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Video transcript

let's see how to approach an intramolecular aldol condensation so over here on the Left we have a two five hexane diode and if we add sodium hydroxide and we heat things up we would form this compound over here on the right so we would actually form a ring let's go ahead and number our our diamond over here it's symmetrical so we could start from either side so I could say this is carbon one carbon two three four five and six so we have two five hexane down here and if we add sodium hydroxide we're going to deprotonate at the alpha carbon so I need to find my alpha carbons here so that's the carbon next to the carbonyl so carbon one could be an alpha carbon carbon three it could be an alpha carbon and then over here right these would also be alpha carbons to this carbonyl but since it's symmetrical we don't need to worry about those so we just need to focus in on the possible alpha carbons at one and three let's let's first do this thinking about deprotonating at alpha carbon one here at alpha carbon which is carbon one so I put a my alpha proton on here and I think about hydroxide taking this alpha proton right leaving these electrons behind on this carbon and so here I have I have the diode drawn in a different conformation and let's go ahead and number this one so this would be carbon one two three four five and six this is just going to help us when we're thinking about forming our product and so if we're deprotonating at carbon one alright so these electrons in here would end up on carbon one to form a carb anion so I'm showing those electrons in magenta on carbon one so that's a negative one formal charge on this carbon and for this mechanism going to show the carb anion functioning as a nucleophile so it's going to attack this carbonyl over here we know this carbonyl has a partial negative oxygen a partial positive carbon so this carbon right here is electrophilic so our nucleophile is going to attack our electrophile right so these electrons in here going to attack here pushing these electrons off onto our oxygen so let's go ahead and show the result of that we would actually form our ring here so I'm going to draw our five membered ring we would have our carbonyl right here and we would form an alkoxide then we could we can think about protonating to form our alcohol so I'm just going to hand draw the aldol here and then we would have a methyl group right here so let's let's follow some of those some of those electrons right so the electrons in magenta moved in here to form our bond and then let's number our carbons too so this would be carbon one carbon two right three four five and six in terms of the numbering system that we used over here on the left so it just helps us to to follow those carbons along here and so this would be our aldol and then we know that this is our alpha carbon right here so we could deprotonate that so I'm going to go ahead and draw an alpha proton right here and so we could think about hydroxide coming along right functioning as a base right and taking that proton so the hydroxide takes this proton these electrons move in here to form your double bond and you have hydroxide as your leaving group so when we draw the product right we have our ring right we have our carbonyl right here we would have a double bond now and then we would have our methyl group like that and so that's obviously our product let's follow those electrons right we already had a carbon-carbon bond formed in here but if we take the proton right these electrons in blue can move in here to form our double bond and then we have our product and so this would be the the major product for this reaction so so that's that's showing deprotonation at the alpha carbon which we labeled one right here let's so let's show what would happen if we deprotonated the Alpha carbon that we numbered three so we'll do that and the next area we have down here alright so the same conformation we're starting with before all right so one two three four five six if we deprotonate at carbon three this time then that's where we'd form our carb anion so let me go ahead and show the lone pair of electrons on this carbon three right here that gives that carbon a negative one formal charge and so we could think about this being our nucleophile and attacking right here at this carbon so that would push these electrons in here off onto the oxygen so let's go ahead and draw the results of that this time we would actually form a three membered ring so let's go ahead and draw everything in and then let's let's follow everything around here so we have a methyl group right here and then we would have an O and that'd be our alkoxide which we could protonate to form our aldol as our intermediate here so let's once again number these carbons even though this isn't how you would actually number it just helps us to follow everything along here so let me go ahead an umber in red so one two three four five and this is where carbon six again using these numbering system over here on the left just to make it a little bit easier for us so the electrons in magenta right are the ones that form this bond between carbon three and carbon five like that and once again we know that carbon three is an alpha carbon so there's an alpha proton on here and so we could think about hydroxide coming along and functioning as a base right so hydroxide is going to take this proton leave these electrons behind and then we have hydroxide as a leaving group and so another possible product would be this three membered ring right here and then we would have a methyl group and then we would also have our ketone over here like that and so we had already formed a carbon-carbon bond with electrons in magenta and then if we take this proton right these electrons in blue or I could move in here to form our double bond and so it turns out that on this small ring here it's possible to form this but it's not isolated in large yield and in this reaction because there's too much angle strain for this three membered ring and so this isn't really isolated as your product your product is this one there's less angle strain to form this five membered ring but it is possible to get a three membered ring in this so but the five membered ring is going to be your major product here so let's do out one more intramolecular aldol condensation all right so very similar to the last one actually so it looks a little bit more complicated but we can analyze it the same way so if we're looking for our alpha carbons right we know those are the ones next to our carbonyl so let's go ahead and number this compound this would be carbon one two three four five and six and so for our alpha carbon all right so we know carbon one would be an alpha carbon and we know carbon three would be an alpha carbon carbon four and then carbon six all right so those are our possibilities so carbons 3 & 4 if you think about these being alpha carbons users give us a ring with a little bit too much angle strain so we can kind of rule those out and so now now it's down to thinking about carbon 1 or carbon 6 and for that we need to look at the base right so we're using potassium hydroxide here so we're using a non sterically hindered base and it isn't very strong and so it's going to favor the formation of the thermodynamic enolate which is the more stable one because it's more substituted so we talked about this in an earlier video and so if you look at an alpha if you look at deprotonation at at alpha carbon which we have labeled one here that would give you the kinetic enolate which is form the fastest was not the most stable if you deprotonate at the alpha carbon that six that would give you the thermodynamic enolate so it's more substituted than the kinetic enolate so it's more stable so if you think about the fact that on this alpha carbon right there are two alpha protons right here and if your base takes one of these alpha protons right then these electrons would move into here push these electrons off onto your oxygen so let's do the same the same thing down here we have a different conformation so very similar to how we did the last one so one of these acidic protons right here on our alpha carbon I'll go ahead and draw it in right here so we have our hydroxide come along and function as a base right so hydroxide anion all right so negative one charge is going to take this proton right leaving these electrons into here pushing these electrons off onto the oxygen so I'm going to draw the oxyanion this time all right so this is going to be in equilibrium so let's go ahead and show first I'll show what we would get and then we'll follow some electrons around right so I have my carbonyl here and then we're going to form our oxy anion so I'm going to go ahead and draw negative one formal charge on this oxygen there's now a double bond right here and then we go ahead and draw on the rest of our carbons so this is the this is the thermodynamic enolate right so if you look at this double bond this is going to be the most stable enolate forms because of the substitution of this double bond here so following those electrons let's go ahead and make those electrons in here blue so these electrons in here all right move in here like that and then you could think about your electrons your PI electrons in here moving off onto your oxygen to form your oxy anion like that and so once again this is the thermodynamic enolates because this reaction is under thermodynamic control because of our choice of base next we're going to show our oxy anion functioning as a nucleophile so once again our carbonyl is polarized right partial negative partial positive alright so we could think about these electrons right in red moving into here and then we could think about these electrons pretend these are our PI electrons in here attacking at our carbonyl right pushing these electrons off onto our oxygen so let's go ahead and draw all right the result of our nucleophilic attack right here all right so we would form our ring and be five membered rings let's go ahead and draw our five membered ring here we would form a carbonyl all right and then let's see at this carbon we would have this group coming off here all right and then we would form an alkoxide which you could protonate to form our aldol and then we have a methyl group right here so let's let's follow those electrons so a lot happened here so if the oxyanion acts as your nucleophile all right so these electrons in here which I'm saying are your PI electrons right are going to form this bond right here so very similar to how we did it in the previous video except we're using an oxy and on this time as their nucleophile these electrons in here move in to reform your carbonyl alright and if we're thinking about carbons alright so this carbon right here is this carbon and let's identify some other carbon so this carbon right here is this carbon and then this carbon is this carbon just so it's not it's getting a little bit confusing in terms of what we have so if we go back to our alpha carbon here right so we there's an acidic proton on that alpha carbon so I could go ahead and draw that in alright and then our base could come along so hydroxide could come and take that acidic proton so let's go ahead and show that so negative one formal charge takes this proton right these electrons move in here and you lose hydroxide and then you have your products let's go ahead and draw on our products we have a five membered ring right we have a ketone here we have a double bond here and then we retain the stereochemistry of this group that's coming off and then we have a methyl group so we had already formed a carbon-carbon bond in here all right and then let's let's go ahead and make these electrons in here let's make them make the magenta all right so these electrons in here we're going to move in to form your PI bond to give you your products so the product is called sis Jasmine and it's found in jasmine flowers and so it gives its scent so this is a pretty cool reaction an intramolecular aldol condensation which gives you a product that's used in the perfume industry and a pretty good yield