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Current time:0:00Total duration:11:23

Video transcript

a la condensations are extremely important reactions in organic chemistry and let's look at first an aldol addition where if we take something like acetaldehyde and add some sodium hydroxide we form this aldol product right so the term aldol comes in the fact that we have an aldehyde here and in the O L because we have an alcohol as well let's look at the mechanism for the formation of this so if we start with our acetaldehyde and go ahead and draw it out here and we look for our alpha carbon right it's the one right next to our carbonyl so here is our alpha carbon there are three alpha protons on that alpha carbon I'm just going to show one here and then we can have sodium hydroxide come along as act as a base so we have the hydroxide anion so negative one formal charge on the oxygen it's going to take this proton right leaving these electrons behind on our alpha carbon so let's go ahead and show the carb anion that would result alright so we have our carbonyl and then we have our hydrogen on the right side and the left side we have a lone pair of electrons on this carbon so we have a carb anion so the electrons in magenta here moved off on to this carbon we can draw a resonance structure right these electrons move into here these electrons go off onto our oxygen and we would form the oxyanion right so we have a double bond here and then our oxygen up here would have three lone pairs of electrons so negative one formal charge and then a hydrogen so the electrons in magenta moved in here to form our double bond and so we know that the negative charge is best supported on an oxygen so the oxyanion is probably the best way to show an enolate anion however for the aldol condensations i'm always going to use the carb anion as our enolate anion it's actually a little bit easier for students to look at the mechanism that way so we're going to start with our our carb anion so I'm just going to redraw it down here so here's our carb anion and once you form your enolate anion your email aid anion can act as a nucleophile all right this carbon here is negatively charged and because we use something like sodium hydroxide sodium hydroxide is not basic enough to completely form enolate anions so only a small out of your enolate anion is formed and so this can act as a nucleophile and attack another molecule of acetaldehyde so go ahead and draw in another molecule of acetaldehyde and I'm going to draw it in a particular way I'm going to leave off the hydrogen and I'm going to put it right next to this it just helps me when I'm doing an aldol reaction here alright so if I know the carbonyl of my other molecule of acetaldehyde is it polarized right the e oxygen is partially negative and the carbonyl carbon here is partially positive so the carbonyl carbon is electrophilic it wants electrons and it's going to get electrons from our nucleophilic enolate anion so these electrons here are going to attack our carbon right push these electrons off onto our oxygen so let's go ahead and show the result of that so on the left side right we would have our oxygen with three lone pairs of electrons and then we form a carbon-carbon bond here so that's it's extremely important and then we also have over here on the right our aldehydes so let's let's show the movement of electrons right so these electrons right here right this is our our carb anion attack our carbonyl carbon form this bond right a carbon-carbon bond forming reaction to form an alkoxide intermediate and so we could protonate our alkoxide with water right so water is right here so we could show a lone pair of electrons picking up this proton and we can go ahead and draw our aldol product now so when we draw our aldol right here like that so the formation of a carbon-carbon bond makes an aldol addition and the next an aldol condensation a very important reaction in organic chemistry here so there's our aldol product all right let's look at an aldol condensation which is kind of a continuation of what we were just talking about and you can you can form different products depending on reaction conditions so for example here we're starting off with acetaldehyde again and studying hydroxide in water we're adding Heat so we're changing the reaction conditions and so we're going to get a different product and so we call this an aldol condensation so you can see the product that we get is an e now alright so we get an email we get a double bond here for the and then we have an aldehyde right so we also hear it referred to as I alpha beta unsaturated because this carbon right here is the alpha carbon and this carbon is the beta carbon it's unsaturated because we formed a double bond here so this is an aldol condensation and to to form this email product we're going to start with the aldol that we just made in the previous reaction so acetaldehyde and right with sodium hydroxide can form an aldol as we just saw so let's go ahead and draw in the aldol product alright so the same mechanism that we just discussed and I should point out that we're doing a base catalyzed aldol condensations here so so we have a hydrogen here and then our aldol had an oxygen and then a hydrogen right here so there's our aldol and if we look at our aldol right the carbon right next to the carbonyl is our alpha carbon and there's still two acidic protons on that alpha carbon so i'm just going to draw one in there and then we can think about sodium hydroxide acting as a base and forming another enolate anion so i'm going to have sodium hydroxide come along so OS it's going to function as a base alright it's going to take this proton right leave these electrons behind on this carbon so let's go ahead and draw that right so we would have our oxygen right here with two lone pairs of electrons right we would have our carbon now with a lone pair of electrons making a carb anion and then we have our carbonyl over here like that all right so let's once again show those electrons these electrons right in here right forming our carb anion and then to form our email product we can just think about and go ahead and draw an arrow here we could just think about these electrons moving in here to form a double bond and kicking these electrons off onto the oxygen to form hydroxide as a leaving group all right so we can say that hydroxide leaves and then we form our double bond so I can go ahead and show that you know let's say these electrons right here were the magenta ones to form our email product and so once again I'm using the carb anion a form of our resonance structure instead of the oxyanion so again it's probably more appropriate to use the oxyanion but just easier from a student perspective to use a cheese a carb anion to go ahead and form your email product that way so in terms of in terms of another product right this is the major product that we would form here right this is the trans products everything about this double bond these hydrogen's are on opposite side so this is the trans products that's the major one you would form a small amount of the cysts product as well so let me go ahead and draw that in but the cysts product would be the minor product because it has more steric hindrance than the trans product so this is the major product and and and I'm sure the one that that your professor wants you to put on an exam question so the trans product has decreased steric hindrance alright let's do let's do one more of these aldol condensations and so here i've drawn out the reaction and so so the question be like write a mechanism right right a plausible mechanism for the formation of this ino n-- alright so we have an e known this time we're starting with a ketone and so we have a double bond and then the ketone so an e known product so we're going to use sodium hydroxide and we're going to heat things up here so let's think about what would happen right so we have our ketone and we have two alpha carbons which are symmetrical so doesn't really matter which one we take a proton from so I'm just going to say take a proton from this one alright so we take a proton from here to form our enolate anion and once again I'm going to form the the carb anion just to make things simpler so we have our carb anion now and then we have another molecule cyclohexanone is going to come along so we go ahead and draw that in alright and we know we have a polarized carbonyl situation partial negative partial positive so our nucleophilic enolate anion is going to attack right here all right so we can show these electrons attacking right here pushing those electrons off onto your oxygen so just going a little bit fast through this mechanism here and so we have now a bond that formed right and now this oxygen right here is going to have three lone pairs of electrons around it negative one formal charge and I can go ahead and draw on the rest of my ring like that so let's follow some of those electrons here so I'll make these electrons in blue this time forming the all-important carbon carbon bonds right here alright and then these electrons came off onto our oxygen ghin forming our alkoxide alright so we could think about water alright protonating our alkoxide intermediate so lone pair takes that proton leaves those electrons behind and hopefully we have enough room to draw the aldol alright so over here we would have our aldol intermediate right we have our Oh H we have our ring like that okay thinking about what happens next right the carbon next to the carbonyl is the Alpha carbon so that's our alpha carbon and so there's still a proton on that alpha carbon and I don't really have a lot of room here but let me see if I can draw in an alpha proton like that and so we could have hydroxide once again functioning as a base alright so picking up this proton alright and then leaving these electrons behind on that carbon in red so let's get some more room here so now what we would have is our our ring alright we have our our a carbonyl there and then we have a lone pair of electrons on this carbon at negative one formal charge over here we have our Oh H like that and then we have our ring okay so in the next step once again I'm not drawing an oxyanion so so I'm not going to draw the the oxyanion form of this enolate anion here but we could show we could show these electrons moving into here right and pushing these electrons off on to the oxygen and so let's go ahead and draw our product once we do all of that so we would have our ring right we would have our carbon eel here and then we would form a double bond and then now we can see that we have our products so let's follow those electrons so electrons in in blue here alright these electrons move in here to form our double bond right and then let's let's use red to show these electrons in here coming off and so we get hydroxide as our leaving group and we form our conjugated enone as our product and so this is a one way to represent the mechanism and and once again the reaction conditions will determine what kind of product you get our enon here is conjugated right of an alternating single double bond situation so that favors that fit that favors this product here so that shifts the equilibrium once again to make more of your n own so practice doing a lot of these aldol condensations