Oxymercuration - Online Tutor, Practice Problems & Exam Prep

The oxymercuration–demercuration reaction adds alcohols to alkenes via a two-step mechanism. Initially, a double bond reacts with mercuric acetate (Hg(OAc)₂) and water, forming a bridged ion known as a mercurium ion. This intermediate allows for anti addition, resulting in trans products. The subsequent reduction step, typically using NaBH₄, replaces the mercury with a hydrogen atom, yielding a Markovnikov alcohol. This reaction is significant for its regioselectivity and stereochemistry, avoiding carbocation rearrangements and ensuring the formation of stable products.

General properties of oxymercuration-reduction.

All right, guys. So we just talked about how you can use aqueous acid to add alcohols to double bonds through acid-catalyzed hydration. But it turns out that there's actually more than one way to add alcohols to double bonds and that's actually a huge focus of this section. So what I'm going to do now is I'm going to show you guys an alternate reaction that we could use that also adds alcohols to double bonds. It's called oxymercuration-demercuration. So first of all, what you're going to notice is that this is just a really long name for a reaction, so I personally like to shorten it. I always call it, instead of that long name, I usually just call it oxymerc and that's fine. If you just say oxymerc, that just means it's oxymercuration, demercuration or reduction is fine. So in general, what do we see? Like, what's the general reagents that we use for this? What we're going to find is that you have a double bond of course and then we're going to react it with this weird reagent that's a mercury with 2 acetate or acetyl groups on it. So that's what OAc means. It means acetyl groups and I'll show you what that looks like in a second. So you have that and you have water. So, this is really the first step and this is called the oxymercuration step. Now if you're confused about how you would remember this, well, eventually, you are going to need to memorize this. But one way that I like to maybe make it simpler is that notice that the word oxymercuration has mercury in it, "Mercury". And notice that the reagent of oxymercuration is Hg, which is mercury. So whenever you just see the Hg, you automatically know this is an oxymercuration. In fact, we won't see any other reactions with mercury until organic chemistry 2. So you're pretty safe that if it's a mercury, that's oxymerc. Okay? Then the second step of this is to react with NaBH₄ and some kind of base like NaOH. This is called the reduction step or the demercuration step. Now typically, that's the way things work. When you have something above the arrow and something below it, that could either mean that you have a reagent and its solvent or that you just have a 2 step reaction. Sometimes you're going to see these written as separate steps like 1, 2, but sometimes you'll just see it written without the numbers and you're just supposed to know that what's on top of the arrow is the oxymerc part. What's at the bottom is the reduction part. Regardless, we're going to learn these reagents. Don't worry. But regardless, look what happens. We still get an alcohol. So let's go ahead and look at the general feature of this mechanism. So basically, the intermediate for oxymerc is not going to be the same as hydration. Instead of being a carbocation, what it's going to be is what we call a bridged ion. That's going to be a big deal when it comes to predicting products. That's actually going to matter a lot. The stereochemistry here is actually going to matter. It's going to be decisively anti. So what that means is that anti stereochemistry is the same way of saying that at the end, you're going to get trans products. So if you ever hear me say anti, that just means that at the end, you're going to expect your alcohol and your to be trans to each other. And like I just told you guys, the products are alcohols. So now let's come to the last two facts. Would we expect there to be rearrangements in this mechanism? Remember that rearrangements happen when we have carbocations. Do we have a carbocation? No. So it turns out that this reaction is actually not going to have any rearrangements because it doesn't have any carbocations. I'm just going to write that right here. No carbocation. So there's actually no way for it to rearrange at all. Finally, is it going to follow Markovnikov regiochemistry? Yes, it is. Markovnikov's rule is still going to apply even though it's not a carbocation, it's still going to apply. So now when we look down at our general product, it actually makes sense what it looks like. Notice that what I have oops, just a second. Notice that what I have is that once again, this is my Markovnikov location and I attach my alcohol to it. So my alcohol is going to go Markovnikov. On top of that, notice that the H that I added on the other side is going towards the dash and the Oh is going on the wedge. What that means is that I have a Markovnikov anti alcohol. Does that make sense? Because basically, it's Markovnikov because it went to the most stable location. It's anti because my H and the Oh that I added are trans to each other and it's an alcohol. So even if you didn't know the full mechanism, you could still predict the products just based on these facts. But obviously, we need to know the mechanism, so let's go ahead and get started with that.

Acid-catalyzed oxymercuration-reduction mechanism

So for this mechanism, the first step, just like the first step of all these mechanisms, really is electrophilic addition. I'm going to take my double bond, and I'm going to try to find something that's electrophilic. So in this case, that's actually going to be the mercury because when I drew out HgOAC₂, the OAC₂s are these things right there. There are 2 of those acetyl groups. So those groups are pulling electrons away from the mercury giving that a very, very strong partial positive. So what this means is that I'm actually going to get an interesting series of arrows. I'm going to get that this double bond attacks the Hg because it's positive, and that in order to make that bond, I break a bond. So I'm going to break off one of these acetyl groups and it's going to leave. But on top of that, there's one more arrow that's going to form, which is that the Hg instead of just attaching to one side or the other like I would usually do for an H, the Hg is actually going to go ahead and grab back. So I'm actually going to get 2 arrows. You're going to get one arrow going to the Hg. I'm going to get another arrow from the Hg back to the double bond. What this is going to do is it's going to give me a bridged ion that looks like this, like step 2. In step 2, what we see is that now the Hg from here is attached right there to both of the atoms. It's attached partially to the top one and the bottom one. This is unlike hydration because remember hydration, I would always attach the H just to one, and then I would get a carbocation on the other. In this case, I don't get a carbocation. I get a bridged ion.

Now notice that the OAC here just has to do with the fact that one of the OACs is still attached. I'm also going to get ± OAC- that just left. Just so you guys know. So now I've got this ion. In general, it's called a bridged ion. Specifically, it's called a mercurium ion. It’s an ion made out of mercury. What you're going to notice is that there's a positive charge distributed throughout these 3 atoms. Why is there a positive charge? Because actually, since these are partial bonds, they're not a full bond. What that means is that these carbons don't have enough total bonds. Basically, both of them are kind of like making a half bond, so it adds up to a full positive deficit. There’s basically one bond missing. So my question is now how do I get rid of this intermediate because it sucks. It's really unstable. It turns out that the nucleophile in this case is going to be water. Why is my nucleophile water? Because if you look up towards my reagents, water is the second part that's given to me for oxymercuration. Now that nucleophile could change, but in oxymerc, it’s always going to be water. So now we go down here and we see that water has electrons. It has plenty of electrons to give, and I've got this positive charge so inside the ring. So this water is really attracted to that positive charge.

Now the question is which side of the ring is it going to attack? Is it going to attack the top part or is it going to attack the bottom part? Because both of these are different. It could either attack the top part and become tertiary or the bottom part and become secondary. And it turns out the way we judge which side it attacks is by the side that's going to have the most positive character. Let’s think about it. I just told you that there's basically partial positives on all of these atoms. Partial positive, partial positive, partial positive. It's being distributed throughout. But it turns out that one of these atoms is going to have the most positive charge. Why? Because it's going to be the one that's the most stable with a positive charge. So between the two carbon atoms, which one do you think is going to be the one that's more stable? The one that's a secondary carbon or the one that is a tertiary carbon? Which one's going to like to have a positive charge more? Tertiary. So that means that this one's actually going to be more positive and this one's actually just going to be a little bit less positive. So you can imagine that there's a little bit more positive density at the top. And what that means is that the water since it's negative, since it has electrons, it's going to go for the one that's the most positive, which is the top one. So you're going to go for the most substituted. This mechanism, just so you know, if you've already learned it, it's an SN2 mechanism. It’s actually a backside attack. So if you don't know that, that's fine. But anyway, what's going to happen here is that the water attacks there, and then it's just going to erase some of the stuff that we're not using.

So the water attacks the top part and now I've got too many bonds to that carbon because that carbon is now going to have 4 bonds. 1, 2, 3, 4. That's the new bond that's being created. And it has that partial bond to the mercury. So if I make a bond, I have to break a bond, and I'm going to break the bond to the mercury. Now this brings us to an interesting pattern that we're going to see all throughout this section, which is that any time you have a 3 membered ring or a bridge or anything like that and you break it open, that ring is highly, highly strained. Bond angles do not want to be at 60 degrees like that. So when you break open part of the ring, and it snaps open, the groups are going to end up facing opposite directions. And the reason is you can think of it almost like a loaded spring. Once you break it open, both groups are going to go opposite directions because they want to get as far away from each other as possible. What that means is that if my water, let's say, attacks from the front from the top. Let’s say that the water attacks from the top of this 3 membered ring and they break open the part with the mercury, that means the mercury is now going to go down because the mercury is going to try to snap open to the opposite side of where the water attacked. What this gives us now is the reduction step. So now what I have is I have water that attacks from the top. Remember that I said that let’s say it attacked from the top. If it attacked from the top, that means that the methyl group that used to be here, CH₃, got pushed down to the bottom. So now it’s at the bottom. Well, another thing that happened was that the mercury was part of that 3 member ring and it broke open. So that means if the water is on the top, the mercury has to be at the bottom and that's what we have right here. The mercury is now going to face towards the bottom side.

So that's the end of our oxymercuration step. Now, we have to get to the demercuration or the reduction step. Now the fun part about this is that you actually don't need to know the mechanism. So I'm just going to write here don't need MEK. Why is that? Because it's really complicated and it has to do with stuff that you haven't learned yet. In fact, reduction in general, you don't have a good grasp on yet. Oxidation-reduction is like its own topic in organic chemistry, and you haven't really gotten there fully yet. So we're not going to really bother with that too much. All you need to know is that basically you have a base that's going to wind up taking away a proton from my water to become alcohol. And then you have a reducing agent. By the way, NaBH₄, get used to it now, is a reducing agent. We'll talk about this more later in the semester. That's going to reduce basically a reducing agent adds hydrogens to things. So that's going to reduce my mercury into an H. So at the end, what I'm going to wi

Predict the product of the following reaction.

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Here’s what students ask on this topic:

What is the mechanism of oxymercuration-demercuration?

The oxymercuration-demercuration mechanism involves two main steps. First, the alkene reacts with mercuric acetate (Hg(OAc)₂) and water, forming a bridged mercurium ion. This intermediate allows for anti addition, resulting in trans products. In the second step, sodium borohydride (NaBH₄) reduces the mercurium ion, replacing the mercury with a hydrogen atom. This yields a Markovnikov alcohol, where the hydroxyl group attaches to the more substituted carbon. This reaction avoids carbocation rearrangements, ensuring stable product formation.

Why is oxymercuration-demercuration preferred over acid-catalyzed hydration?

Oxymercuration-demercuration is often preferred over acid-catalyzed hydration because it avoids carbocation rearrangements, leading to more predictable and stable products. The reaction proceeds via a bridged mercurium ion intermediate, which ensures anti addition and trans product formation. Additionally, it follows Markovnikov's rule, adding the hydroxyl group to the more substituted carbon. This makes it a reliable method for synthesizing alcohols from alkenes with high regioselectivity and stereochemistry control.

What reagents are used in the oxymercuration-demercuration reaction?

The oxymercuration-demercuration reaction involves two main reagents. In the first step, mercuric acetate (Hg(OAc)₂) and water are used to form the bridged mercurium ion intermediate. In the second step, sodium borohydride (NaBH₄) is used as a reducing agent to replace the mercury with a hydrogen atom, completing the formation of the Markovnikov alcohol. Sometimes, a base like NaOH is also used in the reduction step.

Does oxymercuration-demercuration follow Markovnikov's rule?

Yes, oxymercuration-demercuration follows Markovnikov's rule. During the reaction, the hydroxyl group (OH) is added to the more substituted carbon of the double bond, while the hydrogen (H) is added to the less substituted carbon. This regioselectivity is due to the formation of a bridged mercurium ion intermediate, which directs the nucleophilic attack of water to the more stable, more substituted carbon.

What is the stereochemistry of the products in oxymercuration-demercuration?

The stereochemistry of the products in oxymercuration-demercuration is anti, meaning the substituents added across the double bond are trans to each other. This occurs because the reaction proceeds through a bridged mercurium ion intermediate, which forces the nucleophile (water) to attack from the opposite side of the mercury. As a result, the final product has the hydroxyl group and the hydrogen atom added in a trans configuration.