Oxidation of alcohols | Chemical Processes | MCAT | Khan Academy


Let’s see what happens
when you oxidize alcohol. So in the top left
here, we’re starting with a primary alcohol. And the carbon that’s
attached to the OH group is your alpha carbon. To oxidize an alcohol, you
must have alpha hydrogens. You must have hydrogens attached
to that alpha carbon in order for the mechanism to work. So in that mechanism,
you’re actually going to lose one of
those alpha hydrogens. And we’ll take a look at the
mechanism in a few minutes. So if I were to oxidize
this primary alcohol, I’ll add something to oxidize
my primary alcohol, like that. One way to think about the
oxidation of an alcohol is to think about the number
of bonds of carbon to oxygen. On the left side here, we have
one bond of our alpha carbon to this oxygen. In the mechanism, we’re going
to lose a bond of carbon to hydrogen, and we’re going
to gain another bond of carbon to oxygen. So you’re increasing the number
of bonds of carbon to oxygen. So that would, of
course, give me two bonds of carbon to oxygen,
if I oxidize my alcohol one time. And I’m going to lose
one of those hydrogens. So one of those
hydrogens is still left. And my alkyl group
is still attached. So obviously this would give me
an aldehyde functional group. So if you oxidize a
primary alcohol one time, you’ll get an aldehyde. Let’s take a look
at the oxidation states of my alpha carbon
and see what happened to it. So if I want to
assign an oxidation state to my alpha
carbon on the left, once again, I have to
put in my electrons. Each bond consists of
two electrons, like that. And I need to think about
electronegativity differences. Oxygen is more
electronegative than carbon, so it’s going to take
those two electrons. Carbon versus carbon is
a tie, so each carbon will get one of those electrons. Carbon actually is slightly more
electronegative than hydrogen, so carbon will win and take
those electrons right there. Carbon normally has
four valence electrons, and in this
instance, it is being surrounded by five electrons. So four minus five will
give me an oxidation state of negative one for
my alpha carbon. So let’s look and see what
happened to that alpha carbon after we oxidized it. Right? So over here on the right, if
I wanted to assign an oxidation state to what is now
my carbonyl carbon, once again I think about my
electronegativity differences. And I know that oxygen
is going to beat carbon. Carbon versus carbon is a tie. And carbon versus
hydrogen– carbon will win. So the oxidation
state of that carbon– normally four valence electrons
surrounded by three this time. So 4 minus 3 will
give me plus 1. So I can see that my oxidation
state went from negative 1 to plus 1. So an increase in the oxidation
state is, of course, oxidation. So if you oxidize a
primary alcohol one time, you will get an aldehyde. What about if you keep going? Right? So if you form an aldehyde–
and sometimes it’s hard to stop the reaction
mixture from continuing to oxidize. So if you oxidize
an aldehyde, you think about what functional
group you would get. Well, again, a
simple way of doing it would be to think
on the left side, I have two bonds of
carbon to oxygen. Is there any kind
of functional group where carbon is bonded
three times to an oxygen? So that, of course, would
be a carboxylic acid. So if I think about the
structure of a carboxylic acid, I can see that carbon is
actually bonded three times to an oxygen, if you will. Three bonds of carbon to oxygen. And over here, I have my
alkyl group, like that. So if you oxidize
an aldehyde, you’re going to get a carboxylic acid. Let’s look again at
the oxidation state of my carbonyl carbon. So once again, I put
in my electrons here and I think about
electronegativity. So oxygen, of
course, beats carbon. Right? Tie between these two carbons. And oxygen beats carbon again. So in this case, normally
four valence electrons. Now there’s one. So 4 minus 1 gives us an
oxidation state of plus 3. So once again, an increase
in the oxidation state means oxidation. If you oxidize an aldehyde,
you’ll get a carboxylic acid. Let’s look at a
secondary alcohol now. So we’ll go down here to
our secondary alcohol. And once again, identify
the alpha carbon, the one attached
to your OH group. We need to have at least one
hydrogen on that alpha carbon, and so we have one right here. So if we were to oxidize
our secondary alcohol– so we’re going to oxidize
our secondary alcohol. Once again, a
simple way of doing it is thinking my alpha
carbon has one bond to oxygen, so I could increase
that to two bonds and that should be an
oxidation reaction. In the process,
I’m going to lose a bond to my alpha hydrogen. So I’m now going to have two
bonds of carbon to oxygen, and I’m going to lose the bond
that that alpha carbon had with the hydrogen there. So that leaves my two
alkyl groups, like that. So now I have two alkyl groups. And of course, this would be
a ketone functional group. So if you oxidize a
secondary alcohol, you’re going to end
up with a ketone. I can assign oxidation states. So once again, let’s
show that this really is an oxidation reaction here. And I go ahead and put in my
electrons on my alpha carbon, and think about
electronegativity differences. So once again,
oxygen beats carbon. Carbon versus carbon is a tie. Carbon versus
hydrogen, carbon wins. And then carbon versus carbon,
of course, is a tie again. So normally, four
valence electrons. In this example, it’s
surrounded by four. So 4 minus 4 gives us
an oxidation state of 0 for our secondary alcohol. And when I oxidize it, I’m going
to get this ketone over here on the right. So let’s take a look
at the oxidation state of the carbon
that used to be our alpha carbon
on the left, which is now our carbonyl carbon. So once again, we
put in our electrons and we think about
electronegativity difference. Right? So oxygen is going
to beat carbon. So we go like that. Carbon versus carbon is a tie. Carbon versus carbon
is a tie once again, [? it’s ?] normally
4, minus 2 this time around that carbon giving us
an oxidation state of plus 2. So to go from a secondary
alcohol to a ketone, we see there’s an increase
in the oxidation state. So this is definitely
an oxidation reaction. Let’s look now at
a tertiary alcohol. So here is my tertiary alcohol. And when I find
my alpha carbon, I see that this time,
there are no hydrogens bonded to my alpha carbon. So according to the mechanism
which we’ll see in a minute, there’s no way we can
oxidize this tertiary alcohol under normal conditions anyway. So if we attempted
to oxidize this, we would say there’s
no reaction here since we are missing
that alpha hydrogen. Let’s take a look
at the mechanism and see why we need to
have that alpha hydrogen on our alpha carbon. So if I were to start my
mechanism here with an alcohol, remember, this must be either a
primary or a secondary alcohol in order for this
oxidation to work. So I’m going to go ahead
and show my alcohol there. All right. So again, either primary
or secondary, like that. And when we have our primary
or secondary alcohol– and it’s going to be
reacting with chromic acid. So here is the dot structure
for chromic acid, like that. So I’ll just simplify
it right here. I won’t worry too much about
my lone pairs of electrons. And chromic acid can come from
several different reagents. Probably the most common reagent
would be sodium dichromate. So Na2Cr2O7. Sulfuric acid- H2SO4. And water. And all of this
together is usually referred to as
the Jones reagent. So a mixture of sodium
dichromate, sulfuric acid, and water is called
the Jones reagent. And that will mix
together to give you chromic acid in solution. OK. So another way to
do it would be, you could start from
chromium trioxide. So you could also use
a different reagent which consists of CrO3,
chromium trioxide, and H3O plus, and acetone, and that
will also generate chromic acid in solution. So whichever one you
would like to use. The first step of
the mechanism is similar to the formation
of nitrate esters that we saw in the
previous video. OK. So this is going to be
a reaction equilibrium, or it’s reversible. And if you remember, in the
formation of nitrate esters, it’s a similar mechanism
for the formation of all inorganic esters here. And we’re going to lose
this hydrogen and this OH, and those are going
to produce water. And we can stick those
two molecules together. And so we would get this as
the initial product here. We’re going to have the end
result of putting that oxygen bonded to that chromium
atom like this. So this is a chromate
ester intermediate. All right. So this is what we would make. All right. So in the next step
of the mechanism, we need something to
function as a base. And water is going
to do that for us. So water comes along like this. Two lone pairs of electrons. One of those lone pairs
can function as a base, and it’s going to take
that alpha proton. Remember, this is our alpha
hydrogen on that carbon. And over here, we’re going
to take just the proton, just the nucleus of
that hydrogen atom. And so this lone pair
of electrons in here could take that proton. That’s going to leave the
electron that hydrogen brought to the dot structure behind,
and these two electrons are going to move
into here to increase the number of bonds
of carbon to oxygen. At the same time,
that is going to kick these electrons in this
bond off onto the chromium. So let’s go ahead
and draw the result, the product of
that reaction here. So let’s see if we can get
some space– so right here. Well, we’re going to
lose that alpha hydrogen. Right? So now our carbon still is
bonded to two other things. We lost that alpha
hydrogen, and now it’s double-bonded to that oxygen. So that would be the mechanism. We went from one bond
of carbon to oxygen on our primary or
secondary alcohol. We’ve now increased it to two
bonds of carbon to oxygen. So the other products here, we
would make H3O plus, of course. So we’ll go ahead
and put H3O plus when water picks up that proton. We would form HCrO3 minus
as our other product. Now, if the alpha carbon
is the one being oxidized, so if this carbon is
oxidized to this carbon– it’s the same carbon, but
this carbon is being oxidized. Something must be being reduced. So this is a redox reaction. If you oxidize something,
something else is reduced. And that something
else is chromium. So if you were to assign an
oxidation state to chromium in the sodium dichromate
over here– so in this guy over here– chromium has an
oxidation state of 6 plus. So when we look at
our products and we find chromium in
our products here, if you were to
assign an oxidation state to this chromium,
you’d get 4 plus. So Cr4 plus. And there’s some
other chemistry that goes on, which
ends up converting the chromium from
4 plus into 3 plus. And so overall, you
can see that you’re starting out with
6 plus over here, and you’re ending up
with 3 plus over here. That’s a decrease in
the oxidation state. So chromium is being reduced. So that alpha carbon
is being oxidized, and chromium is being reduced
in this redox reaction. In the next video,
we’ll take a look at several examples involving
primary and secondary alcohols.

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