Concept: Concept: General Reaction13m
Hey, guys, so now we're going to talk about a topic that scares a lot of students, but it's really not that bad. Honestly, I think it's just the name that scares people. And the topic is organometallics. Now you're thinking this is going to be terrible. Organometallics sound very confusing. But they're not. They're very easy actually. All they are is they're alkylating agents. What is an alkylating agent?
It means that I'm going to be able to use this reagent to put an alkyl group on something else. That's the first thing. What it's going to consist of is usually a group IA or IIA metal, so I'm talking about the first and the second column of the periodic table, those metals are going to be directly bonded to a carbon structure.
Now if you think about it, that's going to give this a very interesting molecular property because typically we usually see electronegative atoms attached to carbon. There have been several instances this semester where you've been exposed to an electronegative atom bonded to carbon.
For example, this functional group right here, C with an X on it, remember X stands for alkyl halide, I'm sorry, stands for halogen. So this would be an alkyl halide. And what would be the typical dipole of an alkyl halide? Do you guys remember? Where would I draw the dipole towards? Towards the X, right? So I would get a dipole towards the X. That's going to give me a partial negative here, a partial positive here. What kind of charge do I have on that carbon? Well, we have a positive charge meaning that this carbon is going to be a great electrophile. That's typically the way that carbon acts. A lot of reagents that we've seen this semester, carbon is going to react as an electrophile.
But organometallics are different because notice that the X, X stands for halogen. What group is halogen in? Let's start off with carbon. Maybe this is a better explanation. Carbon, if I was just to draw my ghetto periodic table that I do so often, carbon is in group IV. Are you guys cool with that? Halogen is all the way over here in Group VII. I'm just going to draw VII. Which one is more electronegative? Carbon or halogen? Obviously, the halogen because as you move to the right in the periodic table, you get more electronegative. Are you guys getting this so far? Halogen, fluoride, fluorine is at the very top, so it's obviously more electronegative.
Now let's think about these organometals that we're using for organometallics. Well, I told you that these metals are usually in group I or in group II, so that means are they going to be on the right side or the left side of carbon? They're going to be on the left side. Actually, my metals – I'm just going to use the letter M to stand for them in general because I don't know exactly which metal it is. It could be lithium. It could be magnesium, something like that. These are going to be in group I or in group II.
So which one is going to be actually more electronegative? The carbon or the metal? The answer is the carbon will. Organometals like I have drawn here. I just put M in general because I don't know if it's lithium or magnesium, whatever. We'll get to the exact metals in a second. The dipole goes in the opposite direction. That means that instead of this carbon being a great electrophile, organometallics are examples of where your carbon is in an amazing nucleophile. That's going to change everything. I wrote nucleophilic, I was trying to write nucleophile.
The carbon is an amazing nucleophile because it has a negative charge now and that means that now I'm going to be able to use this in a unique set of reactions because normally we don't have carbon with a negative charge, but now we do.
So there's four types of organometallics that you guys will be responsible for. There's four types that you need to know and they're commonly talked about in organic chemistry.
The first one is sodium alkynides. These are the easiest type because most likely, if you've been doing your homework, you've already seen one of these at least in another chapter. The way that organometallics work is we usually use a terminal alkyne to react with a strong base. Now that strong base, there's a lot of different strong bases we could use, but typically it's NaH2 or NaH. These bases have the ability to pull off – let's just say I'm using NH2-, they have the ability to pull off the most acidic hydrogen on that molecule.
When we learn in the acids and base chapter how to predict acidity, we would have found that the H at the very end of that terminal alkyne is very acidic compared to the others. So I would grab that H with my base and I would give a negative charge to my carbon. What does that mean? Well, that means that my final product is not going to have a negatively charged carbon and is there a spectator ion? Yes. Remember that there was also a Na+ present. We just dissociated it. The Na+ can wind up making a bond to my negatively charged carbon.
Now this bond right here that I'm drawing and this is going to be true with all of the bonds that we're going to draw today, this is an ionic bond. Do you guys remember what the definition of ionic means? That's really easy. This is going back to chapter one. But an ionic bond was simply a bond that had such a great difference in electronegativity that there's essentially no sharing. Do you remember that? Basically, you can draw a bond, but really there's very little sharing going on of electrons.
Since this bond is ionic, you can draw it two different ways. You can either draw it as a bond, like we've done here, or you could also choose to draw it as ions with a negative there and with a positive there. Both of these representations are perfectly accurate because one of them shows that there is a bond, ionic. But the other one shows how the bond is very weak, meaning that – weak in terms of that it can be easily dissociated because I have almost full charges on both. Does that make sense guys? And this is obviously a carbon here. That is a carbon, I just didn't draw it, but it is a carbon.
The reason we call this an organometallic is because sodium is my metal. Remember that I said the group I or II metals are typically your metals, so I could actually just boil this down to being R-M and R-M is going to be the general structure that we're going to use for all organometallics. So R simply means I have some kind of carbon group with a negative and M means I have some kind of group I or II metal with a positive. Does that make sense? This one is one that hopefully you've already seen by this point in the course. This is your first organometal.
Let's go onto the next one. How about if we want to make a Grignard reagent. Now I know I might have just lost a few of you guys because that does not look like it says Grignard. It looks like it says Grignard. But, it's pronounced Grignard. Just got to live with that one. Just take it from me. I've been doing this for a while.
So you've got a Grignard reagent. And how do we make one? What we're going to do, you don't need to know the whole mechanism. It's fine. But what you do need to do is you need to recognize how to make it and what it looks like. Well, you start off with an alkyl halide. Start off with an alkyl halide. Then what do we do? We add elemental, meaning by itself, magnesium, elemental magnesium and what is this? Et2O. What is that? That's an ether because all it means is that I have an O in the middle with two ethyl groups coming off of it. That would be diethyl ether.
You put in elemental magnesium, your diethyl ether. That's all going to complex together and what you're going to wind up getting is a Grignard reagent. Now, why is a Grignard reagent considered an organometallic? Well, because once again we have the same situation where I have a carbon that has a group I or II metal attached to it. If I were to draw that dipole, it would really give almost all of its electrons to that carbon, so I'd get a negative here and I would get a positive there.
Another way to write a Grignard is to draw it ionically. So I'd have a negative charge here and then I would put MgBr or I said Br because usually we use Br, but it could be any X with a positive. And these are associated together as an organometallic. Is that cool?
That's another common one. Just so you guys know X could stand for any halogen. Are you cool with that? Usually we don't use fluorine. Fluorine isn't used very commonly in these, but definitely iodine, chlorine, bromine are all fair game. Usually, it's bromine, but you can also use the other ones as well. Cool.
So let's keep going. So organolithium. My organolithium compound is going to be super similar to my Grignard reagent. I'm just going to use slightly different reagents. I'm going to start off with an alkyl halide again. No difference there. And now I'm going to react it with two equivalents of elemental lithium instead of magnesium and ether. What winds up happening is that now I just get my lithium directly attached to my carbon. That's it There's no X on the other side. This can be written the other way which would be ionic.
Once again, this counts as an organometallic because I have an R group with a negative charge and one of the group I or II metals with a positive charge.
Then finally we have Gilman reagents. Now, if the phrase Gilman doesn't really ring a bell maybe it's because, in your textbook, they may not call it a Gilman reagent, they may call it a lithium dialkylcuprate. Really, they're the same exact thing. They just have – one is named after the scientist and one is just the general name. But whatever, regardless, these are the same exact thing. If you look online, let's say you're trying to find practice problems online, and you could totally look up lithium dialkylcuprate and they'll give you Gilman reagents and vice versa. Cool.
You start off once again with an alkyl halide. And by a mechanism that you don't need to know, please don't learn this right now. This is not what the point of this topic is. We're just trying to figure out how to make them and what they look like. What we're going to wind up getting is you combine two lithiums in ether. That's the same as, really the same exact step as the organolithium.
But then you have a second step. The second step is the giveaway that it's Gilman. You have copper iodide. You have CuI, copper iodide. And what that's going to do is it's going to take two of those organolithiums together and complex them to one copper.
So what you're going to wind up getting is now this structure right here. This is your lithium dialkylcuprate or you Gilman reagent and this still follows the general structure of R-M, organometallic because I have an R group that has a negative charge and then this copper would have a positive charge. Now I know that copper isn't group I or II, but there's a lithium there as well, so that's why I count it as group I or II.
All right guys. So what do you need to know here? What's the point of the story? The point of the story is know your reagents to make these – know the preparation step. Know exactly which reagents you need to make your organometallics. Then know how to recognize the names of each of these organometallics. I haven't taught you how they react yet. We're going to get there. But for right now just know how to name it, how to make it. Cool guys. Let's go ahead and move on to the next topic.
Concept: Concept: Ruining Organometallics3m
Really quickly I just want to talk about this blank space that we didn't fill out. I just want to let you guys know that because organometallics are strong nucleophiles that means that they're also going to be able to react as strong bases. When you're a strong nucleophile, you can usually react as a strong base and that means you're good at pulling off what? Protons. Remember that nucleophile is the Lewis definition. That means you're a good electron pair donor. But when I say base, that's Bronsted-Lowry. What I'm saying is that you're going to be a great proton acceptor.
So how does that play into things? Why is that important? Because you have to be aware of cross-reactions with acidic hydrogen. Let's say I have an organometallic R-M. Remember that we've been using R-M all day to mean organometallic and I introduce it to a molecule with an alcohol. I introduce these to each other.
Let's say I want the organometallic to react at some other part of the molecule. Well, this would be a problem for me. The reason is because, as you guys might remember, the hydrogen on an alcohol is acidic. That's going to be a problem because my R is basic. It has a negative charge. So what you're going to wind up getting is a mechanism where your negative winds up pulling off the H and attaching a negative charge to the O.
And what you wind up getting is something that we consider a ruined organometallic, where what you're going to get is an O-. And then that's going to be attached to basically, it's just going to be RH. All we did here is we just protonated the R group. Usually, we don't want to just protonate the R group. We want it to react with something else like an electrophile. This would be a really bad idea – bad, sad face because I don't want just to get a negative charge and a protonated R.
What I prefer to get is to react with an electrophile, E+. All I'm trying to say here is you have to be very careful about alcohols, water, in particular, and then obviously carboxylic acid, stuff like that. All of these things, you have to be very careful about because they have acidic hydrogens that can ruin, that can mess up your organometallic and this is what we consider ruined. Even in your book, it's going to talk about how you can ruin an organometallic by introducing it to one of those three things or just some kind of acidic proton, some kind of proton that has a plus charge.
I just wanted to go over that really quick. Let's move on to the next topic.
Which of the following compounds are suitable solvents for Grignard reactions?
What organometallic compound will be formed from the reaction of excess methylmagnesium chloride and GaCl3? (Hint: See Table 12.1.)
Which statement below about organometallic compounds is false?
a. organometallics contain metal-carbon bonds
b. the metal-carbon bond is highly polarized.
c. they are important sources of carbon nucleophiles.
d. they are important sources of carbon electrophiles
e. they are used to form new carbon-carbon bonds.
Explain why the following reaction will fail.
Which of the following would be a good solvent for doing Grignard reactions (i.e., which would not react with RMgBr)?
Which would be a suitable solvent for the preparation of ethylmagnesium bromide from ethyl bromide and magnesium?
Which synthesis of a Grignard reagent would fail to occur as written?