Formation of imines and enamines

Voiceover: Let's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone. And you add an amine to it and you need an acid catalyst and over here on the right if your Y is equal to hydrogen or an alkyl group which is a R double prime you will form an imine. So let's go ahead an write that this would be an imine. And the Y can be other things, we'll talk about those other things in the next video. You also form water and so if you, since it reacts as [that] equilibrium if you wanted to shift equilibrium to the right, you could remove water as it's formed and equilibrium will shift this way and give you more of your imine product. So let's look at the mechanism to form imines. And I've seen two different ways to start off, the mechanism and so I'll present both ways and you can choose which one you want to use. And so one way would be to think about an acid being present right, so we'll say it's H-A plus the generic acid and you think about this as protonating your imine. So you have an imine and protons, you're protonating your imine to form this generic acid here. And then your carbonyl oxygen is going to be protonated. A lone pair of electrons on the oxygen takes this proton, leaves these electrons behind, so let's go ahead and show the results of that. So now we have our oxygen, with a plus one formal charge and we still have our prime group over here, so let's show those electrons. So this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here. We've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic. And so we have a good electrophile, this carbon here is a good electrophile and the imine can be a good nucleophile. So a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that's going to kick these electrons in here off onto the oxygen, so let's draw the results of that. So now we're going to have our nitrogen is forming a bond with our carbon, so let's show those electrons so let's make those blue. So this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here. This nitrogen is also bonded to two hydrogens. So let's draw those in. And still our Y group like that. That gives our nitrogen a plus one formal charge. And for our carbon, it's also bonded to this oxygen, this oxygen now has two lone pairs of electrons. So let's show the movement of those electrons so I'll use green here. So these electrons in green moved out onto our oxygen. The carbon is still bonded to our alkyl group so R and R prime, so I'm saying we start with a ketone here. And so that's one way to start off your mechanism and of course there is another way to do it. So let's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly. So partial negative oxygen, partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly. Right, attack this carbon push these electrons off onto your oxygen. So let's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen. Our nitrogen is bonded to two hydrogens and our Y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge. So negative one formal charge here. We have our alkyl groups down here, so let's once again show our electrons. So if nitrogen attacks directly, This lone pair in blue forms this bond right here. And then we can say that these pi electrons in here kick off onto our oxygen so it doesn't really matter which lone pair you make them as long as it states that one right there. And then we can show this intermediate going to the one we already talked about, if we just protonate our negative one formal charge on our oxygen here. So let's show that, so if we have our generic acid so H-A plus, we can show a lone pair picking up this proton, leaving these electrons behind. And so now that would give us this intermediate the same one that we had before. So whichever way you would like to start off your mechanism, once we've reached this intermediate we can think about the base. The imine here, a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens. So let's show what we would make from that. Let's get a little more room down here alright. So we deprotonate and then we form this structure, a carbon bonded to a nitrogen and then we still have our Y group, we still have a hydrogen on this nitrogen. We have a lone pair of electrons on this nitrogen, so that lone pair came from right in here. So you take that proton and leave those electrons in magenta behind on your nitrogen. So we also have an O-H group bonded to our carbon here, and then we also have our alkyl group, so R and R prime. So this intermediate is called a carbinolamine. So let me go ahead and write that. So this is called a carbinolamine. And for the next step we can protonate the O-H group. So we take our generic acid once again so H-A plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let's go ahead and show that. So let's get once again some more space in here. So if we protonate our carbinolamine, we would have our carbon here, our nitrogen, lone pair of electrons on our nitrogen, our Y group, our hydrogen here, our alkyl groups. And then if we protonate the O-H, we would form water as a good leaving group. So right in here, this would be a plus one formal charge on our oxygen and let's show those electrons. So in here, lone pair of electrons on the oxygen, pick up this proton and so forming this bond right here and now you can see we have a good leaving group. So we have water as a good leaving group. So if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen. So let's go ahead and show that. So our next step here, this is where we lose water. So we're going to minus H2O and let's go ahead and show our next structure here. We have our carbon double bonded to our nitrogen this time, our nitrogen is still bonded to our Y group. And our hydrogen over here. That gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups And so let's show those electrons here on our nitrogen. So I'm going to go ahead and make those magenta. So these magenta electrons move in here to form our double bond between our carbon and our nitrogen. And this is an important structure, this is called an iminium ion, so let's go ahead and draw that. So an iminium ion and then we lose water of course. So minus water at this stage. So we're almost to our final product, we would just have to deprotonate our iminium ion. And so we can do that with our imine. Could take that proton and leave these electrons behind on the nitrogen. So that's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our Y group here. A lone pair of electrons on that nitrogen and then we have our alkyl groups and so let's show those electrons here in blue moving in here, off onto our nitrogen. And then once again, if our Y is a hydrogen an alkyl group we have formed an imine. Alright, so if that's formation of an imine, let's look at an example. So here is our reaction, we're going to start with a ketone here, so cyclohexanone and react it with a primary amine. So the nitrogen is bonded to one carbon. So this is a reaction with a primary amine and then we're going to use sulphuric acid as our catalyst here. So to figure out the product of this reaction since it's kind of a long mechanism so it doesn't really makes sense to run through the entire mechanism but we could think about part of the mechanism here. We know that the nucleophile is going to be our amine and it's going to attack our carbonyl carbon here. So we know that in our mechanism we lose one of these protons on our amines so let's say it's that proton right there. And we know we're going to lose water so minus H2O and this going to think about taking us to our iminium ion steps, so we're going to have our ring. And so we've already lost water so now we're going to have a double bond to our nitrogen. And our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen. So we still have the hydrogen attached here at this point. Which gives us a plus one formal charge on our nitrogen. So this is our iminium ion right here. And then in our last step we know that our amine can come along and act as a base. So a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products. So we go over here and we draw our product, double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen. So if I follow those electrons, so these electrons in here in blue ended up on our nitrogen to form our imine. So a few things that I want to point out here for the iminium ion, we still have a proton attached to it. So deprotonation for our last step will form an imine. And this is once again done with a primary amine right here. And so that's a little bit different from what we're going to talk about next, we're going to talk about formation of an enamine, and formation of an enamine is, it starts off the same way in terms of the mechanism but it's this iminium ion step that changes a little bit. So let's look at reacting a ketone with a secondary amine this time. And so let's look at this one right here, so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time. And once again, we can kind of run through the mechanism. just thinking a little bit about what happens without going through each individual step, We know the nucleophiles going to attack here and we know we're going to lose a proton. So this is the only proton we have left to lose and so we lose that one, we're going to lose water in the mechanism to form our iminium ion, so let's go ahead and draw that out We're going to have our ring, we're going to have a double bond to this nitrogen, we've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here. And so in the previous reaction, let me go back up to here to the previous reaction again. Once again we still had a proton on our iminium ion at this step and so we're able to deprotonate here. But that's not the case with what we have for this ion. We don't have a proton on our nitrogen here. And so we can't deprotonate in the same place, we have to pick a carbon adjacent over here. So here's a proton over here on this carbon and our base could come along. So let's go ahead and draw out our amine base with a lone pair of electrons right here. And a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products, so let me go ahead and draw this out here. So now we're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen. So let's follow some electrons. So these electrons in here once you deprotonate, move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product. So this called an enamine, so let me go ahead and write this out here, an "enamine." And the name come from "-en" come from the double bond. So we form double bonds, that's the "-en" portion and then the "amine" portion comes from this part up here. We have an amine so this is an enamine. And so formation of an enamine happens because once again we don't have the same iminium ion that we had before because of the fact that we started with a secondary amine. So that's why it's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone. And if it's secondary, you're going to form an enamine here. And enamines are useful synthetic intermediates, So we'll talk much more about them in later videos. But once again the mechanism is the same until you get to this last step here.