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Topic: Balancing an equation with sodium borohydride  (Read 12347 times)

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Offline Alexis7

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Balancing an equation with sodium borohydride
« on: March 03, 2013, 05:39:09 PM »
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.

Offline Hunter2

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Re: Balancing an equation with sodium borohydride
« Reply #1 on: March 04, 2013, 02:12:39 AM »
No. Write two equations. The Reduction from Keton to Alcohol.

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

Offline Alexis7

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Re: Balancing an equation with sodium borohydride
« Reply #2 on: March 04, 2013, 02:35:24 AM »
C13H8O + BH4- ==> C13H9O- + BH3

BH3 + CH3CH2OH ==> CH3CH2OHBH3

C13H9O- + CH3CH2OHBH3 ==> C13H9OH + CH3CH2OBH3-

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

Offline discodermolide

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Re: Balancing an equation with sodium borohydride
« Reply #3 on: March 04, 2013, 03:05:47 AM »
Not quite correct I think. What about the H2 that is produced?
The reaction of BH3 with ethanol will give CH3CH2OBH2 + H2
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Offline orgopete

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Re: Balancing an equation with sodium borohydride
« Reply #4 on: March 05, 2013, 08:37:24 PM »
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.
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Offline discodermolide

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Re: Balancing an equation with sodium borohydride
« Reply #5 on: March 05, 2013, 10:15:07 PM »
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
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Offline orgopete

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Re: Balancing an equation with sodium borohydride
« Reply #6 on: March 06, 2013, 01:53:33 AM »
@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.
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Offline discodermolide

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Re: Balancing an equation with sodium borohydride
« Reply #7 on: March 06, 2013, 01:59:24 AM »
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
« Last Edit: March 06, 2013, 02:10:27 AM by discodermolide »
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Offline Dan

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Re: Balancing an equation with sodium borohydride
« Reply #8 on: March 06, 2013, 02:37:38 AM »
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)nBH4-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).
 
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