reduction of alkynes

In this video, we're going
to take a look at two ways

to reduce alkynes.

The first way is a
reaction we've seen before.

This is the
hydrogenation reaction.

And we saw it before when
we hydrogenated alkenes

to form alkanes.

Here we're going to hydrogenate
an alkyne to form an alkene.

And to do a
hydrogenation reaction,

we need some hydrogen gas,
so some H2 right here.

And then a metal
catalyst, so we're

going to use Lindlar
palladium, which

is a special type of catalyst.

It will catalyze the
reduction of the alkyne

on the left, the
alkene on the right.

However, the reduction of
the alkene to the alkane

down here is slow.

So, so slow that we can
stop it if our goal is

to just make an alkene.

So this reaction will
form a cis-alkene.

And it was a syn edition
of our hydrogens.

So we're going to get
the two hydrogens adding

on to the same
side, and this has

to do with the mechanism of
a hydrogenation reaction.

So you can check out the
earlier video on hydrogenation

of alkenes to see more details.

So Lindlar palladium,
a poison catalyst,

it will reduce an
alkyne to an alkene.

It will produce a cis-alkene.

All right, so that's how
to make a cis-alkene.

Let's take a look at how
to make a trans-alkene.

So how do we reduce an alkyne
to make a trans-alkene.

So here is our alkyne So we
have our triple bond like that.

And we're going to
add sodium metal,

and we're also going to add
liquid ammonia like that.

So we're going to
form a trans-alkene.

So I'm going to put-- this
time my two hydrogens are going

to be on opposite
sides of each other.

So this is formation of
a trans-alkene like that.

And it does this by an anti
addition of hydrogens, right?

So these are adding from
opposite sides like that.

Let's take a look
at the mechanism

to form a trans-alkene.

So I start with my alkyne.

So I'll go ahead and
put in my carbons

there and put an R
group on the left side.

And I'll make this
an R-prime group

to distinguish it from
the R group over there.

So we start with sodium, which
we know, being in Group 1,

has one valence
electron like that.

And in the first step
of the mechanism,

this sodium atom is going to
donate its valence electron

to the alkyne.

So when we're showing the
movement of one electron,

we use a half-headed arrow.

So I'm going to show this
electron moving over here,

but it's only one
electron so I'm only

going to do a half-headed
arrow like that, not

a full-headed arrow.

So one of these bonds
here between the carbons

is going to break.

And one of the
electrons is going

to move over here to
this carbon like that.

And one of the
electrons is going

to move over to the
carbon on the left.

So let's go ahead
and draw the result

of all those electrons
moving around.

So we have an R group here.

And we had a triple bond,
but now we only have a double

between our two carbons, and
then we have R-prime over here.

So the carbon on the right
picked up an electron

from sodium, and it also
picked up an electron

from the breaking of
that one bond there.

So now it has two
electrons around it

like that, which
gives us a negative 1

formal charge on this carbon.

So it's a carbanion.

It's an anion here.

The carbon on the left
picked up one electron

for the breaking of
that bond like that.

So that's a radical
that's something

we haven't talked about before.

So we actually form what's
called a radical anion here.

So let's go ahead
and write that.

This is a radical
anion, so radical

because there's an
unpaired electron there.

And then it also has a carbanion
in the same molecule like that.

So we have these electrons
that are pretty close together,

at least how I've
drawn them, right?

So we know that electrons
are all negatively charged,

so all these electrons are
going to repel each other.

So this isn't the most
stable way for this molecule

to have in terms
of a conformation.

These electrons
are going to repel,

and they're going to
want to try to be as far

away from each other
as they possibly can.

So what's going to
happen is, we have

our two carbons right here.

And let's say that these
two electrons stay over here

on this side.

This one electron's going to
go over to the opposite side.

They're going to try to get
as far away from each other

as they possibly can.

And same thing with these
R group here, right?

So this R group is going
to try to get as far

away from this R-prime
group as it possibly can.

So this trans conformation
is the more stable one.

So this is our negatively
charged carbanion right here.

So in the next step
of the mechanism,

we remember ammonia is present.

So let's go ahead and
draw an ammonia molecule

floating around like that.

So here is our ammonia molecule.

And the carbanion is
going to act as a base,

and it's going to take a proton
from the ammonia molecule.

So this lone pair
of electrons is

going to form a new
bond with this proton,

and these electrons are going
to kick off onto the nitrogen.

So let's go ahead
and draw the results

of that acid-base reaction.

So now we have our two
carbons, with an R group

right here, R-prime right here.

And now this carbon on the
right is bonded to a proton.

It bonded to a
hydrogen like that.

And then we still have
our radical down here,

so there is one electron
on that carbon as well.

All right, so the next
step of our mechanism?

Well, there's plenty
of sodium present.

So here's a sodium atom
with one valence electron.

The sodium is going to donate
this electron to this carbon.

So just use a
half-headed arrow to show

the movement of one electron.

So if that sodium atom donates
that one valence electron

to that carbon, let's go ahead
and draw the results of that.

So we have two carbons double
bonded, an R group over here,

a hydrogen, and an R-prime.

And this carbon had
one electron around it.

It just picked up one
more from a sodium atom.

So it's like that,
which would give it

a negative 1 formal charge.

So this carbon has a
negative 1 formal charge.

So let's go ahead and draw
that negative 1 formal charge.

It's a carbanion.

And once again, ammonia
is floating around,

so let's go ahead
and draw ammonia

right here, so NH3 like that.

And the same thing is going to
happen as did before, right?

The negative charge is
going to grab a proton.

It's going to act as a base.

And these electrons are going
to kick off onto the nitrogen

here.

And so we protonate
our carbanion,

and we have completed our
mechanism because now we

have our two R groups
across from each other.

And we added on two hydrogens
across from each other

as well like that.

So we formed a trans-alkene.

All right, so
that's the mechanism

to form a trans-alkene.

Let's look at a few examples.

Let's start with this
alkene right here.

OK, so carbon triple
bonded to another carbon,

and we'll put a methyl group
on each side like that.

OK, so let's do a few
different reactions

with the same substrate here.

So our first reaction will
just be a normal hydrogenation

with hydrogen gas, and let's
use platinum as our catalyst.

So this is not a
poison catalyst.

This is a normal catalyst.

So what's going to
happen is, first

you're going to reduce
the alkyne to an alkene.

And then since there's
no way of stopping it,

it's going to reduce
the alkene to an alkane.

So this is going to
reduce the alkyne

all the way to an alkane.

So if we go back up here
to beginning, remember,

we said that a poison catalyst
will stop at the alkene,

but if it's not a
poison catalyst,

it's just going to hydrogenate
your alkene to an alkane

down here.

So this reaction is going
to produce an alkane.

Let's go ahead and
draw the product.

So we know that there are
four carbons in my starting

materials.

There's going to be four
carbons when I'm done here,

so these two carbons in the
center here are going to turn

into CH2's.

And then on either side,
we still have our CH3's.

So this is going to form
butane as the product.

All right, this time
let's use a hydrogen gas,

and let's use a
Lindlar palladium here.

This is our poisoned catalyst.

So it's going to reduce
our alkyne to an alkene,

and then it's going to stop.

And you have to think, what
kind of alkene will you get?

You will get a cis-alkene.

So if we draw our two hydrogens
adding on to the same sides,

so now we have our methyl
groups going like that.

So our methyl groups
will be going-- this

and this would be our
product, a cis-alkene.

All right, let's do one
more, same starting material.

So this one right
here, except this time

we're going to add
sodium, and we're

going to use ammonia
as our solvent.

And remember, this will reduce
our alkyne to an alkene,

but it will form a
trans-alkene as your product.

So when you're drawing
your product down here,

you want to make
sure that your two

hydrogens are trans
to each other.

So they add on the mechanism,
and then your two methyl groups

would also be on the
opposite side like that.

So look very closely as to what
you are reacting things with.

Is it a normal
hydrogenation reaction?

Is it a hydrogenation reaction
with a poison catalyst,

which would form a cis-alkene?

Or is it reduction with
sodium and ammonia,

which will give
you a trans-alkene.