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reduction of carboxylic acids using BH3-THF


lila

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

 

 

 

 

 

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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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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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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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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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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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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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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