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Topic: Diastereoselective Reduction of Enone with Ipc2BCl  (Read 6178 times)

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Offline noda

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Diastereoselective Reduction of Enone with Ipc2BCl
« on: June 14, 2006, 08:41:25 AM »
In the next scheme there is reduction of enone how is the stereoselectivity achieved?
I'm not sure about the answer I know that The hydride is transferred to carbonyl carbon
through the six-membered ring transition state. does it has to be in planar ring transition state? and I think that the boron coordinate with the carbonyl and because of the large groups on the borane they causing steric resistance and that cause stereoselectivity.
am I right?

Offline Dan

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Re: Diastereoselective Reduction of Enone with Ipc2BCl
« Reply #1 on: June 14, 2006, 11:42:56 AM »
think that the boron coordinate with the carbonyl and because of the large groups on the borane they causing steric resistance and that cause stereoselectivity.
am I right?

Yes, this explains why the recation is stereoselective. But as to rationalising which reducing agent results in which product, you will have to draw transition states. This is really difficult, and I have never been able to draw a decent diagram to explain the stereoselectivity achived with each isomer of this B reagent. I wonder if anyone can, it's an absolute nightmare!
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Offline noda

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Re: Diastereoselective Reduction of Enone with Ipc2BCl
« Reply #2 on: June 14, 2006, 12:59:47 PM »
is it because of the bridgedhead methyls on the cyclohexane? someone can draw the transition state in the two reagents so I can understand the stereoselectivity

Offline movies

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Re: Diastereoselective Reduction of Enone with Ipc2BCl
« Reply #3 on: June 15, 2006, 01:48:12 PM »
Here's a guess at the TS using the - enantiomer.

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