Concept: Concept: Crossed Aldol10m
Hey guys in this video we're going to explore, what happens when you try to run an aldol condensation on a mixture of different aldehydes. So, whenever you run a condensation on two different ketones or two different aldehydes mixed products are going to be difficult to avoid because what's going to happen is that we've got two different enolates that are possible, not within the same ketone or aldehyde but on different aldehydes, so that opens up a big can of worms because, for example, let's say, we have aldehyde A that can form an enolate here, an aldehyde B that can form an enolate here, we've got an issue because it turns out that once you form a enolate A, A can react with itself to give you A-A, right? So, A negative could attack A the actual protonated version of A to give you A-A, A negative could also attack B so the non enolate version of B to give you A-B, condensation product, B negative, so the enolate of B could attack A to give you BA, which by the way A-B and B a look like they're similar but they're not there are going to be different compounds because you have different numbers of carbons in each and finally B negative could react with B to give you B-B. So, what's going to happen guys, you're going to get this terrible like punnett square of Aldol products and it's just going to be a disaster, okay? Now, for the sake of teaching purposes I am in going to fill in this punnett square. So, you guys can see all the possibilities, but typically this is not something that we would ever do in a classroom, you would never draw all four aldol products because it's just it's a synthetically useless reaction, no one wants that high of a mixture of products but for the sake of learning, let's just see what each of these combinations look like, and always remember, that the first letter I'm using as the enolate. So, I'm doing enolate, enolate, enolate. So, doing my whole left right thing, let's start off with A-B and I am going to, you might need to grab some extra paper for this because it's going to be a lot, I'm going to minimize this and keep this in the corner, I know I'm cheating, okay? Alright, so let's start off with A-B. So, A-B you would look like this and I'm going to try to draw small, it would be that my enolate looks like this. Remember, that I told you guys that you should draw your R groups facing down and your anion am facing towards the right plus B, which I would have to actually draw my H facing towards the left and my CH3 facing towards the right. Remember, that you want your smallest group towards the anion. So, what I'm going to get is a product that looks like this where I have H carbon, alcohol carbon, okay? Now, for the sake of just drawing these quickly, I'm going to be drawing the beta hydroxy carbonyls but you guys know that you could also dehydrate these. So, let's put that one here and again the hydration product would probably be the more accurate one, okay? Now, to make sure we did it, right? We should count up our carbon to see if the number of carbons in A and B add up, so notice that A has three carbons and B has two carbons, so I just have a five carbon product, do I? 1, 2, 3, 4, 5, good. So, at least I'm right with the numbers, let's do A-A. So, A-A should be this enolate, same enolate plus this unreacted. Now, notice really minor little detail here. Notice that on the electrophile side I kept the R group facing up but on the enolate it faced down, guys it doesn't really matter which way you fix the electrophile side, I'm just facing it up because that's what I'm normally used to seeing, the only one you have to be careful about is the one that's facing down because notice that my product actually made sense because I made it face down originally, okay? So, I'm actually saving myself a ton of time by doing these things ahead of time. So, I'm going to do my mechanism, again remember, these are all reversible arrows guys. So, again my apologies, I'm kind of overlooking a few things here, but what we would get is H carbon attached to OH and now two carbons coming off the chain, okay? See how that changed things. So, this product would be this was A-A, right? So, this was H carbon, OH carbon, carbon. So, now we should count up our carbons again to see if it's right.
This one was a five carbon chain, let's see the other one. Now, if it's A-A that should be three plus 3, that should be six 1, 2, 3, 4, 5, 6, yep, six carbons, count up yourself, okay? So, we've got two of these. Notice how they're different, I mean, they look kind of similar but there's carbons in different places. So, B plus A. So, guys B looks like this, H methyl plus A, which looks like this, that's going to give me a compound that looks like this, carbon alcohol carbon, so notice that this one, let's just draw it first. So, I've got my aldehyde and I got my alcohol and I got my ethyl, because. So, this one B plus A should also be 5, is it 5? 1, 2, 3, 4, 5, yes it is, Now guys, notice that now I got A plus B and B plus A together, are they the same compound? no, they both have five carbons but they're slightly different, these are the kinds of terrible mixtures that you get when you do a cross aldol, okay? So, it's just like, that's what this is called cross aldol, it sucks, you don't want to do this, finally B plus B is going to give me, that is actually the simplest one to draw, and that I'm going to do my mechanism, I'm going to get H, alcohol, carbon and that's going to give me a molecule, looks like this. Alright, so I should have four carbons, do I? 1, 2, 3, 4, yes because it's B plus B, 2 plus 2, excellent guys. So, as you can see, this is literally a synthetic chemists nightmare, they do not want to see this, this is like what keeps them awake at night, okay? Not even a nightmare, they can't go to sleep, so how do we avoid this from happening? Well, one, if you have multiple enolates that are possible it's just going to happen. So, don't do it, but that's what we try to do, if you're going to do a mixed reaction or cross reaction, then only work, make sure that one of the enolates are, sorry, make sure that one of the aldehydes or ketones is going to be nonenolizable, okay? What does that mean? nonenolizable means it doesn't have an alpha proton, I'm going to put it here, non alpha proton, if it doesn't have an alpha proton then you can't do anything about it, you can't make an enolate out of it. So, these are some great examples, formaldehydes, does that have any alpha hydrogens? no guys, it has aldehydes hydrogens but you can't remove those, so this is, this would be formaldehyde, benzaldehyde, another great example, see how benzaldehyde doesn't have an alpha hydrogen and I mean you could name this one, I guess it would be 2, 2 dimethylpropanal, right? I don't really care that you memorize all the different good nonelizable carbonyls but you should just be able to visualize that, hey, if it doesn't have an alpha carbon or, I'm sorry, an alpha hydrogen then that's a good candidate for a cross why? Because then these are always going to be forced to be always the electrophile, that means that they're always going to get attacked. So, you will only get one product because, what you're going to get is your enolate, let's say it's A minus attacking this or attacking this and you're not going to worry about B minus as well, it would just be your enolate attacking the electrophile, see how that simplifies things? Cool guys. So, hope that made sense, hope that didn't remind you too much of genetics, let's move on to the next video.
Problem: Give the structure of the aldehydes or ketones used to create the product prepared by a crossed aldol condensation reaction.3m
Problem: Give the structure of the aldehydes or ketones used to create the product prepared by a crossed aldol reaction.3m
Problem: What product can be isolated from the following aldol condensation reaction?3m
Propose mechanisms for the following base-catalyzed condensations, with dehydration.
(a) 2,2-dimethylpropanal with acetaldehyde
(b) benzaldehyde with propionaldehyde
What two molecules produced the molecule below under heating in aqueous solution with catalytic base?
The following mixed aldol reaction would produce which of the products shown?
1) A and B only
2) A only
3) B and D only
4) C and D only
5) a mixture of all 4
Which of the following is not a product of the following reaction?
Complete the following reactions by drawing the structures of the principal major products in the box shown. Indicate relative stereochemistry where necessary. If there is no reaction, write NR.