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reduction of carboxylic acids using BH3-THF


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#1 lila

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Posted 26 June 2010 - 12:32 PM

Please I need the detallied mechanism of the reduction of carboxylic acids using BH3-THF ?

Thanks for any help


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#2 Horza2002

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Posted 26 June 2010 - 06:53 PM

The mechanism is attached below.

First, the borane coordinates to the carbonyl activating it with the electronpoor boron atom. After that, the oxygen lone pair then donates to the boron atom followed by the transfer of the hydride to the carbonyl. Elimination of the oxyboronyl intermediate followed by attack of a second hydride attack and quenching then give the alcohol.
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#3 mississippichem

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Posted 26 June 2010 - 11:02 PM

I wonder if this reaction might be carried out asymetrically, perhaps with a bulky, chiral, organo-substituted borane-THF adduct.

Edited by mississippichem, 26 June 2010 - 11:06 PM.
incorrect structure posted

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#4 Horza2002

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Posted 27 June 2010 - 09:59 AM

Yes you might be able to do it chirally, but that would strongly depend on the groups of the carboxylic acid. They would need to be fairly large to get any decent amount of enantioselectivity.
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#5 mississippichem

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Posted 27 June 2010 - 05:02 PM

Yeah that's what I figured. Check this out (p.86).

[attachment=1709:organic.pdf]



This chemistry is in the same ball park stereo-wise (not a reduction) but with a thioester and a boron enolate; really cool chemistry. You're right, even in this case bulkier groups on the substrate encourage enatioselectivity.
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#6 Horza2002

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Posted 27 June 2010 - 06:28 PM

Yer those reactions are typical of chiral bororeagents. Theres a chapter on them in Clayden Greeves Warren and Wothers (the ungrad organic chemistry bible) that talks about them.

The main driving force for the enantioselectivity is the steric interaction between the ligands on the boron and the other groups present on the substrate. If you could improve boryl reagents theyd make a marked improvement over metal based chiral reducing catalysts.
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#7 mississippichem

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Posted 27 June 2010 - 06:47 PM

Yes they would. Cheaper too, LDA can put a dent in the research budget, and its a bitch to handle. I really love Hydrazine reductions of carbonyls that proceed through the hydrazone intermediates, they are beautiful, and N2 is probably one of your easier groups to remove.
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#8 Horza2002

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Posted 27 June 2010 - 09:16 PM

Yer N2 is wonderfully easy to remove isn't! Ive never done that reaction personally; favoured method is NaBH4 for ketone/aldehyde reduction....nice and simple. Hydrazine is not the nicest reagent in the world to use though is it?
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#9 mississippichem

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Posted 27 June 2010 - 09:37 PM

It's not so bad in under 30% concentration (aqueous). Keep away from anything remotely oxidizing of course.
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#10 lila

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Posted 28 June 2010 - 07:01 PM

thanks but how the control in enantioselectivity during the transition state. I think the mechanism is still unclear.:confused::rolleyes:

thankssssss
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#11 mississippichem

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Posted 28 June 2010 - 07:38 PM

thanks but how the control in enantioselectivity during the transition state. I think the mechanism is still unclear.:confused::rolleyes:

thankssssss


This reaction is not enatioselective unless a chiral borane reagent is used. It depends on what borane reagent is used. In the case of BH3 the reaction will result in a racemate or an achiral product depending on the substrate. In the case of a chiral borane reagent, addition of the hydride cannot occur on the side that is sufficiently sterically blocked.
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#12 Horza2002

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Posted 28 June 2010 - 09:06 PM

If you have a look at the mechanism I said ealier, the second stage is when the hydride is transfered to the carbonyl. If you have a chiral boran substrate, then one of the faces will have more of a steric interaction than the other. Thats means that one face os favoured for the hydride transfer
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#13 lila

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Posted 29 June 2010 - 11:33 AM

it usually the BH3 -THF is generated by using NaBH4/I2 and many papers confirmed this method give only one enantiomer....so I was trying to know the transition state which lead to this enantioselectivity.:-(
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#14 Horza2002

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Posted 1 July 2010 - 09:24 AM

Do they give an ee value for that process. If its an low ee then its not worth metntioning really
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