How does Sodium Borohydride reduce 9-Fluorenone to 9-Fluorenol with methanol as the solvent?

Question

I'm talking about mechanism here. I think a hydride ion from NaBH4 attacks the carbonyl carbon as one of the C=O bonds becomes a lone pair on the oxygen, but then what happens? Does the oxygen's lone pair attack the hydroxyl hydrogen from the MeOH or is it a hydrogen from NaBH4?

Answer 1

Good place to look: http://www.organic-chemistry.org/namedreactions/luche-reduction.shtm

Ignore the CeCl3, it just makes the reaction go faster, but the mechanism is exactly the same (i.e. without the cerium just ignore the first step of the activated methoxide attacking the boron and start for the B-H bond attacking the carbonyl carbon).

Answer 2

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To obtain a balanced equation you need to be somewhat familiar with the mechanism of the reaction. It can be broken down into 3 steps: 1) Nucleophilic addition of hydride (by borohydride) to the carbonyl carbon of fluorenone C13H8O + BH4- ==> C13H9O- + BH3 2) Complexation of BH3 with ethanol BH3 + CH3CH2OH ==> CH3CH2OHBH3 Note that in a Lewis structure the product should appear with a formal positive charge on O and a formal negative charge on B. 3) Formation of product (proton transfer) C13H9O- + CH3CH2OHBH3 ==> C13H9OH + CH3CH2OBH3- Next, the ethoxyborohydride formed here can react with 3 additional equivalents of fluorenone, thereby picking up additional ethoxy groups in the process. Adding these steps together and canceling like terms, we arrive at: 4C13H8O + BH4- + 4CH3CH2OH ==> 4C13H9OH + B(OCH2CH3)4- You may also write the equation with counterions (Na+) as appropriate. We see that there is a 4:1 molar ratio between fluorenone and borohydride in the reactants. That is, 1.32 mmol of borohydride require 4 x 1.32 mmol = 5.28 mmol of fluorenone to react completely. However, there are only 1.11 mmol of fluorenone present. Therefore, fluorenone is the limiting reagent.

Answer 3

Reduction Of 9-fluorenone Mechanism