[narrator] Let’s have another look at hydroboration oxidation, including the mechanism but also
predicting the products. We know a couple of things about this process that we’ve
learned already. Number one is that it’s a syn addition so both atoms of borane add
from the same side of the double bond and also, perhaps more importantly, the process
is anti markovnikov. When we predict the product, let’s draw our cyclohexane ring first, we
are going to--let’s say we add from the top, we’re going to push the methyl away
from us while adding the OH toward us after the mechanism is complete. We know for these
processes if we add from the other side, since we formed two chiral centers, we could also
add from the bottom and we would have the enantiomer. But, only these two possible stereoisomers
and none of the other two stereoisomers. Predicting the products is one thing but then drawing
the mechanism of this beast is another. Let’s do it, let’s not be afraid. If we start
here we know that we approach anti markovnikov
and we are going to dump the pi electrons into the empty orbital on boron. We know there’s
a p orbital that’s empty there so we can fill that and then the BH bond is going to
go where the markovnikov position would be. We see the famous square transition state.
Where we have the vertical bonds that are forming and the horizontal bonds that are
breaking. This is not necessary to show in the mechanism unless I request it. What is
required are the intermediates and electron movements, so all the arrows and all the intermediates
that we draw. Now, after adding the BH3 and the THF this part is complete. But now we
bring in the deprotonated peroxide anion that will also dump its electrons into the p orbital.
As an aside I would like to point out that this can happen twice more to make a compound
where boron has three R groups. In this problem, R is going to be equal to our cyclohexane.
Basically where our alkene was. Three of those can happen at once. Really it’s three equivalents
but to simplify the mechanism we are just going to show it happen once, make it easier
to draw for us. Dump the electrons from the lone pair into the boron. Now we have an intermediate
with a formal charge of -1 on the boron. Boron makes four bonds and so what we do now is
we invert the bond. We need to have a CO bond here instead of a CB, so we’d like to have
a CO instead. What we do is we break the CO bond--or CB bond, form a CO bond, and break
this weak peroxide bond forming a small molecule. We get an enthalpy and an entropy gain there
that drives this. We basically invert where the boron is sitting. It’s a tricky part,
okay, it’s a tricky part. Hard to remember. But, we can kind of guide ourselves through
it. Kind of break the CB, invert into the oxygen, break the weak peroxide bond, and
here’s what we’re left with. Now our boron is ready to be attacked again and so we generally
show that with a hydroxide that we just formed. Although others are surely floating around
and we make a complex that, again, has a boron with a -1 formal charge. In equilibrium, then
our complex can then dump its electrons from boron’s p orbital, I guess it’s from SP3
as it rehybridizes back to P, dumps those electrons back on to oxygen forming an alkoxide.
Our alkoxide can then, in equilibrium, also protonate in an acid base reaction with water
solvent to form one of our predicted products. The
other would come from addition to the opposite side of the alkene.