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Biotech / Medical : Sepracor-Looks very promising -- Ignore unavailable to you. Want to Upgrade?

To: tommysdad who wrote (4591)	10/20/2000 10:20:46 AM
From: jayhawk969	Read Replies (1) \| Respond to of 10280

Slightly off topic: It has been over 30 years since I took Organic Chemistry. I am not sure I want to go back and relearn it. There are some reactions referred to enantioselective reactions where one enantiomer is produced in excess of the other primarily because one side of the reactant molecule or intermediate was hindered so that side of access was less likely than the less hindered side. Such a mixture of two enantiomers will show an optical rotation. We can determine the specific rotation of the mixture and then knowing the specific rotation of the enantiomer in excess we can determine the enantiomeric excess. enantiomeric excess = (specific rotation of the mixture/ specific rotation of the pure enantiomer in excess) 100 The enantiomeric excess can be determined in another way if we know the amount of each enantiomer produced. If one knows the moles of each enantiomer produced then: enantiomeric excess = (moles of major enantiomer - moles of other enantiomer / Total moles of both enantiomers) 100 Let's work a problem determining enantiomeric excess. Let's suppose that we have a reaction mixture which we measure for rotation, and the observed specific rotation of the reaction mixture is measured to be +8.52 degrees of rotation. The specific rotation of the S-configured enantiomer was said to be -15.00 degrees of rotation. Determine the enantiomeric excess. The sign on the specific rotation of the mixture tells us which enantiomer is in excess. Since the pure S-Configured isomer was measured to be -15.00 degrees, then the R-configured isomer would be +15.00 degrees. The positive sign on the observed specific rotation of the mixture tells us that the R-configured isomer is in excess. So the enantiomeric excess would be: enantiomeric excess = (+8.52 / +15.00)(100) = 56.8 % in excess of R-isomer This means that 56.8% of the entire mixture is in excess of the R-configured isomer.