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Are the mechanism and the product for this given reaction right or is there anything wrong with it? 1,2 - Epoxybutane has originally the 2R configuration and the product got an R configuration as well.

I was not sure about, where the nucleophile attacks. Under acidic conditions steric hindrance will be ignored, right? In this case the hydrogen atom will be added to the less substituted carbon atom and the ring will be opened.

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  • $\begingroup$ Aluminium binds to the oxygen first $\endgroup$
    – Waylander
    Sep 29, 2020 at 17:08
  • $\begingroup$ Al loses hydride to form Alane, (AlH3 + H-). It binds to the oxygen as Alane freeing up the hydride to do the nucleophilic attack. Whether they fully separate or remain partially associated is not afaik fully established $\endgroup$
    – Waylander
    Sep 29, 2020 at 18:01
  • $\begingroup$ Well, I wager that hydride transfer happens after coordination - hydride hardly ever is alone and Al can coordinate up to 6 groups. $\endgroup$
    – Mithoron
    Sep 29, 2020 at 18:55
  • $\begingroup$ ...and LiAlH4 has aluminum impurities in it that can act as Lewis acids to get reduction going. This process frees up AlH3. $\endgroup$
    – user55119
    Sep 29, 2020 at 19:38

1 Answer 1

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The free electron pairs of the oxirane may act as a Lewis base and coordinate with a Lewis acid like $\ce{Li+}$. This form of activation facilitates the attack of the alanate $\ce{[AlH4]-}$ as a $\sigma$-nucleophile at the less substituted side of the epoxide.

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Once opened, the nucleophilic oxygen attacks at the electron deficient centre of $\ce{AlH3}$ (Lewis acid). But the $\ce{Al-O}$ bond hydrolyzes upon the workup, yielding the alcohol and the hydroxides of lithium and aluminum.

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