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Intramolecular aldol condensation

How to determine the product of an intramolecular aldol condensation. Created by Jay.

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  • blobby green style avatar for user samiryusuf15
    At , when the alpha hydrogen is taken away by the OH(-), why does the bond from that alpha hydrogen go in between C5 and C6 to form a double bond? Whereas in the previous examples, when the alpha hydrogen is taken away, a carbanion is formed at the alpha carbon. Just wondering why there is a difference.
    (9 votes)
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    • male robot hal style avatar for user Yogeshwar
      He only drew a resonance structure of the carbanion formed on C6. This structure was more stabilised (due to resonance delocalization of electrons), but it does not affect the reaction mechanism in any way.

      As you can see, at , he shows that the pi-electrons of C6 attack the carbonyl carbon C2.
      This is exactly the same as the formation of a carbanion on C6, and its subsequent attack on the carbonyl carbon C2.
      (7 votes)
  • blobby green style avatar for user Max Myers
    At , how is the alpha hydrogen getting deprotonated by the hydroxide if there is no heat to drive that reaction? I learned that if there is no heat then that OH on the 3 carbon is part of your product.
    (6 votes)
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  • male robot hal style avatar for user jonathantullock
    At , the speaker said that KOH is not a strong base. But aren't all Group 1 & 2 metal hydroxides considered strong bases?
    (3 votes)
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  • leafers sapling style avatar for user yasminhooman
    At roughly....I thought that NaOH would lead to a Thermodynamic removal of the H at C3 based on the other videos, while larger bases like LDA removed H from C1. What am I missing here?
    (3 votes)
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    • leaf green style avatar for user dffsfs
      I thought the same at first, but the final product isn't actually thermodyanmically stable because of 3 carbon ring, so only kinetic 5 carbon ring product will have high yield. In this case, BOTH the kinetic AND thermodynamic product is the 5 member ring. The 3 member ring is neither. If you were to draw out an energy diagram, it would look something like this. http://i.imgur.com/9N4CICm.png
      (3 votes)
  • leaf green style avatar for user Courtney Smith
    At , why would the 3-membered ring even form? Isn't there a lot of steric hindrance keeping it from forming?
    (2 votes)
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  • orange juice squid orange style avatar for user dylandevens
    Wouldn't the section continue to make a diccclic compound? It seems to me that in the presence of excess KOH that the double bonded carbons would attack the ketone but he didn't mention this. Can anyone elaborate as to why the reaction does not continue?
    (2 votes)
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  • blobby green style avatar for user farazarastu
    Why can't you form a homopolymer from an intermolecular reaction instead of intramolecular? Kinetic reasons? How would you favor intermolecular to make this happen?
    (2 votes)
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  • blobby green style avatar for user namansiddique
    At , shouldn't you protonate the OH because its such a bad leaving group? Or is it okay because the entire reaction is happening in a basic solution?
    (2 votes)
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  • blobby green style avatar for user cpopo9106
    for the last reaction, how come the carbonanion did not attack the carbon of the carbonyl at the bottom instead to form the same product?
    (2 votes)
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  • blobby green style avatar for user Ahszha
    at , why could the alpha 3 or alpha 4 carbon be used?
    (1 vote)
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Video transcript

Voiceover: Let's see how to approach an intramolecular aldol condensation. So over here on the left, we have a two five hexanedione. And if we add sodium hydroxide and we heat things up. We would form this compound here on the right. So we would actually form a ring. Let's go ahead and number our, our uh,dion over here. It's symmetrical so we can start on either side. So I can say this is Carbon one, Carbon two, three, four, five and six. So we have two five hexanedione 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 also carbon one, could be an alpha carbon. Carbon three, 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, an alpha carbon which is carbon one. So I put put 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 dione drawn in a different conformation. And let's go ahead and number this one. So this will be carbon one, two, three, four, five and six. This is just gonna help us when we're thinking about forming our product. And so if we're deprotonating at carbon one, all right, so these electrons in here would end up on carbon one to form a carbanize. So I'm showing those electrons in magenta on carbon ones, so that's a negative 1 formal charge on this carbon. And for this mechanism, I'm gonna show the carbanion 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's going to attack our electrophile, right? So these electrons in here, are going to attack here, pushing these electrons off onto our oxygen. So let's go ahead and show the results of that. We would actually form our ring here, so I am gonna draw our five membered ring. We would have our carbonyl right here, and we would form an alkoxide and then we could we could think about protonating to form our aldols, so I go ahead and 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 bonds. 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 gonna 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 leafing group. So when we draw the product, right, we have our ring. All 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 let's show what would happen in we de-proteinated the alpha carbon that we numbered three, so we'll do that in the next area we have down here, all right. So the same confirmation we're starting with before, 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 and ions. Let me go ahead and show the one pair of electrons on this carbon three right here. That gives that carbon a negative 1 a formal charge. And so we can think about this being our nuclei file and attacking right here at this carbon. So that would push these electrons in here off to 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 alc oxide, 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. It just helps us to follow everything along here, so let me go ahead and number in red. And 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 key tone over here like that. And so, we had already formed a carbon, carbon bond with electrons and magenta. And then if we take this proton, right, these electrons in blue, right, can move in here to form our double bonds. And so, it turns out that on this small ring here, all right it's, it's possible to form this. But it's not isolated in, in large yield in this reaction because there is too much angle stream for this three membered ring. And so this isn't really isolated as your product. Your product is this one, there is 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, is gonna be your major product here. So lets do 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 carbonyls. Let's go ahead and number this compound. This'll be carbon one, two, three, four, five and six. And so, for our alpha carbons, right, 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, right? So those are our possibilities. So carbons three and four, if you think about these being alpha carbons. These will give us a ring with a little bit too much angle strain. So we can kind of rule those out. And so now, now its down to thinking about carbon one or carbon six. 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 gonna 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 I had labeled one here, that would give you the kinetic enolate, which is one of the fastest, but it's not the most stable. If you deprotonate at the alpha carbon at 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, the same, the same thing down here where we have a different confirmations. 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 and ion. All right so negative 1 charge is gonna take this proton, right, leaving these electrons in here pushing these electrons off onto the oxygens. I'm gonna draw the oxy anion this time, right? So this is going to be in equilibrium, so let's go ahead and show first off I'll show what we would get, and then we will follow some electrons around, right. So I have my carbonyl here and then we're gonna form our oxyanion, so I'm gonna go ahead and draw negative one formal charge on this oxygen. There's now a double bond right here, and then we'll go ahead and draw on the rest of our carbons. So this is the, this is the thermodynamic enolates, right? So if we look at this double bond, this is going be the most stable enolate that forms, because of the substitution of this double bond here. So, following those electrons, lets go ahead and make those electrons in here blue. So these electrons in here. All right we move in here, like that, and then you can think about your electrons, your pi electrons and 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 gonna show our oxyanion functioning as a nucleophile. So once again, our carbon yield is polarized, right? Partial negative, partial positive. All right, so we could think about these electrons, right, in red, moving into here. And then we could think about these electrons again, these are, are pi electrons in here. Attacking at our carbonyl, right, pushing these electrons off onto our oxygen. Lets go ahead and draw right, the results of our nucl, nuclei philsacrate here, right, so we would form our ring, it be five membered rings. So lets go ahead and draw our five membered ring here. We would form carbonyl, right? And then, let's see at this carbon, we would have this group coming off here. All right, and then we'll 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 oxy anion acts as your nucleophile, all right, so these electrons in here, which I'm saying are your pi electrons, right, are gonna form this bond right here. So, very similar to how we did it in the previous video. All right, except we're using an oxion on this time as a nucleophile, these electrons in here move in to reform your carbonyl. All right, and we're thinking about carbons, all right, so this carbon right here, is this carbon. And, let's identify some other carbons. 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, all right. So we, there's an acidic proton on that alpha carbon. So, I can go ahead and draw that in. All right, and then our base could come along, so then hydroxide could come along and take that acidic proton. So let's go ahead and show that. So negative 1 formal charge, takes this proton, right, these electrons move in here, and you lose hydroxide. And then, you have your products, so let's go ahead and draw in our products. So, we have a five membered ring, right? We have a key tone 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'd already formed a carbon, carbon bond in here, right? And then let's let's go ahead and make these electrons in here let's make them, let's make them magenta, all right? So these electrons in here are gonna move in to form your pi bonds to give you your product. So, the product is called, cis jasmone, it's found in jasmine flowers and so it gives it its scent. So this is a, a pretty cool reaction in intramolecular outlaw condensation which gives you a product that's used in the perfume industry in a pretty good yield.