Below is a picture of the mechanism: enter image description here

I understand the mechanism but what I don't understand is why it's $\mathrm{S_N2}$. The solvent used is polar and protic, there is a tertiary carbon that would be suitable for $\mathrm{S_N1}$ as well. For what reason does this reaction occur exclusively via $\mathrm{S_N2}$?

  • $\begingroup$ unfavourable leaving group $\endgroup$ – Mithoron Feb 4 '15 at 14:12
  • $\begingroup$ Its exactly the same with either mechanism $\endgroup$ – RobChem Feb 4 '15 at 14:13
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    $\begingroup$ unfavorable for Sn1 $\endgroup$ – Mithoron Feb 4 '15 at 14:14

Quotation from Clayden:

Without protonation, the epoxide oxygen is a poor leaving group, and leaves only if ‘pushed’ by a strong nucleophile: the reaction becomes pure SN2. Steric hindrance becomes the controlling factor and methoxide attacks only the primary end of the epoxide.

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  • $\begingroup$ The nucleophile isn't strong in this solvent though is it? $\endgroup$ – RobChem Feb 4 '15 at 14:18
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    $\begingroup$ I'm afraid you're strongly overestimating effects of solvation. $\endgroup$ – Mithoron Feb 4 '15 at 18:31
  • $\begingroup$ Does this mean only in the presence of a strong acid (protonation of oxygen), in SN2, the nucleophile will attack the more substituted side? Otherwise (neutral/basic conditions), it will attack the less substituted side as normal SN2 does? [Due to the hydroxyl group is a better leaving group (more stable than its conjuagte base)]? $\endgroup$ – 234ff yesterday

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