Carbonyl Condensation Reactions

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Carbonyl Condensation Reactions
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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: 00:20 - Malonic Ester Example 01:32 - Reaction of Enols and Enolates with Other Carbonyls 04:24 - Aldol Addition and Aldol Condensation: Base Catalyzed 17:43 - Special Points about the Aldol Condensation 27:40 - Aldol Addition and Aldol Condensation: Acid Catalyzed 37:57 - Crossed Aldol Reactions
Hydrocarboxylierung Computer animation Nitrogen fixation Aldol Carbon (fiber) Condensation reaction Claisen-Kondensation
Reaction mechanism Oxygenierung Hydroxyl Alpha particle Electron Sodium hydroxide Halogenation Kohlenstoffgruppe Beta sheet Condensation reaction Claisen-Umlagerung Hydrocarboxylierung Aldol Ketone Veresterung Carbon (fiber) Aldehyde Carbonylverbindungen Protonation Chemical reaction Aldol reaction Wine tasting descriptors Lone pair Water Hydrogen Acid Computer animation Enol Dehydration reaction Base (chemistry) Mixing (process engineering) Cobaltoxide Claisen-Kondensation
Schwermetall Resonance (chemistry) Collision Alpha particle Molecule Electron Data conversion Elimination reaction Condensation reaction Concentrate Aldol Walking Chemical reaction Aldol reaction Alcohol Hydroxide Electronic cigarette Death Food additive Lone pair Acid Materials science Enol Chemical compound Acetone Cobaltoxide Medical history Tannin Thermoforming Reaction mechanism Hydroxyl Wursthülle Pharmacokinetics Lactitol Beta sheet Exotherme Reaktion Conjugated system Motion (physics) Process (computing) Aldehyde Hydrocarboxylierung Setzen <Verfahrenstechnik> Ketone Asset Carbon (fiber) Aldehyde Water Volumetric flow rate Hydrate Hydrogen Gesundheitsstörung Computer animation Functional group Covalent bond Dehydration reaction
Sense District Left-wing politics Reaction mechanism Hydroxyl Alpha particle Molecule Electron Sodium hydroxide Acid dissociation constant Addition reaction Lactitol Colourant Beta sheet Elimination reaction Condensation reaction Aldehyde Hydrocarboxylierung Concentrate Aldol Ketone Walking Alkoxide Carbon (fiber) Aldehyde Chemical reaction Hydroxide Water Hydrate Acid Gesundheitsstörung Computer animation Functional group Enol Acetone Dehydration reaction Base (chemistry) Cobaltoxide Deuterium
Physical chemistry Chemical property Activity (UML) Benzaldehyde Hydroxyl Wursthülle Aromaticity Reactivity (chemistry) Alpha particle Molecule Beta sheet Condensation reaction Aldehyde Hydrocarboxylierung Aldol Ketone Chemical reaction Kupplungsreaktion Hydrogen Acid Computer animation Enol Chemical compound Covalent bond Acetone Dehydration reaction Base (chemistry) Separator (milk)
Activity (UML) Plant breeding Hydroxyl Chemistry Molecule Wasserstoffion Data conversion Beta sheet Hydrocarboxylierung Aldol Veresterung Carbon (fiber) Aldehyde Carbonylverbindungen Protonation Chemical reaction CHARGE syndrome Aldol reaction Water Hydrogen Sodium Acid Computer animation Enol Covalent bond Chemical compound
good afternoon we're gonna get started alright have you guys noticed that like somewhere in the middle of this class it's just like time is just flying by so fast we're starting our last carbonyl chapter and oh my gosh okay I made a mistake last time of some several people pointed out to me so let's fix that I did not count carbons so it makes a really good point I have the right number of carbons here 1 2 3 4 5 but this is 1 2 3 4 5 6 I've inadvertently added another carbon right very easy mistake to make I'd like to say I did that on purpose to see if you caught it but I didn't just like that 1 2 3 4 5 ok very easy to make that mistake so that makes a good point alright we're going to start chapter 24 we're going to learn about the aldol condensation and the claisen condensation
we'll start with the we're going to start with the aldol first but let me just introduce the idea behind this chapter so we know that needles and a delights from chapter 23 are electron rich nucleophiles they come react with a number of different electro files and so we have our carbonyl compound which is an electrophile because we have the partial positive charge on carbon if we remove the beta hydrogen at the alpha hydrogen we get this electron rich nucleophile and then when we add electrophiles that will attack the electrophile so we incorporate an electrophile into the Alpha position and so of course our electrophile czar d2o d3o + bromine BR 2 + RX so that's what we talked about in chapter 23 so in this chapter we're going to add to those lists of electrophiles and say what if the electrophile was another carbonyl compound rather than d2o d3o + bromine Rx would if we had that electron rich carbon group attack another carbonyl and that's the focus of this chapter here so if it does that you have a condensation reaction and we're going to talk about two broad categories of these reactions if you have an enol or an enolate plus an aldehyde or ketone it's so the enol or enolate of course is the nucleophile and the aldehyde or ketone is the electrophile then you have an aldol addition or an aldol condensation depending on what products you get out of the audition or of aldol condensation if you have an enolate plus an ester so enol eight of course being your nucleophile ester being your electrophile if you have an enolate plus an ester that's a claisen condensation so we'll start with aldol addition and aldol condensation and then we'll talk about claisen and then you'll see once we get through these sections we're going to kind of mix it up because you can have you can mix the two of these up and get you and get unique products alright so we'll start with the aldol will do base catalyzed and then we're gonna do acid catalyzed your book does not have acid catalyzed but i'm going to show you how sit catalyzed all right so here's base catalyzed and here's the product that you get and then we're gonna go through the mechanism very highly likely that this mechanism will be on midterm two very highly likely you'll see an aldol very highly likely you'll see a claisen condensation on midterm two it's kind of usually how I roll with that material okay that's aldol addition and our product is a beta hydroxy carbonyl so let's let's see where we get that here's alpha position here's beta position this is a beta hydroxy carbonyl so you want beta hydroxy carbonyl to be on your radar if you see a beta hydroxy carbonyl you can make that potentially by an aldol addition and so then but what can also happen is is this can this can go further more sodium hydroxide and typically he'd not necessarily though and you could do a dehydration and if you do a dehydration you get an alpha beta unsaturated carbonyl and that's called an aldol condensation product and you'll see that sometimes we can sometimes we stop at the beta hydroxy and sometimes we go all the way but again this is our alpha beta unsaturated carbonyl so a nice way to make alpha beta unsaturated carbonyl so that's the idea I'm going to show you conceptually what's happening and then we'll go through the complete mechanism so here's the what's going on here which is not the complete mechanism if I ask you for this mechanism on the test and that what I want you see is the complete mechanism all right this is where everything we've learned in the carbonyl chapters comes together in this chapter here all right so we have electron rich nucleophile its electron rich nucleophilic carbon and then over here we've got the partial positive on the carbonyl carbon partial negative here so this is electron poor or electrophilic carbon all right so what we have here is the going to have the lone pairs coming from the electron-rich the lone pairs always come from electron rich and go-to electron poor I would attack the carbonyl we're going to kick electrons up onto oxygen so here's mixing up everything we learned in chapter 23 with everything we learned in the rest of the chapters all right and then so then you can imagine we do protonation this is looking like chapter 20 now we do protonation and then usually heat although sometimes you don't need heat as we'll see coming up usually heat you lose water so that's a dehydration reaction and that takes you all the way to the aldol condensation product alright so let's go through the complete mechanism questions so far conceptually for that reaction anybody too fast ok we'll leave that there all right complete mechanism alright so
hydroxide ion now as we've been doing throughout these chapters you can when we go when we make the enolate ion you can leave the electrons on carbon or you can push them onto oxygen I'm going to try to go back and forth here and do it both ways what I find is that most students like to leave it on carbon in chapter 24 it makes it a little easier to draw products but let's do draw it both ways and I'll show you the alternative way right underneath here so for fair coverage I like to go back and forth here so and then the alternative is to go on to oxygen I'll put that one in green so there's our nucleophile so all we're drawing here is just the resonance structure for the one above and that that's what makes it perfectly fine to do this either way so again these guys are resonance structures so it's perfectly fine to draw it that way and now so now we've made our nucleophilic carbon and now we're going to attack electrophilic carbon so that's going to mean that we're gonna attacked it the aldehyde that has not been deprotonated because once we deprotonate the aldehyde you turned into a nucleophile we're not gonna have a nucleophile attacking nucleophile so we want we want to have we want to attack carbonyl that has not been deprotonated and so electrons are gonna come from the lone pair on carbon we're going to attack the carbonyl carbon we're going to kick electrons up onto oxygen all right if you draw it the the alternative way here then this is what it will look like okay and and and again it's your choice you do not have to draw both ways but you can that's going to it's going to give you the same product so let's draw the product here notice our reversible arrows this is a reversible reaction and then we're going to protonate and we're going to protonate with water and when we protonate with water we're going to regenerate our catalyst that's aldol addition beta hydroxy carbonyl is your product if you do just the aldol addition all right so this guy right here is going to take you to the same thing so we'll just do these arrows right here and I'm just going to put this in a box as the alternative way to draw this so from here on out sometimes I'm going to push electrons onto oxygen sometimes I'm going to push them onto carbon when I make the enolate all right so let's keep going here because we want to go all the way to the dehydrated product so what I'm going to do is scribble this out right here and draw one of those hydrogen's in because it is a two-step elimination two-step elimination we make the enolate ion first and then we kick off hydroxyl there is a heavy temptation to just go directly let me show you what I'm talking about we're gonna we're gonna actually remove this acidic hydrogen right here make an enolate first we're gonna push electrons onto oxygen and then we're gonna kick off hydroxyl there is a heavy tannin temptation to just have the hydroxyl come in and remove this hydrogen push electrons here and kick off the hydroxyl in one step it has not been shown to be a one-step process so if you do it as a one-step process you will miss four points because when you're combining steps into one you miss the points for both steps so it is a two-step elimination that being said I'm a bunch of you are going to do it as one step on the test and well maybe that won't happen this time so you make the enolate first and then here I like to when I make the enolate here I like to motion push electrons onto oxygen because I just like the flow of the whole thing you'll see here make the enolate first and of course we're going to be kicking off of hydroxyl as a leaving group going to something that's really stable so as we've seen we can have hydroxyl leave if we're going to something that's really stable all right and then the electrons come down from oxygen here and then we move this over and then we kick off hydroxyl so we're going to do this in the forward direction and in the reverse direction will be doing the reverse direction in discussion next week and you can see this will be our product here all right so let's let's do it let's put a note for ourselves here important two-step elimination step one making late and when you make the EULA you can leave electrons on carbon but you do have to make the enolate first and then in step two you eliminate hydroxide questions on that mechanism anybody yes oh that is not irreversible let me fix that I don't know why I did that thank you yeah okay more questions yes yes no it's not boy I'm having history of this now I normally don't let me fix that yes that is the incorrect number of carbons okay so now I can't do that anymore for the rest of the quarter right okay so everybody fixed that I've got one too many carbons I'm doing that a lot lately all right 2-step elimination everybody have this yeah so hydroxides
Olivia group conjugation makes the double bond particularly stable this allows hydroxide to removed in a very exothermic reaction alright I want to make up some points about this algal condensation here on the next page and that is aldol reactions in equilibrium process as you saw with all my reversible arrows especially the ones I fixed in the last step and the concentration of aldol product at equilibrium depends on the substrate use with the aldehydes the aldol product is favored but not by a huge amount so 55% at equilibrium and we're gonna have some ways to tweak this a little to get the product that we do want so 55% at equilibrium is not a good ratio so that means we still have 45% that is not that has 45% of starting material with ketones starting materials are favored and so I'm here with the product and what you want to do their first learning curve for al doll is to be able to draw the correct product so you want to go back today and make sure that I'm drawing these products correctly here by going through the mechanism for both of these reactions and for this one here is only 1% at equilibrium so this kind of kind of goes along with what we talked about with hydrates and with asset Al's the more electrophilic the more hydrate the more asset al you form and in this case the more aldol product you form and so when you try to do this with a ketone the carbonyl is more hindered and less electrophilic and so that's why we see only 1% of product and so if you want to drive the equilibrium and we're gonna again talk about some other ways to do this also but if you want to drive the equilibrium then you want to go all the way to the aldol condensation product both reactions can be forced to completion by dehydration dehydration drives the alcohol eager learner to the right because performing such a stable product so all the product wants to be in that form and so basically if we take this same and acetone and we drive the equilibrium and usually by heating again we're gonna see some examples where we don't even have to heat usually you just heat that up and you're gonna form the alpha beta unsaturated carbonyl and then because this is such a stable product that's what's going to drive the equilibrium so we get a hundred percent conversion why exothermic and the answers right here exothermic the product ketone is conjugated with the double bond which makes it more stable product key tone is conjugated with the double bond conjugation it makes something more stable lower energy and that's it so it remember at equilibrium when we have an equilibrium process we always favor the most stable on compound alright and that helps us to understand a little bit what the next point I want to make about the aldol condensation and I actually got some students asking me this on Wednesday conditions for Alvo reaction are the same as for hydrate formation right exact same conditions how do we know we're not going to form hydrate what do you think well how do we know we're not going let's let's show the product for hydration let's do reversible arrows here hydration was chapter 20 so here's the hydrate and again we can form the aldol addition product or we can form the aldol condensation product in this case we would heat why do we why don't we form hydrate instead anybody it's well no not total immersion what do we know about the stability of a hydrate it's not stable right so why we're in equilibrium conditions this is unstable and in equilibrium conditions you always form the more stable product and this is much more stable at under equilibrium conditions under equilibrium conditions the most stable product is favored you absolutely will form hydrate when you're doing an aldol reaction you will form hydrate but it's a reversible process and as soon as your form it it's going to be like okay that's a dead end it's going to go back again so remember the molecules don't have brains you know they don't they don't know oh I have to go already that that's more stable they're gonna collide and if they collide the right orientation with the right energy that we react but if it's a reversible processing and it's not something that's stable it will just go right back again so they're pretty much good to try out all the possible things until they hit upon something that's more stable once they hit upon something that's more stable that's low energy they don't go back again okay so it's just we like to think of them as being smart but they're not all right and I'm supposed to write that down there but we can just write it up there anyway all right so the third point I want to make look at the P K's for enol 8-iron formation in the above reaction all right sir certainly this is not favored for me the enol with hydroxide let's look at how much we have this is where that little thing that I do about how much do we have at equilibrium is be actually very helpful really comes into play well here there's the acid-base reaction we know not we know how to know the direction of equilibrium and acid-base you look for the acid
conjugate acid-base pairs and we look up the pKa for the acid there's one acid on the right hand side and there's one acid on the left so that's approximately 15 and this one is approximately 20 so now we have an idea the reaction is favored to the left but how much does it favor to the left at equilibrium favored to the left at equilibrium we have 10 to the third to 1 no 10 to the fifth to 1 so the equilibrium only contains a small amount of the you know light yeah go ahead did I draw that wrong yes I did oh okay okay now really I'm not gonna do that anymore okay I swear okay and and here's a good example of where we actually want only a small amount of enol aid because remember and then the aldol the enolate attacks an aldehyde that hasn't formed an enolate if we completely deprotonated the aldehyde we it would have nothing to attack does that make sense if we used Lda and we completely formed the enolate then all of the aldehydes converted to the enolate now we have nothing left to attack so the fact that we're only forming a tiny amount of the enolate is good because that tiny amount of you know later we're forming that one enolate now has ten to the fifth of these two attack okay so it's got a lot of those to attack and so that's that's important in this reaction that we have a weak base here if we use two Lda it wouldn't we wouldn't get that we wouldn't get we used one equivalent of Lda we would have nothing left to attack so that's our actually very helpful in this reaction all right so that's what I say here even though the equilibrium concentration of enol eights small it still serves as a useful reactive intermediate once the enol light reacts with an electrophile it's removed from the equilibrium and then now we're going to form another one that can now attack another of aldehyde so eventually all reacts via a low concentration of the enol 8i on all right let's look at the acid catalyzed mechanism and and notice what I say here the acid catalyzed mechanism goes all the way to the dehydrated product and you can't isolate the addition product remember that okay in base the the dehydration you have to kick off hydroxyl it's not as favorable so you can sometimes stop at the beta hydroxy in acid that leaving groups protonated and it comes off so much more easily so let's go through the acid catalyzed mechanism you get the same product sigh product is water here alright so we know that in acid catalyzed mechanism enol is the nucleophile not enolate ion we don't want to be making enolate ions in an acid catalyzed mechanism so let's go through this here I have put this on exams before also the acid catalyzed reaction alright so just to slow me down a little bit when I slow down I make less mistakes I'm going to do the changing colors thing because I'm already over my limit of mistakes for the day so that slows me down a little bit and that always helps me so we're going to make an enormous so notice even though I'm stopping it starting with the acetone and this is going to go all the way to the dehydrated product and let's change this into a two and I'm going to remove this what do I want to use to deprotonate that with water not hydroxide and we don't want a hydroxide ion appearing in an acid catalyzed mechanism so why do I okay you didn't see that all right so we're gonna make the enol that's going to be our reactive intermediate so there's the you know the enol is going to attack a protonated carbonyl because that's how we do acid catalyzed mechanisms okay so I already showed how to make a protonated carbonyl I did that in this step right here so I'm just going to have my protonated carbonyl I don't need to show it twice there is protonated carbonyl in this reaction mixture so that's ready to go and then I'm gonna have the electrons on oxygen come down and we're going to attack the carbonyl so big difference with the acid catalyzed we always attack a protonated carbonyl and we always attack from the enol not the enolate ion and notice um we don't have a we don't have an alkoxide because we attacked a protonated carbonyl so that's already protonated for us we have our protonated carbonyl we are still going to have our two-step elimination well actually it's going to be three steps right we're gonna make the enol first acid catalyzed reactions always have more steps so we're going to make the enol again now what do we know about leaving groups in an acid catalyzed mechanism what's something we always do to leaving groups in an acid catalyzed mechanism protonate them first so I'm going to protonate that carbonyl and you're not going to protonate that hydroxyl make it a better leaving group so I'm in the acid catalyzed mechanism we are not kicking off water we're not kicking off
hydroxide we're kicking off water so does it look like we're ready to go to take off our leaving group electrons on oxygen come down and we kick off our leaving group so always a few extra steps in our acid catalyzed mechanisms and you can see we are one step away all we need to do is deprotonate the carbonyl now and we'll be all set with our product deprotonate with water that regenerates our acid catalyst so the kind of mistakes I see on mechanisms like this on the exam or when students mix like half acid catalyzed half base catalyzed so that would that would be the biggest mistake so in other words like deprotonating with hydroxide when you're not going to have two hydroxide appearing in an acid catalyzed mechanism so kind of mixing up the two and that's the thing you want to get straight before them in four before midterm two questions on the acid catalyzed mechanism yeah electrons on the oxygen attack the carbonyl directly Oh like you're talking about say where'd I put that so right here have water attack of the carbonyl their first step right here one step below that oh the oxygen attacked the carbonyl so this coming in attacking the carbon he'll not so much yes yes it absolutely will but that goes to a hydrate which is not stable you are going to form a lot of hydrate in this reaction you won't isolate hydrate and how do we know that how can we know that if we take if you take labeled water so Oh 18 water and you do this reaction you will have a teen in the carbonyl and the only way you can do that is if you are attacking the carbonyl remember we had the mechanism where we took acetone and then we treated it with hydroxide and water that was labeled and we got labeled acetone so if you did this reaction if you did this reaction with labeled water you will see Oh 18 in the oxygen so we know we're forming hydrate there's another reaction that we do with these same conditions did you remember what that is in chapter 23 right if we so we make the enol okay so we make the enol here and the you know just protonates back again right reverse that reaction that's also going to happen so we don't the molecules don't necessarily do a straight line like this so we can form the enol on that we could go right back to starting material how do we know that's actually happening if we use d2o with deuterium and oxygen labeled you will start seeing deuterium is incorporated into these alpha positions that's how we know all of these things are happening but again at equilibrium you're going to form the more stable product so all these things will happen but once you've formed a more stable product it stays and doesn't go backwards okay more questions that was a really good question by the way anybody all right okay so that's acid catalyzed now we want to talk about crossed aldol so this would be more of a self condensation where we take our aldehyde we add base aldehyde or ketone we add base and it condenses with itself what if we wanted to use two different aldehydes and condense them together so we're gonna have to be really careful if we try to do that here's a bad example we'll call this aldehyde a and we'll call this aldehyde B and we treat that with sodium hydroxide and h2o so you get four different products this would be enol eight of a adds to a and if I were you what I would go home and do today is I would prove to myself that these are the four possible products make sure you know how to do that because that's the first learning curve for aldol okay this
one is nina late of a adds to b and this one would be enol eight of b adds to a and enol eight of B adds to be all right so that's not something we would want to do because if we want there's if there's a specific product out of that four that we want we want to make just that product we don't want to have to separate those four compounds that would not be a very easy separation they're going to have very similar physical properties so we're gonna have a couple ways around this first way around this is choose one of the reagents so that choose their agents so that one of the participants has no alpha hydrogen so that you can only form one enolate and then you choose the other participant to be more electrophilic and then you can get it clean reaction so here's an example carefully choose your reagents all right so if you notice that we have benzaldehyde as our first reagent cannot form an e no wait so no alpha hydrogen's therefore no enolate and this is where all of that reactivity comes into play here so you'll notice here our key till we can form the enolate so we're gonna form the enolate it's on it's symmetrical so there's only one side where we can you can go on either side we'll get the same product so we don't have to worry about kinetic versus thermodynamic enolate so we're going to form the enolate here and it will add two benzaldehyde why will that selectively attack benzaldehyde which one which compounds the most electrophilic benzaldehyde so this is going to work really well when we take a two carbonyls one you form the enolate and the other one's more electrophilic it's going to be a good reaction so this is more electrophilic so we formed the you laid here and it will add two benzaldehyde them because aldehydes are more electrophilic than ketones all right and so what's the product you get and in this case you cannot isolate the alpha beta hydroxy carbonyl because the product is so stable it goes directly to this at room temperature don't have to heat so formed at room temperature can't isolate the beta hydroxy carbonyl and that's because the product is so stable notice the double bond is not only conjugated with the ketone it's also conjugated with the aromatic ring and so that's what we talked about underneath here you don't even need to heat the reaction to get to dehydration dehydration to occur once the enolate of acetone is formed it will more readily add to benzaldehyde then another molecule acetone so this is a very good example of a an aldol condensation unmixed aldol where you can get one single product questions on this example anybody so careful choice of reagents the other thing is maybe you maybe you have two compounds that can form elates but one of them is so much more acidic so all that acidity we learned about the relative acidity of carbonyls in chapter 23 that comes into play here so here we have active methylene compound way more acidic than a non active methylene compound so when we add base we're going to exclusively form the enolate on the active methylene compound we are not going to form the enolate on the regular carbonyl and so this is going to be another great reaction here is going to work well so let's go
through that most acidic hydrogen in the active methylene position which carbonyl is more electrophilic we have an aldehyde and we have a diaster what's more what's more electrophilic alright so we're going to tack the aldehyde exclusively when we have one inna late that's going to exclusively form in one of the reagents and it's going to attack selectively the other reagent you're going to have a clean cross crossed aldol reaction so that's the most acidic hydron so what you're going to do is you're going to form this enolate remember with sodium ethoxide we completely form this enol a complete conversion to this enol eight and now we're going to attack the aldehyde so there isn't any Laster left that's it there isn't any active methylene compound left that's not deprotonated you're going to protonate then eliminate i'm not going to show that here we'll do that in discussion protonate then eliminate and once again you can't stop at the beta hydroxy carbonyl because this compound is the product is so stable that double bond that you just formed is conjugated with both esters both both carbonyls of both esters alright so that's going to work really well there's a third technique that we're going to use and that's called a directed aldol and this kind of puts you in the driver's seat so here's what you do you prepare the enolate of one carbon eel with Lda okay so do chapter 23 chemistry completely form the enolate and then add the second carbonyl to the c9 so let's see what that looks like Eldar you gives you the mostly the kinetic enolate so you form this first and then you add your second on carbonyl so we add this next so this is going to attack that carbonyl and rather than letting the molecules do your own their own thing you're actually in charge here what's happening if you know what you're doing and this also lets you this also lets you isolate the beta hydroxy carbonyl compound depending on what you do if you add water here and your k4 careful not to add acid you can isolate the beta hydroxy carbonyl if you add acid what's going to happen you're going to dehydrate aren't you so you're in charge you're you directed the reaction we're going to stop right
there and I hope you guys have a great weekend