If we start with a carboxylic acid that has at least one alpha proton on it, so right here is our alpha carbon, we can see in this reaction we're going to substitute in. So instead of a hydrogen at the alpha position, we're going to have a bromine at the alpha position. So here's your alpha position.
This is alpha substitution of a carboxylic acid, and this is called the Hel-Volhard-Zelinski reaction. In the first step, we add bromine and phosphorus tribromide, and in the second step, we add water. So let's look at these one by one. So we'll start with phosphorus tribromide. So if you add PBR3 to a carboxylic acid, we saw in a previous video that the phosphorus tribromide is going to turn the carboxylic acid into an acyl bromide.
So let me go ahead and draw in the acyl bromide. Still thinking about this alpha proton right here. And so you can see we took the carboxylic acid and we substitute in a bromine for the OH. So we form our acyl bromide.
And if you think about the The next part of this reaction, there's acid presence. We're gonna do an acid catalyzed tautomerization. You can think about our acyl halide, this is being the keto form of our acyl halide.
I'm gonna go ahead and draw the enol form of our acyl halide. We would have our R group, and then we have a double bond here. Then we would have an oxygen bonded to a hydrogen, lone pairs of electrons on this oxygen, and then a bromine right here.
This is an acyl halide enol. And so once again, we've talked about acid catalyzed tautomerization in an earlier video. So just to simplify things, you have a double bond right here between this carbon and this oxygen.
You can move that double bond between this carbon and this carbon. So here's our double bond now. And then we can also think about moving a proton. So moving a proton from here over to here.
And it's not necessarily the same proton. It's just a very simple way of thinking about this. So converting a keto form, a keto form of our acyl acyl bromide into an acyl bromide enol over here on the right. So once we form our enol, we can think about the bromine, right?
So we have Br2 present as well. So let's go ahead and draw in bromine here. So a molecule of bromine is normally non-polar, right? So we have atoms of the same electronegativity, and so it's normally a non-polar molecule. However, you can induce a dipole on the bromine molecule.
So things like pylons. electrons can help induce a dipole. So these electrons in red right here could move closer to the bromine on the right, giving it a partial negative charge.
And if those electrons in red are moving to the right, they're moving away from the bromine on the left, giving the bromine on the left a partial positive charge. And so we talked about this when we talked about reactions of alkenes. So the pi electrons in here, we saw that pi electrons could function as a nucleophile. And those pi electrons in blue, could attack the electron deficient bromine, which is gonna function as an electrophile.
So this bromine on the left here could function as an electrophile. So we have a nucleophile attacking our electrophile. So if these electrons in the oxygen move into here to form the carbonyl, these electrons in blue could attack this bromine, and then these electrons in red would come off onto the bromine on the right to form the bromide anion. So let's go ahead and draw what we would make.
So we would have our R group, we would have this carbon right here, we would have. We would have a carbonyl like that, and then we have a bromine right here. And then we would have a bromine here as well.
So let me show those electrons. So the electrons in blue, so these pi electrons in blue are gonna function as a nucleophile, attack this bromine, and so that's these electrons right here, and this bromine. And then we're gonna lose the bromide anion over here on the right. So we make the bromide anion over here on the right, and then we also could deprotonate, so we lose this proton.
So if you think about loss, of HBr at this step, so minus HBr. So I'm not going through the entire mechanism here, just a shortened way of thinking about it. So we've converted our acyl halide enol over here on the left to this molecule over here on the right, which has a bromine at the alpha position, and it's still an acyl bromide. So we call it an alpha bromo acyl bromide. And then we're ready for the second step of the HVZ reaction, which is where we add water.
So we could think about adding water, so let's go ahead and draw that in. So we have water right here, and water has two lone pairs of electrons, and so water can function as a nucleophile. We know this carbon right here is electrophilic, so this oxygen is more electronegative, withdrawing some electron density away from this carbon, and so water functions as a nucleophile, it attacks this carbon right here. These electrons kick off onto this oxygen, and then when the electrons move in, And to reform your carbonyl, these electrons in here are going to come off onto bromine to form your bromide anion as a leaving group. And then you could also think about losing a proton from water after it bonds to that carbon.
And so. So the OH is going to replace the bromine and then we're gonna once again lose HBr is just a simple way of thinking about it, right? So loss of HBr at this step.
And that converts our alpha-bromoacyl bromide into our final product, right? result is to get back our carboxylic acid, but we've substituted a bromine at the alpha position now. And so that's the HVZ reaction.
Again, I didn't go through the complete mechanism, just a way of thinking about the reaction. It would take just a little bit too long for one of these videos to go through every single step. So one use for the HVZ reaction is to synthesize amino acids, and so that's one of the classic uses here.
And so we could make an amino acid starting from this carboxylic acid. So this is a three carboxylic acid. carbon carboxylic acid, so this is propanoic acid.
And if we react propanoic acid with, once again, bromine, phosphorus tribromide, and then water, this is the HVZ reaction, which we know allows us to substitute the alpha position. So right here is the alpha position, and we know there are two hydrogens bonded to this carbon at the alpha position. We're gonna substitute a bromine for one of those hydrogens.
So we already know what the product looks like for the HVZ reaction, so we draw in our carboxylic acid. and we know there's gonna be a bromine, right, substituting at the alpha position, and there's still a hydrogen bonded to this alpha position here. And so thinking about this alpha carbon, right, thinking about this alpha carbon right here, there are four different substituents attached to this alpha carbon.
And so you're thinking about stereochemistry. This is a chiral sensor, and so you're gonna get a mixture of enantiomers for your product here. So once you form that, to make an amino acid, in the second, In another reaction, you could add excess ammonia.
So let's go ahead and draw in ammonia right here. And so let's think about what the ammonia might do. Well, one thing the ammonia would do is function as a base and deprotonate the acid.
So this is a carboxylic acid. You could think about deprotonating the carboxylic acid, which would form an ammonium salt. So that's one thing the ammonia could do. Something else the ammonia could do is function as a nucleophile and attack this carbon ray.
here, so this is an SN2 type mechanism. So the ammonia could function as a nucleophile and attack the carbon in red. And therefore, these electrons would come off onto bromine to form the bromide anion. And so you get a substitution reaction.
You substitute an amino group for the bromine. So let me go ahead and draw the results of the nucleophilic substitution reaction. So I'm going to show the carboxylic acid. And I'm going to show an amino group has sub... substituted in for the bromine.
So once again, I'm not concerned with the exact mechanisms here. And I'm also not concerned with exactly what the structure would look like at a certain pH. So this would be deprotonated with the use of ammonia. I'm just more interested in showing you that this is the dot structure for alanine, which is an alpha amino acid. So right here is the alpha carbon.
And we have an amino group that's substituted in for the bromine at the alpha position. And then we have a carboxylic acid. So an amino acid.
as just a simple way of thinking about it. Once again, in terms of stereochemistry, this alpha carbon is a chiral center, so we would make a mixture of enantiomers. But this is one of the classic ways of making alanine, using the HVZ reaction, and doing substitution at the alpha position. And once again, I'm not concerned with exact pH and what the structure would look like at different pHs.
I just want you to think about the usefulness of substitution at the alpha position of carboxylic acids. Use. using the HVZ reaction, and then using ammonia to produce amino acids.