Leah here from leah4sci.com and in this video we'll look at the oxidation and reduction reactions for alkenes and alkynes. You can find this entire video series along with my redox practice quiz and cheat sheet by visiting my website leah4sci.com slash redox. In the last video, we looked at the definitions for oxidation as the loss of an electron or a hydrogen atom which is a proton coming along with that electron.
The gain of an oxygen atom or an additional bond to oxygen, so if you have an oxygen with a single bond and you create a double bond or going up in charge meaning the oxidation number of that atom increases or goes up spelled with an O to remind you that it's oxidation. We define reduction as gaining an electron or a hydrogen atom when a proton comes along with that electron, losing an oxygen atom or a bond to an oxygen atom. So for example, if you have a double bound oxygen to make that single bound, that's a loss of a bond, not an atom and it's still reduction. And finally, to reduce the oxidation number or to reduce the charge of that atom. Now let's see how this applies to alkenes and alkynes.
We'll start with the oxidation of alkenes. Remember it's not just gaining an oxygen atom but the entire molecule has to gain oxygen to be oxidized. So we'll start with the oxidation of alkenes.
That means if I have a reaction like acid catalyzed hydration which is an alkene reacting with H2SO4 and H2O, you'll also see this as H3O+. The final product does gain an oxygen atom. We get an alcohol on one of the former pi bound carbon atoms and a hydrogen on the other one. If you look at the atoms individually, yes, this carbon got oxidized.
But this carbon here got reduced and having one atom oxidized and another atom reduced on the same molecule meaning there's no net change in the oxidation state of the molecule. So you want to keep in mind that if you're adding something that's oxidizing to one carbon and reducing to the other carbon, it's not oxidation. Examples would include adding something like H2O because it'll give you an OH in one atom and a hydrogen on the other. This hydrogen doesn't count because it's on oxygen, we're specifically looking at the bond between oxygen and carbon.
Adding an alcohol is a very similar reaction, you add an OR to one carbon, hydrogen to the other. But it's not just oxygen, any highly electronegative atom can be considered oxidation. So if you add something like HX, this is your hydrohalogenation reaction where you're adding an X. chlorine or bromine to one carbon and hydrogen to the other, same thing. They all cancel out and that means there's no net oxidation on the molecule.
So how do you get oxidation for an alkene? You have to add oxygen without adding the same number of hydrogen or you have to increase the number of bonds to oxygen without having an equal number of new bonds to hydrogen to cancel that out. A good example of this is the dihydroxylation reaction where you're adding two alcohols, two OH groups to the carbon atoms. That means while this carbon atom is getting oxidized, this carbon is also getting oxidized, we have no reduction to cancel it out. The two reactions you've probably covered for this are reacting an alkene with KMnO4 in cold dilute conditions, this will add an OH to each carbon or osmium tetroxide, OSO4 Followed by H2O2 or NaHSO4 and H2O.
Both of these will give you a syn diol which is why it's called dihydroxylation. You can get a similar product in the two-step reaction that starts out with something like MCPBA which is a per acid to give you an epoxide followed by opening in acidic or basic conditions you'll typically see this as H3O+. You have an epoxide intermediate but the final product is going to be a trans dial which is a good reaction to know if you're looking for the trans instead of cis product. Now notice we defined this oxidation without looking at any of the math and calculations. In organic chemistry you do not want to waste your time with tricky or complex math if you can simply ask yourself which atom did I gain or lose bonds to and in this case it's oxygen.
I'll run through the math really quick just to prove it to you but I don't want you doing this on your exams. Carbon here is bound to one hydrogen atom. Hydrogen has an oxidation number of plus one.
If the molecule is neutral, carbon has to have a minus one to balance. Same thing here, hydrogen is plus one, carbon is minus one. In the product, Oxygen is the more electronegative atom and so it gets an oxidation number of negative one.
But let's not forget the invisible attached hydrogen atom with an oxidation number of positive one. If we have a negative and a positive attached to carbon, the other carbons are each zero when attached to carbon while negative and positive cancel out making the carbon itself zero. The change in oxidation number from reactant to product. goes from negative one to zero. If we go from negative one up to zero we have to increase by one, that means we have the oxidation number going up by one, spelled with an O, showing us that it's an oxidation.
Again, if this is something that confuses you, forget the oxidation numbers and simply look at the fact that carbon didn't have oxygen, now it has oxygen, that's an oxidation. The other carbon is the same thing, it doesn't have oxygen, it gets oxygen. that's an oxidation. There are two reactions you likely studied for oxidative cleavage of alkenes. The first one is ozonolysis which uses ozone in step one but it can be followed by a reductive workup such as DMS or an oxidative workup such as H2O2.
The word oxidative workup should imply that the products will be much more oxidized. The second reaction is using KMnO4. in hot basic conditions.
The trick is to redraw the skeleton exactly as the starting molecule but separate the pi bond components and we'll draw the same thing for each one. The only difference is in the final product. Reductive conditions for ozonolysis will give me a ketone on a secondary carbon and an aldehyde on a primary carbon. Oxidative conditions will give me a ketone again for the secondary carbon but a carboxylic acid for the primary and KMnO4 is a stronger reagent so we go straight to the carboxylic acid for the primary carbon and a ketone for the secondary. Oxidative cleavage on a terminal carbon will give you a single carbon fragment for one of the products.
The secondary carbon is gonna be the same, a ketone regardless of what you use but the terminal carbon will be different depending on the reagents that we use. Ozone analysis under reductive conditions will give us formaldehyde for that single carbon product and formic acid which is a carboxylic acid under oxidative conditions. KMnO4 will oxidize it all the way giving us carbon dioxide for the product.
If you're looking at the three products in comparison, you'll notice that the CO2 is the most oxidized and formaldehyde is the least oxidized. How do we see that? Count the number of bonds between carbon and oxygen.
The more bonds to oxygen the higher the oxidation number. Formaldehyde has 2, formic acid has 3, and carbon dioxide has 4. We didn't do any math here, we simply counted the number of oxygen bonds to that central carbon atom. Compared to the oxidation of alkenes, reduction is very simple and very straightforward.
If we have a pi bond on the molecule and we react it with H2 in a metal catalyst for example palladium carbide, the metal will grab the pi bond, the metal is also holding on to hydrogen and the hydrogens will slide in. Since the metal holds the alkene from the same face of the pi bond, this is considered a syn addition. So for example if we had two green hydrogen atoms to start with, we can show them on dashes and the incoming hydrogen on wedges or the other way around.
Now it doesn't matter for this molecule because it's achiral but if you had something like a ring as you're starting alkene, you will have to show the stereochemistry for the incoming hydrogen atoms. We have very similar patterns for the oxidation reactions for an alkyne. For example you learned two different reactions for the hydration of an alkyne, oxymercuration and hydroboration where the product starts out as an enol.
where you add an alcohol and a hydrogen and then tautomerizes through ketoenol tautomerization to give you a carbonyl. In this case the final product is a ketone but if it's terminal and you have hydroboration you will get an aldehyde. At first glance it looks like this should be an oxidation reaction because we have two new bonds to hydrogen but don't forget what happened. The first product is just the addition of water. The carbon that is oxidized by the alcohol is cancelled out by the carbon reduced with the hydrogen atom so there is no net redox reaction.
When the molecule undergoes Ketoenol Tautomerization, once again there is no net gain or loss of hydrogen or oxygen or electrons. Just imagine that this hydrogen atom gets moved to the other carbon so we have the hydrogen that was added in the first step and the hydrogen that was added by moving atoms around in Ketoenol Tautomerization. The final product has two new bonds to an oxygen atom and two new bonds to a hydrogen atom.
for no net oxidation or reduction. In order to have an oxidation reaction, you have to add more bonds to oxygen than you do to hydrogen in your final product. So let's take a look at a couple of reactions.
If we react an alkyne with K-Metaphor under weak conditions, in this case it's H2O neutral as opposed to acidic or basic, we're going to get a diketone as the product. We only break the pi bonds but not the sigma bond. So you get a carbonyl on the first carbon that was sp and a carbonyl on the second former sp carbon. Each carbon atom that reacted gained bonds to oxygen which tells us it's oxidation and we have no hydrogen bonds to cancel it out making this a definite oxidation reaction.
If you carry out the same reaction on a terminal carbon, it'll start out trying to give you a diketone but K-amino 4 is a strong oxidizing reagent and it'll wind up turning that terminal carbon into a carboxylic acid making it even more oxidized because of that extra third bond to oxygen. Like alkenes, alkynes can undergo oxidative cleavage using the same trick where you slice the molecule between the pi bonds and add an oxygen atom for every pi bond you break. Notice that for an alkyne, we add two oxygens per carbon so we know it's going to be much more oxidized compared to its alkene counterpart.
For oz analysis we only need water in the final step. So we'll draw the product as the two carbon fragments that we had initially but then make sure to add two oxygen atoms to each fragment for two carboxylic acids. We get the same product for K-mino4 except that since we're using hot basic conditions, KOH, we can't form a carboxylic acid.
In a basic solution, the base will deprotonate. So for step one, we would just show a carboxylate with an O-for each one. But then in step two, we add an acid like HCl and that protonates the carboxylate to give us that same carboxylic acid product.
A terminal alkyne will have different products compared to an internal alkyne, specifically for the terminal carbon that is attached to one carbon and one hydrogen atom. The trick is still the same, just slice it down the middle and add in the two oxygen atoms. For the internal fragment, that's the portion of the molecule that has two carbons, we're going to get a carboxylic acid for both. But the terminal fragment, that single carbon atom, will be fully oxidized with four bonds to oxygen as carbon dioxide.
A fun fact to note is that if carbon dioxide is your product, it won't stay dissolved in solution but instead it'll start bubbling out. So one way to know if you have a terminal alkene or alkyne is to cause a reaction that results in carbon dioxide and watch for any bubbles escaping from solution. Unlike alkene reduction, since the alkyne has two pi bonds, we have more options for the product. The first reduction we saw with alkenes was reacting with H2 in a metal catalyst like palladium carbide. This reaction will grab that carbon chain, break every single pi bond and put the hydrogen atoms where the pi bonds used to be and the alkyne is no different.
This is a complete reduction since every single pi bond was broken and we maximize the number of hydrogen atoms that we can add. With an alkyne you can also do a partial reduction where you break just one of the two pi bonds. If you react it with H2 in a Lindlar's catalyst, the product is going to be a cis alkene. Basically one of the pi bonds was broken.
and the hydrogens are added to the same side of that molecule. This is because the Lindler's catalyst can be thought of as a poison metal catalyst. It acts the same way as a metal catalyst but it doesn't go all the way. It does one reduction and then stops.
The final reduction is reacting the alkyne with a neutral metal catalyst in liquid ammonia. A neutral metal like sodium. has one valence electron making it very very reactive.
Electrons like to be paired either in a bond or a lone pair. The sodium atom donates its electron to the alkyne Forming a radical intermediate you also get a carbanion intermediate and I'll discuss this in the alkyne tutorial series But the idea is that if you have electrons and electrons on the intermediate They want to go as far away from each other as possible and trans puts them further apart Which is why you're going to get a trans product in the final product. You'll have hydrogen donated from ammonia to each of the two former sp carbons making them sp2 with one hydrogen atom and a pi bond in the trans configuration.
When you think of aromatic compounds you think of benzene which is a six member drink with three pi bonds. But just because it has pi bonds does not mean it's going to react the same way. In fact, if you try to react benzene with any of the standard alkene or alkyne reduction reagents you're going to get no reaction. Benzene is aromatic making it too stable to react unless you expose it to extreme conditions.
If we take that benzene ring and we react it with H2 in a palladium carbide catalyst but then we add extreme conditions, high temperature and high pressure, you're making the reaction extreme enough to reduce benzene all the way down to a cyclohexane. We'll show the initial hydrogens in light blue, this reaction breaks every pi bond and adds a second hydrogen atom to each carbon making it sp3 hybridized and fully reduced. Be sure to join me in the next video where we look at the mechanism for a unique aromatic reduction, the Birch reduction.
You can find this entire video series along with my redox practice quiz and cheat sheet by visiting my website leah4sci.com slash redox.