Alternatives to Direct Alkylation of Enolate Ions

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Alternatives to Direct Alkylation of Enolate Ions
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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: 00:21 - Base Promoted Halogenation of Aldehydes and Ketones 11:46 - Alpha Halogenated Carbonyl Compounds in Synthesis 17:55 - Direct Alkylation of Enolate Ions 27:44 - Kinetic vs Thermodynamic Enolate 43:45 - Alkylation and Acylation of the Alpha Carbon via an Enamine Intermediate
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all right we saw that when you did there's an acid-catalyzed and a base catalyzed reaction in the acid catalyzed reaction you get how many bromine is incorporated acid catalyzed one base catalyzed you get all grow all out all you know Liza behide Rajan's replaced with bromine so that means that if you have here we have one two three four analyzable hydrogens they're all replaced with bromine if you have a methyl ketone where there's a methyl on one side all three hydrogen's will be replaced with bromine you form this intermediate that is attacked by hydroxide so we're going to incorporate three bromine boom-boom-boom we were teach that two more times then hydroxide ion attacks the carbonyl and now the electrons come down and run kick off CBR 3 so remember when we do that attack and we form that tetrahedral intermediate the best leaving group is going to leave so which is a better leaving group so one of my students looked up the pKa of this conjugate acid all right pKa of the conjugate acid here is 13.8 pKa of the conjugate acid here is what 15 approximately fifteen fifteen point seven so that is a better leaving group all right and so that's going to leave and then you get the carboxylic acid and as soon as the carboxylic acid forms then the CBR 3 comes in deprotonates so it's making that CBR 3 more acidic is the fact that it has the three bro means that are pulling electron density away from the carbon and making it more negative less negative let's less well less electron density on carbon that stabilizes it and so that's why it's a better leaving group let's go ahead and label PK's here all right this guy here better leaving group pKa of the conjugate acid which is HCV our three equals thirteen point eight and then we go over here and this one pKa of the conjugate acid this is worse leaving group I just doesn't want to let me do that let's see what's going on here let's try that worse leaving group pKa of the conjugate acid which is water equals approximately fifteen or fifteen point seven if you want to be exact so the better leaving group leaves and that kicks off so why doesn't that happen over here up at the top so so so why doesn't the hydroxide come and attack this carbon eel and then we brake on one side here here all right well I can't why doesn't it why does that only happen when it's a methyl ketone what do you think all right so thinking about this same thing here hydroxide ion attacking them o and adding kicking off that CBR to that side of the Ring what do you think why doesn't it always it only happen with a methyl when there's three bro means they're two bronies is not enough to stabilize the carbon I and you need three of them okay so so we need to have all three bro means they're to stabilize the carbon ion right here here's our leaving group must have three to stabilize the carbon anion enough for it to leave if you only have two that's not enough so you've got to have all three and bro means with their collective inductive effect stabilizing that group enough to make it leave all right so you'll see this reaction dealt with bromine you'll see this reaction done with the iodine also okay so if it's bromine it's a bromo form reaction if it's iodine it's an iota form reaction and and back in the day when they didn't when they didn't have spectroscopy so readily available and even before spectroscopy that's how you were able to tell if you had a methyl ketone you would do an iodophor reaction because when the I fly see I the HCI 3 leave that that CI 3 leaves they forms a yellow precipitate so that was a chemical test for methyl ketones so um you'll see iodine used you'll see bromine used you won't really see chlorine used all right so now you're I'm sure you're wondering maybe not but I hope you're wondering why do we only get one bromine incorporated with acid and we get multiple bro means incorporated with base right you're wondering about that ok or if you're not you're hopefully you're wondering about it right now all right so let's look at this so once we have one bromine incorporated here this would be our product right and if you're looking at the acidity of this hydrogen we changed colors in case you can't see that we're looking at the acidity of this hydrogen right here if we in ordered in the base in order to put another bromine on we have to remove the the that hydrogen to form this intermediate right so hydroxide ion has to come in and we have to remove that hydrogen to form that intermediate that's that's how we're gonna get another bromine on there right what effect does that bromine have on that is that going to stabilize that conjugate base or destabilize what's gonna do it's going to stabilize its pulling electron density way making that less negative okay so so this hydrogen is more acidic once one bromine is incorporated because the conjugate base is stabilized by the inductive effect of the bromine so what is the inductive effect of the bromine it's pulling electron density away making that less negative and let's fix that word there the R pulls electron density away from the cart negatively charged carbon making it less negative less negative is more stable less negative therefore the acid is more acidic what happens when we put a second bromine on there that hydrogen is even going to be more acidic because now it has two bromine stabilizing the conjugate base okay so it makes good sense so therefore multiple halogenation is favored all right so that's makes good sense with the base now let's talk about the acid alright in acid in order to cut his foot a second bromine on remember we have we have a probe we have to protonate the carbonyl right and that protonated carbonyl has a resonance structure so let's draw the two resonance structures so different
mechanism with the acid catalyzed we're protonating the carbonyl we have that resonance structure what is that bromine doing to that resonance structure especially you can see it on the second resonance structure what is that stabilizing or destabilizing it's destabilizing it's pulling electron density away from a carbon that's already electron deficient bromine destabilizes the protonated carbonyl intermediate necessary for formation of the enol and remember the enol is what our reactive intermediate is an asset and if we're destabilizing the intermediate that leads to the enol then it's not going to happen it's going to be less likely to happen and that's because I'm bromine makes carbon more positive and therefore less stable so only on only mono halogenation occurs so all we have going on here is this bromine pulling electron density the bromine doesn't care that that carbons are already electron deficient it's still gonna pull electron density way making it more positive that's going to destabilize so as you can see before this quarter you might have had the misconception that inductive effects are always stabilizing that's not the case inductive effects stabilized a negative charge but they don't stabilize a positive charge all right questions on that reasoning anybody all right so let's talk about alpha alpha halogenated carbonyls and synthesis so we're gonna use these in synthesis which is your of course your favorite part of organic chemistry right okay so we see that when we make an enolate or we make an enol we have a nucleophilic alpha carbon don't we we turn that alpha carbon into a nucleophile when we put a bromine there we change it back to an electrophile right so now it's an alkyl halide just pulls electron density away so we have a partial negative charge on the bromine and partial positive so we're back to electrophilic carbon which we're much more familiar with because we've been doing that more of the quarter so we have an electrophilic alpha carbon so now rather than the elect the alpha carbon being nucleophilic and attacking electrophiles the alpha carbon is electrophilic so it's going to attack nuclei and so nucleophiles are going to attack it so nucleophile comes in from tax the alpha carbon we kick off bromine as a leaving group so we've dramatically added to the range of things that we can do with two carbonyls in this chapter so that's something that we haven't been able to do until now and so that so of course compare that with the nucleophilic alpha carbon that we've been looking at when we take a base to deprotonate right we make our enol 8i anion and then we can push electrons onto oxygen right we can draw both resonance structures so there's both resonant structures and so now we can attack an electrophile electrons on oxygen come down and we can attack an electrophile and incorporate an electrophile into the Alpha position so nucleophiles in the Alpha position not a problem electrophiles in the Alpha position also not a problem so you see the two two ways to do this okay so that's why that's a very helpful reaction so one example of using this is to use cyanide ion to tack the alpha carbon kick off bromide as a leaving group this is an sn2 reaction so backside attack sn2 reaction and we just sweet so we put a cyano group in the Alpha position that's also converts that into an active methylene compound so this is a active methylene compound so we we have enhanced acidity of the hydrogen that's in between those two activating groups okay so enhanced acidity on that methylene carbon that's right in between the two activating groups alright the other thing we can do after we incorporate a one bromine so what do we want to use if we want to incorporate one bromine acid or base acid right the other thing that we can do is we can turn this into an alpha beta unsaturated carbon eel right so these are the reagents the lithium carbonate lithium bromide DMF is the solvent and so what ends up happening here as we remove this carbon right here oh that's co 3 2 - sorry ah you know what we're just gonna draw I'll come back and erase that so we do an elimination now we confirm an alpha beta unsaturated carbonyl compound and so now what can we do with alpha beta unsaturated carbonyl compounds one two addition and one for addition right all right so let me draw the arrow pushing properly for that ok 1 2 1 4 addition so in 1 2 addition we can incorporate we can attack the carbonyl with the nucleophiles that do irreversible addition we can also have nucleophiles come in and attack in the beta position so now we can incorporate nucleophiles into the Alpha position electrophiles into the Alpha position nucleophiles into the beta position and nucleophiles at the carbonyl carbon so quite a range of things that we can do here alright and that's all i want to say about alpha halogenated carbonyl compounds are there any questions on that anybody all right so we've talked about how to incorporating protons into the Alpha position deuteriums into the Alpha position bro means into the Alpha position and we can do iodine also and now what about incorporating carbon groups into the Alpha position alright so direct alkylation so we would call that alkylation of enolate ions so here's an example right here of a general reaction we take an ester which has a pKa of 25 and we take LD a very
strong hindered base it's so strong that the enolate is formed completely at low temperature alright so we're going to form the enolate let me draw the enolate here or the ester so naked will be charged carbon in the Alpha position and then we can certainly push those electrons up onto oxygen let me draw the second resonance structure and now rather than adding a proton source or rather than adding bromine we can add an alkyl halide all right so the electrons on oxygen come down and we backside attack check the carbon bonded to the leaving group kick off the leaving group and now we've incorporated an alkyl group into the beta position pretty powerful thing to be able to do and so remember I've said this before when we make new carbon-carbon bonds those are going to be very important reactions so this is a new carbon-carbon bond forming strategy so hopefully you started the carbon-carbon bond forming reaction page that I recommended that we're going to add to a bunch of new reactions in chapter 20 that are carbon-carbon bond forming right with the green URI agents with sodium Alka knives with blue theme reagents and now we're going to add this alright so this is an sn2 reaction so if we think back to what's important for an sn2 reactions is subject to the same considerations so primary alkyl halides sort oscillates work great with the very best being ch3 X so methyl halides or methyl tosylates alo benzyl these are good these guys work great all have no beta hydrogen's that can be eliminated why is that important why do we have to worry about beta hydrogen's so we've got to worry about elimination we just said that but what is it about the nucleophile that we're talking about is that a nucleophile that that is thatis is that a strong nucleophile and also a really strong base I mean a pKa it's 25 so when you deprotonate that that's a really strong base and really strong bases favored eliminations so we got to stick with things and then aren't going to do a lot of em elimination or things that aren't going to do elimination at all so those are three examples if you have a secondary alkyl halide sort oscillates that's not going to work well because the nucleophile is a really strong base a deprotonated ester is a strong base therefore e 2 is going to predominate and how do we know that because we know the pKa of the ester is 25 therefore it's a weak weak acid and therefore the conjugate base is strong so steer clear of secondary alkyl halides in this reaction they're not going to work well tertiary alkyl halides don't work at all so that's a really bad idea we're only gonna get elimination so here's what it would look like just to remind you because this is all going all the way back to chapter 8 which was a while ago so tertiary alkyl halide - hindered for backside attack okay I'm really working on this here that's that always means that I need to slow down so I will slow down how about that okay so remember elimination anti-periplanar remember all of that here's our base I'm gonna grab that beta hydrogen we're going to break the carbon hydrogen bond we're going to form a double bond here we're going to kick off iodide as a leaving group so we have not gotten any alkylation in the Alpha position whatsoever all we've gotten is elimination product so no tertiary at all all right so this reaction works well for Esther's and night trials let's show what it looks like with the nitrile sew with a nitrile Lda THF so strong race necessary strong base necessary for complete deprotonation followed by a methylbromide can do some elimination but not too much because it's primary and so what we would form is here this is our we eliminate the alpha hydrogen so that's our that's our enolate ion and then we can and that that is resonance stabilized so let's draw the resonance structure structure not really a needle a but it's a resonance stabilized anion okay and then we have ethyl bromide come in this is going to do backside attack kick off the leaving group attacking the carbon bearing the leaving group and we kick off bromide ion and so you can see what we haven't what we've done is we've incorporated a ethyl group into the Alpha position all right questions on direct alkylation of esters and nitriles all right annoyed some ketones can also react with alkyl halides but the reaction is a little bit more problematic little more khalat more
complicated okay so let's talk about that so Esther easy nitrile easy the problem here is that that we have two places where the enol a can form are we going to remove for this proton on this side are we going to remove one of these protons on that side so we have a regio selectivity issue okay so we can move this proton here or we could remove one of the two over there and we got a name for each of those so we've removed this proton right here this is what our enolate ion would look like I'm going to draw the residence structure where we push the electrons on oxygen here we can do that on that side or we can go and remove one of the acidic protons on the other side and let me draw that you know a tie on so one of those is kinetic is formed faster and one of those is the thermodynamic product which one's the thermodynamic product one on the left or one on the right thermodynamic product is going to be the more stable so look for the most substituted alkene here so everybody see that the one on the right has the more substituted alkene so that is what we call the thermodynamic enolate which is the most stable and why because the alkyne is more substituted so that means the other one must be the kinetic enolate and it makes sense that it would be easier to remove a proton on the left-hand side because it's less hindered right we don't have a methyl in the way so this is our kinetic enolate which means it's formed more rapidly why removal of less hindered proton is faster especially when we're using a hindered base like LD a right doesn't that make sense it's a hindered base we specifically chose that as a base because we don't want it attacking the carbonyl it's a non-nucleophilic base but it makes sense that we would want it to be it would make sense that that hindered base is going to more easily remove a proton on the left-hand side because there's no methyl on the way by just in conditions it's possible to establish either kinetic or thermodynamic control and so I'm going to do this a little bit differently than the book departing from the book a little bit because I don't agree with what the book is saying on this I'm so and I'll show you what I'm talking about we definitely agree on kinetic enolate to form a kinetic enolate LDH EHF -78 with a slight excess of Lda I'll explain that in just a minute and so what we do here is the ketone is added slowly to the LDA solution so there is never excess on ketone so the book doesn't talk about the slight XSL da but that's going to give you more of the kinetic enolate and what you get if you do that and you add that ketone to the elds stirring LD a solution slowly you get 99% of the kinetic enolate and you get 1% of the thermodynamic enolate and so your major product after you add an alkylating agent after you add an alkyl halide you get 45 to 47 percent I looked these fields up online notice the poor yield so even though you get high selectivity you get a poor yield in this reaction all right and so the product composition here is determined by the rate of formation relative rates of formation of each product so since it's easier to remove a proton on the on the least substituted side um you're going to get most of that product here product composition is determined by relative rates of formation of each product so that's what that's that we favor the kinetic enolate how do we favor the thermodynamic enolates sometimes we're going to want the thermodynamic enolate let's talk about that on the next page all right so we know that when we have a thermodynamic conditions for a reaction we have conditions that allow equilibration all right and once once the equilibrium is formed will have a thermodynamic mixture it doesn't does in so if we live we allow equilibrium to reach will have a thermodynamic mixture and we have but we do have to allow equilibration to form and and to do the kinetic enolate we're deliberately not allowing the equilibrium to form so ideal conditions which anyways can be entered to rapidly strong base protic solvent or slight excess of ketone and warmer temperature so remember when we had kinetic and thermodynamic products last quarter we had the kinetic forming at low temperature we had the thermodynamic at a higher temperature so if you do this at a higher temperature and this time we're going to have a slight excess of ketones so we have a real a proton source here to allow for equilibrium then we're going to be able to form the thermodynamic product and so in this case here the LDA is added to the ketones so ketone is always in excess LD a added to the ketone so opposite here so um ketone is always in excess so Lda THF 20 degrees slight excess of ketone on the other conditions our sodium hydride room-temperature also with a slight excess of ketone so you can do that also and so let's do it let's divide these guys up let's have blue for these conditions here we'll call those conditions 1 or we'll have let's see nice a bunch of clicking pens here those would be our second conditions here so do you mind art so if you choose method 1 if it will let me choose method 1 it's not letting
me oh no what just happened what do I do here what is it is that it right there okay
cool it's like hanging off the edge there it doesn't want to let me do this hmm all right what color we have here this is conditions for number 1 even though it's it's green it's not gonna let me change it maybe you want to get lower in this page it well 22 percent kinetic and 78 percent thermodynamic not great is it 78% 22% for the second condition using sodium hydride a little bit worse 26% and 74% that is the thermodynamic mixture there's nothing you can do to change that you can't make it all thermodynamic no matter how hard you try that's the kinetic mixture they're very close to each other okay so now let's say we add and this great lq alkylating agent that has no no beta hydrogens so no elimination this will be our major product but low yield you will get product from the third the kinetic enolate you definitely will so I would call this poor selectivity for both of these poor selectivity for both of these and I also want to say that the product composition is determined by the relative stability z' of the products in contrast to the kinetic conditions where the product composition was determined by how fast the reaction took place okay so completely different here so the question is what's the big deal with the excess ketone why do we have excess ketone how does that allow a collaborating conditions to take place let's look at that so let's say we have some excess ketone here we've got the methyl here we got the hydrogen right here so we've really formed the enolate let's say we form the kinetic the kinetics going to form faster but now we can equilibrate and so what happens is electrons come down this is going to come and grab that proton and then we can push electrons up onto oxygen so with a little bit of excess ketone presence present we can now form the thermodynamic enolate from the kinetic enolate this can also be done using a proton source as solvent but neither one of these solvents LD AR and a sodium hydride you can you can't use either one of those with a protic solvent so that's a completely different story altogether and then we have that and then we have there's our little bit of excess ketone back again so just a little bit of excess ketone is enough to allow this to ??quilibre to get our thermodynamic mixture which is approximately 3/4 the thermodynamic enolate and will end 1/4 the kinetic you know age all right so the book uses sodium ethoxide in an ethanol to form the thermodynamic enolate and this is the way I learned it but it doesn't really work very well and I'll show you why so near my sodium ethoxide it's not strong enough to quantitatively remove the alpha hydrogen we know we looked at the direction of equilibrium here already but let's do it again so this is what the reaction would look like we have our acid on this side which is PKA about 20 and we have our acid on this side PKA about 15 so as you can see at equilibrium in this so this is where this really comes in handy this at equilibrium thing at equilibrium we have 10 to the 5th ketone to everyone annoyed alright so that means that if you want to alkylate that position you have to kind of throw everything together like this so these things are all together so not as the book will do step 1 Naoe tea and ethanol followed by our alkylating agent because we really we've only formed a small amount ok so what's the problem with that with everything altogether we're only forming a small amount of being late the problem is is that while we're all this enol aid is trying to form we also have side products that can happen the alkyl halide can just as easily react with the alkoxide iron right
you can do an ex sn2 reaction backside attack kick off the leaving group so that's one side product that can happen that's why we get poor yields second side of a product that can happen is an aldol reaction which we're gonna learn about in chapter 24 so you'll see that coming up and the other thing that can happen is reaction with the kinetic annoying so you know we are still forming kinetic enolate faster it's equilibrating but if the kinetic enolate forms in the presence of our alkyl halide then it can react with it so we're going to have all these possible things happening so it's not a very good reaction so I'm going to give you some alternatives that would rather you use instead and I will ask you for alternatives on the exam so I want you to be able to do this so alternative number one is to rather than making an e lights from a ketone you make an enamine okay so remember any means chapter 21 so here's a the very common secondary amine that's used to make an enemy this is called parola diene I would say that's the most common one okay so there's your enamine right and if you if you look at the structure of analyte is it you mean isn't it kind of reminding you of an enolate ion right so on this guy can isn't this guy can attack electrophiles let's show you what the arrow pushing looks like and she'll see the similarity here to have an electrophile here are generic electrophile electrons on nitrogen come down then we attack the electrophile and that's very similar to an enol a tie on doing this very same thing so here's our generic electrophile coming in electrons come down we come and grab the electrophile do you see the similarity there here and there you know in the left hand case here we have the electrons on nitrogen coming down and then we have this grabbing the electrophile here we have the electrons on oxygen coming down and grabbing the electron it's going to still do the same thing but it actually works a lot better in much higher yields so here's the typical reaction we know how to make a need to me h3o plus catalytic or you can write pH 5 remember that pH 5.0 so you make an enemy so I'll be using any means in this chapter we'll also be using ena means in chapter 25 so expect to see ena means on this next test coming up okay so here's our we're using ethyl bromide electrons on nitrogen come down we're going to backside attack detect the carbon bonded to the leaving group we're gonna kick off the leaving group this isn't a minium ion here that forms similar to a carbonyl we now have an ethyl in the Alpha position and now we had HCL h2o and we go from the iminium ion back to a ketone we did the mechanism for that we did we did the Emmy and this is going to look just the same for going from an iminium ion back to a ketone we did that with an amine and so I could possibly ask you that mechanism on the test so advantages better yields on symmetrical ketones can be selectively alkylated at the least substituted side and ena means can also be a slated so if you have a an unsymmetrical ketone then the ena mean is going to form you're going to form that what is equivalent to the kinetic enolate it's going to go exclusively on the less substituted side so it's an alternative for a kinetic enolate and then when we add we add this we add our alkyl halide and then we get rid of the iminium ion that forms we get a ketone very much much much more cleanly much higher yield ethyl on the left hand side that's all on the right hand side all right so where does the selectivity come from the end the less least substituted you mean is more sterically favorite conjugation between the nitrogen and the pi bond means that the bolded bonds must all be in the same plane and you can see that if we try to form it on the other side here all these bonds bolded bonds are in the same plane if we try to form the eena meena on this side can you see the problem between the methyl and these hydrogen's here so that's why we're going to exclusively form the uni mean on this side so we'll stop right there
and we will continue this some on Wednesday