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From what it looks like, Gilman reagents are good nucleophiles, and will attack at the less-substituted side of an asymmetric alkane because Gilman reagents don't necessarily need a substantial amount of partial positive character at the site of attack.

However, don't strong nucleophiles open epoxides in an SN2 like manner? Where's the inversion?

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In the first reaction, the reaction doesn't take place at a stereocenter, so there's no way to observe the inversion.

In the second reaction, the stereochemistry in the product is R, R. The Gilman reagent approaches opposite the C-O bond. After bond rotation, we see the product in the indicated conformation.

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    $\begingroup$ First one or second one? @jerepierre $\endgroup$ – Dissenter Oct 19 '14 at 1:31
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    $\begingroup$ @Dissenter I've clarified my answer to address both examples. $\endgroup$ – jerepierre Oct 19 '14 at 2:04
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    $\begingroup$ but how do Gilmans open epoxides? From the less substituted side ... but does opening proceed with inversion or retention of stereochemistry? $\endgroup$ – Dissenter Oct 19 '14 at 7:25
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    $\begingroup$ @Dissenter The reaction goes with inversion. $\endgroup$ – jerepierre Oct 19 '14 at 16:39
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    $\begingroup$ @Dissenter You're right, it's RR. However it's still inversion. $\endgroup$ – jerepierre Oct 21 '14 at 20:29

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