Before we get to preparation
of alcohols using sodium borohydride, let's take a look
at a few of the other ways to make alcohols
that we've already talked about in earlier videos. For example, you could
make alcohols from alkenes, and you can add the OH on
in a Markonikov fashion, or you can add the OH on in
an anti-Markovnikov fashion. So that's one type of
way to make your alcohol. So you could also do
so from alkyl halides. The process could
be an SN1 process, or it could be in SN2 process. And then, what we
haven't talked about yet is how to prepare alcohols
from carbonyl compounds, and there are a couple
different ways to do that. You could use sodium
borohydride, NaBH4, which is what we're going
to talk about in this video. On the next video
we're going to talk about the use of lithium
aluminum hydride, so LiALH4. And you could also use an
organometallic, something like a Grignard reagent. So I'll put that down
here as well too. And we'll talk about that
in a future video as well. So all we have time for this
video is sodium borohydride. So let's check out
the general reaction for the use of sodium
borohydride to form an alcohol. And so you can see
over here on the left, we're starting with either
an aldehyde, or a ketone. So there's either a hydrogen
attached to your carbonyl, or an R prime group like that. We're going to add
sodium borohydride, and then we're going
to add a proton source it could be
just about anything. So H+ tends to work. And we're going to form either a
primary or a secondary alcohol, depending on our
starting materials. This is going to be a primary
or in a secondary alcohol. Let's look at the mechanism
for the preparation of alcohols using sodium borohydride. Let's go ahead and start
with a ketone here. So we're going to
start with a ketone. So we'll just do the
reaction like that. So I have my carbonyl
like that, put in my loan pairs of
electrons, and then make this an R prime
to make it our ketone. And sodium borohydride
comes along. So let's go ahead and draw the
structure for that-- so Na+, positive formal charge. And then we have boron, bonded
to four hydrogens like that. And there's a negative 1
formal charge on our boron, so we'll go ahead and put
that in there as well. So the mechanism I'm
going to show you is a simplified mechanism. The actual mechanism is a
little more complicated, but this mechanism works. So let's just go with
the simplified version. All right, so the first
thing you have to think about is this carbonyl here,
this carbon double bonded to an oxygen. There's an
electronegativity difference between the carbon
and the oxygen. So right here, this carbon
and this oxygen, the oxygen is more electronegative, meaning
that those bonds, the electrons in the double bond
between the carbon oxygen, are going to be pulled
closer to the oxygen. Therefore, oxygen has a
partial negative charge, right? So increased electron
density around it. Whereas the carbon
here is losing some of the electrons around it. So it's partially
positive, like that. So if carbon is
partially positive, it's our electrophile, right? It wants electronns. So where can we
get our electrons? We can get our electrons
from right here. So the two electrons
on this hydrogen here are going to move out
and attack this carbon. So that's going to be
the nucleophilic portion of our molecule, like that. And that would be too many bonds
to a carbon, so one of these, this pi bond here, is going
to kick off onto the auction. So they're going to put
two electrons on that. So let's go ahead
and draw the results of that nucleophilic attack. So we have R, and
then we have R carbon, and then R prime over here. And this top auction
had two lone pairs of electrons that just picked up
one more, for a total of three. And that gives it a negative
1 formal charge, like that. And we added on a
hydrogen, like that. So let's go ahead and color
code some electrons here. All right, so we can see
where everything went. So I'm saying this hydrogen and
the two electrons in this bond became this hydrogen and these
two electrons right here. So really, what you're
doing is, sodium borohydride is a source of hydride anions. So a hydride anion
would be hydrogen with two electrons around it
and a negative 1 formal charge. So that's the way to think
about these reagents. Hydride itself is not
the best nucleophile, because it isn't
polarizable enough. Such a small atom that it
doesn't really work very well as a nucleophile by itself. So this, of course, is
our simplified version. All right, in the next
step of our mechanisms, get some acid-base chemistry. So we have protons
floating around, right? So we added some
H+ to our solution, and the loan pair of
electrons picks up the proton, and we are done. So let's go ahead
and draw the product. It would be our
alcohol, like that. And let's go ahead and put
those lone pairs of electrons. And there's only one hydrogen
added onto our carbon. So if we start with
the ketone, we're going to end up with a secondary
alcohol, as in the situation. All right, let's do an
actual reaction here. And let's start with vanillin. All right, so vanillin is a
very nice smelling molecule. And if you look at the
structure of vanillin, there's an aldehyde functional
group right there on our ring. So there's some other
functional groups. So go ahead and put in the
rest of the vanillin molecule, like that. So we're going to add sodium
borohydride in our first step. We're going to add
sodium borohydride, and then, once
that's reacted, we're going to add a
source of protons. HCL works pretty well
for this reaction, so we'll just say a
source of H+, like that. So when you're trying to figure
out the product on a test, it's not even necessary do that
really simplified mechanism. You can pretend
like the hydride ion is going to act
as a nucleophile, even though we've already
covered that doesn't quite do that. So here we have our
hydride nucleophile. So it's going to attack
this carbon right here. This is our carbonyl
carbon, right? So it's going to be
partially positive. That's going to kick these
electrons off onto our oxygen, like that. So let's go ahead and draw the
intermediate for this reaction here. So we'll have our
benzene ring, like that. I'll put in our other
functional groups over here, which don't
participate in the reaction. And let's go ahead and
draw what we have here. So we had one hydrogen
already bonded to that carbon. We just added another
hydrogen onto that carbon. And then we have an oxygen still
there with three loan pairs of electrons around it, giving
it a negative 1 formal charge. So when we add our
source of protons, a loan pair will pick
up that proton there to form our alcohol. So we'll go ahead and draw
our products right over here. So once again we have an
OH coming off of our ring. And we have this part portion
of the molecule, this ether portion right here. And then we have an
alcohol, like that. So this molecule, since it
has such a similar structure to vanillin, also has
a nice vanilla smell. It's very pleasant. So this is a very good
undergraduate organic chemistry lab to do. And let's classify
the type of alcohol we got as either primary,
secondary, or tertiary. So remember how to do that,
you look at the carbon that's directly attached to the
OH-- that's this carbon-- and you see how many carbons
that carbon is attached to. That carbon is attached
to one other carbon. So this is an example
of a primary alcohol. So if you start with an
aldehyde, as we did over here-- this is our aldehyde--
you're going to end up with a primary
alcohol as your product. Let's do another one. This time let's do a ketone. All right, so let's go ahead
and draw a ketone over here. So we'll do cyclohexanone. So here's our ketone, like that. And we're going to add sodium
borohydride, like this-- NaBH4. And we'll add methanol. And we'll do this
on one step here. So once again, think
about sodium borohydride being a source of
hydride anions. All right, so again even
though this isn't technically the mechanism, it's
our simplified version. So if you're taking a test, this
is the easiest way to do it. Attacks that carbon, kicks these
electrons off onto your oxygen. So we have our intermediates
right over here. So let's go ahead and draw that. So we have the
hydrides anion attack to form a new bond to what
used to be our carbonyl carbon. Now this oxygen has
three lone pairs of electrons around it and
a negative 1 formal charge. It's going to pick up
a proton from methanol as its protons source this time. And therefore, our
final product, if we're going to go ahead and draw it. I'll keep the hydrogen in there,
and you'll see why in a second. And then I have
my OH, like that. So I form cyclohexanol
as my product. This is a reduction reaction. So we are reducing the
molecule to form our alcohol. So let's take the reactants and
the product from this reaction and let's assign some
oxidation states, so we can see how this is
an example of a reduction reaction. So I'm just going to redraw
our starting ketone here. So let's go ahead and put in
some of the atoms this time. So I'm just going to redraw
that starting ketone, like this. And I'm going to
go ahead and draw the product that
we got, as well. All right, so I'll put
in these carbons here. And then we added on
a hydrogen, like that. And then we forms our
alcohol over here, like that. So let's go ahead and think
about oxidation states. So when we're assigning
oxidation states, we need to go ahead and draw
in these electrons here, all right. So let me go ahead and put in
these electrons on these guys right in here, like that. All right, so let's think
about how we did this in some of the earlier videos. It's actually thinking
about electronegativity. And if I'm thinking about the
four electrons between carbon and oxygen, oxygen is
more electronegative. So oxygen is going to get
all of those electrons. So we go ahead and go like that. Carbon versus carbon, fighting
over these two electrons, it's an equal
electronegativity, obviously, since it's the same element. So each carbon's going to
get one of those electrons, all right. And the same thing
for over here. So when we assigned
our oxidation states, we said that's the number
of valence electrons that carbon normally has,
which, of course, is four. And from that we subtract the
number of electrons around it when account for
electronegativity, which is two in this case. So the oxidation state
of that carbon is plus 2. So we have a plus 2
oxidation state on the left. Let's think about the right. Let's go ahead and
put in our electrons. Each bond consists
of two electrons. So we go ahead and put the
fact that each bond consists of two electrons
in here, like that. And then once again we
think about differences in electronegativity. So again, we know carbon
versus carbons is a tie, so we can go like that. And carbon versus oxygen--
oxygen's more electronegative, so it's going to get
those electrons there. But carbon verses
hydrogen-- carbon is a little bit more
electronegative. So now carbon's going to
take those two electrons. So carbon normally has four. In this case, carbon
has four around it after we count for
electronegativity. So the oxidation state of
that carbon atom is zero. So let's think about what
happened to the oxidation state. We started off with a
plus 2 oxidation state, and that number was reduced
to an oxidation state of zero. So this is a reduction
reaction, right? This is a reduction. So several different
definitions you could use. You could think
about reduction being a decrease in the
oxidation state, or a reduction in
the oxidation state. You could also think about
that carbon gaining electrons-- so it picked up
two more electrons. And yet another way to
think about it is carbon lost a bond to oxygen,
Over here on the left, carbon had two bonds to oxygen. Over here in the right, carbon
only has one bond to oxygen. And it formed a
bond with hydrogen. So increased the number
of bonds to hydrogen, decreased the number
of bonds to oxygen is another way to think
about this being a reduction reaction. But assigning your
oxidation states is probably the
best way to do it. In the next video,
we'll take a look at how to prepare alcohols using
lithium aluminum hydride, which is very similar to using
sodium borohydride.