Videos > Ring opening reactions of epoxides: Acid-catalyzed
Ring opening reactions of epoxides Acid-catalyzed Video Lecture
Text Transcript from Video
Here's the general reaction
for a ring opening of epoxides
when everything
is acid-catalyzed.
So if I first start by looking
at my epoxide over here
on the left, I can
classify this carbon,
and I can see this carbon is
attached to two other carbons,
so this carbon
would be secondary.
Or I could think about a
hydrogen replacing the R group,
and then it would
be a primary carbon,
so either primary or secondary.
If I look at the carbon
on the right side,
so this carbon right
here, it's attached
to three other carbons, so
this carbon on the right
would be tertiary.
So if I react this epoxide
with a weak nucleophile
so H-Nu is going to refer
to my weak nucleophile,
and I make it an
acid-catalyzed reaction,
I can see the product
over here on the right.
The nucleophile has added to
the most substituted carbon,
so here i can see
my nucelophile.
I add it to my tertiary
carbon over here on the right.
So that's the regiochemistry,
a nucleophilic attack
of the most substituted carbon.
And I can see, when
the ring opens,
the OH group is going to be
anti- to the nucleophile.
Let's take a look at a reaction.
And we can run through the
entire mechanism for this ring
opening.
So if I start with
my epoxide, so I'll
go ahead and make it like this.
I'm going to have my epoxide
coming out at me in space.
And then, at this
top carbon here, I'm
going to have a methyl group
going away from me in space.
So here's my methyl group.
I'm going to react this
epoxide with ethanol.
So I'm going to go ahead
and put ethanol in here.
Ethanol is going to function
as my week nucleophile.
And it's acid-catalyzed, so I'll
make sure and put my protons
in there like that.
OK, let's go ahead
and redraw the epoxide
so we can see it a
little bit better.
So we're going to be
looking down on our epoxide,
like this, so, therefore,
the oxygen is going
to be up relative to
the plane of the ring.
A lone pair of
electrons on the oxygen,
like that, and then,
at this carbon,
there's going to be a methyl
group going down in space.
The first step of the mechanism
is an acid-base reaction.
This is an
acid-catalyzed reaction,
so they're H plus
protons floating around.
Lone pair of electrons
on oxygen are
going to pick up that proton.
So I'm going to
protonate my epoxide.
So when I draw the product
of that acid-base reaction.
And I now have, on
my oxygen, I still
have one lone pair of electrons.
The other lone pair formed
a bond with that proton,
so now it is a
protonated epoxide,
which gives the oxygen a
plus one formal charge.
I still need to
have my methyl group
down relative to
the ring, like that.
Now when I think
about the next step,
I need to think about the
classification of the carbon
atoms.
So if I look at this
top carbon right here,
I can see that this one is
attached to three other carbon,
so this is my tertiary carbon.
The one down here
is only secondary.
And I know the
nucleophile is going
to attack the more highly
substituted carbon, which
would be this one down
here, like that, so this
is the carbon that's
going to get the attack.
If I think about that oxygen
being positively charged,
oxygen doesn't like to
be positively charged,
oxygen is very electronegative.
So it's going to do its best
to pull some of these electrons
and the bond between that carbon
and that oxygen a little bit
closer to the oxygen there.
So I'm going to
withdraw some electron
density from this carbon.
So this carbon is
going to end up
having a partial, positive
charge, like that.
And if I think about what
kind of a carbo cation
would that be, it has a little
bit of partial cargo cation
character, and I can
think about the fact
that that carbon is attached
to those three other carbons--
it's a little bit easier
to see here --there's
one, two, and three
other carbons.
And, in earlier videos, we've
seen that tertiary carbo
cations are the most
stable, much more stable
than secondary carbo cations.
And if I have a weak
nucleophile attacking,
it's going to attack the
most substituted carbon,
so it's going to
attack the one that
would be the most
stable carbo cation.
So again, we have more
of a partial carbo cation
character here.
But when the weak
nucleophile comes along,
that lone pair of
electrons is going
to attack this
carbon right here.
And that is going to
kick these electrons
in here, the magenta ones,
off onto your oxygen.
So let's go ahead
and draw the results
of that nucleophilic attack.
So I have my ring right here.
And I've opened my epoxide.
So now this oxygen is going to
swing over to the left here,
and it's bonded to a hydrogen.
It had one lone
pair of electrons.
It just picked up
one more lone pair
of electrons, the magenta ones.
And, in the course of
that nucleophilic attack,
the methyl group that's down
relative to the ring here
gets pushed up.
So the methyl group will
end up relative to the ring.
And I have a lone pair of
electrons from the ethanol
molecule that are now
bonded to this carbon.
So I can go ahead and
show those electrons.
And there's an oxygen, a
hydrogen, and then two carbons,
and then there's
still a lone pair
of electrons left in
that oxygen, which
gives that oxygen a
plus one formal charge.
So let's just go
ahead and highlight
those electrons really fast.
So, let's see, I'll
make them red here.
So these electrons
right here, those
are the ones that
attack the carbon,
and those are the
ones that formed
these covalent bonds, like that.
So we're almost done.
The last step will be
an acid-base reaction
to get rid of that plus one
formal charge on the oxygen.
So another molecule of
ethanol comes along.
This one's going
to act as a base.
So I go ahead and draw
another molecule of ethanol.
And it has lone pairs
of electrons on it.
One of those lone
pairs of electrons
are going to pick up that
proton which kicks these two
electrons off onto this oxygen.
And we're going to end
up with our products.
So let's go ahead and draw
the product right up here.
And, like usual, we're
going to be looking down
on our molecule this
way, and so we're
going to look down right
here and draw what we see.
So we're going to have
our ring, like that.
And, at the top carbon, which
would correspond to this one
right here, I can
see that there's
a methyl group coming out at me.
So I can go ahead and
put my methyl group
coming out at me like that.
And then, going away from me,
I'm going to have my oxygen,
so I'm of course talking
about this oxygen
down here like that.
So I can go ahead and draw
that oxygen going away from me.
And connected to that oxygen
of course would be two carbons.
So I can go ahead and do that.
And when I look at the carbon
on the left over here--
now I'm talking
about this carbon,
I can see there's an OH group
coming out at me in space,
so I can go ahead and put
that OH group coming out
at me in space right there.
So that's going
to be our product.
So, again, I went
ahead and showed--
I'm going to go ahead and put
my two lone pairs of electrons
on the oxygen, because I showed
that proton being removed
in the last step of
that mechanism there.
So let's look at
the stereochemistry.
And let's first start
with this top carbon here.
So I go back to
my reactants and.
I look at this top carbon.
And I think about the fact that
I have a wedge to an oxygen,
so oxygen's coming out
at me, and a carbon's
going away from me in space.
When I look at my
products-- this
would be the carbon that
corresponds to that carbon,
it's the opposite.
Now I have a carbon
coming out at me in space,
and an oxygen
going away from me.
So that's inversion
of configuration,
so at that top carbon there,
so at this carbon right here,
I'm going to get inversion of
absolute configuration here.
So let me go ahead and
write that-- inversion
of configuration.
So there's some
stereochemistry involved.
And when I look at
the other carbon--
so when I look at
this carbon down here,
I can see that there's an oxygen
coming out at me in space.
And when I look at that
carbon for my products,
I see there's still an oxygen
coming out at me in space.
So it's the same absolute
configuration at this carbon.
So we only see inversion
of configuration
at the carbon
which is undergoing
the nucleophilic attack, which
is the most substituted carbon.
for a ring opening of epoxides
when everything
is acid-catalyzed.
So if I first start by looking
at my epoxide over here
on the left, I can
classify this carbon,
and I can see this carbon is
attached to two other carbons,
so this carbon
would be secondary.
Or I could think about a
hydrogen replacing the R group,
and then it would
be a primary carbon,
so either primary or secondary.
If I look at the carbon
on the right side,
so this carbon right
here, it's attached
to three other carbons, so
this carbon on the right
would be tertiary.
So if I react this epoxide
with a weak nucleophile
so H-Nu is going to refer
to my weak nucleophile,
and I make it an
acid-catalyzed reaction,
I can see the product
over here on the right.
The nucleophile has added to
the most substituted carbon,
so here i can see
my nucelophile.
I add it to my tertiary
carbon over here on the right.
So that's the regiochemistry,
a nucleophilic attack
of the most substituted carbon.
And I can see, when
the ring opens,
the OH group is going to be
anti- to the nucleophile.
Let's take a look at a reaction.
And we can run through the
entire mechanism for this ring
opening.
So if I start with
my epoxide, so I'll
go ahead and make it like this.
I'm going to have my epoxide
coming out at me in space.
And then, at this
top carbon here, I'm
going to have a methyl group
going away from me in space.
So here's my methyl group.
I'm going to react this
epoxide with ethanol.
So I'm going to go ahead
and put ethanol in here.
Ethanol is going to function
as my week nucleophile.
And it's acid-catalyzed, so I'll
make sure and put my protons
in there like that.
OK, let's go ahead
and redraw the epoxide
so we can see it a
little bit better.
So we're going to be
looking down on our epoxide,
like this, so, therefore,
the oxygen is going
to be up relative to
the plane of the ring.
A lone pair of
electrons on the oxygen,
like that, and then,
at this carbon,
there's going to be a methyl
group going down in space.
The first step of the mechanism
is an acid-base reaction.
This is an
acid-catalyzed reaction,
so they're H plus
protons floating around.
Lone pair of electrons
on oxygen are
going to pick up that proton.
So I'm going to
protonate my epoxide.
So when I draw the product
of that acid-base reaction.
And I now have, on
my oxygen, I still
have one lone pair of electrons.
The other lone pair formed
a bond with that proton,
so now it is a
protonated epoxide,
which gives the oxygen a
plus one formal charge.
I still need to
have my methyl group
down relative to
the ring, like that.
Now when I think
about the next step,
I need to think about the
classification of the carbon
atoms.
So if I look at this
top carbon right here,
I can see that this one is
attached to three other carbon,
so this is my tertiary carbon.
The one down here
is only secondary.
And I know the
nucleophile is going
to attack the more highly
substituted carbon, which
would be this one down
here, like that, so this
is the carbon that's
going to get the attack.
If I think about that oxygen
being positively charged,
oxygen doesn't like to
be positively charged,
oxygen is very electronegative.
So it's going to do its best
to pull some of these electrons
and the bond between that carbon
and that oxygen a little bit
closer to the oxygen there.
So I'm going to
withdraw some electron
density from this carbon.
So this carbon is
going to end up
having a partial, positive
charge, like that.
And if I think about what
kind of a carbo cation
would that be, it has a little
bit of partial cargo cation
character, and I can
think about the fact
that that carbon is attached
to those three other carbons--
it's a little bit easier
to see here --there's
one, two, and three
other carbons.
And, in earlier videos, we've
seen that tertiary carbo
cations are the most
stable, much more stable
than secondary carbo cations.
And if I have a weak
nucleophile attacking,
it's going to attack the
most substituted carbon,
so it's going to
attack the one that
would be the most
stable carbo cation.
So again, we have more
of a partial carbo cation
character here.
But when the weak
nucleophile comes along,
that lone pair of
electrons is going
to attack this
carbon right here.
And that is going to
kick these electrons
in here, the magenta ones,
off onto your oxygen.
So let's go ahead
and draw the results
of that nucleophilic attack.
So I have my ring right here.
And I've opened my epoxide.
So now this oxygen is going to
swing over to the left here,
and it's bonded to a hydrogen.
It had one lone
pair of electrons.
It just picked up
one more lone pair
of electrons, the magenta ones.
And, in the course of
that nucleophilic attack,
the methyl group that's down
relative to the ring here
gets pushed up.
So the methyl group will
end up relative to the ring.
And I have a lone pair of
electrons from the ethanol
molecule that are now
bonded to this carbon.
So I can go ahead and
show those electrons.
And there's an oxygen, a
hydrogen, and then two carbons,
and then there's
still a lone pair
of electrons left in
that oxygen, which
gives that oxygen a
plus one formal charge.
So let's just go
ahead and highlight
those electrons really fast.
So, let's see, I'll
make them red here.
So these electrons
right here, those
are the ones that
attack the carbon,
and those are the
ones that formed
these covalent bonds, like that.
So we're almost done.
The last step will be
an acid-base reaction
to get rid of that plus one
formal charge on the oxygen.
So another molecule of
ethanol comes along.
This one's going
to act as a base.
So I go ahead and draw
another molecule of ethanol.
And it has lone pairs
of electrons on it.
One of those lone
pairs of electrons
are going to pick up that
proton which kicks these two
electrons off onto this oxygen.
And we're going to end
up with our products.
So let's go ahead and draw
the product right up here.
And, like usual, we're
going to be looking down
on our molecule this
way, and so we're
going to look down right
here and draw what we see.
So we're going to have
our ring, like that.
And, at the top carbon, which
would correspond to this one
right here, I can
see that there's
a methyl group coming out at me.
So I can go ahead and
put my methyl group
coming out at me like that.
And then, going away from me,
I'm going to have my oxygen,
so I'm of course talking
about this oxygen
down here like that.
So I can go ahead and draw
that oxygen going away from me.
And connected to that oxygen
of course would be two carbons.
So I can go ahead and do that.
And when I look at the carbon
on the left over here--
now I'm talking
about this carbon,
I can see there's an OH group
coming out at me in space,
so I can go ahead and put
that OH group coming out
at me in space right there.
So that's going
to be our product.
So, again, I went
ahead and showed--
I'm going to go ahead and put
my two lone pairs of electrons
on the oxygen, because I showed
that proton being removed
in the last step of
that mechanism there.
So let's look at
the stereochemistry.
And let's first start
with this top carbon here.
So I go back to
my reactants and.
I look at this top carbon.
And I think about the fact that
I have a wedge to an oxygen,
so oxygen's coming out
at me, and a carbon's
going away from me in space.
When I look at my
products-- this
would be the carbon that
corresponds to that carbon,
it's the opposite.
Now I have a carbon
coming out at me in space,
and an oxygen
going away from me.
So that's inversion
of configuration,
so at that top carbon there,
so at this carbon right here,
I'm going to get inversion of
absolute configuration here.
So let me go ahead and
write that-- inversion
of configuration.
So there's some
stereochemistry involved.
And when I look at
the other carbon--
so when I look at
this carbon down here,
I can see that there's an oxygen
coming out at me in space.
And when I look at that
carbon for my products,
I see there's still an oxygen
coming out at me in space.
So it's the same absolute
configuration at this carbon.
So we only see inversion
of configuration
at the carbon
which is undergoing
the nucleophilic attack, which
is the most substituted carbon.
About this Video
Apr 21, 2026 Last updated
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