Concept: Nucleophilic Addition9m
Now I'm going to introduce a new mechanism that's going to be really important for this chapter. This mechanism is really completely unique. You haven't seen anything like it in this course so far. The name of it is nucleophilic addition. What is nucleophilic addition? Well, nucleophilic addition is actually one of the most important ways that carbonyl compounds can participate in organic reactions.
How does this work? Basically, if you look at a typical carbonyl, what you always have is you have a carbon attached to an oxygen and two groups on both sides. This is always going to produce a very predictable partial charge. And that partial charge you should be looking for is a partial positive. Notice that if I were to assign a positive and a negative charge to this molecule, I would say that the O is going to have my negative charge and the carbon is going to have my positive.
Nucleophilic addition is going to be the addition of nucleophiles or negatively charged species to that electrophilic carbon. So the reason this thing is so reactive and the reason that carbonyls are so good at this is because the carbonyl carbon is electrophilic. Electrophilic means that it's going to react with negatively charged things very, very well. Does that make sense so far? Let's go ahead and look at the general mechanism for nucleophilic addition so you guys can get a better idea of what's going to happen.
Nucleophilic addition, in general, involves a carbonyl. It could really be any carbonyl compound, and a nucleophile that has a negative charge. It doesn't always have to have a negative, but most of the time it will. Sometimes it may be neutral just with a lone pair. As long as it has the ability to donate electrons, it can be a nucleophile.
In our first step, if we were to try to predict the first arrow here, where do you think we would start the arrow from? Would we start from the carbonyl? Would we start from the nucleophile? Do you guys remember these rules for mechanisms? Pretty sure you guys remember that you always start from the area of high density to the area of low density, meaning that your electron flow is going to start from your nucleophile to somewhere because that's the thing with the most electrons.
Now where are we going to go? Are we going to want to go to a partial negative? Hell, no. We want to go to the partial positive. So we're going to go ahead and that arrow didn't work out. Let's try that again. We're going to go ahead and attack the bottom carbon, the electrophilic carbon. That makes sense because now I have a negative and a positive interacting.
But we've got a problem. Does carbon like to have five bonds? Remember that each mechanism arrow represents a new bond that's going to be created. No. So if we make a bond, we have to break a bond. Can you think of the easiest bond to break here? It wouldn't be to kick out one of those R groups. Notice that right now I'm dealing with a ketone. These are R groups, but it could also be an H, like an aldehyde. Regardless, do we have something that's easy to kick out? No. R groups hate to be broken off and get a negative charge. They're terrible leaving groups.
So this is the bond that we're going to break. We're going to break the double bond to the O making a negatively charged species with a nucleophile on it. Now this intermediate is going to be very important throughout the course of this topic because you have to pass through this stage every single time. This is called a tetrahedral intermediate. This is actually an intermediate we're going to see a lot in orgo two. In orgo two we're going to deal a lot with this mechanism. But for right now, you don't have to know it super in-depth, you just need to know the basics.
We've got this negative charge. What do you think we're going to do after that? Well, your final product never wants to have a formal charge on it. So our last step here is going to be to protonate. I'm going to go ahead. I'm going to use some kind of acid. There always will be some kind of protonating agent that you can extract a proton from.
Now I've got my proton. What do I get at the end? Well, what I have is I still have this nucleophile. But now I have an alcohol. So nucleophilic addition produces what we call substituted alcohol. Why? Because there's an alcohol on the final product and you have one extra substituent afterwards. Sometimes that's going to be an R group, but it really depends on whatever your nucleophile was. Whatever your nucleophile is, that's going to be what's attached. Makes sense so far? Cool.
Now what I want to do is I want to show you guys some general – you know the general mechanism now, this is the mechanism that we're going to follow pretty much throughout all of the topics that include nucleophilic addition. But now I want to show you guys some specific additions that proceed through this mechanism.
Let's talk about this one. What if my nucleophile is H-, hydride? What would you expect? Well, what you would expect is that your H-, I'm just going to draw this mechanism and then the other ones I won't draw. The H- attacks here, kicks this up. What I wind up getting is an O- that protonates and I wind up getting an extra H at the bottom. Would you guys agree that this H here is the same as this H here? Cool.
What did we just do? What we just did was we added a hydrogen on one side and a hydrogen on another. Now you might have learned this already, you might have not. But just so you guys know, anytime that we're adding two hydrogens across a double bond, that reaction is called a reduction.
Just so you guys know, if you've already learned about reduction, then you know that this is the mechanism that reduction proceeds with. If you don't know about it, that's fine. Just know that reduction is the name of a reaction that adds two hydrogens across a double bond. Either way, it's cool that we can do a reduction through nucleophilic addition. Makes sense so far?
What if we use other nucleophiles, though. How about if it's not H-. Well, this is just a random one. What about CN-. Well, you can image that CN-, same thing. We're going to wind up attacking my carbonyl and I'm going to wind up getting a CN here. Once again, substituted alcohol. Why? Because now I have an alcohol with one extra substituent on it. This would be a cyano group. Cool?
Let's try another one. What about if you used this nucleophile. Now I know that looks a little complicated, but you should know this nucleophile by now. This is a nucleophile called an alkynide. If it doesn't ring a bell by now, you probably haven't been doing enough homework because you're probably going to see an alkynide come up by now, at this point of the game. An alkynide is a strong nucleophile. It's also a strong base. You'd get the same thing.
What you would wind up getting is a negative charge attacking, kicking up the O. And what you wind up getting is C, triple bond CH. Now you might be wondering, “Johnny, why did you draw it at an angle?” Because some of you might remember that triple bonds are always drawn in a linear fashion. I kind of cheated right here. But the reason I didn't draw it straight down is because it's literally going to go off the screen if I draw it straight down.
But just so you guys know, if this was my organic chemistry test, if I was writing it for my professor, I definitely would have aimed the triple bond facing down so that it would be in a linear fashion because it's sp hybridized, 180 degrees, yadda, yadda , yadda. Remember that comes from chapter one or two. Cool.
I just want to show you guys these are different nucleophilic additions. By the way, this alkynide also falls into another category of molecules called an organometallic. I'll explain what that means in a second. Why am I giving it such a big name? Well, because I have a carbon and I have a lithium together. Lithium is a metal. Carbon is an organic compound, so let's call it an organometallic. And just so you guys know, later on when we talk about organometallics or if you already have talked about organometallics, organometallics also follow the same nucleophilic mechanism.
So what's the point of this? I'm just trying to tell you guys there's a very important mechanism. It's also a very easy mechanism. So there's no reason not to really understand it. And we're going to be repeating this several times throughout the semester.
I hope that makes sense. Let's go ahead and move on to the next topic.
Draw the product of each of the following reactions.
Is the following statement true or false?
If the newly added group in the tetrahedral intermediate is a stronger base than the group attached to the acyl group in the reactant, then formation of the tetrahedral intermediate is the rate-limiting step of a nucleophilic addition–elimination reaction.
Predict the organic product(s) of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product, if stereoisomers are produced in the reaction be sure to indicate the relationship between them. Draw the answer in skeletal form.
When nucleophilic addition to a carbonyl group occurs, the carbon attacked undergoes this hybridization change:
Write the structure of the major organic product formed in the reaction of 1-methylcyclohexene with each of the following:
Write the structure of the major organic product formed in the reaction of 2-methyl-2-butene with each of the following:
Write the structure of the major organic product formed in the reaction of 1-pentene with each of the following: