- Oxidation of alcohols I: Mechanism and oxidation states
- Oxidation of alcohols II: Examples
- Biological redox reactions
- Protection of alcohols
- Preparation of mesylates and tosylates
- SN1 and SN2 reactions of alcohols
- Formation of nitrate esters
- Preparation of alkyl halides from alcohols
Oxidation of primary alcohols to aldehydes (and then carboxylic acids); oxidation of secondary alcohols to ketones. Created by Jay.
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- As Sal would put it, the way Jay discusses the reduction of Chromium starting at about11:50is kind of "hand wavy." How would I be able to determine on my own that the Oxidation State of the Chromium in HCrO3 goes from +4 to +3? "There's some other chemistry that goes on..." does not help me understand this process. It may be beyond the scope of this video, but "Curious minds want to know."(7 votes)
- Disproportionations and single electron transfers lead to chromium (V) acid and stable Cr(III) hydroxide. The chromium (V) acid promotes a two-electron oxidation of an alcohol and becomes Cr(III).
In simple terms, redox reactions continue to occur until the Cr reaches the +3 stable point. If you look at the periodic table, you will see that the Cr+3 has a filled s orbital which increases stability and is the driving force for these continued reactions.(8 votes)
- at07:43I thought when you use Jones's oxidation, CrO3, H2SO4, acetone you make a carboxylic acid from a primary alcohol?(3 votes)
- By using jones reagent , we get RCHO group ie , an aldehyde. Jones reagent is a relatively mild oxidising agent. Only a strong oxidising ahent such as chromic acid (H2CrO4) could oxidise an alcohol to carboxylic acid.
The oxidising order is as follows -
alkanes -> alcohols -> aldehydes -> carboxylic group(11 votes)
- Can anyone please explain why Na2CrO7 have Cr6+ and when oxidize HCO3 you get Cr4+ then convert to 3+?(3 votes)
- The rules for oxidation numbers are:
1. H is +1.
2. O is -2.
3. Group 1 metals are +1.
3. The sum of all the oxidation numbers must add up to the overall charge.
Thus, in Na2Cr2O7, 2Na → +2, and 7O → -14. These add up to -12, so the 2Cr must be +12, and one Cr must be +6.
In HCrO3-, H → +1 and 3O → -6. These add up to -5. Since the overall charge is 1-, the Cr is +4.
So oxidation number of Cr changes from +6 to + 4. We are not saying that Cr6+ ions are converted to Cr4+ ions.
The HCrO3- later disproportionates to form HCrO4- and Cr3+ ions. You can calculate that the oxidation number of Cr changes from +4 to +3 (and+7). It is the Cr3+ ions that we see as the final product..(10 votes)
- The first example of oxidization from alcohol to aldehyde, we lose two hydrogens (the one on the alcohol and the other one attached to the alpha carbon itself.) I thought it was only supposed to lose one?(5 votes)
- In order to form a double bond you have to lose two hydrogens. Remember 1 degree of unsaturation (a double bond) equals a lose of 2 H's from the chemical formula.(3 votes)
- In class, I heard that Ni, Pd, or Pt needed to be used as a catalyst to oxidize the alcohol, is that true?(3 votes)
- Yes, in fact, these are common catalysts is organic synthesis, and keep in mind that there are probably some other requirements such as temperature or pressure(3 votes)
- In oxidation of a primary alcohol; where dose the extra oxygen comes from to form a carboxylic acid?(2 votes)
- In oxidation reactions, what happens with the hydrogen attached to the hydroxyl group when the oxidising agent is introduced? It seems that two hydrogens are lost in the reaction, not one? The best explanation I have heard at is that it "floats off"..as in this video http://www.youtube.com/watch?v=vUQUVoTV5Ys at5:55. I need a better explanation please.(3 votes)
- what is catalytic oxidation? plz help me frndz n thnkz in advnce:)(1 vote)
- Catalytic oxidation is a reaction with oxygen that occurs more rapidly and at a lower temperature in the presence of another substance (called a catalyst) than it would in the absence of the catalyst. The catalyst only speeds up the reaction. It doesn't get used up in the process. It can be used over and over again.
A good example is the catalytic converter in your car. It catalyzes the oxidation of unwanted byproducts of combustion in the exhaust of your car engine to more friendly substances. Two of the main pollutants in automobile exhaust are carbon monoxide and unburned gasoline. These are oxidized to carbon dioxide and water before they leave in your exhaust.(4 votes)
- At9:12, when the hydrogen is lost from the alcohol, does it take one electron with it? Similarly, does the -OH also take one electron from the bond it is sharing with the Cr with it when it leaves?
I'm assuming that is how they are able to form the water, but I wanted to know if I understood the flow of electrons correctly.
Thank you!(2 votes)
- Yes, in effect, when the hydrogen atom leaves, it takes one electron with it. And when the OH leaves the Cr, in effect, it also takes one electron with it.(2 votes)
- why on a carbon two -OH groups are not stable(1 vote)
- In some cases they are stable, but in most cases a 1,1-diol can become more stable by forming a carbonyl group, as in RCH(OH)₂ ⇌ RCH=O + H₂O.(3 votes)
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. I'll 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. Obviously, this would give me an aldehyde functional group. If you oxidize a primary alcohol one time, you will get an aldehyde. Let's take a look at the oxidation states of my alpha carbon and see what happened to it. All right. So if I wanted to assign an oxidation state to my alpha carbon on the left, once again, I have to put in my electrons. Right? 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 4 minus 5 will give me an oxidation state of negative 1 for my alpha carbon. So let's look and see what happened to that alpha carbon after we oxidized it. Over here on the right, I want to assign an oxidation state to what is now my carbonyl carbon. And 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. 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. If you oxidize a primary alcohol one time, you will get an aldehyde. What about if you keep going? 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. 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 will get a carboxylic acid. Let's look at a secondary alcohol now. All right. 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. We have one right here. If we were to oxidize our secondary alcohol-- so we're going to oxidize our secondary alcohol. Once again, a simple way of doing 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 it 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 tha 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. 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. Once again, oxygen beats carbon. Carbon versus carbon is a tie. Carbon versus hydrogen, carbon wins. And carbon versus carbon, of course, is a tie again. Normally, four valence electrons. In this example, it's surrounded by four. 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. 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 hydrogen bonded to my alpha carbon. 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. 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. 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. Again, either primary or a secondary. Like that. And when we have our primary or secondary alcohol, it's going to be reacting with chromic acid. So here is the structure for chromic acids. 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-- you could start from chromium trioxide. So you could also use a different reagent, which consists of CrO3-- chromium trioxide-- and H3O+, 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, there'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. 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. 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. And 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 results of 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. 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+, of course. So we'll go ahead and put H3O+ when water picks up that proton. We would form HCrO3- 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. 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+. 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. 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.