“…The ethyl ester (2) (5.64 g) was converted to the free P-chloro-L-alanine hydrochloride (3) by refluxing 1 h in 2 N HCI (60 ml) (13). The mixture was repeatedly evaporated to dryness with the addition of water and finally ethanol, and the residue was crystallized from methanolether; yield, 4.0 g (go%), [a], -10.8" (3.2, water containing one equivalent of NaOH), corresponding to [a], -14.0" (2.5, water) for the free acid, lit.…”