Electrophilic Addition Reactions of Type 2 Carbonyl Compounds

Video thumbnail (Frame 0) Video thumbnail (Frame 5820) Video thumbnail (Frame 10198) Video thumbnail (Frame 12321) Video thumbnail (Frame 24589) Video thumbnail (Frame 36857) Video thumbnail (Frame 49946) Video thumbnail (Frame 62685) Video thumbnail (Frame 75423)
Video in TIB AV-Portal: Electrophilic Addition Reactions of Type 2 Carbonyl Compounds

Formal Metadata

Electrophilic Addition Reactions of Type 2 Carbonyl Compounds
Title of Series
Part Number
Number of Parts
CC Attribution - ShareAlike 3.0 USA:
You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this license.
Release Date

Content Metadata

Subject Area
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: 00:22 - Electrophilic Addition Reactions of Type 2 Carbonyl Compounds: Acyl Substitution 03:53 - Mechanism and Reactivity in Acyl Substitution Reactions 06:34 - The Stability of the Carboxylic Acid Derivative 08:15 - The Leaving Group Ability 10:30 - The Rate of Reaction 27:03 - Reactions of Acyl Halides 41:44 - Reactions of Anhydrides 49:33 - Reactions of Esters
Setzen <Verfahrenstechnik> Walking Alkoxide Chemical reaction Elektrophile Addition Alcohol Electronic cigarette Reactivity (chemistry) Water Lone pair Computer animation Functional group Addition reaction Generic drug Carbonylgruppe Grignard-Reaktion Stuffing
Sense District Asymmetric induction Weakness Carboxylate Resonance (chemistry) Chloride Reactivity (chemistry) Electron Carbonylgruppe Addition reaction Elimination reaction Derivative (chemistry) Process (computing) Chemical reactor Aldehyde Setzen <Verfahrenstechnik> Concentrate Octane rating Carbanion Ketone Walking Carbon (fiber) Chemical reaction Acid Computer animation Functional group
Activation energy Ester Carboxylate Hydrogen chloride Resonance (chemistry) Acetic acid Action potential Chloride Wursthülle Chlorine Methoxygruppe Reactivity (chemistry) Ethanol Molecule Shear strength Electron Addition reaction Carbonylgruppe Tetraederstruktur Elimination reaction Mixture Conjugated system Setzen <Verfahrenstechnik> Acyl Octane rating Induktiver Effekt Walking Veresterung Carbon (fiber) Chemical reaction Hydroxide Water Acid Computer animation Metabolic pathway Functional group Orders of magnitude (radiation) Proteinkinase A Iron Base (chemistry) Cobaltoxide Carboxylierung
Ester Activation energy Carboxylate Sunscreen Substitutionsreaktion Chloride Chlorine Reactivity (chemistry) Electron Addition reaction Carbonylgruppe Ballistic trauma Elimination reaction Conjugated system Hydrophobic effect Setzen <Verfahrenstechnik> Acyl Thionylchlorid Hydride Octane rating Walking Veresterung Sulfur Chemical reaction Acid Gesundheitsstörung Computer animation Acid anhydride Functional group Proteinkinase A Chemical compound Zunderbeständigkeit Base (chemistry) Cobaltoxide Methanol
Übergangszustand Activation energy Ester Hydrogen chloride Carboxylate Substitutionsreaktion Plant breeding Chloride Hydrochloric acid Vancomycin Leak Wursthülle Atomic number Ammonium Lead Electron Addition reaction Amine Carbonylgruppe Lactitol Data conversion Tetraederstruktur Elimination reaction Pyridine Hydride Octane rating Walking Veresterung Methylamin Protonation Solution Chemical reaction Water Acid Gesundheitsstörung Computer animation Acid anhydride Functional group Etomidate Proteinkinase A Iron Amination By-product Base (chemistry) Cobaltoxide Thermoforming Stuffing
Computer animation
good afternoon we're gonna get started so um today we're gonna talk about some reversible reactions of type two carbonyls and i will let you know the stopping point for Friday's midterm so I opened up chapter 22 later on this afternoon and I'm going to write I'm gonna send you an email telling you the relevant problems from Southland chapter 22 and I'll also tell you the relevant problems from the end of the chapter first Beth okay questions anybody about upcoming midterm all that kind of stuff No all right all right so here's where we start talking about the electrophilic addition reactions of the reversible but it will first a quick review of the irreversible addition reactions of type 2 that will be very relevant for you for midterm one right so here we have strong nucleophiles I'm using grignard as an example here and here's our generic type 2 carbonyl with the group L that could be a leaving group also has a lone pair we make our tetrahedral intermediate let me fix that we make our tetrahedral intermediate irreversible addition the leaving group leaves boom we kick that we kick the leaving group off and then we have to make a decision about whether the addition is going to happen again and it turns out that the addition is going to happen again here because well we don't know we can't compare reactivity here when we have the the generic version but the greener it is going to add again until we end up getting an alcohol so this is a great way to synthesize alcohols so we end up getting the alkoxide and when we add water in a second step we protonate so we'll just pump put water over here so that's actually a possible mechanism for midterm one that we've already done all right so there's our generic example of addition of a reversible reagent now here's the deal for this one here we have only one leaving group possible and that's this leaving group al and that's because this group that just came on can't come back off again as a leaving group when you use weaker nucleophiles we're actually going to have a choice of leaving groups so we have two no leaving group ability okay so here we only have one possible leaving group L down here when we have these guys there we're going to have other potential leaving groups so we're gonna have to be in a position to make decisions about that so that's why I want to go into a little bit more in depth on the reactivity of the the type two carbonyls we've already
talked about one aspect we did not talk about leaving group ability so we're gonna just do a little bit more in depth here so here's our G here's our reaction here one more time no one's going to miss down on the test because I've done it so many times right our particular reaction so a nucleophile attacks the carbonyl that's addition which I'll abbreviate ad that's an addition here's our tetrahedral intermediate and then in the second step elimination electrons come down and kick off the leaving group and of course the overall process is a shil substitution now in Smith they say the rate of the reaction has only to do with the leaving group ability and that's not really an accurate way of looking at it because the second step depends on leaving group ability the first step does not depend on leaving group ability does it we first we have to do the first step first when we look at just the leaving group we're talking about the second step we still have to look at the first step because if the first step doesn't happen then the second step can't happen so it's a little bit more a little bit more more of an explanation than that that's what we're going to be focusing on than this this next part of the chapter all right so the rate of the reaction is going to depend on the stability of the carbonyl derivative that's because of that's that's going to affect the rate of addition the first step which is step one and the rate of the reaction also depends on the leaving group ability of L that would be for the rate of elimination the leaving group ability is going to affect the rate of elimination which would be step two so we may be able to do step one and then if we can't do step two then it will just go backwards okay so we have to talk about both steps if we're going to talk about the rate of the reaction so let's talk about the first one the stability of the carbanion we've already done that we know about the stability of the carboxylic acid derivative or the type 2 stability depends on the concentration of positive charge on the carbonyl carbon and we talked about the inductive effect and the resonance effect to explain that and so you could go to page 12 to 14 of your notes to do
that I don't need to talk about that again and here's our order of reactivity that everybody knows for the the midterm on Friday right so acid chlorides have a strong inductive which destabilizes and a weak resonance resonance is the thing that stabilizes inductive destabilizes resonance stabilizes and so that makes sense that acid chlorides are the are the least stable least stable or most reactive and then we saw carboxylate has the Cork's carboxylate anion is the most resonance stabilized of all of these guys so this is the best resonance and resonance stabilizes and so it makes sense that this is going to be the most stable least reactor notice I don't have a lights and ketones here because in this particular chapter we're talking about just to type tooth so I left them out entirely aldehydes and ketones don't do acyl substitution so on that's why I left them out all right so the second thing we need to talk about is leaving group ability best
leaving groups are weak bases right we learned that back in 51 a so this is this is sort of that the adage I told you at the beginning class everything you've got learned this year is gonna all come back to you in 51 C so that was back in 51 ad that we learned that so let's talk about leaving group ability here so leaving group how do we measure a leaving group ability weakest base so we look at the pKa of the conjugate acid right so here's our leaving groups I'm gonna circle all of them hydroxide ethoxide NH 2 minus CL minus okay pKa of the conjugate acid the conjugate acid here would be HCl that's -7 pKa of this conjugate acid that would be just acetic acid we know that's about five pKa of Hydra leaving group ability of hydroxide the pKa of the conjugate acid the conjugate acids water and that's about 15 and likewise here we have ethanol as our conjugate acid for that leaving group and that's also about 15 and then NH 2 - we'd look at the pKa of the conjugate acid which be ammonia that's about 35 so most of these we know rounded to the nearest 5 on those few of PKS that I want you to know route it to the nearest 5 so what we can see here is that we have increasing base strength as we go this way and then in the other direction we have increasing leaving group ability so the arrow would go the other way so notice an acid chloride that's our most reactive and it's also has the best leaving group these things tend to work together for the type 2 carbonyls alright so let's go back to the rate of the reaction so the first step is addition second step is elimination if we have we decrease the stability of the carbonyl so as that carbonyl as that type two carbonyl becomes less real s stable you're going to increase the addition rate right the more stable the carbonyl the slower the addition rate so that's the first step and then when we have the second step this is all about leaving group ability we M for elimination if we increase the leaving group ability we increase elimination and so if we have a poor leaving group that's going to slow down that second step and may not even enable that second step to take place all right so when it looks a look at some energy diagrams for the extreme cases I'm going to start off with carboxylic acids and esters and and they have about they are very similar they're very similar in their reactivity both of these guys have resonance stabilization and they have some inductive effect and resident stabilization but what we're going to say for this class is that this ester in this carboxylic acid be about the same electro Felicity so they they're about the same electro Felicity and we looked at leaving group ability and using our PKA rounded to the nearest five these are not exactly the same but they're about the same about PKA 15 so energy level these guys are about the same so that's why I have them here right there and there that should be at the same energy level and and and and and that's what it's going to look like and so if you look at here we have energy of activation for the first step is right here taking the ester and going to a carboxylic acid here's the energy of activation for the first step we have our tetrahedral intermediate is going to be right here in the energy diagram let's see what that looks like what our tetrahedral intermediate is going to look like all right so what's gonna happen is the hydroxides gonna attack the carbonyl we're going to give us that tetrahedral intermediate so we can actually even draw that down here let's just draw that down here boom that's gonna attack the carbon you know we're gonna kick electrons up onto oxygen just like we've been doing and that's what our tetrahedral intermediate would look like and notice all right we have two leaving groups now we have two potential leaders in groups so that's what's different about this chapters we're gonna have to choose leaving groups both of these leaving groups have about the same leaving group ability so similar leaving group ability so this tetrahedral intermediate is going to want to go back to a carbon u if this methoxy group comes off it'll go here so it starts off here it will go here and if the hydroxide comes up it will go right back to starting material and because those two leaving groups are about the same we have the same height for this this energy barrier here for both of these so what we're gonna have to do is when we're trying to take a an ester and convert it into a carboxylic acid if we don't drive the equilibrium we're going to get a 50/50 mixture right so we will have to drive the equilibrium here and we're going to talk about that reactive reaction coming up but we will just definitely have to drive the equilibrium so and that's because the energy of activation of the reverse reaction is approximately equal to the energy of activation of the forward reaction so therefore we will get a 50/50 mixture at equilibrium unless you drive the reaction so we're gonna have to worry about that when we get to Esther's talk about converting esters to carboxylic acids and the reverse reaction converting carboxylic acids into esters we're gonna have to drive the equilibrium okay so that's why that's when the two are about the same let's look at in the second energy diagram what happens when we have a carbonyl compound stabilized by residents and it can't Anna containing a poor leaving group that would be a namet see what happens here so here's our amma dried here and we want to try to convert an dammit into an acid chloride alright
so chloride ion attacks the damid what would the tetrahedral intermediate look like let's say let's draw the tetrahedral intermediate over here all right so we have nh-2 here we have chloride very large energy of activation for that first step why is that so large energy of activation because Emma's are pretty stable aren't they they have very strong resident stabilization not as strong as carboxylates but they're very strongly resonance stabilized okay so that's because Amma's are pretty stable so when things are stable they're not going to want to react so it really arts energy of activation for the first step and then if you look at the energy of activation for the second step once we come up once the molecules have enough energy to overcome that first energy barrier then they have to come up with this much more okay so those so so e8 so so the molecules have to come up with this much extra energy in order to to go to the product so what's the rate-determining step for this reaction first step or second step the second step because you need to come up with this the molecules need to cope with this much energy and this much more so that the rate determining step is the highest and highest mountain to climb there and that's the highest mountain to climb so this is so large this energy of activation and want an energy of activation to that the reaction doesn't happen and when your daughter tetrahedral intermediate you can see why this reaction doesn't happen which one is the best leaving group here that's the best leaving group pKa of the conjugate acids -7 pKa of the conjugate acid of NH 2 minus is 35 so what's the best leaving group by far it's gonna be chlorine so what this guy wants to do is those electrons come down and kick the chloride ion right back off again so in other words even if you can get this first step to happen this next the second step is not going to happen because we want the little best leaving group to leave we can't have this chloride stay and have this as a leaving group because this is by far orders of many many many orders of magnitude better leaving group so that's why too high of energy so can't you can't take an abbot and convert it into an acid chloride directly by acyl substitution it doesn't happen all right so this is the best leaving group so it's easier for the tetrahedral intermediate to collapse back to starting material so what's going to happen is that those electrodes are going to come down so by collapsing back to starting material I mean these electrons are going to come down and kick that chloride iron right off so you're gonna just go right back here anyway okay so there's there's another extreme case now let's look at the other the other extreme here where we have an energy diagram where the carbonyl is very reactive contains a great leaving group and this is scrolling all weird on me sorry about that so our best example of that is an acid chloride we're taking acid chloride and converting it into a carboxylic acid alright so let's see what that looks like hydroxides going to attack I put these two close together but you get the idea hydroxides going to attack the carbonyl here so this is our tetrahedral intermediate let's draw it what's the best leaving group cards the best leaving group that's the best leaving group and so that's what's going to happen so best leaving group loss of this therefore loss of this leaving group is faster so it has a smaller energy of activation if we wanted the hydroxide to leave then that would be a larger energy of activation and so we're going to follow the lowest energy pathway here so we have two things going on here we've got a great leaving group so that's going to be a small energy of activation we also have a small energy of activation for the first step so this is EA one this is EA two so EA one is small so therefore we have a fast first step why acid cards are very electrophilic that's it and ea2 is also small because we have a great leaving group acid chlorides have a great leaving group questions on the energy diagrams anybody alright so what you've noticed is if the reaction goes readily we have a lower energy product it's exothermic overall over here this is endothermic not going
to be easy right doesn't happen and then we also have where we have similar reactivity in this energy diagram and those are the ones where we're going to have to drive the equilibrium okay so going uphill is a problem going downhill
is not so we're going to summarize these this thing does not scroll very well these factors can be summarized and that's at the top of your next page so here's our here's our order of reactivity increasing stability of the carbonyl and we also have increasing basicity of the leaving group so the road as a result of that is that if we have increasing stability that we're going to have a faster reaction rate this direction increasing reactivity and we're also going to have so that's for the first step one and we also have increasing leaving group ability and that's going to affect step two and this one here is step two all right so that's the reactivity of this reaction in a nutshell that's going to actually help you make decisions here coming up when we look at all of these type 2 reactions with reversible reagents ok so what we're gonna do is we're gonna we're going to start with this order here we're gonna start with the most reactive here acid chlorides and we're going to talk about n hydride so we're going to talk about carboxylic acids esters and then Amex and we're gonna do it in that order so we're going to start with reactivity of acid chlorides and pretty much it's very easy to take any of the acid core an ESO card and convert it into any of the other type 2 carbonyls very easy so we start here this is an energy level here's our acid chloride and so then we these are the various other different energy levels this is really not to scale but here we have an hydride energy level we have ester and carboxylic acid so ester in acid and then we have a m''d and it's easy to go downhill so if we have an acid chloride you can have the small energy of activation here and then small energy of activation for the second step and then likewise we can go down here easily and we can go down here easily if we try to go the other direction which would be uphill reaction so if you try to go uphill you run into a problem so for example let's say you try to take an ester which would be right here and go to an acid chloride large energy of activation for the first step and the second step would be even larger energy of activation than the first step so can't go the opposite direction so we're gonna put a big X through that so in general less reactive acyl compounds can be easily synthesized for more reactive ones forming a more reactive acyl compound from a less reactive one requires special reagents or forcing conditions I want you to remember that special reagents or forcing conditions so that would be this example right here and we're gonna see both of those types we're gonna see special reagents that we're going to see forcing conditions all right so let's look at some easel substitution reactions with acid chlorides we can easily make an ID we can easily make an acid chloride into anhydride we would use conjugate base of a carboxylic acid so it looks like this so any one of these acid chloride mechanisms is a possibility for a midterm - but they're all the same very much the same anyway that would be addition and the problem with if you try to or if you try to memorize the reagents and products you're gonna get completely overwhelmed if you actually do the arrow pushing you draw you look at the tetrahedral intermediate and make decisions about what's going to happen then it's going to be much much easier for you so I really highly recommend that so let's look here we have this what's the best leaving group chloride ions the best leaving group pKa of the conjugate acid it's -7 it's gonna come off like a shot that's really going to be a fast reaction so here's our best leaving group and that's what's going to happen in these reactions best leaving group is going to win best leaving group is going to leave so then we get elimination and there we have our anhydride all right now let's make an ester from an acid chloride this is probably going to remind you of the thionyl chloride mechanism from 51b chapter 9 I'm thinking so anyway remember pyridine non-nucleophilic base let's draw that our favorite non-nucleophilic base from chapter 9 so imagine that this is sign of chloride we're gonna be doing the exact same thing so just so this is addition let's draw the intermediate that we get all right what's our best leaving group so methanol what if methanol leaves pKa of the conjugate acid of methanol is protonated methanol that has a pKa of about minus 5 or minus 6 pKa of the conjugate base of chlorine is minus 7 so what's the best leaving group chloride ions the best sleeping group and so that's what's going to happen best leaving group is going to leave so are you seeing that the correlation here with the thionyl chloride we attack the sulfur we kicked electrons up onto oxygen and electrons on oxygen came down and kicked off chloride exactly the same here second step is elimination and
again this is best leaving group oops and here's where the pyridine comes in the pyridine comes in to deprotonate that oxygen so Esther from an acid chloride another possible mechanism for the term one alright acid chloride plus two equivalents of methylamine we got that too there I want to talk about that in just a second but let's go through this we need two equivalents here I know you hate that when we got to worry about two equivalents but you'll see why here what I'm talking about so yeah this is going to cause a problem I guess I'll go on to the next page here so here's edition I'll cop side leaving group what's our best leaving group always chloride iron how about that huh so -7 for HCl about 10 for the ammonium ion which with the pKa for that amine if it leaks ok so then the second step is elimination so we've already used one equivalent of our mean and now we need a second equivalent to deprotonate so acid chloride to Amman all right so let's let's take a look here this is our first student first equivalent and here's our second equivalent sorry about that so an extra equivalent of the amine is needed in this reaction why ch3 and H 2 will be protonated reaction and CH 3 NH 3 plus is not nucleophilic if we don't have a second equivalent of a mean in there then we will protonate the rate remaining amine and we will only get 50% conversion okay so reaction will stop at 50% so that's the problem with that one so why not just you protonate with the chloride ion generated in the reaction that's really not going to help us anyway because if we generally do you protonate with chloride ion will form HCl what's HCl going to do with the remaining Amitha mean that's in there I'm just going to protonate it right so just going to protonate it so let me show you what I mean here so we do the addition we get to this point you say okay why do I need excess I mean I can just do this instead you protonate with chloride iron there's my base and of course if you do that you will produce hydrochloric acid what's the fate of the hydrochloric acid well it's going to react with the amino I'm gonna do long arrow to the right short arrow to the left and we're going to prove that we have that plus we have CL minus so we look for our acid on each side of the equation hydrochloric would be here on this side and the protonated amine would be our acid here and we look up the pKa this is PKA minus seven and this is a PKA about 10 so at equilibrium we have one hydrochloric acid for every 10 to the 17th protonated amine so it really doesn't matter what we use we do need either two equivalents of amine or there's another thing that you can do and we did this back in chapter 18 and I'll just draw it right up here and you can draw it in the margin of your cage you can also use one equivalent of a mean and one equivalent of pyridine the non-nucleophilic base instead of two equivalents of rain all right so remember from last quarter if you had any last quarter if you don't it doesn't matter I'll show you right now remember from last quarter when we had antolín and we had we run into all sorts of problems with annalen last quarter when we're trying to do electrophilic aromatic substitution and so a solution to many of our problems with a med was to take with annalen was to take annalen and convert it into acid annelid remember that is it all just running back to you here hopefully we had that in pyridine right we made a sedan allayed and acid analysts much more well behaved in an electrophilic aromatic substitution so that's what this reaction is I promised you we would be covering at this quarter and there it is and we'll be doing the mechanism for that in discussion this week okay questions on reactions of acid chlorides anybody okay and so that ends midterm 1 material return 1 this direction yeah so you know here's the deal I was tempted I was thinking about this last night and I was tempted you know let's just not have chapter 2 if I buy fear saying that because I'll get it Emo's did you say that it wasn't good to be on there I was thinking about where to end the material for the first midterm and I thought it's better to have a little bit more on the first midterm because I think most students find overwhelming amount of material for midterm 2 don't you think that it usually ends up being that way so that's why we're stopping at this point it will actually make it easier for you for the rest of the quarter when things start to get really intense God didn't do anything did you see how that did that just whenever I said intense what does it mean okay stuck in our reactions to van
hydrated of course this will not be on the test so as we've seen here's an hydrides right here it's easy to go downhill so we can easily convert an N hydride into an ester or carboxylic acid we can also easily convert it into a namet and that's because we have the energy diagram looking like this where we have the second step is lower in energy here boom-boom-boom-boom-boom best leaving group is going to leave and that accounts for the lower energy second step for that reaction well we can't do I gotta stop emphasizing words here something's going on here we can't do an hydride up to acid chloride that would be an uphill reaction and it would look like this we have a large energy of activation for the first step and it's even larger energy of activation for the second step because if the if we're trying to make an acid chloride into an anhydride we have to have a worse leaving group leave and that's the problem so that's why we have a really large energy of activation for that second step alright so we want to put a big X through that right that's not going to happen uphill doesn't happen unless you use fortunate conditions are special reagents and this is not one of them that we're going to be able to do that with all right so let's show why anhydride can't go to acid chloride so you can see from the energy diagram you can also see it by actually I'm doing the arrow pushing for this reaction so chloride is going to tack one of the two carbonyls of the anhydride does not matter which I will always have symmetrical anhydride so you don't have to decide which carbonyl to attack alright and now we're going to draw our tetrahedral intermediate look at what we have and then make a decision what's the best leaving group chloride iron in order to go from an anhydride to an acid chloride you have to have a worse leaving group lead and that's not going to happen so what is going to happen is his electrons are going to come down and kick chloride right back off so it goes back to anhydride right let's look at some more examples do you want me to include in hydrides on the test I know okay it is easy to go downhill let's look at some down hills here water water attacks and anhydride tetrahedral intermediate this reaction actually bet goes better when it has a little bit of acid catalyst in it and you'll see kind of why here our best leaving group is water which would reverse and go back to reach the starting material so it's so in this one I'm going to deprotonate and this action actually works better if you add a little bit of an acid or base catalyst so it looks like this if we deprotonate that first now it's very clear what the best leaving group is and that's what we're going to have we so this reaction will go without acid or base but it works a lot better when you do use acid or base catalysis and so now we definitely have our best leaving group is going to be here that's our best leaving room so electrons come down and we kick off the best leaving group then that leaves us with a carboxylic acid same thing with an avid nose we're using two equivalents here yeah deprotonate intramolecular protonate intramolecularly you can pronate and deprotonate into like really if it makes a five or six membered transition state so if you count all the atoms involved and you count five or six that can happen but I don't believe we have that here so it would be you're having this coming and grab that yeah one two three four four it can't reach it has to be five or six great question by the way alright so let's look at the amide here also need two equivalents or one equivalent of this and pyridine so it depends what if you would you choose two equivalent if it's a very inexpensive cheap a mean that you don't mind you losing half of it if it's an important amine that's something that you've had to synthesize you don't want to waste half of it so you will use one equivalent of amine and one equivalent to pyridine yes yeah no it actually works without it but as you can see we have this one's a little bit different and it actually helps it with you don't have to write small amount of acid or base okay yeah the reaction will go but it will be incomplete it's kind of like with the Amman yeah alright so I'm not going to go through this one is this going to look just like the previous example here there's all your side products a in case sapling ask for that we want to keep track of that kind of stuff which I don't usually have you do on the test all right so that's Anna hydrates questions on anti drives anybody talk about reactance investors in the last couple last minute okay so we already know we don't already know what we can do here and what we can't do so these guys we can convert back and forth it's a reversible reaction that we're gonna have to draw it so ester we can convert to a carboxylic acid and we concurred a carboxylic acid to an ester so we can since we're talking about esters we can easily go to the right we can't go this way cross that out we can't go this way cross that out but we can easily go this way so this will be easy and these can't do can't go uphill but we can go downhill easily we'll talk
more about this on Wednesday