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In the reduction of esters to alcohols, the RO group is pushed off. However, the negative charge on the O atom is destabilized by the electron donating R group. Why is the RO group able to function as a leaving group given that the resultant alkoxide is unstable?

Link to reaction mechanism.

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    $\begingroup$ This is simplified mechanism, in reality this would be probably complex with Al not bare ethanolate. $\endgroup$ – Mithoron May 7 '17 at 15:44
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The intermediate before forming the alkoxide is a tetrahedral intermediate which is also unstable. You're going from a high energy species to another high energy species.

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  • $\begingroup$ Sorry I don't understand what you mean by "going from a high energy species to another high energy species". $\endgroup$ – Jonathan Smith May 7 '17 at 2:59
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    $\begingroup$ You're saying that the alkoxide is unstable. How that is different from the intermediate before. That is also an alkoxide except that it has an oxygen bonded to the central carbon. On collapse of the tetrahedral intermediate, you're going to form a new alkoxide and a carbonyl group, plus frequently two species. This is entropically and energetically favorable. You're focused on the alkoxide being "unstable", but the thing that comes before is also "unstable" and more so. $\endgroup$ – Zhe May 7 '17 at 15:02

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