hydrogen that's being added onto will be added onto the same side or opposite side? And the answer is it will be the same side.

So a concerted process will always give you a syn reaction pathway, or what we call a, oops, ah. And of course I can't erase it. Oh, using the wrong one. There it is.

We can use a syn reaction pathway or we form a cis product between the boron and the hydrogen that was added in. Let's go over that one more time. As the double bond attacks the open p orbital in the boron, the electrons in the hydrogen-boron bond attack the opposite carbon of the double bond at the same exact time.

So if the boron added from the top face, the hydrogen adds from the top face. If the boron adds from the back face, the hydrogen adds to the back face. And we form these two products right here. Now, you're going to get to know that I hate having black box mechanisms. But I'm going to have you have one mechanism right now that's a black box.

We do not need to know the chemistry here. If you want to know this chemistry, ask me in the discussion section or over email, and I will show it to you. happily, but I don't want to put it in a video where it's testable material.

So we're not going to go through the mechanism for the oxidation. I'm going to give you the basic trend. When you treat it with H2O to an OH-, the BH2 is pulled off and OH- is put in its place. And we form these two products right here.

And if I shrink this down, ah, it's the same products that we see up above, but But still, this is our major, this is our minor. That seems weird. Tertiary carbocations are more stable than secondary carbocations, so this should be our major.

Now what I just said is still true. Tertiary carbocations or tertiary carbocation-like systems are going to be more stable than secondaries and primaries and stuff like that. Let's look at this reaction a little bit more.

As the double bond attacks the p-orbital, I know you're like, oh dear gosh, not again. As it attacks the p-orbital, this carbon can build up a partial positive, or this carbon can build up a partial positive. Which one's better holding a partial positive charge, the tertiary or the secondary? And the answer is the tertiary, right?

So the bond is going to form between the secondary carbon and the boron to create this partial charge here, that which the hydrogen adds in onto that carbon, forming this. as our major intermediate or major product of the first step of the hydroboration step itself. We're still governed by the carbocation-like systems. The reason why we're getting the opposite regioselectivity is because the hydrogen, in the form of a hydride, is attacking that carbon last. Very important distinction, right?

Now that is hydroboration. Let's go ahead and open up another page. I know you think this is never going to end, but we are almost there. The next one that we want to talk about is called hydrogenation.

A hydrogenation reaction is going to take a carbon-carbon double bond, and we're going to reduce it with hydrogen gas, and we're going to use either palladium or platinum. And what we end up forming here is an alkane. What is the role of the metal? The metal's job...

The palladium or the platinum, you can never draw lead. Lead PB does not work. The palladium or the platinum, its whole job is to hold that hydrogen gas in place. a gas and it's moving around like crazy. So if the hydrogen gas is being held, it allows the double bond to attack one of those hydrogens, and then the other electrons and the hydrogen attack the opposite side of the carbon, just like we saw in hydroboration.

And that is how we form our alkane from an alkene. And now with that, we are done with reactions for alkenes. There are many other reactions for alkenes. There are.

I'm just covering these basic ones right here, and this is basic even for full-year organic chemistry. We don't need the more detailed ones. None of your majors require it, nor any application into biological chemistry.

I know this was a long video. I highly recommend they'll go back and watching it again. Practice the problems in your textbook.

Read the textbook. Come to office hours and discussion sections, and we can talk about this. I hope each of you are doing well.

And I look forward to seeing you in a session sometime this semester.

Transcript for:Understanding Alkene Reactions and Mechanisms

Transcript for:
Understanding Alkene Reactions and Mechanisms