Grignard Reaction - Video Tutorials & Practice Problems

Reactions of Organometallics

Now I want to discuss some common reactions of organometallics. 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. 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. We're going to start off with the simplest electrophile. It's an electrophile that you should already know at this point in 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, right? Carbon can make four bonds, so this is the side that I look at to determine if it's an electrophile or a 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 and nucleophile because that one doesn't matter to me. I care about the one with the highest bonding preference. So, 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, RM stands for in general an organometallic where I have a negative charge on the R and I have a positive charge on the M. So 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 anything. Oh, you might be wondering, "Johnny, well, 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 an 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 is 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. So 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. Well, 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. So 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 to 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 addition? You've 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 would 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. So this actually turns out to be a tertiary alcohol. This one in particular would be a tertiary alcohol. Why? Because I start 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. But I'm just telling you in this case, since it started off with a ketone and since I'm 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 would determine if it's secondary or tertiary. Cool guys. So now I want to look at another mechanism. Another mechanism and this is one that you're not supposed to know up until this point in this course, so I'm just going to let you know, is a nucleophilic acyl substitution. This is normally an organic 2 mechanism. Usually, we learn this in orgo 2. But there are other schools that teach it in Orgo 3 differently. It just depends. But 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 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 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. Sorry, I forgot to write, just like this was my tetrahedral intermediate. They're 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. Okay? And what you're going to wind up getting is something that looks like this. A carbonyl again with R1 and R. 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 2. 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 2. That's my second one. I'm gonna go ahead and I'm going to attack again. What I'm gonna wind up getting eventually after protonation, I'm just gonna skip the protonation step, is you're going to wind up getting O at the top. Remember that it protonates. It makes an O- then it protonates. Then you're going to get R1 on one side. You're going to get R and you're going to get another R 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 are 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 through 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 by using the same 2 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, "Woah, woah, woah. I don't even know what an epoxide is." Cool. Cool. It's fine. I'll explain. An epoxide is simply a cyclic ether. Right? Cyclic ether. It's an ether. R-O-R. 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 with anything, but cyclic ethers do. They actually like to react and like to open up. So, when using organometallic on an epoxide,erot.