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Organic chemistry
Course: Organic chemistry > Unit 7
Lesson 5: Synthesis and cleavage of ethersWilliamson ether synthesis
Two-step synthesis of an ether from an alcohol, A strong base is a dded to deprotonate the alcohol, which then attacks an alkyl halide. Created by Jay.
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Isn't it NaOH instead of NaH?(10 votes)
I don't think it has to be... You can, I believe, use NaOH but NaH is still a strong base and it drives the reaction forward due to the hydrides bonding to each other and forming H2 gas that bubbles out.(21 votes)
- what are the possible side reactions in williamson ether synthesis?(3 votes)
The major side reaction is elimination from the alkyl halide. You are using a strong base, an alkoxide ion. So you cannot use a tertiary alkyl halide at all -- it will undergo preferential elimination of HX. Primary alkyl halides are best, because they give primarily substitution reactions (ether formation). You have to juggle the conditions carefully with secondary substrates in order to get good yields of ether formation.(12 votes)
How can the attacked carbon on the alkyl halide at2:28be tertiary and SN2 still occur?(3 votes)- At0:15he specified that the halide must be a 1° halide. He was a little careless at2:28. He should have shown the H atoms explicitly to show that the halide is primary.
I agree that it is confusing. Without the H atoms, the compound looks like tert-butyl halide, but it is really methyl halide, so the reaction really is SN2.(9 votes)
- this a general question...is there a time where it is more appropriate to use NaH+ as oppossed to KOH...or is it ok to use NaH as the strong base for all reactions of williamson ether synthesis?(4 votes)
- Yes, NaH could be used in all of the williamson synthesis, but if the situation demands like if the alkaoxide ion so formed is stabilised by resonance then there is no need of a strong base. Then reaction can also be carried out with a slightly weaker base.(6 votes)
I've seen Williamson-Ether synthesis done with Bu4N+, Br- included with CH3I (alkyl halide) and NaOH (the strong base). What do the former two compounds do for this reaction? Also, do they do anything in the laboratory setting?(2 votes)- They are highly important in a laboratory synthesis.
You are referring to a process called reverse phase catalysis.
The NaOH converts the alcohol to the alkoxide:
ROH + Na⁺OH⁻ ⇌ RO⁻Na⁺ + H₂O
Then the alkoxide reacts with the alkyl halide to form the ether.
RO⁻ + CH₃I → ROCH₃ + I⁻
The overall reaction is CH₃I(l) + ROH(l) + NaOH(aq) → CH₃OR + NaI + H₂O
The problem is that ROH is soluble in CH₃I (and water), but NaOH and the alkoxide are soluble only in water.
The alkoxide is in the water layer, but the methyl iodide is in the organic layer. The reaction between them will be quite slow.
Enter a phase transfer catalyst, the quaternary ammonium bromide, Bu₄N⁺Br⁻ (let's call it Q⁺Br⁻). Its function is to bring the OR⁻ into the organic layer so it can react with the methyl iodide.
Despite being a salt (and water-soluble), the four nonpolar butyl groups make it soluble in the organic layer as well.
We have the equilibrium, Q⁺Br⁻(org) ⇌ Q⁺Br⁻(aq)
In the aqueous layer, Q⁺Br⁻(aq) + OR⁻(aq) ⇌ Q⁺OR⁻(aq) + Br⁻(aq)
The quaternary alkoxide moves into the organic layer, Q⁺OR⁻(aq) ⇌ Q⁺OR⁻(org)
In the organic layer, CH₃I(org) + Q⁺OR⁻(org)→ CH₃OR(org) + Q⁺I⁻(org)
Summary: The Q⁺ goes from the organic layer into the water layer, where it grabs the alkoxide and pulls it into the organic layer to react in the Williamson synthesis.(5 votes)
- if i use methyl fluoride or chloride is there any problem
And among the three which will react fast?(1 vote)- There are certainly health concerns, and they are both highly inflammable gases. The boiling points of CH3F, CH3Cl, CH3Br, and CH3I are -75°C, -24°C, 3.5°C, and 42.5°C. Reactions of the first three would have to be conducted at low temperatures, and this would make the reaction too slow to be practical.
No matter the temperature, the order of reactivity is
F < Cl < Br < I. So you should use CH3I. This is easier to handle, because it is a liquid at room temperature and is the most reactive of the four halides.(5 votes)
I thought very encumbered bases were more likely to perform an E2 mecanism on a primary carbon... At the end of the video, I understand that there can't be any E2 because we are using an alkyl group. But what if we used a haloalcane with more than two carbons? Would we still have a SN2 mecanism and thus the formation of an ether?(1 vote)
If the haloalkane is secondary, we can get some E2 products, especially if the base is bulky (but all of the bases used in this example are small and not sterically hindered). Williamson ether synthesis cannot take place on tertiary carbons because it is a SN2 mechanism.(2 votes)
- @2:25, the sodium alkoxide gets converted into an ether. The sodium alkoxide has an ionic bond. Why would it want to lose its ionic bond and form an ether with a covalent bond ??. I thought ionic bonds were much more stable so the sodium alkoxide should also be more stable right ?! :/(2 votes)
- Yes, you're right about ionic stability but since they involve bondage of charged species hence they remain disassociated {i.e, separate} in the aqueous polar solvent in which the reaction is being carried out.Thus the alkoxide anion is more susceptible to covalent bonding.(0 votes)
- I'm pretty sure at2:17the C has only 2 methyl groups on it; the third one on the right is supposed to be the bond that was connected to X, right?(0 votes)
Don't they have to be hydrogens? SN2 can't occur with a tertiary carbon, or occurs so little to be negligible.(4 votes)
- Is Williamson Ether always Sn2 (primary halide attack)?(1 vote)
- Yes, a Williamson ether synthesis almost always uses the SN2 attack of an alkoxide on a primary halide.(1 vote)
2:28Video transcript
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.