HO_ch19_2e
有机化学综合测试题
有机化学综合测试题综合测试题⼀、命名或写结构（每题1分，共10分）1. 2. C CH 6H 5CON(CH 3)23C=CHCHOHCH 2CH 2CH 2CH 2CH 3CH 33.4.5. 6. (E)-3,4-⼆甲基-3-庚烯7. (R)-甲基仲丁基醚 (Fischer 投影式) 8. 2-氨基-3-硝基吡啶9. 反-1,4-⼆甲基环⼰烷(最稳定的构象式) 10. δ-⼰内酰胺⼆、完成下列反应，写出主要产物（每题2分，共30分）1、2、3.4.5. 6. 7.8. 25CH -O 3--N=N-H C 6H 5CH 2CH 3 + Br 2+ HBr OH H 2SO 4BrCH 2CH=CHCl + H 2O -OH 2CH 2COCH 3 + ( )H +CH 2－C －CH 3O OH 2N--OH + CH 3-C-O-C-CH 3 O O(1:1)CH 3CONH 2NaNO 2+ HCl BrCH H 2C －CH 2BrCH 2( )①CH 33+( )CrO 3( )-CH=CH 2 + HBr ( )9. (CH 3)2C-CH 2OH + HIO 4H 2 OH 10. C 3C =CH11.12.13.14. 15. 三、选择题(含多选题。
每题1分, 共30分)1. 属于烃基的是 A.C 6H 5CO- B.-CH 2COOH C.-CH=CH 2 D.-OCH 32. 所有碳原⼦处于同⼀平⾯的分⼦是A. CH 3CH=CHCH 2CH 3B. CH 2=CHC ≡CHC. CH 2=CH-CH 2CH 3D. 3. 当丁烷从最稳定构象旋转240°时其间经过⼏次最⾼能量状态？A 、1次B 、2次C 、3次D 、4次4. 在苄基正离⼦中，带正电荷的碳原⼦杂化类型是A.sp 杂化B.sp 2杂化C.sp 3杂化D.不杂化5. 化合物CH 2=CHCH ＝CHCH 3分⼦中存在A. π-π共轭效应B. p-π共轭效应C. σ-π超共轭效应D. σ-p 超共轭效应6. 下列共价键伸缩振动所产⽣的吸收峰的波数最⼤的是A. C －OB. C —CC. C －ND. C －H7. 酯的碱性⽔解机制属于A.亲核加成－消除B.亲核取代C.亲电取代D. 游离基取代8. 构象异构属于A.构型异构B.互变异构C.构造异构D.⽴体异构H 2( )3+( )△( )NH 22( )－－CHO + CH 3CHO 3CH －CH 2O + CH 3OH H +CHCH 3(CH 2)12CH=CHCH-CH-CH 2-O-P-O-CH 2CH 2N +(CH 3)3 + H 2O OH NH 3OO -OH -HOOC-CH 2-CHCOO - + 1mol NaOH+NH 3-CH 39. D-葡萄糖和D-半乳糖的关系是A.对映体B.⾮对映体C.异头物D.差向异构体 10. 不属于S N 2机制的说法是A.产物的构型完全转变B.增加氢氧化钠浓度，卤代烷⽔解速度加快C.反应不分阶段⼀步完成D.反应速度叔卤代烷明显⼤于伯卤代烷11.下列叙述正确的是A.皂化值越⼤，油脂平均分⼦量越⼤。
微波消解-冷原子荧光光谱法测定V_(2)O_(5)-WO_(3)TiO_(2)系脱硝催化剂中汞的含量
doi: 10.11857/j.issn.1674-5124.2020030019微波消解-冷原子荧光光谱法测定V 2O 5-WO 3/TiO 2系脱硝催化剂中汞的含量闫月娥(攀枝花学院钒钛学院，四川 攀枝花 617000)摘 要: 该文采用酒石酸+氢氟酸+硝酸体系微波消解处理V 2O 5-WO 3/TiO 2脱硝催化剂试样，以氢化物发生-冷原子荧光光谱法(HG-AFS)测定其中汞的含量。
文中探讨原子化器炉温、原子化器高度、载气及屏蔽气流量及硼氢化钾-氢氧化钠浓度等关键参数对汞测定的影响，并通过系列试验确定上述关键参数的最优解：还原剂浓度为0.2 g/L （硼氢化钾）–2.0 g/L （氢氧化钠），载气和屏蔽气流量分别为600 mL/min 和800 mL/min ，原子化器炉温200 ℃，原子化器高为10 mm 。
共存元素干扰试验表明，在该实验条件下，少量As(Ⅲ)的干扰可加入高锰酸钾（1.0 g/L ）将其氧化成As （Ⅴ）得以消除，Pb （Ⅱ）的干扰可加入1.0 mL 硫酸（1+5）生成硫酸铅沉淀消除。
汞的浓度在0~2.0 μg/L 范围内线性相关系数为0.999 7，方法检出限为0.02 μg/L 。
方法用于实际样品分析，相对标准偏差（RSD ，n =7）小于4.0%，加标回收率为95%~104%。
关键词: 脱硝催化剂; 汞; 氢化物发生-冷原子荧光光谱法中图分类号: O659.2文献标志码: A文章编号: 1674–5124(2021)04–0072–05Determination the content of Hg in V 2O 5-WO 3/TiO 2 denitrification catalyst bymicrowave digestion-cold atomic fluorescence spectrometryYAN Yuee(College of Vanadium and Titanium, Panzhihua University, Panzhihua 617000, China)Abstract : The sample of V 2O 5-WO 3/TiO 2 denitrification catalyst was treated by microwave digestion in tartaric acid+hydrofluoric acid+nitric acid system, the mercury content was determined by hydride generation-cold atomic fluorescence spectrometry (HG-AFS). The influence of the key parameters such as the furnace temperature of atomizer, the height of atomizer, the flow rate of carrier and shielding gas, and the concentration of potassium borohydride-sodium hydroxide on the determination of mercury was discussed in this paper, and then, the optimal solution of the above key parameters was obtained after a series of experiments: the concentration of reducing agent is 0.2 g/L (potassium borohydride) –2.0 g/L (sodium hydroxide), the flow rates of carrier gas and shielding gas are 600 mL/min and 800 mL/min respectively, the furnace temperature of atomizer is 200 ℃, and the height of atomizer is 10 mm. The interference test of coexistence elements show that: the interference of a amount of As (Ⅲ) and Pb (Ⅱ) can be eliminated by added potassium permanganate (1.0 g/L) which can make the As (Ⅲ) oxidized to As (Ⅴ) and sulphuric acid (1+5) which can make the Pb (Ⅱ)收稿日期: 2020-03-04；收到修改稿日期: 2020-04-12基金项目: 攀枝花市平台建设项目（20180816）作者简介: 闫月娥（1981-），女，河南周口市人，工程师，硕士，主要从事钒钛产品检验。
有机人名反应大全
索引：Arbuzov 反应Arndt-Eister 反应.Baeyer-Villiger 氧化Beckmann 重排Birch还原Bischler-Napieralski 合成法Bouveault・Blanc 还原Bucherer 反应Cannizzaro 反应Chichibabin 反应Claisen酯缩合反应Claisen・Schmidt 反应Clemmerisen 还原Combes合成法Cope重排Cope消除反应Curtius 反应Dakin反应Darzens 反应Demjanov 重排Dieckmann 缩合反应Elbs反应Eschweiler-Clarke 反应Favorskii 反应Favorskii 重排Friedel—Crafts 烷基化反应Friedel—Crafts 酰基化反应Fries重排Gabriel合成法Arbuzov（加成仮应Gattermann 反应Gattermann・ Koch 反应Gomberg-Bachma nn 反应Hantzsch合成法Haworth反应Hell-Volhard-Zelinski反应Hinsberg 反应Hofmann烷基化Hofmann消除反应Hofmann重排（降解）Houben・Hoesch 反应Hunsdiecker 反应Kiliani孰化增碳法Knoevenagel 反应Knorr反应Koble反应Koble-Schmitt 反应Leuckart 反应Lossen反应Mannich 反应Meerwein・Ponndorf 反应Meerwein・Ponndorf 反应Michael加成反应Norrish I和II型裂解反应Oppenauer 氧化Paal-Knorr反应Pictet・Spengler 合成法Pschorr 反应Reformatsky反应Reimer-Tiemann 反应Reppe合成法Robinson缩环反应Rosenmund 还原Ruff递降反应Sandmeyer 反应Schiemann反应Schmidt 反应Skraup合成法Sommelet-Hauser 反应Stephen 还原Stevens 重排Strecker氨基酸合成法Tiffeneau・Demja nov 重排Ullmann 反应Vilsmeier 反应Wagn er-Meerwei n 重排Wacker反应Williamson 合成法Wittig反应Wittig-Horner反应Wohl递降反应Wolff-Kishner-黄鸣龙反应Yu rev反应Zeisel屮氧基测定法亚磷酸三烷基酯作为亲核试剂与卤代烷作用，生成烷基麟酸二烷基酯和一个新的卤代烷:R*(RO)3P +亚髓酸三烷荃酯R1夏------- - (RO)2P=O 4 RX烷基瞬酸二诜基酯卤代烷反应时，其活性次序为：Rl>RBr>R'Clo 除了卤代烷外，烯丙型或烘丙型 卤化物、 卤代醯、或卤代酸酯.对甲苯磺酸酯等也可以进行反应。
考研经典合成题第2套带答案
2 CH 3
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考点：1. 由醇制备烯烃；2. 由格氏试剂和酮制备醇；3. 片呐醇重排； 4. 氯甲基化反应；5. 邻二醇合成；6. Williamson 法醚的合成。
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Me Me3CC CH(CH2 )4CH3
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O HO O α OH (1). HCN OH (2). H3O CO 2Et
NaOEt BrCH2CH2Br
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37. 用指定的原料和其它必要的试剂合成目标化合物
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2023重庆二诊(教科院卷)高2023届学业质量调研抽测(第二次)化学试卷及答案
高2023届学业质量调研抽测（第二次）高三化学试卷（化学试题卷共8页，考试时间75分钟，满分100分）注意事项z1.答卷前，考生务必将自己的学校、姓名、考号填写在答题卡上。
2.回答选择题时，逃出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。
如需改动，用橡皮擦干净后，再逃涂其他答案标号。
回答非逃择题时，将答案写在答题卡上，写在本试卷上无效。
3考试结束后，将本试卷和答题卡一并收回。
可能用到的相对原子质量：Hl C12016Na 23Mn 55一、选择题z本题共14个小题，每小题3分，共42分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1.民以食为天。
下列说法错误的是A.硬化油不易被空气氧化变质，是制造人造奶油的原料8.亚硝酸纳具有一定毒性，不能用作食品防腐剂C.铝制餐具不直用来长时间存放酸性食物或碱性食物D.酱油中添加铁强化剂，可减少缺铁性贫血问题的发生2.SOCb是一种液态化合物，其与水反应的化学方程武为：SOCb Hb O =S 02↑＋2HC1↑，下列说法错误的是＼αOH A，。
为式构S 结矶的U P2UH M 比角F D山UAA户L（、飞＼B氯离子的结构示慈因为也）｝JJD.S02与时）的空间结构相同3.氟他脑是一种抗肿瘤药，其结构简式如下。
下列关于氟他股说法正确的是A .第一电离能：C <N <O <FB.所有元索者I ！处于元素周期表的p 区C.1个分子中采取sp 3杂化的原子数为4D.基态氟原子最高能级的轨道形状为哑铃形臼N心女×4.加入下罗1J 物质后，不会影响溶液中N矿、Al 3•、Fe 3•、S03－、c1－大量共存的是A.苯盼B.Mg(OHhC.Na乞C030.NH,Cl高三化学试卷第1页（共8页）5.设NA为阿伏加德罗常数的值，下列说法正确的是A.39g0含有1键的数目为1.5NAB. 18gD 20和l8gH 20中含有的质子数均为lONAC.22. 4LS02和11.2L02完全反应生成so l 的分子数为NA D.15. 6gNa 202与足量C02完全反应，转移电子数为0.2N A 6.下列叙述正确的是A.工业上用98.3%的浓硫酸吸收S03$1J 备硫酸B.乙烯、乙醋、乙酸均能与氢气发生加成反应C.Zn-N键中离子键成分的百分数大于Zn-0键．0.氧氟酸溶蚀破璃与Fe C h溶液腐蚀覆铜板均为氧化还原反应7.下列实验装置与操作正确，且能达到实验目的的是』Fe(OH)3肢体制备B. NH3制备4 ：主翩翩氯－II2在＿.-J 水的精捧f／YI _pH试纸c.液液分离D.pH测试8.凯芙拉是一种耐高温、耐酸碱的高分子材料，司由脐IJY在一定条件下反应制得，相应结构简式如下。
人教版八年级上册英语 Unit4 习题课件 Period 3 Section B (1a-2e)
1. 南北方的菜肴有共同之处。【中考·铁岭】
Thedishesinthenorthandsouthhavesomething_
__________________. incommon
2. IlikesportsandLilylikessports, too.
A
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5. Thetalentshows________(提供) give
peopleawaytomaketheirdreamscometrue.
二、根据汉语意思完成句子, 并背记英语句子
6. 所有的这些节目都有一个共同点。
Alltheseshows________onething_____________
A. have; incommonB. take; atcommon
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【点拨】用固定结构法。have…incommon 意为“……相同”。
3. Thereare_____B___flowersonthehill. A. kindsB. allkindsof C. kindofD. allkindof
5. 每个人都应该在环保方面发挥作用。(play)
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6. 赫本在《罗马假日》中扮演一位公主, 其堪称经典。 【中考·无锡】
1 talent
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STERIS Vaporized Hydrogen Peroxide (VHP) 生物消毒系统及服务
STERIS provides a wide range of vaporized hydrogen peroxide (VHP®) bio-decontamination systems and services, utilizing Vaprox® Sterilant for broad-spectrum efficacy against viruses, bacteria, yeasts, and bacterial spores. Vaporized hydrogenperoxide bio-decontamination is crucial, not only for pharmaceutical and biotechnology production, but also for agricultural industries and healthcare facilities. The STERIS bio-decontamination systems use a “dry process” hydrogenperoxide vapor distribution, which eliminates the risk of condensation on surfaces.The advantages ofdecontamination with vaporized H 2O 2 include:• Easy to use• Effective against biological contaminants• Ideal for low-temperature processes• Processes can be validated • Compatible with a wide variety of materials• Environmentally friendly and safe for operators• Leaves no toxic residue, only water vapor and oxygenA trusted partnerFor several years, STERIS has used a trusted sensing technology from Vaisala. In 2018, STERIS became interested in a new solutionfrom Vaisala: the HPP270-series hydrogen peroxide vapor probe. The probes feature PEROXCAP® sensing technology. The sensors provide accurate measurements for hydrogen peroxideconcentration or ppm (parts per million) and several other parameters, most importantly: relative humidity, temperature, and a new parameter: relativesaturation — which indicates when condensation will occur.Validating bio-decontaminationIn DSVA (surface disinfection by airways) the aim is to prove that bacteria and microorganisms have been eradicated and the results must demonstrate maximumeffectiveness throughout the bio-decontaminated area. To validate a cleanroom bio-decontamination, it is essential that STERIS use a highly accurate sensor that can provide stable, repeatable data on the concentration of hydrogen peroxide vapor ppm. Vaisala’s unique technology met STERIS’s requirements for measurement reliability and repeatability. 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For STERIS’s dry method, it is necessary to avoid condensation that can lead to equipment deterioration.In the absence of hydrogen peroxide vapor, the relativehumidity of the air is equal to the relative saturation (1). When vaporized H 2O 2 is introduced, the relative saturation is greater than the relative humidity (2).During H 2O 2 vapor bio-decontamination processes, there is always water vapor in addition to hydrogen peroxide vapor. To control condensation, you need to know both the humidity of the air caused by water vapor and by hydrogen peroxide vapor.Relative saturation, which indicates the concentration of vaporizedhydrogen peroxide and water vapor in the air, is the only value thatrepresents both vapors. Monitoring the relative saturation value during a process is therefore crucial, because it indicates saturation point of the combined vapors: water and hydrogen peroxide.Reliable measurements mean reliable processesSTERIS systems required a probe capable of providing accurate measurements for hydrogen peroxide ppm, temperature, relative humidity and relative saturation. Using the unique Vaisala PEROXCAP ® hydrogen peroxide sensor technology, the HPP272 probe can also measure two other parameters: dewpoint and vapor pressure, which can also be useful parameters in bio-decontamination. The probe guarantees reliable and precise hydrogen peroxide measurements throughout thebio-decontamination cycle, even in high humidity.The reliable and reproducible measurements of Vaisala’s vaporized hydrogen peroxide probes allow STERIS to achieve a high degree of confidence in their bio-decontamination procedures, success during annual audits, and a high level of product quality.Please contact us at/contactusScan the code formore informationRef. B212075EN-A ©Vaisala 2020This material is subject to copyright protection, with all copyrights retained by Vaisala and its individual partners. All rights reserved. Any logos and/or product names are trademarks of Vaisala or its individual partners. The reproduction, transfer, distribution or storage of information contained in this brochure in any form without the prior written consent of Vaisala is strictly prohibited. All specifications — technical included — are subject to change without notice.。
有机化学综合复习题及解答
有机化学综合复习题及解答1．写出下列反应的主要产物：OH OH+(1).H SO (CH 43)23CCH 2OH(2).(CH 3)2CC(CH 3)H 2(3).H 2SO 4(4).OHNaBr,H 2SO 4OH(5).OHHBr(6).OHPCC CH 2Cl 2CH 3CH 3(7).PBr HOH 3(8).O ()C MgBr C acetone12H 52H 5CH 3(2)H 3O +CH 3OCHCH CH 2OH(9).(10).CH H 5IO 6CH 33OHCH 3(11).H SO 4(A1)O 3H CH 23B(2)Zn,H 2OOH解答：O(1).(CH 3)2CCHCH 3(2).CH 3CC(CH 3)3(3).Br(4).CH 3(CH 2)3CH 2Br(5).(6).CH 3(CH 2)5CHOCH 3CH 3(7).HBr (8).OH +antiomer C 2H 5C 2H 5CH 3OHH 3C CH 3OO(9).(10).CH 3C(CH 2)4CCH(11).ACH 33CH 3CHCH CH 2CH 3CH 3BO CH 3O 2．解释下列现象：（1）为什么乙二醇及其甲醚的沸点随分子量的增加而降低？b.p.CH 2OH CH 2OH197CCH 2OCH 3CH 2OH125CCH 2OCH 3CH 2OCH 384C（2）下列醇的氧化（A）比（B）快？(A)OHMnO 2O(B)MnO 2OHO（3）在化合物（A ）和（B ）的椅式构象中，化合物（A ）中的-OH 在e 键上，而化合物（B ）中的-OH 却处在a 键上，为什么？(A)OH(B)OOOH解答：2.（1）醇分子中的羟基是高度极化的，能在分子间形成氢键，这样的羟基越多，形成的氢键越多，分子间的作用力越强，沸点越高。
甲醚的形成导致形成氢键的能力减弱，从而沸点降低。
（2）从产物看，反应（A ）得到的是共轭体系的脂芳酮，而（B ）没有这样的共轭体系。
WiFi Extender 用户手册说明书
1. Coloque las antenas.Gire las antenas en el sentido de las agujas del reloj para colocarlas en el conector. Doble y gire las antenas hasta su posición. Compruebe que las antenas están bien apretadas.2. Encienda el extensor de rango.Coloque el extensor de rango cerca del router WiFi. Conecte el extensor de rango a una toma de corriente. Pulse el botón de encendido si fuera necesario.3. Conéctelo al router WiFi.Pulse el botón WPS del extensor de rango. El indicador de tasa deconexión y el indicador de estado de conexión entre el dispositivo y elextensor de rango parpadean en verde.Espere un máximo de 2 minutos y pulse el botón WPS en el router WiFi.Tras unos segundos, el indicador de tasa de conexión de 2,4 GHz seilumina en verde, lo que indica una buena conexión entre el router WiFi y el extensor de rango. Si el indicador de tasa de conexión de 2,4 GHzno se ilumina o el indicador parpadea en ámbar, inténtelo de nuevo. Si el indicador continúa sin iluminarse, siga las instrucciones que aparecen a la derecha sobre “No (o no estoy seguro)” a la derecha.4. Añada una banda de WiFi.Si el router WiFi es compatible con la banda de 5 GHz (no todos losrouters lo son) y desea ampliar esa banda, repita el Paso 3 y compruebe que el indicador de tasa de conexión de 5 GHz se ilumina en verde fijopara confirmar la conexión.5. Elija una ubicación y compruebe la intensidad de la señal.Ahora que el extensor de rango está conectado al router WiFi, colóquelo en una ubicación que aumente alcance de WiFi. La ubicación elegida debe encontrarse dentro del alcance de la red del router WiFi existente.El indicador de tasa de conexión le ayudará a elegir un lugar en el que la calidad de la conexión entre el extensor de rango y el router sea óptima.Consulte la sección “Indicadores” al dorso para saber cómo muestran la mejor conexión los indicadores.Si no logra conectarse o la calidad de la conexión es mala, sitúe elextensor de rango más cerca del router WiFi y vuelva a intentarlo hasta que el indicador de 2,4 GHz o 5 GHz se ilumine en verde.6. Conecte los dispositivos WiFi.Coloque el dispositivo WiFi en una ubicación con mala cobertura delrouter WiFi. Busque el nuevo nombre de red del extensor de rango:Nombre de la red existente. NombreDeMiRedNuevo nombre de la red ampliada. MyNetworkName_2GEXToMyNetworkName_5GEXTSeleccione la nueva red y utilice la contraseña del router WiFi paraconectarse. Ahora está utilizando la red WiFi ampliada.1. Coloque las antenas.Coloque las antenas en los conectores y gírelas en el sentido de lasagujas del reloj para fijarlas. Doble y gire las antenas hasta su posición.Compruebe que las antenas están bien apretadas.2. Encienda el extensor de rango.Coloque el extensor de rango cerca del router WiFi. Conecte el extensor de rango a una toma de corriente. Pulse el botón de encendido si fuera necesario.3. Conecte el extensor de rango.Abra el administrador de conexión WiFi desde su un ordenador, tableto smartphone, y conéctese a la red del extensor de rango denominadaNETGEAR_EXT. Cuando esté conectado al extensor de rango, losindicadores de conexión entre el dispositivo y el extensor de rango seiluminarán en verde fijo.4. Configure con un navegador web.Abra un navegador web. Aparecerá directamente la pantalla de inicio.Si no se abre la ventana de inicio, vaya a .Para configurar el extensor de rango, realice los siguientes pasos:a. Haga clic en el botón NEW EXTENDER SETUP (Configuración delnuevo extensor de rango).b. Rellene los campos en la página Create Account (Crear cuenta) yhaga clic en el botón NEXT (Siguiente).c. Haga clic en el botón de WiFi RANGE EXTENDER (Extensor derango WiFi).d. Seleccione la red WiFi cuya cobertura desea ampliar y haga clic en elbotón NEXT (Siguiente).Si no desea ampliar ambas bandas WiFi, desactive las casillas2.4 GHz WiFi Networks (Redes WiFi de 2,4 GHz) o 5 GHz WiFiNetworks (Redes WiFi de 5 GHz).e. En el campo Password (network key) [Contraseña (clave de red)],escriba la contraseña de la red WiFi existente y haga clic en el botónNEXT (Siguiente).f. Establezca el nombre de red (SSID) y la contraseña de su extensorde rango y haga clic en el botón NEXT (Siguiente).g. Utilice un administrador de redes WiFi del ordenador o dispositivo WiFipara conectarse a la red WiFi del extensor de rango que acaba de crear.5. Elija una ubicación y compruebe la intensidad de la señal.Ahora que el extensor de rango está conectado al router WiFi, colóquelo en una ubicación que aumente alcance de WiFi. La ubicación elegida debe encontrarse dentro del alcance de la red del router WiFi existente.El indicador de tasa de conexión le ayudará a elegir un lugar en el que la calidad de la conexión entre el extensor de rango y el router sea óptima.Consulte la sección “Indicadores”al dorso para saber cómo muestran la mejor conexión los indicadores.Si no logra conectarse o la calidad de la conexión es mala, sitúe elextensor de rango más cerca del router WiFi y vuelva a intentarlo hasta que el indicador de 2,4 GHz o 5 GHz se ilumine en verde.Modelo EX6200Guía de inicio rápidoIntroducción¿Tiene el router WiFi un botón WPS?SíNo (o no estoy seguro)Modelo EX6200Guía de inicio rápidoHardwarePanel frontal y lateralPanel posteriorDeberá registrar su producto para poder hacer uso de nuestro servicio telefónico de soporte técnico NETGEAR. NETGEAR recomienda registrar su producto a través del sitio web de NETGEAR.Podrá encontrar actualizaciones del producto y asistencia técnica en .Encontrará el manual del usuario online en o a través de un vínculo en la interfaz del GEAR le recomienda que utilice solo los recursos del soporte técnico oficial de NETGEAR.Para consultar la declaración de conformidad actual de la UE, visite: /app/answers/detail/a_id/11621/. Para obtener información sobre el cumplimiento de normativas, visite /about/regulatory/. Consulte el documento de cumplimiento normativo antes de conectar la fuente de alimentación.© NETGEAR, Inc., NETGEAR y el logotipo de NETGEAR son marcas comerciales registradas de NETGEAR, Inc. Cualquier marca comercial distinta a NETGEAR que se utilice, se usa únicamente a modo de GEAR, Inc. 350 East Plumeria Drive, San Jose, CA 95134 (EE. UU.)Julio de 2015。
有机化学第四版答案_高鸿宾主编
第一章 习 题(一) 用简单的文字解释下列术语：（1） 有机化合物：碳氢化合物及其衍生物。
（2） 键能：形成共价鍵时体系所放出的能量。
（3） 极性键：成鍵原子的电负性相差为0.5~1.6时所形成的共价鍵。
（4） 官能团：决定有机化合物的主要性质的原子或原子团。
（5） 实验式：能够反映有机化合物元素组成的相对比例的化学式。
（6） 构造式：能够反映有机化合物中原子或原子团相互连接顺序的化学式。
（7）均裂：共价鍵断裂时，两个成鍵电子均匀地分配给两个成鍵原子或原子团，形成两个自由基。
（8） 异裂：共价鍵断裂时，两个成鍵电子完成被某一个成鍵原子或原子团占有，形成正、负离子。
（9） sp 2杂化：由1 个s 轨道和2个p 轨道进行线性组合，形成的3个能量介于s 轨道和p 轨道之间的、能量完全相同的新的原子轨道。
sp 2杂化轨道的形状也不同于s 轨道或p 轨道，而是“一头大，一头小”的形状，这种形状更有利于形成σ键。
（10） 诱导效应：由于成键原子的电负性不同而引起的电子云的转移。
诱导效应只能通过σ键传递，并且随着碳链增长，诱导效应迅速减弱。
（11） 氢键：由氢原子在两个电负性很强的原子之间形成“桥梁”而导致的类似化学键的分子间或分子内作用力。
氢键具有饱和性和方向性，但作用力比化学键小得多，一般为20~30kJ/mol 。
（12） Lewis 酸：能够接受的电子的分子或离子。
(二) 下列化合物的化学键如果都为共价键，而且外层价电子都达到稳定的电子层结构，同时原子之间可以共用一对以上的电子，试写出化合物可能的Lewis 结构式。
(1) C H 3N H 2 (2) CH 3O C H 3 (3) CH 3C OH O(4) C H 3C H =C H 2 (5) C H 3C C H (6) CH 2O 解：分别以“○”表示氢原子核外电子，以“●”表示碳原子核外电子，以“★”表示氧原子核外电子，以“△”表示氮原子核外电子，题给各化合物的Lewis 结构式如下：(1)HH H H。
semi-syntheticar...
Semi-synthetic aristolactams—inhibitors of CDK2enzymeVinod R.Hegde *,Scott Borges,Haiyan Pu,Mahesh Patel,Vincent P.Gullo,Bonnie Wu,Paul Kirschmeier,Michael J.Williams,Vincent Madison,Thierry Fischmann,Tze-Ming ChanSchering Plough Research Institute,2015Galloping Hill Road,Kenilworth,NJ 07033,USAa r t i c l e i n f o Article history:Received 7August 2009Revised 23December 2009Accepted 4January 2010Available online 7January 2010Keywords:Semi-synthetic analogs Aristolactams IC 50SARa b s t r a c tSeveral analogs of aristolochic acids were isolated and derivatized into their lactam derivatives to study their inhibition in CDK2assay.The study helped to derive some conclusions about the structure–activity relation around the phenanthrin moiety.Semi-synthetic aristolactam 21showed good activity with inhi-bition IC 50of 35nM in CDK2assay.The activity of this compound was comparable to some of the most potent synthetic compounds reported in the literature.Ó2010Elsevier Ltd.All rights reserved.In the preceding Letter we have reported on the isolation of a potent CDK2enzyme inhibitor SCH 546909,a natural product aris-tolactam analog with an inhibition IC 50of 140nM.This prompted us to undertake a semi-synthetic study of different analogs from this class.Many total syntheses of aristolactam analogs have been reported in the literature,1,2however sub-structure literature searches revealed that these compounds could be easily prepared from naturally occurring,aristolochic acids.HO H 3CONHOHOSCH 546909Several publications and reviews have been published on the occurrence,synthesis and biological activities of aristolochic acids.Aristolochic acids and aristolactams are classified as aporphinoids because of their basic skeleton which bears a distinct similarity to that of aporphins.Aristolochic acids exhibit tumor inhibitory activ-ity against the adenocarcinoma 755test system but in mice they induced papiloma.3They are also known to form covalent DNA adducts by enzymatic reductive activation of aristolochic acids in the presence of DNA.4They are also shown to induce mutagenicity in mice.5Aristolochic acid is commercially available from SigmaChemical Co.and ACROS.The commercially available aristolochic acid is a complex mixture of several analogs,with the major com-ponents being aristolochic acids II &I in 1:4ratio.We have sepa-rated commercial aristolochic acid mixtures on a preparative HPLC using YMC ODS-A C-18,10l m,5Â50cm HPLC column,eluting with 0.05%trifluoroacetic acid and acetonitrile (60:40)to obtain compounds 1–8.A typical 600mg of commercial aristolo-chic acid afforded 27.7,4.6,7.8,71.6,5.4,6.8,315.8,and 3.2mg of aristolochic acid C (5),6aristolochic D (7),77-hydroxy aristolo-chic A(6),8aristolochic acid II (1),aristolochic acid IV (4),97-meth-oxy aristolochic acid A (3),10aristolochic acid I (2),2and aristolochic acid III (8).11In our semi-synthetic modifications to prepare aristolactam analogs,the aristolochic acids were first converted to their lactams.The purified aristolochic acids were hydrogenated in ethanolic solution under 40psi hydrogen in presence of Pd/C catalyst,over-night at room temperature.The amino compound produced on reduction of nitro group,on further ring closure results in lactam.After separation and derivatization to the resulting lactam,the aromatic phenol ether derivatives were deprotected with BBr 3in methylene chloride solution.A typical demethylation 12involved stirring the aristolochic methyl ethers (15mg)in CH 2Cl 2(50ml)at 0°C with the dropwise addition of BBr 3(7.5ml,1M)in CH 2Cl 2at 0°C and then continue stirring overnight at room temperature.The reaction mixture was quenched in ice,extracted with ethyl acetate,and dried.The demethylated product was purified by HPLC.The Methylenedioxy group was removed by stirring aristolac-tams in CH 2Cl 2at 0°C and dropwise addition of a solution of PCl 5(1:1ratio).The reaction mixture was slowly allowed to attain room0960-894X/$-see front matter Ó2010Elsevier Ltd.All rights reserved.doi:10.1016/j.bmcl.2010.01.007*Corresponding author.Tel.:+19088203871;fax:+19088206166.E-mail address:**********************(V.R.Hegde).Bioorganic &Medicinal Chemistry Letters 20(2010)1384–1387Contents lists available at ScienceDirectBioorganic &Medicinal Chemistry Lettersj o ur na l h om e pa ge :w w w.e lse v ie r.c om /lo c at e/bm c ltemperature during2h and then quenched with ice,extracted with CH2Cl2and dried.The O-dihydroxy compound formed was purified by HPLC.3,4-Dihydroxy-12-chloro aristolactams were prepared from methylenedioxy containing derivatives via treatment with the dropwise addition of PCl5(1:2.5ratio)in CH2Cl2at0°C and slowly allowing the reaction mixture to attain room temperature during 3h.The reaction mixture was quenched in ice,extracted with CH2Cl2and dried.The halogenated product was further purified by HPLC.O OCOOHNO2OONR1R2R31. R1 = R2 = R3 = -H2. R1 = -OCH3, R2 = R3 = -H3. R1 = R2 = -OCH3, R3 = -H4. R1 = R3 = -OCH3, R2 = -H5.R1 = R2 = -H, R3 = -OH6. R1 = -OCH3, R2 = -OH, R3 = -H7. R1 = -OCH3, R2 = -H, R3 = -OH8. R1 = -OH,R2 = R3 = -H9. R1 = R2 = -OCH3, R3 = -HOR1R2R3R4R515. R1 = R2 = R3 = R4 = R5 =-H16. R1 = -OCH3, R2 = R3 = R4 =R5 = -H17. R1 = R2 =-OCH3 ,R3 =R4 = R5 = -H18. R1 = R3 = -OCH3, R2 = R4 = R5 = -H19. R1 = -OCH3, R2 = -OH, R3 = R4 =R5 = -H20. R1 = -OCH3, R3 = -OH, R2 = R4 = R5 = -H21. R1 = -OH, R2 = R3 = R4 = R5 = -H22. R1 = R2 = -OH, R3 = R4 = R5 = -H23. R1 = R3 = -OH, R2 = R4 = R5 = -H24. R1 = R2 = R4 = R5 = -H, R3 = -OH,25. R1 = -OCH3, R2 = R3 = R4 = -H, R5 = -CH326. R1 = -OH, R2 = R3 = R4 = -H, R5 = -CH3ARISTOLOCHIC ACID ANALOGSHO HOCOOHNO2R4R1R2R310. R1 = R2 = R3 = R4 = -H11. R1 = R2 = R3 = -H, R4 = -Cl12. R1 = -OCH3, R2 = R3 = R4 = -H13. R1 = -OH, R2 = R3 = R4 = -H14. R1 = -OCH3, R2 = R3 = -H, R4 = -Cl 27. R1 = R2 = R3 = R4 = -H28. R1 = R2 = R3 = -H, R4 = -Cl29. R1 = -OCH3, R2 = R3 = R4 = -H30. R1 = -OH, R2 = R3 = R4 = -H31. R1 = -OCH3,R2 =R3 = -H, R4 = -ClHOHONHOR4R1R2R3The aristolochic acid analogs prepared were tested in CDK2as-say13with the resulting inhibition IC50s are tabulated in Table1. Many analogs showed CDK2activity>10l M,however compounds 13,16,19,21,and24exhibited CDK2inhibition under10l M. Compound21showed a CDK2inhibition IC50of35nM,potency similar to the most potent CDK2inhibitor reported in the litera-ture14Compound13,having a hydroxyl group at C-9also showed activity in the l M range.Several natural products,like aporphinoids,morphine,and fused berberine classes of compounds,were also tested to evaluate importance of the lactam ring in the CDK2activity.All these com-pounds excepting sinomenine,sinoacutine,and tetrahydroberber-ine,have tetrahydro pyridine ring attached to phenanthrine moiety.Sinomenine and sinoacutine have morphine like ring sys-tem but tetrahydroberberine has two tetrahydro-isoquinoline ring system.All these compounds failed to show inhibition in CDK2as-say at50l M.Only compound21,displayed strong CDK2inhibition,about threefold better than the natural product SCH546909.Based on the activity profile of the different aristolochic acid and aristolactam analogs,it appears the lactam ring is essential for potent CDK2inhibition.This has been shown to be true for sev-eral potent inhibitors reported in literature.31,32Hydroxyl groups at C-7or C-9positions also appear to enhance CDK2inhibition. Additionally,theprotection of the dihydroxy groups atthe C-4 and C-5positions contributes toward the potency.However,pro-tection of amide–NH by a methyl group or substitution by a halo-gen at C-10results in reduced activity.The observations are only empirical and a detailed study would be necessary to evaluate a complete structure–activity relationship.Protection increasesthe potency in CDK2assay-CH3 in this positiondecreases activityhalogen in this positiondecrease in potencyincreases the potencyAristolochic acids and aristolactams have phenanthrin aromatic moiety similar to another class of natural product that includes staurosporine,isolated from fungus.Staurosporins are also potent kinase inhibitors and have been extensively studied as antitumor compounds.Like staurosporine,these compounds are also planar molecules and are sparingly soluble in various solvents including water.Increasing the solubility properties by salt formation or by Table1CDK2inhibition IC50s of aristolochic acids and aristolactam analogs Compound CDK2IC50(l M)1>202>203>204>205306257258159>201013.4111812>3013 5.71416.51516151616 1.217>151815>1519 2.92018>3521180.03522>302318>502420,21 2.152519,21>352617>35274>3528>2529>353016>353110Dicentrine22>50Crebanine23>50Roemerine-HBr24>50Isocorydine25>50Corydine26>50Corytuberine27>50Sinoacutine28>50Sinomenine29>50Stephanine,25>50Tetrahydro-berberine30>50V.R.Hegde et al./Bioorg.Med.Chem.Lett.20(2010)1384–13871385forming inclusion compounds with b -cyclodextrin appear to improve cellular activity.Aristolactam 21was further tested in a kinase counter screen assays,along with the natural product SCH 546909and 3233,as shown in Table 2.The results indicate that the inhibitors share a similar activity in the CDC2(cyclin-A dependent kinase,$90%homology)assay,and a lesser selectivity in other kinases assays like CDK4,AUR2(Aurora kinase),MAPK (mitogen-activated protein kinase),and AKT (ATP kinase).Cellular activities:Compound 21,the most potent and selective CDK2inhibitor from this series,was evaluated in two cellular pro-liferation assays:a colony forming assay and a soft agar growth as-say.In the soft agar growth assay compound 21showed comparable activity to 32,although compound 32appeared to lose some potency in this assay format compared to the clonogenicity assay (Table 2).In the clonogenicity assay using MCF-7cells,all three compounds inhibited growth at similar micromolar pound 21inhibited proliferation of tumor cells,with IC 50values consistent with CDK2inhibitors that are competitive with respect to ATP.The anti-proliferative activity of the com-pounds was up to eightfold selective for the tumor cells relative to the HFF normal cell line.The anti-proliferative activity of 21ar-rests the tumor cells and protects the normal cells from chemo-therapy-induced toxicity.32These data are consistent with an anti-proliferative mechanism expected for inhibition of CDK2and revealed that the aristolactam class of compounds have potential for treating proliferative disorders,including chemotherapy-in-duced alopecia.NN HNORPyrazoloquinolines32. R = -OCH 3 (SCH47089)Computer based interaction design of 21with CDK2enzyme:Thedocking experiments on CDK2enzyme with 21(SCH535270)and staurosporine were performed and are shown in Figure 1A and B.The computer docking model suggests the lactam of 21(SCH 535270)interacts with CDK2enzyme active sites in a manner anal-ogous to that observed for compounds of staurosporine class of inhibitors bound to fibroblast growth factor receptor kinase.34Two hydrogen bonds were formed between the c -lactam moi-ety of 21and CDK2.Specifically,the amide nitrogen was hydrogen bonded to the backbone carbonyl of glu-81of the CDK2enzyme and the lactam carbonyl oxygen was hydrogen bonded with the backbone NH of leu-83amide.Staurosporin also binds to the CDK2enzyme in a similar fashion.The C-9hydroxy group also ap-pears to stabilize the binding at some other sight of enzyme core.35SCH 535270,like staurosporin,is a planar molecule and exhibits similar biological properties.X-ray crystallography:Our attempts to determine the X-ray structure of inhibitor SCH 535270bound to CDK2have failed.The crystals were prepared by soaking the compound in presence of CDK2enzyme and cyclin A.The parameters like compound concentration and duration of soak were screened.Examination of the electronic density maps did not reveal the binding mode of the compound.In some cases,X-ray crystallogra-phy has failed to determine the co-structures of a compound bound to the CDK2protein even in the case of potent inhibitors.A possible explanation is that inhibitor binding requires the com-plete CDK2-cyclin-A complex,which is protocol in our screening assay.CDK2is activated by complexing with cyclin-A that induces conformational changes in the protein that affect the ATP binding site to some degree.The most significant effect involves a rotation of the C-helix,which alters the active-site geometry in the region of the triad of catalytic active-site residues Lys-33,Glu-51,and Asp-145.The amino group of Lys-33can be potential interaction site for inhibitors.The amino nitrogen appears to hydrogen bond with the oxygen of methylenedioxy group.Efforts to grow crystalsTable 2Inhibition (IC 50)of SCH 546909,21and 32in different kinases CompoundActivity IC 50(nM)Selectivity (nM)Cellular activity (l M)CDC2CDK4AUR2MAPK AKT SAG MCF7Clonogenicity SCH5469091402141420214035,335———32(SCH47089)2020020005000>50,000—>10 3.021(SCH535270)352009000350012,00011,4002–2.53.5Figure puter modeling of binding of 21(SCH 535270)and staurosporin with CDK2enzyme.1386V.R.Hegde et al./Bioorg.Med.Chem.Lett.20(2010)1384–1387of the activated CDK2-cyclin-A protein complex are in progress and will be reported in future publications.The aristolactam class of compounds represents a novel class of CDK2inhibitors.Exploration into semi-synthetic analogs provided a potent CDK2inhibitor from this class.Binding interactions by docking experiments suggested carbonyl of glu-81and NH of leu-83amide of the CDK2enzyme are involved in hydrogen bond-ing with the lactam functionality of aristolactams.CDK2inhibition causes an arrest of the cell cycle and exhibits a selective killing effect on several tumor cell lines.36AcknowledgmentsAuthors gracefully acknowledge Dr.E.Lees and Dr.R.Doll for their helpful discussion on CDK2inhibitors.References and notes1.Mix,D.B.;Guinaudeau,H.;Shamma,M.J.Nat.Prod.1982,45,657.2.Shamma,M.;Monit,J.C.Isoquinoline Alkaloids Research1972–1977;PlenumPress:New York,1978.3.(a)Kupchan,S.M.;Wormser,.Chem.1965,30,3792;(b)Kupchan,S.M.;Merianos,.Chem.1968,33,3735.4.(a)Stiborova,M.;Frei,E.;Breuer,C.A.;Schmeiser,H.H.Cancer Res.1990,50,5464;(b)Cinca,S.;Voiculetz,N.;Schmeiser,H.;Wiessler,M.J.Med.Biochem.1997,1,3.5.Pistelli,L.;Nieri,E.;Bilia,A.R.;Marsili,A.;Scarpato,R.J.Nat.Prod.1993,56,1605.6.Hong,L.;Sakagami,Y.;Marumo,S.;Xinmin,C.Phytochemistry1994,37,237.7.Nakanishi,T.;Iwasaki,K.;Nasu,M.;Miura,I.;Yoneda,K.Phytochemistry1982,21,1759.8.Wu,T.-S.;Leu,Y.-L.;Chan,Y.-Y.Chem.Pham.Bull.1999,47,571.9.De Pascual,T.J.;Urones,J.G.;Fernandez,A.Phytochemistry1983,22,2745.10.Wu,T.-S.;Chan,Y.-Y.;Leu,Y.-L.Chem.Pharm.Bull.2000,48.11.Urzua,A.;Salgado,G.;Cassels,B.K.;Eckhardt,G.Planta Med.1982,45,51.12.Gerecke,M.;Borer,R.;Brossi,A.Helv.Chim.Acta1976,59,2551.13.Guzi,T.J.;Paruch,K.;Dwyer,M.P.;Doll,R.J.;Girijavallabhan,V.M.;Mallams,A.;Alvarez, C.S.;Keertikar,K.M.;Rivera,J.;Chan,T.-Y.;Madison,V.S.;Fischmann,T.O.;Dillard,L.W.;Tran,V.D.;He,Z.;James,R.A.;Park,H.;Paradkar,V.M.;Hobbs,D.W.;Kirschmeier,P.;Bannerji,R.U.S.Patent Appl.Publ.2008,pp.387.14.(a)Babu,P.A.;Narasu,M.L.;Srinivas,K.ARKIVOC2007,2,247(Gainesville,FL,USA);(b)Ruetz,S.;Fabbro,D.;Zimmermann,J.;Meyer,T.;Gray,N.Curr.Med.Chem.:Anti-Cancer Agents2003,3,1;(c)Dumas,J.Exp.Opin.Ther.Patents2001, 11,405.15.Priestap,H.Phytochemistry1985,24,849.16.Coutts,R.T.;Stenlake,J.B.;Williams,W.D.J.Chem.Soc.1957,4120.17.Chakraborty,S.;Nandi,R.;Maiti,M.;Achari,B.;Bandyopadhyay,S.Photochem.Photobiol.1989,50,685.18.Eckhardt,G.;Urzua,A.;Cassels,B.K.J.Nat.Prod.1983,46,92–97.19.Achari,B.;Bandyopadhyay,S.;Chakravarty,A.K.;Pakrashi,.Magn.Reson.1984,22,741.20.Mizuno,M.;Oka,M.;Tanaka,T.;Yamamoto,H.;Iinuma,M.;Murata,H.Chem.Pharm.Bull.1991,39,1310.21.(a)Chakraborty,S.;Nandi,R.;Maiti,M.;Sur,P.Indian J.Phys.B.1991,65B,587;(b)Priestap,H.A.Magn.Reson.Chem.1989,27,460.22.Chen,C.C.;Huang,Y.L.;Ou,J.C.;Su,M.J.;Yu,S.M.;Teng,C.M.Planta Med.1991,57,406.23.Shamma,M.;Slusarchyk,W.A.Chem.Rev.1964,64,59.24.You,M.;Wickramaratne, D. B.;Silva,G.L.;Chai,H.;Chagwedera,T. E.;Farnsworth,N.R.;Cordell,G.A.;Kinghorn,A.D.;Pezzuto,J.M.J.Nat.Prod.1995,58,598.25.Roblot,F.;Hocquemiller,R.;Cave,A.;Moretti,C.J.Nat.Prod.1983,46,862.26.Manske,R.H.F.Can.J.Res.1932,7,258.27.Wang,C.-C.;Kuoh,C.-S.;Wu,T.-S.J.Nat.Prod.1996,59,409.28.Kunitomo,J.;Ju-Ichi,M.;Yoshikawa,Y.;Chikamatsu,H.J.Pharm.Soc.Jpn.1974,94,97.29.Terui,Y.;Tori,K.;Maeda,S.;Sawa,Y.K.Tetrahedron Lett.1975,33,2853.30.Chen,C.Y.;MacLean,D.B.Can.J.Chem.1968,46,2501.31.Dumas,J.Exp.Opin.Ther.Patents2000,11,405.32.Davis,S.T.;Benson,B.G.;Bramson,H.N.;Chapman,D.E.;Dickerson,S.H.Science2001,291,134.33.Afonso,A.;Kelly,J.M.;Chackalamannil,S.U.S.Patent5459146,1995,pp13.34.Mohammadi,M.;McMahon,G.;Sun,L.;Tang,C.;Hirth,P.Science1997,276,955.35.Bramson,H.N.;Corona,J.;Davis,S.T.;Dickerson,S.H.;Edelstein,M.;Frye,S.V.,;Gampe,R.T.,Jr.;Harris,P.A.;Hassell,A.;Holmes,W.D.;Hunter,R.N.;Lackey,K.E.;Lovejoy,B.;Luzzio,M.J.;Montana,V.;Rocque,W.J.;Rusnak,R.D.;Shewchuk,L.;Veal,J.M.;Walker,D.H.;Kuyper,L.F.J.Med.Chem.2001,44, 4339.36.Walker,D.H.;Luzzio,M.;Veal,J.;Dold,K.;Edelstein,M.Proc.Am.Assoc.CancerRes.1999,40,A4783Physico-chemical properties:Aristolochic acid C(5):UV k max:225,256,308,410nm;FABMS328(M+H)+,350 (M+Na)+,366(M+K)+,1H NMR(DMSO-d6)d:10.63(COO H),8.48(9-H),8.46(d, J=4Hz,5-H),8.10(d,J=17Hz,8-H),7.75(s,2H),7.29(dd,J=17,4Hz,7-H),6.48(s,12-H2).13C NMR(DMSO-d6)ppm:168.0(11-C),159.8(6-C),145.8(3-C),145.5(4-C),143.1(10-C),132.5(8-C),131.0(4b-C),126.4(9-C),123.7(1-C),121.5(8a-C),118.8(7-C),117.2(10a-C),116.2(4a-C),111.9(2-C),111.1(5-C), 102.8(12-C).7-Hydroxy aristolochic acid A(6):UV k max:224,271,318,384nm;ESMSÀve mode,m/z356(MÀH)À.Aristolochic acid D(7):UV k max:224,243,333,408nm;ESMS m/z358(M+H)+.Compound(21)18:UV k max:214,242,258,294,328,398nm;ESMS:m/z280 (M+H)+;1H NMR(DMSO-d6)d:10.72(s,NH),10.2(s,–OH),8.03(d,J=15Hz,5-H),7.63(s,2-H),7.37(t,J=15Hz,6-H),7.36(s,9-H),7.06(d,J=15Hz,7-H),6.46(s,12-H2).13C NMR(DMSO-d6)ppm:168.1(11-C),153.8(8-C),148.8(3-C),147.1(4-C),134.0(10-C),125.8(6-C),125.3(4b-C),125.3(10a-C),123.2(8a-C),119.3(1-C),117.5(5-C),112.3(7-C),111.3(4a-C),105.4(2-C),103.2 (12-C),98.7(9-C).Compound(16)16:UV k max:225,239,258,295,329,394nm;ESMS:m/z294 (M+H)+;1H NMR(DMSO-d6)d:10.67(s,NH),8.22(d,J=16Hz,5-H),7.70(s,2-H),7.50(t,J=16Hz,6-H),7.35(s,9-H),7.20(d,J=16Hz,7-H),6.48(s,12-H2),4.0(s,13-H3).13C NMR(DMSO-d6)ppm:168.1(11-C),155.3(8-C),148.8(3-C),147.1(4-C),134.7(10-C),125.7(6-C),125.0(11a-C),124.8(4b-C),124.0(10a-C),119.3(1-C),118.7(5-C),111.0(4a-C),108.3(7-C),105.7(2-H),103.3(12-C),97.9(9-C)add–OCH3value.Compound(29):ESMS:m/z316(M+H)+;1H NMR(DMSO-d6)d:9.12(d, J=16Hz,5-H),7.52(t,J=16Hz,6-H),7.50(s,2-H),7.22(d,J=16Hz,7-H),3.92 (s,–OCH3),13C NMR(DMSO-d6)ppm:NOE from–OCH3to proton doublet at d7.22due to7-H and no NOE from–OCH3.All the new compounds were purified by HPLC and identified by MS.V.R.Hegde et al./Bioorg.Med.Chem.Lett.20(2010)1384–13871387。
新人教版七年级上册英语 Unit 7 Period 1 Section A(1a-2e) 重点习题练习复习课件
5. —____B____? —I want to buy a pair of sports shoes.【中考•云南】
A. What's wrong
B. What can I do for you 【C解. H析o】w 交mu际ch法i。s i根t 据答语句意“我想买一双运动
鞋。”可知问句是提供服务的用语，可用“Can I
D. Take it easy
【解析】上句句意为“年轻的女士，我可以帮助你 吗？”。所以答语应用“Yes，please.”来回答，故选C。
7. The boy is too fat, so he needs __t_o__ea_t__(eat) less meat.
8. —Must I clean the room now?
watch观看，watch TV看电视，排除A, D; look
at强C.调w看at的ch动; L作oo;kseaet看; s见ee，强调看的结果。Look at Dth.elpoiocktuarte;看Lo这ok张a图t; 片se，e 是祈使句，强调“看”
这一动作，看到花是强调“看”的结果。故选C。
is Mike's.
8. Sally's yellow ___so_c_k_s__ the sofa.
are on
9. What color is your brother's __sw__e_a_te_r_ , Alan?
10. That green ___s_k_i_r_t _
is only 15 dollars.
A. Here you are
B. They are for you
C. It doesn't matter
爱德华斯信号设备-视觉信号灯系列说明书
V I S U A L S I G N A L S1AC voltage frequency is 50/60 Hz2Calculated at operating power to 75% efficiency.NOTE: Mounting options notincluded (ordered separately)Edwards 116DEXMST-FJ and 116EXMST Series Xenon strobe beacons are explosionproof, signaling devices suitable for use in hazardous indoor or outdoor applications requiring NEMA Type 3R or 4X installations. The housing is cast aluminum with a corrosion resistance epoxy powder coat, and includes a dome guard. The fluted, high-impact glass dome provides even light distribution.The 116DEXMST-FJ Series is Diode Polarized for use in electrically supervised circuits. Both versions can be bracket, ceiling, pendant or stanchion mounted.Features and Specifications• Xenon strobe light source • Flash rate 65 fpm• High impact glass dome, dome guard included • Quick connect for easy assembly and installation• Diode Polarized for use in electrically supervised circuits (116DEXMST-FJ Series only)• Suitable for indoor and outdoor hazardous applications • B racket, ceiling, pendant or stanchion mountingoptions (ordered separately)• NEMA Type 3R and 4X enclosure • E xplosionproof: Class I, Div 2, Groups A and B, Class I, Div 1 and 2, Groups C and D; Class II and III, Div 1, Groups E, F and G; Class II andIII, Div 2, Groups F and G1Stanchion 116EX-SPendant 116EX-P3/4" Conduit EntryCeiling/Wall 116EX-C (4) 3/4" HubCeiling/Wall 116EX-C (4) 3/4" HubsWall BracketElbow 116EX-B116DEXSTM-FJ Series116EXMST-N5 Series*Letter in this position designates lens color: A - amber, B - blue, C - clear, G - green, M - magenta, or R - redV I S U A L S I G N A L S1Stanchion Mount116EX-Sdr_vs_explodeproof_wallMt_01.epsCeiling Mount 116EX-Cdr_vs_explodeproof_ceilingMt_01.eps6 3/4"4"Pendant Mount116EX-P。
《有机化学》第三版 王积涛 课后习题答案 南开大学 无水印版本
CH2CH2CH2CH2CH2CH3 |
CH3C| HCH2CH2CH2CH3
Cl
15.8%
Cl
42.1%
CH3CH2C| HCH2CH2CH3 Cl
42.1%
1H：2。H=1：4
6 ×1 = 6 8 × 4 32
32 38
× 100%=84.2%
2）异已烷
CH3 CH3|CHCH2CH2CH3
CH3
||
|CH3 CH3-|C-CH2|CH2
Cl CH3
45%
H3C Cl
40%
H3C Cl
15%
5、写出下列各取代基的构造式：
1) Et-
CH3CH2-
2) i-Pr-
CH3 CHCH3
3) i-Bu4) s-Bu5)
CH3-—CH-CH2CH3
— CH3-CH2-CH-CH3
CH 3 — CH 3-CH-CH 2-CH2-
根据杂化后形成的杂化轨道有着最佳的空间取向，彼此做到了最大程度
的远离，可想像 2 个 sp 杂化轨道之间的夹角应互为 1800，sp 杂化的碳原子
应为直线型构型。
8、 二氧化碳的偶极矩为零，这是为什麽？如果 CO2 遇水后形成 HCO3-或 H2CO3,
这时它们的偶极矩还能保持为零吗？碳酸分子为什麽是酸性的？
Br
2- -2-
(6) HBr(
)
CH3 HBr CH3-C=CH2
CH3 Br CH3CH-CH2
1- -2-
(7) HI
CH3 HI
CH3-C=CH2
CH3 CH3C-CH3
I
2- -2-
(8) HI (
)
2024届高考化学一轮复习：《第19讲 化学能与热能》
高考化学一轮复习： 《第19讲 化学能与热能》【真题再现·辨明考向】1．(2023年江苏卷) 氢元素及其化合物在自然界广泛存在且具有重要应用。
11H 、21H 、31H 是氢元素的3种核素，基态H 原子1s 1的核外电子排布，使得H 既可以形成H +又可以形成H －，还能形成H 2O 、H 2O 2、NH 3、N 2H 4、CaH 2等重要化合物；水煤气法、电解水、光催化分解水都能获得H 2，如水煤气法制氢反应中，H 2O(g)与足量C(s)反应生成1molH 2(g)和1molCO(g)吸收131.3kJ 的热量。
H 2在金属冶炼、新能源开发、碳中和等方面具有重要应用，如HCO －3在催化剂作用下与H 2反应可得到HCOO －。
我国科学家在氢气的制备和应用等方面都取得了重大成果。
下列化学反应表示正确的是A. 水煤气法制氢：C(s)＋H 2O(g)＝H 2(g)＋CO(g) △H ＝－131.3kJ∙mol -1B. HCO －3催化加氢生成HCOO －的反应：HCO －3＋H 2=======催化剂 HCOO －＋H 2OC. 电解水制氢的阳极反应：2H 2O －2e －＝H 2↑＋2OH －D. CaH 2与水反应：CaH 2＋2H 2O ＝Ca(OH)2＋H 2↑2. (2023年江苏卷)二氧化碳加氢制甲烷过程中的主要反应为：CO 2(g)＋4H 2(g)＝CH 4(g)＋2H 2O(g) △H ＝－164.7kJ∙mol -1CO 2(g)＋H 2(g)＝COg)＋H 2O(g) △H ＝41.2kJ∙mol -1在密闭容器中，1.01×105Pa 、n 起始(CO 2)∶n 起始(H 2)＝1∶4时，CO 2平衡转化率、在催化剂作用下反应相同时间所测得的CO 2实际转化率随温度的变化如题图所示。
CH 4的选择性可表示为n 生成(CH 4) n 生成(CO 2)×100%。
2023【人教版】七年级英语下册教学课件-第1课时(A 1a-2e)
Conversation 2
Peter: How do you get home from school, Jane? Jane: I take the bus. Peter: How long does it take? Jane: Oh, about one hour and 30 minutes. Peter: Wow! That’s a long time! Jane: Yes, I only go home on weekends. Peter: How far is it from your home to school? Jane: It’s about 60 kilometers.
Talk about how to get to places. Improve your listening skills through listening training.
Let’s watch a video
交通工具 What transportation is mentioned in the video?
Unit 3 How do you get to school?
Section A 1a-2e
R·七年级下册
Learning objects
Learn curriculum words and useful expressions: train, bus, subway, ride, sixty, seventy, eighty...
ninety one hundred
2a
Listen and repeat. Then write the correct numh切bu数nedr目ren时de数，x词t其，t后o意不t为加h“e-s一，w百也o”不rd。和. 当of连前用面。有具体数字表示确
【精品课件】有机化学第十五章缩合反应CondensationReactions
15
2.1 Claisen Ester Condensation
O 2 CH3CH2COC2H5
1) C2H5ONa, EtOH
2) HOAc, H2O
O
O
CH3CH2C CH C OC2H5
CH3
81%
O (1) NaOR
2 RCH2COC2H5 (2) H+
OO RCH2CCHCOC2H5
R
β-酮酸酯
+
C2H5OH
C2H5ONa 180 oC
催化量的C2H5ONa
O 2 CH3COC2H5
其它含有活泼氢的化合物也可发生类似的反应
O RCA (I) +
I: 酰氯、酐、酯
O C C (II) H
O
O
RC C C
II: 含活泼氢的羰基化合物
18
2.2 Crossed Ester Condensation
O
CH2CO2H
CH2CHO CH2CHO
+
NH2CH3
+
C
O
N CH3
CH2CO2H
CH3 N
-CO2
H2/Ni
pH = 5
N CH3 CO2H
O CO2H
O
OH
9
Mannich Reaction
H NHCH3 O HCH
CH3 N+ H2C
H
CH3 N
O H
O
H
H
O
应用：(2) 用作合成中间体
O
+
C6H5CCHCH2NMe2
CH3 CHCH2NMe2
+ CH3
MgX
芴甲氧羰基的结构
芴甲氧羰基的结构一、引言芴甲氧羰基（Fluorescein-methyl ester，简称FME）是一种有机化合物，具有独特的化学结构和优异的物理、化学性质。
在生物学、药物化学等领域具有广泛的应用。
本文将对芴甲氧羰基的结构、物理性质和化学性质进行详细阐述，以期为相关领域的研究提供有益参考。
二、芴甲氧羰基的化学结构1.分子式和结构式芴甲氧羰基的分子式为C20H11NO4，结构式中包含一个苯环、一个甲氧基和一个羰基。
2.原子间的键合关系芴甲氧羰基中，苯环上的碳原子与羰基的氧原子通过一个双键相连，形成一个稳定的芳香环结构。
甲氧基与苯环相连，使分子具有较好的溶解性和反应活性。
3.空间构型芴甲氧羰基分子呈扭曲的平面结构，苯环、甲氧基和羰基在空间上呈现出一定的排列。
这种结构有利于分子间的相互作用，从而影响其物理和化学性质。
三、物理性质1.颜色和溶解性芴甲氧羰基的颜色为橙红色，随着溶剂极性的增加，颜色逐渐加深。
在水中的溶解度较低，但在有机溶剂如甲醇、乙醇、丙酮等中溶解性较好。
2.熔点和沸点芴甲氧羰基的熔点为111-113℃，沸点为292℃（0.2kPa）。
四、化学性质1.反应活性和稳定性芴甲氧羰基具有一定的反应活性，可与醇、胺等发生酯化、酰化等反应。
在酸、碱条件下，容易水解生成相应的醇和羧酸。
2.降解途径芴甲氧羰基在环境中的降解主要受光、热和微生物作用，降解产物包括芳香族化合物、醇和羧酸等。
3.反应类型及其应用芴甲氧羰基作为一种有机染料，广泛应用于生物学实验、荧光探针、药物制剂等领域。
此外，还可作为合成其他有机化合物的中间体。
五、结论芴甲氧羰基作为一种具有独特结构的有机化合物，具有丰富的物理和化学性质。
在生物学、药物化学等领域具有重要应用价值。
本文对芴甲氧羰基的结构、物理性质和化学性质进行了综述，为其在相关领域的研究和应用提供了一定的理论基础。
1-乙基-(3-二甲基 氨基丙基)碳化二 亚胺盐酸盐纯度
1-乙基-(3-二甲基氨基丙基)碳化二亚胺盐酸盐纯度1-乙基-(3-二甲基氨基丙基)碳化二亚胺盐酸盐（简称EPAH）是一种有机化合物，化学式为C9H19N2·4HCl，分子质量为288.99 g/mol。
它是一种常用的生物荧光探针，广泛应用于细胞生物学和生物医学研究领域。
EPAH是一种固体粉末，易溶于水和一些有机溶剂，如甲醇和乙醇。
它的溶解度在室温下大约为10 mg/mL。
纯度高的EPAH通常以白色结晶的形式存在。
在实验操作中，我们通常会按照所需用量，将EPAH溶解在适量的溶剂中，以制备工作液。
EPAH具有较强的荧光特性，其荧光峰位在波长为427 nm左右，发出蓝色荧光。
这使得EPAH被广泛用作细胞标记剂，能够通过其特定的荧光特性在显微镜下追踪和观察细胞的运动和行为。
在细胞实验中，需要将EPAH添加到培养基中，浸泡细胞一段时间，使其进入细胞内并与分子靶标结合。
通过激发特定波长的光源，观察细胞产生的荧光信号，可以获得有关细胞的许多重要信息，如细胞的形态、分裂、运动和死亡等。
这些信息对于细胞研究和生物医学研究具有重要意义。
然而，纯度对于EPAH的应用至关重要。
高纯度的EPAH可以提供更可靠和准确的结果。
低纯度的EPAH可能含有其他杂质，这些杂质可能干扰细胞实验的结果，降低实验的可重复性和准确性。
因此，在使用EPAH进行实验之前，需要选购纯度较高的EPAH。
供应商通常会提供有关EPAH纯度的相关证明和数据。
在实验室中，也可以通过某些分析方法，如质谱分析、红外光谱分析等，对所购买的EPAH样品进行鉴定和检测。
总之，EPAH是一种常用的生物荧光探针，广泛应用于细胞生物学和生物医学研究中。
高纯度的EPAH可以提供更可靠和准确的结果。
在使用EPAH进行实验之前，需要选购纯度较高的EPAH，并通过相关分析方法对EPAH样品进行鉴定和检测，以确保实验结果的准确性和可靠性。
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Microeconomics The study of how households and firms make choices, how they interact in markets, and how the government attempts to influence their choices. Macroeconomics The study of the economy as a whole, including topics such as inflation, unemployment, and economic growth.
Personal Consumption Expenditures, or “Consumption”
Consumption Spending by households on goods and services, not including spending on new houses.
Gross Private Domestic Investment, or “Investment”
FIGURE 19-1
The Circular Flow and the Measurement of GDP
© 2008 Prentice Hall Business Publishing Economics R. Glenn Hubbard, Anthony Patrick O’Brien, 2e.
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Learning Objective 19.1
Gross Domestic Product Measures Total Production
Chapter 19: GDP: Measuring Total Production and Income
Production, Income, and the Circular Flow Diagram
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Learning Objective 19.1
Gross Domestic Product Measures Total Production
Chapter 19: GDP: Measuring Total Production and Income
Components of GDP
© 2008 Prentice Hall Business Publishing Economics R. Glenn Hubbard, Anthony Patrick O’Brien, 2e.
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Learning Objective 19.1
Gross Domestic Product Measures Total Production
Learning Objective 19.1
Solved Problem
Calculating GDP
Chaproduction and Income
19-1
PRODUCTION AND PRICE STATISTICS FOR 2007
Chapter 19: GDP: Measuring Total Production and Income
Measuring Total Production: Gross Domestic Product
Gross domestic product (GDP) The market value of all final goods and services produced in a country during a period of time, typically one year.
GDP Includes Only Current Production GDP includes only production that takes place during the indicated time period.
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© 2008 Prentice Hall Business Publishing Economics R. Glenn Hubbard, Anthony Patrick O’Brien, 2e.
Investment Spending by firms on new factories, office buildings, machinery, and additions to inventories, and spending by households on new houses.
Don’t Let This Happen to YOU!
© 2008 Prentice Hall Business Publishing Economics R. Glenn Hubbard, Anthony Patrick O’Brien, 2e.
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GDP: Measuring Total Production and Income
Chapter 19: GDP: Measuring Total Production and Income
GDP Includes Only the Market Value of Final Goods Final good or service A good or service purchased by a final user.
Intermediate good or service A good or service that is an input into another good or service, such as a tire on a truck.
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Learning Objective 19.1
Gross Domestic Product Measures Total Production
Chapter 19: GDP: Measuring Total Production and Income
Measuring Total Production: Gross Domestic Product
(1) PRODUCT (2) QUANTITY (3) PRICE PER UNIT
Eye examinations
Pizzas Textbooks Paper
PRODUCT (1) QUANTITY
100
80 20 2,000
(2) PRICE PER UNIT
$50.00
10.00 100.00 0.10
Transfer payments Payments by the government to individuals for which the government does not receive a new good or service in return.
© 2008 Prentice Hall Business Publishing Economics R. Glenn Hubbard, Anthony Patrick O’Brien, 2e.
Business cycle Alternating periods of economic expansion and economic recession.
© 2008 Prentice Hall Business Publishing Economics R. Glenn Hubbard, Anthony Patrick O’Brien, 2e.
GDP Is Measured Using Market Values, Not Quantities The word value is important in the definition of GDP.
© 2008 Prentice Hall Business Publishing Economics R. Glenn Hubbard, Anthony Patrick O’Brien, 2e.
Increases in GDP Help Revive American Airlines
Learning Objectives
19.1 Explain how total production is measured.
19.2 Discuss whether GDP is a good measure of well-being.
Remember What Economists Mean by Investment
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© 2008 Prentice Hall Business Publishing Economics R. Glenn Hubbard, Anthony Patrick O’Brien, 2e.
Learning Objective 19.1
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Learning Objective 19.1
Gross Domestic Product Measures Total Production
Chapter 19: GDP: Measuring Total Production and Income
Production, Income, and the Circular Flow Diagram
Recession The period of a business cycle during which total production and total employment are decreasing.
Economic growth The ability of an economy to produce increasing quantities of goods and services. Inflation rate The percentage increase in the price level from one year to the next.