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Is it possible, chemically, to change a dextrorotatory compound to its levorotatory counterpart? e.g. Is it possible to change levodopa to dextrodopa?

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Related, but not a duplicate:… – jonsca Feb 11 '13 at 5:37
up vote 6 down vote accepted

Yes,this process is called stereoconversion. Both SN1 and SN2 reactions can invert the chirality of a carbon center. In the case of the SN1 mechanism, departure of a leaving group generates a prochiral intermediate which is then open to nucleophilic attack on either face. If both faces are equally acessible to the nucleophile this will result in racemisation of the molecule. In the case of the SN2 reaction the nucleophile will enter a trigonal bipyramidal intermediate with the molecule. Assuming the nucleophile has the same stereochemical precedence as the leaving group, the stereochemistry will be inverted.

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Doesn't "dextrorotatory compound to levorotatory counterpart" mean we have to preserve the groups on the chiral carbon, rather than substituting one of the groups with another. – Aditya Sriram Jan 14 '13 at 3:12
@AdityaSriram - Yes, but if the leaving group and nucleophile are the same species, you're in the clear. This can happen in the case of keto-enol tautomerism at a stereocenter. – Richard Terrett Jan 14 '13 at 4:03
Adding an example to your answer will really help a lot. – Aditya Sriram Jan 14 '13 at 5:29
@RichardTerrett I think Aditya Sriram is right you need to add an example not just of how you would change the chirality from dex to levo but how you would preserve the groups on the chiral carbon. – Brenton Horne Jan 14 '13 at 5:42

The following example(s) might help to illustrate the stereoconversion described by Richard Terrett:

stereoconversion of alcohol

In the case of alcohols, the "classical" method is the Mitsunobu reaction, using triphenylphosphine ($\ce{Ph3P})$ and diethyl azodicarboxylate (DEAD, $\ce{EtO2C-N=N-CO2Et})$ as reagents, and water as a nucleophile to quench the intermediate with inversion of configuration.

In the course of the reaction, $\ce{Ph3P}$ adds to DEAD in the presence of a proton donor to form an intermediate.

Mitsunobu intermediate

The latter adds to the alcohol, forming a stable $\ce{P-O}$ bond.

alcohol Ph3P adduct

The addition of a nucleophile proceeds according to $S_N2$ with inversion of configuration, the formation of triphenylphosphine oxide renders the reaction irreversible. enter image description here

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