2022学年高考化学模拟题汇编专题18选修5有机化学基础【含答案】

专题18选修5有机化学基础/r/n1．（福建省福州高三/r/n模拟/r/n）β-/r/n羰基酯化合物是重要的化工合成原料，在有机合成工业和制药工业具有广泛的用途。某/r/nβ-/r/n羰基酯化合物/r/nG/r/n的合成路径如图：/r/n已知：/r/nI.CH/r/n3/r/nCOOC/r/n2/r/nH/r/n5/r/nCH/r/n3/r/nCOCH/r/n2/r/nCOOC/r/n2/r/nH/r/n5/r/nII./r/n回答下列问题：/r/n(1)B/r/n的化学名称为/r/n_______/r/n。/r/n(2)C/r/n到/r/nD/r/n的反应类型为/r/n_______/r/n。/r/n(3)C/r/n中的官能团名称/r/n_______/r/n。/r/n(4)/r/n写出反应/r/nE/r/n到/r/nF/r/n的化学方程式/r/n_______/r/n。/r/n(5)/r/n写出/r/nG/r/n的结构简式/r/n_______/r/n。/r/n(6)/r/n化合物的/r/nF/r/n的同分异构体/r/nH/r/n能同时满足以下三个条件，/r/nH/r/n的结构简式/r/n_______/r/n。/r/n(/r/nⅰ/r/n)/r/n不能发生银镜反应；/r/n(/r/nⅱ/r/n)/r/n易水解，且/r/n1molH/r/n和/r/n3molNaOH/r/n反应；/r/n(/r/nⅲ/r/n)/r/n苯环上二取代，且核磁共振氢谱峰面积比为/r/n1/r/n：/r/n1/r/n：/r/n2/r/n：/r/n2/r/n：/r/n6/r/n。/r/n(1)/r/n苯甲醚/r/n(/r/n2/r/n)/r/n取代反应/r/n(/r/n3/r/n)/r/n氯原子、醚键/r/n(/r/n4/r/n)CH/r/n3/r/nOH+/r/nH/r/n2/r/nO+/r/n(/r/n5/r/n)/r/n(/r/n6/r/n)/r/n/r/n【分析】/r/n根据给定信息/r/nII/r/n结合/r/nC/r/n的结构简式可逆推知/r/nB/r/n为/r/n，所以可知/r/nA/r/n为苯酚，/r/nC/r/n与/r/nNaCN/r/n发生取代反应生成/r/nD/r/n，/r/nD/r/n在酸性条件下可生成/r/nE/r/n（/r/n），其分子式为/r/nC/r/n9/r/nH/r/n10/r/nO/r/n3/r/n，/r/nE/r/n与甲醇在浓硫酸作用下可发生酯化反应生成/r/nF/r/n，根据给定条件/r/nI/r/n可知，/r/nG/r/n的结构简式为/r/n，据此分析解答。/r/n根据上述分析可知，/r/n(1)B/r/n为/r/n，其化学名称为苯甲醚；/r/n(2)C/r/n与/r/nNaCN/r/n发生反应，/r/n-CN/r/n替换原来的/r/n-Cl/r/n转化为/r/nD/r/n，所以其反应类型为取代反应；/r/n(3)C/r/n的结构简式为：/r/n，其中的官能团名称为氯原子、醚键；/r/n(4)E/r/n与甲醇在浓硫酸作用下可发生酯化反应生成/r/nF/r/n，其化学方程式为：/r/nCH/r/n3/r/nOH+/r/nH/r/n2/r/nO+/r/n；/r/n(5/r/n根据上述分析可知，/r/nG/r/n的结构简式为：/r/n；/r/n(6)F/r/n为/r/n，其分子式为：/r/nC/r/n10/r/nH/r/n12/r/nO/r/n3/r/n，不饱和度为/r/n5/r/n，其同分异构体/r/nH/r/n不能发生银镜反应，则没有醛基；易水解，且/r/n1mol/r/n该和/r/n3molNaOH/r/n反应，说明分子中可能含酯基或羧基，结合/r/nO/r/n原子个数为/r/n3/r/n，可推知分子结构中含/r/n；又分子中核磁共振氢谱峰面积比为/r/n1/r/n：/r/n1/r/n：/r/n2/r/n：/r/n2/r/n：/r/n6/r/n，则说明含有/r/n2/r/n个甲基，所以推知/r/nH/r/n的结构简式为：/r/n。/r/n2．（广东珠海市/r/n高三/r/n二模）/r/n化合物/r/nG/r/n是合成一种心血管药物的中间体，其合成路线如下：/r/n已知：/r/n+/r/n/r/n+H/r/n2/r/nO/r/n回答下列问题：/r/n(/r/n1/r/n)/r/nB/r/n的化学名称是/r/n______/r/n，/r/nC/r/n反应生成/r/nD/r/n的反应类型是/r/n______/r/n，/r/nF/r/n的含氧官能团名称是/r/n_________/r/n。/r/n(/r/n2/r/n)物质/r/nG/r/n的结构中有两个六元环，其结构简式为/r/n_______________/r/n(/r/n3/r/n)写出/r/nB/r/n反应生成/r/nC/r/n的化学方程式/r/n____________/r/n。/r/n(/r/n4/r/n)/r/nD/r/n反应生成/r/nE/r/n时用/r/nNa/r/n2/r/nCO/r/n3/r/n水溶液而不用/r/nNaOH/r/n溶液的原因是/r/n________________/r/n。/r/n(/r/n5/r/n)/r/nH/r/n是/r/nE/r/n的同分异构体，写出符合下列条件的/r/nH/r/n的结构简式(写出一种即可)/r/n_______________/r/n。/r/n①/r/n1/r/n/r/nmol/r/n/r/nH/r/n可以消耗/r/n1/r/n/r/nmol/r/n/r/nNa/r/n生成氢气/r/n②/r/n1/r/n/r/nmol/r/n/r/nH/r/n可以和/r/n3/r/n/r/nmol/r/n/r/nNaOH/r/n反应/r/n③核磁共振氢谱共有/r/n4/r/n个吸收峰，峰面积之比为/r/n6/r/n:/r/n2/r/n:/r/n1/r/n:/r/n1/r/n(/r/n6/r/n)设计由甲醇和乙酸乙酯制备丙酸乙酯(/r/n)的合成路线(无机试剂任选)/r/n_______________/r/n。/r/n（1）/r/n邻甲基苯酚(或/r/n2/r/n-甲基苯酚)/r/n/r/n取代反应/r/n/r/n酯基、醛基/r/n/r/n（2）/r/n/r/n/r/n（3）/r/n/r/n+/r/nCH/r/n3/r/nCOCl/r/n→/r/n+/r/nHCl/r/n（4）/r/n防止酯基在/r/nNaOH/r/n条件下被水解/r/n/r/n（5）/r/n/r/n或/r/n/r/n（6）/r/nCH/r/n3/r/nOH/r/nHCHO/r/nCH/r/n2/r/n=/r/nCHCOOC/r/n2/r/nH/r/n5/r/nCH/r/n3/r/nCH/r/n2/r/nCOOC/r/n2/r/nH/r/n5/r/n/r/n【分析】A/r/n和溴甲烷发生取代反应生成/r/nB/r/n，根据/r/nC/r/n的分子式和/r/nD/r/n的结构简式可知/r/nC/r/n的结构简式是/r/n，/r/nC/r/n发生取代反应生成/r/nD/r/n，/r/nD/r/n水解生成/r/nE/r/n，/r/nE/r/n发生催化氧化生成/r/nF/r/n，/r/nF/r/n发生已知信息反应生成/r/nG/r/n为/r/n，据此解答。/r/n(/r/n1/r/n)根据/r/nB/r/n的结构简式可知/r/nB/r/n的化学名称是邻甲基苯酚(或/r/n2/r/n-甲基苯酚，/r/nC/r/n反应生成/r/nD/r/n是甲基上的氢原子被溴原子取代，反应类型是取代反应，根据/r/nF/r/n的结构简式可知/r/nF/r/n分子中含氧官能团名称是酯基、醛基。/r/n(/r/n2/r/n)物质/r/nG/r/n的结构中有两个六元环，其结构简式为/r/n。/r/n(/r/n3/r/n)根据以上分析可知/r/nB/r/n反应生成/r/nC/r/n的化学方程式为/r/n+/r/nCH/r/n3/r/nCOCl/r/n→/r/n+/r/nHCl/r/n。/r/n(/r/n4/r/n)由于酯基能和氢氧化钠反应，所以/r/nD/r/n反应生成/r/nE/r/n时用/r/nNa/r/n2/r/nCO/r/n3/r/n水溶液而不用/r/nNaOH/r/n溶液的原因是防止酯基在/r/nNaOH/r/n条件下被水解。/r/n(/r/n5/r/n)①/r/n1/r/n/r/nmol/r/n/r/nH/r/n可以消耗/r/n1/r/n/r/nmol/r/n/r/nNa/r/n生成氢气，说明含有/r/n1/r/n个羟基或羧基；②/r/n1/r/n/r/nmol/r/n/r/nH/r/n可以和/r/n3/r/n/r/nmol/r/n/r/nNaOH/r/n反应，说明含有酚羟基以及酚羟基形成的酯基；③核磁共振氢谱共有/r/n4/r/n个吸收峰，峰面积之比为/r/n6/r/n:/r/n2/r/n:/r/n1/r/n:/r/n1/r/n，则满足条件的结构简式为/r/n或/r/n；/r/n(/r/n6/r/n)首先甲醇发生催化氧化生成甲醛，甲醛和乙酸乙酯发生已知信息反应生成/r/nCH/r/n2/r/n=/r/nCHCOOC/r/n2/r/nH/r/n5/r/n，然后加成即可得到/r/nCH/r/n3/r/nCH/r/n2/r/nCOOC/r/n2/r/nH/r/n5/r/n，合成路线为/r/nCH/r/n3/r/nOH/r/nHCHO/r/nCH/r/n2/r/n=/r/nCHCOOC/r/n2/r/nH/r/n5/r/nCH/r/n3/r/nCH/r/n2/r/nCOOC/r/n2/r/nH/r/n5/r/n。/r/n3．（广东汕头市·金山中学高三三模）/r/n扁桃酸衍生物是重要的医药中间体，以/r/nA/r/n和/r/nB/r/n为原料合成扁桃酸衍生物/r/nF/r/n路线如下：/r/n(1)A/r/n与/r/nB/r/n的反应为加成反应，且/r/nA/r/n的分子式为/r/nC/r/n2/r/nH/r/n2/r/nO/r/n3/r/n，可发生银镜反应，且具有酸性，则/r/nB/r/n的结构简式是/r/n____/r/n。/r/n(2)/r/n试写出/r/nC/r/n与/r/nNaOH/r/n反应的化学方程式：/r/n___/r/n。/r/n(3)E/r/n是由/r/n2/r/n分子/r/nC/r/n生成的含有/r/n3/r/n个六元环的化合物，/r/nE/r/n的分子中不同化学环境的氢原子有/r/n___/r/n种。/r/n(4)/r/n下列说法正确的是/r/n___/r/nA．F/r/n常温下易溶于水/r/nB．1molF/r/n在一定条件下与足量/r/nNaOH/r/n溶液反应，最多消耗/r/nNaOH/r/n的物质的量为/r/n3mol/r/n。/r/nC．D→F/r/n的反应类型是取代反应/r/nD．/r/n一定条件下/r/nC/r/n可以发生缩聚反应/r/n(5)/r/n符合下列条件的/r/nF/r/n的同分异构体/r/n(/r/n不考虑立体异构/r/n)/r/n有/r/n___/r/n种。其中氢原子个数比为/r/n1:1:2:2:3/r/n的结构简式为/r/n____/r/n①属于一元酸类化合物；②苯环上只有/r/n2/r/n个取代基且处于对位；③遇氯化铁溶液发生显色反应。/r/n(6)/r/n己知：/r/n+CH≡CH/r/n，请设计合成路线以/r/nB/r/n和/r/nC/r/n2/r/nH/r/n2/r/n为原料合成/r/n(/r/n无机试剂任选/r/n)_____/r/n。/r/n（1）/r/n/r/n（2）/r/n/r/n+2NaOH→/r/n+2H/r/n2/r/nO/r/n（3）/r/n4/r/n/r/n（4）/r/nBCD/r/n（5）/r/n4/r/n/r/n（6）/r/n/r/n【分析】/r/nA/r/n与/r/nB/r/n的反应为加成反应，且/r/nA/r/n的分子式为/r/nC/r/n2/r/nH/r/n2/r/nO/r/n3/r/n，可发生银镜反应，且具有酸性，则/r/nA/r/n为/r/nOHCCOOH/r/n，/r/nB/r/n为/r/n；/r/nC/r/n与甲醇发生酯化反应得到/r/nD/r/n，/r/nD/r/n和/r/nHBr/r/n发生取代反应得到/r/nF/r/n；由于/r/nC/r/n中既含醇羟基，也含羧基，因此会得到副产物如两个/r/nC/r/n分子生成的环酯等。/r/n(1)/r/n由分析可知/r/nB/r/n为/r/n；/r/n(2)C/r/n为/r/n，羧基、酚羟基均能与/r/nNaOH/r/n按/r/n1:1/r/n反应，则/r/nC/r/n与/r/nNaOH/r/n反应的化学方程式为/r/n+2NaOH→/r/n+2H/r/n2/r/nO/r/n；/r/n(3)E/r/n是由/r/n2/r/n分子/r/nC/r/n生成的含有/r/n3/r/n个六元环的化合物，则/r/nE/r/n为/r/n2/r/n分子/r/nC/r/n生成的环酯，/r/nE/r/n为/r/n，则/r/nE/r/n分子中不同化学环境的氢原子有/r/n4/r/n种；/r/n(4)A/r/n．/r/nF/r/n含酯基和/r/n-Br/r/n、酚羟基，常温下不易溶于水，/r/nA/r/n错误；/r/nB/r/n．/r/n1molF/r/n含/r/n1mol/r/n酚羟基、/r/n1mol/r/n酯基、/r/n1mol/r/n溴原子，则在一定条件下/r/n1molF/r/n与足量/r/nNaOH/r/n溶液反应，最多消耗/r/nNaOH/r/n的物质的量为/r/n3mol/r/n，/r/nB/r/n正确；/r/nC/r/n．对比/r/nD/r/n和/r/nF/r/n的结构可知/r/nD→F/r/n的反应类型是取代反应/r/n(H/r/n被/r/nBr/r/n取代/r/n)/r/n，/r/nC/r/n正确；/r/nD/r/n．/r/nC/r/n含一个羧基和一个醇羟基，一定条件下/r/nC/r/n可以发生缩聚反应生成聚酯类物质；/r/nC/r/n含酚羟基，且酚羟基邻位有/r/nH/r/n，一定条件下可发生酚醛缩聚，/r/nD/r/n正确；/r/n选/r/nBCD/r/n；/r/n(5)F/r/n为/r/n，/r/nF/r/n的同分异构体/r/n(/r/n不考虑立体异构/r/n)/r/n：/r/n①属于一元酸类化合物，则含/r/n1/r/n个/r/n-COOH/r/n；/r/n②苯环上只有/r/n2/r/n个取代基且处于对位；/r/n③遇氯化铁溶液发生显色反应，则含酚羟基；/r/n符合条件的有/r/n、/r/n、/r/n、/r/n，共/r/n4/r/n种；其中氢原子个数比为/r/n1:1:2:2:3/r/n的结构简式为/r/n；/r/n(6)/r/n以/r/nB(/r/n)/r/n和/r/nC/r/n2/r/nH/r/n2/r/n为原料合成/r/n，进行逆合成分析：/r/n可由/r/n和/r/nHC≡CH/r/n合成，/r/n可由/r/n催化氧化得到，/r/n由/r/n与氢气加成得到，因此合成路线为/r/n。/r/n4．（湖北武汉市汉阳一中高三二模）/r/n有机化合物/r/nH/r/n可用来制备抗凝血药，可通过如图路线合成。/r/n已知：/r/n①RCOOH/r/n(R/r/n为烃基/r/n)/r/n②/r/n酚羟基一般不易直接与羧酸酯化/r/n请回答：/r/n(1)F/r/n的名称为/r/n_______/r/n。/r/n(2)C+F→G/r/n的反应类型为/r/n_______/r/n；/r/nH/r/n中含氧官能团的名称为/r/n_______/r/n。/r/n(3)/r/n在/r/nA→B/r/n的反应中，检验/r/nA/r/n是否反应完全的试剂是/r/n_______/r/n。/r/n(4)/r/n写出/r/nG/r/n与过量/r/nNaOH/r/n溶液共热时反应的化学方程式/r/n_______/r/n。/r/n(5)/r/n化合物/r/nD/r/n的同分异构体有多种，其中能与/r/nFeCl/r/n3/r/n溶液发生显色反应的结构有/r/n_______/r/n种/r/n(/r/n不包括/r/nD/r/n本身/r/n)/r/n，其中核磁共振氢谱有/r/n4/r/n组峰，且峰面积比为/r/n2/r/n：/r/n2/r/n：/r/n1/r/n：/r/n1/r/n的结构简式为/r/n_______(/r/n任写一种/r/n)/r/n。/r/n(6)/r/n苯甲酸苯酚酯/r/n(/r/n)/r/n是一种重要的有机合成中间体。请根据已有知识并结合相关信息，试写出以苯酚、甲苯为原料制取该化合物的合成路线/r/n(/r/n无机试剂任选/r/n)/r/n：/r/n_______/r/n。/r/n（1）/r/n邻羟基苯甲酸甲酯/r/n/r/n（2）/r/n取代反应/r/n/r/n羟基、酯基/r/n/r/n（3）/r/n/r/n银氨溶液/r/n(/r/n或新制氢氧化铜悬浊液/r/n)/r/n（4）/r/n/r/n+3NaOH/r/nCH/r/n3/r/nCOONa+CH/r/n3/r/nOH+H/r/n2/r/nO+/r/n/r/n（5）/r/n11/r/n/r/n/r/n（6）/r/n/r/n【分析】/r/n物质/r/nA/r/n为乙醛，乙醛催化氧化生成物质/r/nB/r/n乙酸，乙酸和三氯化磷发生取代反应生成乙酰氯，根据/r/nG/r/n的结构简式可推知/r/nE/r/n为甲醇，甲醇和物质/r/nD/r/n发生酯化反应生成物质/r/nF/r/n：邻羟基苯甲酸甲酯，/r/nC/r/n和/r/nF/r/n发生取代反应生成物质/r/nG/r/n：/r/n，最后/r/nG/r/n生成/r/nH/r/n，据此分析答题。/r/n（/r/n1/r/n）根据物质/r/nG/r/n：/r/n可推知/r/nE/r/n为甲醇，/r/nE/r/n和/r/nD/r/n发生酯化反应生成/r/nF/r/n，/r/nF/r/n的名称为：邻羟基苯甲酸甲酯；/r/n（/r/n2/r/n）/r/nC+F→G/r/n发生取代反应，生成/r/nG/r/n和/r/nHCl/r/n；/r/nH/r/n的结构为：/r/n，其中含氧官能团为：酯基、羟基；/r/n（/r/n3/r/n）/r/nA/r/n为乙醛，可以和银氨溶液发生银镜反应，也可以和新制氢氧化铜悬浊液产生砖红色沉淀，利用此性质检验乙醛的存在，故银氨溶液/r/n(/r/n或新制氢氧化铜悬浊液/r/n)/r/n；/r/n（/r/n4/r/n）/r/nG/r/n为：/r/n，/r/nG/r/n中含有酯基，与过量/r/nNaOH/r/n溶液共热时会发生水解反应，故/r/n+3NaOH/r/nCH/r/n3/r/nCOONa+CH/r/n3/r/nOH+H/r/n2/r/nO+/r/n；/r/n（/r/n5/r/n）物质/r/nD/r/n为：/r/n，其中能与/r/nFeCl/r/n3/r/n溶液发生显色反应的结构有：间羟基苯甲酸、对羟基苯甲酸、/r/n、还可以将羧基拆成一个醛基和一个羟基，两个羟基一个醛基同时连在苯环上，共有/r/n6/r/n种结构，所以同分异构体一共有/r/n11/r/n中，故/r/n11/r/n；/r/n（/r/n6/r/n）根据已知②酚羟基一般不易直接与羧酸酯化，所以设计合成路线时先将甲苯氧化为苯甲酸，苯甲酸发生取代反应生成酰氯，再与苯酚发生取代反应制得目标产物，设计合成路线如下：/r/n。/r/n5．（江苏高三/r/n模拟/r/n）/r/n有机物/r/nE/r/n是一种常用的药物，可通过如图路线合成：/r/n已知：/r/nRCOOH/r/nRCOCl/r/n(1)D→E/r/n的反应类型为/r/n___/r/n。/r/n(2)A/r/n的分子式为/r/nC/r/n4/r/nH/r/n6/r/nO/r/n3/r/n，写出/r/nA/r/n的结构简式/r/n___/r/n。/r/n(3)/r/n写出同时满足下列条件的/r/nC/r/n的一种同分异构体的结构简式/r/n___/r/n。/r/n①不能发生银镜反应，但能与金属钠反应/r/n②分子中只有/r/n3/r/n种不同化学环境的氢/r/n③该分子不能使紫色石蕊溶液变红色/r/n(4)/r/n有机物/r/nE/r/n中，含有/r/n___/r/n个手性碳原子。/r/n(5)/r/n写出以甲苯、/r/nCH/r/n2/r/n(COOC/r/n2/r/nH/r/n5/r/n)/r/n2/r/n和/r/nCH/r/n3/r/nONa/r/n为原料制备/r/n的合成路线流程图/r/n___/r/n。/r/n(/r/n无机试剂任用，合成路线流程图示例见本题题干/r/n)/r/n。/r/n（1）/r/n取代反应/r/n/r/n（2）/r/n/r/n/r/n（3）/r/n/r/n/r/n（4）/r/n1/r/n/r/n【分析】/r/n由合成路线图可知，乙酸甲酯（/r/n）和甲酸乙酯在碱的作用下加热生成/r/nA/r/n，又因为/r/nA/r/n的分子式为/r/nC/r/n4/r/nH/r/n6/r/nO/r/n3/r/n，所以/r/nA/r/n为/r/n，/r/n在氧气的催化氧化下可把醛基转化成羧基，/r/nB/r/n与甲醇酯化可生成/r/n，/r/n与溴乙烷发生取代反应生成/r/nC/r/n（/r/n），/r/n与/r/n在甲醇钠的作用下发生取代反应生成/r/nD/r/n（/r/n），/r/n与/r/n发生取代反应生成/r/nE/r/n（/r/n），据此分析解答。/r/n(1)D→E/r/n为/r/n与/r/n发生取代反应生成/r/n，反应类型为取代反应，故取代反应；/r/n(2)/r/n的分子式为/r/n，根据前后物质推断，/r/n+/r/n+/r/n，则/r/nA/r/n的结构式为/r/n；/r/n(3)/r/n根据要求不能发生银镜反应和不能使紫色石蕊溶液变红色，该物质不能含有/r/n和/r/n官能团/r/n,/r/n能与金属钠反应，说明含有/r/n，同时分子中只有/r/n3/r/n种不同化学环境的氢，则该结构式为/r/n，故/r/n；/r/n(4)/r/n四个键上连接不同的原子或原子团的碳原子称为手性碳原子，由/r/nE/r/n的结构可知，含有/r/n1/r/n个手性碳原子，即/r/n，故/r/n1/r/n；/r/n(5)/r/n以甲苯、/r/nCH/r/n2/r/n(COOC/r/n2/r/nH/r/n5/r/n)/r/n2/r/n和/r/nCH/r/n3/r/nONa/r/n为原料制备/r/n，可以先将甲苯用高锰酸钾氧化成苯甲酸，再用/r/nSOCl/r/n2/r/n将苯甲酸转化成/r/n，/r/n与/r/nCH/r/n2/r/n(COOC/r/n2/r/nH/r/n5/r/n)/r/n2/r/n反应生成/r/n，/r/n在酸性条件下酯基水解生成/r/n，/r/n在加热的情况下生成/r/n，所以合成路线为：/r/n。/r/n6．（天津高三二模）/r/n某药物中间体/r/n(W)/r/n的一种合成路线如下：/r/n

/r/n已知：/r/n回答下列问题：/r/n(1)W/r/n所含官能团名称是/r/n_______/r/n；/r/n的反应类型是/r/n_______/r/n。/r/n(2)Y/r/n的结构简式为/r/n_______/r/n。/r/n(3)M/r/n分子式为/r/n_______/r/n(4)/r/n在/r/nX/r/n的同分异构体中，苯环上含/r/n1/r/n个取代基的结构有/r/n____/r/n种/r/n(/r/n不含/r/nX)/r/n，其中核磁共振氢谱有/r/n4/r/n组吸收峰的结构简式为/r/n_____/r/n。/r/n(5)/r/n化合物/r/nM/r/n含有手性碳原子的数目为/r/n_______/r/n，下列物质不与/r/nM/r/n发生反应的是/r/n_______/r/n。/r/nA．/r/n氢气/r/nB．/r/n溶液/r/nC．/r/n酸性高锰酸钾溶液/r/nD．/r/n银氨溶液/r/n(6)/r/n以甲苯、/r/n为主要原料，合成/r/n(/r/n无机试剂自选/r/n)/r/n，设计合成路线。请写出/r/n“□”/r/n内物质结构简式、/r/n“→”/r/n上反应试剂及条件/r/n_______/r/n。/r/n（1）/r/n羧基/r/n/r/n取代反应/r/n/r/n（2）/r/n/r/n/r/n（3）/r/n/r/n（4）/r/n3/r/n/r/n（5）/r/n1B/r/n（6）/r/n/r/n、/r/n、/r/n/r/n【分析】/r/n根据/r/nX/r/n、/r/nZ/r/n的结构简式以及已知反应可知，/r/nX/r/n与/r/nCH/r/n3/r/nCOCl/r/n发生取代反应，取代的位置为/r/nX/r/n苯环取代基的对位，则/r/nY/r/n的结构简式为/r/n，其他物质已知，根据流程图分析即可。/r/n(1)/r/n根据/r/nW/r/n的结构简式可知，/r/nW/r/n所含官能团名称是羧基；由分析知，/r/n的反应类型是取代反应。/r/n(2)Y/r/n的结构简式为/r/n。/r/n(3)/r/n由/r/nM/r/n的结构简式可知，/r/nM/r/n的分子式为/r/n。/r/n(4)X/r/n苯环上取代基为某丁基，丁基/r/n(-C/r/n4/r/nH/r/n9/r/n)/r/n有/r/n4/r/n种，则在/r/nX/r/n的同分异构体中，苯环上含/r/n1/r/n个取代基的结构，即取代基的种类数决定了这种/r/nX/r/n的同分异构体的种类数，应有/r/n3/r/n种/r/n(/r/n不含/r/nX)/r/n，其中核磁共振氢谱有/r/n4/r/n组吸收峰的结构简式为/r/n。/r/n(5)/r/n手性碳原子是指连接/r/n4/r/n个不一样的原子或原子团的碳原子，则化合物/r/nM/r/n含有/r/n1/r/n个手性碳原子，如图所示/r/n，标注/r/n*/r/n的碳原子为手性碳原子，/r/nA/r/n．/r/nM/r/n含有碳碳双键和醛基，可与氢气发生加成反应；/r/nB/r/n．/r/n溶液不能与/r/nM/r/n发生反应；/r/nC/r/n．/r/nM/r/n含有碳碳双键和醛基，可被酸性高锰酸钾溶液氧化；/r/nD/r/n．/r/nM/r/n含有醛基，可被银氨溶液氧化；故选/r/nB/r/n。/r/n(6)/r/n甲苯与/r/n发生取代反应生成/r/n，/r/n被酸性高锰酸钾氧化生成/r/n，/r/n被氢气还原生成/r/n，所以合成路线为：/r/n。/r/n7．（福建厦门外国语学校高三/r/n模拟/r/n）/r/n辣椒素又名辣椒碱/r/n(capsaicin)/r/n，是常见的生物碱之一、辣椒素/r/nH/r/n的合成路线如图。/r/n请完成下列问题/r/n(1)B/r/n的键线式是/r/n_______/r/n。/r/n(2)E/r/n中官能团的名称是/r/n_______/r/n。/r/n(3)C→D/r/n中反应/r/n1)/r/n的化学方程式是/r/n_______/r/n，反应类型是/r/n_______/r/n。/r/n(4)F/r/n与/r/nG/r/n反应生成/r/nH/r/n时，另一产物为/r/n_______(/r/n填化学式/r/n)/r/n。/r/n(5)/r/n的同分异构体中，同时符合下列条件的有/r/n_______/r/n种/r/n(/r/n不含立体异构/r/n)/r/n。/r/n①/r/n具有四取代苯结构，且核磁共振氢谱显示，其苯环上只有一种化学环境的/r/nH/r/n②/r/n红外光谱测得其分子结构中含有/r/n和/r/n-OH/r/n（1）/r/n(2-/r/n甲基/r/n-7-/r/n溴/r/n-3-/r/n庚烯/r/n)/r/n（1）/r/n/r/n碳碳双键、羧基/r/n/r/n（3）/r/n+2KOH/r/n+2C/r/n2/r/nH/r/n5/r/nOH/r/n取代/r/n(/r/n水解/r/n)/r/n反应/r/n/r/n（4）/r/nHCl/r/n（5）/r/n4/r/n【分析】/r/n物质/r/nA/r/n与/r/n中的溴原子发生取代反应得到物质/r/nB/r/n，物质/r/nB/r/n和/r/n发生取代反应得到物质/r/nC/r/n：/r/n，/r/nC/r/n发生水解反应生成/r/n，再将/r/n酸化得到二元羧酸类物质/r/nD/r/n，/r/nD/r/n发生脱羧反应得到物质/r/nE/r/n，/r/nE/r/n与二氯氧硫发生取代反应的到酰氯类物质/r/nF/r/n，/r/nF/r/n与/r/nG/r/n发生取代反应的目标产物/r/nH/r/n，据此分析答题。/r/n（/r/n1/r/n）/r/nA/r/n中的醇羟基与/r/n中的溴原子发生取代反应，生成/r/nB/r/n，所以/r/nB/r/n的键线式为：/r/n；/r/n（/r/n2/r/n）/r/nE/r/n的键线式为：/r/n，官能团为：碳碳双键、羧基，故碳碳双键、羧基；/r/n（/r/n3/r/n）物质/r/nC/r/n为酯类物质，在碱性条件下发生水解，反应方程式为：/r/n+2KOH/r/n+2C/r/n2/r/nH/r/n5/r/nOH/r/n，故/r/n+2KOH/r/n+2C/r/n2/r/nH/r/n5/r/nOH/r/n；水解反应或取代反应；/r/n（/r/n4/r/n）/r/nF/r/n与/r/nG/r/n反应发取代反应生成/r/nH/r/n时，同时生成小分子/r/nHCl/r/n，故/r/nHCl/r/n；/r/n（/r/n5/r/n）/r/n的同分异构体中符合/r/n①/r/n具有四取代苯结构，且核磁共振氢谱显示，其苯环上只有一种化学环境的/r/nH②/r/n红外光谱测得其分子结构中含有/r/n和/r/n-OH/r/n条件的有以下四种：/r/n、/r/n、/r/n、/r/n，故/r/n4/r/n。/r/n8．（辽宁铁岭市/r/n高三/r/n二模）/r/n瑞德西韦是一种核苷类似物，研究表明它对治疗新冠病毒具有一定的治疗效果，其中/r/nK/r/n为合成瑞德西韦过程中重要的中间体，其制备方法如下图所示。请回答：/r/n已知：①/r/n/r/n/r/n②/r/n(1)A/r/n的结构简式为/r/n___________/r/n；/r/nC→D/r/n的反应类型/r/n___________/r/n。/r/n(2)/r/n写出化合物/r/nG/r/n的名称为/r/n___________/r/n；/r/nJ/r/n中所具有的官能团名称为/r/n___________/r/n。/r/n(3)/r/n写出/r/nE→F/r/n的化学方程式/r/n___________/r/n。/r/n(4)X/r/n是/r/nC/r/n的同分异构体，满足下列条件的/r/nX/r/n的同分异构体有/r/n___________/r/n种，其中苯环上的等效氢最少的结构为/r/n___________/r/n。/r/n①苯环上含有硝基；/r/n②/r/n1molX/r/n可以与足量/r/nNaHCO/r/n3/r/n反应生成/r/n1molCO/r/n2/r/n。/r/n（1）/r/n/r/n取代反应/r/n/r/n（2）/r/n/r/n甲醛/r/n/r/n酯基、氨基/r/n/r/n（3）/r/n/r/n+/r/n+HCl/r/n（4）/r/n13/r/n种/r/n/r/n/r/n【分析】/r/n根据/r/nA/r/n的化学式及/r/nD/r/n的结构简式可知：/r/nA/r/n为/r/n；/r/nA/r/n发生取代反应生成/r/nB/r/n为/r/n，/r/nB/r/n发生硝化反应得/r/nC/r/n为/r/n，/r/nC/r/n发生取代反应得/r/nD/r/n，根据/r/nE/r/n的化学式可知，/r/nD/r/n再发生信息/r/nii/r/n的取代生成/r/nE/r/n为/r/n；结合/r/nF/r/n的化学式可知，/r/nE/r/n和/r/nA/r/n发生取代生成/r/nF/r/n为/r/n，根据/r/nF/r/n和/r/nK/r/n的结构简式可知/r/nJ/r/n为/r/n。/r/nF/r/n和/r/nJ/r/n发生取代反应生成/r/nK/r/n，根据信息/r/ni/r/n可逆推得/r/nG/r/n为/r/nHCHO/r/n，/r/nH/r/n为/r/nHOCH/r/n2/r/nCN/r/n，/r/nI/r/n为/r/nNH/r/n2/r/nCH/r/n2/r/nCOOH/r/n，/r/nG/r/n发生加成得/r/nH/r/n，/r/nH/r/n取代后再水解得/r/nI/r/n，据此分析答题。/r/n根据上述分析可知：/r/nA/r/n为/r/n，/r/nB/r/n为/r/n，/r/nC/r/n为/r/n，/r/nE/r/n为/r/n，/r/nF/r/n为/r/n，/r/nG/r/n为/r/nHCHO/r/n，/r/nH/r/n为/r/nHOCH/r/n2/r/nCN/r/n，/r/nI/r/n为/r/nNH/r/n2/r/nCH/r/n2/r/nCOOH/r/n，/r/nJ/r/n为/r/n。/r/n(1)A/r/n结构简式为/r/n；/r/nC/r/n为/r/n，/r/nC/r/n与/r/nH/r/n3/r/nPO/r/n4/r/n发生取代反应产生/r/nD/r/n：/r/n，故/r/nC→D/r/n的反应类型为取代反应；/r/n(2)/r/n化合物/r/nG/r/n结构简式为/r/nHCHO/r/n，名称为甲醛；化合物/r/nJ/r/n结构简式为/r/n，其中所含官能团名称为酯基、氨基；/r/n(3)/r/n化合物/r/nE/r/n为/r/n，/r/nE/r/n与苯酚发生取代反应产生/r/nF/r/n为/r/n和/r/nHCl/r/n，故该反应的化学方程式为：/r/n+/r/n+HCl/r/n；/r/n(4)/r/n化合物/r/nC/r/n为/r/n，/r/nX/r/n是/r/nC/r/n的同分异构体，满足下列条件：①苯环上含有硝基；②/r/n1molX/r/n可以与足量/r/nNaHCO/r/n3/r/n反应生成/r/n1molCO/r/n2/r/n，说明物质分子中含有/r/n1/r/n个/r/n-COOH/r/n。若取代基为/r/n-NO/r/n2/r/n、/r/n-CH/r/n2/r/nCOOH/r/n，二者在苯环上的位置有邻、间、对三种情况，共有/r/n3/r/n种；若取代基有/r/n-NO/r/n2/r/n、/r/n-CH/r/n3/r/n-COOH/r/n，三种取代基都处于邻位，有/r/n3/r/n种；若有/r/n2/r/n个取代基相邻，/r/n1/r/n个相间，有/r/n3×2/r/n种/r/n=6/r/n种；若三个取代基都相间，只有/r/n1/r/n种情况，故符合条件的同分异构体共有/r/n3+3+6+1=13/r/n种；其中苯环上的等效氢最少，说明含有处于对位的取代基，该物质的结构为/r/n。/r/n9．（山东潍坊市/r/n高三/r/n三模）/r/n苯巴本妥主要用于治疗焦虑、失眠及运动障碍等。一种合成苯巴本的路线如下图所示。/r/n

/r/n已知：①/r/n②/r/n回答下列问题：/r/n(1)/r/n路线中生成/r/nA/r/n的反应类型为/r/n_______/r/n，/r/nD/r/n的结构简式为/r/n_______/r/n。/r/n(2)/r/n写出/r/nB→C/r/n的化学方程式：/r/n_______/r/n。/r/n(3)/r/n同时符合下列条件的/r/nB/r/n的同分异构体有/r/n_______/r/n种/r/n(/r/n不包括立体异构/r/n)/r/n①属于芳香族化合物②遇/r/nFeCl/r/n3/r/n溶液不变紫色③可以发生银镜反应/r/n其中能与金属钠反应生成/r/nH/r/n2/r/n，且核磁共振氢谱显示苯环上有四种不同化学环境的氢原子的同分异构体的结构简式为/r/n_______/r/n。/r/n(4)/r/n设计以/r/nCH/r/n2/r/n=CH/r/n2/r/n和/r/nCO(NH/r/n2/r/n)/r/n2/r/n为原料合成/r/n的路线/r/n_______(/r/n无机试剂任选/r/n)/r/n。/r/n（1）/r/n取代反应/r/n/r/n/r/n（2）/r/n+CH/r/n3/r/nCH/r/n2/r/nOH/r/n+H/r/n2/r/nO/r/n（3）/r/n12/r/n、/r/n/r/n（4）/r/nCH/r/n2/r/n=CH/r/n2/r/nCH/r/n2/r/nBrCH/r/n2/r/nBr/r/nCH/r/n2/r/nCNCH/r/n2/r/nCN/r/nHOOC-CH/r/n2/r/nCH/r/n2/r/n-COOH/r/n/r/n【分析】/r/n与/r/nNaCN/r/n发生取代反应生成/r/nA(/r/n)/r/n，由已知①反应原理知/r/nB/r/n为/r/n，/r/nB/r/n与/r/nCH/r/n3/r/nCH/r/n2/r/nOH/r/n发生酯化反应生成/r/nC(/r/n)/r/n，/r/nC/r/n中酯基的/r/nα-H/r/n与乙二酸二乙酯发生取代反应生成/r/n，/r/n水解生成/r/nD(/r/n)/r/n，根据已知②反应原理推得/r/nE/r/n为/r/n，/r/nE/r/n中羧基的/r/nα-H/r/n与溴乙烷发生取代反应生成/r/nF(/r/n)/r/n，/r/nF/r/n中羧基与/r/n脱水生成苯巴本妥。/r/n(1)/r/n由分析知，生成/r/nA/r/n的反应类型为取代反应；/r/nD/r/n的结构简式为：/r/n；/r/n(2)/r/n由分析知，/r/nB/r/n→/r/nC/r/n为酯化反应，对应方程式为：/r/n+CH/r/n3/r/nCH/r/n2/r/nOH/r/n+H/r/n2/r/nO/r/n；/r/n(3)/r/n由题意知，该同分异构中含有苯环、醛基（或甲酸某酯）等结构，一定不含酚羟基，满足要求的结构如下：/r/n、/r/n、/r/n、/r/n（邻间对三种）、/r/n（邻间对三种）、/r/n（邻间对三种），故共有/r/n12/r/n种结构满足要求；能与金属钠反应，说明有羟基，且苯环上含有四种氢的结构为：/r/n、/r/n；/r/n(4)/r/n由目标产物知，需合成丁二酸，丁二酸可由乙烯先与/r/nBr/r/n2/r/n加成，再与/r/nNaCN/r/n取代，再水解生成，最后利用流程最后一步信息得到目标产物，具体合成路线为：/r/nCH/r/n2/r/n=CH/r/n2/r/nCH/r/n2/r/nBrCH/r/n2/r/nBr/r/nCH/r/n2/r/nCNCH/r/n2/r/nCN/r/nHOOC-CH/r/n2/r/nCH/r/n2/r/n-COOH/r/n。/r/n10．（河北沧州市/r/n高三/r/n三模）/r/n有机物/r/nF(/r/n)/r/n是一种镇痛药，它的一种合成路线如图：/r/n

/r/n(1)/r/n的名称为/r/n_______/r/n。/r/n(2)/r/n的反应类型是/r/n_______/r/n反应。/r/n(3)F/r/n中的含氧官能团的名称为/r/n_______/r/n。/r/n(4)C/r/n的结构简式为/r/n_______/r/n。/r/n(5)/r/n写出反应/r/n的化学方程式：/r/n_______/r/n。/r/n(6)H/r/n为/r/nE/r/n的同分异构体，且/r/nH/r/n分子中含有/r/n结构，则符合条件的/r/nH/r/n有/r/n_______/r/n种/r/n(/r/n不考虑立体异构/r/n)/r/n，写出其中核磁共振氢谱显示有/r/n5/r/n组峰且峰面积之比为/r/n的/r/nH/r/n的结构简式：/r/n_______(/r/n任写一种/r/n)/r/n。/r/n(7)/r/n已知：/r/n。写出以/r/n、/r/n为原料制取/r/n的合成路线流程图/r/n_____/r/n。/r/n(/r/n无机试剂和有机溶剂可任选，合成示例见本题题干/r/n)/r/n（1）/r/n环氧乙烷/r/n/r/n（2）/r/n/r/n取代/r/n/r/n（3）/r/n酯基/r/n/r/n（4）/r/n/r/n/r/n（5）/r/n+CH/r/n3/r/nCH/r/n2/r/nOH/r/nH/r/n2/r/nO+/r/n/r/n（6）/r/n23/r/n/r/n（7）/r/nCH/r/n2/r/n=CH-CH=CH/r/n2/r/nCH/r/n2/r/nBr-CH=CH-CH/r/n2/r/nBr/r/n/r/n【分析】/r/nCH/r/n3/r/nNH/r/n2/r/n与环氧乙烷反应生成/r/n，/r/n与/r/nSOCl/r/n2/r/n发生取代反应得到/r/nC/r/n，则/r/nC/r/n为/r/n，/r/n与/r/n在/r/nNaNH/r/n2/r/n作用下发生取代反应生成/r/n，/r/n发生水解得到/r/n，/r/n与乙醇发生酯化反应合成产物/r/n。/r/n(1)/r/n的名称为环氧乙烷；/r/n(2)/r/n与/r/nSOCl/r/n2/r/n反应得到/r/n，/r/n-OH/r/n被/r/n-Cl/r/n取代，故/r/n的反应类型是取代反应；/r/n(3)/r/n中的含氧官能团的名称为酯基；/r/n(4)B/r/n的分子式为/r/n，结合结构分析可知，由/r/nB/r/n到/r/nC/r/n是氯原子取代了羟基，/r/nC/r/n的结构简式为/r/n；/r/n(5)/r/n与乙醇发生酯化反应生成/r/n，化学方程式为：/r/n+CH/r/n3/r/nCH/r/n2/r/nOH/r/nH/r/n2/r/nO+/r/n；/r/n(6)/r/n除/r/n外，还含有两个甲基或一个乙基，若为两个甲基，采取定一移一法，共有/r/n18/r/n种，若为乙基，则有/r/n5/r/n种，共/r/n23/r/n种；其中核磁共振氢谱显示有/r/n5/r/n组峰且峰面积之比为/r/n的/r/nH/r/n的结构简式为/r/n；/r/n(7)CH/r/n2/r/n=CH-CH=CH/r/n2/r/n和/r/nBr/r/n2/r/n发生/r/n1/r/n，/r/n4/r/n加成反应生成/r/nCH/r/n2/r/nBr-CH=CH-CH/r/n2/r/nBr/r/n，/r/nCH/r/n2/r/nBr-CH=CH-CH/r/n2/r/nBr/r/n和/r/nNCCH/r/n2/r/nCN/r/n发生取代反应生成/r/n，/r/n水解得到/r/n，/r/n受热分解生成/r/n，/r/n和乙醇发生酯化反应生成/r/n，故合成路线为：/r/nCH/r/n2/r/n=CH-CH=CH/r/n2/r/nCH/r/n2/r/nBr-CH=CH-CH/r/n2/r/nBr/r/n。/r/n11．（湖南高三/r/n模拟/r/n）/r/n曲美布汀是一种消化系统药物的有效成分，能缓解各种原因引起的胃肠痉挛，可通过以下路线合成。/r/n

/r/n已知：/r/n回答下列问题：/r/n(1)/r/n曲美布汀分子中含氧官能团的名称是/r/n_____/r/n，①的反应条件是/r/n_____/r/n。/r/n(2)/r/n用星号/r/n(*)/r/n标出/r/nG/r/n的手性碳原子/r/n_____/r/n。/r/n(3)/r/n反应③的反应类型是/r/n_____/r/n。/r/n(4)/r/n反应⑦的化学方程式为/r/n_____/r/n。/r/n(5)/r/n化合物/r/nM/r/n是/r/nE/r/n的同分异构体，已知：①/r/nM/r/n为苯的二元取代物，②/r/nM/r/n能发水解反应且水解产物之一能与/r/n发生显色反应，则/r/nM/r/n的可能结构有/r/n_____/r/n种/r/n(/r/n不考虑立体异构/r/n)/r/n；其中核磁共振氢谱为五组峰且能发生银镜反应的结构简式为/r/n_____/r/n。/r/n(6)/r/n参照上述合成路线，以/r/n和/r/n为原料，设计制备/r/n的合成路线/r/n(/r/n无机试剂任选/r/n)_____/r/n。/r/n（1）/r/n酯基、醚键/r/n/r/n光照/r/n/r/n（2）/r/n/r/n/r/n（3）/r/n取代反应/r/n/r/n（4）/r/n+/r/n+HBr/r/n（5）/r/n12/r/n/r/n（6）/r/n。/r/n/r/n【分析】/r/n由/r/nA/r/n的分子式及后续流程可知/r/nA/r/n为/r/n，/r/nB/r/n为/r/n，/r/nB/r/n和/r/nNaCN/r/n发生取代反应生成/r/nC/r/n，对比已知信息可知/r/nD→E/r/n发生/r/n-CN/r/n水解，/r/nD/r/n为/r/n；由/r/nF→G/r/n可知/r/nF/r/n为/r/n；由反应⑧可知/r/nH/r/n为/r/n，据此解答。/r/n(1)/r/n由曲美布汀分子的结构可知含氧官能团的名称是酯基、醚键；反应①为苯环侧链烷基上的/r/nH/r/n被/r/nCl/r/n取代，反应条件是光照；/r/n(2)/r/n与/r/n4/r/n个不同的原子或原子团相连的/r/nC/r/n为手性碳，则用星号/r/n(*)/r/n标出/r/nG/r/n的手性碳原子为/r/n；/r/n(3)/r/n反应③为/r/n和/r/nC/r/n2/r/nH/r/n5/r/nBr/r/n发生反应生成/r/n和/r/nHBr/r/n，属于取代反应；/r/n(4)/r/n反应⑦为/r/n和/r/n反应产生/r/n和/r/nHBr/r/n，化学方程式为/r/n+/r/n+HBr/r/n；/r/n(5)E/r/n为/r/n，其同分异构体/r/nM/r/n满足：/r/n①/r/nM/r/n为苯的二元取代物，即苯环上有/r/n2/r/n个取代基；/r/n②/r/nM/r/n能发水解反应且水解产物之一能与/r/n发生显色反应，则为酚酯；/r/n取代基为/r/nHCOO-/r/n和/r/n-C/r/n3/r/nH/r/n7/r/n(2/r/n种/r/n)/r/n时有/r/n3×2=6/r/n种，取代基为/r/nCH/r/n3/r/nCOO-/r/n和/r/n-C/r/n2/r/nH/r/n5/r/n时有邻间对/r/n3/r/n种，取代基为/r/nC/r/n2/r/nH/r/n5/r/nCOO-/r/n和/r/n-CH/r/n3/r/n时有邻间对/r/n3/r/n种，共/r/n6+3+3=12/r/n种；其中核磁共振氢谱为五组峰且能发生银镜反应的结构简式为/r/n；/r/n(6)/r/n逆合成分析：/r/n用/r/n和/r/n合成，/r/n由/r/n和氢气合成，/r/n由/r/n1,3-/r/n丁二烯和溴发生加成反应合成，即以/r/n和/r/n为原料，设计制备/r/n的合成路线为/r/n/r/n。/r/n12．（江苏南通市/r/n高三模拟/r/n）/r/n化合物/r/nH/r/n具有抗菌、消炎、降血压等多种功效，其合成路线如图：/r/n(1)A→B/r/n的反应类型为/r/n_______/r/n。/r/n(2)/r/n可用于鉴别/r/nB/r/n与/r/nC/r/n的常用化学试剂为/r/n_______/r/n。/r/n(3)F/r/n分子中碳原子的轨道杂化类型有/r/n_______/r/n种。/r/n(4)D/r/n的分子式为/r/nC/r/n10/r/nH/r/n11/r/nO/r/n2/r/nC1/r/n，写出/r/nD/r/n的结构简式：/r/n_______/r/n。/r/n(5)E/r/n的一种同分异构体同时满足下列条件，写出该同分异构体的结构简式：/r/n_______/r/n。/r/n①分子中含苯环和碳氮双键，能发生银镜反应；/r/n②分子中有/r/n4/r/n种不同化学环境的氢原子。/r/n(6)/r/n设计以/r/n、/r/n乙烯为原料制备/r/n的合成路线/r/n____________(/r/n无机试剂和有机溶剂任选，合成路线示例见本题题干/r/n)/r/n。/r/n（1）/r/n取代反应/r/n/r/n（2）/r/nFeCl/r/n3/r/n/r/n溶液/r/n/r/n（3）/r/n3/r/n（4）/r/n/r/n/r/n（5）/r/n/r/n/r/n（6）/r/n/r/n【分析】/r/n对比A、B的结构简式，/r/nA/r/n→/r/nB/r/n发生了取代反应，/r/nC/r/n和/r/nB/r/n相比较，/r/nB/r/n结构上的酯基分成了两个取代基，一个是酚羟基，一个酮羰基，/r/nC/r/n→D→/r/nE/r/n过程中，可知在/r/nC/r/n的结构上引入了一个取代基，根据反应条件以及D的分子式，可推测/r/nD/r/n的结构简式是/r/n，/r/nE/r/n→/r/nF/r/n的过程中先发生了加成、后发生了消去，/r/nF/r/n→/r/nG/r/n，酚羟基和双键相连形成环状，/r/nG/r/n→/r/nH/r/n，结构中的氰基水解成了羧基，据此分析解答。/r/n(1)B/r/n可以看成是/r/n取代了/r/nA/r/n结构中酚羟基上的/r/nH/r/n原子，故/r/nA→B/r/n的反应类型为取代反应。/r/n(2)B/r/n的结构中有官能团酯基，/r/nC/r/n的结构中有酮羰基和酚羟基，可利用酚羟基的显色反应来区分/r/nB/r/n和/r/nC/r/n，故鉴别/r/nB/r/n与/r/nC/r/n的常用化学试剂为/r/nFeCl/r/n3/r/n溶液。/r/n(3)F/r/n的结构简式为/r/n，苯环上的碳原子采取的是/r/nsp/r/n2/r/n杂化，甲基上的碳原子采取的是/r/nsp/r/n3/r/n杂化，碳碳双键上的碳原子采取的是/r/nsp/r/n2/r/n杂化，/r/n-CN/r/n上的碳原子采取的/r/nsp/r/n杂化，/r/n-CH/r/n2/r/n-/r/n中的碳原子采取的是/r/nsp/r/n3/r/n杂化，故/r/nF/r/n分子中碳原子的轨道杂化类型有/r/n3/r/n种。/r/n(4)D/r/n的分子式为/r/nC/r/n10/r/nH/r/n11/r/nO/r/n2/r/nC1/r/n，/r/nD/r/n和/r/nNaCN/r/n反应得到/r/nE/r/n，/r/nE/r/n的结构简式相对于/r/nC/r/n来说，在苯环上多了一个取代基/r/n-CH/r/n2/r/n-CN/r/n，通过/r/nD/r/n→/r/nE/r/n反应的条件，取代基/r/n-CH/r/n2/r/n-CN/r/n可以看成是/r/n-CN/r/n取代了/r/n-CH/r/n2/r/nX/r/n上的一个原子，结合/r/nC/r/n→/r/nE/r/n的反应条件和/r/nD/r/n的分子式，可知/r/nX/r/n原子应是/r/nC1/r/n，则/r/nD/r/n的结构简式相对于/r/nC/r/n来说，在苯环上多了一个取代基/r/n-CH/r/n2/r/nCl/r/n，所以/r/nD/r/n的结构简式为/r/n。/r/n(5)E/r/n的一种同分异构体满足分子中含苯环和碳氮双键，能发生银镜反应，说明有取代基/r/n-CHO/r/n，有/r/n结构，结合/r/nE/r/n的结构简式/r/n和分子式/r/nC/r/n11/r/nH/r/n11/r/nO/r/n2/r/nN/r/n，可知其不饱和度为/r/n7/r/n，/r/n/r/n同分异构体的不饱和度是相同的，已知一个苯环中的不饱和度为/r/n4/r/n，一个醛基中的不饱和度为/r/n1/r/n，一个/r/n的不饱和度也为/r/n1/r/n，从结构简式可以看出分子中只有一个/r/nN/r/n原子，也就是说其同分异构体中只能有一个/r/n，根据不饱和度数，其结构中必然含有/r/n2/r/n个醛基，另外分子中有/r/n4/r/n种不同化学环境的氢原子，醛基中有/r/n1/r/n种，若/r/n连接的/r/nH/r/n原子是/r/n1/r/n种，苯环上有/r/n1/r/n种，还有/r/n1/r/n种根据剩余的碳氢原子个数，说明结构中还有/r/n2/r/n个甲基，这样苯环上就有/r/n2/r/n个醛基，/r/n2/r/n个甲基，和一个/r/n，故满足条件的同分异构体有/r/n/r/n，移动甲基又有/r/n/r/n，/r/n移动醛基又有/r/n/r/n。/r/n(6)/r/n由/r/nH/r/n的合成路线/r/nF→G/r/n可知，要合成出/r/n，就要利用原料合成出/r/n，要合成出/r/n，通过/r/nE→F/r/n可知，要利用原料合成出/r/n和/r/nCH/r/n3/r/nCHO/r/n，通过路线/r/nB/r/n→/r/nC/r/n可知，要合成出/r/n，就需要/r/n，要合成出/r/nCH/r/n3/r/nCHO/r/n，可利用我们熟悉的乙烯制乙醇，乙醇制乙醛来合成。故合成路线如下：/r/n。/r/n13．（山东青岛市/r/n高三模拟/r/n）/r/n有机化合物/r/nI/r/n是一种重要的有机合成中间体，合成路线如图：/r/n

/r/n已知：/r/n①/r/n+/r/n②/r/n+/r/n回答下列问题：/r/n(1)A/r/n的名称为/r/n____/r/n，/r/nH/r/n的结构简式为/r/n___/r/n。/r/n(2)/r/n写出/r/nD→E/r/n的化学方程式/r/n___/r/n，/r/nF→G/r/n过程中涉及的反应类型有/r/n___/r/n。/r/n(3)/r/n物质/r/nC/r/n有多种同分异构体/r/n(/r/n不考虑立体异构/r/n)/r/n，符合下列条件的同分异构体有/r/n__/r/n种。/r/n①苯环上有两个取代基/r/n②既能与/r/nNaHCO/r/n3/r/n溶液反应又能与/r/nFeCl/r/n3/r/n溶液显色/r/n写出其中一种含有手性碳原子的结构简式：/r/n____/r/n。/r/n(4)/r/n设计由/r/nCH/r/n3/r/nCHOHCH/r/n3/r/n为原料制备/r/n的合成路线/r/n(/r/n无机试剂任选/r/n)___/r/n。/r/n（1）/r/n1/r/n，/r/n3—/r/n丁二醇/r/n/r/n/r/n（2）/r/n+2NaOH→/r/n+/r/n+H/r/n2/r/nO/r/n加成反应、消去反应/r/n/r/n（3）/r/n24/r/n/r/n（4）/r/n/r/n/r/n【分析】/r/n催化氧化生成/r/n，/r/n与苯酚发生取代反应生成/r/n，/r/n与/r/n发生加成反应生成/r/n，/r/n在氢氧化钠水溶液中加热发生水解反应生成/r/nE/r/n，/r/nE/r/n酸化后脱羧得到/r/nF/r/n，根据/r/nF/r/n的分子式可推出/r/nF/r/n为/r/n，则/r/nE/r/n为/r/n，/r/n在稀氢氧化钠溶液中加热生成/r/n，/r/n与/r/nH/r/n在碱性条件下发生取代反应生成/r/n，根据反应/r/n+/r/n，推出/r/nH/r/n为/r/n。/r/n(1)A/r/n为/r/n，名称为/r/n1/r/n，/r/n3—/r/n丁二醇，/r/nH/r/n的结构简式为/r/n；/r/n(2)D→E/r/n是/r/n与氢氧化钠反应生成/r/n、/r/n和水，反应的化学方程式为/r/n+2NaOH→/r/n+/r/n+H/r/n2/r/nO/r/n，/r/nF→G/r/n是/r/n转化为/r/n，过程中涉及的反应类型有加成反应形成环、消去反应脱去/r/n1/r/n个水分子形成碳碳双键；/r/n(3)/r/n物质/r/nC/r/n为/r/n，符合条件的同分异构体：①苯环上有两个取代基；②既能与/r/nNaHCO/r/n3/r/n溶液反应又能与/r/nFeCl/r/n3/r/n溶液显色，则含有羧基和酚羟基，根据不饱和度可知应该还含有一个碳碳双键，苯环上除酚羟基外，另一取代基可以为/r/n-CH=CHCH/r/n2/r/nCOOH/r/n、/r/n-C(COOH)=CHCH/r/n3/r/n、/r/n-CH=C(CH/r/n3/r/n)COOH/r/n、/r/n-CH/r/n2/r/nCH=CHCOOH/r/n、/r/n-CH(COOH)CH=CH/r/n2/r/n、/r/n-CH/r/n2/r/nC(COOH)=CH/r/n2/r/n、/r/n-C(CH/r/n3/r/n)=CHCOOH/r/n、/r/n-C(CH/r/n2/r/nCOOH)=CH/r/n2/r/n，两取代基在苯环上的位置有邻、间、对/r/n3/r/n种位置，故符合条件的同分异构体共有/r/n8/r/n3=24/r/n种；/r/n其中含有手性碳原子的结构简式为：/r/n；/r/n(4)CH/r/n3/r/nCHOHCH/r/n3/r/n催化氧化生成丙酮，丙酮发生类似反应①/r/n+/r/n生成/r/n；/r/nCH/r/n3/r/nCHOHCH/r/n3/r/n在浓硫酸中发生消去反应生成丙烯，丙烯与氯气发生加成反应生成/r/nCH/r/n2/r/n=CHCH/r/n2/r/nCl/r/n，/r/n与/r/nCH/r/n2/r/n=CHCH/r/n2/r/nCl/r/n反应制备/r/n，合成路线如下：/r/n。/r/n14．（山东烟台市/r/n高三/r/n三模）/r/n化合物/r/nG/r/n是一种有机光电材料中间体，其合成路线如下：/r/n已知：/r/n(1)/r/n(2)R/r/n1/r/nCH/r/n2/r/nCOOCH/r/n2/r/nCH/r/n3/r/n+R/r/n2/r/nCOOCH/r/n2/r/nCH/r/n3/r/n+CH/r/n3/r/nCH/r/n2/r/nOH/r/n(1)/r/n化合物/r/nA/r/n的结构简式是/r/n_____/r/n。/r/n(2)B→C/r/n和/r/nE→F/r/n的反应类型分别是/r/n____/r/n、/r/n____/r/n。/r/n(3)E/r/n中官能团的名称为/r/n____/r/n；/r/nC→D/r/n的反应方程式为/r/n____/r/n。/r/n(4)/r/n写出两种符合下列条件的/r/nG/r/n的同分异构体的结构简式/r/n____/r/n。/r/n①/r/n具有六元碳环结构/r/n②/r/n能与/r/nNaHCO/r/n3/r/n溶液反应/r/n③/r/n能发生银镜反应/r/n④/r/n核磁共振氢谱显示六组峰，且峰面积之比为/r/n3/r/n:/r/n1/r/n:/r/n4/r/n:/r/n4/r/n:/r/n1/r/n:/r/n1/r/n(5)/r/n以乙酸乙酯和/r/n1/r/n，/r/n4-/r/n二溴丁烷为原料合成/r/n，写出能获得更多目标产物的较优合成路线/r/n(/r/n其它试剂任选/r/n)____/r/n。/r/n（1）/r/n/r/n（2）/r/n/r/n氧化反应/r/n/r/n取代反应/r/n/r/n（3）/r/n/r/n酯基、羰基/r/nHOOCCH/r/n2/r/nCH/r/n2/r/nCH/r/n2/r/nC(CH/r/n3/r/n)/r/n2/r/nCOOH+2CH/r/n3/r/nCH/r/n2/r/nOH/r/nCH/r/n3/r/nCH/r/n2/r/nOOCCH/r/n2/r/nCH/r/n2/r/nCH/r/n2/r/nC(CH/r/n3/r/n)/r/n2/r/nCOOCH/r/n2/r/nCH/r/n3/r/n+2H/r/n2/r/nO/r/n（4）/r/n/r/n或/r/n或/r/n/r/n（5）/r/n/r/n【分析】/r/nA/r/n与乙烯反应生成/r/nB(/r/n)/r/n，结合已知反应可知/r/nA/r/n应为/r/n，/r/nB/r/n与高锰酸钾反应断裂碳碳双键生成羧基得到/r/nC/r/n，/r/nC/r/n为/r/nHOOCCH/r/n2/r/nCH/r/n2/r/nCH/r/n2/r/nC(CH/r/n3/r/n)/r/n2/r/nCOOH/r/n，/r/nC/r/n与乙醇发生酯化反应得到/r/nD/r/n，/r/nD/r/n为/r/nCH/r/n3/r/nCH/r/n2/r/nOOCCH/r/n2/r/nCH/r/n2/r/nCH/r/n2/r/nC(CH/r/n3/r/n)/r/n2/r/nCOOCH/r/n2/r/nCH/r/n3/r/n，/r/nD/r/n在/r/n条件下发生取代反应生成/r/nE/r/n，/r/nE/r/n在乙醇钠条件下与一溴甲烷发生取代反应得到/r/nF/r/n，/r/nF/r/n中的酯基碱性条件下发生水解反应，再酸化得到/r/nG/r/n，据此分析。/r/n(1)/r/n根据以上分析可知/r/nA/r/n为/r/n，故/r/n；/r/n(2)B/r/n到/r/nC/r/n是/r/nB/r/n中的碳碳双键被高锰酸钾氧化断裂生成羧基，/r/nE/r/n到/r/nF/r/n发生的是酯基所连碳上的氢原子被甲基取代的反应，故氧化反应；取代反应；/r/n(3)/r/n由/r/nE/r/n的结构简式可知，其中含有酯基和羰基两种官能团，/r/nC/r/n与乙醇发生反应生成/r/nD/r/n，反应的方程式为/r/nHOOCCH/r/n2/r/nCH/r/n2/r/nCH/r/n2/r/nC(CH/r/n3/r/n)/r/n2/r/nCOOH+2CH/r/n3/r/nCH/r/n2/r/nOH/r/nCH/r/n3/r/nCH/r/n2/r/nOOCCH/r/n2/r/nCH/r/n2/r/nCH/r/n2/r/nC(CH/r/n3/r/n)/r/n2/r/nCOOCH/r/n2/r/nCH/r/n3/r/n+2H/r/n2/r/nO/r/n，故酯基、羰基；/r/nHOOCCH/r/n2/r/nCH/r/n2/r/nCH/r/n2/r/nC(CH/r/n3/r/n)/r/n2/r/nCOOH+2CH/r/n3/r/nCH/r/n2/r/nOH/r/nCH/r/n3/r/nCH/r/n2/r/nOOCCH/r/n2/r/nCH/r/n2/r/nCH/r/n2/r/nC(CH/r/n3/r/n)/r/n2/r/nCOOCH/r/n2/r/nCH/r/n3/r/n+2H/r/n2/r/nO/r/n；/r/n(4)①/r/n具有六元碳环结构/r/n②/r/n能与/r/nNaHCO/r/n3/r/n溶液反应说明存在羧基；/r/n③/r/n能发生银镜反应，说明含有醛基，/r/n④/r/n核磁共振氢谱显示六组峰，且峰面积之比为/r/n3/r/n:/r/n1/r/n:/r/n4/r/n:/r/n4/r/n:/r/n1/r/n:/r/n1/r/n，可知结构中应含有一个甲基，且为对称结构，则符合条件的结构有：/r/n或/r/n或/r/n，故/r/n或/r/n或/r/n；/r/n(5)/r/n可