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Current time:0:00Total duration:9:25

Video transcript

in the previous video we looked at aldol condensations with the same molecule you'd call those a simple aldol condensation in this video we're gonna look at mixed or crossed aldol condensations so no longer are you starting with the same molecule here we don't have two aldehydes of the same we have different aldehydes right we have benzaldehyde and left and propanil on the right and so we need to figure out what sort of enolate anion that we're going to form so when we add our sodium hydroxide as our base what is going to be our enolate anion to do that we need to look for our alpha carbons right so we'll start with propanol we know the Alpha carbon is the one next to the carbonyl and so right here this would be an alpha carbon and there are two protons on that alpha carbon so we have two alpha protons ears and we go ahead and draw in those and so that's a possibility to form our enolate anion this aldehyde hydrogen right here is not we're not going to form an enolate from that so we don't have to worry about the right side of our aldehyde let's look at benzaldehyde now so if we think about this carbon right here it is the carbon next to a carbonyl however this carbon already has four bonds to it and so it doesn't have an alpha proton so benzaldehyde does not have an alpha proton so we don't need to worry about it forming an enolate anion and so we know the emailing anion is going to form from pro finale' here and so let me just go ahead and redraw these in a way that makes it easier for me to see what the product is so I'm just gonna draw my ring here I'm gonna draw my my carbonyl and I like to leave off this hydrogen because it just gets in the way when I'm thinking about my mechanism and for my propanol over here I prefer to draw a different conformation of propanol to make it easier to see the product so I like to I like to put my carbonyl up here like this and then I like to draw this carbon down like that and it's gonna make it easier to come up with the final result alright so if I think about deprotonation alright if I think about sodium hydroxide taking one of these alpha protons here right leaving electrons behind on my alpha carbon I can think about the structure for the carb anion enolate anion right so there's a negative 1 formal charge on this carbon and that's my nucleophilic enolate anion once again I prefer to draw the carbon on as opposed the oxy in I just makes it easier for me when I'm trying to do a quick mechanism like on a test to figure out a product here so now we have a nucleophile all right so I'm gonna go ahead and put these electrons in magenta so this is going to be our nucleophile and our nucleophile is going to attack the electrophilic portion of our benzaldehyde molecule all right so if we think about that the oxygen is partial negative and this carbonyl carbon here is partially positive so that's the electrophilic portion of the molecule and so that lone pair of electrons here on this carbon is going to attack this carbon right pushing these electrons off on to the oxygen and that would form an alkoxide intermediate so let me just go ahead and sketch in here all right so we would form we would form an oxygen right there with three lone pairs of electrons giving it a negative one formal charge and I'll come back to that the important carbon-carbon bond that's formed is right here and then we have go ahead and draw in everything else so we have our aldehyde and then don't forget about this down here so in magenta right these electrons in magenta right it formed our new bond right here and then we would have an alkoxide intermediate so we can think about just protonating it so I'm not concerned with an exact mechanism here I'm more concerned about figuring out the product right so we go ahead and protonate our alkoxide to form our aldol intermediate and so in the next step right we add Heat you're you're most likely going to form your conjugated products so let's uh let's think about what would happen next right so we still have an alpha carbon so this carbon right here next to our carbonyl is an alpha carbon it still has an alpha proton on it all right so we could show a proton right here and we can think about hydroxide once again acting as a base so hydroxide acting as a base coming along and taking this proton right and then we could even just go ahead and think about the final product we could think about these electrons moving into here and then these electrons moving out onto here again I'm not so concerned with the exact detailed mechanism I'm more worried about figuring out how to figure out our product here so when I draw our product alright we have our ring and then we now have a double bond that forms here all right we have this going down and then we of our aldehyde so we form a conjugated product we have an email as our product here so let's say let's follow those electrons right so let's say let's make these electrons in blue here alright so these electrons in blue moved in here to form our double bond and we are already formed this bond in magenta right the carbon-carbon bond forming part to form our aldol and then we lost hydroxide and so this would be our final product are our conjugated products so again the goal is just to to figure out how to draw a product right from these reactants and thinking about where is the alpha carbon and thinking about what is the enolate that forms alright let's let's do another one where I just kind of walk you through my thought process here so let's let's look at this reaction so say you had this on an exam and your task was to draw the product alright so we have once again some different possible alpha carbon so let's first focus on cyclohexanone all right so we know that this could be an alpha carbon and we know that this could be an alpha carbon alright and then for this compound over here we have this these two carbonyls right and this molecule so we know that this could be an alpha carbon and we know that this could be an alpha carbon and so now we have to figure out which one of those alpha carbons is going to be deprotonated when we add sodium ethoxide right as our base right here and if you remember in one of the earlier videos we talked about an alpha carbon between two carbonyls as having the most acidic proton so this alpha carbon right here has two protons on it and it's easy to deprotonate alright because of the the resonance stabilization that we can draw because of those carbonyls and so these protons are the most acidic right much more acidic than than the protons on our ketone here and so these are the ones that are going to be one of these protons could be deprotonated when we add sodium ethoxide here so sodium ethoxide is going to come along right take one of these protons here leave these electrons behind and so let's go ahead and redraw what we would form here so we're gonna form our enolate anion so let me go ahead and draw that so we would now have a lone pair of electrons on our carbon giving that carbon negative one formal charge so let me follow those electrons these electrons in magenta are now on our carbon forming our carb anion and I'm not going to take the time to draw the the resonance structures with the oxyanion because I'm just concerned about figuring out the product here so now we have a nucleophilic enolate anion and we know that's going to attack the carbonyl of our ketone right so going back over here to our ketone right the oxygen is partial negative this carbon is partially positive and so our nucleophile is going to attack our electrophile so you can think about these electrons right attacking here pushing those electrons off onto your oxygen so if we were to draw the intermediate tier all right we would have our ring and we're going to we're going to form an alkoxide so I'm not going to draw all the lone pairs on that oxygen right now I'm more concerned right now is showing the formation of this bond so let me go ahead and draw in everything and then we'll follow some electrons so I've put in my carbon eels and then I have these guys over here like that so the electrons in magenta right these electrons formed our carbon-carbon bond so they formed this bond right here and then we would form an alkoxide and then we would just go ahead and protonate that to form our aldol so once again just to save time not an exact mechanism but thinking about our intermediate as being this aldol alright so because we have heat we're probably going to I once again form a conjugated product here and keep going for the complete aldol condensation and so next we can think about we could think about this alpha carbon right here right still having an acidic proton on it alright so there's still a proton attached to that alpha carbon and so sodium ethoxide come along so we have Anna thought side and line let's go ahead and draw that in here which could function as our base all right so it could take that proton right and think about these electrons moving into here to form our double bond we could think about hydroxide as a leaving group and then that just allows us to figure out our product so we have our ring and then we now have a double bond and then we have a carbonyl over here and and a carbon eel over here and then our oxygen and our ethyl like that so let's uh let's think about those electrons so the electrons in blue here move in to form our double bonds and we had already formed a carbon-carbon bond before so let's say it's these electrons right here and then we have a stable conjugated product and so once again I'm just thinking about how to draw the product for a reaction on an exam and hopefully hopefully this helps