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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 $\mathrm{S_N2}$?

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  • $\begingroup$ unfavourable leaving group $\endgroup$
    – Mithoron
    Feb 4, 2015 at 14:12
  • $\begingroup$ Its exactly the same with either mechanism $\endgroup$
    – RobChem
    Feb 4, 2015 at 14:13
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    $\begingroup$ unfavorable for Sn1 $\endgroup$
    – Mithoron
    Feb 4, 2015 at 14:14

1 Answer 1

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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, 2015 at 14:18
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    $\begingroup$ I'm afraid you're strongly overestimating effects of solvation. $\endgroup$
    – Mithoron
    Feb 4, 2015 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
    Aug 2, 2020 at 13:23
  • $\begingroup$ @234ff Things are more complicated when the epoxide gets protonated. The mechanism is somewhere between SN1 and SN2. Often the SN1 nature prevails and the more susbtituted end gets attacked, but there are numerous examples where the attack occurs on the less substituted. This depends on the nature of the nucleophile as well as sterics of the epoxide. $\endgroup$
    – EJC
    Aug 4, 2020 at 4:16

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