If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Preparation of alcohols using LiAlH4

How to prepare a primary or secondary alcohol from an aldehyde, ketone, carboxylic acid, or ester using lithium aluminum hydride. Created by Jay.

Want to join the conversation?

Video transcript

In the last video, we saw that sodium borohydride will reduce aldehydes or ketones to form primary or secondary alcohols. And if we look at this general reaction, this is either an aldehyde or a ketone over here. If we add lithium aluminum hydride in the first step and then a source of protons in the second step-- which is water-- we will form either a primary or a secondary alcohol, depending on our starting materials. So in that respect, lithium aluminum hydride will react in the same way as sodium borohydride. However, sodium borohydride will only reduce aldehydes or ketones. It won't reduce carboxylic acids or esters. And that's what lithium aluminum hydride does. So we can see that-- if this is an OH right here-- that would be a carboxylic acid functional group. And if we take off the H and put in an alkyl group in our prime group there, we would have an ester. So lithium aluminum hydride, not only reduces aldehydes and ketones, it also reduces carboxylic acids and esters since it's more reactive, which is also why we have to separate these two. We can't have water in the same reaction vessel as our lithium aluminum hydride because it will react with that faster. And so once again, our product will depend on what our starting material is. So the mechanism for the reduction of aldehydes or ketones with lithium aluminum hydride is just like the one for sodium borohydride. So we'll move on to a mechanism for the reduction of an ester. So let's go ahead and do that. So let's start with an ester down here. So we have our carbonyl. Like that. So we'll put in our lone pairs. And down here, we have our R prime group. Like that. So there's our ester. And we add lithium aluminum hydride in excess. So in terms of molar equivalence, let's go ahead and put lithium aluminum hydride down here, Li+. And then, we have Al bonded to 4 hydrogens. Like that. And that's going to give the aluminum a negative 1 formal charge. Like that. So the first step in the mechanism is just like the one we did in the previous video. All right. We think about the carbonyl up here and the difference in electronegativity between this carbon and this oxygen-- oxygen being more electronegative, pulling these electrons closer towards it and the double bond, giving it a partial negative charge. Whereas the carbon down here is losing some electron density, becoming partially positive. So the carbonyl carbon is an electrophile. It wants electrons. And of course, it's going to get electrons from these 2 electrons in here. So these 2 electrons are going to attack this carbon. All right. So a nucleophilic attack. And kick these electrons off onto our oxygen. So that's our first step-- the nucleophilic attack portion. So now, we have R. And it used to have 2 bonds to carbon and oxygen. Now, it has only 1 bond because those electrons moved off onto the oxygen, giving the oxygen a negative 1 formal charge. Like that. So we added on our hydrogen. Like that. And then, we still have our O and our R prime group over here. So it will look like that. In the next step of this reaction, the carbonyl's going to reform. So the electrons in here are going to kick back into here to reform our carbonyl. That would mean 5 bonds to carbon-- which we know never happens-- so that these electrons are going to have to break and come off onto the oxygen. So let's go ahead and draw the results of that. So now, we have R bonded to carbon. Now, we reformed our carbonyls. Now, we have only two lone pairs of electrons on that oxygen now. And then, we still have a hydrogen. Right here. All right. So we lost an oxygen with an R prime group. And that oxygen is going to have three lone pairs around it with a negative 1 formal charge. Oxygen being relatively electronegative, it can handle that formal charge fairly well and be relatively stable. So now we have an aldehyde. And we know that this reaction can occur again with an aldehyde. So since you have extra lithium aluminum hydride floating around, what's going to happen is another reaction. Right? So another reaction just like the one we just did. We're going to have our lithium aluminum hydride again. It's so reactive, you can't stop this reaction from occurring a second time. And same idea. Right? Exact same idea. Partial negative oxygen, partial positive carbon. The carbon wants electrons. It's going to get these electrons right in here. So these electrons are going to attack this carbon, bring the hydrogen along with it, and then these electrons kick off onto the oxygen. So let's go ahead and draw the product of that nucleophilic attack. Right now, we have an R group here and we have our carbon bonded to an oxygen. Once again, it used to have two lone pairs. Now it has three, giving it a negative 1 formal charge. And this hydrogen over here on the right is going to stay here. And we added on another hydrogen. Like that. All right. So in the second step of this reaction, we're going to add water. So we go back up here to refresh our memory. In the second step of this reaction, you add water as a proton source. So let's go ahead and draw water floating around there. So here is our H20 molecule. And we're going to get an acid-base reaction. We're going to get a lone pair of electrons attacking and grabbing that proton there, kicking these electrons off onto the oxygen. So we're going to protonate that alkoxide anion. So you form our alcohol finally. So we're finally done. Now, we have r and then we have oxygen bonded to that hydrogen there. And now only two lone pairs, the formal charge goes away. And you can see what we've done here. We've actually added on two hydrogens to our original carbonyl carbon. All right. This is my original carbonyl carbon. And if I look at it over here, there's actually no hydrogen attached to it over here. Both of those hydrogens came from our lithium aluminum hydride. So this one and this one. So the reaction happened twice. All right. So if you're doing this reaction with a carboxylic acid, it's a similar mechanism. We don't have time to go through it. But you're going to end up with the same product. You're going to add on two hydrogens on to that original carbonyl carbon. Like that. So let's look at the chemoselectivity of this reaction. OK. So now that we've covered sodium borohydride and lithium aluminum hydride, let's see how you can choose which one of those reagents is the best to use. So if I start here with our reactants-- so let's make it a benzene ring. Like that. And let's put stuff on the benzene ring. OK. So let's go ahead and put a double bond here. And then, we'll make this an aldehyde functional group on one end. And then over here on this end, I'm going to put an ester. Like that. And let's look at how we can selectively transform different parts of this molecule using different reagents. So let's say we were to do a reaction wherein we add on sodium borohydride. And then, the proton source in the second step. So we need to think about what's going to happen. Sodium borohydride is selective for aldehydes and ketones only. It will not reduce carboxylic acids or esters. So it's only going to react with the aldehyde at the top right portion of this molecule. So let's see if we can draw this in here. So it's going to react with the aldehyde in the top right portion. All right. So we are still going to have our double bond here. And the aldehyde's going to go away to form a primary alcohol. So we're going to get primary alcohol where the aldehyde used to be. Sodium borohydride has reduced that carbonyl. And the second step-- the protonation of the alkyloxide-- forms your primary alcohol as your product. And the rest of the molecule's going to stay the same. So this ester is going to remain untouched down here. So it's the chemoselectivity of that reaction. All right. Let's say we start with the same starting material. And the first step-- this time we add lithium aluminum hydride. Like that. And the second step-- we add some water. Well, lithium aluminum hydride will reduce aldehydes and ketones, and it will also reduce esters. So lithium aluminum hydride in excess-- so let's just assume this is at an excess here-- it's going to react with this aldehyde portion of the molecule. It's also going to react with this ester portion of the molecule. So it's going to reduce both of those and form alcohol. So let's go ahead and try to draw the product here. All right. So we have our benzene ring, which is untouched. And up here, we know that lithium aluminum hydride is going to form a primary alcohol as well. Like that. And then, down here, what used to be our ester functional group, we're going to add two molar equivalents of hydrogen to that carbonyl. And we're going to end up breaking that bond between the carbonyl carbon and that oxygen. All of this over here is going to go away as a leaving group in our mechanism. And we're going to add on two hydrogens to that carbon. And then, that's going to form our alcohol. So we're going to add two hydrogens onto that carbon, forming a primary alcohol down here as well-- just like in the mechanism that we just discussed. So reduction of esters using lithium aluminum hydride. What about if we were to add a hydrogen gas and palladium as our metal catalyst here? Well, this is also a reduction reaction that we talked about earlier. Hydrogenation is an example of a reduction reaction. And it's going to be chemoselective. If you just used the normal conditions for a hydrogenation reaction, the only thing the hydrogenation reaction is going to touch is this double bond. It's going to reduce this double bond. So let's go ahead and draw the product. It's not going to touch the aldehyde. It's not going to touch the ester. And it's not going to touch the benzene ring. So let's go ahead and draw the product. The benzene ring is not hydrogenated under normal conditions, but we're going to add on two hydrogens across that double bond. And the aldehyde is untouched. And down here, the ester is going to be untouched as well. So that would be our product from a hydrogenation. So three different reductions, three different products. Now, hydrogen will reduce carbonyls under the right conditions. Usually, if you have increased pressure and increased temperatures, you actually can reduce those carbonyls. But again, you can control those conditions. So you can control what part of the molecule is reduced. So that sums up the ways to reduce carbonyl compounds using sodium borohydride and lithium aluminum hydride. In the next video or two, we'll take a look at organometallics and particularly the Grignard reagent.