Oxidation of alcohols

Let's see what happens when you oxidize alcohol. So in the top left here, we're starting with a primary alcohol. And the carbon that's attached to the OH group is your alpha carbon. To oxidize an alcohol, you must have alpha hydrogens. You must have hydrogens attached to that alpha carbon in order for the mechanism to work. So in that mechanism, you're actually going to lose one of those alpha hydrogens. And we'll take a look at the mechanism in a few minutes. So if I were to oxidize this primary alcohol, I'll add something to oxidize my primary alcohol, like that. One way to think about the oxidation of an alcohol is to think about the number of bonds of carbon to oxygen. On the left side here, we have one bond of our alpha carbon to this oxygen. In the mechanism, we're going to lose a bond of carbon to hydrogen, and we're going to gain another bond of carbon to oxygen. So you're increasing the number of bonds of carbon to oxygen. So that would, of course, give me two bonds of carbon to oxygen, if I oxidize my alcohol one time. And I'm going to lose one of those hydrogens. So one of those hydrogens is still left. And my alkyl group is still attached. So obviously this would give me an aldehyde functional group. So if you oxidize a primary alcohol one time, you'll get an aldehyde. Let's take a look at the oxidation states of my alpha carbon and see what happened to it. So if I want to assign an oxidation state to my alpha carbon on the left, once again, I have to put in my electrons. Each bond consists of two electrons, like that. And I need to think about electronegativity differences. Oxygen is more electronegative than carbon, so it's going to take those two electrons. Carbon versus carbon is a tie, so each carbon will get one of those electrons. Carbon actually is slightly more electronegative than hydrogen, so carbon will win and take those electrons right there. Carbon normally has four valence electrons, and in this instance, it is being surrounded by five electrons. So four minus five will give me an oxidation state of negative one for my alpha carbon. So let's look and see what happened to that alpha carbon after we oxidized it. Right? So over here on the right, if I wanted to assign an oxidation state to what is now my carbonyl carbon, once again I think about my electronegativity differences. And I know that oxygen is going to beat carbon. Carbon versus carbon is a tie. And carbon versus hydrogen-- carbon will win. So the oxidation state of that carbon-- normally four valence electrons surrounded by three this time. So 4 minus 3 will give me plus 1. So I can see that my oxidation state went from negative 1 to plus 1. So an increase in the oxidation state is, of course, oxidation. So if you oxidize a primary alcohol one time, you will get an aldehyde. What about if you keep going? Right? So if you form an aldehyde-- and sometimes it's hard to stop the reaction mixture from continuing to oxidize. So if you oxidize an aldehyde, you think about what functional group you would get. Well, again, a simple way of doing it would be to think on the left side, I have two bonds of carbon to oxygen. Is there any kind of functional group where carbon is bonded three times to an oxygen? So that, of course, would be a carboxylic acid. So if I think about the structure of a carboxylic acid, I can see that carbon is actually bonded three times to an oxygen, if you will. Three bonds of carbon to oxygen. And over here, I have my alkyl group, like that. So if you oxidize an aldehyde, you're going to get a carboxylic acid. Let's look again at the oxidation state of my carbonyl carbon. So once again, I put in my electrons here and I think about electronegativity. So oxygen, of course, beats carbon. Right? Tie between these two carbons. And oxygen beats carbon again. So in this case, normally four valence electrons. Now there's one. So 4 minus 1 gives us an oxidation state of plus 3. So once again, an increase in the oxidation state means oxidation. If you oxidize an aldehyde, you'll get a carboxylic acid. Let's look at a secondary alcohol now. So we'll go down here to our secondary alcohol. And once again, identify the alpha carbon, the one attached to your OH group. We need to have at least one hydrogen on that alpha carbon, and so we have one right here. So if we were to oxidize our secondary alcohol-- so we're going to oxidize our secondary alcohol. Once again, a simple way of doing it is thinking my alpha carbon has one bond to oxygen, so I could increase that to two bonds and that should be an oxidation reaction. In the process, I'm going to lose a bond to my alpha hydrogen. So I'm now going to have two bonds of carbon to oxygen, and I'm going to lose the bond that that alpha carbon had with the hydrogen there. So that leaves my two alkyl groups, like that. So now I have two alkyl groups. And of course, this would be a ketone functional group. So if you oxidize a secondary alcohol, you're going to end up with a ketone. I can assign oxidation states. So once again, let's show that this really is an oxidation reaction here. And I go ahead and put in my electrons on my alpha carbon, and think about electronegativity differences. So once again, oxygen beats carbon. Carbon versus carbon is a tie. Carbon versus hydrogen, carbon wins. And then carbon versus carbon, of course, is a tie again. So normally, four valence electrons. In this example, it's surrounded by four. So 4 minus 4 gives us an oxidation state of 0 for our secondary alcohol. And when I oxidize it, I'm going to get this ketone over here on the right. So let's take a look at the oxidation state of the carbon that used to be our alpha carbon on the left, which is now our carbonyl carbon. So once again, we put in our electrons and we think about electronegativity difference. Right? So oxygen is going to beat carbon. So we go like that. Carbon versus carbon is a tie. Carbon versus carbon is a tie once again, [? it's ?] normally 4, minus 2 this time around that carbon giving us an oxidation state of plus 2. So to go from a secondary alcohol to a ketone, we see there's an increase in the oxidation state. So this is definitely an oxidation reaction. Let's look now at a tertiary alcohol. So here is my tertiary alcohol. And when I find my alpha carbon, I see that this time, there are no hydrogens bonded to my alpha carbon. So according to the mechanism which we'll see in a minute, there's no way we can oxidize this tertiary alcohol under normal conditions anyway. So if we attempted to oxidize this, we would say there's no reaction here since we are missing that alpha hydrogen. Let's take a look at the mechanism and see why we need to have that alpha hydrogen on our alpha carbon. So if I were to start my mechanism here with an alcohol, remember, this must be either a primary or a secondary alcohol in order for this oxidation to work. So I'm going to go ahead and show my alcohol there. All right. So again, either primary or secondary, like that. And when we have our primary or secondary alcohol-- and it's going to be reacting with chromic acid. So here is the dot structure for chromic acid, like that. So I'll just simplify it right here. I won't worry too much about my lone pairs of electrons. And chromic acid can come from several different reagents. Probably the most common reagent would be sodium dichromate. So Na2Cr2O7. Sulfuric acid- H2SO4. And water. And all of this together is usually referred to as the Jones reagent. So a mixture of sodium dichromate, sulfuric acid, and water is called the Jones reagent. And that will mix together to give you chromic acid in solution. OK. So another way to do it would be, you could start from chromium trioxide. So you could also use a different reagent which consists of CrO3, chromium trioxide, and H3O plus, and acetone, and that will also generate chromic acid in solution. So whichever one you would like to use. The first step of the mechanism is similar to the formation of nitrate esters that we saw in the previous video. OK. So this is going to be a reaction equilibrium, or it's reversible. And if you remember, in the formation of nitrate esters, it's a similar mechanism for the formation of all inorganic esters here. And we're going to lose this hydrogen and this OH, and those are going to produce water. And we can stick those two molecules together. And so we would get this as the initial product here. We're going to have the end result of putting that oxygen bonded to that chromium atom like this. So this is a chromate ester intermediate. All right. So this is what we would make. All right. So in the next step of the mechanism, we need something to function as a base. And water is going to do that for us. So water comes along like this. Two lone pairs of electrons. One of those lone pairs can function as a base, and it's going to take that alpha proton. Remember, this is our alpha hydrogen on that carbon. And over here, we're going to take just the proton, just the nucleus of that hydrogen atom. And so this lone pair of electrons in here could take that proton. That's going to leave the electron that hydrogen brought to the dot structure behind, and these two electrons are going to move into here to increase the number of bonds of carbon to oxygen. At the same time, that is going to kick these electrons in this bond off onto the chromium. So let's go ahead and draw the result, the product of that reaction here. So let's see if we can get some space-- so right here. Well, we're going to lose that alpha hydrogen. Right? So now our carbon still is bonded to two other things. We lost that alpha hydrogen, and now it's double-bonded to that oxygen. So that would be the mechanism. We went from one bond of carbon to oxygen on our primary or secondary alcohol. We've now increased it to two bonds of carbon to oxygen. So the other products here, we would make H3O plus, of course. So we'll go ahead and put H3O plus when water picks up that proton. We would form HCrO3 minus as our other product. Now, if the alpha carbon is the one being oxidized, so if this carbon is oxidized to this carbon-- it's the same carbon, but this carbon is being oxidized. Something must be being reduced. So this is a redox reaction. If you oxidize something, something else is reduced. And that something else is chromium. So if you were to assign an oxidation state to chromium in the sodium dichromate over here-- so in this guy over here-- chromium has an oxidation state of 6 plus. So when we look at our products and we find chromium in our products here, if you were to assign an oxidation state to this chromium, you'd get 4 plus. So Cr4 plus. And there's some other chemistry that goes on, which ends up converting the chromium from 4 plus into 3 plus. And so overall, you can see that you're starting out with 6 plus over here, and you're ending up with 3 plus over here. That's a decrease in the oxidation state. So chromium is being reduced. So that alpha carbon is being oxidized, and chromium is being reduced in this redox reaction. In the next video, we'll take a look at several examples involving primary and secondary alcohols.