all right we're gonna get started we're going to talk about retrosynthesis for the claisen so we can figure out how to make compounds by claisen condensation anybody have questions for me before we get started all right all right so we're gonna take apart if I give you a compound and I say show me how you can make this using a claisen condensation we had steps to do that with an aldol we have steps to do this with a claisen so here's the steps right here so we're gonna start with start counting from the ester because we have two carbonyls here I encourage you to start counting from the ester so we have alpha and we have beta if we start counting from the ester break the alpha beta bond okay so that's step two add Oh a T to the beta position that's the oet that we lose when we're doing the claisen if you think about the mechanism that's the OEt that we lose at OE t to the beta position put the negative charge in the alpha position so putting the negative the alpha position lets us know where we're making our enolate so let's draw the two compounds that you would use that's just a self condensation isn't it so that's going to work really well so the structural feature that you want to look for is fate Aikido Esther that would be for classic claisen condensation where we have an ester combining with an ester but we also have some cross license that kind of are marginally claisen marginally aldol and so that would be classic claisen or just a 1:3 dicarbonyl so I may tell you show me how you make this by placing or I may not tell you with it's from a claisen and you should recognize oh okay that's a beta key to ester I can make that by claisen and then be able to take it apart and show me how to make that alright so let's look at an example here a target molecule that we're gonna try to synthesize here following the same steps start counting from the ester alpha beta break the alpha beta bond put in it and the Foxxy in the beta position or whatever matches the ester that's already there negative charges in the Alpha position so if we take that apart gonna have to be really careful about counting carbons with these guys and I'm gonna I'm going to count these carbons as soon as I draw this just to make sure because remember I made a mistake the other day doing this same similar thing so let's number that 1 2 3 4 5 so this is 1 2 3 4 5 so that looks good so if I'm going to make that compound on this would be if we'd have to be an intramolecular reaction we know that 5 & 6 member rings form really easily in an intramolecular reaction this strategy uses an intramolecular claisen condensation also known as aka a deacon condensation so in this carpet no chemistry we do have a lot of name reactions so the dieckmann condensation is an intramolecular claisen so what our synthesis would look like we would not need a directed claisen at all this is going to work really well so let's write out our synthesis in the forward direction one two three four five yep so that matches Naoe th o ET ethanol is just the solvent and and what what you should do is when you get home prove this to yourself go through the arrow pushing and make sure that this is what you get remember formation of this deprotonated active methylene compound is what drives the equilibrium and so then in the second step you add acid to protonate questions on that claisen condensation example anybody so definitely will be asked to take apart claisen product to show how to make it just like taking a part in an aldol to show how to make it I you should expect both types of problems on this upcoming midterm all right well a unified look at claisen look at not claisen but any condensation here we're not going to get too caught up in the names we've seen examples of conversations using aldehydes and aldehydes aldehydes and ketones esters and esters festers and al dyes or ketones and probably we should add ketones and ketones here I can't think I left out ketones and ketones so a large variety of reactions are possible

between enol eight ions and compounds containing carbonyl groups how'd you decide the type of action that will occur you can make this decision by considering four questions question number one is there an enol eyes double hydrogen atom on one or both of the reactants include hydrogen's output to carbonyl nitro nitrile or cyano groups I guess that would the nitrile would be the cyano group I just realized that's a little bit redundant here but we're talking about here that hydrogen a pKa about 10 so alpha 2 in activating group alpha 2 ketone PKA about 20 we can also remove alpha the Alpha hydrogen from an ester that is pKa 25 all right so we should consider those and so you want to keep in mind on these PKS so that if you have a mixture of these things you'll know which is the one that's going to be deprotonated okay so will cross out or cyano groups because that's what we're the nitrile is what we're talking about there for that one all right so that's question number one look for you know lights that conform if you are multi

multiple enolates that conform you you're going to probably have a problem with the multiple products is there a carbonyl that can be attacked by the enolate and this is where you need to keep in mind reactivity here so if you think about these guys here you want to know which is the most reactive carbonyl so that you'll be able to make good decisions about the products that you will get so this was material that we covered on midterm 1 and you know the things that you're going to have to consider when you're doing midterm 2 all right so is there a carbonyl that can be attacked by the enolate so you look for that also

hopefully you have if you have a crossed claisen or a crossed aldol you have one enol 8 that can form and then you have the other compound more electrophilic that means you're going to get a good outcome it's the carbonyl in the same molecule as the enol 8 5 or 6 membered rings form

easily if they're not if you can't make a 5 or 6 membered ring we don't want to consider that for this class so for not good 7 not good 8 not good 9 not good 5 and 6 is what we're looking for is the carbon attached to a good leaving group so this is a really is the carbonyl a type 1 or a type 2 answering that question is going to be help you to predict whether we get it claisen or a claisen like condensation or an aldol or an aldol like condensation alright so at the leaving group is present if it is a type 2 carbonyl that you're attacking then you're gonna lose the leaving group so once again we get to the point where we're gonna do the attack we're gonna draw the product we're gonna look at the tetrahedral intermediate and see if it has a leaving group so all those skills that we've been working all quarter all pay off and come back into play in this chapter here so we're attacking the carbonyl that is a type 1 carbonyl so we do have a leaving group and so what that means is that leaving group is going to leave and if you're attacking a type 2 carbonyl that has a leaving group that's going to be a claisen or a claisen like condensation so of course there's our leaving group alright so electrons on oxygen are going to come down kick off the leaving group and then don't forget the last step of this mechanism this is going to be our claisen or a claisen like this is this one here is pure claisen remember the last step we're going to deprotonate that active methylene position that we just created and that's going to drive the equilibrium to the right have to have that key step in order to drive the equilibrium all right so again key point here reaction driven to completion by formation of the stabilized enolate all right so that's what happens if you attack a type type to carve anneal so this is a type 2 if you tak a type 1 then you're not going to get placing because in the claisen we need to kick off a leaving group in the al dog you don't kick off a leaving group we're what we're doing in the AL doll is protonating and then possibly eliminating making that decision so here's our enolate this time we're going to tech an aldehyde we draw our tetrahedral intermediate then we make decisions about what's going to happen next all right so we don't have a leaving group the only leaving group where you have is for the reaction to work reverse itself and do a retro aldol we're looking at those in discussion this week that's but as far as a leaving group that's going to be a productive pathway that's not going to go right back to starting materials we don't have a leaving group so so no leaving group and so what we're going to do is we're going to do protonation and then depending on the conditions potentially elimination so whether you whether you eliminate is going to depend on the structure of the compound whether you're using heat or not sometimes you don't even need heat and sometimes you can't isolate this beta hydroxy if you're condensing two ketones you can't isolate this except in a very very small amount one one two percent something like that so and I'm missing my oxygen here I've dropped my oxygen so I'm going to go back and throw that on there to drive this equilibrium we dehydrate so here reaction is driven to completion by dehydration to form the alpha beta unsaturated carbonyl all right so hopefully this summary on gives you enough skills that you're going to be able to predict any conversation products that I might put on midterm two and/or the final yes it's

call it's gonna be all conjugated so when you draw the resonance structures here this is highly resonance stabilized yeah so this is not these yeah they're not resonance stabilized before you deprotonate those two carbonyls are isolated here they're actually connected by deprotonating okay all right yes did everybody here's question how do you decide whether you're going to go all the way to the alpha beta unsaturated one of two conditions are going to look for heat when you do the reaction or if the double bond that you make is going to be conjugated with something else on the other side so this double bond if this is like a benzene ring or some sort of aromatic ring or extended conjugation then you will get this without even heating it it will happen at room temperature all right more questions we have a few more named reactions to cover so I guess all these guys that we're doing this chemistry wanted to have reactions named after themselves so the next we actually want to talk about is the michael reaction submit the

the micro reaction is a dissonant of an enolate to the double bond of an alpha beta unsaturated carbonyl this reaction works especially well when the enolate is to stabilized enol a form from a beta dicarbonyl compound so let's see what this looks like first let me draw you the product that's the product and what we did what we're doing only just show you what we're doing and then we'll do then we'll draw the mechanism we're deprotonating this guy right here which is a completely deprotonated by sodium ethoxide and this is doing one for addition okay so that's where this is coming in the is there's the alpha beta here's alpha here's beta this is attacking the beta position so it's going to do 1 4 addition rather than 1 2 addition and so let's look at what the mechanism looks like another possible mechanism gotta be hard to decide what Connecticut isn't to put on the test huh I know you have no sympathy for me but it is difficult you know what they found is the more choices you have the more stressful your life is so maybe that's why I get so stressed out writing tests in a different way than you guys I'm we could push we could push these electrons onto carbonyl I'm gonna just push it onto carbon here just for fun we're gonna remove that acidic hydrogen and I'm gonna actually flip this around in a more logical way so I can show the attack a little better I should have I should have drawn it the other way we're going to go this way here and now we're gonna have electrons come and do one for addition we already know one for addition so really none of these steps are new here nucleophile is going to attack in one four position we're going to move these electrons over and up on to oxygen okay and then scroll down a little bit here this is a reaction where you're going to have to be very very careful to count carbons just going to tell you that right now very easy to make a mistake so you notice that once we attack that when we do a 1/4 addition we attack the carbonyl we're going to get an enol 8 ions so now we just need to protonate the enolate ion all right questions on that mechanism anybody she's making a stabilizing wait you see well you know I just attacking one for we've already seen that reaction a couple times so the electrophile is accepting the electron pair from the nucleophile so let's label this is donating this is the nucleophile but this guy Michael did was not was not satisfied to have the reaction named after him he wanted to react he wanted to name the reactions after him also so rather than nucleophile this is the Michael donor okay and this is the electrophile which is really correctly called the Michael acceptor all right so the attacking nucleophile donates electron pair of electrons it's a Michael acceptor and so and this is our stabilized enolate this reaction works really well with stabilizing lights and this right here is 1/4 or a conjugate addition one for addition or conjugate addition is also called that

of enolate - an alpha beta unsaturated carbonyl and that reaction is the michael alright so the product is a 1 5 dicarbonyl so this is another structural feature that you want to look for where do we mean by 1 5 dicarbonyl if we count from this carbonyl this is 1 2 3 4 5 if we counted to this carbonyl it would also be 5 1 2 3 4 5 yes excuse me I wrote one for why did I do that 1 5 so when you're drawing these you're gonna have to be on the lookout for counting carb counting carbons here the one way to do this so you don't make a mistake in the Michael is if you put this on a zigzag the two carbonyls are gonna be up okay if you have one going down you know you've made a mistake the most common mistake is to lose a carbon here and have a one for dicarbonyl like I just drew and I must have been anticipating yeah thinking that is it reversible it doesn't tend to reverse but let's see it is technically reversible yes so we could worry about reversible arrows but we're not going to worry about reversible arrows alright so some of them are common Michael donors and acceptors are on the next page let's take a look at them so best stabilize Deena lights here are some common Michael acceptors alpha beta unsaturated all sorts of compounds including nitriles conjugated nitriles nitro ethylene so where the Nitro group is conjugated and you know what if you see if the nitro looks a little strange to you if you draw it out you can see how this is in the same family here okay so here we have this night and the nitrogen oxygen bond which behaves kind of like a carbonyl in some ways if you think of it that way and seen with the nitrile so all of these are Michael acceptors no notice we have no alpha beta unsaturated carbonyl ik acid why would that be what would happen if you mix the Michael donor with a alpha beta unsaturated carbon carboxylic acid what would it do first they would you protonate the acid okay so we don't have that that doesn't work so I'm not a Michael acceptor why Michael doner we'll just deprotonate the carboxylic acid would app but what happens if our Michael donor is not a stabilized anyway what's gonna happen what's a competing reaction why do we want to use stabilizing weights rather than unstabilized Andales just regular enolates anybody know why that is well wherever we always have that competition between the 1/2 and the 1/4 right if we if we have the stronger that nucleophile gets the more it's going to attack 1/2 right the stronger the nucleophile the more it attacks 1/2 so we're talking about if we just have the enol 8 of a ketone here versus this one mutilated the ketones ketones about PKA 20 that's a pretty strong that's a pretty strong nucleophile and so you're gonna get a lot more attacking 1/2 if you use a stabilizing lay you pretty much get exclusive 1/4 attacks so that's why we want to use stabilized anyways in the micro reaction stronger the based the the more one to attack and let me let me look at your notes and I can show you what page I'm talking about it helps to be able to go look at the page so this would be the end of chapter 20 page 31 oh wait no page 32 so all the strong nucleophiles attack 1 2 and so the stronger we do this I messed up your page hopefully you can find anything ok all right stronger the nucleophile the more one to attack so that's why we want to use stabilize the product of the micro reaction may be treated like any other substituted malonic ester hydrolysis of the ester and

decarboxylation leads to a gamma keto acid or a 1 5 dicarbonyl what would that look like let's draw that Koh water is going to do saponification of the ester so we get the die carboxylate and then if we protonate h3o plus and heat and decarboxylate we can get a gamma gamma keto gamma keto i'm acid let's draw that for you and we're going to also always double check whenever we do the michael to make sure that we have the right number of carbons okay so we're going to count here from one carbonyl to the next doesn't matter which way we start 1 2 3 4 5 1 5 dicarbonyl and you can see where we get the gamma keto acid alpha beta this is gamma o that's Delta sorry Alpha Beta Gamma Delta Delta keto acid I've been saying that wrong the whole time alright questions on Michael reaction more questions anybody alright Michael addition and synthesis retrosynthetic analysis so I might say on the test show me how you would make this compound using a michael reaction so we have steps to take it apart but since it's michael and it's a 1 5 dicarbonyl we can start counting from either carbonyl we're gonna do both and we're gonna pick the best synthesis so if I said provided synthesis of following compound using a michael reaction using the route that would give the best yield that has been a previous test question will it be on midterm 2 i don't know something like it maybe I'm not sure but could be so break beta-gamma bond to see do you see how we could start counting from either one and we're gonna do both and then we're gonna make decisions about what we can do here all right so we'll start counting the undestroyed here starting from the second key carbonyl alpha beta gamma and then down here we'll start from the other one alpha beta gamma all right so here's the steps break the beta gamma bond put a double bond in the alpha beta position you're gonna put a double bond here put a negative charge in the gamma position yeah and then we're gonna look at it and see if these are actually gonna work or how we're gonna make them work okay so once we do that let me just turn make this a solid line I think that looks a little better just make that a solid line let's take that apart so that would be that just because you can take it apart doesn't mean it's something that's going to work so we're gonna take a look I want to draw both of these and then we'll take a look and see which we think is going to work so that would be that one if we did this strategy it would be this plus and if you are a person who tends to lose carbons I recommend putting carbons at the ends of these PI systems rather than using skeletal so something like that will make you less likely to do that so let's just throw a little ch2 on here all right and we'll call the first strategy strategy a and we'll call the second strategy strategy B let's see if we can make a work first things first this looks like a perfectly decent Michael acceptor uh can we make this compound can we make the enolate of a carboxylic acid no that's an impossible compound okay so even though we took it apart we can't use that however we couldn't use a synthetic equivalent to that so impossible I don't know where I'm pointing there impossible compound

so we have a synthetic equivalent something we can use instead that we can manipulate and turn into essentially that it's going to work just like that but it's actually something we can make not an impossible compound all right so if we started with this instead that would be a stabilized enolate that's going to add really nicely to here and then after we do the one for addition we can hydrolyze the esters of decarboxylate and we have essentially added on this piece okay so this is what we call a synthetic equivalent when you label that so if you have something that's impossible to make you want to try something to make something that's that's gonna behave similarly that would you won't you won't have to make something that's impossible to me all right what is the what do we see a problem with be strategy be what's the problem with B I'm seeing a problem with this what do you think did we say in the previous page that you couldn't have an alpha beta unsaturated carbonyl Cassatt in the micro reaction yeah so this is just going to deprotonate the carboxylic acids rather than adding 1/4 it's just going to deprotonate the carboxylic acid all right so this guy right here will deprotonate the carboxylic acid so what week one thing we could do to make them to make strategy be work is that we could convert this into an ester then we could do the one for addition and then we could hydrolyze the ester that would work but is this going to be a good michael reaction no why not that's not a stabilizing wait okay so we want the strategy that's going to work the best so this is the strategy that we're going to use instead of using this compound we're going to use a synthetic equivalent which is an active methylene stabilized enol eight we're going to do one for addition then we're going to hydrolyze both esters decarboxylate and we'll be able to get make good yield of the compound our desired compound so let's see what that would look like all right and and and the and we're going to throw everything together the sodium ethoxide the fastest reaction that can occur is for deprotonation of that active methylene position so we don't have to do this even stepwise we can just throw everything together if you want to deprotonate completely and then add the michael acceptor you can also do that both ways would give you full credit and here's we're going to have to get really careful here about too many carbons one two three four five so it should be five from both of these carbonyls one two three four five one two three four five okay so then we're going to hydrolyze and decarboxylate h3o plus h2o and heat or you can do it two-step Koh water heat and then h3o plus he is the second step and that will take you to the desired product let's draw that every step of the way we're going to make sure that we count carbons 1 2 3 4 5 1 5 dicarbonyl question why is that not say that one more time hydrolysis of an ester I always have heat if you do acid catalyzed hydrolysis of an ester or bathed base you need heat heat for both okay all right more questions that would be a tricky test question huh all right we have another

name reaction just what we wanted and that's the michael reaction followed by aldol I guess the L dog I was pretty calm bull or maybe his name was Aldo I doubt it though he didn't want a reaction named after him he should have though all right Robinson annulation is a michael reaction followed by an aldol alright and pretty remarkable the product that we're going to achieve in all-in-ones in one pot here that's the product that she'll get so let's go through the back of this and we'll do the Michael part and then we're going to do the aldol in the next page so if I I have put this mechanism on test before this would be like sort of two mechanisms in one it would be a Michael mechanism and then an outlaw of course this is going to work really nicely because we have an active methylene compound for our Michael donor and just to go back and forth here I'm going to move electrons onto oxygen this time again you have your choice oxygen or carbon what I find is that most students when they get to this chapter push electrons onto carbon because it's just a little bit easier to see products that you're forming okay so we're going to do one for addition here electrons on oxygen come down we're going to attack one four so it's going to look like that and every step of the way I'm going to check the number of carbons all right so we still have one two three four five but now we're going to protonate with water and we're gonna check one more time 1 2 3 4 5 1 5 dicarbonyl and notice we have and we're gonna throw heat here and I want to make a point about this how I'm going to clue you and knowing whether we should stop with the michael or continue with the aldol if I want to do the Michael but I want to do the Robinson annulation which is Michael followed by Aldo I'm going to show heat there so and we're going to say that you can without heat can stop here all right yes oh I did the wrong thing here thank you let's fix that I was making a very nice enol there but that's not what we want to do okay like that thank you for pointing that out all right so that's the Michael edition now let's do the aldol intramolecular aldol all right so what we're going to do is we're going to remove the civic hydrogen here the reason we remove that hydrogen is because that will give us a six membered ring do you want to go onto carbon or oxygen I heard more carbons okay so I drew that wrong now look at that it just put it right back there how about that the files getting really large and it doesn't like it when it gets really large okay that's awesome all right so we have that electrons on carbon let's draw that we're gonna attack one of the two carbonyls does not matter which I'm going to probably have to stick with one color now cuz it doesn't like to change after it gets really large okay yes let's draw the intermediate and then we'll stop right there because it's time let's just draw that so we I don't forget to draw it next time okay so that's what it looks like after that step well we will continue at that point on Friday