[化学---选修5:有机化学基础]有机物A为芳香烃类化合物,质谱分析表明其相对分子质量为104,分子中含有碳 【化学——选修5:有机化学基础】(15分)已知某有机物A为烃...
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有机物A为芳香烃类化合物,其相对分子质量为104,分子中C原子最大数为
不饱和度为5,分子中含有碳碳双键,故A为 ,A发生加聚反应生成高分子化合物聚苯乙烯,故G的结构为 .D能与氢气发生加成反应,能发生银镜反应,结构A的结构可知,D为 ,E为 ,C为 ,C与HBr酸,发生取代反应生成B,故B为 ,B发生消去反应生成A.F为有香味的有机化合物,故F为苯乙酸苯乙酯, (1)由上述分析可知,D为 ,所含官能团为-CHO,故答案为:-CHO; (2)反应Ⅱ是 HBr酸发生取代反应生成 , 由上述分析可知G的结构简式为 ,故答案为:-CHO; ; (3)反应I方程式为: +KOH
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