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Course: Organic chemistry > Unit 7
Lesson 7: Ring-opening reactions of epoxidesRing opening reactions of epoxides: Acid-catalyzed
How acid-catalyzed reactions can open the highly strained ring of an epoxide. Created by Jay.
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Wouldn't the steric hinderance of the methyl group stop the weak base from attacking the anti-bond of the tertiary carbon. The acid seems to be very weak and bulky making it more likely to attack the secondary carbon for an SN2 reaction?(12 votes)- The reaction is like SN1. It is NOT SN1 but it is carbocationic character driven. Think of it like this, The strong nucleophile will attack vigorously and will not be selective ie. it will attack the carbon which is easily available to it, the one less hindered. Whereas the weak nucleophile which won't be vigorous would be selective and would attack on carbon which is carbocationic driven so it gets more stability and the reaction proceeds even with the weak nucleophile.
In a nutshell, strong doesn't need carbocationic character's stability as it is strong enough and reacts quickly whereas the weak one needs it to compensate its weakness.(15 votes)
- How do you distinguish that a reaction is going to be a strong nucleophile ring-opening reaction or an acid-catalyzed ring-opening reaction?(2 votes)
If it was an acid-catalyzed ring opening reaction, H+ would be one of the first reagents rather than added later to protonate the alkoxide first so that a strong nucleophile isn't required.(4 votes)
- primary alkyl halides react with alkoxide via E2 mechanism to form an alkene (here alkoxide is a strong base) and in Williamson reaction they again react to give an ether by SN2 mechanism. when what happens?(2 votes)
In both cases I believe there will be significant side reactions by the other mechanism leading to unwanted products.
One way to bias the results is by choice of solvent. For example, using a polar aprotic solvent (e.g. an ether) to favors SN2, while a protic solvent favors E2 (since protic solvents interfere with nucleophilicity, but not basicity).(2 votes)
- At3:28, why won't oxygen break the bond from the carbon giving it a full carbocation character. Why do we only see a partial carbocation character.(2 votes)
these videos and intro organic chemistry textbooks are confusing. I feel as if the mechanisms we are given are simplified to the point that when we try to follow the rules that we learn from them universally, we get confused because we get taught more mechanisms that violate those rules. anyone else feel the same way?(2 votes)
same here
i feel (organic) chemistry is highly empirical than theory-based
which means when one found an exception breaking a grounded rule, it tends to add another rule to it as an exceptional rule rather than to explain with the given rule by digging it deeper or replace it with a new one
this way of thinking or practicing seems to give a birth of rows of rules and their exceptional rules, i feel(1 vote)
@5:42, why did the ethanol get a plus charge after it attacked the epoxide? Is it because it gave an electron away to the epoxide?(2 votes)
The Oxygen on the ethanol molecule has a plus-1 charge on it because it is bound to 3 other atoms: the Hydrogen, the Carbon belonging to the Benzene ring, and the remainder of the ethanol molecule. Hope that clears things up!(1 vote)
- how to do mechanism of base catalyzed reaction of epoxide(1 vote)
The hydroxide ion attacks one of the carbon atoms in the epoxide in an SN2 displacement reaction. The C-O bond breaks to form an oxide ion on the adjacent carbon atom. The epoxide ion then picks up a proton from the water. The result is a 1,2-diol.(1 vote)
- If the methyl group was cis to the epoxide on a different carbon of the ring, what happens to the stereochemistry of that methyl group after the ring-opening hydrolysis occurs?(1 vote)
- Can this type of ring opening occur with H2O and H2SO4? Would H2O be a strong enough nucleophile since it would relive ring strain??(1 vote)
In short, yes, but it depends on the substitution of the ring carbons. With greater subsitution, the ring can open (because the carbocation will be stabilised), and then water will attack(1 vote)
- I didn't understand why did the epoxide oxygen atom undergoes protonation first unlike other epoxide reactions.....
The fact is clear that since there is a weak nucleophile, it will attack the most substituted carbon atom.... hence favouring SN1 mechanism.........
But what does it explain on oxygen atom undergoing addition of hydrogen atom first.....?
(In other reactions it was protonated in the second step, ). plz help(1 vote)
This is an acid-catalysed reaction, which means that there will be protons in solution that can participate in an acid-base reaction with the oxygen (from epoxide), and therefore protonate the epoxide.(1 vote)
3:28Video transcript
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.