Reactions at the Alpha-Carbon

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Video in TIB AV-Portal: Reactions at the Alpha-Carbon

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Reactions at the Alpha-Carbon
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14
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27
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CC Attribution - ShareAlike 3.0 USA:
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2015
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English

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Abstract
This is the third (and final) quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. Index of Topics: 00:21 - What Determines the Percentage of Enol at Equilibrium? 05:30 - How would you completely deprotonate acetone? 10:27 - Mechanism of Keto-Enol Tautomerization: Base Catalyzed 15:55 - Mechanism of Keto-Enol Tautomerization: Acid Catalyzed 24:07 - Reactions at the Alpha-Carbon 26:39 - Racemization 31:38 - Deuterium Exchange 36:54 - Acid Catalyzed Halogenation of Aldehydes and Ketones 41:07 - Base Promoted Halogenation of Aldehydes and Ketones
Setzen <Verfahrenstechnik> Activity (UML) Resonance (chemistry) Ketone Carbon (fiber) Carbonylverbindungen Organische Verbindungen CHARGE syndrome Alpha particle Acid Computer animation Electron Functional group Acid dissociation constant Enol Covalent bond Chemical compound Chemical structure Base (chemistry) Cobaltoxide Conjugated system
Sense District Alkyne Activity (UML) Nitrogen fixation Resonance (chemistry) Butyllithium Alpha particle Electron Sodium hydroxide Acid dissociation constant Lithium Conjugated system Stickstoffatom Hydroborierung Ketone Peroxide Hydrogen bond Carbon (fiber) Carbonylverbindungen Protonation Chemical reaction Lone pair Hydrate Hydrogen Tautomer Acid Computer animation Chemical compound Amination Acetone Base (chemistry) Cobaltoxide Thermoforming Methanol
Alkyne Sense District Ionenbindung Kohlenhydratchemie Resonance (chemistry) Hydroxyl Ether Reactivity (chemistry) Anomalie <Medizin> Alpha particle Electron Aldehyde Ketone Hydrogen bond Carbon (fiber) Carbonylverbindungen Protonation Chemical reaction Hydroxide Hydrogen Tautomer Acid Computer animation Resin Enol Covalent bond Chemical compound Chemical structure Base (chemistry) Cobaltoxide Thermoforming
Enantiomere Toluene Resonance (chemistry) Chiralität <Chemie> Alpha particle Electron Lactitol Mixture Setzen <Verfahrenstechnik> Selenite Walking Carbon (fiber) Carbonylverbindungen Protonation Chemical reaction Hydroxide Hydrate Water Tautomer Hydrogen Drop (liquid) Acid Computer animation Enol Chemical compound Base (chemistry) Cobaltoxide Deuterium
Ethylgruppe Ionenbindung Kohlenhydratchemie Carboxylate Resonance (chemistry) Carbonate Chlorine Methylgruppe Alpha particle Electron Halogenation Protein Transformation <Genetik> Elimination reaction Body weight Hydride Ozonolyse Bromide Chloroform Induktiver Effekt Ketone Walking Carbon (fiber) Carbonylverbindungen Katalase Protonation Chemical reaction Hydroxide Water Hydrogen Hypobromite Tea Acid Gesundheitsstörung Computer animation Functional group Enol Base (chemistry) Cobaltoxide Carboxylierung Thermoforming
Computer animation Ketone Methylgruppe
all right we're gonna keep going on chapter 22 everybody got their exams in the Dropbox I will be posting the key on the fifth floor of Roland Hall across from the O chem stockroom on this afternoon later this afternoon are there any questions before we get started anybody all right all right we were talking about acidity of the alpha carbon for carbonyl compounds we have regular carbonyl compounds and then we have active methylene compounds so an irregular carbonyl compound here like this ketone this has a pKa of about 20 if we added a second activate so this is an activating group if we had a second activating group the pKa it becomes much more acidic so we were showing how the direction of equilibrium has completely changed based on which type of compound you have so this has a pKa of about 20 and so we showed the ratio equilibrium 10 to the fifth to one and now let's do the same thing using the same base with an active methylene compound all right so when when we made the base when we when we made the conjugate base here we had one oxygen to push electrons on to and here we have we have one aqua we have one carbon eel here we have another carbon here so we have two different oxygen to remove to to put that charge on to so let's show the resonance structures for this let's go over here to the right so that would look like that notice in this residence structure the other carbonyl is conjugated with the double bond of the enol 8 and that we can now go ahead and move those electrons all the way over on to the other carbonyl and in that resonance structure we also have conjugation with the other carbonyl and the double bond of the enolate ion so as you can see we have something that's much more resonant stabilized so
highly resonance stabilized PKA is nine that makes good sense PK's nine so it's much more acidic because our conjugate base is much more stable what's the ratio at equilibrium let's see let's take a look here all right our acid on this side of the equation is a pKa of nine let's draw our conjugate acid over here that would be methanol PKA 15 and as you can see the equilibrium is favorite favored in the
opposite direction now we're favored to the right ratio I'd equilibrium what would it be 1 to the 10th not 1 to 16 ah I was saying it by writing the opposite thing want to tend to this 6th okay so we've completely switched the equilibrium so with it with a ketone you can't completely remove the acidic alpha hydrogen with methanol you're only going to form a very small amount of the product 1 for every 10 to the fifth of the of the untouched ketone if however you have an active methylene compound so essentially two ketones then now we've completely flipped things we have now 10 to the 6 deprotonated compounds for every non deprotonated so ratio completely switches using the same exact base alright so what if we wanted to completely deprotonate acetone we would have to use a stronger base so active methylene compound don't need a stronger base with non active methylene compound we need a stronger base and the most common base that we're going to use is something called LD a you are feel free to use that abbreviation lithium diisopropyl emit is our base and how you make this is you take an amine and you deprotonate with n butyl lithium so this is n butyl lithium and so what we have here just an acid-base reaction in order to deprotonate Amin you need a really strong basis so that's our really strong face and so there is our Lda all right so here's a typical reaction here on the next page ah let's just change this to a two and draw out that carbon hydrogen bond and we need two lone pairs don't we there's two lone pairs on that nitrogen so let's add another lone pair all right so we're gonna grab those we're gonna grab that hydrogen now where are we going to move those electrons we can move them onto carbon or we can move them all the way up onto oxygen we can move them onto carbon and then draw the resonance structure where they're pushed all the way up onto oxygen I'm just gonna and I'm going to go back and forth because you don't have to draw both resonance structures every time and in fact that you can go straight to pushing electrons up onto oxygen if you like and let's just make sure that equilibrium is favored so we're going directly to the resonance structure where the negative charges on oxygen and then you get Dyess approp amine and i drew that one wrong fix that and now let's determine the direction of equilibrium and make sure that that base is strong enough PKA is about 24 the ketone for this for this particular I mean this is our acid on this side of the equation this is a pKa of 40 so as you can see we go from stronger acid to weaker acid very long arrow to the right very short arrow at equilibrium we have what do we have one one acetone for every 10 to the 20th deprotonated acetone so why do we use Lda instead of other bases it is a strong strong nucleophilic base strong non-nucleophilic base sorry about that it is to hinder to act as a nucleophile and that's important because we know all the reactions that can happen when you when you attack carbonyls so this is what you would form if it attacked as a nucleophile but it's too hindered and so you don't get that so we'll put a big X through that you don't form that so you see how you you wouldn't want to use you have to be really careful with the bases you choose you wouldn't want to use butyl lithium to remove that alpha proton would you what would happen instead it's gonna it would attack the carbonyl so you're gonna have that competition there okay so let's talk about the mechanism for keto-enol tautomerization and you already know a little bit of this right you know the acid catalyzed mechanism if you go all the way back to chapter 10 where we died alkynes hydration of alkynes and that was one of the mechanisms where I said this is guaranteed to be on the midterm and that's why because it could she prepares you for this material here so we'll do base catalyzed first and base catalyzed is what you get when you do hide when you do hydroboration of an alkyne and then you follow that with peroxide sodium hydroxide remember that you've got tautomerization that would be
base catalyzed when you hydrate and alkyne that's acid catalyzed so we're going to talk about both of them here all right so do base catalyzed first we'll use her doctor hydroxide as our base so notice that that time I left electrons on carbon I could also push them all the way up onto oxygen and we'll be doing that also and so unless I tell you to draw both resonance structures you don't need to worry about doing that so here I'm going to go boom here push electrons up on the oxygen to draw the second resonance structure so we're going back to the other direction aren't we and it's back in chapter 10 we went we went from the enol to the keto form now we're going from the keto to the enol form so keto and enol tautomerization just back the other direction so as we start already for most simple ketones the keto form is going to be more stable than the enol form nevertheless the enol is a crucial intermediate in a variety of reactions with ketones and aldehydes so even though it's not favored as long as we're forming a small amount of it we can do reactions with it okay doesn't have to be completely favored we can form a small amount of it and do do reactions which we'll see coming up all right and so here's our resonance stabilized enolate ion all right so that's point number one point number two is the Eataly ions a powerful nucleophile this is the reactivity that we're going to see and really this reactivity you can show from either resonance structure I've got two resonance structures here you can draw from either resonance structure and since that's nucleophilic carbon I'm going to have my generic electrophile here or we can draw it from the other resonance structure and what I do is I leave it up to you to decide which one you like better and so as we're doing mechanisms in this part of the in this chapter are going to be going back and forth from using either one of those resonance structures so they are pushing is going to look a little different depending on which resonance structure that you're going to work from here we're going to attack just directly from carbon if you draw this resonance structure you have the electrons on oxygen come down and then we attack the electrophile like that so those are the two that's the two reactivity their activity that we'll see from either resonance structure so notice what we've done we've incorporated an electrophile into the Alpha position and that's the key thing here electro file and I'm off the page here to the Alpha position by the way I did post those podcast likes last night that you are really looking forward to watching for a carbonyl nomenclature okay okay that's base catalyzed questions on the base catalyzed mechanism anybody all right let's do acid catalyzed what do we almost always do first in an acid catalyzed mechanism pretty the carbonyl first yep protonate the carbonyl and I'm going to also draw this two ways and you get to pick which way you think looks better that's resonance structure number one and we can move electrons onto oxygen to draw a resonance structure number two all right so that's a resonance structure those are two resonance structures for the protonated ketone and now we need to remove the alpha hydrogen our base is very weak but we've enhanced the reactivity by protonating oxygen so I'm going to draw this from ether resonance structure and you can choose which one you like the best in this one here we're going to remove the alpha hydrogen here electrons from the carbon hydrogen bond move over here to make a new double bond and then we push electrons up onto oxygen and that looks a little bit as some students don't like that method because it looks like we're removing a you know a hydrogen from a
carbon and with a really really weak base and that that bothers some students okay so the other way to do this and this will probably look for more familiar to you as we just remove this and you can see what what reaction is this that we saw in chapter 8 that's just an e one right and an e one you can use a weak base so this is you can see it may be so so a lot of students like the second way better because that's an e one and now it looks it's not so bothersome that we're taking and removing a hydrogen from carbon because we have that carpet we have the carbo cation the adjacent carbon so but you get to choose so acid-catalyzed again I literally just touch this thing and it Scrolls it like a huge amount alright so a needle has electron-rich double bond with electrons from this double bond attack and electrophile the resulting cation has stabilized through resonance with the oxygen atom and so this is the reactivity that we will see all right so we've electrodes on oxygen like tons in oxygen are going to come down we're going to grab the electrophile which is going to go into the beta position so there's our electrophile in the beta position and certainly that is resonance stabilized so it looked like that the other way that you can do this well well let me draw the product first and then we'll show the other way you can do this so well now I'm not gonna draw the product what are you doing the next step and we'll see that coming up is deprotonate but the other way that you can do this that you might like better if you're remembering 51b okay so it looks like this you could have the we could just have the arrow come from the PI bond so we did this when we were hydrating alkynes electrophile goes on the side where you're going to get the best carbo cadion the best carbo cation is where the carbo cadion is on the carbon that's bonded to the hydroxyl and then remember we did so in so in chapter 10 our electrophile was an acid which was a proton okay and if you do it that way maybe that makes more sense for a lot of students and let me fix that that should be electrophile here so that's how we're gonna incorporate an electrophile in the Alpha position with an acid catalyzed mechanism let's take a little just do a little show of hands for these last two who likes this one better who likes this one better more of you yeah so you can draw it either way okay excuse me I'm missing a positive charge which purple resin that one oh yeah I am a missing a positive charge will do it green we'll do a green positive charge there okay question yeah you know what we changed this again whatever let's let's yeah you want it would you like this better is that what you're talking about all my green Corrections let's just draw them in here okay are we good now all right so reactions at the alpha carbon we're gonna have acid catalyzed we're going to have base catalyzed we have different reactive intermediates in an acid catalyzed reaction are our reactive compound is an enol in base it's an enol eight ion okay so you got to remember that in base this is what we're going to see if you use that resonance structure or if you use this resonance structure it's going to look like this so enolate enolate ion is going to be our reactive intermediate in acid it's going to be an enol so let's draw the two ways to show that so we can have the electrons on an oxygen come down grab the electrophile like that or we can do the electrophile the double bond attacked the electrophile make the resonance structure that has the positive charge on the carbon that's bonded to the hydroxyl and then of course we can draw the resonance structure there okay so those are your choices but you want to keep it straight if you're in an acid catalyzed mechanism you don't want to make an enolate ion if you're in a base catalyzed mechanism you won't need to make the enol right I mean you could but you don't need to make that you know because the reactive intermediate is the
you know I mean it's the enolate ion okay let's talk about randomization so because when we have formation of eating halls and then we go back from enoki tol tautomerization notice we have reversible arrows that reactions going back and forth and back and forth and back and forth even if you're only forming a tiny amount it's still going back and forth and back and forth so what that means is if you have a chiral alpha carbon you have a really big problem you're gonna Rasim eyes your chiral alpha carbon so let's take a look at that all you need is a trace of acid or base and you literally don't even need to add it just the inside of glassware is acidic enough to catalyze this reaction so if you're working with compounds like this where we have a chiral alpha carbon you actually have to treat the glassware to get rid of all the acidic groups so that it won't erase okay it's it's a really it's a really big issue so you start off optically pure even if you have really pure solvent just the inside of the glassware is enough so your to get rasterization nobody wants to get rasterization I'll tell you that so we're gonna start I'll drop both enantiomers here racemic mixture so this could be base-catalyzed or acid-catalyzed i'm going to show you another reaction that's similar so for this one I'm for Assam ization I'm going to choose base and then for the Deuter putting a do a deuterium in the Alpha position I'm going to use acid and those are we could those mechanisms will be the same for both all right so let's show how this happens in base all right so if we have let's say yeah let's say we have a little tiny bit of hydroxide around I'm going to go directly to the resonance structures where we move electrons up onto oxygen so you can see the problem what we have here
now we know this is the most important resonance structure so let's see what happens here what just happened to that stereocenter gone boom okay so we go from sp3 carbon 2sp to planar SP 2 and therefore planar and that's not going to stay like that we're going to reverse that reaction and so we but we can week but now that we have a planar intermediate it can reap rota Nate from the top or it can reap rota Nate from the bottom so the water can convince you have this planar intermediate water could come in from the top and it can come in from the bottom if it comes in from the top you form one an answer or if it comes in from the bottom you form the other enantiomer so electrons come down here is it coming in from the top or electrons come down and it can't come in from the bottom so here's the green arrow so protonation on top and bottom gives racemic mixture all right so that's something that would you have to worry about if you have carbonyl compounds that have Al chiral alpha carbons another thing that you can do is you can do deuterium exchange to make deuterated compounds so the Alpha hydrogen's can be exchanged for deuterium and so that's going to look like let's see depending on how long you let this go you can get to tea rooms incorporated into all alpha positions so all what we call enolizable hydrogen's eventually replaced with deuterium let's see how that mechanism works so I did added base-catalyzed it's really the same mechanism except now we're just using a deuterium instead I did that I did the rasterization in base-catalyzed now I'm going to do acid catalyzed for this one here I am not going to show incorporation of all four deuteriums I'm going to show incorporation of one and sorry if I ask you this mechanism on the test do I want to see all four or do I want to see just one just one okay so acid catalyzed mechanism and enol is our nucleophile so we need to make the enol so we wanted we want to do the deprotonation with water then that will regenerate our catalyst notice that our catalyst is going to start to get hydrogen's incorporated into it so it's going to look like that so remember in the acid catalyzed mechanism the enol is the nucleophile so we can use that or we can use some acid that doesn't already have our hydrogen incorporated lecterns on oxygen are going to come down we're going to grab deuterium and we certainly could grab hydrogen just as easily and so basically you have to get this you have to let this go for a long time and you you periodically have to replace your acid if you want it all completely converted to deuterium's and when you do that we have one hydrogen we have one deuterium so we got to keep going this is going to go for a long time and so we're going to repeat three more times so all four hydrogen's converted to deuterium's and then in the last step we deprotonate
so same thing going on here the only thing is we can actually detect what's going on by noticing that we have two Terrans incorporated into the Alpha position in the first reaction we did that was base catalyzed you notice because you lose your optical purity questions on either one of those reactions anybody yeah absolutely I don't want you to do it three times yeah we don't want it great three of that done four times in a row because like what if you do it like three times right one time wrong or two times right one time wrong you know what I'm saying my tea eyes wouldn't you not like me just Dimitri says give you no points okay all right so you can incorporate proton in the Alpha position we've just done that twice that's an electrophile right a proton we can also incorporate a bromine into the Alpha position so let's look at that so here V plus equals bromine in the last two examples that a plus was an acid okay a proton we're gonna do an acid catalyzed mechanism for this so we're gonna do a base catalyzed mechanism we're actually gonna get different products based on acid or base okay you've seen that before right chapter xxi we got different products depending on whether we used acid or base remember that acid tail hemiacetal right so this is like this you get it you get a different product all right so we're gonna do the acid catalyzed first protonate the carbonyl first all right we probably to the carbondale first now we're gonna remove an alpha hydrogen so let me draw in the alpha hydrogen I'll just do ch2 and put a hydrogen here what are we gonna use to deprotonate that with we're gonna use water we do we want to use hydroxide no because hydroxide would be half acid has half cat half acid catalyzed half base cows don't want to do that so I'm going to remove this I'm going to go all the way up onto oxygen to make the enol because we're you know that enol is our nucleophile so that's what we want to make bro means our electrophile so now bromine is ready to come in so it looks like that right same as a proton accept all we're doing is you're grabbing bromine we're breaking the bromine bromine bond our new electrophile has been incorporated into alpha position that's our bromine and then in the last step we can use water or bromide ion to deprotonate your choice questions on the acid catalyzed mechanism so in the acid catalyzed mechanism let's write the product up here now that we've just drawn it let's write it replacement of 1-alpha hydrogen in acid and the acid catalyzed mechanism so got to remember that all right now so what about base-catalyzed own otis I say base promoted rather than base catalyzed so now you know something different is going to happen well sometimes it's catalyzed sometimes it's protein it's promoting we'll see you'll see coming up all right something differents happening here all alpha positions are converted all enolizable hydrogens are converted to bro means and if that's what you're doing on value that actually can be catalytic in bass but if you have a methyl ketone if you have a methyl ketone you get a different product so if a methyl ketone is use you need one equivalent of hydride higher Oh h- so this here can be catalytic you need one equivalent weights - so what's going to happen is the same sort of thing on this right hand side you're going to get CDR three but that's unstable under the reaction conditions and so it's going to react with the stoichiometric hydroxide and then when you follow that with acid you get a carboxylic acid that's a big surprise there and H CBR 3
bromoform so you've heard of chloroform chloroform would have three chlorines in hydrogen this is broma form and so with bromine it's called a bromo form reaction with iodine it also works with iodine it's called what do you think iota form yeah you got it all right we've got lovely we have time to do the mechanism so I know you're really curious how you get a carboxylic acid out of that okay so let's do that right now I don't want to make you wait for that base catalyzed mechanism do we protonate the carbonyl first no we remove our alpha hydrogen and we go directly to our reactive intermediate which is an enol a tie on you can leave your electrons on carbon or you can push them all the way up on down to oxygen your choice you don't have to draw both resonance structures so what I'm going to try to do is I'm going to try to go back and forth sometimes I'll leave the electrons on carbon sometimes they're going to push them up onto oxygen but you have the choice this time I'm wanting to push electrons on the oxygen here so enol eight ion in base is our reactive intermediate that's our nucleophile so what you'll notice is the base catalyzed mechanisms are always a little bit shorter than the acid catalyzed mechanisms so now your bromine comes in electrons on oxygen come down now we're gonna grab a bromine and we're gonna break the bromine bromine bond and you're done see how much shorter that is now that's going to repeat two more times all right so now we've got three bro means incorporated and we're kind of leaving our a mystery but once you have one bromine incorporated the bromine to get to go on to that same side this is unstable so what happens is the hydroxide ion is going to attack the carbonyl so now we're going back into comfortable material that you've been seeing all quarter and we have two potential leaving groups we have hydroxide and we have CBR three - and it turns out that the electrons on oxygen come down and kick off CV R 3 - whoops and that CBR 3 - that you just kicked off can't resist deprotonating that carboxylic acid that just formed so you get a carboxylate you get HCV r3 and then in the second step you add acid at h3o plus in second step and that's how you isolate the carboxylic acid so notice what we've done we turned a ketone into a carboxylic acid we know no other way to do that we've talked about you know I know that you guys are creative and you can think of some ways to do that maybe take that ketone do an elimination then do ozonolysis yeah you can do that this is a one step way to do that that gives you the carboxyl so remember that transformation keep methyl ketone two carboxylic acid if it's an ethyl ketone will this work no it has to be a methyl ketone all right so you're I know you're dying to know how we can have a carbo cation as an intermediate right not a carbo-cation a carb and i what am I even saying here carb anion okay I can see you're really upset about that okay it turns out that if you think about this particular carbon ion those bro means are actually removing electron density from the carbon pulling electron density away and that stabilizes a carbon ion so this is a stabilized carb anion inductive effect of the three bro means the three bro means makes carbon less negative and it turns out that you need to have all three of them if you only incorporate two halogens it's not a good enough leaving group so that's why you
need to have a methyl ketone we'll talk more about this next time
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