①A+Na→慢慢产生气泡,说明A中含有-OH; ②A+CH3COOH(加热条件下)有香味的物质,说明A和乙酸发生酯化反应,结合分子式可知A应为丙醇,可为CH3CH2CH2OH 或CH3CH(OH)CH3, (1)由以上分析可知A可为CH3CH2CH2OH 或 CH3CH(OH)CH3,故答案为:CH3CH2CH2OH或CH3CH(OH)CH3; (2)CH3CH2CH2OH 或 CH3CH(OH)CH3在浓硫酸作用下发生消去反应生成的有机产物都是CH3CH=CH2, 故答案为:CH3CH=CH2; (3)A中含有羟基,可与乙酸发生酯化反应,生成产物可为CH3COOCH2CH2CH3 或 CH3COOCH(CH3)CH3, 故答案为:CH3COOCH2CH2CH3或CH3COOCH(CH3)CH3. |