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Does the epoxide ring open in this case? I guessed it does, but it really doesn't. The answer given (among the four options) is another epoxide, with some sort of substitution. Could someone please explain the mechanism for this reaction?

Also, what exactly is the condition for ring opening of epoxides? Thanks for the help!

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marked as duplicate by Mithoron, NotEvans., M.A.R. ಠ_ಠ, Buttonwood, pentavalentcarbon Jul 1 '17 at 18:48

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  1. Epoxide gets opened at the least hindered side by the ethoxide (acting as a nucleophile rather than as a base in this case).
  2. The resulting alkoxide intermediate can itself act as a nucleophile. In the middle intermediate could either kick out ethoxide (which would give back the starting material), or kick out the chloride (to give a different epoxide). Since the chloride is a better leaving group, it is this one that probably leaves (theres an element to which the whole thing is reversible away, and if it does kick out ethoxide, the ethoxide can just attack again to re-form the middle alkoxide intermediate)

The product on the right hand side is probably what the mixture would converge upon, given that chloride is a better leaving group than ethoxide.

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    $\begingroup$ Is there a name for this? I could've sworn this was called the Payne rearrangement, but Wikipedia tells me it's something different. $\endgroup$ – orthocresol♦ Jul 1 '17 at 16:59
  • $\begingroup$ @orthocresol- it's only a Payne rearrangement (I think...) where its a 2,3-epoxyalcohol and the epoxide is 'walked' to the end of a chain before being intercepted by a nucleophile (or at least thats the common textbook variant, using a bulky thiol nucleophile to trap) $\endgroup$ – NotEvans. Jul 1 '17 at 17:14
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    $\begingroup$ It's technically not rearrangement, just standard epoxy making reaction. $\endgroup$ – Mithoron Jul 1 '17 at 17:18
  • $\begingroup$ @Mithoron You guys fon't seem sure. What's the real answer? $\endgroup$ – Mockingbird Jul 2 '17 at 8:14
  • $\begingroup$ "2,3-epoxyalcohol and the epoxide is 'walked' to the end of a chain before being intercepted by a nuclueophile " Nucleophile or electrophile? $\endgroup$ – Mockingbird Jul 2 '17 at 8:17

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