hello everyone and welcome to the lecture on enals and emailings in the third edition of klein this is chapter 21. okay so we're going to start talking about what an enol and enolate is but before we can do that we need to go over some definitions so i want you to call your attention to the carbons in this ketone the carbons that are directly attached to the carbonyl carbon are called the alpha carbons carbons directly attached to those are the beta carbons and then the gamma carbon and then the delta carbons so the reactive carbon that forms an enol or an enolate are the alpha carbons and so those are the ones we're going to spend the most time talking about so yeah i want you to recall that i said that there were two alpha carbons on the ketone in the previous slide and so oftentimes there will be two alpha carbons particularly with a ketone but they won't necessarily have hydrogens right for example if we take a look at this ketone the alpha carbon on the right has two hydrogens but the alpha carbon on the left does not have any hydrogens you cannot form an enol or an enolate from an alpha carbon without any protons attached to it and the reason will be evident when we start looking at the mechanism form an enol or an enolate here's another ketone in this case it's cyclic again there's two alpha carbons with ketones this position has two hydrogens so it's reactive to form an enol or enolate the other alpha carbon does not have any hydrogen's attack attached the position is blocked with methyl groups so this would be a non-reactive alpha carbon for this chapter let's look at a aldehyde as an example aldehydes have that characteristic carbon hydrogen bond between the carbonyl carbon and a hydrogen directly attached so they only have one alpha carbon and as long as we have at least one hydrogen in this position there are actually two only one is drawn in and we will be able to form an enol or an enolate from this substrate let's look at the difference in structure between an enol and an enolate notice that the enol has a alcohol directly attached to a carbon-carbon double bond or an alkene the name is derived from alkene all and we form it from a ketone or an aldehyde that along with the alpha position that has at least one hydrogen attached using catalytic acid or base if we treat a ketone or an aldehyde with an alpha hydrogen that has at least one hydrogen with a strong base instead we will form what is known as an enolate enolates have a full negative charge that is shared between the oxygen and the alpha carbon either way the alpha carbon has now become a nucleophile and it can react with a variety of electrophiles that we'll see in this chapter for example a halogen and for example introducing this alpha-halogenated ketone type structure for the product so if you just have a ketone even a just a sample of the pure ketone if there's even a catalytic amount or acid of acid or base around it will exist in equilibrium with an enol and again recall that an equilibrium right is a distinct conversion of one structure to another in this case these are related as constitutional isomers the ketone and the enol and just a catalytic acid or base will convert the keto form to the enol form these two structures are called tautomers the definition of a tautomer is rapidly interconverting constitutional isomers that differ from each other in the placement of a proton in the position of a double bond so recall what constitutional isomers are those are structures that have the same molecular formula that have a different atom connectivity so notice that the hydrogen from this alpha position has been transferred to the oxygen and the placement of the double bond has also moved from between the carbonyl or carbon oxygen bond to between the carbonyl carbon and the alpha carbon you can see tautomers right exist even in cyclic structures with the equilibrium heavily favoring the keto form typically it is more stable because carbon oxygen double bonds are very stable so those will dominate the equilibrium and we'll have a majority of the keto form in solution there are some exceptions to the keto form being the dominant form at equilibrium specifically here's one exception when we have a 1 3 diketone then if the keto form is actually a minor form a minor tautomer right at equilibrium and that's because in this case the enol form is particularly stable it's stable due to intramolecular hydrogen bonding and also the fact that this pi bond is now in conjugation this carbonyl so the series of overlapping 4p orbital system here is very stable it actually additionally extends up to this oxygen or one of these lone pairs is also in a p orbital so it's really a five atom conjugated system and that's a very stable situation compared to two isolated pi bonds here's another example where the enol actually dominates the equilibrium and it is the case of phenol phenol as you might note is typically drawn this way and there's a reason for that it's because 99.99 of the molecules will exist in the enol form not the keto form there's a few reasons for that once we have a pi bond here right the major reason is now we have an aromatic ring if you look over here this hydrogen has this carbon excuse me has two hydrogens attached to it and therefore it does not have a p orbital and this constitutional isomer is non-aromatic so in addition to being aromatic it's also resonance stabilized when you have the fully conjugated system extending up to the oxygen with a series of p orbitals let's go over the mechanism for the tautomerization of a ketone or aldehyde to a enol the mechanism is drawn differently depending on whether it's acid catalyzed or base catalyzed recall that in mechanisms that are acid catalyzed we typically have intermediates that are either neutral or have a full positive charge base catalyzed reactions typically have intermediates that are either neutral or have a negative charge so when you're drawing a mechanism make sure that they can structures that you draw are consistent with the conditions being employed so if we have an acid in solution it makes sense that the acid will donate a proton to the ketone and it can do so by protonating one of the lone pairs on the oxygen of the carbonyl that gives us this resonance stabilized cation where the positive charge is delocalized between the oxygen and the carbonyl carbon from here water can take the proton that is located at the alpha carbon and electrons can flow to form a pi bond satisfying the carbon's electron deficient state in this resonance structure to form a pi bond between those two carbons notice the equilibrium arrows this process of going from a ketone to an enol is entirely reversible and you could draw a mechanism going back from the enol to the keto form let's look at the mechanism of tautomerization of a ketone to an enol under basic conditions so if we have sodium hydroxide that's a base and it's very likely that it can pick up a proton from the alpha position eventually we'll talk about how acidic these protons are that can leave an anion at the alpha carbon that's resonance stabilized and partially delocalized up to the oxygen atom and after protonation with water we get to the enol structure now you might ask why doesn't this hydroxide which is a strong base but is also a strong nucleophile why doesn't it attack the carbonyl carbon we saw that before in a previous chapter where we formed a hydrate and the first step was nucleophilic attack of hydroxide to the carbonyl carbon of a ketone and that of course that can happen but that is also a reversible reaction whether in some cases right we will have nucleophilic attack on the electrophilic carbon uh leading eventually to a hydrate that process is reversible that gets us back to the keto form and sometimes hydroxide can act as a base instead of a nucleophile and when it does that it will take the hydrogens on the alpha carbon because they're fairly acidic and again we're going to talk about that in a little bit so it's safe to assume that if you have a ketone or aldehyde some fraction of those molecules will exist in the enol form and if we look at the resonance structures of an enol we have two resonance contributors and one shows us that there is a partial negative charge on the alpha carbon so essentially we've created a nucleophile at the alpha carbon and we can utilize that nucleophile to react with a variety of electrophiles which we're going to see with the reactions in this chapter this is an electrostatic potential map showing you that while the oxygen is electron rich the alpha carbon actually is also electron rich so let's practice in the mechanism of the reverse process right so from an enol going back to the keto form let's draw the mechanism that would be acid catalyzed and a separate mechanism that would be base catalyzed again to reverse the process and go from an enol back to the keto go ahead and pause the video to draw it and then we'll go over it okay let's go ahead and go over this problem so i'm going to do the acid catalyzed conditions first so whenever you're drawing the acid catalyzed conditions go ahead and draw your starting material along with some hydronium ion now we know that under acid present it's very likely that the acid will donate a proton and protonate our substrate let's think about where we should proteinate should we protonate the oxygen right or is there another position that we might want to protonate well it's possible to protonate the oxygen that doesn't actually lead us anywhere and that process for proton transfer is reversible so it doesn't actually help us to protonate the oxygen but if you recall looking at the resonance contributors of an enol we saw that the other resonance contributor of an enol shows us that the alpha carbon is has a partial negative charge and is nucleophilic so it makes sense to also protonate the alpha carbon so if we use this pi bond as our nucleophile we can go ahead and pick up a proton at the alpha carbon that's going to leave a positive charge at the carbon attached to the oxygen this is resonant stabilized notice the resonance arrow that i'm drawing here and if we want to show resonance we should join our resonance arrow the pattern is lone pair adjacent to a positive charge okay there's the other resonance contributor where now the oxygen has the formal charge so we know that the true structure is that the oxygen and the carbon attached to it share the positive charge we're very close to the ketone all we have to do is do a final proton transfer to remove this proton water is the best base that we have in solution go ahead and draw a water molecule to come and pick up that proton notice that we also regenerate hydronium ion in that step and therefore all we need is catalytic acid because it gets used and regenerated let's look at the mechanism we have a base catalyzing i'm going to go ahead and draw out the starting email again and some hydroxide with reaction with a base it's likely that the base will deprotonate our substrate the most acidic hydrogen is the one attached to the oxygen so it's very likely that the base will remove that proton notice that that actually gives us an enolate and enolates are resonant stabilized so we can go ahead and draw out the resonance contributor the pattern i'm going to draw is the allylic lone pair this shows us that the negative charge is localized between the oxygen and the alpha carbon and both of those have a partial negative charge as a result notice that we're very close to the keto form in this tautomerization and all we need to do is protonate the alpha carbon to get to the geo form water is the best acid in solution so i'm going to go ahead and draw some water in that will provide a proton to get to the product which is the ketone notice that we also form some hydroxide in this stuff so again only catalytic base is needed because it gets used and regenerated in the mechanism let's go back to our slides and move on let's do another practice problem the question is how many enols can be formed from 3-methyl to butanol so this is partially a question of nomenclature right so we have to remember uh how to draw ketones and notice that it is a ketone from the own in the name and then once we have the correct structure drawn to think about how many enols might be formed from that structure recall that ketones have two alpha carbons so that should give you a clue as to potentially drawing more than one enol go ahead and pause the video and try it for yourself and then we'll talk about it okay so again the first part of this problem is figuring out how to draw this structure by the way this is a very common thing to see on the acs final exam they'll often combine some iupac nomenclature into the problem when they're actually asking you for a reaction or something else involved with that structure okay so first i'm going to look at the parent which is a 2-butanone so i'm going to start by drawing four carbons right and a ketone at carbon number two then i'm going to look at the substituent which is a three methyl group if i counted from this side one two three put the muscle up there okay so hopefully you got to this structure and maybe you identified that there's more than one alpha carbon ask yourself do they do both alpha carbons have at least one hydrogen attached the answer is that they do this one has one hydrogen attached and this structure or this alkyl carbon has three hydrogens attached so we can form enolates by transferring these one of these protons to the oxygen and forming a double bond this direction we're transferring one of the oxygens on this alpha carbon to the oxygen and forming a double bond in this direction so the two enolates sorry enols that can be formed you form an enol with the alpha carbon on the right make a double bond between those two carbons and an oh group where the carbonyl was and also we can go the other direction with our double bond placing it between the alpha carbon on the left and the carbonyl carbon along with an oh group so this is one enol and two enols that can be formed eventually when we talk about reactions we're going to need to think about which substrates can potentially form more than one enol or enolate intermediate and what the conditions favor right if one or the other enolane is favored in those conditions okay moving on let's talk about if you're in the presence of a full equivalent of a strong base then an enolate forms you might be confused because we just talked about a strong base forming an enol right with hydroxide ion but remember that that was only a catalytic amount of the strong base when you have a full equivalent of strong base then you get full conversion of the aldehyde or ketone to an enolate intermediate okay recall that an enolate is different from an enol because of the full negative charge on the intermediate where the enol was neutral and enolate has a negative charge it's delocalized between the alpha carbon and the oxygen because of the full negative charge the alpha carbon is more electron rich and a stronger nucleophile than an enol's alpha carbon so that negative charge as i mentioned is delocalized between the oxygen and the alpha carbon so when you treat an enolate with an electrophile the bond between the substrate and the electrophile might occur between the oxygen and the electrophile or the alpha carbon and the electrophile we call this oxygen attack versus carbon attack and we're calling it attack because this is the nucleophile you say that the nucleophile attacks the electrophile so which process is typically favored as it turns out enolates typically undergo carbon attack which is can be drawn in one of two ways mechanistically if you use this resonance contributor and you need to show two bonds or sorry two arrows excuse me to form the product and if you have drawn this resonance contributor only one arrow is needed to form the product either one of these resonance contributors is acceptable and that's because of course we typically just have to draw one or the other resonance contributor with the full knowledge that the lone pair in question is delocalized the majority of the reaction in this chapter involve enolates so it's going to be very common to show a nucleophilic attack of an enolate recall that we talked about the alpha carbon the beta carbon and the gamma carbon all of those carbons may have hydrogens attached to them however only the alpha protons are acidic and that's because removal of one of these alpha protons results in an enolate which is a resonance stabilized conjugate base recall that resonance stabilizes structures so because we're able to form a resonance stabilized conjugate base we'd be protonating one of these two hydrogens attached to the alpha carbon that makes these alpha protons more acidic than the protons at the beta carbon or the gamma carbon a removal of one of these does not would lead to a localized lone pair which is not as stable as a delocalized lone pair that exists in an alien let's do a practice problem the question says draw both resonance forms of the enolate that can be formed from this aldehyde go ahead and work on this problem pause the video and then we'll go over okay so recall that there are two resonance forms of phenolines and it doesn't matter which one we form first because we can always convert one to the other drawing resonance so if we first start with the resonance contributor where the lone pair is on the carbon alpha carbon that would be one resonance contributor and the other resonance contributor right draw move the lone pair and the negative charge to the oxygen so these are the two resonance contributors and it doesn't matter which one you use when drawing mechanisms right the difference is when you're doing a nucleophilic attack with this one it would only involve one arrow from this right substrate and then we would need two arrows to do a nucleophilic attack on the substrate from this resonance contributor all right let's look at the acidity of these hydrogens at the alpha carbon here we have three different structures two are ketones the last one is an aldehyde all of them have carbonyls where there's an alpha carbon with a hydrogen highlighted in red we look at the pka value ranges between about 17 and 19. so recall that that is about as acidic as water in this case and slightly less acidic for the alpha position of a ketone it's good to keep those pka values in mind as we start talking about the appropriate base that needs to be used to form an enolate let's look at acetaldehyde as an example recall from the people on the previous slide that the pka of the alpha proton of acetaldehyde is 16.7 one common mistake is that students think that the hydrogen attached to the carbonyl is acidic on an aldehyde but the hydrogen attached to the carbonyl of an aldehyde is not acidic like the hydrogens attached to the alpha carbon of an aldehyde and the reason why is if we deprotonate this and put a lone pair and a negative charge on this carbon it is not resonance stabilized and delocalized so the carbon has a full um localized lone pair and negative charge and is not stable and as a result this hydrogen is not acidic so be careful not to deprotonate the aldehyde hydrogen okay let's look at the choice of ethoxide as a base for this proton transfer if we draw out the products of the proton transfer they've done just a simple proton transfer delivering the electrons from this bond as a lone pair on the alpha carbon so they've chosen to draw this resonance contributor of the enolate but it wouldn't have mattered if they had moved electrons all the way up to the oxygen and drawn the other resonance contributor would not change the values in this acid-base equation at all which resonance contributor is drawn here the conjugate acid in this reaction is ethanol and ethanol has a pka of about 15.9 notice that these pka values are not significantly different and so what that tells me is that sodium methoxide is not a strong enough base to fully deprotonate the alpha position of an aldehyde and in fact with a very similar pka value the approximate concentration of the enolate and the aldehyde in solution right is about equal amounts or at least there's a significant amount of both the enolate and the aldehyde in solution even if it's not 50 50 we know that both are present still in solution so it's a partial conversion to the enolate instead of a full conversion to the enolate and that's going to become important because later on in section 21.3 we're going to talk about a reaction called the aldol reaction and that reaction involves the enolate attacking another molecule of aldehyde into form of product so when doing the aldol reaction one would choose a base such as sodium hydroxide or sodium ethoxide because the goal is to only partially convert the aldehyde into enolate and have a significant amount of aldehyde remaining at equilibrium such that these two can react together to form a product let's look at some other choices of base instead of ethoxide let's look at hydride that might be supplied by for example sodium hydride hydride is a very strong base and it can deprotonate the alpha position of an aldehyde or ketone to form an enolate when it does so the conjugate acid is hydrogen gas which if allowed to bubble out of solution will be will drive the reaction forward to form more phenolly and there's a significant enough difference in pka value between h2 and the alpha proton of aldehyde or ketone such that it's essentially exclusive and full conversion to the enolate and there is no more aldehyde in solution so we say that it's essentially irreversible another base that we could use is a base that's called lda lda stands for lithium diisopropyl amide and it's formed when di-isopropyl amine is treated with n-butyllithium this forms a very strong and bulky face which is commonly used in as a base in reactions one of the great things about lda is the fact that the nitrogen atom while very strongly basic with a full negative charge also contains these bulky isopropyl groups this is going to make this nitrogen atom non-nucleophilic and it will only function as a base and not as a nucleophile which might lead to undesired side reactions again lda is commonly used it's a strong bulky base and in the presence of an aldehyde that has an alpha hydrogen the acid-base reaction can take place where the nitrogen atom takes a proton from the alpha position we form the enolate and we form the protonated amine we look at the pka difference between the alpha proton of acetaldehyde and the amine there is about 19 pka units difference right between 17 and 36 so this is essentially an irreversible reaction so if we want to fully convert a aldehyde with at least one alpha proton to an enolate we might choose sodium hydride or lda as the base instead of a weaker base such as sodium hydroxide or sodium methoxide or ethoxide let's compare the pka of the alpha hydrogen of a c aldehyde with the pka of the hydrogens the interior to two carbonyls and a 1 3 diketone these hydrogens are actually more acidic than typical alpha hydrogens they are attached to the alpha carbon of a ketone but notice how they are attached to the alpha carbon of two ketones so that makes them more acidic than typical hydrogens in the alpha position and the pka is more like nine we can see the stability of the conjugate base by drawing the resonance structures where the lone pair and negative charge are delocalized between the alpha carbon and both oxygens having two oxygens contribute to the resonance stability makes this structure more stable than a typical enolate and as a result lowers the pka of these protons that are interior to two carbonyls instead of one because these hydrogens are more acidic sodium hydroxide methoxide or eth oxide is actually a strong enough base to almost completely and irreversibly deprotonate the alpha position but not just any alpha position recall that the alpha position that has the hydrogens that are most acidic are the interior to two carbonyl so we would not want to show deprotonation of one of these terminal methyl groups well they are alpha to a carbonyl they are only alpha to one carbonyl the pka of these hydrogens at the end of the molecules is more like again 18 19. so the hydrogens inside are much more acidic sodium ethoxide would selectively protonate a hydrogen in the center of this molecule forming what we would call a doubly stabilized phenolene with a pka difference between the acid and the conjugate acid of about seven that is significant enough that there are significantly more products than reactants in solution there would be no need to use a strong base like sodium hydride or lda to form this phenoline essentially it would be overkill because a much more simple base like hydroxide or ethoxide is sufficient to produce the enolate without need for stronger base so let's review in summary if you're dealing with a simple aldehyde or ketone and treating it with a straw a stoichiometric amount of a strong base such as sodium hydroxide or sodium ethoxide right we're going to convert part of the ketone to an enolate but a significant amount of the ketone or aldehyde will still remain in solution the reason why is even though this is considered a strong base it's not a strong enough base to completely deprotonate the alpha protons because the alpha protons are not that acidic and this is not as strong a base as some other choices if we wanted to irreversibly and completely deprotonate the alpha position and only have the enolate in solution a stronger base like sodium hydride or lda would be used sodium hydroxide or sodium ethoxide in contrast does completely deprotonate the alpha position of a 1 3 diketone to form the doubly stabilized phenolate in this case a strong base like sodium hydride or lda is simply not necessary to get full converting to the animal let's do a practice problem the question says draw the enolate that is formed when each of the following compounds is treated with sodium ethoxide in each case predict whether a substantial amount of starting ketone will be present together with the enolate at equilibrium go ahead and pause this so you can work on it okay let's go ahead and go over that when we are treating with oxide recall that the hydrogen's interior alpha to two different uh carbonyls are much more acidic than the alpha position where the conjugate base is only localized up to one oxygen so with these hydrogens being more acidic with the pka of 9 we should selectively deprotonate one of the two hydrogens on this central carbon showing the mechanism there i'm just going to go ahead and leave the lone pair on the alpha carbon and if i wanted to i could have shifted the electrons all the way up to the oxygen but it's optional just because i'm drawing a mechanism i'm going to go ahead and add in all the lone pairs as well so recall that the pka difference is about 7 the starting material and ethanol which is the conjugate acid so this is essentially irreversible right and we would not expect a substantial amount of the starting ketone together with the enolate at equilibrium so this we're going to say is not pregnant let's look at this structure where we have one carbonyl notice that it is symmetrical so the alpha positions are equivalent you don't need to worry about selecting hydrogen on either side they're essentially going to produce the same thing let's draw in some ethoxide base and in this case i'm going to show you how to push electrons all the way up to the oxygen drawing the other resonance contributor of the enolate directly you protonate the alpha position and then instead of depositing these electrons as a lone pair on the alpha carbon form a double bond between the two carbons and then push the electrons from the pi bond the carbonyl up to the oxygen so that goes directly to this resonance contributor so the question is right what is going to be present at equilibrium i am going to draw a reverse arrow here because we're not going to exclusively get only enolates at equilibrium because the pka difference between the conjugate acid ethanol and the alpha proton of this ketone is not a significantly different right so we're going to say both the enolate and the ketone are present at equilibrium you just for clarity i'll go ahead and join the conjugate acid so in each case we form ethanol right let's do another practice problem in this case it's a question and it asks can you think of any aldehydes or ketones which cannot form an enolate so go ahead and pause it and see if you can draw a structure of an aldehyde or ketone that is uh enolate would not be able to form okay let's go ahead and go over this question so recall that aldehydes and ketones have alpha carbon and in order to form an enolate we need to have at least one hydrogen attached to the alpha carbon so if we come up with a substrate where there is not a hydrogen attached to the alpha carbon then you will not be able to form an enolate from that specific aldehyde or ketone so for example perhaps i put in benzene ring here right now the alpha carbon does not have any hydrogens attached and no enolate can be formed using this aldehyde what about a ketone start with a simple ketone as long as there is a hydrogen at one of these two alpha carbons and enolate will be possible you would need to come up with a structure where there are not any hydrogens attached to these two alpha carbons so perhaps you might have thought of like a tert-butyl group maybe we would put two charcoal groups now the two alpha carbons neither one has a hydrogen attached so no enolate can be formed from this ketone either there's a variety of other potential answers you might have come up with and if you have any question about whether or not it would work go ahead and let me know in office hours or send me and i'll be happy to take a look all right let's move on to section 21.2 alpha halogenation under acidic conditions so recall that under acidic catalytic acid conditions ketones and aldehydes where there's at least one hydrogen attached to the alpha carbon can form an enol and also recall that that enol is nucleophilic at the alpha carbon so because it becomes a nucleophile in its enol form it can react with electrophiles one such electrophile is bromine we've seen bromine act as an electrophile before back in the first semester of this course we saw bromine react with um alkenes to form anti-dibrominated products so in this case bromine will act as an electrophile and then nucleophilic alpha carbon will attack it and will end up within alpha bromo ketone this works also with chlorine and iodine but fluorine is not a polarizable enough electrophile and the reaction does not work well again acidic conditions mean that an enol can be formed and the alpha carbon of the enol is the reactive intermediate so let's look at the mechanism starting with the ketone under catalytic acid conditions we saw the reaction before where we convert the ketone to its enol tautomer once it's in the enol tautomer we have the exposed the nucleophilic alpha carbon so the nucleophilic alpha carbon can then attack one of the bromines kicking off the other bromine as bromide that leaves this intermediate where there's a proton attached to the carbonyl which can be easily removed with water and the product then is an alpha bromo ketone what happens if you use an asymmetric ketone let's look at this cyclohexanone derivative where we have one alpha carbon has two hydrogens and the other alpha carbon only has one and there's a methyl group attached there so the question then is where does the bromine end up does it end up on this more substituted position alpha carbon or does it end up on the less substituted alpha carbon well it turns out that the major product that places the bromine at the more substituted position and that's because it goes through the more stable enol intermediate recall that alkenes are stabilized by substituents this alkene is tetra-substituted and this alkene is only tri-substituted because there's one hydrogen here so that doesn't count so as a result the more substituted more stable alkene is going to be formed in a greater amount and that ends up with the nucleophile placed at this alpha carbon leading to the bromine at that position in the major product we can actually expand this to synthesize an alpha beta unsaturated ketone so the reason why i call this an alpha beta unsaturated is because this is the alpha carbon to this carbonyl this is a beta carbon this carbonyl and there's a pi bond so pi bonds as you know are units of unsaturation so one of the ways we can name this is calling this an alpha beta unsaturated ketone to recall the first step with catalytic acid and bromine we'll place a bromine regioselectively at this alpha carbon and then pyridine will do an e2 reaction removing a proton at this beta carbon and kicking out the bromine as a leaving group generating an alkene between those two carbons if you remember the structure of pyridine right it's essentially a benzene ring where one of the carbons is replaced by nitrogen that nitrogen has a lone pair that nitrogen's lone pair is considered a non-nucleophilic base so it's great to do e2 reactions where we want the substrate added to act as a base and not a nucleophile other bases that can be used for the elimination are bases like potassium tributoxide again a non-nucleophilic base or lithium carbonate same thing let's do a practice problem this is a synthesis problem so it's going to involve multiple steps so go ahead and try to identify the reagents that can be used to accomplish the following transformation don't forget that we just talked about alpha halogenation to see if you can incorporate that into your synthesis again go ahead and pause the video and when you're ready okay so recall that we have been talking about alpha halogenation right and if we have a carbonyl you'll be able to halogenate on this alpha carbon and then form a pi bond through an e2 elimination reaction so right now we only have the alcohol but we of course know that we can oxidize this alcohol to an aldehyde and aldehydes and ketones are the starting material to form phenols and enolates which are the intermediate that we're studying for the reactions in this chapter so again oxidation of a primary alcohol to an aldehyde you can use pcc just uranium for acromi and from here we can do a two-step process where we alpha halogeny catalytic acid bromine followed by either pyridine potassium terpentoxide or and that will again place a bromine at the alpha carbon and then an elimination lead to the pi bond that we see here okay let's move on to a named organic reaction it's called the hell bolhard zielinski or the hvz reaction and it brominates the alpha carbon of a carboxylic acid the conditions used are in the first step bromine and pvr3 and in the second step just simply water or a hydrolysis notice that the yields of these reactions are quite good let's start with a carboxylic acid this is drawn as just a generic carboxylic acid but notice they included the alpha carbon because we're going to need to utilize the alpha carbon to introduce a bromine at that position in this reaction so while we're not going to go over the entire mechanism it's thought that the tbr3 converts the carboxylic acid to an acid bromide in this case and this is an equilibrium with its keto to enol form creating a nucleophile at the alpha carbon nucleophile could then react with bromine similar to the last reaction we just talked about to place a bromine at the alpha carbon and then basil bromides are not stable in the presence of water so once water is added just like an acid chloride an acid bromide or acyl bromide will be hydrolyzed back to a carboxylic acid halogenation can also be achieved under basic conditions sodium hydroxide in the presence of bromine the n we're going to use a stoichiometric amount of sodium hydroxide so an enolate is the reactive intermediate hydroxide will deprotonate the alpha position of a ketone creating an enolate which then re um does a nucleophilic attack on the bromine kicking out bromide and generating the alpha brominated ketone product this reaction under basic conditions is actually not as ideal as the reaction under catalytic acid conditions because under base um basic conditions poly halogenation typically results right so instead of getting just one bromine on the alpha carbon unfortunately all of the alpha hydrogens that are present will likely be brominated in this case the alpha carbon on the left is blocked by two methyl groups so we don't expect to put any bromines here but there are still two hydrogens attached to the alpha carbon on the right and that can lead to a dibrominated product so if this wasn't the product that you were looking for you probably wouldn't want to use the reactions um under the strong base and instead use the acid catalyzed conditions that we saw previously and the reason why poly halogenation occurs is because once one bromine is added there's still a acidic alpha hydrogen and the deprotonation of that hydrogen is even faster than the deprotonation of the substrate because that new enolate where one bromine is already present is even more stable due to the electron withdrawing nature of the bromine helping to stabilize the enolate let's talk about the halo form reaction a halo form reaction is the reaction of a methyl ketone notice that we have a ketone here and on one side we have a methyl group so that's what i am calling a methyl ketone the methyl ketone gets converted to a carboxylic acid notice that the conditions are exactly what we were just talking about for alpha halogenation under the base presence so recall that this reaction has a hard time stopping by adding just one bromine and that all of the alpha hydrogens will eventually get brominated because the product is even more reactive in the reaction than starting substrate so as a result we're going to end up putting three bromines on this methyl group then sodium hydroxide will come in and do a nucleophilic attack kicking off that tri-brominated methyl anion as a leaving group and eventually leading to a carboxylic acid let's look at that mechanism so again we're going to have three brominations of the alpha position of the methyl leading to this tribrominated alpha carbon and then sodium hydroxide will also will then act as a nucleophile instead of a base doing a nucleophilic attack on the carbonyl carbon electrons will get kicked up to the oxygen to avoid exceeding the octet on this carbon that gives us this tetrahedral intermediate well it turns out that once there are three bromines attached to this carbon we can actually kick this out as a leaving group so we can reach from this intermediate we can regenerate the carbonyl and lose the carb anion as a leaving group it's a reasonable leaving group because it's fairly stable where the carbon anion electron density is being pulled by the three electronegative bromines attached to the carbon that's going to reveal a carboxylic acid product remember that carboxylic acids have a pka of 5 so they're pretty acidic and even though this carbine anion is fairly stabilized by induction of the um of the bromines right it is still basic enough to deprotonate a carboxylic acid so the products before the acidic workup are the carboxylate and bromoform and after the acidic workup the carboxylate will be converted back to a carboxylic acid and potentially isolated at that point so for the for this reaction the haloform reaction you can use bromine chlorine or iodine and it is a useful reaction because essentially we're able to convert a methyl ketone to a carboxylic acid notice that we are losing a carbon in our carbon scaffold so if you ever synthetically need to lose a carbon right one way to do that is potentially use this heliform reaction um some of you might have heard of chemical tests and this is actually a chemical test which was a test for certain functional groups that was widely used before other methods were brought forth that allowed for structure determination so before the advent of nmr and other useful instruments that can help us determine structure we sometimes determine structure through a variety of what are known as chemical tests so the way this works is you could test for the presence of a methyl ketone by adding in sodium hydroxide and actually iodine instead of bromine followed by an acidic workup the products would be a carboxylic acid and instead of bromine form you would get iota form which is h c right and one of the reasons why iodine is used instead of bromine is because iota form is uh will precipitate out of an organic solvent as a nice bright yellow solid so the iota form or the presence of a yellow solid in the reaction was a positive indication that we had a methyl ketone in solution the reaction works best if there aren't any other alpha protons so if this r group was for example like a phenyl or a tert-butyl where there aren't any other hydrogens attached to another alpha-carbon because of course if there was hydrogens attached to a second alpha carbon those hydrogens would also be replaced with bromine which might not be what is synthetically desired okay let's do another synthesis problem as a practice problem so i'd like you to go ahead and think about how you might convert this alkene to this ketone right and of course whenever you're doing these practice problems within the chapter think about how we can use the reaction that we just talked about specifically potentially a halo form reaction as a key reaction to in this synthesis again go ahead and pause and go over when you're ready okay let's go ahead and go over this so you might be thinking about the fact that we actually don't need this carbon right it's not present in the product instead there's a oxygen there right so we definitely do probably want to cleave this carbon carbon double bond to form a carbonyl and you might recall that we can use those analysis to do that that might be a good starting place to think about if there's any way we could proceed after getting a carbonyl instead of an alkene so then ozonolysis of an alkene to produce a ketone utilizes ozone followed by a reducing agent such as zebra and methyl sulfide once we have this carbonyl we're almost there to the product but we do need to introduce another carbon somehow we talked about the fact that we were just talking about alpha halogenation in the haloform reaction specifically so maybe there's a way to utilize that in this synthesis so let's think about what if we use the halo form reaction to convert this methyl ketone to a carboxylic acid would that be helpful to get closer potentially to the product might seem counterintuitive to remove the methyl group and then reintroduce an ethyl group but it actually can be a synthetically fairly straightforward process so again the conditions for the haloform reaction are sodium hydroxide and bromine chlorinator id doesn't matter which one you choose followed by an acidic workup in order to isolate the neutral carboxylic acid this brings us back to our previous chapter where we studied reactions of carboxylic acid derivatives so recall that if we want to react a carboxylic acid to form a ketone it would be best to first convert the carboxylic acid to an acid chloride and acid chlorides are much more reactive carboxylic acids and we do that using thionyl chloride once we're at the acid chloride we can convert that to a ketone using a gilman reagent so because we want to add an ethyl group we're going to go ahead and add a gilman reagent where the copper is attached to ethyl verbs and that allows us to directly from an acid chloride to a ketone with whatever r group that is present in these parentheses if you have studied uh this chapter in um in more detail and perhaps you're looking at this um for a second time you might realize that there are other ways to convert the methyl ketone to an ethyl ketone specifically through what's called the alpha alkylation reaction i specifically didn't use alpha alkylation because we haven't covered that yet in section 21.2 but as you study this reaction this uh chapter further you will note that there are other ways to to go from a methyl ketone to an apple keto but this is one pathway that would work out and it utilizes some reactions from this chapter and some review reactions from previous chapters okay we're moving on to section 21.3 which is aldol reactions so remember when we were talking about acid-base reactions and we were treating aldehydes with hydroxide or an alkoxide like sodium ethoxide an equilibrium formed where significant amounts of both the enolate and the aldehyde are present and when that occurs the enolate can act as a nucleophile and attack the aldehyde and the product of that reaction is a beta hydroxy ketone or aldehyde and it's called the aldol reaction named after the product which has both an aldehyde and an alcohol and the reason it's not called a ketol is because it's much more common to do the aldol addition reaction with aldehydes and ketones because a greater yield of products is obtained with aldehydes compared to ketones so again the reaction involves starting with an aldehyde it should have at least one alpha hydrogen conversion with base to an enolate that can then attack um another molecule of aldehyde and we'll see the mechanism eventually forming beta-hydroxy aldehydes so the mechanism um first involves the base deprotonating the alpha proton to form an enolate and then we mentioned of course that the enolate is nucleophilic at the alpha carbon and we know that aldehydes are electrophilic at their carbonyl carbon so a nucleophilic attack can occur between those along with the resonance arrow to avoid exceeding the octet on this carbon from here we can add protonate the alkoxide that results with the best acid in solution which is water to form the beta hydroxy aldehyde product notice the equilibrium arrows that are being used in the mechanism all of these steps are reversible and it's very possible to go back from the product to the reactants in this reaction so it's in a reaction that needs to be controlled through an equilibrium process and just note that two of these aldehydes are required to form one aldol product so with most organic reactions the substrate is one to one with comparing the substrate in the product right in terms of mole ratio but in this case uh it is a two to one reaction which is fairly uncommon for organic reactions okay so for most simple aldehydes for example eutenel the equilibrium favors the aldol product um you can note about three to one for the energy so a good yield of the beta hydroxy aldehyde product can be isolated but for most ketones the equilibrium favors the reverse reaction or the retro reaction and that's because the product is not quite as stable due to increased steric hindrance here as a result of using a ketone instead of an aldehyde so again we talked about the fact that the reaction is reversible and in reverse we would call it a retro-aldol reaction so the mechanism in reverse is simply forming the same intermediates as you would form in the forward direction so again the retroaldol is favored for ketones so we're going to take go ahead and take a product that's formed from reaction of two ketones in this case acetone to draw the reverse reaction so in reverse hydroxide which again is both a strong nucleophile and a strong base it can do several things and to this substrate right one thing it can do is it can do a nucleophilic attack on this carbonyl and maybe eventually form a hydrate this carbonyl and while that's happening in solution that is not actually leading us back to two acetone molecules so that's not what we're going to draw in our mechanism what else could hydroxide potentially do well there's a ketone here and it has alpha carbons those alpha carbons have hydrogens attached to them as we know those hydrogens are fairly acidic and can be deprotonated with a base such as hydroxide so it's quite possible that hydroxide might be protonate one of the hydrogens attached to these alpha carbons but again that process is also reversible again act as a base but instead of deprotonating the alpha position of the ketone it could deprotonate the hydrogen attached to the oxygen in the product that leads us back to this alkoxide intermediate which can then essentially reform a carbonyl and kick out an enolate as the leaving group so notice that we formed we essentially broke this carbon-carbon bond forming two molecules from one and the leaving group is an enolate and the reason why we can kick out an enolate because it's again it's fairly stable due to the resonance uh delocalization and stabilization of the lone pair and negative charge if the enolate picks up a proton from water we get back to the other keychain so this is entirely a reversible process both the forward and reverse directions let's do a practice problem at this point i would like you to predict the major product of change when the following aldehyde is treated with sodium hydroxide so always look at the section that we're in we've been talking about aldol addition reactions so we're going to be drawing the aldol addition product and two of these aldehydes are combined right to form one aldol product go ahead and pause okay so we are going to be drawing the aldol addition product and again you know that we need to utilize two aldehydes so i'm just going to go ahead and draw a second substrate if it's a two to one uh reaction one of the substrates will form the enolate so it'll be deprotonated sodium hydroxide i'm not going to draw the full mechanism here because i wasn't asked to draw the mechanism i was only asked to predict the major product so i'm just going to go to one of the resonance structures of the enolate and just go ahead and directly uh form the enolate and then i know that i'm going to be doing a nucleophilic attack between the alpha carbon of the enolate and the carbonyl carbon of the aldehyde pushing electrons up to this oxygen after the acidic workup we'll get an oh or an alcohol at that carbon so what we can do here is just draw one all right draw our new carbon-carbon bond that we're forming in this reaction and i can join the rest of the carbons and draw an o-h so if we just line these two aldehydes up essentially one on top of the other form an enolate from the top one imagine the nucleophilic attack that takes place right and then draw the product that results it's fairly easy to visualize make sure that you are getting a beta hydroxy aldehyde as your product because that is the product of an aldol addition reaction it's not always drawn this way so sometimes it might be drawn in a different fashion but don't get too confused by that for example if i wanted to i could redraw the structure and have it look like this this is a more traditional beta hydroxy aldehyde where functional groups are sort of a horizontal more traditional way to look at it right but those are really the same structure okay so just sometimes we draw it this way because it's more easy to visualize how that forms using two aldehydes okay so now let's talk about the aldol condensation reaction recall we were just talking about the aldol addition reaction which produced the beta hydroxy aldehyde so an aldol condensation forms when you heat the reaction mixture so note that before when we were doing the aldol reaction addition reaction we just used sodium hydroxide without any heat if we want to convert the product we can just continue with the sodium hydroxide treatment and heat the reaction up or we could isolate the product and then switch to acid and heat that up so either catalytic acid or base will convert the aldol addition product to the aldol condensation product it's a condensation reaction because we have expelled a small molecule in this case water and form a double bond so how would we classify this product right recall that the product would be called an alpha beta unsaturated aldehyde or ketone in this case albin right so um if we want to get the aldol addition product we would use sodium hydroxide at room temperature we wanted to go all the way to the condensation product we would heat the reaction mixture up and and expel water okay so just uh to make sure everybody understands the aldol addition reaction utilizes two aldehydes or ketones but usually aldehydes right to form the out beta hydroxy aldehyde product that can react further to the alpha beta unsaturated aldehyde and expel a water when the reaction typically when the reaction is heated but there is an exception to that that we'll talk about in just recall the mechanism of aldol addition so the first step is proton transfer where the sodium hydroxide forms phenolate by deprotonating the alpha position of the aldehyde the enolate then attacks another molecule of aldehyde that's still present solution at equilibrium after the proton transfer that leads to an a beta alkoxide to this aldehyde which can be protonated using water resulting in the beta hydroxy aldehyde product the condensation reaction just reacts the out addition reaction product further by utilizing an additional hydroxide to then deprotonate the alpha position between the aldehyde and the alcohol forming a resonance stabilized enolate that then expels hydroxide in an elimination process as the leaving group generating an alkene so i want you to be careful here that when drawing the second part of the mechanism to get to the condensation product make sure you draw this as a two-step process because it is not thought to be an e2 reaction an e2 reaction would involve hydroxide deprotonating the alpha carbon forming in pi bond or alkene directly and using the leaving group all in one step if the reaction were to be e2 we would see second order kinetics for the reaction and we do not see that instead we see first order kinetics so it is thought that hydroxide and also the concentration of hydroxide has no effect on the rate so it is thought that hydroxide is not involved in the redeterminant step and it is involved in proton transfer just to form enolate and then the redetermining step is the elimination step kicking out hydroxide and forming the conjugated product the mechanism is going back excuse me the mechanism here is not again it's not e2 right and it is first order right so it's a type of e1 but we don't call it just by itself an e1 reaction because any one reaction that we talked about last semester involves the formation of a carbocation by loss of a leaving group so notice that we did not form a carbocation intermediate this is not a traditional e1 reaction either so because it has first order kinetics and if this is the molecule by which the rate depends on right and it's not a carbocation we call it an e1cb reaction and the cb stands for conjugate base so it's a first order reaction which is where the one comes from an e1 and the cv stands for the conjugate base this is the conjugate base of the reactant which is acid in this acid-base reaction okay so when it's possible to form more than one stereoisomer as a product for example an e-alkene or z-alkene you can also think of this as a trans alkene and a cis alkene then the more stable less sterically hindered product is typically the major product so traditionally when we draw the product we'll draw the e or the trans alkene more stable than the z or the cis alkene remember i talked about there would be an exception to forming the condensation product typically in order to go from the aldol addition product which would be this one to the condensation product which is this one we would need to heat the reaction this is an exception where it's actually difficult to isolate the aldol addition product in this particular aldol reaction because the aldol addition product very easily converts to the condensation product through loss of water and the reason why it's so easy to do the second part of the mechanism and go to the condensation product is because in this case due to the phenyl rings we have a extended conjugated pi system right so we have p orbitals at every single carbon all the way around this aromatic ring there's p orbitals on the carbon and oxygen here neighboring p orbitals on these two carbons and then it's also in conjugation this second aromatic ring so there's an extended conjugated pi system which is very stable and as a result of this being so stable this readily converts to the condensation product even without heating the reaction remember we talked about the fact that the yields are low for the aldol reaction addition reaction with starting from ketones right because it's an equilibrium process and the retro aldol is favored with ketones due to the product being not very stable due to steric hindrance however if we heat the reaction to go past the elbow addition product all the way to the condensation product the yields increase uh substantially and that's because by eliminating this alcohol group we actually get rid of a lot of the steric hindrance in the product making this much more stable and as a result the equilibrium tends to form fat okay let's take a break to do a practice problem draw the condensation product obtained when the following compound is heated in the presence of sodium hydroxide so again we're in the aldol condensation section so we're going to be drawing an aldol condensation product which is an alpha beta unsaturated aldehyde or ketone in this case we're starting with a ketone so it will be an alpha beta unsaturated ketone right we're using a base and we're heating it so it will go all the way to the condensation product so go ahead and pause and see if you can come up with the structure okay let's go ahead and go over it so one of the there's a little trick to drawing the products as well again we're only expected to draw the product we're not expected to draw the entire mechanism because it didn't ask for that the one thing you can do is line up the alpha carbon right with another molecule and essentially point the carbonyl of the second ketone directly at that alpha carbon right so then it would look like this and the reason i've done that is so that we can essentially circle the carbonyl and alpha carbon and we're going to be forming a double bond between these two carbons and eliminating this hydrogen i'm sorry this oxygen it will eventually become water so we can we can visualize it that way it's actually pretty easy to draw the product we'll just go ahead and draw an alkene there and then connect it to the rest of the molecule so that is out the aldol condensation product you can know how it's an alpha beta unsaturated ketone and that is formed when we heat in the presence of hydroxide or you could use acid all right let's start talking about what we call crossed aldol reactions cross-dial reactions utilize two different aldehydes instead of two of the same aldehyde to produce the aldol product so it's also sometimes called a mixed aldol reaction so in this case we've mixed two different aldehydes so this is known as acetaldehyde or also means ethanol and then this aldehyde is known as propionaldehyde also known as propanol and if we mix these two together with some sodium hydroxide we're going to end up with a complex mixture of products and that's because both of these aldehydes could potentially form enolates because they both have alpha carbons and hydrogens attached to them so we have the possibility of either forming this enolate or this enolate and then each this these two enolates can then react with either the starting aldehyde that form that enolate right or the other aldehyde and so we have this situation where the two enolates form react with two different aldehydes and there's four unique situations here where the outlaw reaction can occur forming four different products this would not be the desired reaction to gene or an organic chemist because we would never want a complex mixture of products because they would likely be very difficult to separate so this is not a synthetically useful process but there are ways to do synthetically useful crossed or mixed aldol reactions so what are the ways we can make this process practical where we use two different substrates instead of forming a dimer with one substrate one way to do it is if one of the substrates is relatively unhindered meaning it makes a great electrophile and it does not have any alpha protons notice that formaldehyde does not have an alpha carbon with hydrogen's attached to it so this is another example of an aldehyde which cannot form enolate right and that's an advantage for this crossed aldol reaction because the only aldehyde that can form enolate is the propional or also known as propanol so only one enolate is possible in this solution and then of course that enolate could attack another propanol but recall that formaldehyde is also present and because there's two hydrogens and no carbon-based groups this carbonyl carbon is not hysterically hindered and more exposed in addition it's more reactive so this is more electrophilic because carbon-based groups are stabilizing the positive charge through hyperconjugation so without one of those carbon-based groups stabilizing the positive charge is carbonated more electrophilic so it's successful then to form an enolate with this aldehyde and have it then preferentially attack the other aldehyde leading to a successful cross aldol product here's another example where we can perform a successful crossed aldol reaction here we have two different aldehydes being used right notice that benzaldehyde again does not have any alpha protons so we cannot form an enolate from the substrate which limits the enolate only forming from the blue aldehyde and then benzaldehyde is uh pretty sterically unhindered because this might look like a big bulky group but recall that the entire thing is cleaner so it's actually not very sterically hindering for a nucleophile to come in and attack this carbon so as a result we can form a successful cross aldol addition reaction however this is placed in brackets because again with the fennel group here and it's hard to isolate the aldol addition product because it directly goes to the aldol condensation product because of the stability of the extended conjugated system here is another example of a successful crossed aldol reaction notice that in the first example we were using sodium hydroxide and in this case we're going to end up using a stronger base we'll call lda which is lithium dioxide propylene and lithium diisopropylemia will deprotonate the alpha position of a ketone or aldehyde forming an enolate and then if we form this at low temperature so that it's um it's fairly stable right so it's usually formed at negative 78 degrees very very cold temperature and then we add the second aldehyde in in a drop-wise fashion genetically faster for the enolate to attack the aldehyde than it is for a enolate to deprotonate the alpha position of the other aldehyde essentially forming two enolates in solution so luckily the nucleophilic attack on this carbonyl carbon of the aldehyde is faster and we're able to successfully do this cross-aldol reaction forming the beta hydroxy keto product all right let's do a practice problem this is a beta hydroxy keto product so when you see that you should be thinking i can form that from an aldol reaction and notice that it's not does not look like it was a dimer so we need to do a crossed aldol reaction to form this so think about how you can successfully do a crossed aldol reaction to produce the following compound go ahead and pause okay let's go go ahead and go over it sometimes it's helpful to draw in a little alpha and beta and then think about what carbon-carbon bond we're forming in this reaction so we're actually going to form the carbon-carbon bond between the alpha and beta carbon so then that simplifies right thinking retrosynthetically that simplifies to the left side being the enolate right and then the right side the electrophile so the nucleophile and the electrophilic carbon so we're just going to convert that that alcohol back into an aldehyde breaking this carbon-carbon bond in a retrosynthetic process so we need to be careful because we're using a ketone and an aldehyde and both of them have alpha carbons that have hydrogens attached to them so there's the potential to form two different enolanes here so when that's the case we need to utilize um example two from the previous slides or method two uh using lda instead of using sodium hydroxide so to form this enolate we would take the ketone we would add lda typically at a very cold temperature so sometimes it might be mentioned negative 78 degrees celsius we usually do that from the enolate at a low temperature and then dropwise add in the other organic components and of course there would need to be an acidic workup and that would produce the product now that i think about it i actually think that typically water is shown as the proton source for this so i think the book does not use the acidic workup as part of the reagents to be drawn over the arrow so i'm going to go ahead and delete that okay so we can also do what's known as an intramolecular aldol reaction if we essentially have uh two ketones or aldehydes tied together as a dial so essentially what we can do is treat with sodium hydroxide and heat to get the albo condensation product and in this case you're actually forming a ring so one of these ketones converts to an enolate using base and then there's a nucleophilic attack that occurs intramolecularly and uh five and six membranes form at rate yields using this reaction so this is not really different from any other aldol condensation reaction it's just done intramolecularly as opposed to inter molecules let's go ahead and do a practice problem practice drawing a reasonable mechanism for this transformation so again you're going to work on forming the enolate from one of the ketones and then doing a nucleophilic attack and then continuing the mechanism until you get to the alpha beta unsaturated condensation product go ahead and pause and then zoom when you're ready okay so the goal of this problem is to think about doing the reaction in an intramolecular fashion so we know that technically all of these carbons that i'm pointing to right now are all alpha carbons right so all of the hydrogens attached to them are acidic the hydrogen's interior are not more acidic than the hydrogens that are at the terminal ends of the molecule because this is a not a 1 3 diketone this is a 1 4 diketone right so an enolate formed from deprotonation of an interior methylene is not doubly resonant stabilized so it's not more stable than the versioning of the protons at the terminal end of the molecule if we were to deprotonate here which will definitely happen in solution and then attack the other carbonyl we would be forming a three-membered ring remembered rings are unstable and so that's not a likely reaction to take place what's much more likely is to deprotonate one of the terminal ends and then do an intramolecular nucleophilic attack which will form a five membrane which is a much more stable ring so the first step is going to be deprotonation using sodium hydroxide and again looking at this kind of gives us an idea of where we should do that deprotonation go ahead and draw in one of the hydrogens attached to the terminal carbon i'm going to show a proton transfer again you can leave these electrons as a lone pair on this alpha carbon negative charge where you can move the electrons all the way up to the oxygen notice that they did leave the lone pair the negative charge on the carbon so just to show you that you can do it either way i went ahead and moved the electrons all the way up to the oxygen and it doesn't matter which one of those two you do they essentially mean the same thing forming an enolate and i'm actually going to try to make it look kind of like this so i can um make it easy for you to visualize the time so from here because i chose to draw this resonance form of the enolate instead of this one you need to push electrons starting from the lone pair on the oxygen to then do the nucleophilic attack on the other carbonyl carbon this is our ring forming step right so we are forming a one two three four five member room go ahead and sketch out formed a carbonyl at this carbon in that step so i'm going to go ahead and draw that in and then the carbon that's being attacked is this one right then it contains an oxygen with a negative charge and a methyl group at this point we need to protonate this to get the aldol addition product and then we'll take that aldol addition product and react it further to get the aldol condensation product best acid in solution when we're using base then i can pick up a proton from water and now we're at the aldol edition product and because we're heating it we don't stop there and we go on to do the second part of the mechanism which is the condensation that involves first the protonating the alpha position between the carbonyl and the newly formed alcohol hydroxyl group i'm going to form another enolate by drawing a second hydroxide and doing a deprotonation remember be careful it's tempting because it's faster to draw this is an e2 reaction and a concerted process but that is not uh what is shown to be the likely mechanism in this reaction so instead of drawing h2 we're going to just go ahead and form the conjugate base again i can deliver these electrons to this carbon right or i can move the electrons all the way up to the oxygen if i want i can go all the way up to the oxygen forming an enolate which then kicks out hydroxide and we form the product in that step some hydroxide last step so again this process right it's called an e1 cb mechanism so make sure to draw the elimination as a two-step process where we form the conjugate base and then do the elimination from the conjugates you