The (S)-configured alpha-bromo carboxylate shown is heated in very dilute aqueous base. Assume the carboxylate oxygen participates (neighbouring group participation through a three-membered ring) before hydroxide or water attacks. What is the product?

The compound is (S)-2-bromopropanoate. The carboxylate displaces bromide from the back side (inversion) to give an alpha-lactone. The nucleophile then opens the lactone at the same carbon from the back side (second inversion). Two inversions give net retention: the lactate has the same arrangement. CIP labels: bromo compound S, lactate S (OH > COO- > CH3 > H, same ranking order as Br > COO- > CH3 > H), so the product is (S)-lactate.