Lecture 15: Tautomerization, Oxidation, and Reduction
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King, Susan
This is the second quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. 
 Good afternoon. We're going to get started. Alright. We are going to finish chapter 11 today and start chapter 12. So hopefully everybody has started cataloging their reactions. There's a lot of new reactions in chapter 12. There's a lot of reactions in chapter 12 that you will use a lot the rest of the year of OCCEM. So I also just looked on our website and there's an addition summary sheet that summarizes everything we covered in chapter 10. I looked and it's not there. So I'm going to put that up and you can take a look. Highly recommend you take a look at that handout. Alright. So we left off last time talking about the mechanism that this was guaranteed to be on the midterm, correct? Yes. It should be. When did you look? Somebody else accessed it? Yes? Yes. So try now. Should be it. Okay. Alright. So we said this mechanism is guaranteed to be on the midterm. So let's add some things here. Alright. When we first made our carbocation here, we put the positive charge on the most substituted side. Of course, that one doesn't have a most substituted side, but if it was a terminal alkyne, you would put it at the most substituted side. This is a Markovnikov addition. And then there's a couple, there's some, a little bit of terminology that we're going to be adding here. This is this type of structure where you have a vinyl alcohol, has a special name. It's called an enol. So when you name alcohols, you use an all ending. And when you name alkenes, you use an ene ending. So that's where we get that name from. It's an enol. And generally speaking, 99% of the time, these structures are not favored at equilibrium. So enols are generally unstable. They are, you form them in reactions, but then they're converted into their more stable form. So the more stable form would be this form right here, which is we call the keto form. So we call this one an enol, or we can call it the enol form. And this is the keto form. And when we have an enol form equilibrating, so notice the equilibrium arrows here. When we have an enol and a keto form equilibrating, we call that a tautomerization. So these two structures, the enol and the keto, I made the arrow the wrong direction there. Let's fix that. They interconvert. So long arrow to the keto form, very short arrow to the vinyl form. And the vinyl, so this is called a tautomerization. And the enol form and the keto form are tautomers. So basically as soon as we form this enol form, then it's going to automatically tautomerize that equilibrium to a ketone. So not favored at equilibrium. So therefore it will tautomerize to a ketone. So this whole process here, let me circle everything that's the tautomerization. Going this whole part right here is how you tautomerize. So going between these species, that's the tautomerization. So you want to make sure on the exam that you actually show that exactly how I've shown it here. So to go from, because some students will just draw the vinyl alcohol and then they'll just draw arrows and they'll draw the keto form, but then you're going to miss all the points here. So let's add up how many points this problem is going to be on midterm two. It's the way that I do my mechanisms is this usually two points per step. So we have, let's count it up. Two, four, six, eight. I want to see both of these resonance structures. So two, four, six, eight, ten, twelve. And then I want you to label the enol form and I want you to label the keto form. So that's 14 points on midterm two. So that's the lowest score that's going to be on midterm two, right? Because that's all you have to, if you want 14 points, that's all you have to study. So that's what I was talking about. That will be my lowest score and I say I wish that would be my lowest score. Okay, so the way that we do mechanisms is if you're missing arrows, like say you just do this first arrow and you don't do the second arrow, you don't get points for that step. You have to have all the arrows. If you're missing lone pairs and charges, you miss points and I believe me the points add up. So make sure you take a look at your test and see what you did wrong on your test with the mechanisms on the test. Alright, so you'll get partial credit if you don't have charges. If you're missing arrows, you won't get points for, you'll miss partial credit. If you combine steps, more than one step, then you miss the points for each step. So if you combine two steps into one, then you miss four points. So that just kind of gives you an idea on how that's graded. Questions on this mechanism, anybody? Anyone? Yes? Oh, very, thank you for pointing it out. So when I grade a mechanism, I don't care if you do these reversible arrows, I'm only focused on curvy arrows. That's it. So don't worry yourself about, oh, if I don't do reversible arrows, I don't care about that. I just care about the curvy arrows. Alright, point and points about this reaction. And this is going to come back in 51C. So this terminology, tautomers, enol form, keto form, that's coming back in 51C, big time. So carbonyl compounds with alpha hydrogens, well we don't know what alpha hydrogens are, but they're hydrogens that are adjacent to the carbonyl. So this carbonyl right here has six alpha hydrogens. Carbonic compounds with alpha hydrogens are in equilibrium with vinylic alcohol, isomers called enols, the two isomers that interconvert are called tautomers. So this one would be the keto form, and this is the enol form. And most of the time, we're going to say 99% of the time the keto form is more stable, so that's what you're going to actually isolate from the equilibrium. For this particular one, we have keq equals 7 times 10 to the minus 6. So that certainly tells you that the keto form is more favored. And that's the case. When you get into 51C, it'll show you a couple of exceptions to this rule, that the keto form is more favored. There's a few of them, a few important ones, but nothing we have to worry about now. So the keto form, usually favored. At equilibrium. So when you're doing predicting products on midterm 2, if you draw an enol, it's going to be the wrong answer. Okay, so we're not going to be isolating enols, because they're not stable. And so tautomerization, like we said, conversion of an enol to a ketone by protonation of the carbon of the double bond and deprotonation of the oxygen atom. This tautomerization can happen in acid or base. You do need a little bit of a catalyst here. So the one that we, the mechanism we saw on the previous page was acid catalyzed tautomerization. On the next page, we're going to see base catalyzed tautomerization. So there's two of them. And you're going to need to know both of them in 51C, certainly. Alright, so when an enol is formed in reaction, it will automatically convert to its keto form. The presence of tracerb acids are based, so don't draw an enol as a product. Because it's not. Don't draw an enol as a product. Alright, so the addition of water to all kinds follows Markovnikov's rules. So the hydrogen adds to the carbon with the most number of hydrogens. And that's certainly to give the more stable carbocation. Terminal alkynes are less reactive and require use of an additional catalyst, HgSO4. Okay, so this is an additional catalyst with terminal alkynes. You will want to include that if you're doing a synthesis problem like this. Additional catalyst needed. Now there's a mechanism that goes along with that additional catalyst. I'm not going to show you that. You are not required to know that. So when I put this question on midterm 2, it will not be a terminal alkyn. It will be an internal alkyn. Okay? Don't need to know that other mechanism. Products always a ketone except when acetylene is hydrated. So the only way that you can get an aldehyde with hydration is if you use acetylene. That's it. On symmetrical internal alkynes, we'll give a mixture of products. There's nothing you can do about it. So this will give two products. So if you were, if you wanted to make one of these two products, you would probably do it a different way because these two products would be really difficult to separate from each other. So that's something you have to keep in mind. Questions on hydration of an alkyne? Anybody? Let's talk about addition of bromine and chlorine. As you can see, I'm not going to spend too long on this. One equivalent of halogen gives you transdi-bromoalkene. If you do two equivalents of halogen, you can get tetra-bromoalkene. Rather than draw the products, let's just go through the mechanism here. I'm going to start with my alkyne here. If you know the mechanism for bromination in chapter 10, this will be a very easy mechanism for you. Okay, so the arrow comes from the pi bond. We attack bromine. We break the bromine-bromine bond and one of the lone pairs on bromine comes and attacks the other side. So we make a bromine. This would be, this one's not a bromonium. It's a bromenium ion. So like a bromonium ion, except we have a double bond in our molecule. Same idea though. We have a bromide ion come in, backside attack. This is symmetrical, so it can attack on either side. I'll just attack right here. So there's our transdi-bromoalkene with one equivalent. And then if we add another equivalent of bromine, and you can do that mechanism, that's a chapter 10 mechanism. It will look just exactly like the mechanism for chapter 10. We're going to make a bromonium ion and then we're going to do backside attack. And I'm not going to attempt to show any stereochemistry here, because there is no stereochemistry here. We don't have any stereocenters. Because we're adding two bromines to each carbon, there are no stereocenters. So the product after one equivalent is this. The product after two equivalents is this. Really straightforward. Students tend to do really well on alkenes. Any time I test on alkenes, students tend to do really well, because they're not super difficult. All right, we're going to end with hydroboration and oxidation of alkynes. Like alkenes, alkynes undergo hydroboration, which we did with borane reagents. It's going to be the same idea. Yes. Yeah. Yes, let's change that to an A. Thank you. We'll fix that right now. Thank you for pointing that out. Tetobromoalkane, yes. All right. So the same concepts that we learned for hydroboration in Chapter 10 are going to also apply here. There's three different borane reagents that you will see in this chapter. BH3THF or 9BBN, which I introduced to you in Chapter 10. 9BBN. Or there's another reagent that tends to work really well with alkynes, and that's called disyamylboring, which is similar to 9BBN in that it's more bulky than a typical, it's more bulky than BH3, certainly. So you tend to get better results. So this one here is called disyamylboring. Sin addition, right? Just like in Chapter 10. So there's our reagent there, and so that you notice is that the boron and the hydrogen both came in on the same side. So this is a sin addition. Unsymmetrical alkynes give mixtures better to use 9BBN here, definitely. 9BBN is much better when you have subtle differences in the groups on either side of the alkyne. So here we have the hydrogen. You'd get both of these if you use BH3. If you use 9BBN or disyamylboring, you get your major one is the second one. So major if 9BBN or disyamylboring. Terminal alkynes also undergo hydroboration. BH3 THF does not work very well for terminal alkynes. I'm choosing not to make an issue about it. But really for terminal alkynes, you need to use disyamylboring or 9BBN. So let's cross that out. 9BBN, you still will see answers. You will still see problems on Smith, and you will still see problems in Sapling where they use 9BBN or BH3. The problem with BH3 is that it tends to add twice, and then that causes all sorts of problems in the product that you actually isolate. So I've drawn the product that you get from hydroboration. We're usually not super interested in those products in this class anyway. So what we normally do is we oxidize these, right? Basic hydrogen peroxide. So let's see what happens when we oxidize these on the next page. Now when we oxidize our hydroboration product, we put an hydrogen hydroxyl right where the boron was, right? Same thing here. Let's draw that. And now we draw that, we look at that product. What type of product is that? It's an enol. Do we want to draw enols in any answers on our test? No. So yes, we do form an enol, and I want you to be able to recognize, wow, that's an enol, I shouldn't draw that. So what this is going to do is it's going to tautomerize. So long arrow to the right, short arrow to the left. And when you tautomerize this, where the hydroxyl is, is going to become a carbonyl. So this is, so that's what you get for the first example, and then we're going to actually show the mechanism here, because it's different than the acid catalyzed. So that's also an enol. And again, exactly where the hydroxyl is, the double bond goes away, and the hydroxyl is going to become a carbonyl. So for both of these, this is base catalyzed tautomerization. So what you want to draw in the box is the carbonyl product. Let's see what that mechanism looks like. So I've already showed you acid catalyzed, that's the one that's guaranteed to be on the midterm. Let's look at what the base catalyzed looks like. I'm going to start with this enol right here. And what I'm going to do first is deprotonate oxygen. So I am doing something very different in base than I did in acid. In the acid catalyzed tautomerization, we hydrated the double bond. We added, we protonated the double bond. Here we're deprotonating oxygen. Notice I'm doing reversible arrows here, because this is a reversible reaction. And I drew resonance structures in the acid catalyzed tautomerization, and I'm going to do the same thing here in the base catalyzed tautomerization. So I know that this is resonance stabilized. I can move these electrons here. This goes all the way back to chapter one. So now instead of arrows that go both directions, I'm going to use a resonance arrow. So that's how you get that double bond there. So now we have negative charge on carbon. And now you can see to get to that product we have here, all we need to do is protonate that carbon. So base catalyzed, we're going to protonate that carbon with water. That regenerates the base that we started with. So this really is only catalytic. So let's put brackets around these two resonance structures here. So exact same outcome, just a little bit different mechanism because it's acid catalyzed versus base catalyzed. And I know some of you had trouble with that when we did epoxides. But acid catalyzed and base catalyzed mechanisms are extremely important in 51C. So that's why I'm emphasizing them now. It'll make it much easier for your next quarter. So stepwise mechanisms, you don't want to combine steps. We talked about that. So don't do this. Let me show you what you don't want to do for this mechanism. So I'll draw it and then you're going to tell me what's wrong with it. And just in case you don't read the don't do this when you're studying, we'll put a big X through this mechanism after we draw it. So there's our enol. We want to tautomerize. So basically I'm just going to kind of do the same thing. Only I'm going to do it all in one step. Combine steps here. So I'm going to grab this here. I'm going to have these electrons come over here. And then I'm going to have the double, the pi bond come and grab a hydrogen from water. That would do my thing all in one step. Why is that bad? What am I doing wrong? Why would that be a very unlikely process here? There's three species all coming together simultaneously with the right orientation, exact orientation, all of them colliding with enough energy to react. It's such a rare event that we don't want to presume that that's happening. So no, we call these termolecular reactions. No termolecular reactions. And so when you combine steps on a mechanism, you generally speaking are doing termolecular reactions. You're having too many things happen at the same time. So that means three molecules coming together simultaneously. Alright, so don't want to do that, don't want to suggest that. Alright, so compare the result of T when a termoleucine undergoes acid, catalyzed hydration versus hydroboration. Well, one's anti-Marconikov and the other one's Marconikov. So remember, where the oxygen goes, that's where the carbonyl is going to be. So hydroboration gives you anti-Marconikov and then hydration gives you a Marconikov. So it's really nice that we have these complementary procedures. So sometimes we want one of these and sometimes we want the other and we have, so now we have two ways to do this at our fingertips that we can use for synthesis. Alright, questions, anybody so far? We're going to start at this point getting into more complicated synthesis. I'm going to give you one example here and then by the time we get to the end of chapter 12, you're going to have much more complicated synthesis. So this is the time to get in gear and start practicing a lot of synthesis. So this is the time when you'll see at the end of the chapter, once you do all the predicting the products, there's a synthesis part where we have reactions from multiple chapters. That's what you want to spend your time on and you want to try to solve those without looking at the answers. So in planning a synthesis, depending on what we're trying to plan, we might need to construct the carbon skeleton. That's what we're going to have to do in this example of functional group interconversion. We're doing that in this example. We're taking an alkyne and converting it into a ketone. So that's functional group interconversion. We are going to have to control regiochemistry. So here's a really good example here, this previous example here. We were controlling regiochemistry by choosing the correct reagent. And then some of the examples we're going to have to control stereochemistry, those are going to be coming up at the end of chapter 12. So we've got to think about all of these things. This particular one, we don't have any stereochemistry, so we don't really have to control this. Can stereochemistry, prepare the following compound from acetylene. So it looks enormously complicated. Using our retrosynthetic arrow is going to really help us. So this is acetylene. And I'm not going to bother going in the forward direction. I can see it in the forward direction, and maybe you can too. And certainly do it that way, but I'm just going to work backwards from the product. So I'm going to use my retrosynthetic arrow that can be made from. And I'm going to work backwards. So again, let's label that. Retrosynthetic arrow means this can be made from. And this is where the recognizing those functional groups, we've really only talked about two ways to make a carbonyl so far. So that makes this a little easier. By the time you get to 51C, there'll be 10 ways to make a carbonyl. And then it's a little more complicated because so now we're a little limited, and this is the time to jump in and get lots of good practice here. So this one can be made from, well, if I had, I only know one way to make a carbonyl right now, and that's from an alkyne. So I would have to do it from, if I had this alkyne, a terminal alkyne, this would be H2O. I want the carbonyl on the most substituted side here, so I would have to use hydration. H2O, H2SO4, and because it's terminal, I need to add HGSO4 as a catalyst. All right, so I need two carbon pieces, so I'm going to have to take this down further. So I know I can make this from, gosh, if I had this right here, bromide, I could have chloride, iodide, or tosate, and if I had this, I could do a substitution reaction. That's chapter 7. Just do an SN2 reaction. And then now, and I'm always going back and looking at what I started with, and I certainly can make this from acetylene by deprotonating it, right? What base would I use for that? NH or NH2? Or you can write, I don't care, H minus or NH2 minus. So I'm not particular about you concluding the counter ion. If you write that in a synthesis, that's totally fine with me. Okay, so we've got our plan. Now let's write out the synthesis in the forward direction. So when I grade tests, this would be the kind that's the open-ended kind, and I don't grade retrosynthetic, and it's too hard to grade. I can't see the way people are thinking. So I only grade the synthesis in the forward direction. So if it was an open-ended, what I call open-ended synthesis, where I ask for products from each step, it would look something like this. Definitely want to number those two steps. I'm deprotonating, and then I'm adding propyl bromide. And then in my last step here, it would be H2O, H2SO4, HGSO4. So this is three steps, this is three steps in synthesis. Or on a test, it might look like this. If it's not an open-ended, if it's one where you just fill in reagents, I do have more of that type on the test just because these other kind are much harder to grade, you got to worry about these things when you have a lot of students. Alright, so then I might have it like this. And so that would be, then you would do the reagents and you would number the individual steps. Step one, NH. I'll just use NH. Step two, propyl bromide. Step three, H2O, H2SO4. HGSO4. And that should be H2O, not H2. What would happen if I didn't number that? What would happen if I mixed sodium hydride, propyl bromide, H2O, H2SO4, and HGSO4? A very potentially explosive reaction between the sodium hydride and the sulfuric acid. Or sodium hydride in water. It's an extremely strong base. And what happens when you mix an extremely strong base with an extremely strong acid and you don't cool it, you just throw them together, it explodes. It might not blow out a wall, but it will blow compound all over you, that you don't want all over you. So that's why it's really critical that we number the steps as we're doing them, when they need to be numbered. That is chapter 11. We're going to start chapter 12 now. Any questions while I'm pulling up chapter 12? It takes a little while to save this, so what can I, anybody have questions? Alright, chapter 12. Pretty excited about this. Now, here's the thing I don't like about chapter 12. There's no mechanisms in chapter 12. I like mechanisms, right? It also helps you learn what the reaction is actually doing. So this is as close as you're going to come in organic chemistry to a chapter which is all memorization. I know a lot of people think O-chem is all memorization, but really it's some memorization and this is all memorization. I will show you approximately what's happening in a couple of different things, but I will not ask you any mechanisms in chapter 12. And the reason is the mechanisms are a lot more complicated, more complicated than we're ready for. Okay, so that's where we're coming from. So we're going to talk about oxidation and reduction. So first of all, recognizing oxidation and reduction in organic compounds. And I think this was in like chapter 4 or 5 or 6 or something like that. It was covered then. I'm going to assume you don't remember it, so I'm going to cover it again. But you will see the exact notes in one of those chapters if you look back in 51A. Alright, so it's a little bit different in organic chemistry than general chemistry to recognize oxidation and reduction. Here's the good news. Does everybody remember balancing redox reactions? We don't do that here at all. Okay, so that's brutal. I know. So let me give you a G-chem example. We also aren't going to be assigning oxidation, we're not assigning oxidation numbers at all. We don't do that. You can. We had a textbook one time that did that, but none of the other ones have. And so we just, you don't need it. So we're not going to cover it. So copper, shiny, red, metal reacting with silver plus. The copper loses two electrons, right? So this loses two electrons. And transfers those electrons over to silver. The electrons are transferred completely from one species to the other. So then we get the blue copper two plus salts. So pretty cool color changes. I like these things. But this is going to be gaining. This gains electrons. Okay, so we know also that if one molecule is oxidized and another is reduced, that's from G-chem. We're going to carry that forward. And the electrons are transferred completely from one molecule to another. That's what happens. Not like that in organic chemistry. So in organic chemistry, we're not actually transferring electrons completely. We're changing the direction of the flow of electrons. So oxidation is loss of electron density. Carbon loses electrons by forming bonds with elements that are more electronegative than it is. So what we're going to see for oxidation is loss of a carbon hydrogen and gain of a carbon oxygen. These both have to happen at the same time. Both have to happen for oxidation. That's what we're going to look for. Loss of carbon hydrogen, gain of carbon oxygen. So what are we talking about with electron density? Well, electronegativity carbon is 2.5. Hydrogen 2.2. Carbon is 2.5. And rather than use oxygen there, I'm going to do a generic here. Let's just do X. That's a little better. Carbon X. So 2.5 for carbon X. Greater than 2.5. So what we're doing here is we're shifting electron density. With a carbon hydrogen bond, the electron density is flowing towards the carbon because it's more electronegative. When we replace that hydrogen with an X, it's flowing towards X. So we've changed the direction of the flow of electrons. So for here, X equals oxygen, nitrogen, halogen typically. These are the ones that we're going to see. That means you have an oxidation. And it's important to be able to recognize if you have an oxidation or reduction because if you have an oxidation, then you'll know you have to use an oxidizing agent. If you have a reduction, you'll know you have to use a reducing agent. That helps you to figure out what the rubber agent to use. Reduction is the opposite. Loss of CX, gain of CH. So as you can see here, we have electrons flowing to X now and we're changing it so the electrons are flowing to carbon. And again, here X equals oxygen, nitrogen, halogen. I'll just say it's satyr. Because it could be anything. More electronegative. So the following shows stepwise oxidation of methane. Most reduced form of carbon, all the way up to carbon dioxide, which is the most oxidized form of carbon. So you can see what's happening here. As we go from methane to methyl alcohol or methanol, we've lost a carbon hydrogen and we've gained a carbon oxygen. So both things have to take place. So therefore it's an oxidation. Moving to the right, it's an oxidation. And then we have, we do that one more time. We break another carbon hydrogen minus carbon hydrogen plus carbon oxygen. We've gained a second carbon oxygen bond because now we have a double bond of carbon. So that's also an oxidation. And then same thing here, we lose another carbon hydrogen and gain carbon oxygen because now we have three bonds. Carbon's bonded to three oxygens. So minus carbon hydrogen and plus carbon oxygen. And then we're also, we're losing another carbon hydrogen, gaining a carbon oxygen. So that's oxidation. Stepwise oxidation. So four steps to go from the most reduced form of carbon to the most oxidized form. If we went back the other way, I'm not going to draw that here, but it's the exact opposite. If we go from here backwards over to here, we're losing a carbon oxygen and gaining a carbon hydrogen. So that's reduction. Let's just do one of those. I'm not going to do all of them. We can go the opposite direction and step back, going from here to here. We minus carbon oxygen plus carbon hydrogen. Therefore, reduction. All right. So very helpful to be able to recognize. So can you recognize oxidation reduction in the following examples here? So what's happening here? We have two hydrogens and we're replacing those with two carbon oxygens, right? So loss of two carbon hydrogens, gain of two carbon oxygens. Or oxidation. Definitely oxidation there. Let's look at the second example here. We're losing two oxygens, two carbon oxygens, and we're gaining two carbon hydrogens. Therefore, it's a reduction. So if you want to do this reaction, you're going to need to use a reducing agent. In the first one, you're going to need an oxidizing agent to do that reaction. So let's look at one here that's one we've already looked at. Is this an oxidation reduction reaction? Well, that's something, it's not in this chapter, so it's not. It's something that we've already talked about. It's an addition reaction, right? So we know that by looking at that, that's addition. That's an addition reaction. We know that's from chapter 10. And so let's look at what's happened here. We gain one carbon oxygen and we gain one carbon hydrogen. You have to gain one and lose the other. So it's not. It's neither an oxidation nor a reduction. It's an addition. Questions on recognizing oxidation or reduction? Ready? So want to know how to do that for midterm two. We're going to do reduction first and then we're going to do oxidation. You're going to get a slew of reactions and you're going to go home and make index cards for them, okay? There's three types of reducing agents. There's the first type where we add H2 plus a catalyst. That's called a catalytic hydrogenation. And you will generally see, so it will be H2 plus palladium, platinum or nickel. Those would be the common catalysts that you would use for catalytic hydrogenation. The second type of reducing agents are where you add two H plus and two electrons. We call these dissolving metal reduction. We're going to talk about all of these. And then the third type is where you add one H minus followed by one H plus and that is a metal hydride reduction. So I can't emphasize enough the importance of the reagents in this chapter. You will use them over and over and over again when you're doing synthesis. Alright, we still have two minutes. So we're going to talk about all three of these in the same order that they're appearing right here. So we'll first talk about catalytic hydrogenation. And we're going to talk about addition of hydrogen to alkenes and then we'll talk about addition of hydrogen to alkynes. So what you do when a catalytic hydrogenation is you add the elements of hydrogen on each side across a double bond. And I'm deliberately showing stereochemistry here. What type of addition is that when the two hydrogens come in on the same side? Yes, so this is a stereoselective syn addition. So that means the hydrogens come in from the same side of the alkene. Need to know that. Highly exothermic reaction but requires a catalyst because of the huge energy of activation. The catalyst changes the nature of the transition state thereby lowering energy of activation. We'll talk more about that next time.
 Good afternoon. We're going to get started. Alright. We are going to finish chapter 11  today and start chapter 12. So hopefully everybody has started cataloging their reactions. There's  a lot of new reactions in chapter 12. There's a lot of reactions in chapter 12 that you will  use a lot the rest of the year of OCCEM. So I also just looked on our website and there's  an addition summary sheet that summarizes everything we covered in chapter 10. I looked  and it's not there. So I'm going to put that up and you can take a look. Highly recommend  you take a look at that handout. Alright. So we left off last time talking about the mechanism  that this was guaranteed to be on the midterm, correct? Yes. It should be. When did you look?  Somebody else accessed it? Yes? Yes. So try now. Should be it. Okay. Alright. So we said this  mechanism is guaranteed to be on the midterm. So let's add some things here. Alright. When we first  made our carbocation here, we put the positive charge on the most substituted side. Of course,  that one doesn't have a most substituted side, but if it was a terminal alkyne, you would put it  at the most substituted side. This is a Markovnikov addition. And then there's a couple, there's  some, a little bit of terminology that we're going to be adding here. This is this type of  structure where you have a vinyl alcohol, has a special name. It's called an enol. So when  you name alcohols, you use an all ending. And when you name alkenes, you use an ene ending.  So that's where we get that name from. It's an enol. And generally speaking, 99% of the time,  these structures are not favored at equilibrium. So enols are generally unstable. They are,  you form them in reactions, but then they're converted into their more stable form. So the  more stable form would be this form right here, which is we call the keto form. So we call this  one an enol, or we can call it the enol form. And this is the keto form. And when we have an  enol form equilibrating, so notice the equilibrium arrows here. When we have an enol and a keto  form equilibrating, we call that a tautomerization. So these two structures, the enol and the keto,  I made the arrow the wrong direction there. Let's fix that. They interconvert. So long arrow to the  keto form, very short arrow to the vinyl form. And the vinyl, so this is called a tautomerization.  And the enol form and the keto form are tautomers. So basically as soon as we form this enol form,  then it's going to automatically tautomerize that equilibrium to a ketone. So not favored at  equilibrium. So therefore it will tautomerize to a ketone. So this whole process here, let me  circle everything that's the tautomerization. Going this whole part right here is how you  tautomerize. So going between these species, that's the tautomerization. So you want to make sure  on the exam that you actually show that exactly how I've shown it here. So to go from, because  some students will just draw the vinyl alcohol and then they'll just draw arrows and they'll  draw the keto form, but then you're going to miss all the points here. So let's add up how many  points this problem is going to be on midterm two. It's the way that I do my mechanisms is  this usually two points per step. So we have, let's count it up. Two, four, six, eight. I want to see  both of these resonance structures. So two, four, six, eight, ten, twelve. And then I want you to  label the enol form and I want you to label the keto form. So that's 14 points on midterm two. So  that's the lowest score that's going to be on midterm two, right? Because that's all you have  to, if you want 14 points, that's all you have to study. So that's what I was talking about. That  will be my lowest score and I say I wish that would be my lowest score. Okay, so the way that we do  mechanisms is if you're missing arrows, like say you just do this first arrow and you don't do the  second arrow, you don't get points for that step. You have to have all the arrows. If you're missing  lone pairs and charges, you miss points and I believe me the points add up. So make sure you take  a look at your test and see what you did wrong on your test with the mechanisms on the test. Alright,  so you'll get partial credit if you don't have charges. If you're missing arrows, you won't get  points for, you'll miss partial credit. If you combine steps, more than one step, then you  miss the points for each step. So if you combine two steps into one, then you miss four points. So  that just kind of gives you an idea on how that's graded. Questions on this mechanism, anybody?  Anyone? Yes? Oh, very, thank you for pointing it out. So when I grade a mechanism, I don't care if  you do these reversible arrows, I'm only focused on curvy arrows. That's it. So don't worry yourself  about, oh, if I don't do reversible arrows, I don't care about that. I just care about the curvy arrows.  Alright, point and points about this reaction. And this is going to come back in 51C. So this  terminology, tautomers, enol form, keto form, that's coming back in 51C, big time. So carbonyl  compounds with alpha hydrogens, well we don't know what alpha hydrogens are, but they're  hydrogens that are adjacent to the carbonyl. So this carbonyl right here has six alpha  hydrogens. Carbonic compounds with alpha hydrogens are in equilibrium with vinylic alcohol,  isomers called enols, the two isomers that interconvert are called tautomers. So this one would be  the keto form, and this is the enol form. And most of the time, we're going to say 99% of the time  the keto form is more stable, so that's what you're going to actually isolate from the equilibrium.  For this particular one, we have keq equals 7 times 10 to the minus 6. So that certainly tells you  that the keto form is more favored. And that's the case. When you get into 51C, it'll show you a  couple of exceptions to this rule, that the keto form is more favored. There's a few of them,  a few important ones, but nothing we have to worry about now. So the keto form, usually favored.  At equilibrium. So when you're doing predicting products on midterm 2, if you draw an enol,  it's going to be the wrong answer. Okay, so we're not going to be isolating enols,  because they're not stable. And so tautomerization, like we said,  conversion of an enol to a ketone by protonation of the carbon of the double bond and deprotonation  of the oxygen atom. This tautomerization can happen in acid or base. You do need a little bit  of a catalyst here. So the one that we, the mechanism we saw on the previous page was acid  catalyzed tautomerization. On the next page, we're going to see base catalyzed tautomerization.  So there's two of them. And you're going to need to know both of them in 51C, certainly.  Alright, so when an enol is formed in reaction, it will automatically convert to its keto form.  The presence of tracerb acids are based, so don't draw an enol as a product. Because it's not.  Don't draw an enol as a product. Alright, so the addition of water to all kinds follows  Markovnikov's rules. So the hydrogen adds to the carbon with the most number of hydrogens.  And that's certainly to give the more stable carbocation. Terminal alkynes are less reactive  and require use of an additional catalyst, HgSO4. Okay, so this is an additional catalyst with  terminal alkynes. You will want to include that if you're doing a synthesis problem like this.  Additional catalyst needed. Now there's a mechanism that goes along with that additional  catalyst. I'm not going to show you that. You are not required to know that. So when I put this  question on midterm 2, it will not be a terminal alkyn. It will be an internal alkyn. Okay?  Don't need to know that other mechanism. Products always a ketone except when acetylene is hydrated.  So the only way that you can get an aldehyde with hydration is if you use acetylene. That's it.  On symmetrical internal alkynes, we'll give a mixture of products. There's nothing you can do  about it. So this will give two products. So if you were, if you wanted to make one of these two  products, you would probably do it a different way because these two products would be really  difficult to separate from each other. So that's something you have to keep in mind.  Questions on hydration of an alkyne? Anybody? Let's talk about addition of bromine and chlorine.  As you can see, I'm not going to spend too long on this. One equivalent of halogen gives you  transdi-bromoalkene. If you do two equivalents of halogen, you can get tetra-bromoalkene.  Rather than draw the products, let's just go through the mechanism here. I'm going to start  with my alkyne here. If you know the mechanism for bromination in chapter 10, this will be a very  easy mechanism for you. Okay, so the arrow comes from the pi bond. We attack bromine. We break  the bromine-bromine bond and one of the lone pairs on bromine comes and attacks the other side.  So we make a bromine. This would be, this one's not a bromonium. It's a bromenium ion. So like a  bromonium ion, except we have a double bond in our molecule. Same idea though. We have a bromide  ion come in, backside attack. This is symmetrical, so it can attack on either side. I'll just attack  right here. So there's our transdi-bromoalkene with one equivalent. And then if we add another  equivalent of bromine, and you can do that mechanism, that's a chapter 10 mechanism. It will look just  exactly like the mechanism for chapter 10. We're going to make a bromonium ion and then we're going  to do backside attack. And I'm not going to attempt to show any stereochemistry here, because there  is no stereochemistry here. We don't have any stereocenters. Because we're adding two bromines  to each carbon, there are no stereocenters. So the product after one equivalent is this. The  product after two equivalents is this. Really straightforward. Students tend to do really  well on alkenes. Any time I test on alkenes, students tend to do really well, because they're not  super difficult. All right, we're going to end with hydroboration and oxidation of alkynes.  Like alkenes, alkynes undergo hydroboration, which we did with borane reagents. It's going to be the  same idea. Yes. Yeah. Yes, let's change that to an A. Thank you. We'll fix that right now. Thank  you for pointing that out. Tetobromoalkane, yes. All right. So the same concepts that we learned  for hydroboration in Chapter 10 are going to also apply here. There's three different borane  reagents that you will see in this chapter. BH3THF or 9BBN, which I introduced to you in Chapter 10.  9BBN. Or there's another reagent that tends to work really well with alkynes, and that's called  disyamylboring, which is similar to 9BBN in that it's more bulky than a typical, it's more bulky  than BH3, certainly. So you tend to get better results. So this one here is called disyamylboring.  Sin addition, right? Just like in Chapter 10.  So there's our reagent there, and so that you notice is that the boron and the hydrogen both came in  on the same side. So this is a sin addition. Unsymmetrical alkynes give mixtures better to use  9BBN here, definitely. 9BBN is much better when you have subtle differences in the groups on  either side of the alkyne. So here we have the hydrogen. You'd get both of these if you use BH3.  If you use 9BBN or disyamylboring, you get your major one is the second one.  So major if 9BBN or disyamylboring. Terminal alkynes also undergo hydroboration. BH3 THF does not work  very well for terminal alkynes. I'm choosing not to make an issue about it. But really for terminal  alkynes, you need to use disyamylboring or 9BBN. So let's cross that out. 9BBN, you still will see  answers. You will still see problems on Smith, and you will still see problems in Sapling where  they use 9BBN or BH3. The problem with BH3 is that it tends to add twice, and then that causes all  sorts of problems in the product that you actually isolate. So I've drawn the product that you get  from hydroboration. We're usually not super interested in those products in this class anyway.  So what we normally do is we oxidize these, right? Basic hydrogen peroxide. So let's see what happens  when we oxidize these on the next page. Now when we oxidize our hydroboration product,  we put an hydrogen hydroxyl right where the boron was, right? Same thing here. Let's draw that.  And now we draw that, we look at that product. What type of product is that? It's an enol. Do  we want to draw enols in any answers on our test? No. So yes, we do form an enol, and I want you to  be able to recognize, wow, that's an enol, I shouldn't draw that. So what this is going to do is it's  going to tautomerize. So long arrow to the right, short arrow to the left. And when you tautomerize  this, where the hydroxyl is, is going to become a carbonyl. So this is, so that's what you get for  the first example, and then we're going to actually show the mechanism here, because it's different  than the acid catalyzed. So that's also an enol. And again, exactly where the hydroxyl is,  the double bond goes away, and the hydroxyl is going to become a carbonyl. So for both of these,  this is base catalyzed tautomerization. So what you want to draw in the box is the carbonyl  product. Let's see what that mechanism looks like. So I've already showed you acid catalyzed,  that's the one that's guaranteed to be on the midterm. Let's look at what the base catalyzed  looks like. I'm going to start with this enol right here. And what I'm going to do first  is deprotonate oxygen. So I am doing something very different in base than I did in acid. In  the acid catalyzed tautomerization, we hydrated the double bond. We added, we protonated the  double bond. Here we're deprotonating oxygen. Notice I'm doing reversible arrows here, because  this is a reversible reaction. And I drew resonance structures in the acid catalyzed tautomerization,  and I'm going to do the same thing here in the base catalyzed tautomerization. So I know that  this is resonance stabilized. I can move these electrons here. This goes all the way back to  chapter one. So now instead of arrows that go both directions, I'm going to use a resonance  arrow. So that's how you get that double bond there. So now we have negative charge on  carbon. And now you can see to get to that product we have here, all we need to do is  protonate that carbon. So base catalyzed, we're going to protonate that carbon with water.  That regenerates the base that we started with. So this really is only catalytic. So  let's put brackets around these two resonance structures here. So exact same outcome, just  a little bit different mechanism because it's acid catalyzed versus base catalyzed. And  I know some of you had trouble with that when we did epoxides. But acid catalyzed and base  catalyzed mechanisms are extremely important in 51C. So that's why I'm emphasizing them  now. It'll make it much easier for your next quarter. So stepwise mechanisms, you don't  want to combine steps. We talked about that. So don't do this. Let me show you what you  don't want to do for this mechanism. So I'll draw it and then you're going to tell me  what's wrong with it. And just in case you don't read the don't do this when you're  studying, we'll put a big X through this mechanism after we draw it. So there's our enol. We  want to tautomerize. So basically I'm just going to kind of do the same thing. Only I'm  going to do it all in one step. Combine steps here. So I'm going to grab this here. I'm  going to have these electrons come over here. And then I'm going to have the double, the  pi bond come and grab a hydrogen from water. That would do my thing all in one step. Why  is that bad? What am I doing wrong? Why would that be a very unlikely process here? There's  three species all coming together simultaneously with the right orientation, exact orientation,  all of them colliding with enough energy to react. It's such a rare event that we don't  want to presume that that's happening. So no, we call these termolecular reactions. No  termolecular reactions. And so when you combine steps on a mechanism, you generally speaking  are doing termolecular reactions. You're having too many things happen at the same time. So  that means three molecules coming together simultaneously. Alright, so don't want to  do that, don't want to suggest that. Alright, so compare the result of T when a termoleucine  undergoes acid, catalyzed hydration versus hydroboration. Well, one's anti-Marconikov  and the other one's Marconikov. So remember, where the oxygen goes, that's where the carbonyl  is going to be. So hydroboration gives you anti-Marconikov and then hydration gives you  a Marconikov. So it's really nice that we have these complementary procedures. So sometimes  we want one of these and sometimes we want the other and we have, so now we have two  ways to do this at our fingertips that we can use for synthesis. Alright, questions,  anybody so far? We're going to start at this point getting into more complicated synthesis.  I'm going to give you one example here and then by the time we get to the end of chapter  12, you're going to have much more complicated synthesis. So this is the time to get in gear  and start practicing a lot of synthesis. So this is the time when you'll see at the end  of the chapter, once you do all the predicting the products, there's a synthesis part where  we have reactions from multiple chapters. That's what you want to spend your time on  and you want to try to solve those without looking at the answers. So in planning a synthesis,  depending on what we're trying to plan, we might need to construct the carbon skeleton.  That's what we're going to have to do in this example of functional group interconversion.  We're doing that in this example. We're taking an alkyne and converting it into a ketone.  So that's functional group interconversion. We are going to have to control regiochemistry.  So here's a really good example here, this previous example here. We were controlling  regiochemistry by choosing the correct reagent. And then some of the examples we're going  to have to control stereochemistry, those are going to be coming up at the end of chapter  12. So we've got to think about all of these things. This particular one, we don't have  any stereochemistry, so we don't really have to control this. Can stereochemistry, prepare  the following compound from acetylene. So it looks enormously complicated. Using our  retrosynthetic arrow is going to really help us. So this is acetylene. And I'm not going  to bother going in the forward direction. I can see it in the forward direction, and  maybe you can too. And certainly do it that way, but I'm just going to work backwards  from the product. So I'm going to use my retrosynthetic arrow that can be made from.  And I'm going to work backwards. So again, let's label that. Retrosynthetic arrow means  this can be made from. And this is where the recognizing those functional groups, we've  really only talked about two ways to make a carbonyl so far. So that makes this a little  easier. By the time you get to 51C, there'll be 10 ways to make a carbonyl. And then it's  a little more complicated because so now we're a little limited, and this is the time to jump  in and get lots of good practice here. So this one can be made from, well, if I had,  I only know one way to make a carbonyl right now, and that's from an alkyne. So I would  have to do it from, if I had this alkyne, a terminal alkyne, this would be H2O. I want  the carbonyl on the most substituted side here, so I would have to use hydration. H2O,  H2SO4, and because it's terminal, I need to add HGSO4 as a catalyst. All right, so I  need two carbon pieces, so I'm going to have to take this down further. So I know I can  make this from, gosh, if I had this right here, bromide, I could have chloride, iodide,  or tosate, and if I had this, I could do a substitution reaction. That's chapter 7. Just  do an SN2 reaction. And then now, and I'm always going back and looking at what I started  with, and I certainly can make this from acetylene by deprotonating it, right? What base  would I use for that? NH or NH2? Or you can write, I don't care, H minus or NH2 minus.  So I'm not particular about you concluding the counter ion. If you write that in a synthesis,  that's totally fine with me. Okay, so we've got our plan. Now let's write out the synthesis  in the forward direction. So when I grade tests, this would be the kind that's the open-ended  kind, and I don't grade retrosynthetic, and it's too hard to grade. I can't see the way  people are thinking. So I only grade the synthesis in the forward direction. So if it was an  open-ended, what I call open-ended synthesis, where I ask for products from each step, it  would look something like this. Definitely want to number those two steps. I'm deprotonating,  and then I'm adding propyl bromide. And then in my last step here, it would be H2O, H2SO4,  HGSO4. So this is three steps, this is three steps in synthesis. Or on a test, it might  look like this. If it's not an open-ended, if it's one where you just fill in reagents,  I do have more of that type on the test just because these other kind are much harder to  grade, you got to worry about these things when you have a lot of students. Alright,  so then I might have it like this. And so that would be, then you would do the reagents  and you would number the individual steps. Step one, NH. I'll just use NH. Step two,  propyl bromide. Step three, H2O, H2SO4. HGSO4. And that should be H2O, not H2. What would  happen if I didn't number that? What would happen if I mixed sodium hydride, propyl  bromide, H2O, H2SO4, and HGSO4? A very potentially explosive reaction between the sodium hydride  and the sulfuric acid. Or sodium hydride in water. It's an extremely strong base. And  what happens when you mix an extremely strong base with an extremely strong acid and you  don't cool it, you just throw them together, it explodes. It might not blow out a wall,  but it will blow compound all over you, that you don't want all over you. So that's why  it's really critical that we number the steps as we're doing them, when they need to be  numbered. That is chapter 11. We're going to start chapter 12 now. Any questions while  I'm pulling up chapter 12? It takes a little while to save this, so what can I, anybody  have questions? Alright, chapter 12. Pretty excited about this. Now, here's the thing  I don't like about chapter 12. There's no mechanisms in chapter 12. I like mechanisms,  right? It also helps you learn what the reaction is actually doing. So this is as close as  you're going to come in organic chemistry to a chapter which is all memorization. I  know a lot of people think O-chem is all memorization, but really it's some memorization and this  is all memorization. I will show you approximately what's happening in a couple of different  things, but I will not ask you any mechanisms in chapter 12. And the reason is the mechanisms  are a lot more complicated, more complicated than we're ready for. Okay, so that's where  we're coming from. So we're going to talk about oxidation and reduction. So first of  all, recognizing oxidation and reduction in organic compounds. And I think this was in  like chapter 4 or 5 or 6 or something like that. It was covered then. I'm going to assume  you don't remember it, so I'm going to cover it again. But you will see the exact notes  in one of those chapters if you look back in 51A. Alright, so it's a little bit different  in organic chemistry than general chemistry to recognize oxidation and reduction. Here's  the good news. Does everybody remember balancing redox reactions? We don't do that here at  all. Okay, so that's brutal. I know. So let me give you a G-chem example. We also aren't  going to be assigning oxidation, we're not assigning oxidation numbers at all. We don't  do that. You can. We had a textbook one time that did that, but none of the other ones  have. And so we just, you don't need it. So we're not going to cover it. So copper, shiny,  red, metal reacting with silver plus. The copper loses two electrons, right? So this  loses two electrons. And transfers those electrons over to silver. The electrons are  transferred completely from one species to the other. So then we get the blue copper  two plus salts. So pretty cool color changes. I like these things. But this is going to  be gaining. This gains electrons. Okay, so we know also that if one molecule is oxidized  and another is reduced, that's from G-chem. We're going to carry that forward. And the  electrons are transferred completely from one molecule to another. That's what happens.  Not like that in organic chemistry. So in organic chemistry, we're not actually transferring  electrons completely. We're changing the direction of the flow of electrons. So oxidation is  loss of electron density. Carbon loses electrons by forming bonds with elements that are more  electronegative than it is. So what we're going to see for oxidation is loss of a carbon  hydrogen and gain of a carbon oxygen. These both have to happen at the same time. Both  have to happen for oxidation. That's what we're going to look for. Loss of carbon hydrogen,  gain of carbon oxygen. So what are we talking about with electron density? Well, electronegativity  carbon is 2.5. Hydrogen 2.2. Carbon is 2.5. And rather than use oxygen there, I'm going  to do a generic here. Let's just do X. That's a little better. Carbon X. So 2.5 for carbon  X. Greater than 2.5. So what we're doing here is we're shifting electron density. With  a carbon hydrogen bond, the electron density is flowing towards the carbon because it's  more electronegative. When we replace that hydrogen with an X, it's flowing towards  X. So we've changed the direction of the flow of electrons. So for here, X equals oxygen,  nitrogen, halogen typically. These are the ones that we're going to see. That means  you have an oxidation. And it's important to be able to recognize if you have an oxidation  or reduction because if you have an oxidation, then you'll know you have to use an oxidizing  agent. If you have a reduction, you'll know you have to use a reducing agent. That helps  you to figure out what the rubber agent to use. Reduction is the opposite. Loss of CX,  gain of CH. So as you can see here, we have electrons flowing to X now and we're changing  it so the electrons are flowing to carbon. And again, here X equals oxygen, nitrogen,  halogen. I'll just say it's satyr. Because it could be anything. More electronegative.  So the following shows stepwise oxidation of methane. Most reduced form of carbon, all  the way up to carbon dioxide, which is the most oxidized form of carbon. So you can see  what's happening here. As we go from methane to methyl alcohol or methanol, we've lost  a carbon hydrogen and we've gained a carbon oxygen. So both things have to take place.  So therefore it's an oxidation. Moving to the right, it's an oxidation. And then we  have, we do that one more time. We break another carbon hydrogen minus carbon hydrogen plus  carbon oxygen. We've gained a second carbon oxygen bond because now we have a double bond  of carbon. So that's also an oxidation. And then same thing here, we lose another carbon  hydrogen and gain carbon oxygen because now we have three bonds. Carbon's bonded to three  oxygens. So minus carbon hydrogen and plus carbon oxygen. And then we're also, we're  losing another carbon hydrogen, gaining a carbon oxygen. So that's oxidation. Stepwise  oxidation. So four steps to go from the most reduced form of carbon to the most oxidized  form. If we went back the other way, I'm not going to draw that here, but it's the exact  opposite. If we go from here backwards over to here, we're losing a carbon oxygen and  gaining a carbon hydrogen. So that's reduction. Let's just do one of those. I'm not going  to do all of them. We can go the opposite direction and step back, going from here to  here. We minus carbon oxygen plus carbon hydrogen. Therefore, reduction. All right. So very helpful  to be able to recognize. So can you recognize oxidation reduction in the following examples  here? So what's happening here? We have two hydrogens and we're replacing those with  two carbon oxygens, right? So loss of two carbon hydrogens, gain of two carbon oxygens.  Or oxidation. Definitely oxidation there. Let's look at the second example here. We're  losing two oxygens, two carbon oxygens, and we're gaining two carbon hydrogens. Therefore,  it's a reduction. So if you want to do this reaction, you're going to need to use a reducing  agent. In the first one, you're going to need an oxidizing agent to do that reaction. So  let's look at one here that's one we've already looked at. Is this an oxidation reduction  reaction? Well, that's something, it's not in this chapter, so it's not. It's something  that we've already talked about. It's an addition reaction, right? So we know that  by looking at that, that's addition. That's an addition reaction. We know that's from  chapter 10. And so let's look at what's happened here. We gain one carbon oxygen and we gain  one carbon hydrogen. You have to gain one and lose the other. So it's not. It's neither  an oxidation nor a reduction. It's an addition. Questions on recognizing oxidation or reduction?  Ready? So want to know how to do that for midterm two. We're going to do reduction first  and then we're going to do oxidation. You're going to get a slew of reactions and you're  going to go home and make index cards for them, okay? There's three types of reducing  agents. There's the first type where we add H2 plus a catalyst. That's called a catalytic  hydrogenation. And you will generally see, so it will be H2 plus palladium, platinum  or nickel. Those would be the common catalysts that you would use for catalytic hydrogenation.  The second type of reducing agents are where you add two H plus and two electrons. We call  these dissolving metal reduction. We're going to talk about all of these. And then the third  type is where you add one H minus followed by one H plus and that is a metal hydride  reduction. So I can't emphasize enough the importance of the reagents in this chapter.  You will use them over and over and over again when you're doing synthesis. Alright, we still  have two minutes. So we're going to talk about all three of these in the same order that  they're appearing right here. So we'll first talk about catalytic hydrogenation. And we're  going to talk about addition of hydrogen to alkenes and then we'll talk about addition  of hydrogen to alkynes. So what you do when a catalytic hydrogenation is you add the elements  of hydrogen on each side across a double bond. And I'm deliberately showing stereochemistry  here. What type of addition is that when the two hydrogens come in on the same side? Yes,  so this is a stereoselective syn addition. So that means the hydrogens come in from the  same side of the alkene. Need to know that. Highly exothermic reaction but requires a catalyst  because of the huge energy of activation. The catalyst changes the nature of the transition  state thereby lowering energy of activation. We'll talk more about that next time.
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