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写出下列物质转化的化学方程式,并指出反应类型.(1)溴乙烷→乙烯CH3CH2BrNaOH/醇△CH2=CH2↑+HBrCH3CH2BrNaOH/醇△CH2=CH2↑+HBr,;(2)甲苯→TNT,;(3)乙醛+银氨溶液→CH3CHO+2Ag

题目详情
写出下列物质转化的化学方程式,并指出反应类型.
(1)溴乙烷→乙烯
CH3CH2Br
NaOH/醇
CH2=CH2↑+HBr
CH3CH2Br
NaOH/醇
CH2=CH2↑+HBr
,______;
(2)甲苯→TNT______,______;
(3)乙醛+银氨溶液→
CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3
CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3
,______.

CH3CH2Br
NaOH/醇
CH2=CH2↑+HBr
32
NaOH/醇
CH2=CH2↑+HBr
NaOH/醇
NaOH/醇NaOH/醇
△△22
CH3CH2Br
NaOH/醇
CH2=CH2↑+HBr
32
NaOH/醇
CH2=CH2↑+HBr
NaOH/醇
NaOH/醇NaOH/醇
△△22

CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3
332
CH3COHNH4+H2O+2Ag+3NH3
△△
3423
CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3
332
CH3COHNH4+H2O+2Ag+3NH3
△△
3423
▼优质解答
答案和解析
(1)溴乙烷发生消去反应生成乙烯,该反应为CH33CH22Br
NaOH/醇
CH2=CH2↑+HBr,
故答案为:CH3CH2Br
NaOH/醇
CH2=CH2↑+HBr;消去反应;
(2)甲苯发生取代反应生成TNT,该反应为
故答案为:;取代反应;

(3)发生银镜反应生成乙酸铵、水、Ag、氨气,该反应为CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3↑,
故答案为:CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3↑;银镜反应.
NaOH/醇
NaOH/醇NaOH/醇NaOH/醇
△△△CH22=CH22↑+HBr,
故答案为:CH33CH22Br
NaOH/醇
CH2=CH2↑+HBr;消去反应;
(2)甲苯发生取代反应生成TNT,该反应为
故答案为:;取代反应;

(3)发生银镜反应生成乙酸铵、水、Ag、氨气,该反应为CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3↑,
故答案为:CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3↑;银镜反应.
NaOH/醇
NaOH/醇NaOH/醇NaOH/醇
△△△CH22=CH22↑+HBr;消去反应;
(2)甲苯发生取代反应生成TNT,该反应为
故答案为:;取代反应;

(3)发生银镜反应生成乙酸铵、水、Ag、氨气,该反应为CH33CHO+2Ag(NH33)22OH
CH3COHNH4+H2O+2Ag+3NH3↑,
故答案为:CH3CHO+2Ag(NH32OH
CH3COHNH4+H2O+2Ag+3NH3↑;银镜反应.
△△△
CH33COHNH44+H22O+2Ag+3NH33↑,
故答案为:CH33CHO+2Ag(NH33)22OH
CH3COHNH4+H2O+2Ag+3NH3↑;银镜反应.
△△△
CH33COHNH44+H22O+2Ag+3NH33↑;银镜反应.
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