In aldol condensation, an enolate ion reacts with another carbonyl compound to form a conjugated enone. The process occurs in two parts: an aldol reaction, which forms an aldol product, and a dehydration reaction, which removes water to form the final product. Created by Jay.
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- The OH is not good leaving group, then why are we saying OH leave when we form double bond(28 votes)
- This is an E1cB elimination which uses a strong base and a bad leaving group. The reaction can occur without the -OH being protonated.(30 votes)
- Is this reaction in the presence of an acid or a base?(2 votes)
- In the video, the aldol condensation is taking place in the presence of a base.
But there is also a mechanism in which the aldol condensation takes place in the presence of an acid.(12 votes)
- Wouldn't this be the actual mechanism?
Aldol + NaOH -> Aldol with a carbanion
The OH on the Aldol with a carbanion grabs a proton from water.
Water leaves and a C-C double bond forms at the same time.(4 votes)
- Jay is correct.
You can think about it this way - where is the proton going to come from in a basic solution?
While hydroxyl is a poor leaving group, the product formed has conjugation between the carbonyl and the carbon-carbon double bond (C=C). This stabilizes the product and overcomes the usual barrier to hydroxyl acting as a leaving group.
See for example page 6 of the notes below:
- Who discovered aldol condensation ?(3 votes)
- How does the heat in an Aldol Condensation make it different from the Aldol addition, and why isn't this explained?(2 votes)
- They briefly mention it at3:35in the video below but still don't provide much explanation. Essentially, the difference is that heat is required for aldol condensation, as it just favors the elimination of water. Without heat, you'll primarily get the addition product.
- Isn't OH a bad leaving group? Then how are we eliminating OH- when we are forming he double bond ? Is it possible that it grabs a proton from water to make it a good leaving group?(2 votes)
- The top question to this video is exactly the same as yours.
Here is the answer given by Tim B: "This is an E1cB elimination which uses a strong base and a bad leaving group. The reaction can occur without the -OH being protonated."(2 votes)
- At6:03and again at10:17, why don't the carbanion electrons bump off the carbonyl oxygen (instead of the hydroxy group)?(2 votes)
- As mentioned in the video, the hydroxyl is a decent leaving group whereas a carbonyl isn't. Furthermore, the formation of the enone (ketone + neighboring alkene) is very favorable as conjugation occurs between the carbonyl and alkene.(2 votes)
- At9:37, why do you add more base? Just to get rid of the hydrogen?(2 votes)
- Don't think that the reagents are limited to 1 use..... Look at the overall picture and this is in a "solution" meaning there is a good amount of base in there to help the reaction move to the products.... Normally in reactions with acid and bases like the aldol, michael, malonic ester synthesis, they all use either an acid or base and if you use it, you're gonna reform it somewhere else in the reaction or use another base/acid floating around in the solution.
Try not to limit your mindset to thinking that just because you used something in the reaction doesn't mean it's the only 1 equivalence in the solution....(2 votes)
- why alpha hydrogen is used in aldol condensation ? please answer in easy way/.(2 votes)
Voiceover: Aldol condensations are extremely important reactions in organic chemistry. Let's look at, first, an "aldol addition," where if we take something like acetaldehyde, and that's some sodium hydroxide, we form this aldol product. So the term, "aldol" comes from the fact that we have an "aldehyde" here, and then, the "ol," because we have an alcohol as well. Let's look at the mechanism for the formation of this: So, if we start with our acetaldehydes, let me go ahead and draw it out here, and we look for our alpha carbon, 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 gonna show one here, and then we're gonna have sodium hydroxide come along, and act as a base. So we have the hydroxide anion, so negative one formal charge on the oxygen; it's gonna take this proton, leaving these electrons behind, on our alpha carbon so let's go ahead and show the carbanion that would result: So we have our carbonyl, and then we have our hydrogen on the right side; on the left side, we have a lone pair of electrons on this carbon, so we have a carbanion. So the electrons in magenta, here, moved off, onto this carbon, we can draw a resonance structure, these electrons move into here, these electrons go off onto our oxygen, and we would form the oxyanion. 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 gonna use the carbanion, as our enolate anion; it's actually a little bit easier for students to look at the mechanism that way. So, we're gonna start with our carbanion, so I'm just gonna redraw it down here, so here's our carbanion, and once you form your enolate anion, your enolate anion can act as a nucleophile, 'cause 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 amount of your enolate anion is formed, and so this can act as a nucleophile, and attack another molecule of acetaldehydes, so I'm gonna go ahead a draw another molecule of acetaldehyde, and I'm gonna 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 polarized; the 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, push these electrons off, on to our oxygen, so let's go ahead, and show the result of that: So, on the left side, we would have our oxygen, with three lone pairs of electrons, and then we form a carbon-carbon bond here, so that's extremely important, and then we also have, over here on the right, our aldehyde. So let's show the movement of electrons: So these electrons, right here, this is our carbanion, attack our carbonyl carbon, form this bond, a carbon-carbon bond forming reaction, to form an alkoxide intermediate, and so we can protonate our alkoxide with water. 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 we will draw our aldol right here, like that. So the formation of a carbon-carbon bond makes an aldol addition, and then next an aldol condensation, a very important reaction in organic chemistry, so, there's our aldol product. Alright, let's look at an aldol condensation, which is kind of a continuation of what we were just talking about, and you can form different products, depending on reaction conditions. So, for example, here we're starting off with acetaldehyde again, and sodium hydroxide, and water, we're adding heat, so we're changing the reaction conditions, and so, we're gonna get a different product, and so we call this an aldol condensation. So you can see, the product that we get is an "enal," so we get an enal, we get a double-bond here, for the "en," and then we have an "aldehyde," so you also hear it referred to as an 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 form this enal product we're going to start with the aldol that we just made in the previous reaction. So acetaldehyde, with sodium hydroxide, can form an aldol, as we just saw, so let's go ahead and draw in the aldol product: So the same mechanism that we just discussed, and I should point out, that we are doing a base-catalyzed aldol condensation here, 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, 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 gonna draw one in there, and then we can think about sodium hydroxide acting as a base, and forming another enalate anion, so I'm gonna have sodium hydroxide come along, so, OH minus, it's gonna function as a base: it's gonna take this proton, leave these electrons behind, on this carbon, so let's go ahead and draw that. So we would have our oxygen right here, with two lone pairs of electrons, we would have our carbon now, with a lone pair of electrons making a carbanion, and then we have our carbon EEL over here, like that. Alright, so let's once again show those electrons, these electrons right in here, forming our carbanion. And then, to form our enal product, we can just think about, let me go ahead and draw an arrow here, we can 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, so we can say that hydroxide leaves and then we form our double bond, so I can go ahead and show that: Let's say these electrons right here, were the magenta ones to form our enal product. And so, once again, I'm using the carbanion form of our resonance structure, instead of the oxyanion, so again, it's probably more appropriate to use the oxyanion, but it's just easier, from a student perspective to use a carbanion, to go ahead, and form your enal product that way. So, in terms of another product, this is the major product that we would form; this is the trans product, so if you think about this double bond, these hydrogens are on opposite sides, so this is the trans product, so that's the major one. You would form a small amount of the cis product, as well, so let me go ahead and draw that in, but the cis product would be the minor product, because it has more steric hindrance than the trans product, so, this is the major product, and I'm sure the one that your professor wants you to put on an exam question, so the trans product has decreased steric hindrance. Alright, let's do one more of these aldol condensations: So here, I've drawn out their reaction, and so, the question would be, "Write a plausible mechanism "for the formation of this enone?" So we have an "enone" this time, we're starting with a ketone, and so, we have a double bond, and then a ketone, so an enone product. So we're gonna use sodium hydroxide, and we're going to heat things up here, so let's think about what would happen: so we have our ketone, and we have two alpha carbons, which are symmetrical, so it doesn't really matter which one we take a proton from, so I'm just gonna say, take a proton from this one, so we take a proton from here, to form our enolate anion, and, once again, I'm gonna form the carbanion, just to make things simpler. So we have our carbanion now, and then we have another molecule, cyclohexanone's gonna come along, so let me go ahead, and draw that in, and we know we have a polarized carbonyl situation: partial negative, partial positive. So, our nucleophilic enolate anion is going to attack right here, so we can show these electrons attacking right here, pushing those electrons off, onto your oxygen, so I'm just going a little bit fast through this mechanism here, and so we have, now, a bond that formed, 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 in 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 bond right here, and then, these electrons came off, onto our oxygen, forming our alkoxide. So we can think about water protonating our alkoxide intermediate: so lone pair takes that proton, leaves those electrons behind, and, hopefully we have enough room to draw the aldol, so, over here, we would have our aldol intermediate, we have our OH, we have our ring, like that. Okay, thinking about what happens next, the carbon next to the carbon EEL is the alpha carbon, so that's our alpha carbon, and so there's still a proton on that alpha carbon. 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: so picking up this proton, and then it 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 ring, we have our carbonyl there, and then we have a lone pair of electrons on this carbon, a negative one formal charge; over here, we have our OH, like that, and then we have our ring. Okay, so in the next step, once again, I'm not drawing an oxyanion, so I'm not gonna draw the oxyanion form of this enolate anion, here, but we can show these electrons moving into here, and pushing these electrons off, onto the oxygen, and so let's go ahead, and draw our product, once we do all of that: So we would have our ring, we would have our carbonyl here, and then, we would form a double bond, and then, now we can see that we have our product, so let's follow those electrons: So, electrons in blue here, these electrons, move in here to form our double bond, and then, 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 one way to represent the mechanism, and, once again, the reaction conditions will determine what kind of product you get: our enone, here, is conjugated, we have an alternating single-double bond situation, and so, that favors this product, here, so that shifts the equilibrium, once again, to make more of your enone. So, practice doing a lot of these aldol condensations.