Cl.[CH:2]1([N:8]([CH2:10][C:11]#[C:12][C:13]2[CH:18]=[CH:17][C:16]([CH:19]3[CH2:24][CH2:23][CH2:22][CH2:21][CH2:20]3)=[C:15]([Cl:25])[CH:14]=2)[CH3:9])[CH2:7][CH2:6][CH2:5][CH2:4][CH2:3]1>C(OCC)(=O)C.CO.[Pd]>[ClH:25].[CH:2]1([N:8]([CH2:10]/[CH:11]=[CH:12]\[C:13]2[CH:18]=[CH:17][C:16]([CH:19]3[CH2:24][CH2:23][CH2:22][CH2:21][CH2:20]3)=[C:15]([Cl:25])[CH:14]=2)[CH3:9])[CH2:3][CH2:4][CH2:5][CH2:6][CH2:7]1A solution containing 13.1 g of the compound prepared in Example 1, in the form of the base, and 1.2 g of 5% palladium on barium sulfate in 100 ml of ethyl acetate and 5 ml of methanol is hydrogenated at normal temperature and pressure. The volume of hydrogen absorbed is 650 ml. After filtration and concentration of the solution, the residue is taken up in ethyl ether and the hydrochloride is precipitated by bubbling hydrogen chloride. 5.6 g of the expected compound are obtained after recrystallization from acetonitrile. Yield: 43%.
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