專題18選修5有機化學(xué)基礎(chǔ)/r/n1．（福建省福州高三/r/n模擬/r/n）β-/r/n羰基酯化合物是重要的化工合成原料，在有機合成工業(yè)和制藥工業(yè)具有廣泛的用途。某/r/nβ-/r/n羰基酯化合物/r/nG/r/n的合成路徑如圖：/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回答下列問題：/r/n(1)B/r/n的化學(xué)名稱為/r/n_______/r/n。/r/n(2)C/r/n到/r/nD/r/n的反應(yīng)類型為/r/n_______/r/n。/r/n(3)C/r/n中的官能團名稱/r/n_______/r/n。/r/n(4)/r/n寫出反應(yīng)/r/nE/r/n到/r/nF/r/n的化學(xué)方程式/r/n_______/r/n。/r/n(5)/r/n寫出/r/nG/r/n的結(jié)構(gòu)簡式/r/n_______/r/n。/r/n(6)/r/n化合物的/r/nF/r/n的同分異構(gòu)體/r/nH/r/n能同時滿足以下三個條件，/r/nH/r/n的結(jié)構(gòu)簡式/r/n_______/r/n。/r/n(/r/nⅰ/r/n)/r/n不能發(fā)生銀鏡反應(yīng)；/r/n(/r/nⅱ/r/n)/r/n易水解，且/r/n1molH/r/n和/r/n3molNaOH/r/n反應(yīng)；/r/n(/r/nⅲ/r/n)/r/n苯環(huán)上二取代，且核磁共振氫譜峰面積比為/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取代反應(yīng)/r/n(/r/n3/r/n)/r/n氯原子、醚鍵/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根據(jù)給定信息/r/nII/r/n結(jié)合/r/nC/r/n的結(jié)構(gòu)簡式可逆推知/r/nB/r/n為/r/n，所以可知/r/nA/r/n為苯酚，/r/nC/r/n與/r/nNaCN/r/n發(fā)生取代反應(yīng)生成/r/nD/r/n，/r/nD/r/n在酸性條件下可生成/r/nE/r/n（/r/n），其分子式為/r/nC/r/n9/r/nH/r/n10/r/nO/r/n3/r/n，/r/nE/r/n與甲醇在濃硫酸作用下可發(fā)生酯化反應(yīng)生成/r/nF/r/n，根據(jù)給定條件/r/nI/r/n可知，/r/nG/r/n的結(jié)構(gòu)簡式為/r/n，據(jù)此分析解答。/r/n根據(jù)上述分析可知，/r/n(1)B/r/n為/r/n，其化學(xué)名稱為苯甲醚；/r/n(2)C/r/n與/r/nNaCN/r/n發(fā)生反應(yīng)，/r/n-CN/r/n替換原來的/r/n-Cl/r/n轉(zhuǎn)化為/r/nD/r/n，所以其反應(yīng)類型為取代反應(yīng)；/r/n(3)C/r/n的結(jié)構(gòu)簡式為：/r/n，其中的官能團名稱為氯原子、醚鍵；/r/n(4)E/r/n與甲醇在濃硫酸作用下可發(fā)生酯化反應(yīng)生成/r/nF/r/n，其化學(xué)方程式為：/r/nCH/r/n3/r/nOH+/r/nH/r/n2/r/nO+/r/n；/r/n(5/r/n根據(jù)上述分析可知，/r/nG/r/n的結(jié)構(gòu)簡式為：/r/n；/r/n(6)F/r/n為/r/n，其分子式為：/r/nC/r/n10/r/nH/r/n12/r/nO/r/n3/r/n，不飽和度為/r/n5/r/n，其同分異構(gòu)體/r/nH/r/n不能發(fā)生銀鏡反應(yīng)，則沒有醛基；易水解，且/r/n1mol/r/n該和/r/n3molNaOH/r/n反應(yīng)，說明分子中可能含酯基或羧基，結(jié)合/r/nO/r/n原子個數(shù)為/r/n3/r/n，可推知分子結(jié)構(gòu)中含/r/n；又分子中核磁共振氫譜峰面積比為/r/n1/r/n：/r/n1/r/n：/r/n2/r/n：/r/n2/r/n：/r/n6/r/n，則說明含有/r/n2/r/n個甲基，所以推知/r/nH/r/n的結(jié)構(gòu)簡式為：/r/n。/r/n2．（廣東珠海市/r/n高三/r/n二模）/r/n化合物/r/nG/r/n是合成一種心血管藥物的中間體，其合成路線如下：/r/n已知：/r/n+/r/n/r/n+H/r/n2/r/nO/r/n回答下列問題：/r/n(/r/n1/r/n)/r/nB/r/n的化學(xué)名稱是/r/n______/r/n，/r/nC/r/n反應(yīng)生成/r/nD/r/n的反應(yīng)類型是/r/n______/r/n，/r/nF/r/n的含氧官能團名稱是/r/n_________/r/n。/r/n(/r/n2/r/n)物質(zhì)/r/nG/r/n的結(jié)構(gòu)中有兩個六元環(huán)，其結(jié)構(gòu)簡式為/r/n_______________/r/n(/r/n3/r/n)寫出/r/nB/r/n反應(yīng)生成/r/nC/r/n的化學(xué)方程式/r/n____________/r/n。/r/n(/r/n4/r/n)/r/nD/r/n反應(yīng)生成/r/nE/r/n時用/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的同分異構(gòu)體，寫出符合下列條件的/r/nH/r/n的結(jié)構(gòu)簡式(寫出一種即可)/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生成氫氣/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反應(yīng)/r/n③核磁共振氫譜共有/r/n4/r/n個吸收峰，峰面積之比為/r/n6/r/n:/r/n2/r/n:/r/n1/r/n:/r/n1/r/n(/r/n6/r/n)設(shè)計由甲醇和乙酸乙酯制備丙酸乙酯(/r/n)的合成路線(無機試劑任選)/r/n_______________/r/n。/r/n（1）/r/n鄰甲基苯酚(或/r/n2/r/n-甲基苯酚)/r/n/r/n取代反應(yīng)/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條件下被水解/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和溴甲烷發(fā)生取代反應(yīng)生成/r/nB/r/n，根據(jù)/r/nC/r/n的分子式和/r/nD/r/n的結(jié)構(gòu)簡式可知/r/nC/r/n的結(jié)構(gòu)簡式是/r/n，/r/nC/r/n發(fā)生取代反應(yīng)生成/r/nD/r/n，/r/nD/r/n水解生成/r/nE/r/n，/r/nE/r/n發(fā)生催化氧化生成/r/nF/r/n，/r/nF/r/n發(fā)生已知信息反應(yīng)生成/r/nG/r/n為/r/n，據(jù)此解答。/r/n(/r/n1/r/n)根據(jù)/r/nB/r/n的結(jié)構(gòu)簡式可知/r/nB/r/n的化學(xué)名稱是鄰甲基苯酚(或/r/n2/r/n-甲基苯酚，/r/nC/r/n反應(yīng)生成/r/nD/r/n是甲基上的氫原子被溴原子取代，反應(yīng)類型是取代反應(yīng)，根據(jù)/r/nF/r/n的結(jié)構(gòu)簡式可知/r/nF/r/n分子中含氧官能團名稱是酯基、醛基。/r/n(/r/n2/r/n)物質(zhì)/r/nG/r/n的結(jié)構(gòu)中有兩個六元環(huán)，其結(jié)構(gòu)簡式為/r/n。/r/n(/r/n3/r/n)根據(jù)以上分析可知/r/nB/r/n反應(yīng)生成/r/nC/r/n的化學(xué)方程式為/r/n+/r/nCH/r/n3/r/nCOCl/r/n→/r/n+/r/nHCl/r/n。/r/n(/r/n4/r/n)由于酯基能和氫氧化鈉反應(yīng)，所以/r/nD/r/n反應(yīng)生成/r/nE/r/n時用/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條件下被水解。/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生成氫氣，說明含有/r/n1/r/n個羥基或羧基；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反應(yīng)，說明含有酚羥基以及酚羥基形成的酯基；③核磁共振氫譜共有/r/n4/r/n個吸收峰，峰面積之比為/r/n6/r/n:/r/n2/r/n:/r/n1/r/n:/r/n1/r/n，則滿足條件的結(jié)構(gòu)簡式為/r/n或/r/n；/r/n(/r/n6/r/n)首先甲醇發(fā)生催化氧化生成甲醛，甲醛和乙酸乙酯發(fā)生已知信息反應(yīng)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，合成路線為/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．（廣東汕頭市·金山中學(xué)高三三模）/r/n扁桃酸衍生物是重要的醫(yī)藥中間體，以/r/nA/r/n和/r/nB/r/n為原料合成扁桃酸衍生物/r/nF/r/n路線如下：/r/n(1)A/r/n與/r/nB/r/n的反應(yīng)為加成反應(yīng)，且/r/nA/r/n的分子式為/r/nC/r/n2/r/nH/r/n2/r/nO/r/n3/r/n，可發(fā)生銀鏡反應(yīng)，且具有酸性，則/r/nB/r/n的結(jié)構(gòu)簡式是/r/n____/r/n。/r/n(2)/r/n試寫出/r/nC/r/n與/r/nNaOH/r/n反應(yīng)的化學(xué)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個六元環(huán)的化合物，/r/nE/r/n的分子中不同化學(xué)環(huán)境的氫原子有/r/n___/r/n種。/r/n(4)/r/n下列說法正確的是/r/n___/r/nA．F/r/n常溫下易溶于水/r/nB．1molF/r/n在一定條件下與足量/r/nNaOH/r/n溶液反應(yīng)，最多消耗/r/nNaOH/r/n的物質(zhì)的量為/r/n3mol/r/n。/r/nC．D→F/r/n的反應(yīng)類型是取代反應(yīng)/r/nD．/r/n一定條件下/r/nC/r/n可以發(fā)生縮聚反應(yīng)/r/n(5)/r/n符合下列條件的/r/nF/r/n的同分異構(gòu)體/r/n(/r/n不考慮立體異構(gòu)/r/n)/r/n有/r/n___/r/n種。其中氫原子個數(shù)比為/r/n1:1:2:2:3/r/n的結(jié)構(gòu)簡式為/r/n____/r/n①屬于一元酸類化合物；②苯環(huán)上只有/r/n2/r/n個取代基且處于對位；③遇氯化鐵溶液發(fā)生顯色反應(yīng)。/r/n(6)/r/n己知：/r/n+CH≡CH/r/n，請設(shè)計合成路線以/r/nB/r/n和/r/nC/r/n2/r/nH/r/n2/r/n為原料合成/r/n(/r/n無機試劑任選/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與/r/nB/r/n的反應(yīng)為加成反應(yīng)，且/r/nA/r/n的分子式為/r/nC/r/n2/r/nH/r/n2/r/nO/r/n3/r/n，可發(fā)生銀鏡反應(yīng)，且具有酸性，則/r/nA/r/n為/r/nOHCCOOH/r/n，/r/nB/r/n為/r/n；/r/nC/r/n與甲醇發(fā)生酯化反應(yīng)得到/r/nD/r/n，/r/nD/r/n和/r/nHBr/r/n發(fā)生取代反應(yīng)得到/r/nF/r/n；由于/r/nC/r/n中既含醇羥基，也含羧基，因此會得到副產(chǎn)物如兩個/r/nC/r/n分子生成的環(huán)酯等。/r/n(1)/r/n由分析可知/r/nB/r/n為/r/n；/r/n(2)C/r/n為/r/n，羧基、酚羥基均能與/r/nNaOH/r/n按/r/n1:1/r/n反應(yīng)，則/r/nC/r/n與/r/nNaOH/r/n反應(yīng)的化學(xué)方程式為/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個六元環(huán)的化合物，則/r/nE/r/n為/r/n2/r/n分子/r/nC/r/n生成的環(huán)酯，/r/nE/r/n為/r/n，則/r/nE/r/n分子中不同化學(xué)環(huán)境的氫原子有/r/n4/r/n種；/r/n(4)A/r/n．/r/nF/r/n含酯基和/r/n-Br/r/n、酚羥基，常溫下不易溶于水，/r/nA/r/n錯誤；/r/nB/r/n．/r/n1molF/r/n含/r/n1mol/r/n酚羥基、/r/n1mol/r/n酯基、/r/n1mol/r/n溴原子，則在一定條件下/r/n1molF/r/n與足量/r/nNaOH/r/n溶液反應(yīng)，最多消耗/r/nNaOH/r/n的物質(zhì)的量為/r/n3mol/r/n，/r/nB/r/n正確；/r/nC/r/n．對比/r/nD/r/n和/r/nF/r/n的結(jié)構(gòu)可知/r/nD→F/r/n的反應(yīng)類型是取代反應(yīng)/r/n(H/r/n被/r/nBr/r/n取代/r/n)/r/n，/r/nC/r/n正確；/r/nD/r/n．/r/nC/r/n含一個羧基和一個醇羥基，一定條件下/r/nC/r/n可以發(fā)生縮聚反應(yīng)生成聚酯類物質(zhì)；/r/nC/r/n含酚羥基，且酚羥基鄰位有/r/nH/r/n，一定條件下可發(fā)生酚醛縮聚，/r/nD/r/n正確；/r/n選/r/nBCD/r/n；/r/n(5)F/r/n為/r/n，/r/nF/r/n的同分異構(gòu)體/r/n(/r/n不考慮立體異構(gòu)/r/n)/r/n：/r/n①屬于一元酸類化合物，則含/r/n1/r/n個/r/n-COOH/r/n；/r/n②苯環(huán)上只有/r/n2/r/n個取代基且處于對位；/r/n③遇氯化鐵溶液發(fā)生顯色反應(yīng)，則含酚羥基；/r/n符合條件的有/r/n、/r/n、/r/n、/r/n，共/r/n4/r/n種；其中氫原子個數(shù)比為/r/n1:1:2:2:3/r/n的結(jié)構(gòu)簡式為/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為原料合成/r/n，進行逆合成分析：/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與氫氣加成得到，因此合成路線為/r/n。/r/n4．（湖北武漢市漢陽一中高三二模）/r/n有機化合物/r/nH/r/n可用來制備抗凝血藥，可通過如圖路線合成。/r/n已知：/r/n①RCOOH/r/n(R/r/n為烴基/r/n)/r/n②/r/n酚羥基一般不易直接與羧酸酯化/r/n請回答：/r/n(1)F/r/n的名稱為/r/n_______/r/n。/r/n(2)C+F→G/r/n的反應(yīng)類型為/r/n_______/r/n；/r/nH/r/n中含氧官能團的名稱為/r/n_______/r/n。/r/n(3)/r/n在/r/nA→B/r/n的反應(yīng)中，檢驗/r/nA/r/n是否反應(yīng)完全的試劑是/r/n_______/r/n。/r/n(4)/r/n寫出/r/nG/r/n與過量/r/nNaOH/r/n溶液共熱時反應(yīng)的化學(xué)方程式/r/n_______/r/n。/r/n(5)/r/n化合物/r/nD/r/n的同分異構(gòu)體有多種，其中能與/r/nFeCl/r/n3/r/n溶液發(fā)生顯色反應(yīng)的結(jié)構(gòu)有/r/n_______/r/n種/r/n(/r/n不包括/r/nD/r/n本身/r/n)/r/n，其中核磁共振氫譜有/r/n4/r/n組峰，且峰面積比為/r/n2/r/n：/r/n2/r/n：/r/n1/r/n：/r/n1/r/n的結(jié)構(gòu)簡式為/r/n_______(/r/n任寫一種/r/n)/r/n。/r/n(6)/r/n苯甲酸苯酚酯/r/n(/r/n)/r/n是一種重要的有機合成中間體。請根據(jù)已有知識并結(jié)合相關(guān)信息，試寫出以苯酚、甲苯為原料制取該化合物的合成路線/r/n(/r/n無機試劑任選/r/n)/r/n：/r/n_______/r/n。/r/n（1）/r/n鄰羥基苯甲酸甲酯/r/n/r/n（2）/r/n取代反應(yīng)/r/n/r/n羥基、酯基/r/n/r/n（3）/r/n/r/n銀氨溶液/r/n(/r/n或新制氫氧化銅懸濁液/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物質(zhì)/r/nA/r/n為乙醛，乙醛催化氧化生成物質(zhì)/r/nB/r/n乙酸，乙酸和三氯化磷發(fā)生取代反應(yīng)生成乙酰氯，根據(jù)/r/nG/r/n的結(jié)構(gòu)簡式可推知/r/nE/r/n為甲醇，甲醇和物質(zhì)/r/nD/r/n發(fā)生酯化反應(yīng)生成物質(zhì)/r/nF/r/n：鄰羥基苯甲酸甲酯，/r/nC/r/n和/r/nF/r/n發(fā)生取代反應(yīng)生成物質(zhì)/r/nG/r/n：/r/n，最后/r/nG/r/n生成/r/nH/r/n，據(jù)此分析答題。/r/n（/r/n1/r/n）根據(jù)物質(zhì)/r/nG/r/n：/r/n可推知/r/nE/r/n為甲醇，/r/nE/r/n和/r/nD/r/n發(fā)生酯化反應(yīng)生成/r/nF/r/n，/r/nF/r/n的名稱為：鄰羥基苯甲酸甲酯；/r/n（/r/n2/r/n）/r/nC+F→G/r/n發(fā)生取代反應(yīng)，生成/r/nG/r/n和/r/nHCl/r/n；/r/nH/r/n的結(jié)構(gòu)為：/r/n，其中含氧官能團為：酯基、羥基；/r/n（/r/n3/r/n）/r/nA/r/n為乙醛，可以和銀氨溶液發(fā)生銀鏡反應(yīng)，也可以和新制氫氧化銅懸濁液產(chǎn)生磚紅色沉淀，利用此性質(zhì)檢驗乙醛的存在，故銀氨溶液/r/n(/r/n或新制氫氧化銅懸濁液/r/n)/r/n；/r/n（/r/n4/r/n）/r/nG/r/n為：/r/n，/r/nG/r/n中含有酯基，與過量/r/nNaOH/r/n溶液共熱時會發(fā)生水解反應(yīng)，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）物質(zhì)/r/nD/r/n為：/r/n，其中能與/r/nFeCl/r/n3/r/n溶液發(fā)生顯色反應(yīng)的結(jié)構(gòu)有：間羥基苯甲酸、對羥基苯甲酸、/r/n、還可以將羧基拆成一個醛基和一個羥基，兩個羥基一個醛基同時連在苯環(huán)上，共有/r/n6/r/n種結(jié)構(gòu)，所以同分異構(gòu)體一共有/r/n11/r/n中，故/r/n11/r/n；/r/n（/r/n6/r/n）根據(jù)已知②酚羥基一般不易直接與羧酸酯化，所以設(shè)計合成路線時先將甲苯氧化為苯甲酸，苯甲酸發(fā)生取代反應(yīng)生成酰氯，再與苯酚發(fā)生取代反應(yīng)制得目標(biāo)產(chǎn)物，設(shè)計合成路線如下：/r/n。/r/n5．（江蘇高三/r/n模擬/r/n）/r/n有機物/r/nE/r/n是一種常用的藥物，可通過如圖路線合成：/r/n已知：/r/nRCOOH/r/nRCOCl/r/n(1)D→E/r/n的反應(yīng)類型為/r/n___/r/n。/r/n(2)A/r/n的分子式為/r/nC/r/n4/r/nH/r/n6/r/nO/r/n3/r/n，寫出/r/nA/r/n的結(jié)構(gòu)簡式/r/n___/r/n。/r/n(3)/r/n寫出同時滿足下列條件的/r/nC/r/n的一種同分異構(gòu)體的結(jié)構(gòu)簡式/r/n___/r/n。/r/n①不能發(fā)生銀鏡反應(yīng)，但能與金屬鈉反應(yīng)/r/n②分子中只有/r/n3/r/n種不同化學(xué)環(huán)境的氫/r/n③該分子不能使紫色石蕊溶液變紅色/r/n(4)/r/n有機物/r/nE/r/n中，含有/r/n___/r/n個手性碳原子。/r/n(5)/r/n寫出以甲苯、/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為原料制備/r/n的合成路線流程圖/r/n___/r/n。/r/n(/r/n無機試劑任用，合成路線流程圖示例見本題題干/r/n)/r/n。/r/n（1）/r/n取代反應(yīng)/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由合成路線圖可知，乙酸甲酯（/r/n）和甲酸乙酯在堿的作用下加熱生成/r/nA/r/n，又因為/r/nA/r/n的分子式為/r/nC/r/n4/r/nH/r/n6/r/nO/r/n3/r/n，所以/r/nA/r/n為/r/n，/r/n在氧氣的催化氧化下可把醛基轉(zhuǎn)化成羧基，/r/nB/r/n與甲醇酯化可生成/r/n，/r/n與溴乙烷發(fā)生取代反應(yīng)生成/r/nC/r/n（/r/n），/r/n與/r/n在甲醇鈉的作用下發(fā)生取代反應(yīng)生成/r/nD/r/n（/r/n），/r/n與/r/n發(fā)生取代反應(yīng)生成/r/nE/r/n（/r/n），據(jù)此分析解答。/r/n(1)D→E/r/n為/r/n與/r/n發(fā)生取代反應(yīng)生成/r/n，反應(yīng)類型為取代反應(yīng)，故取代反應(yīng)；/r/n(2)/r/n的分子式為/r/n，根據(jù)前后物質(zhì)推斷，/r/n+/r/n+/r/n，則/r/nA/r/n的結(jié)構(gòu)式為/r/n；/r/n(3)/r/n根據(jù)要求不能發(fā)生銀鏡反應(yīng)和不能使紫色石蕊溶液變紅色，該物質(zhì)不能含有/r/n和/r/n官能團/r/n,/r/n能與金屬鈉反應(yīng)，說明含有/r/n，同時分子中只有/r/n3/r/n種不同化學(xué)環(huán)境的氫，則該結(jié)構(gòu)式為/r/n，故/r/n；/r/n(4)/r/n四個鍵上連接不同的原子或原子團的碳原子稱為手性碳原子，由/r/nE/r/n的結(jié)構(gòu)可知，含有/r/n1/r/n個手性碳原子，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為原料制備/r/n，可以先將甲苯用高錳酸鉀氧化成苯甲酸，再用/r/nSOCl/r/n2/r/n將苯甲酸轉(zhuǎn)化成/r/n，/r/n與/r/nCH/r/n2/r/n(COOC/r/n2/r/nH/r/n5/r/n)/r/n2/r/n反應(yīng)生成/r/n，/r/n在酸性條件下酯基水解生成/r/n，/r/n在加熱的情況下生成/r/n，所以合成路線為：/r/n。/r/n6．（天津高三二模）/r/n某藥物中間體/r/n(W)/r/n的一種合成路線如下：/r/n

/r/n已知：/r/n回答下列問題：/r/n(1)W/r/n所含官能團名稱是/r/n_______/r/n；/r/n的反應(yīng)類型是/r/n_______/r/n。/r/n(2)Y/r/n的結(jié)構(gòu)簡式為/r/n_______/r/n。/r/n(3)M/r/n分子式為/r/n_______/r/n(4)/r/n在/r/nX/r/n的同分異構(gòu)體中，苯環(huán)上含/r/n1/r/n個取代基的結(jié)構(gòu)有/r/n____/r/n種/r/n(/r/n不含/r/nX)/r/n，其中核磁共振氫譜有/r/n4/r/n組吸收峰的結(jié)構(gòu)簡式為/r/n_____/r/n。/r/n(5)/r/n化合物/r/nM/r/n含有手性碳原子的數(shù)目為/r/n_______/r/n，下列物質(zhì)不與/r/nM/r/n發(fā)生反應(yīng)的是/r/n_______/r/n。/r/nA．/r/n氫氣/r/nB．/r/n溶液/r/nC．/r/n酸性高錳酸鉀溶液/r/nD．/r/n銀氨溶液/r/n(6)/r/n以甲苯、/r/n為主要原料，合成/r/n(/r/n無機試劑自選/r/n)/r/n，設(shè)計合成路線。請寫出/r/n“□”/r/n內(nèi)物質(zhì)結(jié)構(gòu)簡式、/r/n“→”/r/n上反應(yīng)試劑及條件/r/n_______/r/n。/r/n（1）/r/n羧基/r/n/r/n取代反應(yīng)/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根據(jù)/r/nX/r/n、/r/nZ/r/n的結(jié)構(gòu)簡式以及已知反應(yīng)可知，/r/nX/r/n與/r/nCH/r/n3/r/nCOCl/r/n發(fā)生取代反應(yīng)，取代的位置為/r/nX/r/n苯環(huán)取代基的對位，則/r/nY/r/n的結(jié)構(gòu)簡式為/r/n，其他物質(zhì)已知，根據(jù)流程圖分析即可。/r/n(1)/r/n根據(jù)/r/nW/r/n的結(jié)構(gòu)簡式可知，/r/nW/r/n所含官能團名稱是羧基；由分析知，/r/n的反應(yīng)類型是取代反應(yīng)。/r/n(2)Y/r/n的結(jié)構(gòu)簡式為/r/n。/r/n(3)/r/n由/r/nM/r/n的結(jié)構(gòu)簡式可知，/r/nM/r/n的分子式為/r/n。/r/n(4)X/r/n苯環(huán)上取代基為某丁基，丁基/r/n(-C/r/n4/r/nH/r/n9/r/n)/r/n有/r/n4/r/n種，則在/r/nX/r/n的同分異構(gòu)體中，苯環(huán)上含/r/n1/r/n個取代基的結(jié)構(gòu)，即取代基的種類數(shù)決定了這種/r/nX/r/n的同分異構(gòu)體的種類數(shù)，應(yīng)有/r/n3/r/n種/r/n(/r/n不含/r/nX)/r/n，其中核磁共振氫譜有/r/n4/r/n組吸收峰的結(jié)構(gòu)簡式為/r/n。/r/n(5)/r/n手性碳原子是指連接/r/n4/r/n個不一樣的原子或原子團的碳原子，則化合物/r/nM/r/n含有/r/n1/r/n個手性碳原子，如圖所示/r/n，標(biāo)注/r/n*/r/n的碳原子為手性碳原子，/r/nA/r/n．/r/nM/r/n含有碳碳雙鍵和醛基，可與氫氣發(fā)生加成反應(yīng)；/r/nB/r/n．/r/n溶液不能與/r/nM/r/n發(fā)生反應(yīng)；/r/nC/r/n．/r/nM/r/n含有碳碳雙鍵和醛基，可被酸性高錳酸鉀溶液氧化；/r/nD/r/n．/r/nM/r/n含有醛基，可被銀氨溶液氧化；故選/r/nB/r/n。/r/n(6)/r/n甲苯與/r/n發(fā)生取代反應(yīng)生成/r/n，/r/n被酸性高錳酸鉀氧化生成/r/n，/r/n被氫氣還原生成/r/n，所以合成路線為：/r/n。/r/n7．（福建廈門外國語學(xué)校高三/r/n模擬/r/n）/r/n辣椒素又名辣椒堿/r/n(capsaicin)/r/n，是常見的生物堿之一、辣椒素/r/nH/r/n的合成路線如圖。/r/n請完成下列問題/r/n(1)B/r/n的鍵線式是/r/n_______/r/n。/r/n(2)E/r/n中官能團的名稱是/r/n_______/r/n。/r/n(3)C→D/r/n中反應(yīng)/r/n1)/r/n的化學(xué)方程式是/r/n_______/r/n，反應(yīng)類型是/r/n_______/r/n。/r/n(4)F/r/n與/r/nG/r/n反應(yīng)生成/r/nH/r/n時，另一產(chǎn)物為/r/n_______(/r/n填化學(xué)式/r/n)/r/n。/r/n(5)/r/n的同分異構(gòu)體中，同時符合下列條件的有/r/n_______/r/n種/r/n(/r/n不含立體異構(gòu)/r/n)/r/n。/r/n①/r/n具有四取代苯結(jié)構(gòu)，且核磁共振氫譜顯示，其苯環(huán)上只有一種化學(xué)環(huán)境的/r/nH/r/n②/r/n紅外光譜測得其分子結(jié)構(gòu)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碳碳雙鍵、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反應(yīng)/r/n/r/n（4）/r/nHCl/r/n（5）/r/n4/r/n【分析】/r/n物質(zhì)/r/nA/r/n與/r/n中的溴原子發(fā)生取代反應(yīng)得到物質(zhì)/r/nB/r/n，物質(zhì)/r/nB/r/n和/r/n發(fā)生取代反應(yīng)得到物質(zhì)/r/nC/r/n：/r/n，/r/nC/r/n發(fā)生水解反應(yīng)生成/r/n，再將/r/n酸化得到二元羧酸類物質(zhì)/r/nD/r/n，/r/nD/r/n發(fā)生脫羧反應(yīng)得到物質(zhì)/r/nE/r/n，/r/nE/r/n與二氯氧硫發(fā)生取代反應(yīng)的到酰氯類物質(zhì)/r/nF/r/n，/r/nF/r/n與/r/nG/r/n發(fā)生取代反應(yīng)的目標(biāo)產(chǎn)物/r/nH/r/n，據(jù)此分析答題。/r/n（/r/n1/r/n）/r/nA/r/n中的醇羥基與/r/n中的溴原子發(fā)生取代反應(yīng)，生成/r/nB/r/n，所以/r/nB/r/n的鍵線式為：/r/n；/r/n（/r/n2/r/n）/r/nE/r/n的鍵線式為：/r/n，官能團為：碳碳雙鍵、羧基，故碳碳雙鍵、羧基；/r/n（/r/n3/r/n）物質(zhì)/r/nC/r/n為酯類物質(zhì)，在堿性條件下發(fā)生水解，反應(yīng)方程式為：/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；水解反應(yīng)或取代反應(yīng)；/r/n（/r/n4/r/n）/r/nF/r/n與/r/nG/r/n反應(yīng)發(fā)取代反應(yīng)生成/r/nH/r/n時，同時生成小分子/r/nHCl/r/n，故/r/nHCl/r/n；/r/n（/r/n5/r/n）/r/n的同分異構(gòu)體中符合/r/n①/r/n具有四取代苯結(jié)構(gòu)，且核磁共振氫譜顯示，其苯環(huán)上只有一種化學(xué)環(huán)境的/r/nH②/r/n紅外光譜測得其分子結(jié)構(gòu)中含有/r/n和/r/n-OH/r/n條件的有以下四種：/r/n、/r/n、/r/n、/r/n，故/r/n4/r/n。/r/n8．（遼寧鐵嶺市/r/n高三/r/n二模）/r/n瑞德西韋是一種核苷類似物，研究表明它對治療新冠病毒具有一定的治療效果，其中/r/nK/r/n為合成瑞德西韋過程中重要的中間體，其制備方法如下圖所示。請回答：/r/n已知：①/r/n/r/n/r/n②/r/n(1)A/r/n的結(jié)構(gòu)簡式為/r/n___________/r/n；/r/nC→D/r/n的反應(yīng)類型/r/n___________/r/n。/r/n(2)/r/n寫出化合物/r/nG/r/n的名稱為/r/n___________/r/n；/r/nJ/r/n中所具有的官能團名稱為/r/n___________/r/n。/r/n(3)/r/n寫出/r/nE→F/r/n的化學(xué)方程式/r/n___________/r/n。/r/n(4)X/r/n是/r/nC/r/n的同分異構(gòu)體，滿足下列條件的/r/nX/r/n的同分異構(gòu)體有/r/n___________/r/n種，其中苯環(huán)上的等效氫最少的結(jié)構(gòu)為/r/n___________/r/n。/r/n①苯環(huán)上含有硝基；/r/n②/r/n1molX/r/n可以與足量/r/nNaHCO/r/n3/r/n反應(yīng)生成/r/n1molCO/r/n2/r/n。/r/n（1）/r/n/r/n取代反應(yīng)/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種/r/n/r/n/r/n【分析】/r/n根據(jù)/r/nA/r/n的化學(xué)式及/r/nD/r/n的結(jié)構(gòu)簡式可知：/r/nA/r/n為/r/n；/r/nA/r/n發(fā)生取代反應(yīng)生成/r/nB/r/n為/r/n，/r/nB/r/n發(fā)生硝化反應(yīng)得/r/nC/r/n為/r/n，/r/nC/r/n發(fā)生取代反應(yīng)得/r/nD/r/n，根據(jù)/r/nE/r/n的化學(xué)式可知，/r/nD/r/n再發(fā)生信息/r/nii/r/n的取代生成/r/nE/r/n為/r/n；結(jié)合/r/nF/r/n的化學(xué)式可知，/r/nE/r/n和/r/nA/r/n發(fā)生取代生成/r/nF/r/n為/r/n，根據(jù)/r/nF/r/n和/r/nK/r/n的結(jié)構(gòu)簡式可知/r/nJ/r/n為/r/n。/r/nF/r/n和/r/nJ/r/n發(fā)生取代反應(yīng)生成/r/nK/r/n，根據(jù)信息/r/ni/r/n可逆推得/r/nG/r/n為/r/nHCHO/r/n，/r/nH/r/n為/r/nHOCH/r/n2/r/nCN/r/n，/r/nI/r/n為/r/nNH/r/n2/r/nCH/r/n2/r/nCOOH/r/n，/r/nG/r/n發(fā)生加成得/r/nH/r/n，/r/nH/r/n取代后再水解得/r/nI/r/n，據(jù)此分析答題。/r/n根據(jù)上述分析可知：/r/nA/r/n為/r/n，/r/nB/r/n為/r/n，/r/nC/r/n為/r/n，/r/nE/r/n為/r/n，/r/nF/r/n為/r/n，/r/nG/r/n為/r/nHCHO/r/n，/r/nH/r/n為/r/nHOCH/r/n2/r/nCN/r/n，/r/nI/r/n為/r/nNH/r/n2/r/nCH/r/n2/r/nCOOH/r/n，/r/nJ/r/n為/r/n。/r/n(1)A/r/n結(jié)構(gòu)簡式為/r/n；/r/nC/r/n為/r/n，/r/nC/r/n與/r/nH/r/n3/r/nPO/r/n4/r/n發(fā)生取代反應(yīng)產(chǎn)生/r/nD/r/n：/r/n，故/r/nC→D/r/n的反應(yīng)類型為取代反應(yīng)；/r/n(2)/r/n化合物/r/nG/r/n結(jié)構(gòu)簡式為/r/nHCHO/r/n，名稱為甲醛；化合物/r/nJ/r/n結(jié)構(gòu)簡式為/r/n，其中所含官能團名稱為酯基、氨基；/r/n(3)/r/n化合物/r/nE/r/n為/r/n，/r/nE/r/n與苯酚發(fā)生取代反應(yīng)產(chǎn)生/r/nF/r/n為/r/n和/r/nHCl/r/n，故該反應(yīng)的化學(xué)方程式為：/r/n+/r/n+HCl/r/n；/r/n(4)/r/n化合物/r/nC/r/n為/r/n，/r/nX/r/n是/r/nC/r/n的同分異構(gòu)體，滿足下列條件：①苯環(huán)上含有硝基；②/r/n1molX/r/n可以與足量/r/nNaHCO/r/n3/r/n反應(yīng)生成/r/n1molCO/r/n2/r/n，說明物質(zhì)分子中含有/r/n1/r/n個/r/n-COOH/r/n。若取代基為/r/n-NO/r/n2/r/n、/r/n-CH/r/n2/r/nCOOH/r/n，二者在苯環(huán)上的位置有鄰、間、對三種情況，共有/r/n3/r/n種；若取代基有/r/n-NO/r/n2/r/n、/r/n-CH/r/n3/r/n-COOH/r/n，三種取代基都處于鄰位，有/r/n3/r/n種；若有/r/n2/r/n個取代基相鄰，/r/n1/r/n個相間，有/r/n3×2/r/n種/r/n=6/r/n種；若三個取代基都相間，只有/r/n1/r/n種情況，故符合條件的同分異構(gòu)體共有/r/n3+3+6+1=13/r/n種；其中苯環(huán)上的等效氫最少，說明含有處于對位的取代基，該物質(zhì)的結(jié)構(gòu)為/r/n。/r/n9．（山東濰坊市/r/n高三/r/n三模）/r/n苯巴本妥主要用于治療焦慮、失眠及運動障礙等。一種合成苯巴本的路線如下圖所示。/r/n

/r/n已知：①/r/n②/r/n回答下列問題：/r/n(1)/r/n路線中生成/r/nA/r/n的反應(yīng)類型為/r/n_______/r/n，/r/nD/r/n的結(jié)構(gòu)簡式為/r/n_______/r/n。/r/n(2)/r/n寫出/r/nB→C/r/n的化學(xué)方程式：/r/n_______/r/n。/r/n(3)/r/n同時符合下列條件的/r/nB/r/n的同分異構(gòu)體有/r/n_______/r/n種/r/n(/r/n不包括立體異構(gòu)/r/n)/r/n①屬于芳香族化合物②遇/r/nFeCl/r/n3/r/n溶液不變紫色③可以發(fā)生銀鏡反應(yīng)/r/n其中能與金屬鈉反應(yīng)生成/r/nH/r/n2/r/n，且核磁共振氫譜顯示苯環(huán)上有四種不同化學(xué)環(huán)境的氫原子的同分異構(gòu)體的結(jié)構(gòu)簡式為/r/n_______/r/n。/r/n(4)/r/n設(shè)計以/r/nCH/r/n2/r/n=CH/r/n2/r/n和/r/nCO(NH/r/n2/r/n)/r/n2/r/n為原料合成/r/n的路線/r/n_______(/r/n無機試劑任選/r/n)/r/n。/r/n（1）/r/n取代反應(yīng)/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與/r/nNaCN/r/n發(fā)生取代反應(yīng)生成/r/nA(/r/n)/r/n，由已知①反應(yīng)原理知/r/nB/r/n為/r/n，/r/nB/r/n與/r/nCH/r/n3/r/nCH/r/n2/r/nOH/r/n發(fā)生酯化反應(yīng)生成/r/nC(/r/n)/r/n，/r/nC/r/n中酯基的/r/nα-H/r/n與乙二酸二乙酯發(fā)生取代反應(yīng)生成/r/n，/r/n水解生成/r/nD(/r/n)/r/n，根據(jù)已知②反應(yīng)原理推得/r/nE/r/n為/r/n，/r/nE/r/n中羧基的/r/nα-H/r/n與溴乙烷發(fā)生取代反應(yīng)生成/r/nF(/r/n)/r/n，/r/nF/r/n中羧基與/r/n脫水生成苯巴本妥。/r/n(1)/r/n由分析知，生成/r/nA/r/n的反應(yīng)類型為取代反應(yīng)；/r/nD/r/n的結(jié)構(gòu)簡式為：/r/n；/r/n(2)/r/n由分析知，/r/nB/r/n→/r/nC/r/n為酯化反應(yīng)，對應(yīng)方程式為：/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由題意知，該同分異構(gòu)中含有苯環(huán)、醛基（或甲酸某酯）等結(jié)構(gòu)，一定不含酚羥基，滿足要求的結(jié)構(gòu)如下：/r/n、/r/n、/r/n、/r/n（鄰間對三種）、/r/n（鄰間對三種）、/r/n（鄰間對三種），故共有/r/n12/r/n種結(jié)構(gòu)滿足要求；能與金屬鈉反應(yīng)，說明有羥基，且苯環(huán)上含有四種氫的結(jié)構(gòu)為：/r/n、/r/n；/r/n(4)/r/n由目標(biāo)產(chǎn)物知，需合成丁二酸，丁二酸可由乙烯先與/r/nBr/r/n2/r/n加成，再與/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標(biāo)產(chǎn)物，具體合成路線為：/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．（河北滄州市/r/n高三/r/n三模）/r/n有機物/r/nF(/r/n)/r/n是一種鎮(zhèn)痛藥，它的一種合成路線如圖：/r/n

/r/n(1)/r/n的名稱為/r/n_______/r/n。/r/n(2)/r/n的反應(yīng)類型是/r/n_______/r/n反應(yīng)。/r/n(3)F/r/n中的含氧官能團的名稱為/r/n_______/r/n。/r/n(4)C/r/n的結(jié)構(gòu)簡式為/r/n_______/r/n。/r/n(5)/r/n寫出反應(yīng)/r/n的化學(xué)方程式：/r/n_______/r/n。/r/n(6)H/r/n為/r/nE/r/n的同分異構(gòu)體，且/r/nH/r/n分子中含有/r/n結(jié)構(gòu)，則符合條件的/r/nH/r/n有/r/n_______/r/n種/r/n(/r/n不考慮立體異構(gòu)/r/n)/r/n，寫出其中核磁共振氫譜顯示有/r/n5/r/n組峰且峰面積之比為/r/n的/r/nH/r/n的結(jié)構(gòu)簡式：/r/n_______(/r/n任寫一種/r/n)/r/n。/r/n(7)/r/n已知：/r/n。寫出以/r/n、/r/n為原料制取/r/n的合成路線流程圖/r/n_____/r/n。/r/n(/r/n無機試劑和有機溶劑可任選，合成示例見本題題干/r/n)/r/n（1）/r/n環(huán)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與環(huán)氧乙烷反應(yīng)生成/r/n，/r/n與/r/nSOCl/r/n2/r/n發(fā)生取代反應(yīng)得到/r/nC/r/n，則/r/nC/r/n為/r/n，/r/n與/r/n在/r/nNaNH/r/n2/r/n作用下發(fā)生取代反應(yīng)生成/r/n，/r/n發(fā)生水解得到/r/n，/r/n與乙醇發(fā)生酯化反應(yīng)合成產(chǎn)物/r/n。/r/n(1)/r/n的名稱為環(huán)氧乙烷；/r/n(2)/r/n與/r/nSOCl/r/n2/r/n反應(yīng)得到/r/n，/r/n-OH/r/n被/r/n-Cl/r/n取代，故/r/n的反應(yīng)類型是取代反應(yīng)；/r/n(3)/r/n中的含氧官能團的名稱為酯基；/r/n(4)B/r/n的分子式為/r/n，結(jié)合結(jié)構(gòu)分析可知，由/r/nB/r/n到/r/nC/r/n是氯原子取代了羥基，/r/nC/r/n的結(jié)構(gòu)簡式為/r/n；/r/n(5)/r/n與乙醇發(fā)生酯化反應(yīng)生成/r/n，化學(xué)方程式為：/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，還含有兩個甲基或一個乙基，若為兩個甲基，采取定一移一法，共有/r/n18/r/n種，若為乙基，則有/r/n5/r/n種，共/r/n23/r/n種；其中核磁共振氫譜顯示有/r/n5/r/n組峰且峰面積之比為/r/n的/r/nH/r/n的結(jié)構(gòu)簡式為/r/n；/r/n(7)CH/r/n2/r/n=CH-CH=CH/r/n2/r/n和/r/nBr/r/n2/r/n發(fā)生/r/n1/r/n，/r/n4/r/n加成反應(yīng)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發(fā)生取代反應(yīng)生成/r/n，/r/n水解得到/r/n，/r/n受熱分解生成/r/n，/r/n和乙醇發(fā)生酯化反應(yīng)生成/r/n，故合成路線為：/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模擬/r/n）/r/n曲美布汀是一種消化系統(tǒng)藥物的有效成分，能緩解各種原因引起的胃腸痙攣，可通過以下路線合成。/r/n

/r/n已知：/r/n回答下列問題：/r/n(1)/r/n曲美布汀分子中含氧官能團的名稱是/r/n_____/r/n，①的反應(yīng)條件是/r/n_____/r/n。/r/n(2)/r/n用星號/r/n(*)/r/n標(biāo)出/r/nG/r/n的手性碳原子/r/n_____/r/n。/r/n(3)/r/n反應(yīng)③的反應(yīng)類型是/r/n_____/r/n。/r/n(4)/r/n反應(yīng)⑦的化學(xué)方程式為/r/n_____/r/n。/r/n(5)/r/n化合物/r/nM/r/n是/r/nE/r/n的同分異構(gòu)體，已知：①/r/nM/r/n為苯的二元取代物，②/r/nM/r/n能發(fā)水解反應(yīng)且水解產(chǎn)物之一能與/r/n發(fā)生顯色反應(yīng)，則/r/nM/r/n的可能結(jié)構(gòu)有/r/n_____/r/n種/r/n(/r/n不考慮立體異構(gòu)/r/n)/r/n；其中核磁共振氫譜為五組峰且能發(fā)生銀鏡反應(yīng)的結(jié)構(gòu)簡式為/r/n_____/r/n。/r/n(6)/r/n參照上述合成路線，以/r/n和/r/n為原料，設(shè)計制備/r/n的合成路線/r/n(/r/n無機試劑任選/r/n)_____/r/n。/r/n（1）/r/n酯基、醚鍵/r/n/r/n光照/r/n/r/n（2）/r/n/r/n/r/n（3）/r/n取代反應(yīng)/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的分子式及后續(xù)流程可知/r/nA/r/n為/r/n，/r/nB/r/n為/r/n，/r/nB/r/n和/r/nNaCN/r/n發(fā)生取代反應(yīng)生成/r/nC/r/n，對比已知信息可知/r/nD→E/r/n發(fā)生/r/n-CN/r/n水解，/r/nD/r/n為/r/n；由/r/nF→G/r/n可知/r/nF/r/n為/r/n；由反應(yīng)⑧可知/r/nH/r/n為/r/n，據(jù)此解答。/r/n(1)/r/n由曲美布汀分子的結(jié)構(gòu)可知含氧官能團的名稱是酯基、醚鍵；反應(yīng)①為苯環(huán)側(cè)鏈烷基上的/r/nH/r/n被/r/nCl/r/n取代，反應(yīng)條件是光照；/r/n(2)/r/n與/r/n4/r/n個不同的原子或原子團相連的/r/nC/r/n為手性碳，則用星號/r/n(*)/r/n標(biāo)出/r/nG/r/n的手性碳原子為/r/n；/r/n(3)/r/n反應(yīng)③為/r/n和/r/nC/r/n2/r/nH/r/n5/r/nBr/r/n發(fā)生反應(yīng)生成/r/n和/r/nHBr/r/n，屬于取代反應(yīng)；/r/n(4)/r/n反應(yīng)⑦為/r/n和/r/n反應(yīng)產(chǎn)生/r/n和/r/nHBr/r/n，化學(xué)方程式為/r/n+/r/n+HBr/r/n；/r/n(5)E/r/n為/r/n，其同分異構(gòu)體/r/nM/r/n滿足：/r/n①/r/nM/r/n為苯的二元取代物，即苯環(huán)上有/r/n2/r/n個取代基；/r/n②/r/nM/r/n能發(fā)水解反應(yīng)且水解產(chǎn)物之一能與/r/n發(fā)生顯色反應(yīng)，則為酚酯；/r/n取代基為/r/nHCOO-/r/n和/r/n-C/r/n3/r/nH/r/n7/r/n(2/r/n種/r/n)/r/n時有/r/n3×2=6/r/n種，取代基為/r/nCH/r/n3/r/nCOO-/r/n和/r/n-C/r/n2/r/nH/r/n5/r/n時有鄰間對/r/n3/r/n種，取代基為/r/nC/r/n2/r/nH/r/n5/r/nCOO-/r/n和/r/n-CH/r/n3/r/n時有鄰間對/r/n3/r/n種，共/r/n6+3+3=12/r/n種；其中核磁共振氫譜為五組峰且能發(fā)生銀鏡反應(yīng)的結(jié)構(gòu)簡式為/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和氫氣合成，/r/n由/r/n1,3-/r/n丁二烯和溴發(fā)生加成反應(yīng)合成，即以/r/n和/r/n為原料，設(shè)計制備/r/n的合成路線為/r/n/r/n。/r/n12．（江蘇南通市/r/n高三模擬/r/n）/r/n化合物/r/nH/r/n具有抗菌、消炎、降血壓等多種功效，其合成路線如圖：/r/n(1)A→B/r/n的反應(yīng)類型為/r/n_______/r/n。/r/n(2)/r/n可用于鑒別/r/nB/r/n與/r/nC/r/n的常用化學(xué)試劑為/r/n_______/r/n。/r/n(3)F/r/n分子中碳原子的軌道雜化類型有/r/n_______/r/n種。/r/n(4)D/r/n的分子式為/r/nC/r/n10/r/nH/r/n11/r/nO/r/n2/r/nC1/r/n，寫出/r/nD/r/n的結(jié)構(gòu)簡式：/r/n_______/r/n。/r/n(5)E/r/n的一種同分異構(gòu)體同時滿足下列條件，寫出該同分異構(gòu)體的結(jié)構(gòu)簡式：/r/n_______/r/n。/r/n①分子中含苯環(huán)和碳氮雙鍵，能發(fā)生銀鏡反應(yīng)；/r/n②分子中有/r/n4/r/n種不同化學(xué)環(huán)境的氫原子。/r/n(6)/r/n設(shè)計以/r/n、/r/n乙烯為原料制備/r/n的合成路線/r/n____________(/r/n無機試劑和有機溶劑任選，合成路線示例見本題題干/r/n)/r/n。/r/n（1）/r/n取代反應(yīng)/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對比A、B的結(jié)構(gòu)簡式，/r/nA/r/n→/r/nB/r/n發(fā)生了取代反應(yīng)，/r/nC/r/n和/r/nB/r/n相比較，/r/nB/r/n結(jié)構(gòu)上的酯基分成了兩個取代基，一個是酚羥基，一個酮羰基，/r/nC/r/n→D→/r/nE/r/n過程中，可知在/r/nC/r/n的結(jié)構(gòu)上引入了一個取代基，根據(jù)反應(yīng)條件以及D的分子式，可推測/r/nD/r/n的結(jié)構(gòu)簡式是/r/n，/r/nE/r/n→/r/nF/r/n的過程中先發(fā)生了加成、后發(fā)生了消去，/r/nF/r/n→/r/nG/r/n，酚羥基和雙鍵相連形成環(huán)狀，/r/nG/r/n→/r/nH/r/n，結(jié)構(gòu)中的氰基水解成了羧基，據(jù)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結(jié)構(gòu)中酚羥基上的/r/nH/r/n原子，故/r/nA→B/r/n的反應(yīng)類型為取代反應(yīng)。/r/n(2)B/r/n的結(jié)構(gòu)中有官能團酯基，/r/nC/r/n的結(jié)構(gòu)中有酮羰基和酚羥基，可利用酚羥基的顯色反應(yīng)來區(qū)分/r/nB/r/n和/r/nC/r/n，故鑒別/r/nB/r/n與/r/nC/r/n的常用化學(xué)試劑為/r/nFeCl/r/n3/r/n溶液。/r/n(3)F/r/n的結(jié)構(gòu)簡式為/r/n，苯環(huán)上的碳原子采取的是/r/nsp/r/n2/r/n雜化，甲基上的碳原子采取的是/r/nsp/r/n3/r/n雜化，碳碳雙鍵上的碳原子采取的是/r/nsp/r/n2/r/n雜化，/r/n-CN/r/n上的碳原子采取的/r/nsp/r/n雜化，/r/n-CH/r/n2/r/n-/r/n中的碳原子采取的是/r/nsp/r/n3/r/n雜化，故/r/nF/r/n分子中碳原子的軌道雜化類型有/r/n3/r/n種。/r/n(4)D/r/n的分子式為/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反應(yīng)得到/r/nE/r/n，/r/nE/r/n的結(jié)構(gòu)簡式相對于/r/nC/r/n來說，在苯環(huán)上多了一個取代基/r/n-CH/r/n2/r/n-CN/r/n，通過/r/nD/r/n→/r/nE/r/n反應(yīng)的條件，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上的一個原子，結(jié)合/r/nC/r/n→/r/nE/r/n的反應(yīng)條件和/r/nD/r/n的分子式，可知/r/nX/r/n原子應(yīng)是/r/nC1/r/n，則/r/nD/r/n的結(jié)構(gòu)簡式相對于/r/nC/r/n來說，在苯環(huán)上多了一個取代基/r/n-CH/r/n2/r/nCl/r/n，所以/r/nD/r/n的結(jié)構(gòu)簡式為/r/n。/r/n(5)E/r/n的一種同分異構(gòu)體滿足分子中含苯環(huán)和碳氮雙鍵，能發(fā)生銀鏡反應(yīng)，說明有取代基/r/n-CHO/r/n，有/r/n結(jié)構(gòu)，結(jié)合/r/nE/r/n的結(jié)構(gòu)簡式/r/n和分子式/r/nC/r/n11/r/nH/r/n11/r/nO/r/n2/r/nN/r/n，可知其不飽和度為/r/n7/r/n，/r/n/r/n同分異構(gòu)體的不飽和度是相同的，已知一個苯環(huán)中的不飽和度為/r/n4/r/n，一個醛基中的不飽和度為/r/n1/r/n，一個/r/n的不飽和度也為/r/n1/r/n，從結(jié)構(gòu)簡式可以看出分子中只有一個/r/nN/r/n原子，也就是說其同分異構(gòu)體中只能有一個/r/n，根據(jù)不飽和度數(shù)，其結(jié)構(gòu)中必然含有/r/n2/r/n個醛基，另外分子中有/r/n4/r/n種不同化學(xué)環(huán)境的氫原子，醛基中有/r/n1/r/n種，若/r/n連接的/r/nH/r/n原子是/r/n1/r/n種，苯環(huán)上有/r/n1/r/n種，還有/r/n1/r/n種根據(jù)剩余的碳?xì)湓觽€數(shù)，說明結(jié)構(gòu)中還有/r/n2/r/n個甲基，這樣苯環(huán)上就有/r/n2/r/n個醛基，/r/n2/r/n個甲基，和一個/r/n，故滿足條件的同分異構(gòu)體有/r/n/r/n，移動甲基又有/r/n/r/n，/r/n移動醛基又有/r/n/r/n。/r/n(6)/r/n由/r/nH/r/n的合成路線/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，通過/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，通過路線/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，可利用我們熟悉的乙烯制乙醇，乙醇制乙醛來合成。故合成路線如下：/r/n。/r/n13．（山東青島市/r/n高三模擬/r/n）/r/n有機化合物/r/nI/r/n是一種重要的有機合成中間體，合成路線如圖：/r/n

/r/n已知：/r/n①/r/n+/r/n②/r/n+/r/n回答下列問題：/r/n(1)A/r/n的名稱為/r/n____/r/n，/r/nH/r/n的結(jié)構(gòu)簡式為/r/n___/r/n。/r/n(2)/r/n寫出/r/nD→E/r/n的化學(xué)方程式/r/n___/r/n，/r/nF→G/r/n過程中涉及的反應(yīng)類型有/r/n___/r/n。/r/n(3)/r/n物質(zhì)/r/nC/r/n有多種同分異構(gòu)體/r/n(/r/n不考慮立體異構(gòu)/r/n)/r/n，符合下列條件的同分異構(gòu)體有/r/n__/r/n種。/r/n①苯環(huán)上有兩個取代基/r/n②既能與/r/nNaHCO/r/n3/r/n溶液反應(yīng)又能與/r/nFeCl/r/n3/r/n溶液顯色/r/n寫出其中一種含有手性碳原子的結(jié)構(gòu)簡式：/r/n____/r/n。/r/n(4)/r/n設(shè)計由/r/nCH/r/n3/r/nCHOHCH/r/n3/r/n為原料制備/r/n的合成路線/r/n(/r/n無機試劑任選/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加成反應(yīng)、消去反應(yīng)/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與苯酚發(fā)生取代反應(yīng)生成/r/n，/r/n與/r/n發(fā)生加成反應(yīng)生成/r/n，/r/n在氫氧化鈉水溶液中加熱發(fā)生水解反應(yīng)生成/r/nE/r/n，/r/nE/r/n酸化后脫羧得到/r/nF/r/n，根據(jù)/r/nF/r/n的分子式可推出/r/nF/r/n為/r/n，則/r/nE/r/n為/r/n，/r/n在稀氫氧化鈉溶液中加熱生成/r/n，/r/n與/r/nH/r/n在堿性條件下發(fā)生取代反應(yīng)生成/r/n，根據(jù)反應(yīng)/r/n+/r/n，推出/r/nH/r/n為/r/n。/r/n(1)A/r/n為/r/n，名稱為/r/n1/r/n，/r/n3—/r/n丁二醇，/r/nH/r/n的結(jié)構(gòu)簡式為/r/n；/r/n(2)D→E/r/n是/r/n與氫氧化鈉反應(yīng)生成/r/n、/r/n和水，反應(yīng)的化學(xué)方程式為/r/n+2NaOH→/r/n+/r/n+H/r/n2/r/nO/r/n，/r/nF→G/r/n是/r/n轉(zhuǎn)化為/r/n，過程中涉及的反應(yīng)類型有加成反應(yīng)形成環(huán)、消去反應(yīng)脫去/r/n1/r/n個水分子形成碳碳雙鍵；/r/n(3)/r/n物質(zhì)/r/nC/r/n為/r/n，符合條件的同分異構(gòu)體：①苯環(huán)上有兩個取代基；②既能與/r/nNaHCO/r/n3/r/n溶液反應(yīng)又能與/r/nFeCl/r/n3/r/n溶液顯色，則含有羧基和酚羥基，根據(jù)不飽和度可知應(yīng)該還含有一個碳碳雙鍵，苯環(huán)上除酚羥基外，另一取代基可以為/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，兩取代基在苯環(huán)上的位置有鄰、間、對/r/n3/r/n種位置，故符合條件的同分異構(gòu)體共有/r/n8/r/n3=24/r/n種；/r/n其中含有手性碳原子的結(jié)構(gòu)簡式為：/r/n；/r/n(4)CH/r/n3/r/nCHOHCH/r/n3/r/n催化氧化生成丙酮，丙酮發(fā)生類似反應(yīng)①/r/n+/r/n生成/r/n；/r/nCH/r/n3/r/nCHOHCH/r/n3/r/n在濃硫酸中發(fā)生消去反應(yīng)生成丙烯，丙烯與氯氣發(fā)生加成反應(yīng)生成/r/nCH/r/n2/r/n=CHCH/r/n2/r/nCl/r/n，/r/n與/r/nCH/r/n2/r/n=CHCH/r/n2/r/nCl/r/n反應(yīng)制備/r/n，合成路線如下：/r/n。/r/n14．（山東煙臺市/r/n高三/r/n三模）/r/n化合物/r/nG/r/n是一種有機光電材料中間體，其合成路線如下：/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的結(jié)構(gòu)簡式是/r/n_____/r/n。/r/n(2)B→C/r/n和/r/nE→F/r/n的反應(yīng)類型分別是/r/n____/r/n、/r/n____/r/n。/r/n(3)E/r/n中官能團的名稱為/r/n____/r/n；/r/nC→D/r/n的反應(yīng)方程式為/r/n____/r/n。/r/n(4)/r/n寫出兩種符合下列條件的/r/nG/r/n的同分異構(gòu)體的結(jié)構(gòu)簡式/r/n____/r/n。/r/n①/r/n具有六元碳環(huán)結(jié)構(gòu)/r/n②/r/n能與/r/nNaHCO/r/n3/r/n溶液反應(yīng)/r/n③/r/n能發(fā)生銀鏡反應(yīng)/r/n④/r/n核磁共振氫譜顯示六組峰，且峰面積之比為/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二溴丁烷為原料合成/r/n，寫出能獲得更多目標(biāo)產(chǎn)物的較優(yōu)合成路線/r/n(/r/n其它試劑任選/r/n)____/r/n。/r/n（1）/r/n/r/n（2）/r/n/r/n氧化反應(yīng)/r/n/r/n取代反應(yīng)/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與乙烯反應(yīng)生成/r/nB(/r/n)/r/n，結(jié)合已知反應(yīng)可知/r/nA/r/n應(yīng)為/r/n，/r/nB/r/n與高錳酸鉀反應(yīng)斷裂碳碳雙鍵生成羧基得到/r/nC/r/n，/r/nC/r/n為/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與乙醇發(fā)生酯化反應(yīng)得到/r/nD/r/n，/r/nD/r/n為/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條件下發(fā)生取代反應(yīng)生成/r/nE/r/n，/r/nE/r/n在乙醇鈉條件下與一溴甲烷發(fā)生取代反應(yīng)得到/r/nF/r/n，/r/nF/r/n中的酯基堿性條件下發(fā)生水解反應(yīng)，再酸化得到/r/nG/r/n，據(jù)此分析。/r/n(1)/r/n根據(jù)以上分析可知/r/nA/r/n為/r/n，故/r/n；/r/n(2)B/r/n到/r/nC/r/n是/r/nB/r/n中的碳碳雙鍵被高錳酸鉀氧化斷裂生成羧基，/r/nE/r/n到/r/nF/r/n發(fā)生的是酯基所連碳上的氫原子被甲基取代的反應(yīng)，故氧化反應(yīng)；取代反應(yīng)；/r/n(3)/r/n由/r/nE/r/n的結(jié)構(gòu)簡式可知，其中含有酯基和羰基兩種官能團，/r/nC/r/n與乙醇發(fā)生反應(yīng)生成/r/nD/r/n，反應(yīng)的方程式為/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具有六元碳環(huán)結(jié)構(gòu)/r/n②/r/n能與/r/nNaHCO/r/n3/r/n溶液反應(yīng)說明存在羧基；/r/n③/r/n能發(fā)生銀鏡反應(yīng)，說明含有醛基，/r/n④/r/n核磁共振氫譜顯示六組峰，且峰面積之比為/r/n3/r/n:/r/n1/r/n:/r/n4/r/n:/r/n4/r/n:/r/n1/r/n:/r/n1/r/n，可知結(jié)構(gòu)中應(yīng)含有一個甲基，且為對稱結(jié)構(gòu)，則符合條件的結(jié)構(gòu)有：/r/n或/r/n或/r/n，故/r/n或/r/n或/r/n；/r/n(5)/r/n可