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Video transcript
Lecturer: In the previous video we looked at aldol condensations with the same molecule, you call those a simple aldol condensation. In this video we're going to look at mixed or crossed aldol condensations, so no longer are you starting with the same molecule. Here we don't have two aldehydes that are the same, we have different aldehydes. We have benzaldehyde on the left and propanal on the right. We need to figure out what sort of enolate anion that we're going to form. When we add our sodium hydroxide as our base, what is going to be our enolate anion? To do that we need to look for alpha carbons. We'll start with propanal. We know the alpha carbon is the one next to the carbonyl, and so right here this would be an alpha carbon, and there are two protons on that alpha carbon, so we have two alpha protons here. Let me go ahead and draw in those. That's a possibility to form our enolate anion. This aldehyde hydrogen right here is not, we're not going to form an enolate from that, so we don't have to worry about the right side of our aldehyde. Let's look at benzaldehyde now. If we think about this carbon right here it is the carbon next to a carbonyl, however this carbon already has four bonds to it, so it doesn't have an alpha proton. Benzaldehyde does not have an alpha proton, so we don't need to worry about it forming an enolate anion. We know the enolate anion is going to form from propanal over here. Let me go ahead and redraw these in a way that makes it easier for me to see what the product is. I'm going to draw my ring here. I'm going to draw my carbonyl. I like to leave off this hydrogen because it gets in the way when I'm thinking about my mechanism. For my propanal over here I prefer to draw a different confirmation of propanal to make it easier to see the product. I like to put my carbonyl up here like this, and I like to draw this carbon down like that, and it's going to make it easier to come up with the final result. If I think about deprotonation, if I think about sodium hydroxide, taking one of these alpha protons here, leaving electrons behind in my alpha carbon, I can think about the structure of the carbanion enolate anion. There's a negative one formal charge on this carbon, and that's my nucleophilic enolate anion. Once again I prefer to draw the carbon as opposed to the oxyanion, it just makes it easier for me when I'm trying to do a quick mechanism on a test to figure out a product. Now we have a nucleophile. I'm going to go ahead and put these electrons in magenta. This is going to be our nucleophile, and our nucleophile is going to attack the electrophilic portion of our benzaldehyde molecule. If we think about that the oxygen is partial negative and this carbonyl carbon here is partially positive, that's the electrophilic portion of the molecule. That lone pair of electrons here on this carbon is going to attack this carbon, pushing these electrons off onto the oxygen. That would form an alk oxyde intermediate. Let me go ahead and sketch in here. We would form an oxygen with three lone pairs of electrons, giving it a negative one formal charge, and I'll come back to that. The important carbon-carbon bond that's formed is right here. Then we have, I'm going to go ahead and draw in everything else. We have our aldehyde, and don't forget about this down here. In magenta, these electrons in magenta, have formed our new bond right here. Then we would have an alkoxide intermediate and I'm going to think about protonating it. I'm not concerned with the exact mechanism here, I'm more concerned about figuring out the product. We go ahead and proteinate our alkoxide to form our aldol intermediate. In the next step, when you add heat you're most likely going to form your conjugated product. Let's think about what would happen next. We still have an alpha carbon. This carbon right here next to our carbonyl is an alpha carbon, it still has an alpha proton on it. We can show a proton right here. And we can think about hydroxide once again acting as a base. Hydroxide acting as a base, coming along and taking this proton. We can even think about the final product, we can think about these electrons moving into here and then these electrons moving out onto here. Again I'm not so concerned with the exact detailed mechanism, I'm more worried about how to figure out our product here. When I draw our product, we have our ring, and we now have a double bond that forms here. We have this going down and then we have our aldehyde. We form a conjugated product, we have an enal as our product. Let's follow those electrons. Let's make these electrons in blue here. These electrons in blue moved in here to form our double bond. We already formed this bond in magenta, the carbon-carbon bond forming part to form our aldol, and then we lost hydroxide. This would be our final product, our conjugated product. Again, the goal is just to figure out how to draw a product from these reactants, and thinking about where is the alpha carbon and thinking about what is the enolate that forms. Let's do another one where I walk you through my thought process. Let's look at this reaction. Say you had this on an exam, and your task was to draw the product. We have once again some different possible alpha carbons. Let's first focus on the cyclohexanone. We know that this could be an alpha carbon, and we know that this could be an alpha carbon. Then for this compound over here we have these two carbonyls in this molecule. We know that this could be an alpha carbon and we know that this could be an alpha carbon. Now we have to figure out which one of those alpha carbons is going to be deprotonated when we add sodium methoxide as our base here. If you remember one of the earlier videos we talked about an alpha carbon between two carbonyls as having the most acidic protons. This alpha carbon right here has two protons on it, and it's easy to deprotonate, because of the resonance stabilization that we can draw because of those carbonyls. These protons are the most acidic, much more acidic than the protons on our ketone here. These are the ones that are going to be, one of these protons could be deprotonated when we add sodium methoxide here. Sodium methoxide is going to come along, take one of these protons here, leave these electrons behind. Let's go ahead and redraw what we would form here. We're going to form our enolate anions. Let me go ahead and draw that. We would now have a lone pair of electrons on our carbon, giving that carbon a negative one formal charge. Let me follow those electrons, these electrons in magenta are now on our carbon, forming our carbanion. I'm not going to take the time to draw the resonance structure, the oxyanions, because I'm just concerned about figuring out the product here. Now we have a nucleophilic enolate anion, and we know that's going to attack the carbonyl of our ketone. Going back over here to our ketone, the oxygen is partial negative, this carbon is partially positive, so our nucleophile is going to attack our electrophile. You can think about these electrons attacking here, pushing those electrons off on to your oxygen. If we were to draw the intermediate here we would have our ring. we're going to form an alk oxide, so I'm not going to draw all the lone pairs on that oxygen right now. I'm more concerned right now with showing the formation of this bond. Let me go ahead and draw in everything and we'll follow some electrons. I put in my carbonyls, then I have these guys over here like that. The electrons in magenta, these electrons formed our carbon-carbon bond. They formed this bond right here. Then we would form an alk oxyde. Then we would go ahead and protonate to form our aldol. Once again just to save time, not an exact mechanism but thinking about our intermediate as being this aldol. Because we have heat we're probably going to once again form a conjugated product here and keep going for the complete aldol condensation. Next we can think about this alpha carbon right here still having an acidic proton on it. There's still a proton attached to that alpha carbon. Sodium methoxide come along. We have a methoxide anion, let's go ahead and draw that in here, which could function as our base. It could take that proton, and think about these electrons moving into here to form our double bond, we can think about hydroxide as a leaving group. That allows us to figure out our product. We have our ring, and then we now have a double bond, and then we have a carbonyl over here, then a carbonyl over here, then our oxygen and our ethyl like that. Let's think about those electrons. The electrons in blue here move in to form our double bond, and we had already formed a carbon-carbon bond before, so let's say it's these electrons right here. Then we have a stable conjugated product. Once again, I'm just thinking about how to draw the product for a reaction on an exam. Hopefully this helps.