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Current time:0:00Total duration:8:54

Video transcript

if you add lithium aluminum hydride to a carboxylic acid and your work up at a source of protons you can reduce your carboxylic acid to an alcohol and so 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 sign that oxidation state here you'll see there's been a decrease in the oxidation state so there's been a reduction so lithium aluminum hydride is one way to reduce a carboxylic acid you could also accomplish this with boring and boring is actually more chemo selective and so we'll talk about that at the end of the video right now let's focus in on a possible mechanism of lithium aluminum hydride reacting with our carboxylic acid so let's go ahead and redraw our carboxylic acid so 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 so I'm gonna go ahead and draw in alright so aluminum with four bonds to hydrogen giving it a negative 1 formal charge then we have our lithium li plus like that so a hydride we know is a hydrogen with two electrons and we know that's a strong base so you could think about these two electrons here taking this proton and leaving these two electrons behind on your oxygen alright so an acid-base reaction is probably the first step of this mechanism so if you take a proton from a carboxylic acid you're left with the conjugate base which is the carboxylate anion so a negative 1 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 alright lithium is is present so it's probably going to bond with that oxygen and we would also form hydrogen gas right so we would form a ch2 so let's show those electrons right so these electrons in blue right can pick up this proton so that forms a CH 2 hydrogen gas and then we took a bond away from the aluminum so the aluminum is now but only bonded to three hydrogen's and that takes away its formal charge so formal charge of zero now on aluminum so now that we've formed our carboxylate anion that's going to react with a la ch3 so let's go ahead and draw in our carboxylate anion here so we have our our carbonyl we have our R group and we have our oxygen with three lone pairs of electrons of a negative one formal charge so next our al a ch3 comes along so this is just one of the possibilities for the mechanism alright and 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 right it's partially positive because the oxygen is withdrawing some electron density from it so these electrons in here could move in to form a bond and at the same time all right these electrons in blue here can move out to form a bond between the oxygen and the aluminum so let's go ahead and show the result of that so we would now have our carbon would be would be tetrahedral right 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 hydrogen's and go ahead and draw in those two hydrogen's 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 and so we would have our oxygen bonded to the aluminum and we have these two hydrogen's here and so the oxygen would now have three lone pairs of electrons giving it a negative 1 formal charge and so the lithium is probably it's probably now going to bond to this oxygen and we just reformed our carbonyl all right so let's go ahead and show that so we would form our carbonyl here like that and then we have this hydrogen so let's uh let's show some those electrons right so if I say that these electrons in green here on this oxygen move in to form our carbonyl alright and then we had the electrons in red right so this hydrogen is this hydrogen and so we form an aldehyde and so we have an aldehyde and we have excess lithium aluminum hydride and the the lithium aluminum hydride is going to transfer a hydride to our aldehyde so we can we can we can go ahead and reduce our aldehyde with another lithium aluminum hydride we could draw that in so we have lithium aluminum hydride alright so negative one formal charge on the aluminum alright our carbonyl is polarized so partially negative oxygen partially positive carbon right here and so and so once again we can think about these electrons so these electrons right here from attacking this carbon right pushing these electrons and green off on to the oxygen so let's get some more room to show what happens here so this is uh this is what we've seen in in earlier videos here alright so now we would have our carbon right bonded to an oxygen this oxygen now has three lone pairs of electrons so one of those lone pairs were the ones in green alright so draw those in here like that giving that a negative 1 formal charge alright we're gonna form a bond between carbon and hydrogen so I'm gonna show that let's use blue here so these electrons right and this hydrogen right our hydride so lithium aluminum hydride acts as a hydride transfer agent and and transfers these two electrons and this hydrogen right to our carbon and then we still had this carbon in red here I mean started this hydrogen right bonded to the carbon in red like that and then we had our R group here alright so in the final step alright we would just protonate our alkoxide alright so we could add something like dilute acid in our work up here so if we add a dilute acid h3o plus alright let me go ahead and draw that in then our alkoxide could pick up a proton right leaving these electrons behind so protonate our alkoxide would yield our alcohol as our products let me and draw those in so we would have those two hydrogen's all right we have an OHA and then we would have an R group and so 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 the one in so this hydrogen and these electrons right and then also this hydrogen and these electrons both came from lithium aluminum hydride and the mechanism is is definitely more complicated than the one I showed you but this is a simple way to think about it all right let's look at a practice problem so if we had this compound over here on the left and we added lithium aluminum hydride and the source of protons in our workup we just talked about it the fact that it would reduce a carboxylic acid and 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 workup so when we draw the product alright we have our benzene ring and we would turn that ketone into a secondary alcohol as we've seen before and then the carboxylic acid here would turn into a primary alcohol so let's show some of these carbons right so this carbon right here is this carbon and then let's let's do this carbon right here in red right is this carbon so we reduced both functional groups losing lithium aluminum hydride if we did this reaction with borane so BH 3 instead borane is actually chemo selective for the carboxylic acid group only so it's only going to reduce this so 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 and so once again this carbon in red is this carbon so boring is considered to be a little bit better sometimes because of its chemo its ability to the chemo selective right it will only reduce your carboxylic acid group in this case and that's very beneficial times when you're you're not looking to reduce other parts of your molecule