Let's look at the
hydrohalogenation reaction of alkynes. We've seen this reaction
before with alkenes, and there are some similarities
and some differences. So let's look and see
what happens here. We start with our
alkyne, our triple bond. And we add our hydrogen halide. And we add either one
or two more equivalents of our hydrogen halide. If we add one equivalent
of our hydrogen halide, the hydrogen and the
halogen are going to add from opposite sides. And the halogen is going to add
to the most substituted carbon. So this halogen
here is going to add to the most substituted
carbon, which, in terms of
regiochemistry-- so go ahead and write here
regiochemmistry-- that was called Markovnikov's rule. We're going to add the halogen
to the most substituted carbon in the reaction. So with two molar
equivalents, you're going to end up with two
halogens on the same carbon, like that. So let's take a look at one
of the proposed mechanisms for the hydrohalogenation
of alkynes. And it closely parallels the
hydrohalogenation of alkenes. And this isn't considered
to be the perfect mechanism for alkynes, but we're going
to start with that, just to show why this is
Markovnikov's, in terms of regiochemistry. So let's say this
was the mechanism. We start with our alkyne. And we have our hydrogen
halide, so our hydrogen halide like that. And when we did this
mechanism for alkenes, we said the first
thing that happens was these electrons
in this bond are going to form a bond
with this proton here. And then these
electrons are going to kick off onto your
halogen, like that. So let's see we can do. So let's say the hydrogen
adds onto the right side. So now there's only a double
bond between my two carbons. And the hydrogen added onto
the right carbon, like that. And so now, this
right carbon has a bond going down like this. And now this carbon on the left
has only three bonds to it. So that gives it a plus 1
formal charge, like that. And then our halogen
is going to have a negative 1 formal charge. And so our halogen would
act as our nucleophile. And our positively
charged carbon right here would act as
our electrophile. And so you would get the
halogen adding onto that carbon. And since a carbocation was
involved in the mechanism, that's the reason why
Markovnikov's rule is seen, because
you want to form the most stable
carbocation possible. So the more
substituted carbocation is the more stable one. So this really isn't quite
the correct mechanism. And the reason why it's not
quite the correct mechanism is because if this was
the right mechanism, the rate of this
reaction would depend on the concentration of
two molecules, bimolecular. So if we were to write the
rate law for this mechanism, we would say, Oh, OK, I
expect the rate of reaction for hydrohalogenation
of alkynes-- we have a rate
constant in there-- to be proportional to the
concentration of your alkyne, and then the concentration
of your hydrogen halide. Two things, because that's what
we started in this mechanism. We had our hydrogen halide right
here, and we had our alkyne. And that is not
observed experimentally. So this mechanism
is not quite right. So this mechanism is
not quite what happens, but it is kind of important
to think about this carbon as being the positively
charged one and Markovnikov's rule applying. So it helps to think
about this mechanism, but this mechanism
isn't quite right, because this is not the
experimental rate law. The experimental
rate law turns out to be second order with
respect to your hydrogen halide and third order overall. So if you put a
superscript 2 there, it's actually dependent upon
two molecules of your hydrogen halide and one molecule
of your alkyne. So since it's actually
third order overall, we need to come up with
a different mechanism. So let's look at the
mechanism that is currently proposed as the mechanism
for the hydrohalogenation of alkynes. We start with our alkyne here. And we need two molecules
of our hydrogen halide. So here's one molecule of our
hydrogen halide, like that. And let's go ahead and draw
the other one down here. OK, so let's put in those lone
pairs of electrons, like that. So in this mechanism, all
three of these molecules are reacting at the same time. So as these electrons
in here, from the bond between the hydrogen
and the halogen, the halogen is going
to take those electrons and they're going
to move and form a new bond with this carbon. At the same time
that's happening, these two electrons are going
to form a bond with this proton. And then these
electrons are going to kick off onto your
halogen, like that. So if we were to
draw the transition state of all this stuff
happening at the same time-- let's go ahead and put
in our brackets here, like that-- so what's
going to happen? Well, we're going to
have carbon double bonded to another carbon, like that. And then we're going to have a
lot of partial bonds in here. So let's use a different
color for partial bonds. We have our hydrogen
and our halogen. And then we have our hydrogen
and our halogen up here. And then let's use
blue for partial bonds. So this halogen is forming
a bond with this carbon here, at the same time this bond
is breaking in here, like that. And we started
with a triple bond. But that triple bond is leaving
to form a bond over here with this proton, like that. At the same time,
this proton's bond is breaking with this
halogen, like that. So this is all one
giant transition state. And if you think about
what's happening, the reason why I wanted to
show this mechanism up here is to show you, if
this happened stepwise, then this carbon
on the left would get your full positive charge. Let me just highlight that. This carbon on the left gets
your full positive charge, this one right here. And so if we think about what's
happening in this mechanism down here, it's not
quite the same thing. But since the bond is leaving
this carbon on the left-- the triple bond is
leaving this carbon on the left-- this
is the one that's going to get some partial
carbocationic character. So it's going to be
partially positive. And it's hard to see
in this mechanism how this carbon could
be partially positive. But up here, it's easy to
see, because the triple bond is breaking from the
carbon on the left. And it's going over here
to the carbon on the right. So that's the reason why
it's important to think about this mechanism. This gives you your partial
carbocationic character, which explains the
Markovnikov's regiochemistry. All right, so now
we're pretty much done. Once those electrons
finish moving, we're going to get-- let's
go back to our yellow color here-- we're going
to get our alkene. So we're going to form
our alkene, like that. And we formed a bond with
our halogen down here. And then we formed a bond
up here with that proton. And so that is going to be
the product of our mechanism here, like that. So let's look at a
couple of reactions. So let's look at the
hydrohalogenation of alkynes here. So we'll start with an alkyne. So go ahead and
draw it like that. And remember, you want to make
it linear around your alkyne, since they are linear. So in the first
reaction, we'll just add one molar equivalent of
our hydrogen halide, like that. So I know I'm going to add
on hydrogen and halogen across my triple bond. I know that's going
to form a double bond. So the first thing
I'm going to do is count how many
carbons I have. Let's see, 1, 2, 3, 4, and 5. So I know that I'm going to
turn a five-carbon alkyne into a five-carbon alkene. So let's go ahead and draw our
five-carbon alkene, so 1, 2, 3, 4, and 5. And the alkene is going to be
between these first two carbons over here on the
right, like that. So that's part of my product. But now I have to
figure out, OK, which one of these two carbons
do I add my halogen to? Do I add it to this
carbon over here on the left side
of my triple bond? Or do I add that
halogen over here to the carbon on the right
side of my triple bond? And to do that, you need to
think about Markovnikov's rule. So if I add it to the right
one, that would give me a primary carbocation--
I mean, you have to think about
which one will give you the most stable carbocation. And the most stable
carbocation would be your secondary one
over here on the left. And so that is the
one that your halogen is going to add over
here on the left side. So again, watch
those earlier videos for much more, in
terms of detail, about carbocations and stability
and Markovnikov's rule. So let's do the addition of two
equivalents of hydrogen halide, or just make it in excess here. So you have to think about the
fact that, OK, so once again, I'm starting with
five carbons, so I'm going to get five carbons
for my product here. And go back up here to
our original reaction. You can see, if you have
two molar equivalents, you're going to end
up adding your two halogens to the same carbon. And I probably should have
drawn this a little bit differently on the
general reaction. It also is going to
exhibit Markovnikov's for regiochemistry. So when you're trying to figure
out which carbon that is, think about which one would
be the most stable carbocation again. So once again, we
have two choices, the carbon on the left side, or
the carbon on the right side. The carbon on the left
side gives us more stable. So we're going to add on two
equivalents of our halogen here. So we actually form a
dihalide as our product. So that's hydrohalogenation of
alkynes with one or two molar equivalents of your
hydrogen halide.