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I understand that in general, when in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons. And when in basic media, the nucleophile attacks the less substituted side of the epoxide due to steric reasons.

What if the hydride anion attacks an epoxide? I'd imagine the hydride anion to be quite small, and able to attack both the more substituted and less substituted sides of an asymmetric epoxide. What are the experimental results?

Also I have my doubts about hydride anion even being able to open an epoxide; I am told by Clayden that hydride anion never acts as a nucleophile because its 1s orbital is too small.

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    $\begingroup$ If I recall my organic hydride is such a poor nucleophile it won't open an epoxide or do many other organic reactions for the reason you stated. $\endgroup$ – Brinn Belyea Aug 13 '14 at 2:08
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    $\begingroup$ Poor nucleophile because of its heavy solvation, right? $\endgroup$ – Dissenter Aug 13 '14 at 2:13
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    $\begingroup$ The hydride ion is actually gigantic. It's hard to get a reliable value because its poorly held electron cloud is very soft and deforms easily, but if I recall correctly it's theoretically the second largest common monoatomic anion known, at ~200 pm radius, losing only to iodide. $\endgroup$ – Nicolau Saker Neto Aug 13 '14 at 11:53
  • $\begingroup$ Huge because of its very poor ability to stabilize electron density, right? $\endgroup$ – Dissenter Aug 13 '14 at 13:06
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in acidic media, the nucleophile tends to attack the more substituted side of the epoxide due to electronic reasons

This is an $\ce{S_{N}1}$ like reaction. First we would protonate the epoxide oxygen. If we examine the case of an unsymmetrical epoxide and draw the resonance structures for this situation (see below) we would expect the secondary carbocation structure to count more towards the actual structure than the primary carbocation structure. Therefore, we would expect a nucleophile to react preferentially at the tertiary epoxide carbon, quite selectively.

enter image description here

in basic media, the nucleophile attacks the less substituted side of the epoxide due to steric reasons.

This would be a traditional $\ce{S_{N}2}$ reaction. Attack at the least substituted (sterically most accessible) carbon would be expected.

What if the hydride anion attacks an epoxide? I'd imagine the hydride anion to be quite small, and able to attack both the more substituted and less substituted sides of an asymmetric epoxide. What are the experimental results?

The hydride anion is not free, but complexed, so it has reasonable steric bulk - enough for stereoselective reactions to occur. Further, the reaction between LAH and epoxides is quite facile, no doubt accelerated by the release of steric strain in the 3-membered epoxide ring system. A traditional $\ce{S_{N}2}$ reaction takes place. In the figure below, the alcohol shown is formed in ca. 90% yield. Because of the "backside" attack nature of the $\ce{S_{N}2}$ reaction axial attack is strongly preferred in cyclic systems.

enter image description here

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    $\begingroup$ That's interesting. My professor commented that the hydride anion was so small that the opening of the epoxide is not stereoselective. $\endgroup$ – Dissenter Aug 13 '14 at 3:47
  • $\begingroup$ "Lithium aluminum hydride usually delivers hydride with excellent selectivity to the less substituted carbon of an epoxide", see page 358, wavefun.com/products/books/Chapter4.pdf a few more examples showing high stereoselectivity in depts.washington.edu/chemcrs/bulkdisk/chem531A_win06/… $\endgroup$ – ron Aug 13 '14 at 13:22
  • $\begingroup$ I was thinking of sodium hydride; would that be the same thing? $\endgroup$ – Dissenter Aug 13 '14 at 13:22
  • $\begingroup$ No, each form of hydride ($\ce{LiAlH_4, NaBH_4, NaH}$, etc.) has noticeably different properties. $\endgroup$ – ron Aug 13 '14 at 13:24
  • $\begingroup$ Oh okay. I was referring to NaH exclusively. $\endgroup$ – Dissenter Aug 13 '14 at 13:25
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There's much detail to it beyond the scope and nit necessary but the final result is:

  1. Bases (or ions with negative charge) attack the sterically unhindered positions

  2. Acid catalyzed ring opening happens from the position best to stabalize positive charge even though it doesn't form

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