Leah here from leah4sci.com and in this video we're going to look at the acid catalyzed
hydration of alkynes reaction and mechanism.
Acid catalyzed hydration of alkynes is a reaction where a starting alkyne will react with some
sort of lewis acid catalyst, I'm showing H+ but you could see H2SO4 to give us not an
alcohol but a ketone.
This is not a very fast reaction for terminal alkynes and that's why you often see H2SO4
instead which we'll talk about in the next video.
One way to learn your alkyne mechanisms is to recognize that an alkyne has two pi bonds,
an alkene has one pi bond but much of the concept on how they react is very similar.
The pi bond is going to attack, as a refresher, if I have an alkene undergoing acid catalyzed
hydration which we'll as an H+ and H2O, you should remember that the mechanism has the
pibond attacking the H+, a carbocation forms and we get an alcohol product.
In the case of asymmetric alkene, it doesn't where I put the alcohol because it will be
the same.
If I have the same reaction with an asymmetric alkene it's going to follow markovnikov;s
rule where the hydrogen adds to the less substituted carbon.
The carbocation and ultimately the nucleophile will happen on the more substituted carbon
giving me in this case 2 propanol.
This is something to keep in mind for the alkyne mechanism as well.Let's start with
the simple alkyne remembering that where the triple bond ends represents two carbons and
then the chain continues giving us 2 more carbons for a 2 butyne.
The reaction begins with an acid catalyst that we can show as an H+ floating in solution.
Remember you have something like H2SO4, the H+ breaks away.
Your professor may show you the first step in two different ways.
You may see one of the pi bonds using its electrons to reach out for and grab that hydrogen.
Not only is this going to break the pi bond, it's also going to change the change the carbons
from linear when holding a triple bond to trigonal planar which will show like this,
when it's only got a double bond.
We'll show the hydrogen that we added over here giving us a somewhat unstable carbocation.
Right now it doesn't matter where we put the carbocation because we started with asymmetrical
alkyne but if this was a hydrogen rather than a carbon we would follow markovnikov's rule
and put the carbocation on the more substituted therefore more stable carbon.
In the next step, water which is acting as a lewis base is going to reach out for and
attack the carbocation giving us an intermediate that now has an oxygen with 3 bonds.
2 to hydrogen, 1 to carbon only 1 lone pair remains with a formal charge of plus 1.
Some professors may show the first 2 steps as one step as follows: they'll show you that
water attacks the pi bond which breaks the pi bond and that is why the electrons will
collapse and grab that hydrogen atom.
Regardless of which option you learn, the intermediate is the same.
We have a water molecule attached to carbon, the triple bond is broken, we now have just
a double bond with a positive charge on oxygen.
In an aqueous solution we have lots and lots of water floating around so another water
molecule will use its lone electrons to grab a hydrogen off of a positive oxygen atom breaking
the bond between oxygen and hydrogen to collapse those electrons back on to oxygen.
This gives me what looks like the expected final product.
We now have an oxygen bound to just one hydrogen, the initial lone pair and another lone pair
that it got from the bond between hydrogen and oxygen.
Somewhere in solution we now have an H3O+ floating around from the H2O that grabbed
the extra hydrogen off of Oxygen.
It's got a lone pair and positive charge regenerating the acid catalyst which we could also think
of as just water and H+.
But the reason we're not done is that the molecule that we formed is not very stable.
If you have a molecule with an alkene which gives us "ene" and an alcohol giving us "ol"
on the same carbon, this is what we call an enol which we'll write as enol.
And the enol is unstable will undergo what's called keto-enol tautomerization which I teach
in the video link below to give us a keto product.
It's very simple, we have yet another water molecule in solution that reaches for the
hydrogen atom attached to oxygen.
Now the bond between oxygen and hydrogen can't simply collapse on to oxygen to form a lone
pair because this would give oxygen a negative charge in an acidic solution.
We have acid everywhere, there's no way an O minus is going to form in this solution
so we have to find a way to break that bond without forming a negative charge.
And the way this happens is remembering that we still have other H plusses in solution
from other hydroniums that are floating around and so the bond of oxygen to oxygen from hydrogen
instead of collapsing onto the oxygen will collapse down to form a pi bond between oxygen
and carbon which put too many bonds on this carbon atom.
That would give carbon five bonds and so we have to kick out the easiest one, the pi bond
and where does it go?
It get kicked out somewhere to grab a hydrogen atom to now form a single bond.
Again we break the bond between these two atoms because these carbon would have too
many bonds but this carbon can still have a bond on either direction.
That carbon is perfectly fine holding on to those electrons.
What does that look like?
We have the same skeleton with an oxygen that had its initial purple lone pair.
The blue electrons that it gone from breaking this bond between oxygen and hydrogen.
And now the black electrons that collapsed down are sitting as a pi bond between carbon
and oxygen.
And the red electrons that used to form the pi bond are now sitting on this carbon here
with a single bond to hydrogen.
Giving us a ketone as the final product rather than an alcohol as we've seen with the acid
catalyzed hydration of alkenes.
Why?
Because we have an alcohol forming but it's an alcohol and an alkene for an unstable enol
causing it to undergo KET or Keto-enol Tautamerazion.
We just worked through the mechanism for an internal symmetrical alkyne.
If there is symmetry it doesn't matter where I place the enol and ultimately the ketone.
If I number this from the left, 1,2,3,4 then I put the ketone on 2 or 3 but if I put the
ketone on 3, notice that I can just turn around and number it from the right putting the ketone
again on carbon 2.
This changes slightly when you have an asmymetrical alkyne.
If you have a terminal alkyne meaning one side of the alkyne is attached to a carbon
chain but the other side is attached to a hydrogen atom, this is where we have to show
the reaction following markovnikov's rule.
The first step of this reaction where the pi bond reaches for an H+ and we have a carbocation
intermediate, we have a choice between placing the carbocation on the primary position or
on a secondary position.
Remember Markovnikov's rule is all about the stability of the intermediate and even though
the carbocation on a pi bond carbon is unstable, that's why we saw two options for the mechanism,
the fact is that it'll be more stable on the more substituted carbon and if that's where
the carbocation forms, that's where the ketone will ultimately wind up.
This will give us an intermediate that has an enol with an OH on the secondary carbon.
The enol will then undergo Keto-Enol-Tautomerization or technically enol keto tautomerization because
we're going towards the keto form to give us a ketone.
In an upcoming video we'll look at hydroboration we see that keto enol tautomerization gives
us an aldehyde if we're attaching to the terminal carbon.
One thing to keep in mind with this reaction is that even though we're showing this steps
and we know the mechanism this is not a very fast reaction and if you're reacting a terminal
alkyne you're better off adding a catalyst like HgSO4 which is what we're going to look
at in the next video but first one more thing about this reaction, if we're starting with
an assymetrical internal alkyne for example 2-butyne, each of the carbon holding the triple
bond is secondary.
In this case, the reaction is not going to be radio selective meaning we're not going
to favor one carbon over the other because they're both secondary so that if we react
this with an acid catalyst in H2O we're going to get two different ketones forming.
Since the alkyne sits between carbons 2 and 3 and both carbon 2 and carbon 3 are secondary,
the products can have a ketone on carbon number 2 and a ketone on carbon number 3 for a mixed
result.
This is very important to keep in mind synthesis if you're trying to do a chain elongation
with an alkyne intermediate and then looking to turn it into a ketone, if you don't want
side products make sure you go after a different reaction.
But before we tackle synthesis let's move to the next mechanism which is oxymercuration
for alkynes and you can find this along with the entire alkyne video series, practice quiz
and cheat sheets first by giving this video a thumbs up, subscribing to my channel so
you don't miss any new videos and then visiting my website leah4sci.com/alkyne.