Concept: Concept: Ketones from acid chlorides6m
Now we're going to talk about a way that you can make ketones from acid chlorides.
Acid chlorides and esters have something in common. They both have a pretty good leaving group next to the carbonyl. If nucleophilic addition takes place on that carbonyl carbon, that leaving group is prompted to leave allowing a double bond to be reformed. This process is called nucleophilic acyl substitution. It's the subject of another set of videos. That can be found in your carboxylic acid derivatives chapter.
But for right now, all I’m trying to say is that acid chlorides and esters when reacted with organometallics are going to react twice instead of reacting once. Let's take a look. First of all, remember that organometallics have a negative charge on the R. The M ionizes. We don't really care about it. The negative winds up attacking the carbon. We form a tetrahedral intermediate. This makes a compound look like this. We’ve got R at the bottom. We've got R1. We’ve got OR. What takes place next is that instead of protonating my O and getting an alcohol, I wind up kicking out my OR group instead. This gives me a ketone for the time being. This is the first step of a typical reaction of organometallics with acid chlorides and esters. At first it seems like you're going to get a ketone. This video is about making ketones. You’re thinking, “Awesome, I just got a ketone.” Again, this mechanism that we just discussed here is called nucleophilic acyl substitution or NAS. We’re not going to go into it too deeply, just acknowledge that that's what's happening here.
The problem is that the organometallic is going to continue to react with this reagent because it's still got a carbonyl. We bring this molecule down, R1 and R. We tend to react with the grignard or the organometallic twice. I’m going to then get another mechanism, another nucleophilic attack this time giving me a tetrahedral intermediate. Notice that this R group that I started with that is on my organometallic has now added twice because I did two separate mechanisms with it. Now at this point, we don't have anything good to kick out. Typically, this is protonated. We don't have a protonating agent here but remember that at the end of organometallics, we usually have some kind of protonating agent. We’ll just use HA for right now and we get what we call a disubstituted alcohol. I’m calling it disubstituted because I don't know if I started off with an R group or not, but what I do know is that I added the same R group twice. This is the way that organometallics are used to working on acid chlorides and esters. It presents a problem when we're trying to make ketones because we're not going to stop in the middle. We never stop here. We actually go through and make the alcohol. That's why here I said that they react twice yielding disubstituted alcohols.
How can we prevent this from happening? It's actually pretty easy. If we really are bent on it making my acid chloride into a ketone, we can use a Gilman reagent. A Gilman reagent is also called a dialkyl cuprate. You see two alkyl groups and a copper group. What this looks like is literally a copper with two Rs coming off of it. Lithium dialkyl cuprates tend to be weaker than other organometallics. What's going to happen is that they’re going to stop after the first nucleophilic addition. In this case, it’s an NAS reaction. It’s going to kick out the Cl. But what we're going to wind up getting is simply one reaction, not two. You’re going to yield a ketone from your acid chloride. Just to make this mechanism really straightforward, you’ve once again got your R group, your R negative. It attacks. You kick up the electrons. You got your O negative, your Cl, your R. Then you kick out your Cl, but then you stop there. You don't generate another equivalent of R minus. This second equipment does not occur because my organometallic is weaker. Because my organometallic is weaker, a Gilman reagent is not going to react with the ketone.
Just keep in mind that if you use any other organometallic, you’re going to go to an alcohol because you’re going to react twice. But if you specifically use a Gilman reagent, you’re going to stop at the ketone which is what we want for this section. Awesome. Let’s move on to next video.
Problem: Provide the major product for the following reaction2m