Electrophilic Alpha-Halogenations

Concept: Concept: Acid vs. Base Catalyzed

6m
Video Transcript

In this video, I want to talk about specific electrophilic alpha substitutions called alpha-halogenations. It turns out that alpha-halogenation can proceed through an acid-catalyzed and a base-catalyzed mechanism. I'm going to show you both but it's important to know that they actually yield different products because of what the intermediates look like. The acid-catalyzed mechanism is always going to yield mono-halogenation. Let me show you what this mechanism looks like.
For an acid-catalyzed mechanism, this is going to proceed through the enol tautomer. We're going to be trying to make the enol tautomer. The way it works is let’s say you’re using just H3O+ and you protonate. What you're going to wind up getting is something that looks like this. Then we can make the enol. The enol would be with the conjugate base. You take off the alpha-proton with water, my conjugate. I would do this, this and this, then I get my enol tautomer. This enol is now a pretty good nucleophile. I can use it to attack my diatomic halogen in a similar way, not really the same way but similar to another nucleophile. You would go ahead, grab one the X's, kick off one of the X’s. What you wind up getting is a molecule that looks like this, OH with now a new X here. The mechanism would do this. You would also bring down the electrons to reform the double bond. What you would get is something like this. That would be the intermediate.
Eventually, this gets deprotonated using water and you get back to your keto form. That's what we call our alpha halogenation. Notice that we add an X to the alpha opposition but we only had one time. You might be wondering, “Johnny, why would you only add once? Why does it not add again?” It turns out that having the X there to pull electrons away from the alpha-carbon is going to make it less likely to make an enol next time because the enol is going to require a positive charge on those carbons. By pulling away extra electrons, you make it less likely the form an enol, so it only reacts once. Cool so far? Awesome.
This is in contrast to the base-catalyzed mechanism. In the base-catalyzed mechanism, you may yield polyhalogenation. The reason is you would basically going to yield as much halogenation as you have alpha protons. The reason being is that in a base-catalyzed mechanism, the X’s being there is just going to make it more reactive the second time. Let me show you. In a base-catalyzed mechanism, imagine that I'm using OH negative and that I'm just forming an enolate right away. The enolate looks like this, negative charge. That negative charge can react with X2 to this, to that. Go here, go there, and give me my halogen.
But now when it sees that second equivalent of base, now we have a dipole pulling away. It's going to be easier to put a negative charge there because the negative charge is going to be stabilized by the halogen. Therefore, we're going to replace every single hydrogen that was there. We’ll replace all of those alpha hydrogens with X's and we’re going to get polysubstitution. The only reason that I put may yield, not will yield, is because obviously you need multiple alpha protons in order for it to get polyhalogenated. If you only have one, then you’re only going to substitute once. But you can basically continue to keep reacting until you’re out of alpha protons. Interesting, right?
In the next video, I want to show you an application of alpha-halogenation called the haloform reaction. 

Concept: Concept: Haloform Reaction

4m
Video Transcript

Alright guys. So the haloform reaction is simply a base catalyzed alpha halogenation of methyl ketones, okay? So, it's the same thing as base catalyzed alpha-halogenation except that you specifically have a methyl ketone, let me show you the mechanism. So, guys the reagents are just the same as we were doing this before the only difference being that I'm starting with at least one methyl group on one side, okay? What's going to happen is after the formation of your enolate and attacking the x's you're going to get a polyhalogenated alpha carbon, okay? Make sense so far, but here's the deal, if you have a methyl ketone you've now made an awesome leaving group because you have cx-3 so the next equivalent of base that reacts instead of just sitting around and being like, hey I've got nothing else to halogenate, I'm done, the next equivalent can actually go through an NAS mechanism and it can form a tetrahedral intermediate. So, what you would get is a tetrahedral intermediate that looks like this, O negative, OH, R, C, X, X, X, and guys carbon is usually not a good leaving group but a carbon with three halogens is an extremely good leaving group. So, then I would kick out the cx-3, okay? And, what that's going to give me is we just draw over here that's going to give me carboxylic acid crazy, right? So, that's going to give me carboxylic acid plus it's going to give me cx-3 negative, okay?

Now, I know my head's in the way a little bit but then to end this reaction, the cx-3 negative is going to deprotonate the acid because it's an acid, right? And you're going to wind up getting a carboxylate and the molecules that direction is named for haloform, haloform is just basically a methyl group with three hydrogen's replaced with halogens, okay? So, the entire idea behind haloform is that the alpha carbon of this reaction is transformed into a good leaving group through successive halogenations and eventually the OH negative just kicks it out entirely, which is something that would not happen in a base catalyzed mechanism without it, because if a base were to attack here, it would have nowhere to go because there's nothing to kick out, it doesn't have a good leaving group, but on this reaction it does have a weak good leaving group, which is why you get the final products, okay? Now guys, just extra facts, haloform depends on what you're using, if it was chlorines going to be chloroform, right? If it's iodine going to be iodoform and some of these end up being tests that you have to use in your chemistry lab to test for methyl ketones. So, a lot of times iodoform will precipitate out of the solution and it will be a, it's like a yellow precipitate, and it's a test to see if you had a methyl ketone because if you have that precipitate that means that you formed iodoform, which means that you had a methyl group on their ketone, interesting, right? Cool Guys. So, let's move on to the next reaction.

Problem: Provide the major product for the following reaction. 

2m

Problem: Provide the major product for the following reaction. 

2m

Problem: Provide the major product for the following reaction. 

5m

Problem: Provide the major product after each step for the following reaction.

4m