A new class of anti-thrombosis hexahydropyrazino-[1′,2′:1,6]pyrido-[3,4-b]-indole-1,4-dions: Design, synthesis, log K determination, and QSAR analysis

Based on the fact that the cyclization of N-[(3S)]-1,2,3,4-tetrahydro-β-carboline-3-carboxyl]-l-lysine in both of acetic acid aqueous (5%) and rat plasma gave the same product and the hypothesis that the cyclization product is antithrombotic active, we report the synthesis, in vitro anti-aggregation, and in vivo anti-thrombosis activity of 20 hexahydropyrazino[1′,2′:1,6]pyrido[3,4-b]indole-1,4-dions (5a–t) as potential anti-thrombosis agents in this study. Two intermediates (tetrahydro-β-carboline-3-carboxy-l-amino acid benzylesters, 2-aminoacyltetrahydro-β-carboline-3-carboxylic acid benzylesters) were prepared and used for the cyclization to form 5a–t. Coupling hydrochloric acid salt of tetrahydro-β-carboline-3-carboxylic acid esters and Boc-amino acids in the reported literature usually generates very low yield products accompanied by racemization. However, in our case, the free base of tetrahydro-β-carboline-3-carboxylic acid benzylester produced the desired products in high yields and without racemization. The anti-thrombosis results from both in vitro and in vivo studies revealed that 5a–t may be a new class of anti-thrombosis agents with potent effective concentration at 0.5 μmol/kg with oral administration. Moreover, a QSAR analysis was performed on these 20 compounds by using molecular descriptors generated by e-dragon server. Although the activities of these compounds are weakly correlated with the log P values, the current QSAR analysis revealed that the anti-thrombotic activity of these compounds can be explained by their steric and electrostatic effects.

In 5a, R1 = CH3; 5b, R = CH2C6H5; 5c, R = CH(CH3)2; 5d, R = CH2OH; 5e, R = CH(OH)CH3; 5f, R = CH2C6H4–OH–p; 5g, R = tetrahydropyrrol-2-yl; 5h, R = CH2SH; 5i, R = CH2CH2SCH3; 5j, R = CH2CH2CO2H; 5k, R = CH2CO2H; 5l, R = 1,3-imidazol-5-methylene; 5m, R = indol-3-yl-methylene; 5n, R = CH2(CH2)2NHC(NH2)NH; 5o, R = H; 5p, R = CH2CH2CH2CH2NH2; 5q, R = CH2CH2CONH2; 5r, R = CH2CONH2; 5s, R = CH2CH(CH3)2; 5t, R = CH(CH3)CH2CH3.Figure optionsDownload as PowerPoint slide