Elfspice incorrectly assumed that the NaOH was added to the solution to freebase the added amine hydrochloride. It is not. The freebase amine is to be used, and the purpose of the NaOH is to make the monosodium salt of the tartaric acid (which is a dicarboxylic acid).
In solution, the monosodium d-tartrate then preferentially forms a water-soluble salt with d-methamphetamine while the l-methamphetamine (still remaining in its freebase form) can be extracted with a non-polar solvent.
The reason all this happens is that because of the molecular shape of d-tartaric acid is such that
in solution it fits better together with d-methamphetamine than with l-methamphetamine, given a choice (but will freely form a salt with both forms if enough acid is present, therefore one carboxylic acid group on each tartaric acid molecule is 'disabled' by the addition of one equivalent tof NaOH).
(06-13-04): A stereoisomeric confusion by me corrected above (thanks for spotting it, bio!).
I incorrectly assumed that the preferentially formed salt was l-meth d-tartrate (as this is the salt isolated in many resolutions through fractional crystallizations). The correct interpretation of Rusznak's procedure has been entered above.
Also, note the following important discrepancy between the isomer designations:
The dextro isomer of tartaric acid is the d, (+),
L or S isomer; the levo isomer is the l, (-),
D or R isomer.
The dextro isomer of (meth)amphetamine is the d, (+),
D or S isomer; the levo isomer is the l, (-),
L or R isomer.