In this Leah4sci video, we’ll review the alkyne oxymercuration demercuration reaction
and mechanism.
Visit leah4sci.com/alkyne for my alkyne cheat sheet, videos, and more.
Oxymercuration of Alkynes is a reaction where we have a starting alkyne reacting with a
mercury sulfate catalyst written as HgSO4 in H2SO4.
To give us initially an enol that will then undergo keto enol tautomerization to give
us a carbonyl, in this case a ketone.
This reaction is also called the mercuric ion catalyst hydration given the mercuric
ion Hg 2+ catalyst that we'll see in the mechanism.
Since my starting alkyne is symmetrical, I have two options for the final product.
To put my ketone on this carbon or this carbon.
However, if I begin with an asymmetric aldehyde as I have over here with a terminal hydrogen,
the reaction follows markovnikov's rule to give us regioselectivity for a ketone rather
than a terminal carbonyl that would give us an aldehyde.
When studying alkyne reactions, I find it helpful to review their alkene counterparts
to understand the difference and similarities in their mechanisms.
If I have an asymmetrical alkene that undergoes oxymercuration, the reagent looks a little
different but I still have the mercury catalyst that helps control regioselectivity of the
product.
My intermediate appears to be a carbocation except that we cannot have a rearrangement
due to mercury holding on to both carbons or what I consider a carbocation babysitter.
The partial positive sits more on the more substituted carbon following markovnikov's
rule which is how we get the more substituted or markovnikov alcohol product.
As I teach in the alkene reactions shortcut video below you simply break the pi bond,
identify the more substituted carbon and that's where you add your alcohol.
Can I do the same thing for the alkyne?
Yes!
But my product won't be complete.
I break one pi bond and redraw the skeleton with the second pi bond still present, identify
the more substituted markovnikov position and place an alcohol.
But we're not done.
Even though it looks like we have a final product what we have to recognize here is
that when we have an alkene and an alcohol sitting on the same carbon, this alkene alcohol
or enol is relatively unstable and will isomerize through keto enol tautomerization or in this
case enol keto tautomerization to give us our final product which is a carbonyl on the
more substituted markovnikov position.
This enol is not stable enough to be isolated.
Let's take a look at the mechanism in detail starting with propyne which is a 3 carbon
alkyne, recognize that these two are carbons and that we have a terminal hydrogen.
Even though the reagent is presented as HgSO4 in H2SO4 I want you to recognize that what
we're looking at here is simply an Hg 2+ cation countered by an SO4 2- anion, a spectator
ion floating around in solution.
The actual catalyst that we're using is the Hg 2+.
And so the reaction begins when one of the pi bonds will use its nucleophilic pi electrons
to reach out for and grab that mercury ion.
This is where it gets interesting because mercury does have lone electrons that it can
use to attack back and form a bridged or ringed intermediate.
When I redraw my skeleton I no longer have to draw it in linear form because the starting
alkyne had Sp hybridized carbons which gave me a linear structure for 180 degrees.
But after breaking the pi bond I get sp2 hybridized carbons which are 120 degrees and so I can
draw it bent like this.
And let's not forget the bridged mercurinium ion, that's mercury with a +1 charge sitting
and holding on to both carbon atoms.
And that other positive charge sits distributed between the carbon atoms.
This is how we get regioselectivity.
Even though both carbons are holding on to that positive charge, the more substituted
carbon holds on to a bigger percentage of positive, the less substituted carbon holds
on to a smaller percentage making this carbon more susceptible to attack which is exactly
what happens in the next step of this reaction when water comes in to attack the most substituted
carbon breaking that partial bond between carbon and mercury leaving mercury to sit
firmly on that other carbon atom.
Wait a minute, where did the water come from?
Didn't we start with a reagent of HgSO4 in H2SO4?
Don't forget sulfuric acid is an aqueous acid.
You’ll never react with pure H2SO4, it'll always be some concentration of H2SO4 in water
or alcohol if specified.
And so if nothing is given assume you have tons and tons of water molecules floating
around in solution able to come in and attack at this stage, this gives us a carbon skeleton
that hasn't changed, we have the one pi bond, mercury sitting on the terminal carbon, now
with a charge of just plus 1, and the water molecule sitting at the more substituted position
with only one lone pair and therefore a positive charge because the second lone pair is now
sitting as a bond between carbon and oxygen.
To get rid of that charge another water molecule in solution will use oxygen's lone electrons
to grab that hydrogen, collapsing the hydrogen's electrons back on to oxygen.
This gives us the expected enol, alkene and alcohol carbon product.
But we still have an Hg+ sitting on the terminal carbon.
When mercury is removed in the next step, we get what appears to be our final product,
a carbon that lost a pi bond and now has an alcohol at the more substituted markovnikov
position except as we already mentioned this product is an enol because it has an alkene
and an alcohol sitting on the same carbon.
This enol is not very stable and will undergo keto enol tautomerization, a reaction that
I teach in the video below at leah4sci.com/KET.
In this case, the acid-catalyzed version because we're already in an acidic solution.
A water molecule in solution will grab the hydrogen off of the alcohol, collapsing those
electrons back not to the alcohol but down towards that sp2 hybridized carbon atom.
This would give carbon way too many bonds and so the pi bond is forced out, not on to
the carbon atom because that would give us a carbanion in acidic solution which is an
absolute no, no.
Never form a negative carbon in a positive solution, instead those electrons will reach
for a proton in solution on one of the many acidic waters or hydroniums floating around.
Grab that proton, collapse the electrons between hydrogen and oxygen back on to the oxygen
atom to give us as follows.
To make sure nothing is missed, pay attention to every atom and the colors of the electrons
moving around.
Our carbon skeleton hasn't changed with that single bond between carbon and the blue oxygen.
Oxygen's two lone pairs haven't changed but the bond between oxygen and hydrogen now sits
as a pi bond between oxygen and the carbon atom.
The blue hydrogen that used to form an alcohol now sits on a water molecule in solution which
became hydronium with one lone pair and a positive charge.
The purple hydronium from solution now sits as a water molecule with its initial lone
pair and that second lone pair from the bond between oxygen and hydrogen.
That third hydrogen atom sits as a single bond on carbon using the former pi electrons
that grabbed it from hydronium.
Given that we don't show hydrogen in skeletal structure we can just erase it.
As a reminder, since catalysts have to be regenerated in solution, we started with an
acid catalyst and we reformed another hydronium acid catalyst in solution.
And of course the ketone which is our final product.
Let's review the regioselectivity of this reaction.
When we started with the terminal alkyne, don't forget that invisible hydrogen.
We followed markovnikov's rule to get an enol with the alcohol on the more substituted carbon
which ultimately gave us an internal carbonyl or a ketone.
This is the opposite of hydroboration which we'll cover in the next video that will give
us a terminal carbonyl or the aldehyde.
When we start with a symmetrical internal alkyne, because of symmetry I can choose either
carbon for this reaction to give me one single ketone product, and that's because if I choose
the carbon on the right that would be carbon 1, 2 and if I choose the carbon on the left,
that would still be carbon 2.
However, if I start with an asymmetrical internal alkyne such as 2-pentyne, the reaction can
happen at either of the secondary carbons to give me 2 different ketone products.
I can form my ketone on carbon 2 or carbon 3.
First, give this video a thumbs up and if you haven’t already subscribe to my channel
so you don’t miss out on any new videos and then let’s take a look at the next hydration
reaction hydroboration oxidation of alkynes which you can find on my website along with
the alkyne reaction practice quiz and cheat sheet.
See the link below or go to leah4sci.com/alkyne.
The link again is leah4sci.com/Alkyne.