professor Dave here let's talk about hydroboration oxidation.
so we're gonna learn another addition reaction here and it is actually another hydration but the
difference here is whereas oxymercuration demercuration was a
markovnikov hydration now we want to learn a way to do an anti markovnikov
hydration. so to just go over the the the general reaction here this happens in
two steps: a hydroboration and an oxidation and so
the first step the hydroboration the way we're going to remember these conditions
if we see them on a test is boron so hydroboration the boration is for
boron. so if you see BH3 it's a pretty good
indicator that we're going to be doing this reaction and so let's say we have
this alkene here these conditions are going to perform an anti markovnikov
hydration and it will do so in syn fashion, so what does that mean? that
means that we are attaching once again any hydration is an H and an OH and if
it is a syn hydration that means they are coming from the same side so we are
adding H and OH to the same side so let's say here's our H and here's our OH
and if we're doing that in anti markovnikov fashion that means that the
OH is going to go to the less substituted carbon and so what that
means is again here's our H here's our OH, anti markovnikov means OH is down
here syn means they're coming from the same side so we can go like this and if
we go like that that means that the OH is ending up out right so these are
going to be like that and that's why the methyl gets pushed back because if the H
and OH are on the same side remember that means that the implied hydrogen
that we put on there is also there so there is often some confusion there
people think that if it's syn that the methyl ought to be sis to the hydroxyl
in the product that's not the case where we are referring to the two groups that
we are adding to the molecule that's the H and the OH so if H and OH went like
that then we would get this product here but once again we can although there is
stereospecificity in the sense that it is syn we still
will get a mixture of stereoisomers because the H and OH can also come from the
back and so which one did I say was H I don't remember H and OH I believe so if
those came from the back and again I can't show that but if they came from
the back that would be this situation here and the implied hydrogen that would
be this those would be cis to one another on the product which means that
methyl would get pushed forward so that is the general overview of hydroboration
oxidation. we're gonna go over the mechanism now but we may not necessarily
need to know it what we definitely need to be able to do is to see these
conditions and look at a substrate and say I know that those conditions are for
anti markovnikov syn hydration so I'm gonna take H and OH I'm gonna put them
on to that molecule in anti markovnikov fashion so OH to the less substituted
carbon of the two and they're gonna go to the same side of the molecule so
that's what we definitely need to know. now just to elucidate why we have this
regiochemistry and stereochemistry built into this reaction
let's just look at the mechanism real quick so where both the regiochemistry
and the stereochemistry occur is in the hydroboration step. so the first step is
the hydroboration, that's going to be the addition of BH3 to the molecule and the
reason that goes like this is this BH3 is a little bit different, boron is
actually a very electropositive element it's in group 3 it's it's more
electropositive than carbon so actually a BH3 bond is going to be very slightly
polarized towards the hydrogen and we've seen this with BF3 and
other molecules but in general that covalent bond is slightly polarized
towards the hydrogen so if this is going to add to that pi bond right if we're
gonna have a hydrogen and a boron add in that way it's gonna do so preferentially
number one it's going to go it's going to go anti markovnikov because this
hydrogen which is the partial negative is going to want to orient to the carbon
that is slightly better at accommodating a partial positive charge and that is
going to be of course the more substituted carbon
so yes the pi bond is acting as a nucleophile in a sense here because it
although it is all concerted this carbon is going to be the one that is
interacting with the boron and then this covalent bond is going to dissipate as
the pi bond dissipates and this sigma this carbon hydrogen sigma bond is going
to be forming so all of that is happening all at once this is the
transition state of that addition and so so this is what's going on and this is
what we get and so this is the reason and then also the stereochemistry is
going to be sin because this is one molecule approaching one face right we
have BH3 approaching from this side or from behind right it's not going to be
able to do an anti addition because just like other reactions where both groups
are coming from the same molecule it's gonna have to add to the same side so
that explains the stereochemistry and the electronics are what are explaining
the the regiochemistry so that's how we get to here that is the product of our
hydroboration and we can see that we've already got the syn and anti markovnikov
going on there in the hydroboration product so now for the oxidation it's a
little bit different so we're running this we've got and also by the way
whenever we see hydrogen peroxide or any peroxide that's usually our tip-off for
anti markovnikov chemistry right we saw that for anti markovnikov
hydrohalogenation where we were doing that in peroxides that's how we knew it
was anti markovnikov so that's the same thing here that's how we know that this
is happening so that's just one thing to remember so we've got we're running this
in base and so base can go ahead and deprotonate our hydrogen peroxide so
we've got this anion here and then this can go ahead and attack the boron atom
like that and so the negative charge is being transferred from the oxygen to the
boron because oxygen is now neutralized it's given an electron to boron but
boron has gained an electron it has four covalent bonds right one to the rest of
the substrate two to the two hydrogens one to oxygen that now has a formal
negative charge so now this is the key step this is where students often get a
little bit tripped up this this carbon boron bond can migrate over here
to this oxygen so it's it's a it's a rearrangement step and so this bond is
going to disconnect from that right we're not going to have a carbon boron
bond anymore this bond is going to go from carbon to
that oxygen and then we can kick off this OH right there and that is going
to get us to this situation here so this may look a little bit confusing at first
but just look a little more closely this bond is gone this bond is still there
and this bond is gone right so we've lost so we right we get an OH minus
right OH minus has left but this bond we've got a carbon oxygen bond and then
this one is still there so that's why BH2 is still coordinated to this oxygen so
we've sort of flipped things around it's a little bit of a rearrangement and then
all that's going to happen now is we've got hydroxide in solution and so that
can go ahead and attack boron and get these get the oxy anion and so now we're
almost done the oxyanion is just going to react with something in solvent just
to protonate so that could be water if we're doing this aqueous and so that is
how we protonate that and we're gonna get to our product so in this case we
have drawn the hydroxyl on the wedge that would be this and so that from here
that was our oxidation so we had our hydroboration which once again is where
the stereochemistry and the regiochemistry both come into play in
the hydroboration and then we had our oxidation with this interesting
rearrangement step to get our product in this case we drew this one however we
would get a racemic mixture in this particular case so that's what we need
to know about hydroboration oxidation
so to put this together let's look at one example and this is something that
you maybe would see on a test and this is something you could be responsible
for we need to see substrate in these conditions and know what to do we need
to know what what molecules going to be formed and so we see this and we
remember right what is the tip off we've got boron BH3 right there that is going
to be the tip-off for hydroboration oxidation so all we have to do is
recognize these conditions and and associate these conditions with
hydroboration oxidation in our minds and we also have to memorize unless we
memorize the mechanism it might be easier to just memorize the fact that
hydroboration oxidation is an anti markovnikov sin hydration and that's all
we need to memorize because we understand what those terms mean so if
we know that hydro bration oxidation is doing an anti markovnikov sin hydration
it means that we need to add an H and we need to act let me do this in blue
actually we need to add an H and we need to add an OH so we've got an H and
we've got an OHA and where are they gonna go so an H and an O H we know that
if it's anti markovnikov that OH has to go there right we have to put the OH
on the less substituted carbon because it's anti markovnikov so this this is
the less substituted this is more substituted an additional methyl group
so we know that that is going to be the regiochemistry on the product we also
know the stereochemistry it's going to happen in syn fashion and actually in
this particular case it doesn't really matter because we're only producing one
stereocenter and not another so let's say we draw our product so we can draw
it like this and let's just do that so this would be as though that implied
hydrogen right this we added the H and the OH in syn fashion and let's say we
did that and so that methyl got pushed back the thing is that this is not a
stereocenter right we could rotate this and we could have this bond this
methyl end up on the wedge bond so that actually doesn't matter but we are
generating one stereocenter here and that will happen in racemic fashion so
actually we do have to show this so once again this the the right it doesn't
matter what's going on here because that's not a stereocenter but even
though we do have stereospecificity in the sense that it is a syn addition
we're still going to get this racemic mixture here and so all again all we
have to do once again is see a substrate recognize these conditions as
hydroboration oxidation and then know that that means anti markovnikov syn
hydration put our hydroxyl group on the less substituted carbon and then assess
stereochemistry from there.