The Claisen Condensation

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Video in TIB AV-Portal: The Claisen Condensation

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The Claisen Condensation
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18
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27
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2015
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English

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Abstract
This is the third (and final) quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. Index of Topics: 00:36 - Synthesis with the Aldol Condensation 21:16 - The Claisen Condensation 32:32 - Crossed Claisen Condensation
Hydrocarboxylierung Dyeing Ionenbindung Oxygenierung Aldol Ketone Walking Hydroxyl Carbon (fiber) Aldehyde Chemical reaction Alcohol Alpha particle Molecule Gesundheitsstörung Computer animation Chemische Synthese Materials science Chemical compound Beta sheet Retrosynthetic analysis Thermoforming Condensation
Hydrocarboxylierung Setzen <Verfahrenstechnik> Separation process Oxygenierung Reaction mechanism Cyclische Verbindungen Ketone Aldol Walking Carbon (fiber) Hydroxyl Chemical reaction Atomic number Alpha particle Gesundheitsstörung Computer animation Electron Sodium hydroxide Enol Beta sheet Thermoforming Isotopenmarkierung Condensation
Stop codon Hydrocarboxylierung Ionenbindung Organische Chemie Aldol Ketone Walking Aldehyde Alpha particle Computer animation Sodium hydroxide Materials science Enol Boron Chemical structure Beta sheet Process (computing) Condensation Enamine
Ester Reaction mechanism Oxygenierung Hydroxyl Alpha particle Sodium hydroxide Oxocarbonsäuren Beta sheet Process (computing) Aldehyde Hydrocarboxylierung Ketone Aldol Walking Aldehyde Chemical reaction Alcohol Water Hydrogen Acid Gesundheitsstörung Computer animation Enol Chemical compound Saturated and unsaturated compounds Cobaltoxide Claisen-Kondensation Condensation
Ester Reaction mechanism Resonance (chemistry) Alpha particle Electron Claisen-Umlagerung Oxide Hydrocarboxylierung Setzen <Verfahrenstechnik> Aldol Carbon (fiber) Aldehyde Protonation Chemical reaction Aldol reaction Water Hydrogen Acid Gesundheitsstörung Computer animation Functional group Enol Cobaltoxide Base (chemistry) Condensation
Ylide Decomposition Oxygenierung Schwefelblüte Alpha particle Ethanol Fire Benzene Condensation reaction Saponification Aldol Chemical reaction Electronic cigarette Systemic therapy Yield (engineering) Azo coupling Enol Chemical compound Claisen-Kondensation Thermoforming Condensation Isotopenmarkierung Ionenbindung Ester Reaction mechanism Activity (UML) Nitrogen fixation Phenyl group Carboxylate Plant breeding Hydroxyl Wursthülle Methylgruppe Reactivity (chemistry) Clay Sodium hydroxide Sand Hydrolysat Beta sheet Mixture Claisen-Umlagerung Process (computing) Aldehyde Hydrocarboxylierung Setzen <Verfahrenstechnik> Alkane Ketone Aldehyde Protonation Water Hydrogen Combine harvester Sodium Computer animation Functional group Covalent bond Chemical structure Base (chemistry)
Ethylgruppe Ester Reaction mechanism Chloride Hydroxyl Carbonate Chemistry Alpha particle Electron Process (computing) Hydrocarboxylierung Setzen <Verfahrenstechnik> Ketone Aldol Chlorameisensäureethylester Walking Aldehyde Protonation Hydrogen Acid Computer animation Functional group Enol Cobaltoxide Thermoforming
all right what do we talk about last time we talked about Elle ball condensation we talked about self condensation where you take an aldehyde it condenses with itself we also talked
about crossed aldol and you got to be really careful when you do crossed aldol you get multiple products so we talked about ways to direct across the alcohol so that you get one project as your one product is your major product and now we're going to talk about synthesis with aldol so if I give you a compound or you recognize I'm a structural feature how do you take that apart to see how you would make it using an aldol condensation so we're going to do a little bit of a retrosynthesis but first of all I want to make sure that you on you have on your radar the two structural features which would suggest an aldol condensation in order to synthesize those and that would be the two things I'm talking about our beta hydroxy carbonyl or alpha beta unsaturated carbonyl if you see that structural feature you should say to yourself I can make that using an aldol condensation and what we're gonna do is we're going to take it apart show the starting reagents that would be used to make that by an aldol and then decide if it's going to work like we want or if we have to use one of the special techniques that we use for crossed alcohol to give us a one exclusively one product all right so that's what's something you're going to be asked to do on midterm two so key structural feature here remember this is alpha this is beta this is alpha this is beta this is a beta hydroxy carbonyl and this is an alpha beta unsaturated carbonyl all right so what we want to do is we want to have a systematic way to take these apart to see what our two reagents would be to make this so what we're going to do is we're going to number what we're going to mark alpha beta we're going to break the alpha beta bond we're going to put a carbonyl in the beta position and the negative charge in the alpha position so these would be the steps we'll call this one we'll call this two and we'll call this three so we've already marked alpha beta what we're going to do is we're going to break the alpha beta bonds so I'm just going to put a line here to cut that bond and break that bond right there we're going to put a carbonyl in the beta position all right so that means this would be a carbonyl this would be a carbonyl and then negative charge in the Alpha position so we're going to put a negative charge here and a negative charge here and when you do that what you end up is with these two starting materials notice the same starting materials here because remember we can sometimes get a beta hydroxy or sometimes we can go all the way to alpha beta unsaturated and so here's our negative charge here in the Alpha position all right so if you follow those steps that will allow you to take it apart to see what you would how you would make it so now and now so we're going to do an example here determine whether the following alpha beta unsaturated ketone can be prepared by an aldol condensation is this feasible will you get only one product so there will be a problem on the test like this and usually the way I say it is show how you would make the following compound by an aldol condensation and then provide the conditions for this reaction so you would have to if it needs directed if you need to make it enamine first you would do that okay so that's I'm going to ask for both of those things so let's follow the steps here break the alpha beta bond so here's alpha here's beta put a carbonyl in the beta position and negative charge in the Alpha position so on this can be made from and notice this is a this would be made by an intramolecular aldol and let's certainly count carbons to make sure we didn't make a mistake we have one two three four five one two three four five all right and so that's what we would that's we would do a self condensation and in the beta position here we have a negative charge and so of course this would come from the corresponding die ketone and especially in the reactions in this chapter you want to always make sure that you're counting carbons okay so because it's easy to add or lose 1 1 2 3 4 5 6 7 1 2 3 4 5 6 7 and you want to check yourself when you do that so here are both ketones in the same molecule so that's how we take it apart now we have to see it with if we take that compound and we condense it using an elbow condensation are we actually going to get that product remember from 51b that five and six membered rings form
very easily intramolecularly okay so we're looking for a five or six member ring let's look at all the possible analytes and see if this is going to
work because we have more than one possible enolate here so let's draw this so when we're forming that so let's look at all the possible analytes over here and see what's going to happen so we have since it's symmetrical we only have to really look at one side we can start we can look at this ketone or we can look at this ketone let's look at this ketone if we form the anally here so that this carbon will be negatively charged and it cycle eise's on that carbonyl how many how many atoms do we have in the ring so we'll count one two three four five six would make a six membered ring that's going to be favorable five or six but you know of course we could also form the enolate right here because under these conditions this would be sodium hydroxide conditions for aldol condensation we could form the enolate on either side what if we form the enolate here and it's cyclized that would be how many membered ring one two three four membered ring that's not going to happen so we will get mostly one product here just by doing an aldol condensation here so let's label those two positions if enolate forms here and adds to the ketone we'll get a six membered ring which is the one we want if it forms on the other side if the EDA late forms here we'll get a format we'll get a four membered ring which is certainly not favored all right so that's actually going to be an aldol that's going to work very very well for us okay let's look at another example here where we're asked to predict the major product that's on the next page all right so we've got three possible enolates let's let's go through this example and see what we would get we can warm the enolate here let's call that a we can form the enolate here let's call that B or we could form the enolate here let's call that C and what we're going to do is we're going to look at the enolate forming on each of those see let's see what ring it could cyclize to and then we're going to find out what the major product is going to be for this reaction okay so let's do a first and I'm not going to draw the arrows for that I'm going to I'm just already formed the enol latency where it's going to go and I'm going to leave the electrons on carbon because I think that's a little bit easier in this chapter anyway all right so if that cyclize is on this ketone after several steps and this you'll want to go home and prove to yourself by going through the steps of the mechanism all right so that's if we use that one let's do B now okay so let's see if the anal it forms at the V position and it attacks the carbonyl again you want to prove all of these products that I've drawn to yourself of course I'm just drawing the beta hydroxy we could heat that up and we could get to go all the way to the alpha beta unsaturated but right now we're just seeing what kind of what's going to
be favored here so that would be the product if we form the enolate at position B all right and so that's actually going to be a ring that's larger than six and then which is not going to be favored so an eight membered ring I know you guys hate drawing eight members rings so it's just like a stop sign no big deal so now let's analyze each of these products all right so let's go back to the first one and students analyze what we have so we're forming the enolate in the position a we're attacking a ketone so ketone is attacked here we're forming the enolate in the B position and we're attacking an aldehyde aldehyde is attacked what's more favorable a ketone being attacked or an aldehyde being attacked aldehyde being attacked okay so ketone attack plus a six membered ring this is aldehyde attacked plus a six membered ring and so aldehydes better so this one's going to be better so ketone attacked would is not going to be favored at equilibrium especially when we have an aldehyde that can be attacked not favorite at equilibrium and then certainly so down here what do we have we have aldehyde attacks that's favored but we're forming an eight membered ring and that's not favored so also not favored so what is the major product we're going to get B B's the major product yeah and if we if we add heat then we would form the condensation product and that's going to drive it all the way to this product right here that will be the major all right questions on that example anybody a lot of thought process going into that that's not going to be something where you can just memorize the answer and that's really the key with the the problems in this particular chapter not going to be able to just memorize answers okay so there is going to be some thought process which we've been building up for well this is our seventh week right we've been building up all of that knowledge and now it all comes into play in this chapter all right I see some I hear something a little bit of like oh no okay this is a classic midterm example you will see one like this only of course I will change the structure so I want what I want you to know what you're going to need to be able to do show the two Organic starting materials that would be used to synthesize the following ketone by across aldol so we're going to just take it apart see what we get next we have two actually this is where we're kind of asking yourself how are we going to actually make this happen do we just throw the two things together with sodium hydroxide like a typical aldol or do we have to do some sort of directed aldol do we have to make an enamine that's what we want to that's what we want to do alright so the very first thing we're going to do is we're going to take this apart using the steps we talked about on the previous page start counting from the carbonyl alpha beta we're going to break the alpha beta bond we're going to put a carbonyl in the beta position and we're going to put the negative charge in the Alpha position and then we're going to draw those two starting materials so be that Plus this
so I'm going to get charged in the a position we can see once the emulator really ten eight and now we're going to
analyze these two and see if this is actually going to work for us in order not to get multiple products we need just one utilizable higher do we have just one and Eliza Byard no on this one
is there's really only one because it's symmetrical so either side would give us the same one but we have a Nina Liza bhai drogyn on the other compound too so here's an enol izybelle hydrogen and we have only one enolate for this ketone so if we just do the classic aldol conditions which is throw these guys together with sodium hydroxide we are not going to get just one major product once we form the enol a whether what is the most likely carbonyl that's going to be attacked what's more electrophilic aldehydes more electrophilic so but we was but since we have two different inlays we're going to get two different products we don't want to get two different products we want to just get one product so when you provide your conditions for the crossed alcohol you want to provide conditions that are going to give you one major product so what we're going to do here is do a directed aldol so rather than throwing both of these things together with sodium hydroxide we're going to form the enolate completely with the compound that we want to form the enolate from then we're going to add the other compound so you're going to have to make that decision do I need a directed aldol or do I need to just use this classic aldol where you just throw everything together so I'll da th f1 is 78 degrees that's going to give us the enolate in order to be able to do this you're going to need some practice so you definitely want to do as much practice as you can here we have of course gotten to the point where you just you can't memorize the answers at this point you're going to have to think your way through them then we add the aldehyde we need to we need to protonate that oxygen what do we want to protonate that with we think do we want to use water or acid what happens if we use acid it's going to go to the alpha beta unsaturated right so the acid catalyzed aldol if we add acid this is going to want it that hydroxyl is going to want to come off we're going to want to go to the alpha beta unsaturated so we definitely want to use water not acid okay h2o not h3o plus because if we use h3o plus you're going to get the alpha beta unsaturated so if you are forming an alpha beta unsaturated compound then then you definitely want to use acid in your second step so once again you're going to have to think for these problems questions on aldol anybody so we'll have some examples next week and discussion that we'll work through so you have a chance to go through that thought process no questions we're going to start placing now so a bunch of named reactions in this chapter so claisen condensation enol a plus an ester and the class equation is when you take two esters and you they condense with each other so notice the two here it's just going to be a self condensation let's draw the product and then let's go through the mechanism claisen condensation has appeared on many of my midterm twos for 51c so that's just just throwing that out there this is a two-step process product is a13 dicarbonyl also it bores be more specific a beta keto ester and we saw beta key to esters in chapter 23 didn't we we use those we alkylated then we hydrolyzed and decarboxylated okay so that's one of the things that you this is one way to do that so notice it's completely different than an aldehyde aldehyde condensation you get a beta hydroxy ketone or aldehyde or alpha beta unsaturated here you get a 1:3 dicarbonyl so so 1:3 dicarbonyl needs to
be on your radar so if you see it you can say to yourself I might be able to make that by placing condensation okay so that's the structural feature you want to look for alright let's look at the mechanism here now notice the claisen is it's only acid it's only bass college we're not going to have an acid catalyzed claisen so that makes life a little easier for you we're going to make the enolate and again we you have a choice you can move the electrons onto oxygen or you can keep them on carbon I'm going to try to go back and forth here so this time I'm going to move them onto oxygen then I knows we have reversible arrows here is the thermodynamic conditions let's try that again okay so we make the enolate out of one of the esters and it's going to add to one of the other esters that has not been deprotonated yet okay so remember the key thing here is once we do protonate that alpha position now we turn this into a nucleophile so that's a nucleophile and so now it's going to attack an electrophile here alright so electrons on oxygen come down and we attack the carbonyl kick electrons up onto oxygen we're going to get a tetrahedral intermediate so take a look at that tetrahedral intermediate and tell me what's going to happen do we have a leading group so tetrahedral intermediate has leading group so if you if you've if you've internalized everything we've learned so far this quarter this is going to be this going to be easy for you to make these decisions we're going to have the electrons come down and kick off with oxide alright we didn't have a leaving group in the aldol did we so that's why we got a different product so I'm compared with the tetrahedral intermediate that we had in the aldol and because we're attacking an aldehyde we don't have a leaving group right because that's a type 1 carbonyl so compare tetrahedral needed for them in the aldol reaction now we have a leaving group so this goes all the way back to chapter 20 alright so that leaving group is going to leave then what happens okay so what do we know about we just kicked off a Sox I died on what do we know about what's going to happen next can we do can't we just isolate that why do we have to add acid we already formed that why can't we just isolate that hmm it's you can't not eat what what's going to happen is and we saw we saw we went through the thermodynamics of this I mean we went to the acid base whatever this proton is going to be removed from of that ethoxide that you just kicked off and that's actually what drives the equilibrium in this reaction so in the aldol we drive the gluon by losing water this actually drives the equilibrium for this reaction so number one you can't stop this from happening because you're forming this in the presence of base you're forming this in presence of base that is strong enough to completely eliminate to completely remove that alpha hydrogen and on this drives the equilibrium because you're creating a very stable intermediate this is highly resonance stabilized so we can draw could push electrons onto this oxygen so now going to draw a resonance structures here we could also go to the other side we can go all the way over here we can move electrons all the way up onto this oxygen
so super important that you draw this formation of this drives the equilibrium if you cannot form this deprotonated active methylene compound if you cannot deprotonate that position maybe there's alkyl groups there you can't do it Clayton you have to be able to form this or you can't do a claisen and so that's what it's talking about underneath here reaction of EE clearing process and thus reversal because of this the claisen condensation does not give a good yield of the beta key to ester unless the product is converted to an enolate in the basic reaction can mixture so this is key right here the inter need is simply going to goes the reverse reaction and decomposes to give the starting enolate and ester if there are no alpha hydrogen's to deprotonate then the reaction won't go and so that's right here so let's say we try to do a let's say we try to do a claisen with this compound here this one only has one alpha hydrogen you will you will form this but i'm putting it in brackets because it's not stable notice there are no alpha hydrogen deprotonate therefore you-you-you can only isolate very small amounts or none at all so when I say very small amounts I mean 1% 2% maybe and so therefore you can't synthesize this compound by claisen condensation Wilson you can't synthesize this compound and I always want to put two S's on that claisen so we'll fix that it looks a little bit when I do that it looks a little bit like Clawson you know like Clausen pickles always want to put two s is there for some reason okay so can't synthesize that calm how could you synthesize that compound but not using a claisen what could you do how would you how would you do that how about making this without the methyls here and alkylating twice like we did at the end of chapter 23 right so that's why you could do it you make this without the methyls and you could make this without the methyls by a claisen and then you can alkylate twice and that's how you would make it you just can't go directly from this to that not going to happen all right questions on that that's a kind of a key point here questions on claisen condensation mechanism all right so let's look at cost crossed claisen all right and so like the Aldo we're going to have to choose our reagents very carefully if we want to get just one product when we are doing across claisen so the mechanism I showed you that wasn't a class cross claisen but if we want to do a cross claisen we got to choose reagents really carefully and with and so that means that we have to know a little bit about reactivity which means that all the work we've been doing for the last six weeks this is our seventh week all pays off now if you crammed for midterm one it's all that's gone and you're going to have to go back and remind yourself so you know when I put all those reactions that you know this structure reactivity problems at the beginning of the exam break the rank the electrophilicity of these all of that's necessary to be able to do chapter 24 so those of you who have kept up to speed this is going to it's all going to pay off right now here here's your reward okay so a couple ways to do this choose reagents so that one of the participants is especially reactive or has no alpha hydrogen's so only one enol a can form doesn't that sound familiar that's what we were doing with the aldol on friday okay so let's look at the combination here only one ela can form so we're going to form the enolate here the other compound has no analyzable hydrogens so so no enolizable hydrogens so already we're off to a good start now once we've formed the enolate on the first compound which which carbonyl is that more likely to attack aldehyde right aldehydes are more electrophilic that's 1/2 1/2 aldehyde for the other ones 1/2 ketone the esters parts the same and so this is going to be this compound here once we form the enolate on the other compound this one is especially reactive because it's a half aldehyde so that means that this is going to be a good reaction you're going to mix these two esters together you're going to get predominantly one product because we're only forming one inna late and that enolate is selectively going to attack the most electrophilic carbonyl which is on the other one alright so that means this is your product and you definitely want to go back and prove this to yourself so sort of the first thing with these condensation is being able to draw the correct product so that's what you're going to get so prove to yourself
so cover-up the answer and see if you could get it see if you get the same answer all right so that's one way we might not always have that luxury though so let's let's see we have some other ones that aren't we're not going to be able to do that with so a second way alibis or ketones can also be condensed with esters if addition to an ester is desired the ester should be especially reactive all right so let's analyze the two of these this is going to be a condensation reaction and it looks like we only have one analyzable hydrogen right here so that means we're going to form the enolate on the ester the aldehyde has no analyzable hydrogens all right so once we form the enolate from the ester what is it more likely to attack a ylides right aldehydes are more electrophilic than ester so that was not wasted all that time you spent learning the relative reactivity again it's all paying off here so no analyzable hydrogens and it's also the most electrophilic so if you're going to form the enolate exclusively for one and it's going to attack exclusively the other then you've got a good placing so let's actually go through I'm not going to do the whole mechanism what we're just going to show how you draw the product so I'm going to show the enolate already formed and I'm going to show it attacking now this is this is really going to be kind of a mixed it's not really pure clay sand in a pier claisen you attack a an ester and we're protecting an aldehyde so this is not a pure claisen but is you can still do condensations when you mix the two of these you just have to know what you're doing so does our there's our tetrahedral intermediate have a leaving group the only leaving group it has this to reverse itself and it can reverse itself but as far as leaving group productive leaving groups that are not going back to starting material we don't have and so this is going to be more aldol like because we're attacking a type 1 carbonyl so no leaving group so we're not going to get too caught up in the is this a pure claisen is this not a pure claisen isn't an aldol we're not going to get we're too worried about that and so what's going to happen is we're going to protonate like we did with an aldol whoops so we're not going to get a carbonyl like I just started to draw here we're going to get this and it turns out that we can't really stop at this position here at this one it's going to keep going all away and lose water even without heat I'm going to draw out the phenyl ring so you can see why why does that lose water without heat we sell them like that already the double bond that you farm when you eliminate water is conjugated with the benzene ring so we have this highly conjugated system we have the carbonyl double bond single bond double bond single bond double bond single all the way around here okay so you can't isolate the beta hydroxy so you have to know that that goes all the way to the alpha beta unsaturated carbonyl questions on that one yes I did I'm not showing the whole mechanism but yeah you're in you you in this case we would protonate with ethanol rather than water okay yeah we definitely you know we definitely want to use if there is an ester involved you want to choose your base to match the ester what if I use sodium hydroxide here and water instead what would happen because this is actually more like an L dot rather than a claisen what would happen if I added sodium hydroxide to that what's going to happen what what chapter 22 reactions going to happen saponification of the ester you won't get condensation you're going to get hydrolysis of the ester to a carboxylate and then that's going to be the end of it so so in this case we're going to protonate with ethanol rather than water all right you want me to show that yeah let's do that let's show that even though this isn't a complete mechanism we're in a protonate here with ethanol rather than water there was another question over that side no yes you can you can yeah if you don't if you didn't have a benzene ring if this was say a methyl you could stop here but remember you can't get all of this remember the when you attack an aldehyde it was like 55% of this so really to get to get it to all go to one thing you want to drive all the way to this product okay alright so beta beta so the other thing we want to do let's get some more examples so you will see something like this on the midterm and when you see some sort of base like this sodium ethoxide ethanol or sodium hydroxide water and heat you know that you're going to know it's a condensation reaction okay so let's see what's going to happen here do we have any analyzable hydrogen's here no we have analyzable hydrogen here fortunately this is symmetrical so there's only one alight we can form from that compound so let's label that one either side I'm going to go here in elate will form here and it turns out if you're comparing reactivity this is a is a formic ester and the forming ester is actually more
electrophilic so if this has no analyzable hydrogens and it's especially reactive because it's a half aldehyde so it turns out that the ketone is more sterically hindered so you're going to favor forming the attacking the formic Esther so ketone more hindered than formic Esther all right so let's go through the thought process here rather than just drawing the product let's go through the thought process this is not the complete mechanism so there's our enolate you can go on either side here both sides are identical and then we're going to attack the form of guster so because we're attacking a type 2 carbonyl we're going to have a leaving group so let's push electrons up onto oxygen here let's take a look at our tetrahedral intermediate and as you can see here's our tetrahedral intermediate is it has a leaving group so this is going to be more claisen like so that means the electrons on oxygen are going to come down and rather than a beta hydroxy we're going to get actually a carbonyl in the beta position and now that ethoxide that we just kicked off is going to come and remove this active methylene proton and that drives the equilibrium questions on that example anybody all right so they're the previous one was an aldol like this one was a clays in like we're not going to get too caught up in the names we're going to be able to put we are going to be able to figure out what the product is by I'm using all the things that we've learned about with carbonyl chemistry alright so third strategy here is going to be reacting enolate with ethyl chloroform eight or diethyl carbonate these guys are especially reactive both of these especially reacted there especially electrophilic and they have no alpha hydrogen's okay so we can take whatever we want we can take an ester we can take a ketone and we can condense here so we're going to form the enolate form the enolate here I'll just do it on this side here any late forms here and the ethyl chloroform a no ena Liza bhai drogyn and it's especially reactive since it's a half acid chloride okay so that means we're going to predominantly form one product let's go through the thought process that we would use for that so again this is what this wouldn't be a mechanism this is just our thought process if we're if you have for predicting the products so then we have the Clos acid chloride very electrophilic push electrons up onto oxygen and then we take a look all right so what happens now does our tetrahedral intermediate have a leaving group it has two leaving groups which leaving groups which leaving group is going to leave chlorides the best leaving group that's the best leaving group right so electrons are kind of down and take off the best leaving group can we stop there no we're going to have to deprotonate long arrow to the left short arrow to the right we're going to deprotonate that's what drives the equilibrium and so in the second step we add acid all right questions on that when anybody I'm going to have you come up because
it's 1250 so come on up we'll stop right there we'll continue this some on Wednesday
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