To hydrogenate an alkene, you
need hydrogen gas and a metal catalyst, something like
platinum or palladium or nickel. And there are many
others, but these are the ones most commonly used. So what happens is those two
hydrogens from the hydrogen gas are added across
their double bond. And they're added
on the same side of where the double
bond used to be. So it's a syn addition. Let's take a look at why this
is a syn addition of hydrogens. So we have our metal
catalyst over here. So let's go ahead and draw
our flat metal catalyst. And these metals are chosen,
because they adsorb hydrogen really well, which means
that if you bubble hydrogen gas through, the
hydrogen is going to be adsorbed to the surface of
that metal catalyst, like that. And then your
alkene comes along. And your alkene is
also flat, right? The portion of the molecule
that contains the double bond, right? So these two carbons, this
carbon and this carbon, these sp2 hybridized,
which means that the stereo chemistry
around those two carbons is going to be flat. So this portion of
the molecule is flat. So you have one
thing that's flat approaching something
else that's flat. So the only way those
hydrogens can add are to add them on to
the same side, right? So if this carbon
and this carbon, if you add this hydrogen
to the carbon on the left and add this hydrogen to
the carbon on the right, and then you go ahead and you
draw the rest of the bonds, right? This would now be a
wedge and then a dash, and then this would be
a wedge and then a dash. You can see those two hydrogens
have added on to the same side. So these two
hydrogens or these two hydrogens for our syn addition. Notice we're also changing
from sp2 hybridization to sp3 hybridization
over here on the right. So we have to think
about stereochemistry for this reaction for
your products as well. So let's take a look at
an actual reaction here. And let's see if we
can follow along. So if this was my
reaction, I want to hydrogenate this alkene. So I would add some
hydrogen gas and I could choose whichever
metal catalyst I wanted to. I would add two hydrogens
on the same side, right? So I could add two
hydrogens on the same side. So just like I did up there. So we would get now everything
changes from sp2 hybridization to sp3. So we have wedges and
dashes to worry about. And usually you wouldn't
see it drawn like this. That's too much
work, quite frankly. It would be much easier just
to say, well, all I have to do is take away the double
bond, and there's my product. So for some of these
reactions, they're very, very, very simple. Just take away the
double bond and you'll end up with your
alkene-like product. Let's take a look at oxidation
states for this reaction. So I'm going to
redraw this reaction. And this time I'm going
to draw in my atoms. And I'm also going to draw in
my electrons here in a second. So I'm just drawing
out all the atoms here. So I have all these methyl
groups to worry about. And then I have electrons
in these bonds, right? Each one of these bonds
consists of two electrons. So I'm going to go ahead and
put in all of my electrons here, like that. Now, let's assign
oxidation states to those two carbons that
formed our double bond, right? So let's look at oxidation
state for the top carbon. Remember, when you're
doing oxidation states, you're worried about
electronegativity. So oxidation states are all
about electronegativity. So go back and watch the earlier
video on oxidation states. So we have here comparing
the electronegativities of carbon and carbon. Well, obviously
they're the exact same. So in the struggle
for these electrons, it gets divided, right? Each one of these
carbons is going to get one of these electrons. So that's the case for
all of these right here. So that carbon has four
electrons around it. It's the exact same thing
for this carbon, right? This carbon has four
electrons around it. To assign an oxidation
state, we take the number of valence electrons
that atom usually has, which carbon normally
has four, of course. From that we subtract
the number of electrons we just drew around it
for our dot structure. So that would be four. Each one of those
carbons has four. So each of these carbons has
an oxidation state of zero. Let's look at the
product, and let's see if we can assign some oxidation
states for the product. So our product over here on
the right, we had a carbon and we had some methyl
groups bonded to that carbon. We added on a hydrogen. And so each one of these carbons
got a hydrogen added onto it. And let's go ahead and fill in
our electrons in these bonds. So once again, each bond
consists of two electrons, like that. And now we have a single
bond between our carbons. And let's assign some
oxidation states. So once again we know
that the two carbons have the same
electronegativity, right? So the tug of war for
these two electrons right here, it's a tie. So it's a tie, it's a tie. What about carbon
versus hydrogen? Carbon is actually more
electronegative than hydrogen. So in the war over
the two electrons in the carbon-hydrogen
bond, carbon wins, because it's a little
bit more electronegative. So we're going to assign this
extra electron here to carbon. And then again,
carbon versus carbon. So that carbon gets
that electron as well. Same thing down here, right? So it's a tie, it's a tie. Carbon beats hydrogen. And over here, it's a tie. So in the dot
structure on the right, the oxidation states
that the normal number of valence electrons
would be four. From that we subtract the number
of electrons in our picture here, which would
be five electrons. Each one of these carbons
has five electrons around it. So it gained electron. And it's a 4 minus 5 will
give us a negative 1. So the oxidation states of
these two carbons is negative 1. And we can look at our original
oxidation states of being zero, went from zero to negative 1. That's a decrease in the
oxidation state, right? A decrease in the oxidation
state means a reduction. So this is a reduction reaction. So the alkene is reduced by
the addition of these two hydrogens. And you'll see other definitions
for oxidation states. You'll see a gain in
hydrogens is reduction. That's another
definition that's often found in organic
chemistry textbooks. And while that's true,
to me it makes more sense to go ahead and assign
your oxidation states and watch the oxidation
states change as you add those hydrogens, as your
molecule gains hydrogens. So this is a reduction. Let's look at the
stereochemistry of the hydrogenation reaction. So let's do an example
involving stereochemistry. So let's say your
alkene-- let's do that ring again, it
wasn't a very good one-- so let's say your alkene
looked something like this. And you're going to
react that with hydrogen and with platinum. All right, well,
your first thought might be, OK, this is simple. All I have to do is take away
that double bond and I'm done. Well sometimes that's true. But in this case, we actually
formed two new chirality centers, right? So this top carbon here
is a chirality center, and this bottom carbon here
is also a chirality center. So sometimes it's not
quite that simple. We need to think about the syn
addition of those hydrogens when you think about the
possible products that would result. So we're going to get
two products here. Let's look at the
one on the left. Well, one possibility is I
can add those two hydrogens on the same side
as a wedge, right? So I have one
hydrogen as a wedge, the other hydrogen as a wedge. That's our syn addition. And that means that this top
carbon here, this ethyl group, must be going away from me. And down here at
the bottom carbon, the methyl group must
be going away from me. So that's one possible product. The other possibility, instead
of having my two hydrogens as wedges, I can have my
two hydrogens add as dashes. So there's a hydrogen
and then here's a dash, and there's a hydrogen. So at this top carbon
here, now my ethyl group is coming out at me. And at this bottom carbon
now my methyl group is coming out at me, like this. So I have two possibilities. And if I look at
these two products, I can see that they
are enantiomers, they are mirror
images of each other. So these two would
be my enantiomers, and these would be the
products of my reaction. So be very careful when thinking
about syn additions here. Let's do one more example
of a hydrogenation reaction. Let's do a bridged
bicyclic compound. So let's look at a famous
bridged bicyclic compound. Let's see if we
can draw it here. And then I'm going to draw
that back carbon a little bit off like that. And my double bond is
going to go right here. And then this is going
to be a methyl group. And then up here there are
going to be two methyl groups, like that. So this is alpha-pinene,
found in turpentine. And you can see there's
an alkene on this. So if I took this
alpha-pinene molecule and I wanted to
hydrogenate it, I could use palladium and
charcoal, palladium and carbon. And if I think about what
happens in this mechanism, I know that my metal
catalyst there, my palladium, is going to be flat, like that. And so, when it has
those hydrogens, when the palladium adsorbs
those hydrogens, it's going to add those two
hydrogens to my double bond, think about this
guy over here, think about the alpha-pinene as
molecules like a spaceship, right? And the spaceship is
approaching the docking station. So the spaceship is
slowly going down. The spaceship is going to
approach the docking station. And there's only one
way the spaceship can approach the
docking station. And that is the way in which
we have drawn it right here. It could not flip upside down
and approach it from the top, because of the steric hindrance
of these methyl groups. Right? So this is the way
that it approaches. In this part of the
molecule, your alkene, is the flat part, right? So it's easiest for the molecule
to approach in this way. The spaceship analogy
always helps my students. So there's only one
product for this reaction. And let's see if we
can draw it here. And let's see what
it would look like. It would look
something like this. So we have our two
methyl groups right here. So the hydrogens are going
to add from below, right? So this hydrogen, let's
say it adds right here. That's going to push
this methyl group up. So that methyl
group gets pushed up when that hydrogen
adds right down here. And then this other
hydrogen is going to add to the opposite side. And so we can show the
addition of that hydrogen. So there's my syn addition
of these two hydrogens. And there was something
else in that carbon. It was another hydrogen. So another hydrogen got
pushed up right here as well. So that is your only
product, the only product of this reaction. The hydrogenation
reaction is very sensitive to steric conditions.