- Carboxylic acid questions
- Carboxylic acid reactions overview
- Carboxylic acid nomenclature and properties
- Reduction of carboxylic acids
- Preparation of esters via Fischer esterification
- Preparation of acyl (acid) chlorides
- Preparation of acid anhydrides
- Preparation of amides using DCC
- Alpha-substitution of carboxylic acids
How to use the Hell-Volhard-Zelinksy (HVZ) reaction to synthesize alpha-amino acids. By Jay. Created by Jay.
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- At7:38, why does :NH3 attacks the alpha carbon instead of the carbonyl carbon which seems more electrophilic?(5 votes)
- Good question! As Jay mentioned, one thing the added ammonia would do, and very rapidly so, is to deprotonate the carboxylic acid at the hydroxyl group. Due to the subsequent negative charge on the oxygen, nucleophiles would be repelled from the carbonyl carbon and therefore not react with it; the electrophilicity of the carbonyl carbon would also decrease because the deprotonated, negatively charged oxygen would participate in increased electron donation via resonance. :)(5 votes)
- Why is that the alpha carbon, I would think the carboxylic acid carbon would be the alpha carbon cause it's attached to the highest priority functional group.(3 votes)
- The alpha carbon is the first carbon that attaches to a functional group, in this case a carbonyl. It is not the carbon that is itself part of that carbonyl.(5 votes)
- Why the optical rotation of (+)-2-phynyl-2-pentanol goes to zero when heated with aqueous acid to give (+-)-2-phenyl-2-pentanol(2 votes)
- This is Gabriel Synthesis correct?(1 vote)
- Would heat be required to make the amine act as a nucleophile? DCC was required for amide synthesis. Otherwise, it seems that the amine would deprotonate the acid.(1 vote)
- At4:57when you describe how adding water causes the alpha bromo acyl bromide to turn into the product, why doesn't water substitute at the alpha carbon as well? Seems like there is nothing stopping that bromide from undergoing a substitution reaction.(1 vote)
Voiceover: We'll start with a carboxylic acid that has at least one alpha proton on it. Right here is our alpha carbon. We can see this reaction we're going to substitute in. Instead of a hydrogen in the alpha position, we're going to have a bromine at the alpha position. Here's your alpha position. This is alpha substitution of the carboxylic acid. This is called the Hell-Volhard-Zelinksy reaction. In the first step we add bromine and phosphorous tribromide and in the second step we add water. Let's look at these one by one. We'll start with phosphorous tribromide. If you add PBr3 to a carboxylic acid, we saw in a previous video that the phosphorous tribromide is going to turn the carboxylic acid into an acyl bromide. Let me go ahead and draw in the acyl bromide, still thinking about this alpha proton right here. You can see we took the carboxylic acid and we substitute in a bromine for the OH. So we form our acyl bromides. If you think about the next part of this reaction, there's acid presence, so we're going to do an acid catalyzed tautomerization. You can think about our acyl halide. This is being the keto form of our acyl halide and I'm going to go ahead and draw the enol form of our acyl halide. So we would have our R group and then we have a double bond here and 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. 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 to be between this carbon and this carbon, so here's our double bond now. We could also think about moving a proton. Moving a proton from here over to here. It's not necessarily the same proton, it's just a very simple way of thinking about this. Converting a keto form of our acyl bromide into an acyl bromide enol over here on the right. Once we form our enol, we can think about the bromine. We have BR2 present, as well. Let's go ahead and draw in bromine here. A molecule of bromine is normally non-polar. We have atoms of the same electronegativity, so it's normally a non-polar molecule. However, you can induce a dipole on the bromine molecule. Things like pi electrons can help induce a dipole. 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. We talked about this when we talked about reactions of alkenes. The pi electrons in here, we saw that pi electrons could function as a nucleophile. Those pi electrons in blue could attack the electron deficient bromine, which is going to function as an electrophile. This bromine on the left here could function as an electrophile. We have a nucleophile attacking our electrophile. 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. Let's go ahead and draw what we would make. We would have our R group, we would have this carbon right here, we would have a carbonyl like that, and then we would have a bromine right here, and then we would have a bromine here, as well Let me show those electrons. The electrons in blue, these pi electrons in blue are going to function as a nucleophile, attack this bromine and so that's these electrons right here and this bromine. Then we're going to lose the bromide anion over here on the right. We make the bromide anion over here on the right. Then we can also deprotonate, so we lose this proton. You can think about loss of HBr at this step, so minus HBr. I'm not going through the entire mechanism here, just a shortened way of thinking about it. We 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. We call it an alpha-bromoacyl bromide. Then we're ready for the second step of the HVZ reaction, which is where we add water. We could think about adding water, so let's go ahead and draw that in. We have water right here and water has two lone pairs of electrons, so water can function as a nucleophile. We know this carbon right here is electrophilic. This oxygen is more electronegative, withdrawing some electron density away from this carbon. 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 to reform your carbonyl, these electrons in here are going to come off onto bromine to form your bromide anion as a leaving group. Then you could also think about losing a proton from water after it bonds to that carbon. The OH is going to replace the bromine and then we're going to once again lose HBr, as just a simple way of thinking about it. Loss of HBr at this step and that converts our alpha-bromoacyl bromide into our final product. The end result is to get back our carboxylic acid, but we've substituted a bromine at the alpha position now. That's the HVZ reaction. Again, I didn't go through the complete mechanism, a way of thinking about the reaction. It would take a little bit too long for one of these videos to go through every single step. One use for the HVZ reaction is to synthesize amino acids. That's one of the classic uses here. We could make an amino acid, starting from this carboxylic acid. This is a three carbon carboxylic acid. This is propanoic acid. If we react propanoic acid with, once again, a bromine phosphorous tribromide and then water, this is the HVZ reaction, which we know allows us to substitute the alpha position. Right here is the alpha position and we know there are two hydrogens bonded to this carbon at the alpha position. We're going to substitute in a bromine for one of those hydrogens. We already know what the product looks like for the HVZ reaction, so we draw in our carboxylic acid and we know there's going to be a bromine substituting at the alpha position and there's still a hydrogen bonded to this alpha position here. Thinking about this alpha carbon right here, there are four different substituents attached to this alpha carbon. You're thinking about stereochemistry, this is a chiral centers. You're going to get a mixture of enantiomers for your product here. Once you form that, to make an amino acid, in another reaction, you could add excess ammonia. Let's go ahead and draw in ammonia right here. Let's think about what the ammonia might do. One thing the ammonia might do is function as a base and deprotonate the acid. This is a carboxylic acid, you think about deprotonating the carboxylic acid, which would form an ammonium salt. That's one thing the ammonia could do. Something else the ammonia could do is function as a nucleophile and attack this carbon right here. This is an SN2 type mechanism. The ammonia could function as a nucleophile and attack the carbon in red. Therefore, these electrons would come off onto bromine to form the bromide anion. You get a substitution reaction. You substitute an amino group for the bromine. Let me go ahead and draw the result of the nucleophilic substitution reaction. I'm going to show the carboxylic acid and I'm going to show an amino group as substituted in for the bromine. 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. 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. 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 is just a simple way of thinking about it. Once again, in terms of stereochemistry, this alpha carbon is a chiral centers, 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. 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 using the HVZ reaction and then using ammonia to produce amino acids.