|Ch. 1 - A Review of General Chemistry||4hrs & 47mins||0% complete|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 18mins||0% complete|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete|
|Ch. 8 - Elimination Reactions||2hrs & 21mins||0% complete|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete|
|Ch. 10 - Addition Reactions||3hrs & 28mins||0% complete|
|Ch. 11 - Radical Reactions||1hr & 55mins||0% complete|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 42mins||0% complete|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 20mins||0% complete|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete|
|Ch. 23 - Amines||1hr & 43mins||0% complete|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete|
|Ch. 25 - Phenols||15mins||0% complete|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete|
|Ch. 26 - Transition Metals||5hrs & 33mins||0% complete|
|Condensation Reactions||6 mins||0 completed|
|Aldol Condensation||18 mins||0 completed|
|Directed Condensations||8 mins||0 completed|
|Crossed Aldol Condensation||17 mins||0 completed|
|Claisen-Schmidt Condensation||5 mins||0 completed|
|Claisen Condensation||17 mins||0 completed|
|Intramolecular Aldol Condensation||16 mins||0 completed|
|Conjugate Addition||9 mins||0 completed|
|Michael Addition||17 mins||0 completed|
|Robinson Annulation||21 mins||0 completed|
|Intramolecular Condensation Retrosynthesis|
|Acid-Catalyzed Michael Reaction|
|Michael Addition Retrosynthesis|
|Alpha and Beta Alkylation|
|Robinson Annulation Retrosynthesis|
|Condensation Texas Two-Step|
|Cumulative Condensation Reactions|
So far condensation reactions seem pretty straight forward. But, let's see what happens when we have an asymmetrical ketone.
Concept #1: Directed Condensations
Hopefully by now, the aldol condensation is starting to make a little bit more sense. But what happens when you have an asymmetrical ketone? That presents a problem. Whenever you're reacting an enolate-mediated reaction on an asymmetrical ketone, two enolates may be possible. We're going to have to use a directed reaction.
Directed reactions are what we use to pick the enolate that we want because if you only have one choice of enolate, then your enolate is going to hit your electrophile and you're done. But what if you have two possible enolates? Which one is the one that attacks the electrophilic carbon? Who knows? That's why we have to use thermodynamic versus kinetic control.
The thermodynamic enolate, we've learned this before, is the more substituted one. It's going to be favored by small bases. Whereas the kinetic enolate is the less substituted once. It’s the one that’s easier to reach and it’s favored by bulky bases. What that means is that if I want to run an aldol reaction, let's say on the right side of my ketone here, I only want to attack with the enolate on the right side. Then I would use a small base, for example NaOH or any other small base. However, if I wanted to react on the less substituted side of the ring making my enolate on the left side and then having that attacking electrophile, then I definitely have to use a bulky base. For that, we’ve got two options. But the most popular for this chapter being LDA because of the fact that it's a non-nucleophilic base. So we don't have to worry that it's going to actually add to anything. It’s just going to remove a hydrogen. But also, tert-butoxide would be a possibility.
Excellent. Now, go ahead and try to predict the product for the following reaction and then I'll jump in.
Example #1: Predict the Products
Okay. So, what base was this? Guys, I hope you didn't get tripped up on that because I've drawn this for you already, this is just another way to represent LDA, right? LDA stands for lithium diisopropyl amide and that's what we have up here, we got the 2 isopropyl groups, they're negatively charged N and your lithium associated with it, okay? So, whereas my enolate is going to form, by the way another question, let's back up, hold up, we know this is in the aldol section. So, you're obviously thinking along that lines but how would you know that this is even an aldol reaction, we've got LDA and a ketone, how do I know what to do. Remember, I told you guys that anytime you have a base plus a ketone, let me rearrange that, anytime that you have a carbonyl like an aldehyde or ketone, it's important guys, plus a base, okay? To make the enolate, plus no other electrophile, okay? That's going to be an aldol, okay? Because, what that means is that you're going to form an enolate but you're not going to have an electrophile to react it with. So, it's just going to react with itself, okay? So, that's exactly what we have here, we have a ketone with LDA, strong base, gonna make an enolate and no other electrophile. So, I would make my enolate on this side, that means is the less substituted side, it's a directed reaction, I'm going to flip this thing in order to use my rules of how to set up an aldol, right? So, let's go for it.
So, remember I told you guys they always want the enolate facing the anion towards the right, okay? So, L'd twist that around set my anion is facing what's going to be the electrophile, my electrophile I want to draw with the smallest group facing towards the anion. So, I would keep it just as is, okay? We're good to go, we're good to do our first mechanism. So, my negative attacks the O, kicks up the electrons, I wind up getting something like this where I have a molecule that looks like this carbon. Now, that is attached to, trying to use green, that is attached to, that's the new bond and that is attached to what? an O negative and this thing and the methyl group, okay? So, what's going to happen here? Well, basically the conjugate of LDA. So, remember LDA deprotonated, right? So, the conjugate could protonate this. So, we're always going to get a protonation, you're never going to be stuck at the tetrahedral intermediate, you can always use at least the hydrogen that you took to replace this. So, I'm just going to put H+ because I know that there's at least one hydrogen hanging around since I took it off, okay? What this is going to give me, and I ran out of room over here. So, by the way these are all reversible arrows guys, what this is going to give me, is it's going to give me a molecule, looks like this, okay? And that is one of the final answers. So, we're done, good, that's the beta hydroxy carbonyl, but guys I told you that I'm going to be spontaneously dehydrating these guys, why? Because that just seems to be the thing to do, okay? It's going to make us very stable product. So, let's just spontaneously dehydrate this thing, be done with it also we have strong base around. So, there's a justification. So, when you dehydrate, you always dehydrate between the Alpha and the beta, okay? to make an alpha-beta unsaturated. So, then I would draw it like this, my final product, okay? This is my alpha-beta unsaturated and it's also called an enone, right? Awesome guys. Now, actually one thing that I didn't get to mention for the general mechanism of aldol that I'm really glad that we're talking about here, okay? Which is that guys aldol is a reversible reaction. So, you can turn a ketone into a beta hydroxy carbonyl and it can go back, just fine, depending on equilibrium conditions, but there's one irreversible step, the irreversible step is dehydration, once you dehydrate you're not going back, you're not getting alcohol and guys that is the reason why another reason why the dehydration product is the predominant product because once you achieve it you're never going back, so it keeps getting removed from the solution. So, you can imagine that over time you're going to have a 100% dehydration product because you keep removing that reagent from the solution as you create it. So, eventually even, if it's a slow reaction, the rate is slow, you're still going to generate a higher and higher yield at the hydration product until everything has dehydrated, okay? So, it's stable because it's conjugated but it also happens to be a final product because it's an irreversible step of the reaction, okay? So, if that makes more sense as to why, I'm going to be dehydrating these compounds, awesome guys, we did great, let's move on to the next topic.
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