Now I want to discuss some common reactions of organometallics. So as you guys know organometallics are very strong nucleophiles, so they're going to react with things that have positive charges. That just makes sense. What do we call it when something has a positive charge? That's an electrophile. So organometallics are nucleophiles that are going to attack different electrophiles. So I'm looking for things that have a positive charge. The whole point of this topic is I want to go through all the different types of molecules that have positive charges that can be attacked.
So we're going to start off with the simplest electrophile. It an electrophile that you should already know at this point of the course because we've worked with it several times and that's alkyl halides. Alkyl halides, I have it written at the top as well, so you don't have to write it. But alkyl halides are really great electrophiles. Why? Because it's a carbon attached to an electronegative atom, so I get a partial negative here. I get a partial positive there. That partial positive charge is very susceptible to nucleophilic attack.
Now I can already tell some of you guys are forgetting how to determine if this is an electrophile or a nucleophile because you're saying, “Johnny, but I see a negative and I see a positive, so how do I know that it's an electrophile? Why isn't it a nucleophile?” Because we had this rule back in our acid and base chapter, so you're kind of forgetting, but it's okay. I can say it again. We have a rule that says you look at the side of the charge with the highest bonding preference.
So which of those two atoms can make more bonds? Carbon or halogen? Carbon. Carbon can make four bonds. So this is the side that I look at to determine if it's an electrophile or nucleophile. What kind of charge is on the carbon? Positive. So that means this is an electrophile. Does that make sense? Even though both charges are present, I ignore this charge in terms of determining electrophile or nucleophile because that one doesn't matter to me. I care about the one with the highest bonding preference. Cool. Let's go ahead and get started with this mechanism.
It's really easy. It's just going to be a substitution or an elimination reaction of an alkyl halide because as you guys remember R-M stands for, in general, an organometallic where I have a negative charge on the R and I have a positive charge on the M. In this first mechanism, I'm just going to use this as a nucleophile. If this was an SN2 reaction, I would kick out the X and what I would wind up getting is something that looks like this. I would get R, with that little circle around it, so you guys can keep track of it. And now that R is attached to a three-carbon chain. Plus, we're going to get our leaving group. So our leaving group would just be X-. Cool.
I don't think I'm forgetting – oh, you might be wondering, “Johnny, where did the M go?” The M also will leave as a leaving group. So you get M+ and many times the M, the metal, and the X will come together and make an ionic bond. Cool. So that was easy substitution or elimination.
You might be saying, “Johnny, how do I know if it's going to do substitution, like SN2,” like I drew in this case. This was an SN2 reaction. Or how do I know that it's going to do elimination? Well, for that I'm going to need you to go to the substitution elimination chapter and I'm going to need you to look at the flow chart that I have. I have a flow chart that explains exactly how to tell if it's substitution, elimination, SN1, whatever. Just letting you know that that's where that information comes from. Cool. So let's move on to the next one.
Nucleophilic addition on ketones and aldehydes. You should already know what the mechanism is for nucleophilic addition. The mechanism for nucleophilic addition is that I have a partial positive here and I have a negative on my R. So my R, once again, is my nucleophile. My M is going to leave. It's not even going to be around. Remember I said this is an ionic bond. So really you can redraw this as just R- and there's an M+ around.
My R- would attack here, push the electrons up. What I would wind up getting is a tetrahedral intermediate with an O-, an R2, an R1 and then a new R. I'm just going t make that dotted line so you guys can tell exactly where that R is.
Do you guys remember what the next step is of nucleophilic edition? You got to protonate. So some kind of protonating agent. It depends on the exact reaction is going to come in and protonate. So you wind up getting at the end of this, you'd wind up getting a substituted alcohol where now you have an extra R group. So you had two R groups before because it was a ketone, but now you have your extra third R group.
This actually turns out to be a tertiary alcohol. This one, in particular, would be a tertiary alcohol. Why? Because I started off with two R groups. I added a third one, so now there's three R groups around that OH. Will you always get a tertiary alcohol? No. You just have to add up your R groups and figure out the degree of your alcohol and that will be it. I'm just saying in this case, since I start off with a ketone and since we're adding one R group, I'm going to get a tertiary alcohol. Does that make sense? Cool.
Just to clarify one more time, if you're confused about how to determine that, just go back and figure out the degree of it. Just say, is it tertiary, is it secondary. That's all. That's how you'd determine if it's secondary or tertiary. Cool guys.
Now I want to look at another mechanism. Another mechanism and this is one that you're not supposed to know up to this point in this course, so I'm just going to let you know, is a nucleophilic acyl substitution. This is normally an organic two mechanism. Usually, we learn this in orgo two. But there's other – some schools teach it in orgo three. It just depends. The whole point is that even if you don't know it, we can still go through it here.
On an ester, in particular, you have a special situation because you still have that partial positive here. You still have the negative here. You can still do your attack. But there's a difference. There's a difference because I have an OR on that bond, so now I actually could kick out that OR group if I wanted to.
So we're going to go ahead and make the electrons go up to the O. And I'm going to get my tetrahedral intermediate of R1, OR and R dots. So that makes sense so far. This is actually the same exact middle step as the one before. This is our tetrahedral intermediate, just like – sorry, I forgot to write. Just like this was my tetrahedral intermediate. They were both tetrahedral intermediates.
The difference is that guess what? Now, I can kick out the OR group. So I'm going to reform my double bond. Kick out my OR. This is really unique to esters. What you're going to wind up getting is something that looks like this, a carbonyl again with R1 and R dot, dot, dot, dot. Does that look familiar? What kind of functional group is that? Well, this is a ketone.
So is that going to be our final product? No. Because ketones, it turns out, can react with organometallics, like we just learned in step two. The second one was that a ketone that looked very similar to this, reacted with the organometallic. So what that means is that now the organometallic has to react again with the product of this reaction.
So now I'm going to go ahead and use my second equivalent of organometallic, so I'm going to say this is my second R-. I'm just going to put here in parenthesis times two. Okay, that's my second one. I'm going to go ahead and I'm going to attack again.
What I'm going to wind up getting, eventually, after protonation – I'm going to skip the protonation step – is you going to wind up getting OH at the top. Remember that it protonates. It makes an O- and then it protonates. Then you're going to get R1 on one side. You're going to get R, dot, dot, dot, dot, dot. And you're going to get another R, dot, dot, dot, dot, dot, dot because – circle, whatever you want to call that. Because we just added two equivalents of the same R. Notice that this one was also like that because it came from the original alkyl halide.
So one of the biggest giveaways that you used an ester to make this reaction happen is that there's two of the same R group in the final product. If you have two of the same R group in the final product, in the final alcohol, what that tells you is that you probably reacted twice. If you were trying to look backwards, you would know that you could create this molecule through using an ester to react once and then to react twice. Is that cool?
By the way guys, you're not supposed to understand everything right now. This is more just an intro to show you how it works. The only way you're really going to get this is through practice. Feel free to ask me any questions, but, for example, what I just said about knowing if it's an ester using the same two R groups, that's something I'm just going to repeat in practice problems as we work on it. Cool.
So then we have our last reaction of organometallics and that is what we call a base catalyzed epoxide ring opening. Now this is challenging because some of you will have never opened an epoxide before when you get to this point. Others of you already know what I'm talking about. But some of you are like whoa, whoa, whoa. Like, I don't even know what an epoxide is. Cool. It's fine. I'll explain.
An epoxide is simply a cyclic ether. It's an ether, ROR. It's in a ring. But the difference about it is that since it's in a small ring, it's very likely to react. Normally ethers don't react as anything. But cyclic ethers do. They actually like to react and like to open up. When you use an organometallic on an epoxide, this organometallic is going to act like a strong base and it's going to want to open up that epoxide.
Now for those of you who have already gone through this part, through epoxide ring opening, you will remember that there is a difference between acid-catalyzed and base-catalyzed ring opening. An acid-catalyzed ring opening would attack the most substituted side. Whereas a base-catalyzed ring opening would attack the least substituted side. Now for those of you that haven't gone through this yet. Don't worry about the acid part because we haven't gotten there yet. For those of you that have, great. This is just a refresher.
So do you think R-, do you think that's going to be more like an acid or more like a base? Come on guys. I already told you that a nucleophile. It's a strong nucleophile and a strong base. So we're going to go ahead and attack there at the less substituted side. That's going to break open my O. So I'm going to break open my O.
What do I get at the end? Well, if this bond is the one that's at the bottom of the epoxide, this one right here, then what I'm going to get is I'm going to get my R dot, dot, dot, dot, dot. And I'm going to wind up getting this R1. And then I'm going to wind up getting – don't draw this yet. I'm going to show you in a second. O- because basically it springs open.
Now something you guys should know about. Anytime you break open a three-membered ring is that – and this applies for anytime you break open a three-membered ring nucleophilically, it's going to snap open and due to the repulsion factors of these atoms, they're going to wind up facing anti to each other. There's going to wind up getting groups that are anti in location.
So what that means is if this group attacked from the front let's say, then my O- will go to the back. Okay, and that's why I told you not to write it yet. Now you can write this or if you wrote it, erase and just make it dashed and make it front because you want to make sure that your products are anti addition.
Now is this the final product? No. usually there's a protonation step. So now my H – my acid, I don't know what it is is going to be available to protonate R. And then this because it's in the front – I don't know. I can just draw it on the stick. That's fine. We don't know. That's my final product.
This does have a chiral center, so you would get plus enantiomer. Plus enantiomer simply means that you would reverse all the S's and R's. If it was S it would be R, R would become S. This is the final product of a base catalyzed epoxide ring opening with a nucleophile. Notice that now I have this extra R group attached to this compound. These are all considered alkylation reactions because of the fact that I'm always adding an R group. I'm always adding a nucleophilic R group.
So I hope that made sense guys. This is just supposed to be a general run through, but really, you're going to need to practice this in order to get it down solid. So let's go ahead and move on to the next topic.