[Alexis7]

Hi,

So I have a equation about balancing equation:

9-fluorenone is reduced to 9-fluorenol, using NaBH4 and ethanol.

I want to know if this is right or not. So far I have this:

C13H8O+NaBH4+ CH3CH2OH → C13H9OH+NaBH3(CH3CH2OH)^-

My question is this correct, and should I used "Na" in the balanced equation? Thanks.

[Hunter2]

No. Write two equations. The Reduction from Keton to Alcohol.

And the Oxidation from Borhydride to Borate. Probably everything in alkaline Condition.

[Alexis7]

C13H8O + BH4- ==> C13H9O- + BH3

BH3 + CH3CH2OH ==> CH3CH2OHBH3

C13H9O- + CH3CH2OHBH3 ==> C13H9OH + CH3CH2OBH3-

4C13H8O + BH4- + 4CH3CH2OH ==> 4C13H9OH + B(OCH2CH3)4- ------- Does that seem right?

[discodermolide]

Not quite correct I think. What about the H₂ that is produced?
The reaction of BH₃ with ethanol will give CH₃CH₂OBH₂ + H₂

[orgopete]

Not quite correct I think. What about the H₂ that is produced?
The reaction of BH₃ with ethanol will give CH₃CH₂OBH₂ + H₂

I thought the poster gave a good accounting of the net stoichiometry. I am interested to see how hydrogen production is part of it.

[discodermolide]

Not quite correct I think. What about the H₂ that is produced?
The reaction of BH₃ with ethanol will give CH₃CH₂OBH₂ + H₂

I thought the poster gave a good accounting of the net stoichiometry. I am interested to see how hydrogen production is part of it.

http://pubs.acs.org/doi/abs/10.1021/ic50119a025

[orgopete]

@Disco
I did not think NaBH4 reductions in ethanol were unusual. I further thought one mole of NaBH4 reduced four moles of aldehyde or ketone. Perhaps you could provide the proper mechanism and stoichiometry for these reductions.

[discodermolide]

I was only responding to this:
BH3 + CH3CH2OH ==> CH3CH2OHBH3
which, in my opinion is not quite correct.
So I made a suggestion as to how to correct it.
BH₃ with ethanol will give CH₃CH₂OBH₂ + H₂

[Dan]

I further thought one mole of NaBH4 reduced four moles of aldehyde or ketone.

Not in practice. Hydrogen gas is always produced to some extent, you will never get all 4 equivs of hydride. The active reducing species are borates of the formula (MeO)_nBH_4-n^- (where n=1,2,3).

The decomposition of borohydride in methanol (and other alcohols or water) competes with reduction of the carbonyl, but decomposition rate slows as n increases. As a rule of thumb, you get about 2 equiv. hydride per borohydride in practice. I think the most I've personally observed in methanol is 2.5 equiv. (complete conversion in the reduction of a carbonyl with 0.4 equiv NaBH₄ in methanol at -15 °C).