Preparation of alcohols using NaBH4

Before we get to preparation of alcohols using sodium borohydride, let's take a look at a few of the other ways to make alcohols that we've already talked about in earlier videos. For example, you could make alcohols from alkenes, and you can add the OH on in a Markonikov fashion, or you can add the OH on in an anti-Markovnikov fashion. So that's one type of way to make your alcohol. So you could also do so from alkyl halides. The process could be an SN1 process, or it could be in SN2 process. And then, what we haven't talked about yet is how to prepare alcohols from carbonyl compounds, and there are a couple different ways to do that. You could use sodium borohydride, NaBH4, which is what we're going to talk about in this video. On the next video we're going to talk about the use of lithium aluminum hydride, so LiALH4. And you could also use an organometallic, something like a Grignard reagent. So I'll put that down here as well too. And we'll talk about that in a future video as well. So all we have time for this video is sodium borohydride. So let's check out the general reaction for the use of sodium borohydride to form an alcohol. And so you can see over here on the left, we're starting with either an aldehyde, or a ketone. So there's either a hydrogen attached to your carbonyl, or an R prime group like that. We're going to add sodium borohydride, and then we're going to add a proton source it could be just about anything. So H+ tends to work. And we're going to form either a primary or a secondary alcohol, depending on our starting materials. This is going to be a primary or in a secondary alcohol. Let's look at the mechanism for the preparation of alcohols using sodium borohydride. Let's go ahead and start with a ketone here. So we're going to start with a ketone. So we'll just do the reaction like that. So I have my carbonyl like that, put in my loan pairs of electrons, and then make this an R prime to make it our ketone. And sodium borohydride comes along. So let's go ahead and draw the structure for that-- so Na+, positive formal charge. And then we have boron, bonded to four hydrogens like that. And there's a negative 1 formal charge on our boron, so we'll go ahead and put that in there as well. So the mechanism I'm going to show you is a simplified mechanism. The actual mechanism is a little more complicated, but this mechanism works. So let's just go with the simplified version. All right, so the first thing you have to think about is this carbonyl here, this carbon double bonded to an oxygen. There's an electronegativity difference between the carbon and the oxygen. So right here, this carbon and this oxygen, the oxygen is more electronegative, meaning that those bonds, the electrons in the double bond between the carbon oxygen, are going to be pulled closer to the oxygen. Therefore, oxygen has a partial negative charge, right? So increased electron density around it. Whereas the carbon here is losing some of the electrons around it. So it's partially positive, like that. So if carbon is partially positive, it's our electrophile, right? It wants electronns. So where can we get our electrons? We can get our electrons from right here. So the two electrons on this hydrogen here are going to move out and attack this carbon. So that's going to be the nucleophilic portion of our molecule, like that. And that would be too many bonds to a carbon, so one of these, this pi bond here, is going to kick off onto the auction. So they're going to put two electrons on that. So let's go ahead and draw the results of that nucleophilic attack. So we have R, and then we have R carbon, and then R prime over here. And this top auction had two lone pairs of electrons that just picked up one more, for a total of three. And that gives it a negative 1 formal charge, like that. And we added on a hydrogen, like that. So let's go ahead and color code some electrons here. All right, so we can see where everything went. So I'm saying this hydrogen and the two electrons in this bond became this hydrogen and these two electrons right here. So really, what you're doing is, sodium borohydride is a source of hydride anions. So a hydride anion would be hydrogen with two electrons around it and a negative 1 formal charge. So that's the way to think about these reagents. Hydride itself is not the best nucleophile, because it isn't polarizable enough. Such a small atom that it doesn't really work very well as a nucleophile by itself. So this, of course, is our simplified version. All right, in the next step of our mechanisms, get some acid-base chemistry. So we have protons floating around, right? So we added some H+ to our solution, and the loan pair of electrons picks up the proton, and we are done. So let's go ahead and draw the product. It would be our alcohol, like that. And let's go ahead and put those lone pairs of electrons. And there's only one hydrogen added onto our carbon. So if we start with the ketone, we're going to end up with a secondary alcohol, as in the situation. All right, let's do an actual reaction here. And let's start with vanillin. All right, so vanillin is a very nice smelling molecule. And if you look at the structure of vanillin, there's an aldehyde functional group right there on our ring. So there's some other functional groups. So go ahead and put in the rest of the vanillin molecule, like that. So we're going to add sodium borohydride in our first step. We're going to add sodium borohydride, and then, once that's reacted, we're going to add a source of protons. HCL works pretty well for this reaction, so we'll just say a source of H+, like that. So when you're trying to figure out the product on a test, it's not even necessary do that really simplified mechanism. You can pretend like the hydride ion is going to act as a nucleophile, even though we've already covered that doesn't quite do that. So here we have our hydride nucleophile. So it's going to attack this carbon right here. This is our carbonyl carbon, right? So it's going to be partially positive. That's going to kick these electrons off onto our oxygen, like that. So let's go ahead and draw the intermediate for this reaction here. So we'll have our benzene ring, like that. I'll put in our other functional groups over here, which don't participate in the reaction. And let's go ahead and draw what we have here. So we had one hydrogen already bonded to that carbon. We just added another hydrogen onto that carbon. And then we have an oxygen still there with three loan pairs of electrons around it, giving it a negative 1 formal charge. So when we add our source of protons, a loan pair will pick up that proton there to form our alcohol. So we'll go ahead and draw our products right over here. So once again we have an OH coming off of our ring. And we have this part portion of the molecule, this ether portion right here. And then we have an alcohol, like that. So this molecule, since it has such a similar structure to vanillin, also has a nice vanilla smell. It's very pleasant. So this is a very good undergraduate organic chemistry lab to do. And let's classify the type of alcohol we got as either primary, secondary, or tertiary. So remember how to do that, you look at the carbon that's directly attached to the OH-- that's this carbon-- and you see how many carbons that carbon is attached to. That carbon is attached to one other carbon. So this is an example of a primary alcohol. So if you start with an aldehyde, as we did over here-- this is our aldehyde-- you're going to end up with a primary alcohol as your product. Let's do another one. This time let's do a ketone. All right, so let's go ahead and draw a ketone over here. So we'll do cyclohexanone. So here's our ketone, like that. And we're going to add sodium borohydride, like this-- NaBH4. And we'll add methanol. And we'll do this on one step here. So once again, think about sodium borohydride being a source of hydride anions. All right, so again even though this isn't technically the mechanism, it's our simplified version. So if you're taking a test, this is the easiest way to do it. Attacks that carbon, kicks these electrons off onto your oxygen. So we have our intermediates right over here. So let's go ahead and draw that. So we have the hydrides anion attack to form a new bond to what used to be our carbonyl carbon. Now this oxygen has three lone pairs of electrons around it and a negative 1 formal charge. It's going to pick up a proton from methanol as its protons source this time. And therefore, our final product, if we're going to go ahead and draw it. I'll keep the hydrogen in there, and you'll see why in a second. And then I have my OH, like that. So I form cyclohexanol as my product. This is a reduction reaction. So we are reducing the molecule to form our alcohol. So let's take the reactants and the product from this reaction and let's assign some oxidation states, so we can see how this is an example of a reduction reaction. So I'm just going to redraw our starting ketone here. So let's go ahead and put in some of the atoms this time. So I'm just going to redraw that starting ketone, like this. And I'm going to go ahead and draw the product that we got, as well. All right, so I'll put in these carbons here. And then we added on a hydrogen, like that. And then we forms our alcohol over here, like that. So let's go ahead and think about oxidation states. So when we're assigning oxidation states, we need to go ahead and draw in these electrons here, all right. So let me go ahead and put in these electrons on these guys right in here, like that. All right, so let's think about how we did this in some of the earlier videos. It's actually thinking about electronegativity. And if I'm thinking about the four electrons between carbon and oxygen, oxygen is more electronegative. So oxygen is going to get all of those electrons. So we go ahead and go like that. Carbon versus carbon, fighting over these two electrons, it's an equal electronegativity, obviously, since it's the same element. So each carbon's going to get one of those electrons, all right. And the same thing for over here. So when we assigned our oxidation states, we said that's the number of valence electrons that carbon normally has, which, of course, is four. And from that we subtract the number of electrons around it when account for electronegativity, which is two in this case. So the oxidation state of that carbon is plus 2. So we have a plus 2 oxidation state on the left. Let's think about the right. Let's go ahead and put in our electrons. Each bond consists of two electrons. So we go ahead and put the fact that each bond consists of two electrons in here, like that. And then once again we think about differences in electronegativity. So again, we know carbon versus carbons is a tie, so we can go like that. And carbon versus oxygen-- oxygen's more electronegative, so it's going to get those electrons there. But carbon verses hydrogen-- carbon is a little bit more electronegative. So now carbon's going to take those two electrons. So carbon normally has four. In this case, carbon has four around it after we count for electronegativity. So the oxidation state of that carbon atom is zero. So let's think about what happened to the oxidation state. We started off with a plus 2 oxidation state, and that number was reduced to an oxidation state of zero. So this is a reduction reaction, right? This is a reduction. So several different definitions you could use. You could think about reduction being a decrease in the oxidation state, or a reduction in the oxidation state. You could also think about that carbon gaining electrons-- so it picked up two more electrons. And yet another way to think about it is carbon lost a bond to oxygen, Over here on the left, carbon had two bonds to oxygen. Over here in the right, carbon only has one bond to oxygen. And it formed a bond with hydrogen. So increased the number of bonds to hydrogen, decreased the number of bonds to oxygen is another way to think about this being a reduction reaction. But assigning your oxidation states is probably the best way to do it. In the next video, we'll take a look at how to prepare alcohols using lithium aluminum hydride, which is very similar to using sodium borohydride.