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Current time:0:00Total duration:11:54

Preparation of amides using DCC

Video transcript

you start with a carboxylic acid and add something like ammonia to it you might think that the ammonia would function as a nucleophile right so you might think it might attack right here and you lose your Oh H and you would form an amide or amat as your product however this is not what happens at room temperature the ammonia is going to function as a base instead it's going to 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 negative 1 formal charge if you add a proton to ammonia you would form ammonium right so NH 4 plus and we have our ammonium salt here if you heat up this salt you actually can sometimes form your 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 we can see what DCC looks like that's an acronym for Dai cyclohexyl carbo Dai image or Die image alright so we have the D the C and the C here and so if this R double prime group right as 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 and eventually form your amide as your product so let's look at the mechanism the first step is for a 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 alright so we're going to go ahead and show taking a proton away from your carboxylic acid right gives you your carboxylate anion so let's go ahead and draw in our carboxylate anion all right so three lone pairs of electrons on our oxygen so negative one formal charge this nitrogen here is going to pick up that proton and so that nitrogen now is a plus one formal charge so let's show some electrons right so the electrons in magenta here are going to take this proton alright so forming this bond after it took this proton here and we could think about these electrons moving off onto our oxygen to form our carboxylate to anion let's go ahead and draw on the rest of this so we have a nitrogen double bonded to a carbon which is double bonded to another nitrogen with lone pair of electrons and we have our R double prime group so the because we just protonated this nitrogen right it really wants electrons they can withdraw some electron density away from this carbon and so this carbon is going to lose some electron density and become a little bit positive and so this carbon is electrophilic now and our carboxylate anion is going to function as a nucleophile in our next step so the nucleophile attacks our electrophile all right pushing these electrons off onto our nitrogen so let's go ahead and show that so now we would have our carbon all right double bonded to this oxygen and then this oxygen down here right has now formed a bond to this carbon so those electrons in blue right 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 all right these electrons move off on to our nitrogen like that and so our carbon is also bonded double bonded to to this nitrogen alright with our double prime group alright so in the next step our amine is going to function as a nucleophile so let's go ahead and draw in our amine alright so we have our nitrogen with two hydrogen's and an R prime group lone pair of electrons on our nitrogen makes makes this amine able to function as a nucleophile so I'm go ahead and put those electrons in magenta okay so right here right this oxygen is partially negative it's going to withdraw some electron density from this carbon so partially positive and so we have a nucleophile that's going to attack our electrophile right 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 on all of those electrons so negative one formal charge on the oxygen so if these electrons in here and green move off onto our oxygen right we get a negative 1 formal charge this carbon is bonded to you in our group right it's also bonded to this nitrogen this nitrogen now has a plus 1 formal charge so plus 1 formal charge on our nitrogen after the electrons in magenta move in here to form this bond all right so we still have our carbon bonded to this oxygen all right and draw on 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 alright lone pair of electrons double bonded to this nitrogen over here alright with our our double prime group alright so a lot of stuff going on here so the the use of DCC gives you a good leaving group alright so if we think about all this stuff over here alright this is an excellent leaving group so if we reform our carbonyl alright 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 that moving over to here to save some time and if there's a proton out in here alright these electrons could pick up that proton we have an excellent leaving group so this actually forms dye cyclohexyl urea over here on the right so if I circle all of this stuff alright we're going to get dye cyclohexyl urea and let's go ahead and show what would happen right so if we reform our carbonyl let's use those electrons in here in red so those electrons in red move in alright 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 alright and then 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 alright a base taking this proton leaving these electrons behind alright so these electrons are going to be left behind on this nitrogen here so I'll draw them in here in blue and so now we have only one hydrogen on this nitrogen right and then we have our our primate group like that so plus Dai cyclohexyl urea as our other products so we formed our amide so once again DCC allows the amine to function as a nucleophile all right so let's let's see some uses for DCC one of the most famous uses for DCC is to react amino acids together to form a to form peptides right so if we have an amino acid over here on the left so with an R group also called r1 and we have an amino acid over here on the right with a different R group we have r2 not going to worry about stereochemistry here we could we could join these amino acids using DCC all right so if we look for our carboxylic acid over here on the left right 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 alright so we'll just go ahead and think about a protecting group of being over here on this side and we could we could change this to a protecting group over here something like o R Prime right 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 O H 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 alright bonded to our nitrogen bonded to this hydrogen here and then we would have our R group alright and then we have our carboxylic acid and I'll go ahead and put an O R prime here so our carboxylic acid was protected over here on the right to form an ester instead over here we would have our R 1 we would have our nitrogen right with a hydrogen and depending on what protecting group you're using I'm just going to go ahead and put that one in here so we would form a dipeptide alright so right here you can see our am I draw our image and then right here is our peptide bond that formed so this was 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 all right we take it Oh H and 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 eventually of course it became the standard for performing peptides and of course there are different versions of this now but this was 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 2 as a coupling agent to form his peptides in his lab he thought that he could also use it to format 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 if we look over here right we can see it an amine right and we can see a carboxylic acid and so if we do our lasso chemistry right so if we take this o H and this H right we think about losing water and and you can see here we have an an amide right and our ring so an amide in a ring is called a lactam so this is a very famous right this is a beta lactam all right so penicillin is in the beta-lactam antibiotics so it's called a beta lactam because the carbon next to your carbonyl here right 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 us together and this beta lactam was extremely difficult for other chemists to make and so in during world war ii there were several labs that were working on this so I think dr. Sheehan wrote in his book it's called the enchanted ring right because making this beta-lactam ring was extremely difficult that it was this was the this is 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 is uh which can be used in you can do this in water you can do this at room temperature you can do this at at relatively normal reaction conditions and so that made it 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 chemists other chemists were simply tired of trying so some pretty some pretty interesting chemistry and I'll talk more about penicillin in a later video