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Mixed (crossed) aldol condensation using a lithium enolate

Video transcript

here's another way to do a mixed or across aldol condensation so this time using a lithium enolate so if we took this ketone and this aldehyde and just mix them together with some base we would get a mixture of products we wouldn't get our desired product this conjugated enone over here so if we do this stepwise we can do a directed aldol condensation and so if we take this ketone and add Lda to it and then we add our aldehyde and then our workup we can add water and toluene sulphonic acid as an acid source we can get our conjugated enone in a decent yield and so let's look at each of these steps one by one so we'll first start off with the addition of Lda which we know is a strong base so the strong base is going to take a proton from our ketone and so let's analyze our ketone here so the alpha carbons are the ones next to the carbonyl and so this is an alpha carbon on our ketone and this is an alpha carbon on our ketone so which one of those alpha carbons will our strong base deprotonate so we saw in a previous video that Lda is going to form the kinetic enolate so it's going to take a proton from the least sterically hindered side and so it's going to take an a proton from this right side over here to form the kinetic enolate right if it took one from the left that would be the thermodynamic enolate so I can go ahead and draw in an alpha proton on our alpha carbon right here so let's say this is the one that Lda is going to take let's go ahead and draw in Lda so I'm going to go ahead and draw my nitrogen in here and then I have my isopropyl groups on my nitrogen and then I'm going to have the nitrogen with bonded to a lithium right here like that so when this deprotonation occurs it's actually a cyclic mechanism so the oxygen is going to be forming a bond with this lithium alright the nitrogen is going to be forming a bond with this proton here so let's go ahead and show the cyclic mechanism alright so if these electrons right here on the nitrogen anion move in and take this proton right we're going to form a bond right here alright these electrons will move in here to form a carbon-carbon double bond and then these electrons would kick off onto your oxygen to form a bond with lithium so let's go ahead and show the product and then we'll show the movement of the electrons here so we have our oxygen alright is now going to be bonded to lithium like that and now we have a carbon-carbon double bond and then we would have nitrogen bonded to hydrogen now and we have an amine over here on the right so let's go ahead and follow some of those electrons so if I'm saying these electrons in blue here right on the nitrogen anion are going to pick up this proton to form our amine like that all right I could say that these electrons in here right once once it's deprotonated right they could form our double bonds right in here and then I could also point out that these electrons right here right could move out to bond with the lithium to form our lithium enolate and so once again this is the kinetic enolate the one that's warmed up fastest and the use of Lda um helps us to to get our kinetic enolate because these bulky is bulky isopropyl groups right would prevent deprotonation at the alpha carbon on the left and so now we've formed our lithium enolate and so let's go to the next step alright so in the next step we're going to add our aldehyde right here so we're going to add a butte now so let's go ahead and look at the formation of our lithium enolate and then let's add butanol to that so let's go ahead and draw in butte now so here we have our carbonyl and then we would have four carbons so 2 3 4 like that we know that aldehydes can function as electrophile so in the second step we're going to add an electrophilic carbonyl compound right we know this is electrophilic because the oxygen is partially negative and this carbonyl carbon right here is partially positive like that and so in in the second step or it's going to be a nucleophile attacking an electrophile so the enol h is going to function as a nucleophile so we're going to form a bond this time between lithium and oxygen over here and then we're going to form a carbon-carbon bond down here all right so let's show the movement of those electrons so the lithium enolate functions as our nucleophile the alda the aldehyde functions is our electrophile so if these electrons move into here right then these electrons are going to bond with that car urban and then these electrons can bond in here so another cyclic mechanism and so let's go ahead and draw the result of that so we would have these carbons we would form a carbonyl right here all right now we'd form a new carbon-carbon bond right here and then our oxygen would now be bonded to our lithium and then we'll draw on the rest of our carbon so we form a lithium alkoxide product following those electrons right so these electrons in here in red right moved into here to form our carbonyl alright the electrons in here in magenta right these are the ones that attacked our our our carbonyl carbon here to form our new carbon-carbon bond and then finally I could say I could say make these electrons in here blue right and then forming a bond between oxygen and lithium to form our lithium alkoxide intermediate here and so to cyclic mechanisms form your form your carbon-carbon bond here so first you would deprotonate alright we talked about this cyclic mechanism up here and then once you add your aldehyde once you've formed your your lithium enolate right another cyclic mechanism will give you your lithium alkoxide so let's look at the next step right the third step you're going to add water in your workup and so here I've drawn the lithium alkoxide a little bit differently right so the electrons in blue that I had over here in the lithium oxygen bond I'm just going to put those on the oxygen this time alright we know the oxygen has two other lone pairs of electrons right which give it a negative 1 formal charge and then lithium would be a plus one charge and so this is just another way to represent our our lithium alkoxide this is how we've usually done it in our videos here and so in the next step in the workup right if you if you do an aqueous workup so you add some water here you would protonate your alkoxide anion so let's go ahead and show that so we take a proton here from water and let's go ahead and draw the products so once again we would have our carbonyl and then we would protonate the oxygen to form our aldol and so here is our aldol product alright so once you form your aldol let's go ahead and look at the next step here we added some toluene sulphonic acid so we added a source of protons and so in the final step to get to our genome we need to dehydrate our aldol and so we've seen we've seen how to form our enon using base and this time we can do an acid catalyzed dehydration so if you add a source of protons here alright we know we still have two lone pairs of electrons on our oxygen alright one of those lone pairs could pick up that proton so we protonate so let's go ahead and draw what we would make here so we have our carbonyl and we have all of these carbons and now we would have this oxygen right with still has one lone pair of electrons so let's go ahead and show those electrons in these electrons in magenta right picked up a proton right here so we have one lone pair of electrons left which gives that oxygen a plus one formal charge and so we have an excellent leaving group here if we think about these electrons moving off onto the oxygen right we have water as a leaving group so loss of water at this step would yield a cation so let's go ahead and draw the cation that would form so we have our carbonyl and we have all of these carbons and so we lost a bond to this carbon right here so this is the carbon that's going to form our cation so let's go ahead and draw a +1 formal charge on our cation like that and so in the next step a base is going to come along and it's going to take a proton from our alpha carbon so once again we think about where our carbonyl is the carbon next to it is our alpha carbon and so there's there's a proton on here so we can deprotonate it with a base so I'm just going to write a generic base but it could be something like the water that that just left in the previous step so a base is going to come along and take this proton right and so these electrons are going to move in here to form your double bond that's going to take away your +1 formal charge so let's go ahead and draw our product all right so we would have our carbonyl and then we would now have a double bond right here and then we draw on the rest of our carbons like that so if we show those electrons alright let's make them let's make them blue this time so these electrons in here so the base takes the proton and the electrons move in to give us our double bond and we now have formed our a known as our product so once again this is an example of a directed aldol condensation