A new bis-zinc(II) receptor (Zn(2)L), which has two macrocyclic 12-membered tetraamine (cyclen) Zn-II complexes connected through a p-xylene bridge, has been synthesized as a novel host molecule to recognize barbiturates (such as barbital (bar)) in aqueous solution. Each of the zinc(rr) ions in the bis-zinc(II) receptor was originally intended to the dianionic barbital anion (bar(2-) with supplementary hydrogen bonds between the cyclen NH's and the three carbonyl oxygens ill complementary positions to yield a 1:1 complex, Zn(2)L-bar(2-). From an aqueous solution of equimolar Zn(2)L and barbital at pH 8, however, a cyclic 2:2 complex, (Zn(2)L-bar(2-))(2), was isolated and characterized by X-ray crystal analysis. The NMR study in 10% (v/v) D2O/H2O has revealed dissociation of (Zn(2)L-bar(2)(2-) solely into the original target 1:1 complex Zn(2)L-bar(2-) and established the dimerization constant For 2Zn(2)L-bar(2-) reversible arrow (Zn(2)L-bar(2-))(2), K-d (=[(Zn(2)L-bar(2-))(2)]/[Zn(2)L-bar(2-)](2)) to be 10(3.4)M(-1). The thermodynamic parameters were evaluated from the NMR measurements at 25, 35, 45, and 55 degrees C: Delta G = -1.9 x 10(4) J mol(-1), Delta H = - 3.3 x 10(4) J mol(-1), Delta S = - 49 Jmol(-1)K(-1) at 25 degrees C, Potentiometric pH titration of Zn(2)L (1 mM) and barbital (1 mM) disclosed extremely facile deprotonation of the two imido groups of barbital at pH less than 7 to form the dianionic barbital-bound Zn-II complexes Zn(2)L-bar(2-) and (Zn(2)L-bar(2-))(2), whereby the barbital binding affinity for Zn(2)L was estimated to be K-bar (=[Zn(2)L-bar(2-)]/[uncomplexed Zn(2)L][uncomplexed barbital]) = 10(5.8)M(-1) at pH 8 and 25 degrees C with I = 0.10 (NaNO3). The significance of the bis-zinc(II) receptor in stabilizing the dianionic barbital is evident by comparison with the interaction of Zn-II cyclen complex (ZnL) with barbital, which yields only a 1:1 monoanionic barbital complex, ZnL-bar(-) (K-bar = [ZnL-bar(-)]/[uncomplexed ZnL][uncomplexed barbital] = 10(4.2) M(-1) at pH 8 and 25 degrees C with I = 0.10(NaNO3)).