# Why does the opening of an epoxide occur via an Sn2 like mechanism when using a methanoate ion as a nucleophile and methanol as a solvent?

Below is a picture of the mechanism:

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 $\mathrm{S_N2}$?

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

• The nucleophile isn't strong in this solvent though is it? – RobChem Feb 4 '15 at 14:18
• I'm afraid you're strongly overestimating effects of solvation. – Mithoron Feb 4 '15 at 18:31