In this lesson we're going to talk about alkene addition reactions in which a pi bond is exchanged for two sigma bonds. We'll talk about 10 or 15 or different alkene addition reactions, we'll talk about the reagents, the two groups that get added across the alkene, we'll talk about the regioselectivity and the stereoselectivity, and we'll talk about what those are in a little bit, and we'll definitely spend some time talking about how you even go about predicting products for these alkene addition reactions. Alright, so to start this off, we've just got to take a look at the anatomy of an alkene addition reaction. So, in an alkene, we're going to be trading a pi bond for two new sigma bonds.
So in this case, I've just got them labeled A and B. So, and as long as you're adding two different things though, we might have to consider them adding in two different ways. And so I've got it written one way here, but maybe B here adds on the other side.
And then A ends up... where B used to be. And so we've got to at least consider the possibility of these two ways, and we'll find out that for most of these reactions, they end up with two different groups and they prefer one of these over the other. So it turns out in the first step of the mechanism of all our alkene reactions the alkene is the nucleophile and it's going to be a coming and attacking either an H plus ion or an electrophile and when it does so it turns out majority of the time that's going to end up on the less substituted side of the alkene.
So notice this is a primary carbon, this is a secondary carbon, and the less substituted primary carbon is typically where the H plus or electrophile that gets attacked in the first step of the mechanism is going to end up. Now it turns out we can kind of see why this happens here. So it turns out the intermediate in a few of these reactions is a carbocation. And let's say we attached a hydrogen on that less substitute side, it leaves you with a carbocation intermediate on the more substitute side. And so had we done this the other way around, put the H on the secondary carbon, we would have ended up with a primary carbocation, but doing it the way we did, we ended up with a more stable secondary carbocation, and that's going to explain why we preferentially get one over the other of these two different regioisomers.
Cool, it turns out in the next step, a nucleophile is now going to come attack the carbocation and attach there. And so B here, it turns out, is going to be some variety of nucleophiles. And we'll leave that generic here for a little bit.
And it turns out as long as you add the hydrogen or electrophile on the less-substitute side and the nucleophile on the more-substitute side, you'll have accomplished what we call Markovnikov. addition. So Markovnikov studied these reactions, and this was the most common result, and we call it Markovnikov addition.
Now you'll have to memorize who the either H plus or electrophile is for every single one of our sets of reagents, and therefore who the nucleophile is, and then you also have to recall if it goes Markovnikov, or if it ends up doing the opposite. So in this case, A here again is H plus or the electrophile, and B over here the nucleophile, and this is way less common. In fact, we're only going to cover two out of our, like, 15 reactions that will end up going opposite of Mr. Markovnikov's rule here.
So, and when it goes opposite Markovnikov's rule, we just call it anti-Markovnikov. We don't get too creative on the name. Cool.
And so when a reaction goes either Markovnikov or antimarkovnikov, we refer to that as the regioselectivity, because it's going to tell you preferentially which regioisomer is your major product. And the vast majority of the reactions, again, are going to go Markovnikov, but two of them are going to go isomer. them we'll study will go antimarkovnikov. Now we got one more thing to talk about in this regard, and that's stereoselectivity, and that is there any relationship in the stereochemistry of the two groups that add.
So, and before we get there though, we actually have to take this one other place to see why that's even important. So if we look at a reactant in an alkene, we've got two sp2 hybridized carbons that are about to turn into sp3 hybridized carbons and that's important because as sp2 carbons they're never going to be chiral centers but in turning into sp3 hybridized carbons they might turn into chiral centers as a chance as being sp3 and tetrahedral as long as they got four different groups they're going to be chiral centers and so in a typical alkene reaction So you have a chance of forming up to two new chiral centers is typically the way it works. Now, if neither one of these turns into a chiral center, we can kind of classify this here. So let's say number of chiral centers. If you don't form any, if those out of the two options you had, neither one turns into a chiral center, then you're just going to form a single achiral product.
And so I'll label this as the number of products. Let's get the 1A chiral product and life is good, life is easy. Now, on the other hand, let's say one of these two becomes a chiral center, so one out of two.
Well, again, if you have one chiral center, it can exist as R or S. And for every single alkene reaction, doesn't matter any of the 15 that we study here, if you get one chiral center, you're going to get two products, R and S. And if that's the only chiral center, well, then those would be enantiomers. But if there are other chiral centers, then these might be diastereomers instead.
But the key is you still get two products. two different stereoisomers. What it gets tricky though is if you form two new chiral centers. If where you add the electrophile and the nucleophile, if both of those turn into chiral centers, we have a problem. So and the problem is that now you have a chance of getting up to as many as four stereoisomers.
So with two chiral centers, two to the n, where two to the two power is four different stereoisomers, it could be RR, SS, RS, or SR. So with four possibilities, it really depends on the reaction now. And so it turns out when we talk about stereoselectivity, so there is in some reactions a relationship between the nucleophile and the electrophile where they have to add to the same side both from the back or both from the front. In which case then you get two options, both from the back or both from the front. And out of the four possible stereoisomers that could exist when you have two chiral centers, you only get two of them.
There are other reactions though that actually involve backside attack and end up with the nucleophile and the electrophile adding from opposite sides, one from the front, one from the back. in one of two possibilities. And in that case, we call that anti-addition.
You only get two possible products. So if they added the same face, both front or both back, we call that syn addition, and you get two possible products. And if they add the opposite faces, you get two possible products. We call that anti-addition. And then you have the option where there's just no stereo selectivity whatsoever.
And you get both the two syn and the two anti-addition products. You get all four possible stereoisomers. Now, if you look at the little table on your handout, that's like this on your handout.
it doesn't actually say number of chiral centers, it says number of chiral centers formed. So it's not the number of chiral centers in the whole molecule, just how many you formed when your sp2 carbons became sp3. And then also, instead of saying the number of products, it says the usual number of products. So the reason we've got to be careful here is that sometimes you form meso compounds.
And so a meso compound and its mirror image are exactly the same, and so you might actually get fewer products than what we would normally predict here. But with two chiral centers, that is the only time the stereoselectivity matters. So if there's no stereoselectivity and you form two chiral centers, you're generally going to get all four products. So if you have syn or anti stereoselectivity and you form two chiral centers, you're generally just going to get two out of the four possibilities forming as a result of your reaction.
Alright, so now we've covered the regioselectivity, which the options are Markovnikov and anti-Markovnikov, and we've covered the stereoselectivity, which is sin versus anti, or just none. And so one thing to note, be very careful that you don't shorten anti-Markovnikov just saying anti. Is it Markovnikov or anti? Because anti is something else. That's stereoselectivity in this context.
And so just be very careful. Anti-Markovnikov is regioselectivity. Just saying plain old anti refers to stereoselectivity.
Alright, let's look at some examples here. So the first alkene addition reaction we're going to look at is hydrohalogenation, where we're going to add a hydrogen and a halogen across the alkene. And you might recall when we studied elimination reactions, we did dehydrohalogenation, where you lost a hydrogen and a halogen and formed an alkene. It's the exact opposite of hydrohalogenation here.
So elimination and addition, exactly inverse reactions here. So instead of forming an alkene, we're going to be consuming an alkene and forming two new sigma bonds. Now with... Hydrohalogenation, your reagents are either HCl, HBr, or HI.
And you can either add an H and a Cl, an H and a Br, or an H and an I. And so in this case it also goes Markovnikov. So here's an example here.
The H ended up on the less substitute side. The bromine ended up on the more substitute side. So we've got our primary carbon here and our secondary carbon here.
The H went on the primary, the bromine went on the secondary. That is Markovnikov addition. So, and finally the stereoselectivity, it turns out, is none, but the only time we would care is if we formed two chiral centers anyways. And in this case, where the bromine added has two identical methyl groups, not four different groups, not a chiral center, and where the hydrogen added has three identical hydrons, not four different groups, not a chiral center.
With no chiral centers formed, we get just the one achiral product. Alright, so this is an example, but we definitely want to take some time to look at the mechanism here. So, if we take a look... In the first step of every mechanism we know, the alkene is the nucleophile, and whatever else you're adding is the electrophile. In this case, HBr is a strong acid.
You learned in Gen Chem it's also a strong electrophile, and it's going to get attacked here. Specifically, that... partially positive hydrogen. Hydrogen can only have one bond, so we're forming a new one here. The old one here has to break.
And normally I wouldn't draw in this hydrogen, but I'm going to draw it in here so we can kind of keep track. That hydrogen, again, bonds on the less substitute side. Notice I didn't draw it in the product up here. And I'll draw it on the first couple here, but eventually I'm going to stop drawing these H's because you're not going to draw them on your exam either.
Cool, but the H adds on the less substitute side, and the more substitute side is now missing a bond. It used to have a pi bond. Now he doesn't. Only three bonds, and he's a carbocation. And then we also form this bromide ion here.
Cool. And that's step one of the mechanism. And because we formed a carbocation here, you typically will have to check for rearrangements. Now, it turns out there's only going to be three of the reactions for all the alkene reactions that form a carbocation, and therefore they're the only three that are subject to rearrangements.
but this is indeed one of them. Whether you use HCL, HBR, HI, you're going to have to check to see if there's rearrangements. Well, we've got a secondary carbocation. You generally only have a chance of rearranging to one of the adjacent carbons, which are both primary in this case, and therefore they would not be more stable carbocation. and there's no favorable rearrangement for this reaction.
Cool. Now this is not rocket science here. So if you can identify who's electron rich and who's electron poor, it will help you immensely in memorizing some of these mechanisms. And in this case with a negative formal charge, bromine's definitely electron rich, and that makes him a likely candidate to be a nucleophile.
So in our carbocation with a positive formal charge, definitely electron poor. makes him a likely candidate to be an electrophile. And in typical nucleophilic attack mechanistic step, the nucleophile attacks the electrophile, i.e. the nucleophile attaches to the electrophile.
And so this bromine is just attaching to that carbon right here, which looks like our final product. Cool, and again we added the H on the less subsude side, the bromine on the more subsude side, and that was Markovnikov addition. Also we didn't form any chiral centers and so we just get that one achiral product. Now it turns out we have a second option with hydrohalogenation. So here we had HCl, HBr, and HI works exactly the same way, but for specifically HBr and HBr only, you have another option.
With HBr you can mix it with what's called a peroxide and ROOR is our abbreviation for a peroxide, where those R's could simply be hydrogen, which would be hydrogen peroxide. HOOH is H2O2. Or they could just be some sort of, more commonly, make this an organic peroxide, make them like methyl groups or benzene rings or something like this. So, but oftentimes, more commonly, you'll just see it written generically just like this. And I like to think that HBR and ROAR is the name of the reagent.
So we'll see why that's convenient in just a second. In this specific reaction, we're going to add an H. and a bromine however it's not going to go markovnikov now it's going to go anti-markovnikov and so i like to think of hbr and roar the roar scares hbr into going anti-markovnikov so just my stupid little way of remembering it there's also no stereo selectivity in this reaction either now the mechanism of this one we are not going to cover just yet. So mechanistically on this one, it goes through a radical intermediate, not carbocation.
So totally different mechanism. So, and we'll study those in the radical chapter a little bit later. So, but suffice it to say, it does go anti-Markovnikov, but not because we're violating the rules of chemistry or trying to get the less stable carbocation or anything like that.
We're not even going through a carbocation, totally different mechanism. So, but if we predict some products here, if we did the exact same reaction. but added some peroxide with our HBr. Now all of a sudden the bromine ends up on the less substitute carbon, the hydrogen would have added the more substitute carbon, and there's our product. So neither one of the carbons that became sp3 hybridized are chiral centers, neither of them have four different groups, and so we get just the one achiral product.
All right, so so far we haven't formed any chiral centers and it makes this really easy. You don't have to show any stereochemistry, get one achiral product, life is good. So I want to give you an example where we actually do form a chiral center.
And so let's do one more here and this time around I'll use... HCl, could have used HCl, HBr, Hi. So still going Markovnikov then, normal hydrohalogenation. So first step, we're gonna go grab an H, the bond to chlorine breaks, and we're gonna form a carbocation here. And I highly recommend we just draw those carbocations out just to keep track.
So the H ends up on the less substitute side of the alkene over here, not drawn in. So, and the more substitute side is gonna be the carbocation. In this case, it's secondary.
The two adjacent carbons, one's primary, one's secondary. Neither one is better. This one's equally stable, but it has to get better to be a favorable rearrangement for one of the adjacent carbons.
So no favorable rearrangement here, and that's where chlorine ultimately is going to attach. And if we take a look, we'll see something different here where the H added is not a chiral center. There's not four different groups.
But where the chlorine added does indeed have four different groups. A chlorine, a hydrogen that's not drawn in, a methyl, and the carbon of an ethyl. Four different groups, that's a chiral center. And this is not sufficient answer.
So with a chiral center, it could exist as R or S. And if you form one chiral center, no matter which alkene reaction you're looking at, all of them will always form R and S. And so you could represent this one of a couple of different ways. So one, and the most common way, is just to draw both different versions out.
One where the chlorine is a wedge, and one where the chlorine's a dash. So draw both R and S. So although some professors will also do this, and We are kind of the kings of laziness in organic chemistry, and the fewer things we can draw, the better.
And so sometimes we don't even show the stereochemistry on that chiral center, but we draw plus minus just to show that you can get both the positive and the minus enantiomer. Another way of saying you get both the R and the S. So, cool, but more commonly you'll see professors do it the top way, and less commonly the bottom way. I will tell you that every professor will accept the top. Not every professor will accept the bottom, so you're better off just drawing out both enantiomers in this case.
Alright, the next reaction we'll take a look at is acid catalyzed hydration. Now we'll call it hydration because we're going to add water across the alkene. We'll add an H and an OH. But it turns out water and alkenes don't react very well together, so water is not the greatest electrophile.
But if you put acid in there to catalyze it, so you get a much better electrophile H3O plus as it turns out. Now the reagent here is just going to be H2SO4 that is dilute. And so sometimes we'll write this a couple of different ways.
Instead of writing H2SO4 that's dilute, some people will write H2SO4 and water, some people will just simply write H3O+. They're all correct. No matter which of these three ways you see it represented, it all means the same thing, acid-catalyzed hydration.
Now it turns out this is not the only way we carry out the hydration reaction. So we add H and OH across the alkene again. There's actually three different reactions.
that we can use to accomplish this. This is the first one and they all have their differences and we'll emphasize all of those here shortly. Cool. But acid catalyzed hydration here so and I'll choose to write H3O plus one of the ways we can represent it and so in this case There's your product. We add an H to the less-substitute side, the OH to the more-substitute side.
It does go through a carbocation, and had you drawn out your carbocation intermediate, so you'd realize it's not going to rearrange and stuff like that. But there's our product. Adds H and OH, H on the less-substitute side, OH on the more-substitute side. That's Markovnikov addition. And in this case, we did not form any chiral centers, which is why we would get just...
the 1-achiral product. Cool. So that's just predicting products.
We haven't even shown the mechanism yet, and you guys know how to predict the product. But we definitely need to understand the mechanism, and it's very similar to the mechanism with HBR here. Alright, so regardless of how you write it, whether you write H2SO4 in water, dilute H2SO4, H3O+, you should realize that H2SO4 being a strong acid dissociates completely to form H3O+.
And water, again, is not a good electrophile, but... H3O plus is. And so in step one of every alkene reaction we know, the alkene is the nucleophile.
You should definitely know that, which makes H3O plus here the electrophile. And so our alkene is going to come and grab an H, which can only have one bond, so the old one breaks. And one more time here, I'll draw that H in. That H ends up on the less substituted side of the alkene.
That way you get the more substituted and more stable carbocation intermediate. Cool. You also formed a water molecule here as well.
which is not the biggest deal in the world because your solution is dilute H2SO4 and is chock full of water molecules. But we did form one. We should keep track of it.
It's also going to be involved in the next step. Now, in the last reaction, in addition to the carbocation, we had a bromide ion. And the carbocation we identified as...
electron poor and therefore a good electrophile. So but we had Br-which we identified as a good nucleophile. Well we got water in this case and water doesn't have a negative formal charge and he's not a great nucleophile as we learned when we're doing substitution reactions.
So but he'll do in this case. He does have a lone pair of electrons so he actually can act as a nucleophile but the carbocation having no filled octet positive formal charge so highly reactive water will do. And we've seen water react as a nucleophile.
The file with carbocations back when we did SN1 reactions and so water is going to come attach here Giving us this lovely species here. And this lovely species with a positive formal charge on oxygen. So it's a very good acid just like H3O plus is a very strong acid. So very analogous.
So as we saw with SN1 reactions, when a neutral nucleophile comes... and attacks, it ends up with a positive formal charge. And just like we often did there when water or alcohol is attacked, you bring in another molecule of your solvent to deprotonate, to do a proton transfer so that you don't end up with a positive formal charge on your product.
And so in this case, we'll just draw in another water molecule. Our solution, again, is chock full of them. Cool, and that forms us our product, again which is achiral, and we also form some H3O+, the good hallmark of a true catalyst. It is not consumed in the reaction.
And so we used up a molecule, or in this case ion, of H3O to begin with, and then we formed one before the reaction was done. It's not consumed, just as a catalyst shouldn't be. Cool, that is our mechanism. It looks very similar to HBr, with just one extra proton transfer step at the end.
Our next alkene addition reaction is oxymercuration demercuration, and it is also a hydration reaction. It still adds H and OH across an alkene, so it also goes Markovnikov. Now you see the big difference though is that it does not go through a carbocation intermediate, and so it's never, you never have to worry about rearrangements, never going to happen.
So with acid catalyzed hydration, if there's a favorable rearrangement, the major products form through it. Oxymercuration demercuration, there are never rearrangements. All right, so your reagents here are mercuric acetate with water in the first step, followed by sodium borohydride in the second step.
And you'll notice we've got this sequential listing of steps, and that's super important. If you forget to write the one and the two, it is technically wrong. And the idea is that it gives us that it's sequential here. So to our alkene, we're first going to add mercuric acetate and water.
We're going to let that react. We're going to purify the result of that first step. And then we're going to take that purified result and add sodium borohydride.
If you forget to write the 1 and the 2, it just means you added all the reagents at the same time to your alkene, in which case you're not going to get the desired product. So in this case, we're still adding H and OH. The H added on the less substitute side of the alkene.
The OH added on the more substitute side of the alkene. And that, again, is Markovnikov addition. All right, so let's take a look at the mechanism of this reaction. So it turns out the name actually comes from the mechanism.
In step one, we do oxymercuration, and it turns out we're going to add an oxygen and a mercury. In step two, we do demercuration, so you lose the mercury, and it turns out it gets replaced with a hydrogen. That's where this kind of comes from. And so let's take a look at step one here.
So it turns out that the mercuric acetate... dissociates to a small extent here to form a couple of ions. So, and you've got a nice positively charged mercury here, which is an amazing electrophile.
Not only does it have a positive formal charge, but it's a metal. It's low in electronegativity, and the alkene, as we've said, is the nucleophile in step one of all the mechanisms we know. And he's got an amazing electrophile to react with.
So, in the solution. Cool. Now, if we look at what we think is about to happen based on everything we've learned, we're going to find out it's going to be a little bit different. But let's just predict what we think should happen based on everything we've done.
So we think we're about to come and attack Mercury. So, and this might be... where you take some notes off set, cause I'm gonna erase this in a little bit. So, and we think, oh, hey, we're gonna bond a mercury here.
I'll draw this off to the side way over here and we'll attach to that mercury on that left substituted side. So, and form. a carbocation but you might remember we said that we don't form a carbocation in this reaction and so this is not apparently what's happening and so mercury's like hey hey slow down let's not form this carbocation because you know what i don't have the limitations hydrogen did when you bonded to hydra mr alkene before he didn't have any electrons but i'm a big fat metal and i got lots of electrons and so i'm willing just because i like you to form another bond to you as well. And so I'll bond to both of you carbons that way everybody's got a filled octet and life is good. And that's actually what happened.
So it turns out our intermediate here is not a carbocation. Our intermediate has a three-membered ring with mercury and mercury still has a positive formal charge. We call this a mercurinium ion. If I can spell that correctly. So when you see a word ending with i-u-m in chemistry here, so know that you've got a cation, a positive charge, like hydronium, H3O+, ammonium, NH4+, mercurinium here with a positive formal charge.
And so our major intermediate, again, not a carbocation, it's a mercurinium ion, and you never have to worry about rearrangements. So, but to draw this, it's probably easiest to show if we redraw that mercury off over here a little bit. We'll show its lone pair of electrons here.
And so your alkene is going to come and attach to the mercury, and your mercury is going to come and attach back, forming that three-membered ring. And so this is how we typically represent it, so that it's a little easier to see exactly what's going on. Cool. So now we've got this lovely three-membered ring intermediate and no carbocation. And in the next step is where water gets involved.
So water is not a great nucleophile. So, and just like an acid catalyzed hydration where water attacks the carbocation, water is going to be attacking this mercurium ion. The question is where? Well, it's going to be one of the two carbons bonded to mercury because they're both sharing a little bit of that positive charge by both induction or resonance you might even consider, but induction is usually how we present it. So the question is...
Which one's more positively charged? Well, whichever one would be more stable as a carbocation also can handle more of the partial positive charge. And that's the secondary carbon here, not the primary one here. Having more partial positive charge, it would have a lower activation energy if water attacks there. He's more attracted there.
Another way to look at it. And so that's where water, it turns out, is going to come and attack. But we've seen something a little bit different here. We're attacking this three-membered ring, and we're attacking that carbon specifically. This is the first time in this.
chapter we've attacked an sp3 hybridized carbon in every other case so far it's been we've been attacking an sp2 hybridized carbocation well when you attack an sp3 carbon as you learned in the chapter on sn2 it's backside attack totally still backside attack here so if it matters it will eventually that's why it's going to end up being anti uh stereo selectivity because we have to do backside attack when we attack this three-membered ring this mercury amion all right so in this In this case, your mercury is still bonded on the less-substitute side, but now you've got a water molecule bonded on the more-substitute side. And just like we did in acid-catalyzed hydration when the water attached to the carbocation, a neutral nucleophile attacks and you end up with a positive formal charge and another molecule of your solvent, in this case water, will come and deprotonate. Alright, technically I guess we formed some H3O+.
Alright, so this is the mechanism for the first synthetic step in the reaction. So, and there's a little confusion on exactly what's happening in the second step, at least the exact mechanism. And so that's a freebie for you.
You get out of having to know the mechanism for the second step. If we're unsure about it, well then you don't have to know it. Life is good. And so for step one, that mechanism is totally on you. And notice we had one, two, three steps in that mechanism.
But step two with sodium bar hydride? That's where demercuration takes place. You have to know what happens, but you don't have to know the arrow pushing for it at all. And so when we add the sodium borohydride, it simply replaces that mercury with a hydrogen. Demercuration.
Cool. And there's our final product. And this example, again, I didn't form any chiral centers.
And so we just get the one achiral product, as we've seen be the case before. Alright, our next alkene addition reaction here is hydroboration oxidation. So here are our reagents here. So BH3 is borane.
So like methane is CH4, BH3 is borane. And it turns out we complex it with tetrahydrofuran here. So BH3, boron's notorious for not having a filled octet. So the tetrahydrofuran actually will bond to it in kind of a quasi-bond and kind of temporarily give it a filled octet to make it a little more stable. Now technically, though, the reaction can be done without tetrahydrofuran.
Just... plain old borane. So but it turns out borane will commonly dimerize and so you typically see that first step written as B2H6 instead. But the second step is going to involve peroxide under basic conditions so H2O2 and NaOH.
Either way. So, and again, the most common way you're going to see it here is BH3 complexed with tetrahydrofurane. But you might see it this way. So just a word to the wise. All right.
So this is also hydration. It still adds an H and an OH. It adds water across an alkene. But this time it goes anti-Markovnikov.
And so in this case, the H ends up on the more substitute side and the OH on the less substitute side backwards of what we saw with both acid catalyzed hydration and oxymercuration demercuration. Cool. Now this is also, again, two mechanistic, I'm sorry, two synthetic steps. So steps in synthesis. And just like with oxymercuration, you're probably going to be on the hook for the mechanism of step one, but not for step two.
Now we do know the mechanism for step two, but it's long. It involves radicals, and it might be a little complicated for the typical OCHEM1 student. So most professors, most textbooks let you off the hook.
So if on the odd chance you're on the hook for this one, it's on my website, chadsprep.com. You can find it there. All right, but let's take a look at the mechanism for step one because that you are on the hook for.
So take a look here. In fact, you know, I'm going to flip this around so it's a little easier to see when we're drawing out this mechanism. Alright, so our alkene here is going to line up with BH3, and like we've said in the first step of every mechanism we know for these alkene addition reactions, the alkene is the nucleophile.
Great. And so BH3 is the electrophile, and a lot of students think, oh, a tachyhydrin. Not. true in this case. You have something better than hydrogen.
You have a highly electron deficient boron with no filled octet, big fat MTP orbital, amazing electrophile. And so your alkene is actually going to come and attack boron. And just like we've seen before, whatever the nucleophile, I'm sorry, whatever the alkene attacks ends up on the less substituted side. It's going to end up on this less substituted carbon, but it turns out simultaneously this boron hydrogen bond breaks and breaks.
and reattaches to the other carbon. Cool. So I'm gonna save a little room and you should too on your notes. But what we end up with here is this lovely species here, where boron, which had three hydrons, now only has two, and its third hydrogen is now attached to this sp3 carbon here.
So if you notice when this carbon was sp2 hybridized, it only has one hydrogen. Now it has two, it gained a new one. Cool. So, and the reason we had you save a little room here, well, this is, I mean, this is, this is the whole mechanism.
I'm like, it's just one step. So, and that's too easy. And so oftentimes we say, well, you know what, let's make the students draw a transition state for this one, just to make it a little longer.
So you're on the hook for drawing a transition state typically. So transition state has all bonds that are being broken and formed as partial bonds. And so in this case we have a partial bond to boron, a partial pi bond being broken, so a partial bond to hydrogen also being broken, but forming a new bond between carbon and hydrogen right here. And so you end up... let's make that a little cleaner.
But you end up with a transition state in which you've got kind of a four-membered ring for all the bonds that are being broken and formed. So cool. We also use this to explain why this ends up as a sin addition though. So it turns out the two things we're adding here, boron and hydrogen, to add to the same side of the alkene, either both from the front or both from the back. So because they're coming from exactly the same molecule, they're right next to each other, they were bonded together, and so they have to add to the same side.
And that's what's going to be characteristic of a couple of our synoditions, is you add two two atoms at the same time to your alkene from the same molecule, and so they have to add to the same side. Cool. So here our alkene has reacted with BH3, and we end up with it bonded to BH2. This is the hydroboration. We added a hydrogen and a boron.
Hydrogen on one side, boron on the other side. Now it turns out boron had three hydrons to begin with, now he's only got two. So every time he reacts with an alkene, he loses a hydrin.
Well, he's still got two left. And so it turns out this process repeats itself two more times and reacts with two more alkenes, forming in this case what's called a trialkyl borane. And that is actually the true intermediate at the end of your synthesis step number one right here. Cool. And this is typically, again, the entire mechanism you're on the hook for, for most OCHEM1 classes.
So for synthesis step number two here with peroxide and sodium hydroxide. So again, I've seen once in 20 years. Years students be on the hook from their professor for this but so I'm not saying it's an impossible It's just not likely and for most of you all you really need to know though is that when we add?
peroxide and Sodium hydroxide here that we're gonna break every single Carbon borane bond carbon boron bond I should say and in this case just replace them all with an OH. But rather than drawing three of these, you're just going to draw one, because they're all the same thing. So regardless of how you look at it.
So we'll just draw one, I'll take the other two away. And we just essentially get three of these, but from three alkenes. Cool, and that's your final product.
We added an H to the more substitute side, the OH to the less substitute side, it went anti-Markovnikov. Where we added the H, so not a chiral center and where we add the OH, not a chiral center with no chiral centers formed, we just get the one achiral product. There's your mechanism. Cool.
So we've covered three different hydration reactions now. So, and I want to compare and contrast. with one strategic example.
All right, our strategic example here is gonna be this lovely alkene here. And we've got acid catalyzed hydration, oxymercuration, demercuration, and hydroboration oxidation. The top two both go Markovnikov.
The bottom one here is antimarkovnikov. But between the top two, the first acid catalyzed hydration is subject to rearrangements because it goes through a carbocation. And so in this case, we should definitely draw, okay, what would the carbocation look like if we did go through one? So, and in this case, it would.
look, we'd add the H right there and end up with the more substituted more stable carbocation. But even so it's still just a secondary carbocation and the adjacent carbon on the right is not more stable but the one on the left was a tertiary carbon and is. And so as a result we are going to do a hydride shift and so we'll draw in the relevant hydrogen and just transfer him over to this carbon.
That way this carbon gets a filled octet but this carbon will now be missing a bond and will now be the carbocation in the next structure. Cool. And there we get our tertiary carbocation. None of the adjacent carbons are any more stable than that. It's not going to rearrange anymore.
And that's ultimately where water is going to attack and where the OH ends up in the product. And so in this case... There's our product, so undergoing a rearrangement. Now, had we done this with mercuric acetate instead, deoxymercuration, demercuration, then we would have never had the chance for a rearrangement. So we would have just simply added the A.
to the less subsuite side, the OH to the more subsuite side. And if we do this with hydroboration oxidation, then it goes anti-Markovnikov, and we'll add the OH to the less subsuite side, the H to the more subsuite side. Cool. But we should take an exam.
these products a little further and see how many chiral centers we formed. So we added an H here, not a chiral center. In the rearrangement, we actually add another H here. He's now newly sp-thribronized, but still not a chiral center. And this is where the OH added, and he's got two identical methyls, and he's not a chiral center.
And since we didn't form any chiral centers, we don't have any stereochemistry to show, we get one achiral product. But for the next one here, so where the H added, it's not a chiral center, but where the OH added, that is indeed a chiral center. And if you form form one chiral center, you do indeed get two products.
And so technically here, we probably should have drawn this out like so. Show that racemic mixture. And then finally with hydroboration oxidation, we didn't form any chiral centers here either. So where we added the H on the more subsude side, not a chiral center.
Where we added the OH, also not a chiral center. And we get just this one achiral product. Cool. So great example to show you the difference. difference between the three different hydration reactions.
So much of the time acid catalyzed hydration and oxymercuration demercuration will give you the same product provided there's no favorable rearrangement. But when there is like this, you can definitely see the difference in the products they produce. Alright, the next reaction we've got to cover is acid-catalyzed addition of an alcohol. So in this case, it's very similar to acid-catalyzed hydration, but instead of water with acid, we add alcohol and acid.
If adding water and acid is acid-catalyzed hydration, maybe we call this acid-catalyzed inebriation, or something like that. But in this case, you can use any alcohol. In this case, I've chosen methanol.
And the idea is that now we're going to add an H and an OR, where the R, in this case a methyl group, just depends on which alcohol you use. So we've added an H on the less-substituted side. and the OR on the more substitute side. And that is Markovnikov addition.
Turns out this also goes through carbocation. In fact, the mechanism is nearly identical to acid catalyzed hydration. And so pretty much everything's the same. So there's no stereoselectivity.
But we do have to worry about potentially carbocation rearrangements, just not in this example. So let's take a look at that mechanism. So I've got the mechanism of acid catalyzed hydration below. So we can see the parallel here.
So and just like when you put H2SO4 in water it dissociates completely to form H3O+. So it does the same thing in alcohol here as well and so... Instead of protonated water, we've got protonated alcohol here. And again, the first step, any alkene reaction, our alkene is the nucleophile.
And in this case, instead of H3O plus being the electrophile, our protonated alcohol is the electrophile instead. But totally analogous here. Cool, that H ends up on the less substituted side, that way we get the more substituted carbocation.
So, and you also form a molecule of your alcohol. And just like we have water attacking the carbocation in the next step, we just have the alcohol attacking instead. And just like with acid catalyzed hydration, when a neutral nucleophile attacks, it ends up with a positive formal charge, and you just have the solvent deprotonate. So in this case, same thing, positive formal charge, and we'll just have the solvent deprotonate. And the solvent in this case is not water, it's just another molecule of the alcohol.
And that takes us to our final product here. And again, like a true catalyst, Ours is not consumed in the reaction and so our CH3OH2 regenerated before the reaction is done. Cool. Totally analogous to acid catalytic hydration.
Same mechanism, not worth spending any more time on. All right, the next alkene addition reaction we're going to look at is alkoxymercuration demercuration, and very analogous to oxymercuration demercuration, but again, instead of adding H and OH, now we'll be adding an H and an OR, all depending on which alcohol used in step one. So the only difference in the reagents is instead of using water here, we'll pick an appropriate alcohol. All right, so it's still Markovnikov, so H on the left subscude side and the OR, in this case, OCH3 on the more subscude side.
and it's an anti-addition in case you form two chiral centers. So when you do backside attack on that mercurinium ion, leads to the two things that have added being on opposite sides. All right, so let's take a look at that mechanism. So I've got the mechanism for step one of oxymercuration-demercuration here, and we'll see it's exactly analogous for alkoxymercuration.
And just like with oxymercuration-demercuration, you do not have to know the mechanism for step two here. So mercuric acetate is going to dissociate to a small extent just like it did before. So and our alkene is the nucleophile in step one and will come and attack the mercury.
But again mercury's got a lone pair and says don't form a carbocation, I will attack back and we'll form a three-membered ring that we'll call a mercurinium ion here. So there's our mercuridium ion and instead of having a water molecule come and attack again this has taken place in methanol instead and that's what actually is going to come and do backside attack here on the more substitute carbon opening up our three-membered ring. Cool, and again, we had a neutral nucleophile attack, so forming a positive formal charge intermediate here, and we'll deprotonate. But instead of water being our solvent, again, methanol is our solvent. All right, cool.
And that's the end of the mechanism of step one. We've accomplished alkoxymercuration. Cool.
And then again, second step when we add NABH4 would just be the demercuration and replace the mercury there. with a hydrogen getting us to our final product. That's alkoxymercuration.
Demercuration, again, mechanism exactly analogous to oxymercuration, but instead of forming an alcohol for a product, we're going to form an ether as a product instead. Our next alkene addition reaction is going to be catalytic hydrogenation. And catalytic hydrogenation occurs when you add H2 with an appropriate metal catalyst, typically palladium, platinum, or nickel.
So here we've got a common example, and with our... Palladium catalyst that's palladium on carbon or palladium on charcoal or something along those lines Common form of palladium we use for these reactions and we're gonna add an H and an H and because we're adding a two of The same thing so there is no regio selectivity to talk about It's not like well which side gets the H and which side gets the H Like when we added HBR we had an H who gets the H and who gets the BR but if you're adding two of the same Thing then there is no Regio selectivity no Markovnikov no intermarkovnikov to talk about So however, there is stereoselectivity here and it is a syn addition as we'll see. Alright, so we don't exactly know the mechanism of this reaction, which is a beautiful thing, because if we don't know the mechanism, then you don't have to know the mechanism.
So we kind of have an idea of, you know, maybe how it goes and stuff, but we're not sure. So it turns out that hydrogen is going to adsorb to the surface of one of our metal catalysts here, and it turns out that weakens the hydrogen-hydrogen bond. So, and our alkene is going to line up to add both hydrons at the same time here from the same molecule of H2. And because we're adding both hydrons at the same time from the same molecule, they have to add to the same face.
So, and that's what makes this a syn addition. So I'm only showing this so that we understand why it's a syn addition. So, but other time we saw syn addition was with BH3 and we added both the boron and the hydrin at the same time from the same molecule.
Which is again the exact same reason they added the same phase. So we can explain why this is syn addition. So let's take a look at a couple examples, and in this case there's no mechanism to show, we're just going to predict the products. So in this first one, if we add an H to both sides, we just simply turn an alkene into an alkane.
And so a lot of students just think, oh yeah, it takes the double bond away. Well again, it adds two H's. We don't draw them in, but they both got added there.
So this is technically also called a reduction. Some people even call the reaction catalytic reduction. So when you're adding hydrogens. So But that'll be more important later on. So if we take a look at a more creative example here.
So we didn't form any chiral centers here. We're about to in this one though. So in this case, H2.
And again, we like to think that made the alkene disappear, but again, we added a hydrogen to both sides. So, but this is not going to be sufficient now. So this carbon's got four different groups. He's got the methyl carbon, these two carbons are different, and he's now got a hydrogen. Same thing, this carbon's got the ethyl carbon, these two carbons, which are unique, and then a hydrogen as well, and we formed two chiral centers.
Now, this is the first example we've looked at where we formed two chiral centers. And again, when you form two chiral centers, you may get as many as four products. So, but if you have stereoselectivity, you're only going to get two out of the four typically. So in this case, this is syn addition.
We're only going to get the two syn products. And what this means is that the two hydrogens we added, added to the same face, the same side of the molecule, both wedges or both dashes. So let's say I made them both dashes. Well, if they're both dashes, that would mean that our methyl and our ethyl are both wedges.
And that would be one of the products. But again, the two hydrons could have been both on the wedge side, which would mean that my methyl and ethyl are both on the dashed positions. And so we get this racemic mixture right here, two products.
Cool, and those would be our two syn products. Now, one thing to be careful of, and this question shows it pretty commonly here, so it's an example worth noting. Instead of having a methyl and an ethyl there on our cyclohexane, let's just say we had two methyls, and we did exactly the same reaction. And so you might be inclined to, again, draw both wedges. And both dashes.
And this time you're gonna lose some points unfortunately. So we did indeed still form two chiral centers. These are still both chiral centers, but see that mirror plane of symmetry?
This is an achiral molecule. Specifically, it is a meso compound. And this and this are exactly the same. meso compound just flipped over but they're exactly the same and so when you form a meso compound in one of these reactions so when you form two chiral centers you're not going to get as many products as you think so normally when you've got syn addition and two chiral centers you get two products but not if you form a meso compound just draw one of the two if you draw them both in all likelihood you're losing points okay one last thing we got to look at with hydrogenation So oftentimes we use this as a way of characterizing alkenes that we're doing the hydrogenation reaction on.
So it turns out that when we hydrogenate an alkene and turn it into an alkane, it is always going to be an exothermic reaction. So that's the first off, but there are varying degrees of how exothermic it's going to be. And we call the amount of heat it releases the heat of hydrogenation.
And so you might have to rank some alkenes to see how... Exothermic are they? And stuff like that.
So here I've got three different alkenes that are all related, that all upon hydrogenation form exactly the same product. So, and I've kind of ranked them in order of increasing energy here as well. So this is our most stable alkene. So the alkene is between what's going to be a tertiary and secondary carbon. So this is the second most stable.
The alkene here is between what's going to be a primary and secondary carbon. And this one's the least stable, the least subsumed. being between what's going to be a primary and secondary carbon.
And so we learned in the chapter on elimination reactions that the more substituted the alkene, the more stable. And so this is the lowest energy alkene, this is the highest energy alkene. But upon hydrogenation, they all form a lower energy alkane. And if we kind of take a look at what this might look like.
So you're going to end at the same place in all three hydrogenation reactions, right at this alkane. But you're going to start at three very different energies. And the most substitute alkene would start at the lowest energy, and therefore it would be the least exothermic reaction. So the next one would be a little more exothermic.
And then the least substituted alkene there would have the most exothermic reaction in being reduced. And so you might have to rank these. And so we'd say that this alkene that was the least substituted, that starts off the highest energy, would have the highest heat of hydrogenation. It would release the most heat upon hydrogenation. The next alkene addition reaction here is halogenation, and we are going to add a halogen to both sides of the alkene, usually chlorine and chlorine, or bromine and bromine.
So typically the way it works, and because we're adding two of the same thing, again, there is no regioselectivity, so no more Kovnikov, no more antimarkovnikov to even talk about. So, but it turns out this one is an anti-addition, and it turns out this is going to involve a three-membered ring, with a halogen this time, not with mercury, like in oxymercuration-demercuration. So, but involving that three-membered ring, we're going to be doing backside attack on an...
sp3 hybridized atom again and that's why it's going to result in anti-addition as we'll see all right so if we kind of take a look at how this works we'll just predict a couple products and then we'll work through the mechanism here and so we'll kind of use our standard alkene here and if we had br2 and ch2 cl2 we would add a bromine to both sides Cool. Now where I added the bromine on the less substituted carbon, that did not turn into a chiral center. It's got two hydrons.
So, but this one did. And again, if you form one chiral center, no matter which alkane reaction you're talking about, you will always form both versions, R and S. And so this would not be a sufficient answer here.
You'd have to draw both enantiomers. So you could get that version of the chiral center or... This version of the chiral center.
Notice, again, this bromine, a lot of students have asked me, well, Chad, why aren't you showing wedges or dashes here? Well, it's not a chiral center. It only can exist in one form, regardless of how you draw it. You can make this a wedge, a dash, or the bond in the plane.
It's the same thing either way. But here at a chiral center, it totally matters. it's a different stereoisomer depending on how you represent it.
And so we had to definitely show that there. Cool, but just two products. Now, on the other hand, what if we formed two chiral centers like we're about to in this one?
So I'm going to add a bromine to both sides yet again. So, and in this case, we do form two chiral centers. And when you form two chiral centers, that's when your stereoselectivity matters. And again, this is anti-addition.
And so in this case, you have to add the two things you add, in this case two bromines, to opposite faces. And what that means is one's going to be a wedge and the other one a dash. But you could do this the other way as well. The top one could have been the dash. and the bottom one, the wedge.
And these are enantiomers. This is not a meso compound. In this case, both bromines would have had to be in both wedges or both dashes for it to be meso. So, not a meso.
These are enantiomers, and you get them both in this example. Cool. I want to also show the mechanism for this last one as well.
Alright, so in this case we know the first step that the alkene is going to be the nucleophile every time and in this case He's gonna come and attack bromine and again you might think well are we gonna form a carbocation and we are not going to form a carbocation because again just like mercury bromine's got electrons and he's like oh no no don't form a carbocation I will also attack you back and form that three-mered ring. Now it turns out when we attack the bromine the other bromine does break away and leave so to speak. Which takes us here.
All right, so we've got this lovely three-membered ring with bromine. When we had a three-membered ring with mercury and a positive charge, we called it a mercurium ion. This one's called a bromonium ion. Had it been with chlorine, it would have been called a chloronium ion.
Or for any generic halogen, we call it a halonium ion. So I just want you to be familiar with those terms. We also formed another bromide ion here as well. Cool. Now I haven't shown any stereochemistry here on my intermediate, but technically those are chiral centers, and so I should be showing some stereochemistry.
Well, they have to be bonded on the same side. I can't have one, you know, bonded out in front of the board, one behind the board, because they're bonded the same thing, the same exact atom. So they're either both wedges or both dashes, and it's really my choice. In fact, it's the same thing either way, because it's a in this case. So I'll make them both wedges, but what I really just want to emphasize is now we've got to do backside attack.
And this bromide here, these two carbons are both equivalent, so it doesn't really matter which one we attack. If one of them was more substituted than the other, though, that bromide would prefer the more substituted one. These two carbons share the partial positive charge, or they share that positive charge of the bromine, so they're partially positive. And whichever one is more substituted will have more of that partial positive charge, and bromide can attack it. with greater attraction and lower activation energy, so to speak.
But in this case, it totally doesn't matter. And so we'll come and do backside attack. Again, when you attack an sp3 carbon, it's got to be from the backside. And it's backside relative to the leaving group.
And in this case, the leaving group is the bromine. Now, I realize the bromine is not leaving the entire molecule, but it is leaving that carbon. And so as a result... So...
So the top carbon had a bromine attached as a wedge, the new one's going to be attaching from the back face as a dash. Cool. Had I decided to attack the bottom one instead, then it would have been the other enantiomer instead.
Cool. And that's again why we do the anti-addition, this backside attack of an sp3 carbon in the intermediate. Alright, so this is halogenation. So, we use an inert solvent with our halogen, and the two most common inert solvents are CCL4 and CH2Cl2. And technically, these are pretty common inert solvents.
In fact, if you did the reaction with HBr, you probably did it with one of these solvents, except we... didn't take the time to draw it out for you. Well, there's a reason we didn't take the time to draw out for you because it didn't matter.
But here it does because we have an option. When you're doing halogenation, you can do it with an inert solvent and exactly as we see, or you can do it with a reactive solvent. which is usually a water or an alcohol is most common. So, and all of a sudden, those actually participate in the reaction.
And instead of adding two halogens, we'll get a slightly different reaction as we're about to see. So this is halohydrin formation. Along with your halogen, which again is typically Cl2 or Br2, we're either gonna use water or an alcohol instead.
And so it turns out instead of adding two halogens, we're gonna add a halogen and either an OH or an OR, depending on if we used water or alcohol. And so in this example here, we're gonna add a chlorine on. the less substitute side and we're going to add an OH on the more substitute side and technically this is going to be Markovnikov. Now a lot of students memorize Markovnikov as the side with more hydrogens gets another hydrogen like the rich get richer or something like that but in this example we don't add any hydrogen across the alkene and so students often get confused on what Markovnikov means here if they memorize that kind of like overly simplistic version. So but it turns out the halogen is going to be adding as it gets attacked by the alkene so adding as an electrophile and so it's gonna end up on the less substituted side for Markovnikov.
So, and then water or the alcohol is gonna be attacking back. So just like bromide attacked here and we'll end up on the more substituted side, as we'll see. So I'm gonna use a slightly different alkene here.
So, and if we predict our products. So our chlorine should end up on the less substitute side, and in this case the OH on the more substitute side, and both of these turn into chiral centers. So we just form two chiral centers, and again, that's when stereoselectivity matters. And so forming two chiral centers, it's anti-addition. And so we're only going to form two out of four possible stereoisomers, the two anti-versions.
And so in this case, anti means that the chlorine and the OH have to end up on opposite sides of the ion. Alkenes, one a wedge and one a dash. And so we could have had the OH as the wedge, which would have meant the chlorine was a dash. But if the OH was a wedge, that also means that this methyl group up here was a dash as well.
And we could have just as well got the enantiomer here also. In which case, the OH would have been the dash and the methyl the wedge. And if the OH was the dash, the chlorine would have to be the wedge on. This side, again, to be anti-addition. And so we get a pair of enantiomeres yet again.
Two chiral centers form, but anti-addition. Let's take a look at that mechanism. All right.
So in this case, we are going to first again with our alkene come and attack chlorine. So that's going to cause the other chlorine to break off and leave. But rather than form a carbocation, this chlorine again is like, I got lone pairs. Don't form a carbocation.
I will come and attach to the other carbon as well. And so in this case, I'm just going to attach the chlorines as wedges. They could just as well be dashes.
So in here, it actually matters. This is not a miso. So, but I'm just going to choose the one where it's wedges.
Just know it's actually a pair of enantiomers as our intermediate here. So, and the idea is that we also formed a chloride ion. And so now, normally, we'd have that chloride come and do backside attack. And we said it would be on the more substituted carbon, not the less.
So, however, we used a solvent that's not inert this time around, water. And when you use water or alcohol, so they are actually... weak nucleophiles.
So, and in this case they're weak nucleophiles. Chloride's a better nucleophile, but there's only one chloride, and he's the solvent. There's millions and zillions of them. And so as a result, the solvent preferentially gets to do the backside attack instead and open up that ring. So chlorine is still going to be attached to the other carbon still as a wedge.
So but on this carbon where we did backside attack, so the chlorine was a wedged bond, the leaving group there, so the OH is going to be where the water attaches as the dash and that flips in takes place, Walden inversion, causing the methyl group that used to be a dash to now be the wedge instead. And as we've seen in the past, when you've got a neutral nucleophile attacking and attaching and ending up with a positive formal charge, you'll have another molecule of the solvent deprotonate. And so we'll bring in another water molecule. Cool, and there is one of our products.
And again, had we showed our lovely intermediate here with the two bonds to chlorine as dashes and the methyl group to the wedge, that's how we would have arrived at our other enantiomer as well. Oh, and just to make sure we're balancing this correctly, she really should for any proper mechanism. We also formed some H3O plus. Cool.
Only difference here, have we used alcohol instead of water? So is everywhere you see an OH here, it would have been an OR, where the R would have just been indicative of what alcohol you use. Use methanol, OCH3. Use an ethanol, OCH2CH3.
Same diff. The next alkene addition reactions we'll look at are kind of come as a pair. Epoxidation and anti-dihydroxyl.
It turns out that anti-hydroxylation, the first step is epoxidation, and so the first is really part of the second, but you might see it independent as well. So it turns out the reagent here is going to be what's called a peroxy acid, or simply a per acid. So it kind of looks like a carboxylic acid with one too many oxygens. I've got one drawn out here, so notice it's got three auctions here instead of normal two. And that auction-auction single bond is a rather weak bond, so it's kind of governed in this reaction, so explaining the reactivity.
So If you just add a peroxy acid, so you get the epoxide. Now one thing you should know is that the most common peroxy acid is called metachloroperoxybenzoic acid. And it's just a special, turns out, peroxy acid. So we've got benzoic peroxy acid and metachloro. That is MCPBA.
I like to think of it as a Canadian who likes peanut butter. Me crave peanut butter. And so the truth is you're more likely to see MCPBA written out as an abbreviation than just to see generically, you know, peroxy acid or to see it drawn out or anything like that.
So MCP is the most common form, but technically you could see it any one of these ways. So I want to make sure you recognize all of them. So cool, if you just add a peroxy acid you get the epoxide.
Cool, done. And notice you don't actually add two things, you just add a single oxygen. And so that's why it's hard to say, what are the two groups added?
Well, you just get one. And also since both carbons here of the alkene, or what used to be the alkene, are bonded to the same atom, they have to be added on the same side, and so it's a syn addition. And since we're only adding one thing, there's definitely no like, who's bonded to what kind of regio selectivity to talk about, no Markovnikov or anti-Markovnikov kind of thing.
Now it turns out, if you take this epoxide and add H3O plus to it, so it carries out what we call acid catalyzed ring opening in epoxide. Reaction will, you know, study in greater depth in second semester. And in this case you end up with an OH on both sides. So one of these OHs was the O of the epoxide and another one comes from the water in H3O plus. So, and because we're adding two of the same thing on both sides, an OH on both sides, so again, there'll be no, you know, Markovnikov or anti-Markovnikov, no regioselectivity to talk about.
So, but it is going through a three-membered ring, and so we'll be doing backside attack on an sp3 carbon, which will lead to anti-addition, and that's why the two OHs end up trans to each other in this case. Cool, in the second example, we did form two chiral centers, and with two chiral centers, so we had to worry about our stereoselectivity, which was anti, and in this case, only got the two anti products both trans so a pair of enantiomers in this case all right so let's take a look at the mechanism a little bit so if we take a look i've lined up our alkene and our peroxy acid here and first step our alkene is the nucleophile as we expect and we're going to come and attack so that oxygen right there so but this bond is going to break this bond between oxygen hydrogen back so that the oxygen ends up bonded to both carbons of the alkene. Simultaneously, as we attack that oxygen, this bond breaks, forming a double bond there, and this bond is used to make a bond to the hydrogen instead. And so what we end up with is, one, our epoxide.
And again, whether these are both wedge bonds or both dash bonds, it is exactly the same thing. And then we're going to end up with... Just a plain old carboxylic acid. So this carbon here is going to have a double bond to this oxygen. This will now be a single bond to an O that now has an H, and it'll be a carboxylic acid.
Cool, so there's the epoxidation step. The second step is the acid catalyzed ring opening with H3O+. So let's see how that plays out.
So this takes a couple of steps. So it turns out the first step, so it's taking place in acid, is that we actually protonate the epoxide first. We'll do that with our H3O+.
Cool, and with a protonated epoxide, it is actually a much better electrophile. than had it not been deprotonated. And so, I'm sorry, not been protonated.
So, and these two carbons are sharing the partial positive charge, and now they have even more partial positive charge, and now that the oxygen's even more withdrawing, having a positive formal charge. Cool, and now water's just going to come do backside attack. So, if it mattered, typically we like to say that it would attack the more subsoil one, but I just kind of left it here alone, and I'll treat that in a little more detail in second semester.
So, but water's going to come and do backside attack there, which is why it ends up on the opposite side. And as we've seen already a few times this chapter, when a neutral nucleophile attacks it ends up with a positive formal charge and you'll come and deprotonate it with another molecule of whoever your solvent is, which in this case is water. And that gets us to our final product here. and a molecule of hydronium.
Cool. So, when you've got these two OHs here, we call that dihydroxylation, adding two hydroxyl groups. And in this case we call it anti-dihydroxylation because they end up on opposite faces of the molecule. So, and that's important because we're about to cover syn-dihydroxylation, where you end up adding two OHs and they end up cis to each other on a ring like this.
They add to the same face rather than opposite. Alright, the last of our alkene addition reactions here is going to be syn-dihydroxylation. We actually have two different sets of reagents that can pull this off. We've got osmium tetroxide, followed by either sodium bisulfite or sodium sulfite.
So, or potassium permanganate, and we say cold and dilute under basic conditions. The reason we have to say cold and dilute is because if it's hot and concentrated it actually does a different reaction. In this case we add two OHs across the alkene, so dihydroxylation, but this time notice they're on the same side.
They add to the same face and we typically don't require students to know the mechanism for this and stuff like that. However, we often show you the major intermediates involved. With osmium tetroxide you get a cyclic osmate ester, with potassium manganate you get a cyclic manganate ester, and the idea is you can see why these two auctions, which are the ones for the product in either case, So you can see why they have to end up being cis to each other into the same phase, because they come from the same molecule of OSO4 or MnO4-in either case. Cool.
So like I said, don't have to know the mechanism here, but I did want to show the intermediates to explain why it ends up being a syn addition. One last thing I want to point out here is that where we added the two OHs, they both are chiral centers. And when you form two chiral centers, that's where your stereoselectivity matters. And so in this case, being a syn addition, So we had to show the OHs as both wedges or both dashes.
But in this case, look at that plane of symmetry. This molecule is achiral. It's a meso compound. And whether you make them both wedges or both dashes, it's the same thing. And once again, we expected two products with two chiral centers in syn addition.
We're only getting one due to actual formation of a meso compound in this case. So finally, I want to finish off these alkene addition reactions by just working a few examples, predicting some products, and showing how we predict the number of products based on how many chiral centers we formed, this sort of thing. So if we look at this first one, we're adding HBr to an alkene, and you should remember that HBr is one of the three reactions that goes through a carbocation intermediate, and you're definitely going to want to draw out that carbocation intermediate. So in this case, we're going to add the hydrogen to the less-substitute side, and end up with a carbocation on the more-substitute side, which in this case is tertiary. And the three adjacent carbons are secondary, secondary, and primary.
This one's not going to rearrange. And that's ultimately where the bromine is going to end up, right? Where that carbocation is.
And so we added the H to the less substitute side, the bromine to the more substitute side, because this is the Markovnikov addition of H and Br. Okay, cool. Now I'll examine where we added the H.
And so it had a chance of becoming a chiral center, but definitely doesn't have four different groups. So and this guy here also doesn't have four different groups. And you might think at first inspection, maybe he does. So, but he's got a methyl, he's got a bromine, and then these two CH2s look similar. They still look similar and they meet in the middle and they're exactly equivalent.
This thing's actually got a mirror plane of symmetry right down the middle here and so no four different groups for him either. He's not a chiral center. So out of a chance of forming two chiral centers we didn't form any and we get one achiral product. We have no stereochemistry to show whatsoever. Now if we do the same reaction but with peroxide now, now it's going to go antimarkovnikov and you're going to see this is going to be a pain in the butt.
And so in this case we don't have to worry about rearrangements or anything, but it's going to go antimarkovnikov. So the more substitute carbon got the hydrogen, which is not drawn in, and the less substitute carbon got the bromine. And now we have a little bit of a problem, because we formed two chiral centers.
Where the hydrogen added, four different groups. Where the bromine added, four different groups. Two chiral centers, that's when your stereoselectivity matters. And for HBr and peroxide, it doesn't have any stereoselectivity whatsoever. And so that being the case, you not only get the two syn products, you get the two anti products and you get all four possible stereoisomers.
And so in this case, this would not be a sufficient answer and maybe I need to do something like this. Let's make sure we get our five member grain correct. Cool, so we get four different stereoisomers here.
So these two are a pair of enantiomers and these two are a pair of enantiomers. These two resulted from syn addition, these two resulted from anti addition. And in case you think I just misspoke, so keep in mind that what's syn and anti is relative to the hydrogen that wasn't drawn in.
And the H and the Br are both wedges here, and they're both dashes here. That's syn addition. And here, so the bromine's on a wedge, the hydrogen's on a dash, that's anti-addition. And here the bromine's on a dash, the hydrogen's on a wedge, and again, that's anti-addition.
So again, it's always relative not to the two groups you can see drawn in, but the two that actually got added across the alkene. So pain in the butt, first example we saw where you get all four stereoisomers. So it's difficult to actually write a question where that happens, but it does show up on occasion, just like it did here.
Cool, this next one, Br2 and CCl4, and you might recall CCl4 is one of your classic inert solvents, and so this is just going to add two bromines across our alkene here. Cool. And unfortunately, where both bromines added, we formed two chiral centers. And again, that's when your stereoselectivity matters.
And for halogenation, so we form that three-member ring with bromine, and the carbons we attack are sp3 hybridized, so we have to do backside attack, and it results in anti-addition. And so in this case, this is not sufficient. We've got to show that these two bromines are on opposite sides here. So if one's a wedge, the other's a dash, but that means we've got to show stereochemistry all the way around. If this bromine's a wedge, that means this methyl group's a dash.
And if here the bromine is the dash, that means this methyl groups the wedge. And then we draw the enantiomer as well. Cool. Pair of enantiomers yet again.
So form two chiral centers, but it's anti-addition, so we only got the two antiproducts. And finally this last one is very similar to one we already kind of worked out. So in this case, we're going to add two hydrons across the alkene. And when we do so, it reduces it simply down to an alkane.
And in this case, where the two hydrons added, both are chiral centers. When you form two chiral centers, we typically need to know your stereoselectivity in that one example. And in this case, it's syn addition. So instead of expecting to get four products, we only expect to get the two syn products. However, due to symmetry here, so being syn, we could think, oh, it's both wedges or both dashes.
So if I make the both methyl groups wedges, that means the two hydrons were on the dashes. But if you think you've got to draw the other, we've got to see, oh, there's a mirror plane of symmetry in this. This is a meso compound. And whether we put the two methyl groups on both wedges or both dashes, it's exactly the same thing.
And so one of those unusual examples, again, where instead of getting two products like we expected with syn addition and two chiral centers formed, we only get a single meso compound. So now we've got one last set of reactions for alkenes to cover and technically these are not addition reactions so it's what's referred to as oxidative cleavage and the most common way to do this is called ozonolysis and involves ozone which is O3 the structure which is right here. So we're not going to show the entire mechanism but we'll show just a little bit of it to get our appetite wet here.
So your alkene actually reacts with ozone here and does something looking like this. So to form what's called a melozynide. This melozynide is going to rearrange to an ozonide.
It takes a couple of steps and again I'm not going to show the whole mechanism, but you get this lovely ozonide and it's this ozonide that actually is going to undergo step two. And so typically for ozonolysis you've got step one which is ozone and then step two you've got some variability there and you might add a reducing agent or you might add an oxidizing agent. And so we have two different sets of conditions for ozonolysis. We've got reducing conditions and we've got oxidizing conditions. And it's all about whatever you add in step two.
Now for reducing conditions, we're typically going to use dimethyl sulfide. which might be written as DMS. That could be step number two.
Or you might see commonly zinc and water used for step two as well. So there's your ozonolysis under reducing conditions. under oxidizing conditions we'll use a mild oxidizing agent in this case for step two we use hydrogen peroxide and this is going to slightly affect what products we get now if you look at what's kind of going on here so i like to think when i look back at my original alkene if i want to predict products of ozonolysis i just cleave that alkene in half hence the name oxidative cleavage so i'll cleave it in half and then i will oxidize both sides to make it oxidative cleavage you And so if we did that here, so cleave both sides and then oxidize both sides, that means give a double bond to oxygen to both sides.
And so if you look, the top product here is a ketone. And the bottom one here is an aldehyde. And for the aldehyde, you typically want to draw in the hydrogen that it's also bonded to. One time we draw a carbon-hydrogen bond, and I don't know why. So, but get a ketone and an aldehyde.
Now, ketone... Ketones are ketones, and they're not oxidizable under any normal means, but aldehydes can be oxidized to carboxylic acids. So when you do this under reducing conditions, either with dimethyl sulfide or zinc, so an aldehyde stays an aldehyde. Your reducing agent keeps it from getting oxidized, and so you get your aldehyde. But if we did this under oxidizing conditions instead, so your ketone again is still a ketone, not going to change, but your aldehyde is going to get oxidized to a carboxylic acid instead.
So again, your ketone is still a ketone, but instead of an aldehyde, you get a carboxylic acid. Cool. So if you're predicting the products of ozonolysis here, again, just cleave your carbon-carbon double bond in half, and then give both sides a double bond to oxygen. If either side is a ketone, great. That's never going to change regardless of the conditions.
But if one of them turns turns out to be an aldehyde or both. So that's when the conditions matter. So if you use reducing agents, aldehydes stay aldehydes. If you use an oxidizing agent peroxide, then your aldehydes are gonna become a carboxylic acid for your product instead.
Cool, we got one other way we can pull this off instead of ozonolysis. Turns out we can do the same cleavage, oxidative cleavage with permanganate. So if you recall we used permanganate a little bit ago to do syn-dihydroxylation, but we used it and it was cold and dilute. Well now we're using it under much harsher conditions.
We're using it with hot concentrated and under acidic conditions here. So and this does the same thing as ozonolysis under oxidizing conditions. So permanganate is definitely a strong oxidizing agent. And so again we get a ketone just like we did before.
And ketones are always going to be ketones for oxidative cleavage, but instead of getting the aldehyde again, we get the carboxylic acid. Cool. Don't have to know the mechanism for the second one.
So the truth is, I've never seen students responsible for the mechanism on ozonolysis either. I'm not saying it's not possible, I'm just saying it's unlikely.