Master Organic Chemistry Reaction Guide

SN2 reaction of alkoxide ions with alkyl halides to give ethers (Williamson synthesis)

Description: Alkyl halides (or tosylates) will react with alkoxy ions to form ethers. This reaction is called the Williamson ether synthesis.

Notes: X here is a halide (Cl, Br, I) or sulfonate (OTs, OMs). The counter-ion on the alkoxy ion can be any alkali metal (e.g. Li, Na, K)


Notes: Note that since this is an SN2 reaction and proceeds via backside attack, there will be inversion of configuration at the carbon (note the last two examples). Note that in the second example that this ether would be difficult to make the opposite way (CH3O- attacking a tertiary alkyl bromide) since SN2 reactions don’t work on tertiary centers.

Mechanism: In the SN2 reaction the nucleophile (RO-) attacks the carbon with the good leaving group, forming a C–O bond and breaking the C–Br bond (Step 1, arrows A and B).

Notes: Again, the Na(+) is not crucial here, it’s just a spectator ion.



Comment section

25 thoughts on “SN2 reaction of alkoxide ions with alkyl halides to give ethers (Williamson synthesis)

    1. True, although that’s a rather large subject. Here, note that it works best for primary (and to some extent secondary) alkyl halides.

  1. He doesn’t need to explain why it’s SN2 :)
    When the halide group leaves, if the alkoxy ion doesn’t atake at the same time, you would be left with a primary carbo-cation. These are very unstable because you only have a carbon giving charge to the primary. If you had two more carbons linked to the carbon with the halide, you could have a SN1 reaction, since the carbons around the tertiary carbo-cation would be estabilizing the positive charge.

    I hope I made myself clear. I’m Portuguese :)

  2. If you’re starting with an ether and trying to determine which starting materials are required, does it matter which halide you use for the leaving group?

    1. It absolutely matters. Since the reaction is SN2, you want to set the reaction up so that your alkyl halide is (preferably) primary. You certainly wouldn’t want to try to get an SN2 to work on a tertiary alkyl bromide.

  3. Hi, great work!
    I have a small doubt, though. If the alkyl halide is tertiary, the reaction will follow SN1 mechanism instead of SN2, and thus a carbocation will be formed.

    Q1) Will the reaction not proceed to give us an ether?
    Q2) My teacher’s answer to Q1 was no. He said that a carbocation always reacts with a base to give an alkene. Is that actually true?

    Thanks in advance!

  4. Can you show me the rxn if I use second halide with sodium alkoxide as I want to make an alkene out of it but I want to know the rxn please

  5. in the last example (a alkyl group attached to the benzene ) the approach of nucleophile is heavily hindered so how would the reaction proceed to the final product ??

    1. It’s not really that hindered. The benzene is flat. Also, benzylic positions are unusually reactive because the aromatic pi system stabilizes partial positive charge in the transition state.

  6. E2 elimination is favoured when the atacking alkoxideion is tertiary but in willamsons’ Synthesis sn2 rxn is favoured in presence of tertiary alkoxide ion how is it possible reply me soooon

  7. why wouldn’t e1 happen with naome and a secondary bromine group on an alkane chain? like why would this proceed with substitution if naome can also be an unhindered base?

    1. E1 is not an option with a secondary bromine. Loss to give a carbocation would not happen without significant heat. On the other hand, with a secondary bromine, E2 is a real issue to watch out for. The substitution reaction would likely be accompanied by elimination products (from E2)

    1. It certainly could, especially with secondary alkyl halides! I tried to highlight the formation of the ether in the case of the secondary alkyl halide but I should note that I’m not sure on the yield.

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