Williamson Ether Synthesis (in detail) Video Lecture - Chemistry for JEE

One way to make ethers is
to use the Williamson ether
synthesis, which is where
you start with an alcohol,
and you add a strong base
to deprotonate the alcohol.
Once you deprotonate
the alcohol,
you add an alkyl halide,
and primary alkyl halides
work the best.
We'll talk about
why in a minute.
And what happens is you end
up putting the R prime group
from your alkyl
halide on to what
used to be your alcohol to
form your ether like that.
So let's look at the mechanism
for the Williamson ether
synthesis, where you
start with your alcohol.
We know that alcohols can
function as weak acids.
So if you react an alcohol
with a strong base, something
like sodium hydride, we know
that the hydride portion
of the molecule is going to
function as a strong base.
This lone pair of
electrons is going
to take that proton,
which is going
to kick these electrons
off onto the oxygen.
So if we're drawing the product
of that acid-based reaction,
we now have an oxygen
with three lone pairs
of electrons around it, giving
it a negative 1 formal charge.
And we call that an
alkoxide anion, which
would interact
with the positively
charged sodium ion
floating around.
So there's some electrostatic
or ionic interaction
between those opposite charges.
And here's where you
introduce your alkyl halide.
So if we draw our alkyl halide,
it would look like this.
And we know that there's an
electronegativity difference
between our halogen
and our carbon,
where our halogen is going
to be partially negative,
and our carbon is going
to be partially positive.
Partially positive carbon
means that that carbon
wants electrons.
It's going to function
as an electrophile
in the next step
of the mechanism.
And a lone pair of
electrons in the oxygen
is going to function
as a nucleophile.
So opposite charges attract.
A lone pair of electrons
on our nucleophile
are going to attack our
electrophile, our carbon.
At the same time, the
electrons in the bond
between the carbon
the halogen are
going to kick off onto
the halogen like that.
So this is an
SN2-type mechanism,
which is why a primary alkyl
halide will work the best,
because that has the decreased
steric hindrance compared
to other alkyl halides.
So what will happen is,
after nucleophilic attack,
we're going to attach our
oxygen to our carbon like that,
and we form our ether.
So if we wanted to, we
could just rewrite our ether
like this to show it as we added
on an R prime group like that.
Let's look at an example of
the Williamson ether synthesis.
So if I start with a molecule
over here on the left,
and it's kind of an
interesting-looking molecule.
It's called beta-naphthol.
And so beta-naphthol has two
rings together like this,
and then there's an OH coming
off of one of the rings,
like that.
So that's beta-naphthol.
And in the first
part, we're going
to add potassium
hydroxide as our base.
Now, potassium hydroxide
is not as strong of a base
as sodium hydride
is, but in this case,
it's OK to use a
little bit weaker base.
So the lone pair of
electrons on the hydroxide
are going to take that proton,
leaving these electrons behind
on the oxygen.
So when we draw the conjugate
base to beta-naphthol--
and we can go ahead
and show that-- we're
going to take off
that proton, which
is going to leave
that oxygen there
with three lone
pairs of electrons,
giving it a negative
1 formal charge.
So this is our alkoxide anion.
And this alkoxide anion
is resonance stabilized.
So a resonance-stabilized
conjugate base
stabilizes the
conjugate base, which
makes beta-naphthol a little bit
better acid than other alcohols
that we will talk about.
So since beta-naphthol is
a little bit more acidic,
that's why it's OK for us to use
a weaker base for this example.
So potassium hydroxide
is strong enough
to take away the acidic
proton in beta-naphthol
because the conjugate
base to beta-naphthol
is resonance stabilized.
So in the second step, once we
have formed our alkoxide anion,
this is where we add
our alkyl halide.
So if I add my alkyl
halide in my second step--
let's see if we can
have enough room
here-- I'm going to use methyl
iodide as our alkyl halide.
So methyl iodide
looks like that.
And once again, we
know this carbon
is going to be the
electrophilic carbon, so
nucleophile, electrophile.
So a lone pair of electrons on
the oxygen attacks the carbon,
kicks these electrons
off onto the iodide,
and we form our product.
So let's go ahead and draw the
ether product that will result.
So these rings are going
to stay the same like that.
And we now are going to have
our oxygen attached to a methyl
group, which came from the
methyl iodide like that.
So we formed our product.
This product is
called nerolin, which
is a fixative used in perfume.
So this has an
interesting smell to it.
So if you ever get a chance
to do this Williamson ether
synthesis, it's just interesting
to see what nerolin smells
like, what it looks like,
and to think about it
as being a component
of some perfumes.
Let's think about
synthesizing an ether.
So if you were given a problem
where the question said
something like, OK,
here is the ether
that you want to synthesize.
What would you need
in order to do so?
So you need to think about,
OK, there's my ether,
and I'm going to make it from
some other things over here.
And if I analyze the alkyl
groups attached to my ether,
and I have a methyl
group over here,
and this would be like a
cyclohexyl group over here.
And one of those
two groups I'm going
to use for my alkyl halide.
You want to use the group that's
the least sterically hindered
since it's an
SN2-type mechanism.
So you want to go
with the methyl group.
So in your second
step, you would
need to add something
like methyl iodide.
That's the least
sterically hindered,
so that's going to improve
your yield on this reaction.
So that's the second step.
And in the first step, you'd
have to add a strong base,
so we'll use sodium
hydride here.
And your alcohol, therefore,
must come from this.
So this must be where
your alcohol comes from.
So if I'm going to show
my starting alcohol,
it would have to look like this.
So if I add that alcohol in
the first step, sodium hydride,
I take off that proton,
form an alkoxide,
that alkoxide nucleophilic
attacks the methyl
iodide to add the
methyl group on,
and to form the
ether on the right.
So that's how to think about
using the Williamson ether
synthesis.
So think about
retrosynthesis and think
about which alkyl
group is the best
one to use for
your alkyl halide.