This is the second of three ways to add alcohol to a double bond. It is similar in acid-catalyzed hydration in terms of products, but the mechanism is WAY different.
Concept: General properties of oxymercuration-reduction.5m
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 and 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 and 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, it's fine.
In general, what do we see? 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 two acetate or acetyl groups on it. So that's what OAc means, it means acetyl group, so 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 how you'd remember this, well, eventually you are going to need to memorize this. But one way the 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 orgo two, so you're pretty safe that if it's a mercury, that's oxymerc.
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 even mean that you have a reagent and its solvent or that you just have a two-step reaction. Sometimes you're going to see these written as separate steps like one and two. 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. And 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 H to be trans to each other. 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 you 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 carbocations. So there's actually no way for it to rearrange at all.
Finally, is it going to follow Markovnikov's 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 out general products, 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 can still predict the products just based on these facts. But obviously, we need to know the mechanisms. Let's go ahead and get started with that.
Concept: A worked-example of the acid-catalyzed oxymercuration-reduction mechanism.10m
So for this mechanism, the first step, just like the first step of all these mechanisms, 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 Hg(OAc)2, the OAc2's are these things right there. There's two 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 two arrows. I'm 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 two.
In step two 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 OAc's is still attached. I'm also going to get plus 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 three atoms, throughout this one, this one and that one. 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 is kind of 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 inside of 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. Well, 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 ones 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.
What that means is that the water, 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 back side attack. If you don't know that, that's fine.
But anyway, what's going to happen here is that the water attacks there. I'm just going to erase some of this 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 four bonds, one, two, three, four. 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 anytime that you have a three-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 we 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.
So what that means is that if my water let's say attacks from the front, from the top. Let's say that the water attacked from the top of this three-membered ring and I 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 a water that attacked 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 three-membered ring and it broke open. So that means if the water's 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 mech. 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 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. And 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 wind up getting is a product that looks like this. Like I said, you don't need to know the mechanism, you just need to be able to predict that it's going to look like this. I would get now an alcohol at the top. I would get my methyl group towards the back because it got pushed there and then finally, I would get the H towards the back. Notice that my mercury now became an H.
And now notice that once again, this is now showing you how I'm getting my Markovnikov or I'm just going to call him Mark anti-alcohol because my alcohol went to the more substituted location which was the water. The water attacks the more substituted Markovnikov. Then it's anti because the bridge snapped open, so it had to face opposite directions. And then it's an alcohol because we used water as the nucleophile. Does that make sense? Cool.
This is a very important reaction for you guys to know. It's basically taught in pretty much every textbook. It might show up on later standardized tests, stuff like that and obviously you need it for your exam. So make sure that you know everything about it, literally.
Let's go ahead and do some practice with oxymerc.
1. Electrophilic Addition
2. Nucleophilic Substitution (SN2)
3. Reduction (demurcuration)
Problem: Predict the product of the following reaction.6m
Complete the following reaction supplying the missing product and showing correct regio- and stereochemistry where applicable. If a racemic or diastereomeric mixture forms show all stereoisomers; if no reaction takes place, write N.R.
Draw the mechanism of the following reaction. If no reaction would occur, say so. Show stereochemistry where relevant. Show the intermediate between steps 1 and 2.
Complete the mechanism for the following reaction. Draw all the arrows to indicate movement of electrons, write all lone pairs, all formal charges, and all products for each step. In the dotted boxes write which mechanistic element is involved in each step. If a racemic mixture is formed in the final product, you must draw both enantiomers and write racemic.
Consider the strucutres below and answer the following questions.
b. Which compound is most likely to form a secondary alcohol by oxymercuration, but a tertiary alcohol by hydration using aqueous sulfuric acid?
Consider the strucutres below and answer the following questions.
a. Which compound will form different products by oxymercuation and hydroboration?
Follows Markovnikov’s Rule
a) Hydrogen goes to the double bonded carbon with _______ hydrogens. ( ______ substituted alkene carbon)
b) Hydroxide ion goes to the double bonded carbon with _______ hydrogens. ( ______ substituted alkene carbon).
Predict the starting material of the following reactions.