Organometallics on Ketones

Now that we understand nucleophilic addition of carbonyls, I want to focus in on one of the most important nucleophiles of all that can react in nucleophilic addition and that's organometallics.
I know organometallics freak a lot of people out. They look weird. They’ve got a lot of atoms on them. They’re metals. But these are some of the easiest compounds to work with in terms of pushing arrows because organometallics are just really strong nucleophiles. We're going to do the same things with them as we would normally do with anything that has a negative charge.
Organometallics do tons of different reactions. If you want to thoroughly review organometallics, you should go back to that section of your text or of Clutch videos. I've got plenty of videos on them. But what I’m going to focus on right now is just really a refresher of what they do to ketones and aldehydes. Remember that we can use the letters R and M to represent any organometallic because there's always going to be some kind of R component, some kind of alkyl group and some kind of metal. The two that are the most common in this section are grignards. Grignards would be RMgBr, and organolithium is RLi.
These molecules are both extremely similar on how they react because they both have that ionic bond with an extremely strong dipole towards the R because remember that in your periodic table, the further you go towards fluorine, the more electronegative you get. Carbon is actually pretty close to fluorine. Whereas these groups 1 and 2 metals are some of the least electronegative atoms possible. It’s going to be a very ionic bond towards the carbon, so ionic in fact that we can write these things ionized. We can actually write it as R negative, MgBr positive. The same thing for organolithium. We could write it as R negative, Li positive.
That being said, these are just both the source of R negative. When that R negative see the carbonyl, we’re just going to do a nucleophilic addition. We're going to attack the partial positive carbon, move the electrons up. What we’re going to get is a tetrahedral intermediate, O negative, R1, R2. But now I have my new R group. I just call it R since that’s the one that I just added. That tetrahedral intermediate would normally stay there but these are usually followed by a protonation step. I don't know exactly what the protonation step would be. But I’m just going to put some kind of HA. Some kind of protonation would occur and you would get your substituted alcohol. You get OH, R1, R2 and the R that we actually added.
That’s just a reminder. You should have worked with organometallics before in this course. In case you forgot, this is really what you need to know for this section. You need to know how to add an organometallic to a ketone and to an aldehyde. You’re always going to get an alcohol product and you’re going to get a new R group attached to that tetrahedral carbon in the middle. That being said, why don’t you take a look back at the molecule below and try to do the whole reaction. Draw all the mechanism as well and then I’ll show you the answer. Go for it.

Alright, so for this reaction the first thing to do is draw your charges. You have to remember that your C 6 H 5 is going to be negatively charged and your M G B R is going to be positively charged. So I know his looks strange but in actuality what we're really dealing with is it's almost like having a benzene benzene nucleophile. We pretty much have a negative charge just on that benzene ring and then M G B R is hanging out like a cation just like a spectator ion. So we're going to get our nucleophilic addition, we're going to attack and what we're going to get is a tetrahedral intermediate. That tetrahedral intermediate will have an O negative, it will have my new benzene ring attached, the fourth group is that there was a hydrogen there, right?

There used to be a hydrogen, it's still there, nothing changed and usually guys if you're stopping at a tertahedral intermediate like let's say that the second reagent wasn't there, then many times the way that you might see it in the text or your professor might write it is that the M G B R was actually attached to the O because one is negative and one is positive so your T I, tetrahedral intermediate, a lot of times will just be written like that but we do have a protonating agent in this case. It's N H 4 C L. This is a very common protonating agent for Grignards and for organometallics. Now at first glance guys it actually looks really weird because you've got N H 4 C L, it almost looks like nitrogen has five bonds but that would break nitrogen's octet, right? It can only have four bonds, how are five things attached to nitrogen?

Well, guys it's not really like that, what it is that you've got your nitrogen with your for H's and that nitrogen is going to have a formal charge because it's unhappy with the number of valence electrons it has, it wants five and has four, so then the C L negative will be hanging out associated with the N H 4 positive but it's not actually going to be attached by a bond. So this protonating agent guys is going to get rid of the negative charge in the O and give us our eventual alcohol, and our alcohol looks it's like this with our benzene ring and you got your N H 3 and I mean it doesn't really matter guys these other by-products are not really of our concern but your M G B R really is just hanging out with a positive charge and you could attach it to the C L if you wanted. The C L has a negative so that would be fine too. Awesome guys, so that was it for that mechanism. Let's move on to the next topic.