I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.
I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.
Stabilization of NaBH4 in Methanol Using a Catalytic Amount of NaOMe. Reduction of Esters and Lactones at Room Temperature without Solvent-Induced Loss of Hydride, More
You need methanol to solve NaBH4, other solvents are not very suitable. The active reducing species can be NaBH4 or NaOCHxHx (in which NaBH4 has first reacted with methanol)
You need methanol to solve NaBH4, other solvents are not very suitable. The active reducing species can be NaBH4 or NaOCHxHx (in which NaBH4 has first reacted with methanol)
I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.
I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.
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You need methanol to solve NaBH4, other solvents are not very suitable. The active reducing species can be NaBH4 or NaOCHxHx (in which NaBH4 has first reacted with methanol)
You need methanol to solve NaBH4, other solvents are not very suitable. The active reducing species can be NaBH4 or NaOCHxHx (in which NaBH4 has first reacted with methanol)
More
VOTE