June 04, 2020, 01:55:21 AM
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Topic: SN2- Synthesize an R enantiomer from a optically pure R 2-bromooctane in 1 step?  (Read 3024 times)

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Offline abstruse

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Hi

I was going over synthesizing specific enantiomers using SN2 reactions and my text talks about using a 2 step method for generating a R enantiomer (Assume I want to create (R)-2-Octanethiol from optically pure (R)-2-Bromooctane). The text follows a backside displacement of the Br by I in the first SN2 reaction and then the product ((S)-2-Iodooctane) reacts with HS- (another backside displacement) to form (R)-2-Octanethiol.

My question is cant we just follow a one step frontside displacement where HS- displaces Br- since frontside displacement reactions leave the product with the same configuration. If this is not possible, why is it not possible?

Any help is appreciated!!
« Last Edit: October 31, 2009, 07:57:11 PM by abstruse »

Offline dunno260

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It has to do with the orbitals the nucleophile interacts with on the carbon atom.  A frontside attack is just simply not possible.  An Sn2 reaction requires a backside attack because the nucleophile binds to the anti-bonding orbital at carbon which sits on the other side of the bond between carbon and the leaving group. 

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