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Current time:0:00Total duration:11:29

Video transcript

if you start with an aldehyde or a ketone and add a catalytic amount of acid or base you'll find the aldehyde or ketone is going to be an equilibrium with this product over here on the right which we call an enol so the name in all comes from the fact that we have a double bond in the molecule so that's where the N part comes in and we also have an alcohol all right you can see the O H over here so that's where the O L comes in so this is the enol form and then over here this is the keto form so the keto form and the enol form and these are different molecules there are isomers of each other so we call them tautomers and they're in equilibrium with each other they're not different resonance structures let's see if we can analyze our aldehyde or ketone to see how to form our enol and so if we look at the carbon that's next to the carbonyl carbon we call this the Alpha carbon and there are two hydrogen's attached to the Alpha carbon in this case let me go ahead and draw those in those are called the alpha protons and so if we think about transferring one of those alpha protons from the Alpha carbon to the oxygen right even though it's most likely not the same a proton just helps to think about doing that we can also think about moving the double bond right so over here on the left the double bond is between the carbon and the oxygen and removing that double bond over here between the two carbons so transferring one alpha proton and shifting your double bond converts the keto form into the enol form and then we also have a hydrogen rights over here alright we still have a hydrogen left on this carbons let me go ahead and draw in that hydrogen so that's this hydrogen in blue here so that's how to think about converting a keto tautomer into an enol one let's look at the acid catalyzed mechanism for this so if we start with our aldehyde or ketone and add h3o plus alright the first thing is going to happen is protonation of our carbonyl and so lone pair of electrons picks up this proton like that so we can go ahead and draw that we would protonate our carbonyl so now our oxygen would have a plus 1 formal charge let me just go ahead and draw on those electrons here and let's say we started with an aldehyde so we'll make this an H so the lone pair of electrons on our oxygen right picked up a proton like that we can draw a resonance structure for this right we can move these electrons off onto our oxygen so let's go ahead and show a resonance structure so we have our R group all right and now we would have our oxygen with two lone pairs of electrons so let me go ahead and draw in those two lone pairs of electrons on our oxygen and then we took a bond away from carbon right so we took a bond away from this carbon so this carbon right here so plus one formal charge on that carbon and then we could show the movement of those electrons so these electrons right here I'm saying moving out on to the oxygen like that and so this is uh this is our intermediate here alright so we we know that our alpha carbon has two protons on it so once again let's find our alpha carbon here it is right here and we know we have two protons attached to it two alpha protons if you will and so in the next step of our mechanism we're going to get a cool of water acting as a base let me go ahead and show a molecule of water here and the water is going to take one of those alpha protons right let's say once again it takes this alpha proton and leave these electrons behind they're going to move in here to form our double bond so let's go ahead and draw let's go ahead and draw our products right so we would have our R group here and now we would have a double bond form between our two carbons and then we would have our oxygen and then we would have two lone pairs of electrons on our oxygen we have our hydrogen and then we would have another hydrogen right here so let's go ahead and follow some of those electrons alright so let's go ahead and make these electrons in here blue alright so these electrons are going to move in in here it doesn't really matter which one you say it is let's just say it's that one right to form our double bond and then the electrons in red moved off onto this oxygen and then we said that these electrons were in magenta and so you can see that we have forms our enol here right so this is our enol right and then we started with our keto form like that so keto enol tautomerization let's look at the base catalyzed version alright so once again we start with our aldehyde or ketone but this time we're going to add a base so something like hydroxide and so we find our alpha carbon so here is our alpha carbon once again with two alpha protons so I'm going to go ahead and draw in those two protons here and the base is going to take one of those protons let's say it takes this one over here on the right that leaves these electrons behind on this carbon so let's go ahead and draw the resulting anion here so we would have our carbon eel like that alright and once again let's say we started with an aldehyde and then we would have a lone pair of electrons on this carbon the carbon in red here so let me go ahead and identify those electrons so these electrons in here and magenta have moved off onto this carbon like that which gives that carbon a negative 1 formal charge right it's a carb anion there's still a hydrogen attached to that carbon in red right this hydrogen right here is still attached to it I'm just not drawing it in so we can see a little bit better alright so this is uh this is one form of the anion that we could have we could draw a resonance structure to show the other form so if we move these electrons in magenta into here and push these electrons off onto the oxygen let's draw the resonance structure so we would have alright on our group here we would have a double bond and then our oxygen would have three lone pairs of electrons giving it a negative 1 formal charge and then we would have our hydrogen over here so the electrons in magenta alright moved in here to form our PI bond and then we can say that these electrons in here right moved off onto our oxygen so we could go ahead and show that and so let me just go ahead and put the other bracket on here and so we have two forms of this anion this is called the enol 8 anion so this is the enol 8 anion this is going to be extremely important in future reactions and you can see the enol 8 an ID has two resonance structures one where we've shown the negative charge on the carbon alright so that would be that would be this one right over here so the negative charge in the carbon so this is our carb anion form so carve an ion and then we also have a resin structure where the negative charge is on the oxygen so we could call this the oxy anion and if you think about which one contributes more to the overall hybrid oxygen is more electronegative than carbon and so it's better able to have a negative 1 formal charge on it so the oxy anion contributes more to the resonance hybrid all right let's think about the last step in our mechanism right to form our enol if we think about our oxy anion all we'd have to do is protonate that oxygen here so we could just go ahead and draw a water molecule all right so we have a water molecule this time water is going to function as an acid it's going to donate a proton so let's say these electrons in blue right take this proton leave these electrons behind and so from our oxy anion that we can go ahead and draw our enol product so we have our R group here we would have our double bond we would have our oxygen all right now protonated like this to form our enol product so let me just go ahead and show those electrons in blue right picked up a proton here to form our enol so that's how to get there using using base catalyzed and once again we will talk much more about the enolate anion in future videos here so let's look at a situation where the alpha carbon is a chiral Center okay so let's look at this right here so here's our alpha carbon right and it's a chiral let's just say it's a chiral Center so if R and R double prime are different from each other we would have four different things attached to this carbon right and so the Alpha carbon here is sp3 hybridized with tetrahedral geometry so let's say it's either the R or the S enantiomers doesn't really matter which one but you can see now we have only one alpha proton right only one alpha proton but because there is an alpha proton right we can form an enol so in either an acid or base catalyzed mechanism right we could think about the proton here in red you can think about transferring one to this oxygen and moving your double bond and then we our enol alright so here is our enol now let's look and see what happened to the carbon in red right here so on the left right the alpha carbon is sp3 hybridized with tetrahedral geometry now this carbon is sp2 hybridized with trigonal planar geometry and so whatever stereo chemical information we had over here on the left right whether it was the R or the S enantiomer it's been lost now that we form the enol the enol is a chiral it's flat it's planar and so when when we reform the keto form right so one of the possibilities is to form the enantiomer that we started with but the other possibility is to form the other enantiomer and so you can see that's what I've shown here I've shown the hydrogen outgoing away from us and our our double prime group coming out at us so this is the this is the enantiomer and so because because we form the enol right we can get a mixture of enantiomers so enol ization can lead to two racing ization we can get a mixture of enantiomers and if we and if we wait long enough right we could get an equal mixture of these guys so this one and this one right would be an equilibrium with our enol form and so that's something to think about if you have a chiral Center at your alpha carbon let's look at two quick examples of keto and enol forms and so over here on the Left we have cyclohexanone and then the right would be the enol version of it right so you can think about one of these as being your this being your alpha carbon right and you can move these electrons in here and push those electrons off and you can see how that would give you this enol form so it turns out that the the keto form is is favored right so the equilibrium is actually far to the left favoring formation of the keto form so even under just normal conditions so not acid or base catalyzed and so there's only a trace amount of the enol presence however there are some cases where the enol is extra stabilized and that's the case for this example down here so we have the keto form and we have the enol form so once again you could think about these electrons moving in here pushing those electrons off giving you your enol form this is a specially stabilized in all right so this is this is phenol right here and we know that phenol has an aromatic ring and so the formation of the enol form is extra stabilized because of this aromatic ring and so this time the equilibrium is actually to the right and much more of it is in the enol form than the keto form so in this case we have some special stabilization