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.
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.