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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: noda on June 14, 2006, 08:41:25 AM

Title: Diastereoselective Reduction of Enone with Ipc2BCl
Post by: noda 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?

Title: Re: Diastereoselective Reduction of Enone with Ipc2BCl
Post by: Dan on June 14, 2006, 11:42:56 AM

Quote from: noda on June 14, 2006, 08:41:25 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!

Title: Re: Diastereoselective Reduction of Enone with Ipc2BCl
Post by: noda 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

Title: Re: Diastereoselective Reduction of Enone with Ipc2BCl
Post by: movies on June 15, 2006, 01:48:12 PM

Here's a guess at the TS using the - enantiomer.