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Let's look at two ways
to prepare alkynes

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from alkyl halides.

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So here I have an alkyl halide.

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So this is a dihalide,
and my two halogens

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are attached to one carbon.

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We call this a geminal dihalide.

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So this is going to be a
geminal dihalide reacting

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with a very strong
base, sodium amide.

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

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

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and this E2 elimination
reaction is actually

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going to occur twice.

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

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

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elimination of a
geminal dihalide.

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So let's start with
our dihalide over here.

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And this time we're
going to put in all

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

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

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and then I have two
hydrogens on this carbon.

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OK.

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Sodium amide is a source
of amide anions, which

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

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So a strong base means that
a lone pair of electrons

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here on our nitrogen is
going to take this proton.

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And these electrons,
in here, are

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

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these electrons kick
off onto our halogen.

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So an E2 elimination mechanism.

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You can watch the
previous videos

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on E2 elimination
reactions for more details.

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

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And our other product
is going to be

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carbon double-bonded
to another carbon.

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

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And over here, in the
carbon on the right,

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

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So we're not quite
to our alkyne yet.

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So we've done one E2
elimination reaction,

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and we're going to do one more.

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

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and it's negatively charged.

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It's going to
function as a base.

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It's going to take
this proton this time.

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And these electrons
are going to move

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in here to form our triple bond.

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

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like that.

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So that is going to finally
form our alkyne here.

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So you always have to
have your base in excess,

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if you're trying to do this.

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

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elimination.

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This time the halogens are
not on the same carbon.

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

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We have two carbons
right here, and we

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have two halogens right here.

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And then hydrogen,
and then hydrogen.

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This time we have two
halogens on adjacent carbons.

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So this is called
vicinal dihalides.

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So let's go ahead
and write that.

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

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So a vicinal dihalide will
react in a very similar way

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if you add a strong
base like sodium amide

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and you use ammonia
for your solvent.

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

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So you're going
to get an alkyne.

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

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Let's look at the mechanism.

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

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

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Lone pairs of electrons on
our halogens, like that.

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

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So we have our amide
anion, and once again,

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functions as a strong base.

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It's going to take a proton.

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

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These electrons are
going to move in

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

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on to our halogen.

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So that's our first E2
elimination reaction.

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So let's just go ahead
and write E2 here

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to remind us this is
yet another E2 reaction.

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

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So now we're going
to have carbon

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double-bonded to another carbon.

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

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

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

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reaction to form our alkyne.

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We're going to get another
E2 elimination reaction.

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

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

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like that.

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It's going to
function as a base.

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Lone pair of electrons
takes this proton.

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These electrons kick in
here to form our triple bond

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at the same time
our halogen leaves.

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And so we form our
alkyne like that.

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So you can produce alkynes
from either vicinal or geminal

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dihalides via a double
E2 elimination reaction.

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

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

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try to make an alkyne
from an alkene.

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

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

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So here's a benzene
ring like that.

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

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

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

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like that.

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So this is
1,2-Diphenylethylene. .

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

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And you could use a
solvent like carbon

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tetrachloride or
something like that.

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And we're reacting an
alkene with a halogen.

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And we've seen this
reaction before

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in the videos on
reactions of alkenes.

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

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

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Our benzene rings
aren't going to react

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as readily as our
double bonds will.

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

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like that.

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And we know that
we're going to add

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a bromine to either
side of our double bond.

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So let's go ahead
and add a bromine

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to either side of
our double bond here.

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And we'll also have
a hydrogen bonded

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to each one of these carbons.

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That hydrogen was
originally there as well,

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over here on the left.

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

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here.

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And now we have a
vicinal dihalide.

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So if we add a strong base
to our vicinal dihalide

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we can prepare an
alkyne from that.

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So if we add an excess of
sodium amide in ammonia

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

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reaction.

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

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elimination reaction
and form a triple bond.

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So we're going to
form a triple bond.

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

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are these two
carbons right here.

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So when you run
through the mechanism,

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you're going to get an alkyne.

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And then, on either
side of that alkyne,

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you're going to get
a phenyl groups.

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

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

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so we'll go ahead and do this.

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So this would be our product.

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

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This will be diphenylacetylene.

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So you can synthesize
alkynes from alkenes,

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or you could synthesize
an alkyne from a dihalide.

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So this is one way to do it.