preparation of alkynes

Note that if a terminal alkyne is produced, the strong base will deprotonate the alkyne to form an alkynide anion. A proton source (like H2O) must be added to regenerate the terminal alkyne.

Let's look at two ways
to prepare alkynes

from alkyl halides.

So here I have an alkyl halide.

So this is a dihalide,
and my two halogens

are attached to one carbon.

We call this a geminal dihalide.

So this is going to be a
geminal dihalide reacting

with a very strong
base, sodium amide.

So this is going to give us
an E2 elimination reaction.

So we're going to get an
E2 elimination reaction,

and this E2 elimination
reaction is actually

going to occur twice.

And we're going to end up with
an alkyne as our final product.

So let's take a look at the
mechanism of our double E2

elimination of a
geminal dihalide.

So let's start with
our dihalide over here.

And this time we're
going to put in all

of our lone pairs of electrons
on our halogen, like that.

So let me go ahead and
put all of those in there,

and then I have two
hydrogens on this carbon.

OK.

Sodium amide is a source
of amide anions, which

we saw in our previous video
can function as a strong base.

So a strong base means that
a lone pair of electrons

here on our nitrogen is
going to take this proton.

And these electrons,
in here, are

going to kick in to form a
double bond at the same time

these electrons kick
off onto our halogen.

So an E2 elimination mechanism.

You can watch the
previous videos

on E2 elimination
reactions for more details.

So we're going to form ammonia
as one of our products.

And our other product
is going to be

carbon double-bonded
to another carbon.

And then we're going to still
have our halogen down here.

And over here, in the
carbon on the right,

we're still going to have
a hydrogen, like that.

So we're not quite
to our alkyne yet.

So we've done one E2
elimination reaction,

and we're going to do one more.

So we get another-- another
amide anion comes along,

and it's negatively charged.

It's going to
function as a base.

It's going to take
this proton this time.

And these electrons
are going to move

in here to form our triple bond.

And these electrons are going
to kick off onto our halogen,

like that.

So that is going to finally
form our alkyne here.

So you always have to
have your base in excess,

if you're trying to do this.

Let's look at a very similar
reaction, a double E2

elimination.

This time the halogens are
not on the same carbon.

So let's go ahead and draw
the general reaction for this.

We have two carbons
right here, and we

have two halogens right here.

And then hydrogen,
and then hydrogen.

This time we have two
halogens on adjacent carbons.

So this is called
vicinal dihalides.

So let's go ahead
and write that.

So this is vicinal, and the
one we did before was geminal.

So a vicinal dihalide will
react in a very similar way

if you add a strong
base like sodium amide

and you use ammonia
for your solvent.

So you're going to form
an alkyne once again.

So you're going
to get an alkyne.

It's going to be via a double
E2 elimination reaction again.

Let's look at the mechanism.

So let's start with our
vicinal dihalide down here.

So let's go ahead and put
our halogens in there.

Lone pairs of electrons on
our halogens, like that.

And then we have hydrogen, and
we have hydrogen right here.

So we have our amide
anion, and once again,

functions as a strong base.

It's going to take a proton.

So it's going to take
this proton right here.

These electrons are
going to move in

to form our double bond the same
time these electrons kick off

on to our halogen.

So that's our first E2
elimination reaction.

So let's just go ahead
and write E2 here

to remind us this is
yet another E2 reaction.

And let's go ahead and
draw the product of that.

So now we're going
to have carbon

double-bonded to another carbon.

And then we're going to
have a hydrogen right here.

And then we're going to have
our halogen up here, like that.

And then we're going to
have-- we need one more

reaction to form our alkyne.

We're going to get another
E2 elimination reaction.

So sodium amide-- another anion
of sodium amide comes along.

So let's go ahead and put in
those lone pairs of electrons,

like that.

It's going to
function as a base.

Lone pair of electrons
takes this proton.

These electrons kick in
here to form our triple bond

at the same time
our halogen leaves.

And so we form our
alkyne like that.

So you can produce alkynes
from either vicinal or geminal

dihalides via a double
E2 elimination reaction.

Let's see how we could use
this in a synthesis reaction.

So let's go ahead and
try to make something--

try to make an alkyne
from an alkene.

OK, so let's start
with an alkene here.

And I'll put some benzene
rings on this guy here.

So here's a benzene
ring like that.

Put in my lone pairs-- sorry,
put in my bonds like that.

And then I'm going to put
a double bond right here.

And then I'm going to put
another benzene ring attached

like that.

So this is
1,2-Diphenylethylene. .

And I'm going to react
this alkene with bromine.

And you could use a
solvent like carbon

tetrachloride or
something like that.

And we're reacting an
alkene with a halogen.

And we've seen this
reaction before

in the videos on
reactions of alkenes.

We're going to add two bromines
across our double bond.

So we're going to draw the
product of this reaction.

Our benzene rings
aren't going to react

as readily as our
double bonds will.

So let's go ahead and draw in
our other benzene ring here,

like that.

And we know that
we're going to add

a bromine to either
side of our double bond.

So let's go ahead
and add a bromine

to either side of
our double bond here.

And we'll also have
a hydrogen bonded

to each one of these carbons.

That hydrogen was
originally there as well,

over here on the left.

And we form
1,2-Dibromo-1,2-diphenylethane

here.

And now we have a
vicinal dihalide.

So if we add a strong base
to our vicinal dihalide

we can prepare an
alkyne from that.

So if we add an excess of
sodium amide in ammonia

we know that we're going to
get a double E2 elimination

reaction.

And those halogens are going
to go away in our double E2

elimination reaction
and form a triple bond.

So we're going to
form a triple bond.

So those two carbons, the
ones that form a triple bond,

are these two
carbons right here.

So when you run
through the mechanism,

you're going to get an alkyne.

And then, on either
side of that alkyne,

you're going to get
a phenyl groups.

So let's go ahead and draw in
our benzene rings, like that.

So it doesn't really matter
how we draw our electrons,

so we'll go ahead and do this.

So this would be our product.

So let's go ahead and
put in those electrons.

This will be diphenylacetylene.

So you can synthesize
alkynes from alkenes,

or you could synthesize
an alkyne from a dihalide.

So this is one way to do it.