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Williamson 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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  • blobby green style avatar for user Zac
    Isn't it NaOH instead of NaH?
    (10 votes)
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  • blobby green style avatar for user Ana Ramos
    what are the possible side reactions in williamson ether synthesis?
    (3 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      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)
  • mr pink red style avatar for user Chris Cameron
    How can the attacked carbon on the alkyl halide at be tertiary and SN2 still occur?
    (3 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      At he specified that the halide must be a 1° halide. He was a little careless at . 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)
  • blobby green style avatar for user brittney weisend
    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)
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    • leaf green style avatar for user Anurag Singh
      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)
  • female robot amelia style avatar for user Bianca Khatibshahidi
    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)
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    • spunky sam blue style avatar for user Ernest Zinck
      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)
  • leaf green style avatar for user gentech
    if i use methyl fluoride or chloride is there any problem
    And among the three which will react fast?
    (1 vote)
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    • spunky sam blue style avatar for user Ernest Zinck
      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)
  • mr pants teal style avatar for user francoiszam
    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)
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    • leafers sapling style avatar for user mir.schwabe
      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)
  • spunky sam blue style avatar for user Aditya Moitra
    @ , 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)
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    • leaf green style avatar for user Anurag Singh
      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)
  • blobby green style avatar for user Justin
    I'm pretty sure at the 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)
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  • blobby green style avatar for user jlee4001
    Is Williamson Ether always Sn2 (primary halide attack)?
    (1 vote)
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Video 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.