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Voiceover: You start with the carboxylic acid and add something like ammonia to it. You might think that the ammonia would function as a nucleophile. So you might think it might attack right here and you lose your OH, and you would form an amide as your product. However, this is not what happens at room temperature. The ammonia is gonna function as a base instead and it's gonna take the acidic proton on your carboxylic acid, leaving these electrons behind on your oxygen. So, you would actually form your carboxylate anion here, so three lone pairs of electrons on your oxygen giving it a -1 formal charge. If you add a proton to ammonia you would form ammonium, so NH4+ and then we have our ammonium salt here. If you heat up this salt you actually can sometimes form your amide. However this is definitely not the best way to make an amide. A much better way would be to use something called DCC, and so let's get some more room down here so that we can see what DCC looks like. That's an acronym for Dicyclohexylcarbodiimides. So we have the D, the C, and the C here. And so if this R double prime group is a cyclohexyl group then we would have DCC. And so if you start with your carboxylic acid and add an amine, the use of DCC allows your amine to function as a nucleophile and eventually form your amide as your product. So let's look at the mechanism. The first step is for DCC to function as a base. And so this lone pair of electrons on the nitrogen take this proton, leave these electrons behind on your oxygen. So let's get some more room here. So, we're gonna go ahead and show taking a proton away from your carboxylic acid gives you your carboxylate anion, so let's go ahead and draw in our carboxylate anion. So three lone pairs of electrons on our oxygen, so -1 formal charge. This nitrogen here is gonna pick up that proton and so that nitrogen now has a +1 formal charge. So let's show some electrons. So the electrons in magenta here are going to take this proton, forming this bond after it took this proton here. And we can think about these electrons moving off onto our oxygen to form our carboxylate anion. Let's go ahead and draw in the rest of this. So we have a nitrogen double-bonded to a carbon, which is double-bonded to another nitrogen with a lone pair of electrons, and we have our R double prime group. So because we just protonated this nitrogen, it really wants electrons, it can withdraw some electron density away from this carbon. And so this carbon is gonna lose some electron density and become a little bit positive. And so this carbon is electrophilic now. And our carboxylate anion is gonna function as a nucleophile in our next step. So the nucleophile attacks our electrophile, pushing these electrons off onto our nitrogen, so let's go ahead and show that, so now we would have our carbon, double-bonded to this oxygen, and then this oxygen down here has now formed a bond to this carbon, so those electrons in blue have formed this bond now, and this carbon is bonded to this nitrogen, bonded to this, and our R double prime group, and if we showed, let's make these electrons in here green, these electrons move off onto our nitrogen, like that. And so our carbon is also double-bonded to this nitrogen, with R double prime group. So in the next step our amine is gonna function as a nucleophile. So let's go ahead and draw in our amine. So we have our nitrogen with two hydrogens, and an R prime group, lone pair of electrons on our nitrogen, makes this amine able to function as a nucleophile, so let me go ahead and put those electrons in magenta. So right here, this oxygen is partially negative, It's gonna withdraw some electron density from this carbon, so partially positive. And so we have a nucleophile that's going to attack our electrophile. So our amine is going to attack this carbon, push these electrons off onto our oxygen. So let's go ahead and show the result of this nucleophilic attack. Alright, so let's get some room down here. We would have our carbon bonded to this oxygen. So let's go ahead and draw in all of those electrons, so, -1 formal charge on this oxygen, so if these electrons in here in green move off onto our oxygen we get a -1 formal charge. This carbon is bonded to an R group, it's also bonded to this nitrogen, this nitrogen now has a +1 formal charge. So +1 formal charge on our nitrogen, after the electrons in magenta move in here to form this bond. So we still have our carbon bonded to this oxygen, and draw in our lone pairs of electrons, and then we have this carbon, we have a nitrogen, we have our R double prime group, we have a hydrogen, lone pair of electrons double-bonded to this nitrogen over here with our R double prime group, so a lot of stuff going on here. So, the use of DCC gives you a good leaving group. So if we think about all this stuff over here, this is an excellent leaving group. So if we reform our carbonyl, let's go ahead and show that, so if these electrons in here move in to reform our carbonyl, these electrons could come off onto our oxygen, and we could even show them moving over to here, to save some time, and if there's a proton out here, these electrons could pick up that proton, and we have an excellent leaving group, so this actually forms dicyclohexylurea over here on the right. So if I circle all of this stuff, we're gonna get dicyclohexylurea, and let's go ahead and show what would happen. So if we reform our carbonyl, let's use those electrons in here in red, so if those electrons in red move in, now we would have our R group, our carbon is now double-bonded to this oxygen with only two lone pairs of electrons, so the electrons in red move in to reform our carbonyl, and then this carbon is still bonded to this nitrogen. So let's go ahead and draw this nitrogen in here. And if we think about a base taking this proton, let me go ahead and change colors, a base taking this proton, leaving these electrons behind, so these electrons are gonna be left behind on this nitrogen here, so I'll draw them in here in blue, and so now we only have one hydrogen on this nitrogen, and then we have our R prime group, like that. So plus dicyclohexylurea as our other product. So we formed our amide. So once again, DCC allows the amine to function as a nucleophile. Alright, so let's see some uses for DCC. One of the most famous uses for DCC is to react amino acids together to form peptides. So if we have an amino acid over here on the left, with an R group, so we'll call it R1, and we have an amino acid over here on the right with a different R group, we have R2, not gonna worry about stereochemistry here, we could join these amino acids using DCC. So if we look for our carboxylic acid over here on the left, and our amine over here on the right, we know that DCC could form an amide. Now for something like this you have to be a little bit more careful because you have an amine over here on this side, and a carboxylic acid over here on this side, and so you would have to add a protecting group, so we'll just go ahead and think about a protecting group being over here on this side, and we could change this to a protecting group over here, something like OR prime, so an ester instead of our carboxylic acid, and when we add DCC, we can think about DCC as being a dehydrating agent, so we can think about losing water. So let's go ahead and show that. So we would lose the OH from our carboxylic acid and the H from our amine. So we can see, that's H2O. So if we think about minus H2O, we can stick those two amino acids together. So let's go ahead and do that. So we would have our carbonyl bonded to our nitrogen, bonded to this hydrogen here. And then we would have our R group, and then we have our carboxylic acid, and I'll go ahead and put in an OR prime here, so our carboxylic acid was protected over here on the right to form an ester instead. Over here we would have our R1, we would have our nitrogen, with the hydrogen, and depending on what protecting group you're using. I'm just gonna go ahead and put that one in here. So we would form a dipeptide. So right here you could see our R-amide, and then right here is our peptide bond that formed. So this was developed in the 1950s at MIT and it was published by Dr. Sheehan's group, somewhere around 1955. And he actually calls this up here, what I've done, we take the OH and the H, and just kind of think about removing them to lose water, as lasso chemistry. So, it's certainly not the best way to think about exactly what's happening, but it's a good way of thinking about sticking these two individual amino acids together to form your dipeptide. And so Dr. Sheehan's group was using DCC in the 50s, and it eventually, of course, it became the standard for forming peptides. And of course there are different versions of this now, but this was a breakthrough in the 1950s. And Dr. Sheehan was also involved in the total synthesis of penicillin in the 50s, and when he used DCC as a coupling agent to form his peptides in his lab, he thought that he could also use it to form penicillin. So let's take a look at the structure of penicillin. So over here on the right is penicillin salt, and this is actually from Dr. Sheehan's Total Synthesis of Penicillin. So he was the first to do so in 1957. So if we look over here we can see an amine and we can see a carboxylic acid, and so if we do our lasso chemistry, so if we take this OH and this H, we think about losing water, and you can see here we have an amide in our ring. So an amide in a ring is called a lactam. So this is very famous. This is a beta-lactam. So penicillin is in the beta-lactam antibiotics. So it's called a beta-lactam because the carbon next to your carbonyl here would be your alpha-carbon, and then the carbon next to that would be your beta-carbon. So here we have a four-membered ring to form our beta-lactam. And so Dr. Sheehan used DCC in his synthesis to join this together, and this beta-lactam was extremely difficult for other chemists to make. And so during World War Two there were several labs that were working on this, so I think Dr. Sheehan wrote in his book, it's called "The Enchanted Ring," because making this beta-lactam ring was extremely difficult, that this was the tricky part. How do you form a beta-lactam ring? It was extremely difficult for most chemists to do so. But the use of DCC, which can be used in, you can do this in water, you can do this at room temperature, you can do this at relatively normal reaction conditions, and so that made him the first one to synthesize penicillin, the first total synthesis of penicillin. So according to him, in his book, he said there was no competition at the time, and he thought that other chemists were simply tired of trying. So, some pretty interesting chemistry. And I'll talk more about penicillin in a later video.