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Reduction of carboxylic acids

How to reduce carboxylic acids using lithium aluminum hydride (and borane). Created by Jay.

Video transcript

Voiceover: If you add lithium aluminum hydride to a carobxylic acid, and then your work up at a source of protons, you can reduce your carboxylic acid to an alcohol. If you think about the oxidation state of this carbon, if you assigned one, that's this carbon over here on your alcohol. And if you assigned an oxidation state here, you'll see there's been a decrease in the oxidation state. So, there's been a reduction. Lithium aluminum hydride is one way to reduce a carboxylic acid. You could also accomplish this with borane, and borane is actually more chemoselective. We'll talk about that at the end of the video. For right now, let's focus in on the possible mechanism of lithium aluminum hydride reacting with our carboxylic acid. Let's go ahead and re-draw our carboxylic acid. I'm gonna go ahead and put in our carbonyl. And then, we know the acidic proton on our carboxylic acid is the one on the oxygen. Lithium aluminum hydride can be a strong base. I'm gonna go ahead and draw it in. So, aluminum with four bonds to hydrogen, giving it a negative one formal charge. Then we have our lithium, [L I] plus, like that. A hydride we know is a hydrogen with two electrons, and we know that's a strong base. You can think about these two electrons here taking this proton and leaving these two electrons behind on your oxygen. So an acid-base reaction is probably the first step of this mechanism. If you take a proton from a carboxylic acid, you're left with a conjugate base, which is the carboxylate anion. So a negative one formal charge on this oxygen. And we could follow those electrons, so these electrons in magenta move on to this oxygen to form our carboxylate anion. Lithium is present, so it's probably going to bond with that oxygen. We would also form hydrogen gas. So we would form H two, so let's show those electrons. So these electrons in blue, or I could pick up this proton, so that forms H two, hydrogen gas. And then, we took a bond away from the aluminum, so the aluminum is now only bonded to three hydrogens, and that takes away it's formal charge. So, formal charge of zero now on the aluminum. Now that we've formed our carboxylate anion, that's going to react with [A L H] three. Let's go ahead and draw in our carboxylate anion here. We have our carbonyl, we have our R group, and we have our oxygen with three lone pairs of electrons, so negative one formal charge. Next, our [A L H] three comes along. This is just one of the possibilities for the mechanism. We're gonna form a bond between the oxygen here and the aluminum. And we're gonna form a bond between the carbon and the hydrogen. So, if we think about these electrons in red right here, this carbon is partially positive because the oxygen is withdrawing some electron density from it. So these electrons in here can move in to form a bond. And, at the same time, these electrons in blue here can move out to form a bond between the oxygen and the aluminum. Let's go ahead and show the results of that. We would now have our carbon, it would be tetrahedral. So let's draw it like this. We would have a bond to this hydrogen right here. So the electrons in red move in here to form this bond, and then that carbon is still bonded to an oxygen with three lone pairs of electrons, so it still has a negative one formal charge, like that. And then we would have our carbon bonded to this oxygen. This oxygen has two lone pairs of electrons on it. And we just formed a bond to the aluminum. So the electrons in blue form our bond here, and the aluminum is still bonded to two hydrogens. We can go ahead and draw in those two hydrogens. Alright, if we take these electrons and move them in here to form our double bond, we would have to push these electrons off onto our oxygen. So we would have our oxygen bonded to the aluminum, and we have these two hydrogens here, so the oxygen would now have three lone pairs of electrons, giving it a negative one formal charge. The lithium is probably now going to bond to this oxygen. And we just reformed our carbonyl, right? So let's go ahead and show that. We would form our carbonyl here, like that. And then we have this hydrogen. Let's show some of those electrons. So, if I say that these electrons in green here, on this oxygen, move in to form our carbonyl, right? And then we had the electrons in red, so this hydrogen is this hydrogen, and we form an aldehyde. We have an aldehyde and we have excess lithium aluminum hydride. The lithium aluminum hydride is going to transfer a hydride to our aldehyde. So we can go ahead and reduce our aldehyde with another lithium aluminum hydride. I'm gonna draw that in. We have lithium aluminum hydride, so negative one formal charge on the aluminum. Our carbonyl is polarized, so partially negative oxygen, partially positive carbon right here. So, once again we can think about these electrons, these electrons right here, from attacking this carbon, pushing these electrons in green off onto the oxygen. Let's get some more room, to show what happens here. This is what we've seen in earlier videos here. Now we would have our carbon bonded to an oxygen. This oxygen now has three lone pairs of electrons. One of those lone pairs were the ones in green. I'll draw those in here, like that, giving that a negative one formal charge. We're gonna form a bond between carbon and hydrogen, so I'm gonna show that. Let's use blue here. These electrons and this hydrogen are a hydride, so lithium aluminum hydride acts as a hydride transfer agent and transfers these two electrons and this hydrogen right to our carbon. Then we still had this hydrogen, in red here, bonded to the carbon in red, like that, and then we had our R group here. The final step, we would just protonate our alkoxide, so we could add something like dilute acid in our work up here. If we add a dilute acid, H three O plus, we go ahead and draw that in, then our alkoxide could pick up a proton, leaving these electrons behind. Protonating our alkoxide would yield our alcohol as our product. Let me go ahead and draw those in. So we would have those two hydrogens, we have an OH, and then we would have an R group. That's one of the possibilities for the reduction of a carboxylic acid, with lithium aluminum hydride. With the end result of transferring two hydrides, right? So this hydrogen and these electrons, and then also this hydrogen and these electrons. Both came from lithium aluminum hydride. The mechanism is definitely more complicated than the one I showed you, but this is a simple way to think about it. Let's look at a practice problem. If we had this compound over here on the left and we added lithium aluminum hydride and the source of protons in our work up, we just talked about the fact that it would reduce a carboxylic acid. We have a ketone present here as well, and we've seen in earlier videos that lithium aluminum hydride will reduce the ketone as well, and turn that into an alcohol on our work up. So when we draw the product, we have our benzene ring, we would turn that ketone into a secondary alcohol, as we've seen before. Then the carboxylic acid here would turn into a primary alcohol. Let's show some of these carbons. This carbon right here is this carbon, and then let's do this carbon right here, in red, is this carbon. We reduced both functional groups using lithium aluminum hydride. If we did this reaction with borane, so [B H] three instead, borane is actually chemoselective for the carboxylic acid group only. So it's only going to reduce this. If we draw the product using borane, we would have our benzene ring, and the borane wouldn't touch the ketone, so that is left here. It would reduce the carboxylic acid, so we would turn that into a primary alcohol. So once again, this carbon in red is this carbon. Borane is considered to be a little bit better sometimes, because of its ability to be chemoselective, right? It will only reduce your carboxylic acid group, in this case, and that's very beneficial sometimes, when you're not looking to reduce other parts of your molecule.