All right, there we go. Okay, wow, got really quiet all of a sudden. TGIF, right? Let's get through this.
Okay, I have quizzes up here for quiz four. Let me run through the announcements. I want to talk a little bit about quiz five because that's going to look a little bit different. So active four, quiz four both had their deadlines now, so we should be moving on into our new content on the ethers and epoxides that we're going to hopefully get through all of and wrap up today.
So active assignment five will cover that chapter 18, ethers and epoxides. content so you can find that that's been available since I think Wednesday afternoon and then I have quiz five ready I just didn't give it my last once over to make sure all the updates that I did over the last couple days were correct on it so right after class I will get that looked at and I'll have it posted for you so quiz five is a little different because I don't want to have it due next Friday because your exam next Friday is on the content includes the content on quiz five you want to know how you did on that and what the right and wrong responses were before then. So the way that our just-before-the-exam quizzes will work is they are due end of day on Wednesday. So that means, like, Wednesday when it turns midnight and it's now formally Thursday, that's your due date.
And so it's due. It's all multiple choice. There's no open-ended things. So you can get the answers right away.
The answers will come out in the form of an announcement. So I'll have an announcement scheduled to release right at midnight. So because of that, absolutely no late responses are allowed because once the answers are public, it's just not fair for me to accept a response. No matter how much you swear that you are honest about it, do your classmates really believe you? We're not going to have that battle.
So absolutely dead cut off at 1159 on Wednesday because answers will be out at 12. I might make it 1201 just in case there's like some discrepancies in clock. So a two minute buffer, but right after midnight, you will have those answers. answers.
So that gives you all of Thursday to kind of digest what's going on. So these are typically a little bit shorter quiz also than what you're going to have. This one is 10 questions, but the last three questions are part of a continuum that are, I don't know, maybe not so exam relevant, but more thinking about how it fits into your career relevant. So that's how usually the right before exam quizzes will go.
Five, seven, eight maybe questions on the content, and then two or three. three that are more like contemplating career. Okay. But then answers released immediately.
Questions about that? You'll see the quiz later today, and then we'll have plenty of time on Monday if you have questions about it and Wednesday as well. All right.
And this is all because our exam is coming week from tomorrow. So we will spend plenty of time next week. The week will be largely review.
Monday and Wednesday will be almost completely review. I do want to go back and hit on the one topic we skipped that was tangential to Grineard. That will take 15, 20 minutes maybe, but the rest of the time will be time to review, get all your questions answered, go over things lingering from quizzes past, etc. As long as there's not lines, I do have queued up first for today a couple of things about quiz three that I wanted to make sure everybody was on the right track in their way of thinking about two of the questions on there before we get back into finishing up our ethers and epoxides. for today.
We'll be going back to the Williamson ether synthesis. So we'll pick up where we left off. I'm going to talk a little bit more about the requirements of the reaction. So what you need to have in the two pieces that go into the Williamson ether synthesis, and then we'll go through the mechanism of that as well. And then we'll move on to epoxides.
I'll remind you of the structure and properties. And thankfully there's no nomenclature here. We will not be naming epoxides at all.
And then we have one type of reaction we'll be talking about doing on epoxides. So that type of reaction has two different flavors. flavors, two different types of reagents that we'll react it with, which is going to have regiochemical and stereochemical outcomes. So there's some terms to review as well as seeing how that looks for the structures involved in those reactions.
So that's our content for today. Well, let's go back to the old stuff for just a minute. Before I do that, Thomas, Cameron, do you guys have anything for any SIs wrapped? Okay.
You guys have good attendance this week, I hope? Yeah. Okay. All right. Next week, right?
Next week's the mad rush. Okay. So don't forget, they keep...
Keep reminding you of all the meeting times. These are also all on the homepage on Canvas. So don't forget about these really valuable resources you have in your SI leader and the TA that can give you a lot of guidance helping you prepare for the upcoming exam. All right. Okay.
So I pulled two quizzes from quiz three that I wanted to talk about. The first one I'll just spend a quick second on. So number seven, everybody got right, no matter what you chose.
Because somehow in my update to this one, there was a typo in it that just made two of the answers nonsense. So A is reasonable-ish. It's as reasonable as it was ever meant to be. D is reasonable-ish. But somehow B and C got scrambled.
And the... Sorry, I don't have the color I want up here. There we go. What happened was this HBR wound up tagged onto C where it doesn't belong. It was supposed to be over here with B.
So that was supposed to be HBR with tert-butyl peroxide and heat. And C was supposed to be just bromine and light, two very different things than what you actually see there. So that was not an answerable question, so I gave everybody full credit for it.
But now that you see the way, and maybe you figured out this is... the way the reagents were supposed to look, but now that you see that, does it make a little bit more sense what the correct answer would be? So it's either B or C.
Obviously, the issue had to be with one of those if I was giving everybody full credit on that, but what we're trying to do here is that anti-Markovnikov type addition across an alkene, right? So this was actually in the radicals chapter. not your alkenes addition chapter.
So these B and C are the two sets of conditions that come in that chapter. So B is the bromination reaction where you start with an alkane, right? So that's actually not going to to be disrupting your double bond at all.
That would not give the product here. So C is not correct. B, where we add HBR, but then we use the tert-butyl peroxide radical to change the regioselectivity of that and make it anti-Markovnikov, this is the one that actually works.
The other reaction that's in that radicals chapter, in case you don't recall, was NBS and heat, which would do something to this starting material, but this is the one that does that a little bit. allylic bromination. So it's going to add a bromine over to this position.
So actually both C and D, we put a bromine here without touching the double bond. Okay. They wind up inadvertently giving you the exact same product.
Okay. No matter which one of these you use. And then A is just nonsense. Just made up stuff. Any questions on that one?
Pretty straightforward once you see the right reagents in the right places. And then I do want to spend a second talking about the synthesis question. So do be sure to go and look at the comments that I made because of the way I've had to reallocate points to make Canvas cooperate with being with you being able to see your running. grade throughout the semester.
I can't really allot the points to this that I would like to. I don't like to make the synthesis and mechanism questions worth a ton of points, but I like to have a little more to play with than just two points. So I wound up not taking off on things that normally I would be taking points off of for.
The main thing that people lose points on, and it wasn't just this semester, it's every semester, the very first time that I give you a mechanism question, is for not following the rules of the question. So typically in bold, you're going to see your rules for what you are allowed to use in doing your synthesis. First of all, if I say it's a hydrocarbon, a hydrocarbon is a substance that is containing only... hydrogen and carbon atoms. Can't have oxygens, can't have halogens, nitrogens, nothing.
Can have double bonds or triple bonds. Doesn't have to be an alkane, but it can't have anything other than hydrogen or carbon. So usually when you see that, you're thinking like alkene, alkyne, because you know how to do all kinds of things with alkenes and alkynes.
So those were the muscles I was trying to get you to flex here is working all that alkene, alkyne chemistry. But I also gave you the opportunity to use alkyl halogen. halides.
Okay. So these can have double and triple bonds. That's fine.
And a halogen, but you just can't put other functional groups. Like you can't throw a ketone plus a halogen on there or something like that. And then most importantly was this two and only two carbons rule. And that's where you might have lost some points. Okay.
So being able to stay within those constraints really lets me fine tune the kind of synthesis strategy. I want you to be looking for here and in this case all this little jibber-jabber I have here about but I gave you an odd number of carbons and I'm telling you you have to have an even number of carbons how do you rectify that situation well not by saying well I'm just gonna use a one carbon piece or a three carbon piece because that doesn't follow the rules okay so you lose points for not following the rules and it kind of becomes compounded because the key thing I'm looking for here is that you're Recognizing ozonolysis, that reaction that cleaves a carbon-carbon bond, as something that has to be employed in this strategy. Okay, that's me pushing you a little too much and like forcing you to have only one reaction.
reaction that gets you the right place in your strategy. It's a little harsh. So I didn't take off quite so many points if you didn't use ozonolosis, but not following the rules did have, like, I think I took one out of two points off for not following.
the rules. So if you don't see ozonolysis, you know, that's fine. I mean, we don't always all see things the same way.
So I'm really just looking for creativity and ingenuity in this. And when people do ozonolysis on this, there's a couple of different ways they go about it. Sometimes what they're getting to prior to getting to the spinal aldehyde is maybe a six carbon alkene.
And then they're able to do the ozonolysis and cleave. that alkene down to the aldehyde in the final step of their synthesis. So they're just losing one carbon out of it.
Some students see it a little bit differently and they don't think about, oh, I can throw a carbon away. And they go for, and I love this one because I didn't even think of this until I saw a student do it. They do a 10 carbon chain.
like this and then when they do the ozonolysis they've got two of the desired product that's actually super cool i think all right and there are people that saw that ozonolysis had to be used and did it in a number of different ways where they you know some sometimes they're throwing away three carbons instead of just one. So there's lots of ways to go about that. Some people even did the ozonolysis early, like started from... an alkene, and then maybe cleave that down to formaldehyde from the get-go.
The only problem is you get kind of stuck there because at the point you were working on this, you didn't formally have all the knowledge of Chapter 17 and reactions like a sodium borohydride or lithium aluminum hydride reduction that would take you to an alcohol that you can make into a one-carbon alkyl halide. So that's where this one kind of falls apart because the other rules are that you're supposed to be using only reactions from Chapters 8 through 11, Chem 235, not Chem 238 reactions. at this stage.
So part of why I have that in there, these quizzes, and they're open book, open note, open internet. And it is really easy to go on Chegg, to go to a tutor, and get a solution to these problems. And I don't feel good about giving full credit to a problem that's been solved using tools that are outside of where you should be in this class. Now I understand some of you, this isn't your first time in this course, and maybe you remember things like sodium borohydride, lithium aluminum hydride reduction. that's what the regrade requests are there for.
So if you think I'm being really unfair and taking off those points because it was your independent work from things you know that maybe the bulk of the class doesn't know or I wouldn't think that you know, please let me know about that. So put in a regrade request for that if I've done that to your grading. Trying to be fair here. Sound good?
Questions about approaching synthesis problems? because there will not be another one that I grade prior to the exam. There are lots and lots of practice synthesis problems in your chapter 17 and chapter 18 workbooks. Those are a lot of old exam questions, so great place to get extra practice on synthesis right there. I will be posting solutions to those as we go.
They're not all up yet, but they will be as we get through the next week. There will also be synthesis practice questions on your practice exam. You will have that no later than Monday. a minimum do that practice exam practice the reactions that are there and for those of you that have more time and have the ability to put in that extra effort definitely those workbook problems year in year out are what students say are the thing that made the difference for them and being prepared for those hard questions on the exam and i'm looking at cameron to see if he's like yeah yeah that usually that's what my si's and tas are saying right good Back to some new chemistry. Oh, one question back here.
Is the exam going to be only two? Yes, it is all open-ended. Sometimes there's a question or two that's multiple choice. You'll see on the practice exams in the oddball ways I add multiple choice just so I can grade faster. That's really the reason I do it is, man, it takes a long time to grade 65 exams.
You guys are a small class for me, so I'm kind of excited about this, but it's still slow. Right, okay, well then, I pulled up exactly where we left off last time. This is just cut and paste right out of our lecture notes from Wednesday's class. Just the example, the generic... version and then a couple of examples I gave you for Williamson ethersynthesis and I just want to elaborate on a couple of bits of this so that you really understand why we're using the reagents we're using and I think maybe we should start with the mechanism of this and I'll explain why why we use these reagents as we go, right?
So for the mechanism, I'm gonna use our second example here. So the cyclohexanol reacting with one bromopropane after we've added sodium hydride into it. All right, so our mechanism is going to start with just our alcohol, and I'm drawing out that alcohol with the OH bond shown explicitly, because we're going to be pulling that proton off in our first step, using the first reagent in the first step. step of this reaction which is sodium hydride.
So sodium hydride is going to dissociate in our organic solvent to give us a hydride. Now we've kind of seen hydride before right, sodium borohydride, lithium aluminum hydride, it has hydride. in the name.
We formerly were using it as a hydride, but we never actually saw this hydrogen with a lone pair and a negative charge on it just kind of floating free in the reaction. Sodium hydride, it is going to dissociate into something that really is a free hydride. And a free hydride is a very reactive hydride. Now, I don't want to say that you should be more afraid of working with sodium hydride than lithium aluminum hydride because this doesn't cause the same problems.
that we have with lithium aluminum hydride, but it does do the side reaction that we are concerned about with lithium aluminum hydride. That free hydride is very good at very rapidly deprotonating something that's fairly acidic, like an alcohol. So our first step is to take that hydride, have it grab that acidic proton. Now, the way that that hydride in sodium hydride behaves as a base, it doesn't behave like a super strong base. but it does behave as an irreversible base.
So it's a fast reaction, but not a super exothermic reaction. So we don't have those same exact hazards that we had with the lithium aluminum hydride getting protonated. We get to an alkoxide, and then we're making Hydrogen gas, right?
The fact that that's a gas is what makes this irreversible. That gas just bubbles on out of solution and we're not going backwards. Not to mention like we're going to be talking about a hydride leaving group if we try and reduce this or try and reverse this hydride leaving group. The worst. Okay, so really not going to reverse at all.
So sodium hydride is a safe way of doing this kind of reaction without the fire hazard. And it really is only used for deprotonating alcohols and not much else. Like it's too weak. Think about like sodium making the...
acetylide ions. You're using sodium azide to make acetylides, right? Back in Chem 235, when we take our terminal alkyne and remove that proton, we would use sodium azide for that.
So PKA-wise, it would look like the sodium hydride would work, but it absolutely does not. Okay, you need a stronger base to do that. So this kind of acts as a weaker base. but irreversible so great for deprotonating and alcohol and not much else and once you've done that now our alcohol doesn't have so many different ways it can react okay it pretty much is a base or nucleophile and in this case we want it to react as a nucleophile the electrophile we're going to add this to is our alkyl halide And this is going to proceed through a simple SN2 process.
So we're just going to add in to the carbon that our halogen is on, kick out the halide ion, throw the lone pairs on that bromine for good measure, and that is going to give us All of the carbons of what used to be our alcohol plus its oxygen and that oxygen has formed a new bond over to all of the carbons of our alkyl halide, right? This was a three carbon unit. I want to make sure all three of those are there.
So it's basically redraw your alcohol, take off its proton, throw all the carbons of the other piece onto it. And that's it. That's the whole mechanism.
Super easy. Deprotonate SN2. You've been doing this for months, right? Okay. Now, one of the things that I pointed out when I showed you kind of the generic reaction, I'm going to scroll up just a hair here, is in the generic reaction here, I'm showing you that that alkyl halide that we're using in the second step must be a primary alkyl halide.
So seeing this mechanism helps us to understand why that needs to be the case. If we want to do an SN2 reaction, in that final step that actually makes the ether. We really need this to have a primary leading group. If you think back to your nucleophilic substitution and elimination chapter, when you're talking about, are we going SN1 versus SN2?
If we're trying to go SN2, what are the chances of it going E2? When does it get messy? As soon as you have a secondary leaving group, right?
Tertiary leaving group, we can start pushing things SN1, but as soon as you have a secondary or tertiary leaving group, elimination becomes a real complication. And it's especially a complication when the nucleophile you're dealing with is not just a nucleophile, but also a pretty strong base. So that strong base is thinking elimination is a lot easier than substitution, except when there's a primary leaving group, right? The sterics of that don't impede it at all.
So the only time a strong nucleophile does well as a nucleophile and does a substitution reaction really is when you have a primary leaving group. So that's what really limits us here to primary alkyl halides. So that means when we're thinking backwards about a reaction, such as, let's just kind of reverse engineer the first example I gave you, making methyl tert-butyl ether.
When you think about synthesis planning. You always have to figure out which route is going to give you the option of a primary leaving group. It doesn't matter if our alcohol is a primary or secondary or tertiary alcohol, that's inconsequential, but that alkyl halide absolutely has to be primary for the Williamson ether synthesis to be efficient.
So if we were to take our methyl tert-butyl ether, and I'll just draw it without any color coding here, and try to figure out how to make this. We have two different pathways we could think about. We can either make our new carbon-oxygen bond of the ether on the left-hand side of the oxygen, we'll call that route A, or we can make it on the right-hand side of the oxygen, we'll call that route B.
In route A, We're asking to have our alkyl halide be tertiary and our alcohol would be Essentially primary. Methanol. Methyl sometimes gets treated as like better than primary because it's even less hindered than something that's formally primary. If we look at path B, we're going to flip those rules around. So the oxygen is staying on the left-hand side with the oxygen.
So now we're talking about tert-butanol. And then our alkyl halide will be on the other side with just the methyl. So we're talking about a methyl bromide primary or better.
So this is the better route because we will not have elimination competing in that route. So in the Chapter 17 workbook, you will see a series of questions, and I think that exact example is one of them, unfortunately. But there are at least three other examples where you can play around with, how would I make this ether? What are the two pieces I need to put together to make this ether most efficiently, most effectively? So you'll play with which of the two possible combinations gives you the primary alkyl halide, and that's the winner to go with.
So look for that for the extra practice on that kind of problem. And that's pretty much everything about ethers. We're good on that. I'm ready to start talking about our specialty ethers, the epoxides.
And normally I like to give you a few more examples before we move on to the next topic, but I just really want you to get all the content now. We'll have plenty of time for examples next week. All right, you guys remember what epoxides are?
You learned how to make them back in the Alkenes chapter, but you haven't really talked too much about them. An epoxide is, well that's not at all a straight line, a three-membered ring with an oxygen in it. So if you think of it as an ether, it is just a very small cyclic ether.
We've seen other cyclic ethers here and there. We've been talking about THF, which is a solvent. That's another cyclic ether. It's just a bigger ring. You can do this in a six-membered ring.
There are even, there's a fairly common solvent called dioxane. It's a six membered ring with two ethers in it. There's all kinds of cyclic ethers out there, but epoxides are very different from all other cyclic ethers.
the way that they react. All of them can kind of do what epoxides do, but not as effectively as epoxides do it. And it's because of their structure.
When you see a three-membered ring, does it give you a little bit of an uncomfortable feeling? Like, should I really be drawing this? What feels weird about that three-membered ring?
Bond angles, that bond strain, right? So it just feels like this thing's going to break. I always tell students when we're learning about epoxides for the first time in Chem 235, let's pull out our model kits and build it, but let's maybe not build that three-membered ring because it's going to break.
break your model pieces right so but they do exist they exist stably in nature there are a lot of compounds there are antibiotic compounds out there like the apothelones it's a class of antibiotics that are isolated from a marine sponge, that part of the active group within it for its antibacterial activity is an epoxide. So this marine sponge makes it. It just exists in nature. We can isolate it and use it as a drug with all that ring strain. That's crazy.
So I'm going to not go on and on about epoxides and their crazy bonding and talk about banana bonds and like, are we really SP2 or SP3 in these things? There's all kinds of cool things about their bonding that it's really unique to epoxides. What you need... to know is how they behave just is different because of that bond angle strain and it makes them far more electrophilic than any other cyclic ether.
So epoxides really are very electrophilic at the carbons that are adjacent to the oxygen. So part of how we can envision this and rationalize this, let's just redraw that epoxide structure and think about what's going on. So we have that electronegative oxygen in it. So the easiest way to think about it is, let's pull the electron density away from those carbons. And that's absolutely what it's doing.
So we can draw it with those polarity arrows showing the polarity of each of the carbon-oxygen bonds. So the way you really need to be thinking about epoxides is that they are partially negative on the oxygen, but then they are partially positive on the carbons adjacent to it. And then part of the way that we rationalize our alkyl halides being good electrophiles, you look at bromine, it's not that electronegative. But we say if we make a carbon-bromine bond, that that's electronegative enough that it's really making the carbon next to it electronegative.
more positive. Do you think that really happens that much? It kind of doesn't. It's more the fact that that bromine can leave that makes that carbon more reactive and electrophilic.
And then that is the same case here. The fact that we can not just have our oxygen leave, but when it does and we open the ring, we alleviate that ring strain. So here it's a little bit of leaving group ability and a lot of ring strain alleviation that makes those carbons really electrophilic. So you'll see that as we go through the reactions we're going to do with our epoxides. Okay, so think of those as the two driving forces behind our epoxides being good electrophiles.
So let's talk about this in the context of the kinds of reactions that our epoxides will do. So if our epoxide is going to behave as an electrophile, we're going to want to react these with nucleophiles. When you think of a nucleophile, what's the first thing you think of?
Sure, hydroxide is a great nucleophile. So we can react this with something like sodium hydroxide. This may be in water as the solvent, but typically if we do a reaction in base, we're going to need to neutralize the product of this.
So we're going to follow this up with little acid to neutralize it. So this is just like what we did with our green yard reagents, right? Followed those up with acid in a subsequent step to protonate at the end. So when we do this, what we will wind up having happen is we will ring open our epoxide because we've added in our nucleophile.
In this case, it's an OH and then eventually protonated. the oxygen that was our epoxide oxygen. So the mechanism of this process, you can probably guess. Take our hydroxide. We're going to add it in and you can see by where I placed it in the product.
We're adding it into this very accessible, less substituted carbon of the epoxide. We're going to break open the carbon oxygen bond on the same carbon we're adding into. Our oxygen then stays attached to the other carbon of the epoxide, where we made our new bond to that less substituted carbon of the epoxide.
Our negative charge winds up on the oxygen that was the epoxide oxygen. And then... Our second step, that hydronium comes along, piece of cake to protonate that alkoxide and get to the product jump.
Now, this made us a diol. You might have seen this back in your Alkenes chapter, by the way, right? MCPBA followed by acid or followed with hydroxide will get you to a diol. This does work, but how do you know which one's which?
So knowing which one's which, we could do with really sophisticated... isotope labeling experiments or we can just say well if we did this with a different slightly different nucleophile let's use sodium methoxide i'm just going to use the exact same substrate here so that we can talk about how we know that we get this particular ether out of the product. Notice it's an ether and an alcohol now.
And you can imagine by NMR, like it's really easy to tell that apart from if the nucleophile had added in over here without even thinking about some sort of an isotopic labeling experiment. Well, these should be okay to tell those apart by spectroscopy. So we know that we're getting that first one through spectroscopy, but the reason why that's happening is simple sterics. Our nucleophile is adding in kind of indiscriminately.
Our epoxide is fairly stable the way it is. Yes, it's reactive. Yes, it wants to alleviate that ring strain. It really doesn't care which of those carbon-oxygen bonds is breaking when it's in this state.
It's really this negatively charged nucleophile. It's saying, I want to get rid of my extra electron density, that negative charge, and I'm going to do that the fastest way possible. And it will meet a lot more resistance to add in to the more substituted side than it will to add in on that less substituted side.
So that's why when we're using a negatively charged nucleophile or a strong nucleophile we will always get this kind of selectivity. The nucleophile in the very first step will opt to go the path of least resistance just based on sterics. Oops and that was supposed to be an OCH3 and that is not what I drew at all. so that your nucleophile winds up in the more accessible, less substituted position at the end of the process.
We can do this with other structures. We can put our epoxide right on. A cyclic system, adding a methyl group to one side of it so that you can have different substitution on either side and have an obviously more and less sterically hindered side.
We can use things that aren't alkoxides as our nucleophiles. You can add in nitrogen containing compounds. You can add in Grignard reagents.
I'm going to add in phenylmagnesium bromide. And I'm using this large of a Grignard reagent just so it's really obvious where it lands. Like if I used a methylmagnesium bromide, which methyl is which will start to become an issue.
In our Grignards, we're also going to follow this up with that aqueous acidic workup. So the mechanism of this one is also going to have our Grignard reagent, which we're thinking of as a carbanion, right? It's going to react like this. So when we think of it as being a carbanion equivalent, it's really easy to see that it's a strong nucleophile.
When it goes to add into our epoxide, it goes for the easy access side. So the oxygen of the epoxide stays on the carbon that had that methyl on it in the beginning. We're making our new bond to the carbons of our greenery agent here, which is just going to be a benzene ring. And then we will protonate in our final step. Acid and we get to an alcohol with some new carbons added on the carbon adjacent to it.
A few notes on this. Especially this Grignard reaction. Notice that we're winding up with an alcohol product. But look at where our new carbon-carbon bond winds up in our alcohol product. We think about a Grignard reacting to an epoxide.
We wind up with our new carbon-carbon bond. One carbon removed from the alcohol that is the product. What happened if we added a Grignard reagent into an aldehyde or a ketone?
You add into an aldehyde or a ketone, right? Those were electrophiles. We're making a new carbon-carbon bond and we're producing an alcohol product. Where was that new carbon-carbon bond relative to the alcohol in the product? It's on the same carbon.
So synthetically, this opens up a little bit different opportunity for us in making carbon-carbon bonds to be able to add our Grignard reagents to epoxides. So let's just take a quick look at what would happen if we had, I'm going to just do a similar starting material. Maybe if we started with this ketone and added that same Grignard reagent to it. We're going to add right into that carbonyl carbon in our first step.
I've got methyl pointing down so our new bond is right here to our benzenes from the greenery agent. And then that O minus is on the same carbon as we just added into. So in the end we also get a new alcohol product, but that alcohol And the carbons we added in are on the same carbon, the same carbon that was, in this case, a ketone to start with.
It's a very different look. So that's factor number one, is that we have that nice juxtaposition. Factor number two, stereochemistry.
So there is stereochemistry to all of these reactions. And I'm just going to rewrite the one we've just been dealing with here. So that five-membered ring. And I'm going to draw our epoxide with stereochemistry to it. Because we have two chiral centers in that starting material we were using.
No matter what nucleophile we add in, I'm going to go generic here because your nucleophile could be your oxygen-centered nucleophile or your carbon-centered nucleophile. No matter which one you're using, the process is going to be the same. How is that nucleophile adding into our epoxide?
Look at the curved arrows that we have for this. Scroll back up. We have one arrow adding in to the carbon and then another arrow breaking that bond from that same carbon.
That's SN2 right? So if this is an SN2 process, what happens stereochemically when that nucleophile adds in to the carbon it's going to add into? You're going to get inversion, exactly.
So pay careful attention to what's happening here. The chemistry is happening not on the side where the epoxide oxygen stays. That epoxide oxygen is staying on the top carbon of that epoxide. So the stereochemistry at the top of the epoxide is exactly the same, only now it's an OH on it instead of the epoxide.
And then on the bottom part of that epoxide where our nucleophile added in, that's where the inversion happened. So whether we're adding in another oxygen-containing nucleophile or making a new carbon-carbon bond with a Grignard reagent, you get inversion only at that carbon where the addition occurred. So, that's the stereochemical course of the reaction.
I mentioned at the very beginning of class, and I'm going to leave you with the beginning of it. of this and we'll wrap this up first thing on Monday with the idea of regiochemistry in these reactions. So regiochemistry is maybe a kind of vague term to you. Regiochemistry means that when have two different sides that we could possibly react with that we have some selectivity there.
You saw this with alkenes and we called it Markovnikov versus anti-Markovnikov selectivity when we decided whether our nucleophile added to the more or the less substituted side. So we have something similar happening with our epoxides here too and everything we've been talking about so far is that when we use a strong nucleophile In a strong nucleophile, we are defining as one that has a negative charge on it. It's actually broader than that, but we're going to keep it super simple.
If it has a negative charge on it, it's a strong nucleophile, but that's going to add to the less substituted side of our epoxide. So as it turns out, if we change our nucleophile strength, A weak nucleophile, which will be a nucleophile that does not have a charge, this is what we'll be able to add to the more substituted side. So I'm going to leave you with an example of this. Go back to the first epoxide that we were looking at, three carbons, propylene oxide is what this is called. If we were to react propylene oxide with a weak nucleophile, a good example of a weak nucleophile, methanol.
Looks just like our sodium methoxide, but without that negative charge. It's neutral. And this can absolutely behave as a nucleophile, but we need to kick it along with a little bit of a catalyst.
So we need a catalytic amount of acid, and specifically a fairly strong acid. So we like things like sulfuric acid for this. If we do that, we can actually flip that regioselectivity, and we can have our methanol add in to the more substituted side, leaving the epoxide oxygen on the less substituted side with the proton it's going to pick up from its catalyst. So on Monday, we'll go through the mechanism of that.
We'll talk a little bit more about the stereochemistry of these processes. We'll have this wrapped up halfway through Monday's class. Good? All right. Enjoy your weekend.
Look at that. I ended at 10 o'clock. Never mind.