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For the acid catalyzed opening of epoxides, the epoxide is first protonated to form the oxonium ion. I read in some textbooks that the $\ce{H2O}$ molecule, which acts as a nucleophile attacks in a $\mathrm{S_N2}$ concerted mechanism. Given that $\ce{H2O}$ is a weak nucleophile, how can a $\mathrm{S_N2}$ concerted mechanism be possible, 8 instead of $\mathrm{S_N1}$ which is expected of $\ce{H2O}$?

Also, for an unsymmetrical epoxide, why will the nucleophile attack at the more substituted $\ce{C}$?

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  • $\begingroup$ Take a look in Organic Chemistry by Clayden & Warren, the whole thing is fully explained (as it probably is in any good organic textbook). $\endgroup$ – NotEvans. Apr 20 '17 at 18:28
  • $\begingroup$ I can only find the base catalyzed epoxide opening, but not the acid catalyzed reaction. $\endgroup$ – Jonathan Smith Apr 21 '17 at 14:52
  • $\begingroup$ chem.libretexts.org/Textbook_Maps/… This link might help. $\endgroup$ – Tyberius Apr 23 '17 at 2:11

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