In this video we're going to look at the hydroboration oxidation of alkynes.
You can find my entire alkyne reaction video series along with the practice quiz and cheat
sheet by visiting the link below or going to leah4sci.com/Alkyne.
Hydroboration Oxidation of Alkynes is the reaction in which an alkyne reacts with some
form of boron, here I’m using disiamylborane followed by NaOH and H2O2 to give us an enol
which is terminal if starting with a terminal alkyne that will then tautomerize to give
us an aldehyde as the final product.
We get an aldehyde because this is an anti-markovnikov reaction and the alcohol therefore the carbonyl
adds to the final carbon where is if you want an internal carbonyl, a ketone, you need to
use the oxymercuration reaction which I teach in the previous video.
I recommend reviewing alkenes when studying alkyne reactions so that you can see the similarities
and differences between their mechanisms.
If I start with a terminal alkene and react it first with BH3 and THF followed by NaOH
and H2O2, I will get an alcohol product as follows; the quick way to find this product
as I teach in the alkene shortcut video below is to break the pi bond, identify your less
substituted carbon and simply add an alcohol.
But for the alkyne, we didn't get an alcohol.
So what happened?
At first, nothing.
We identify and break the pi bond, re-draw the skeleton, in this case with one pi bond
because we broke the second, identify the less substituted carbon and give it an alcohol.
But unlike the alkene the reaction doesn't stop here.
Because we have both an alkene and an alcohol sitting on the very same carbon to give us
an enol, an unstable molecule that we cannot isolate.
And this will automatically tautomerize to give us the keto or technically here the aldehyde
form.
This reaction is stereoselective in that the alcohol and hydrogen add syn to the same face
of the pi bond that broke.
And regioselective in that if we have a terminal alkyne, the alcohol and therefore the carbonyl
will follow anti-markovnikov addition onto the terminal carbon.
Before diving into the mechanism, let's take a look at the boron reagent because in alkene
reactions, you simply used BH3, borane but in alkyne reactions you may see BH3, you may
see BH3 bound to another BH3 in the form of B2H6.
In the above reaction I use disiamylborane and some professors will teach it with 9-BBN.
So what's going on?
When it comes to this reaction in real life, we have to pay attention to things like safety
and percent yield.
This is where the different reagents come in.
Because BH3 being very small will add to multiple alkynes as you’ve seen with a tri-alkylation
for hydroboration of alkenes.
It’s also not a molecule you really want to be around.
But by making it bigger, for example fusing two together or adding other things to it,
it becomes easier to work with in that it only adds once.
B2H6 is just 2 molecules of BH3 also known as diborane.
This thing right here is the shortened version for disiamylborane which looks like this.
A boron bound to a hydrogen and then 5 carbons attached in a three carbon chain with a methyl
group on either carbon.
And since it's di, we have one on either side.
9-BBN is a monster of a molecule known as 9-Borabicyclo[3.3.1]nonane, that looks like
this.
Don't worry, you don't need to know what this looks like and you probably don't even need
to know the full name.
What I want you to take away from here is that we can have multiple reagents for this
reaction so don't get caught off-guard.
Memorize one reagent for when you're asked to provide it.
I recommend disiamylborane and then for the rest, if you're given the reagent just look
for the boron.
If you see an alkyne reacting with some crazy boron reagent, you're probably looking at
a hydroboration reaction.
With this in mind, let's look at the mechanism when propyne reacts with disiamylborane.
As with the alkene reaction, we have the boron and hydrogen situating themselves over the
pi bond where the hydrogen goes towards the internal side and boron external.
In the interest of sanity, I'm just going to put Sia instead of drawing the whole thing
and then show the concerted mechanism that happens when the pi bond attacks the empty
p-orbital in boron to form a bond between itself and carbon simultaneously breaking
the bond between boron and hydrogen so that those electrons come down to the internal
carbon to form a bond between carbon and hydrogen.
This is not a one step and then another step reaction, instead it’s a concerted mechanism
that happens at the same time so that the electrons are simultaneously floating around
in a circle.
Our starting alkyne was linear because the sp hybridized carbon had a hundred and eighty
degree bond angle but once that pi bond is broken, we now have an sp2 carbon at a hundred
and twenty degrees and so we can draw it at an angle.
The terminal carbon now has a bond to boron which still has the two siamyl groups attached
and the internal carbon is now bound to hydrogen.
Notice that boron and hydrogen added to the same face of the pi bond for a syn addition.
Fortunately, you don't have to know the mechanism for the next step of this reaction where NaOH
and H2O2 will replace the boron group with an OH to give us our alcohol product.
If you're in the lucky minority that has to know this, ask your professor exactly which
steps to memorize because I'm telling you, you likely don't have to know this.
This leaves us with what appears to be our final product.
The pi bond is broken, hydrogen at the more substituted position, alcohol at the less
substituted anti-markovnikov position, and everything is happy.
Or is it knowing that an enol, an alkene and an alcohol sitting on the same carbon is not
very stable.
This will tautomerize using the keto-enol tautomerization reaction that I teach linked
below at leah4sci.com/KET.
Using the base catalyzed mechanism, why base?
Given that we already have NaOH in solution, we're already in a basic solution.
And so one of those hydroxide ions will reach out for and grab the hydrogen off of the alcohol
giving oxygen back those electrons.
This is an extra step that we did not see in the acid catalyzed keto-enol tautomerization
that we saw on the oxymercuration mechanism.
The terminal oxygen now has three lone pairs, the two that it had originally and the third
that came from the bond between itself and hydrogen to give it a negative charge.
That purple hydroxide is now floating around in solution as a water molecule, thanks to
the extra hydrogen from alcohol.
The negative oxygen is unhappy and will look for something to attack but it won't be oxygen
because if you look at the resonance you'll see the negative charges distributed between
oxygen and carbon.
If we show the resonance form with the negative on carbon, that's a much more reactive, much
less stable contributing structure and therefore the one that is going to attack.
But we don't show the resonance because the electrons won't sit on carbon instead they’ll
immediately seek out another positive hydrogen in solution, in this case off a water molecule.
How is neutral water positive?
In a basic solution, the partially positive hydrogen on a water molecule is the most positive
compared to all the negativity floating around.
When the hydrogen is grabbed, the electrons between oxygen and hydrogen will collapse
on to oxygen to regenerate the hydroxide catalyst, the one that we used up over here.
And our product looks something like this where I want you to pay close attention to
the atoms and electrons.
We had our carbon skeleton with the red oxygen attached.
We had a black lone pair and a red lone pair that were untouched.
We had this lone pair that collapsed towards the carbon to form a pi bond and we had the
green pi bond that broke away to grab a hydrogen.
We already had the dark blue hydrogen but now these green electrons are sitting on that
same carbon bound to a second hydrogen.
And even though it looks like a mess, this is our final product.
So let's redraw it in the more conventional format where the carbonyl is facing down and
we'll add in the invisible hydrogen just to remind us that yes, this is a terminal carbonyl,
therefore it's an aldehyde.
Let's do a quick review of regioselectivity for this reaction.
If I start with a terminal alkyne and react it with disiamylborane followed by NaOH and
H2O2, the final product without following Markovnikov's rule will give me a terminal
carbonyl, in this case an aldehyde.
If I start with a symmetrical alkyne like 2-butyne, the carbonyl can add to either side
to give me the same product.
If I count from the left, I can put the carbonyl on two, if I count from the right, I can put
the carbonyl on two for a single 2-butanone product.
Notice that when I have an internal alkyne, the product is a ketone even though the molecule
underwent hydroboration, it's only a terminal alkyne that could give me the aldehyde.
Where it gets interesting is if the starting molecule is an internal alkyne but asymmetrical.
In this case I have the option to add the carbonyl to carbon 2 or carbon 3 for two completely
viable yet different products to give me 2-pentanone or 3-pentanone.
This is important to keep in mind when working on synthesis if you're trying to control the
presence of your carbonyl.
If you found this video helpful, first make sure to give this video a thumbs up then subscribe
to my channel so you don't miss out on any new videos then be sure to check out my entire
alkyne reaction series along with the practice quiz and cheat sheet by visiting my website
linked below, leah4sci.com/Alkyne, the link again leah4sci.com/Alkyne.