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1.4,4＇-二硝基二苯甲烷(Ⅰ)與硫酸亚碘醯(Ⅱ)作用,生成一种新型的含碘杂環化合物(Ⅲ_a)。 2.化合物Ⅲ_a與氯化鈉、溴化鈉、碘化鉀或苦味酸作用,置换成为相應的鹵化物(Ⅲ_(b-d))或苦味酸盐(Ⅲ_e)。 3.用鹼性高錳酸鉀溶液氧化化合物Ⅲ_a得2-碘代-4-硝基苯甲酸。 4.碘化物(Ⅲ_d)在它的熔點温度进行熱解,得2,2＇-二碘代-4,4＇-二硝基二苯甲烷。继还原,得相應的二氨基化合物。二者均为新化合物。 5.化合物Ⅲ_a用氢氧化鈉溶液处理,得到蓝绿色物質,溶於乙酸乙酯、丙酮、吡啶中成鲜明的蓝绿色溶液。乙酸乙酯溶液酸化後变成黄棕色,再鹼化復现蓝绿色。将化合物Ⅲ_a在无水吡啶中加熱,亦呈现蓝绿色。 6.在类似的条件下,二苯甲烷與硫酸亚碘醯起作用。
The reaction of 1,4,4’-dinitrodiphenylmethane (Ⅰ) with iodine (Ⅱ) sulfate produced a novel iodine-containing heterocyclic compound (Ⅲ_a). 2. Compound III_a interacts with sodium chloride, sodium bromide, potassium iodide or picric acid to replace the corresponding halide (Ⅲ_ (b-d)) or picrate (Ⅲ_e). 3. With alkaline potassium permanganate oxidation of compound III_a 2-iodo-4-nitrobenzoic acid. 4. Iodide (III_d) is pyrolyzed at its melting point to give 2,2’-diiodo-4,4’-dinitrodiphenylmethane. Subsequent reduction, the corresponding diamino compounds. Both are new compounds. 5. Compound III_a treated with sodium hydroxide solution to give a blue-green material, dissolved in ethyl acetate, acetone, pyridine into a clear blue-green solution. Acidified ethyl acetate solution into a yellow-brown, then alkaline blue-green reappear. Compound III_a was heated in anhydrous pyridine, also showing a blue-green color. Under similar conditions, diphenylmethane works with iodine sulphate.