ch6-3_eng ratioed logic
顶空气相色谱法检测生活饮用水中三卤甲烷
分析检测顶空气相色谱法检测生活饮用水中三卤甲烷付文凯（英德市疾病预防控制中心，广东英德 513000）摘 要：目的：建立顶空气相色谱方法同时检测饮用水中三卤甲烷的方法。
方法：应用顶空气相色谱法对生活饮用水中的三卤甲烷进行分析。
结果：本方法检测饮用水中三卤甲烷的4种组分分离度好，在10 min 内完成检测，且各组分的曲线相关系数均＞0.998。
本方法的最低检出限为三氯甲烷0.015 µg·L-1、二氯一溴甲烷0.012 µg·L-1、一氯二溴甲烷0.014 µg·L-1、三溴甲烷0.015 µg·L-1，方法精密度较好，相对标准偏差为0.8%～4.6%，不同浓度样品的加标回收率为94.2%～103.3%。
结论：该方法检测饮用水中三卤甲烷简单、准确，灵敏度和精密度高。
关键词：顶空气相色谱法；三卤甲烷；生活饮用水Determination of Trihalomethane in Drinking Water byHeadspace Gas ChromatographyFU Wenkai(Yingde Center for Disease Control and Prevention, Yingde 513000, China) Abstract: Objective: To establish a headspace gas chromatography method for simultaneous determination of trihalomethane in drinking water. Method: Headspace gas chromatography was used to analyze the trihalomethane in drinking water. Result: The separation of 4 components of trihalomethane in drinking water was well determined by this method, and the detection was completed within 10 min, and the curve correlation coefficients of each component were＞0.998. The minimum detection limits of the method were 0.015 µg·L-1 for trichloromethane, 0.012 µg·L-1 for dichlorobromomethane, 0.014 µg·L-1 for dichlorobromomethane, and 0.015 µg·L-1 for tribromomethane. The accuracy of the method was good, and the relative standard deviation ranged from 0.8% to 4.6%. The recoveries of samples with different concentrations ranged from 94.2% to 103.3%. Conclusion: This method is simple, accurate, sensitive and precise for the determination of trihalomethane in drinking water.Keywords: headspace gas chromatography; trihalomethanes; drinking water随着生活水平的不断提高，人们对饮用水安全问题越来越重视[1]。
乙酸蒸汽催化重整制氢的研究进展
CHEMICAL INDUSTRY AND ENGINEERING PROGRESS 2017年第36卷第5期·1658·化 工 进展乙酸蒸汽催化重整制氢的研究进展王东旭1，肖显斌2，李文艳1（1华北电力大学能源动力与机械工程学院，北京 102206；2华北电力大学生物质发电成套设备国家工程实验室，北京 102206）摘要：通过生物油蒸汽重整制备氢气可以减少环境污染，降低对化石燃料的依赖，是一种极具潜力的制氢途径。
乙酸是生物油的主要成分之一，常作为模型化合物进行研究。
镍基催化剂是乙酸蒸汽重整过程中常用的催化剂，但容易因积炭失去活性，降低了制氢过程的经济性。
本文首先分析了影响乙酸蒸汽重整制氢过程的各种因素，阐述了在这一过程中镍基催化剂的积炭原理，讨论了优化镍基催化剂的方法，包括优化催化剂的预处理过程、添加助剂和选择合适的载体，最后对乙酸蒸汽重整制氢的热力学分析研究进展进行了总结。
未来应重点研究多种助剂复合使用时对镍基催化剂积炭与活性的影响，分析多种助剂的协同作用机理，得到一种高活性、高抗积炭能力的用于生物油蒸汽重整制氢的镍基催化剂。
关键词：生物油；乙酸；制氢；催化剂；热力学中图分类号：TK6 文献标志码：A 文章编号：1000–6613（2017）05–1658–08 DOI ：10.16085/j.issn.1000-6613.2017.05.014A review of literatures on catalytic steam reforming of acetic acid forhydrogen productionWANG Dongxu 1，XIAO Xianbin 2，LI Wenyan 1（1 School of Energy ，Power and Mechanical Engineering ，North China Electric Power University ，Beijing 102206，China ；2 National Engineering Laboratory for Biomass Power Generation Equipment ，North China Electric PowerUniversity ，Beijing 102206，China ）Abstract ：Hydrogen production via steam reforming of bio-oil ，a potential way to produce hydrogen ， can reduce environmental pollution and dependence on fossil fuels. Acetic acid is one of the main components of bio-oil and is often selected as a model compound. Nickel-based catalyst is widely used in the steam reforming of acetic acid ，but it deactivates fast due to the carbon deposition. In this paper ，the affecting factors for the steam reforming of acetic acid are analyzed. The coking mechanism of nickel-based catalyst in this process is illustrated. Optimization methods for nickel-baed catalyst are discussed ，including optimizing the pretreatment process ，adding promoters ，and choosing appropriate catalyst supports. Research progresses in the thermodynamics analyses for steaming reforming of acetic acid are summarized. Further studies should be focused on the effects of a combination of a variety of promoters on carbon deposition. Catalytic activity and the synergy mechanism should be analyzed to produce a novel nickel-based catalyst with high activity ，high resistance to caborn deposition for hydrogen production via steam reforming of bio-oil. Key words ：bio-oil ；acetic acid ；hydrogen production ；catalyst ；thermodynamics第一作者：王东旭（1994—），男，硕士研究生，从事生物质能利用技术研究。
六水三氯化铁催化合成乙酸异戊酯反应时间的探讨
２１ ０ ０年 ３月
高职教育研究 ・
高 校 实 验 室 工 作
研 究
Ｓ ｒａ ＮＯ． ０ ＮＯ． ｅｉｌ １ ３， １
Ｍ ａ ．２ ０ ｒ ０ｌ
ＧＡ０ＸＩ ＨＩ Ａ０ Ｓ ＹＡＮＳ ＨＩＧＯＮＧＺ ＵＯ ＹＡＮＪ Ｕ Ｉ
・
．
上
／ ＼
水 三 氯化 铁 催化 合成 乙酸 异戊 酯反 应 间 的 讨 时 探
刘 小 忠 ，周 娇
（ 湖南化 工职业 技术 学 院 １实训处 ；２化 学工程 系 ，湖南 株洲 ４２ ０ ） １０ ４
摘 要 ：用 六 水三 氯 化 铁 作 催 化 剂 ，改 变反 应 时 间和 不 用带 水 剂合 成 乙酸 异 戊 酯 。结 果 显 示 ： 当反 应 时 间 为 ５分钟 时 ， 乙酸 异 戊 酯 的 产 率 为 ８ ． ６ ； 当反 应 时 间 为 ６ ５５ ％ ０分钟 时 ， 乙酸 异 戊 酯 的 产 率 为 ９ ． ４ ； ＂ ００ ％ － ３ 反 应 时 间 为 １ ０分钟 时 ， 乙酸 异 戊 酯 的 产 率 为 ８ ．６ 。 实验 得 出 ， 乙酸 异 戊 酯产 率 最 高 的 反 应 时 间 为 ６ ８ ２ １％ ０
折 射 率
１． ９８ ３９
１． ９ ３９ ９ １． ００ｌ ４ １． ＯＯ ４ Ｏ １． ０Ｏ１ ４ １ ３ ９ ． ９９ １ ４ ２ ． ００ １ ４ ． ０Ｏ１ １ ４ ＯＯ ． Ｏ １ ４Ｏ ｌ ． Ｏ １ ４０ ． ０２
２ 实验 部 分
２ １ 实 验 仪 器 与 药 品 ．
●作 者 简 介 ：刘 小 忠 （ ９２一） １７ ，男 ，汉 ， 湖 南 祁 阳 人 ， 高 级 实
胺基保护——邻苯二甲酰亚胺脱保护
New1H-Pyrazole-Containing Polyamine Receptors Able ToComplex L-Glutamate in Water at Physiological pH ValuesCarlos Miranda,†Francisco Escartı´,‡Laurent Lamarque,†Marı´a J.R.Yunta,§Pilar Navarro,*,†Enrique Garcı´a-Espan˜a,*,‡and M.Luisa Jimeno†Contribution from the Instituto de Quı´mica Me´dica,Centro de Quı´mica Orga´nica Manuel Lora Tamayo,CSIC,C/Juan de la Cier V a3,28006Madrid,Spain,Departamento de Quı´mica Inorga´nica,Facultad de Quı´mica,Uni V ersidad de Valencia,c/Doctor Moliner50, 46100Burjassot(Valencia),Spain,and Departamento de Quı´mica Orga´nica,Facultad deQuı´mica,Uni V ersidad Complutense de Madrid,A V plutense s/n,28040Madrid,SpainReceived April16,2003;E-mail:enrique.garcia-es@uv.esAbstract:The interaction of the pyrazole-containing macrocyclic receptors3,6,9,12,13,16,19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene1[L1],13,26-dibenzyl-3,6,9,12,13,16,-19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene2[L2],3,9,12,13,16,22,-25,26-octaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene3[L3],6,19-dibenzyl-3,6,9,12,13,-16,19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene4[L4],6,19-diphenethyl-3,6,9,12,13,16,19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetraene5[L5],and 6,19-dioctyl-3,6,9,12,13,16,19,22,25,26-decaazatricyclo-[22.2.1.111,14]-octacosa-1(27),11,14(28),24-tetra-ene6[L6]with L-glutamate in aqueous solution has been studied by potentiometric techniques.The synthesis of receptors3-6[L3-L6]is described for the first time.The potentiometric results show that4[L4]containing benzyl groups in the central nitrogens of the polyamine side chains is the receptor displaying the larger interaction at pH7.4(K eff)2.04×104).The presence of phenethyl5[L5]or octyl groups6[L6]instead of benzyl groups4[L4]in the central nitrogens of the chains produces a drastic decrease in the stability[K eff )3.51×102(5),K eff)3.64×102(6)].The studies show the relevance of the central polyaminic nitrogen in the interaction with glutamate.1[L1]and2[L2]with secondary nitrogens in this position present significantly larger interactions than3[L3],which lacks an amino group in the center of the chains.The NMR and modeling studies suggest the important contribution of hydrogen bonding andπ-cation interaction to adduct formation.IntroductionThe search for the L-glutamate receptor field has been andcontinues to be in a state of almost explosive development.1 L-Glutamate(Glu)is thought to be the predominant excitatory transmitter in the central nervous system(CNS)acting at a rangeof excitatory amino acid receptors.It is well-known that it playsa vital role mediating a great part of the synaptic transmission.2However,there is an increasing amount of experimentalevidence that metabolic defects and glutamatergic abnormalitiescan exacerbate or induce glutamate-mediated excitotoxic damageand consequently neurological disorders.3,4Overactivation ofionotropic(NMDA,AMPA,and Kainate)receptors(iGluRs)by Glu yields an excessive Ca2+influx that produces irreversible loss of neurons of specific areas of the brain.5There is much evidence that these processes induce,at least in part,neuro-degenerative illnesses such as Parkinson,Alzheimer,Huntington, AIDS,dementia,and amyotrophic lateral sclerosis(ALS).6In particular,ALS is one of the neurodegenerative disorders for which there is more evidence that excitotoxicity due to an increase in Glu concentration may contribute to the pathology of the disease.7Memantine,a drug able to antagonize the pathological effects of sustained,but relatively small,increases in extracellular glutamate concentration,has been recently received for the treatment of Alzheimer disease.8However,there is not an effective treatment for ALS.Therefore,the preparation of adequately functionalized synthetic receptors for L-glutamate seems to be an important target in finding new routes for controlling abnormal excitatory processes.However,effective recognition in water of aminocarboxylic acids is not an easy task due to its zwitterionic character at physiological pH values and to the strong competition that it finds in its own solvent.9†Centro de Quı´mica Orga´nica Manuel Lora Tamayo.‡Universidad de Valencia.§Universidad Complutense de Madrid.(1)Jane,D.E.In Medicinal Chemistry into the Millenium;Campbell,M.M.,Blagbrough,I.S.,Eds.;Royal Society of Chemistry:Cambridge,2001;pp67-84.(2)(a)Standaert,D.G.;Young,A.B.In The Pharmacological Basis ofTherapeutics;Hardman,J.G.,Goodman Gilman,A.,Limbird,L.E.,Eds.;McGraw-Hill:New York,1996;Chapter22,p503.(b)Fletcher,E.J.;Loge,D.In An Introduction to Neurotransmission in Health and Disease;Riederer,P.,Kopp,N.,Pearson,J.,Eds.;Oxford University Press:New York,1990;Chapter7,p79.(3)Michaelis,E.K.Prog.Neurobiol.1998,54,369-415.(4)Olney,J.W.Science1969,164,719-721.(5)Green,J.G.;Greenamyre,J.T.Prog.Neurobiol.1996,48,613-63.(6)Bra¨un-Osborne,H.;Egebjerg,J.;Nielsen,E.O.;Madsen,U.;Krogsgaard-Larsen,P.J.Med.Chem.2000,43,2609-2645and references therein.(7)(a)Shaw,P.J.;Ince,P.G.J.Neurol.1997,244(Suppl2),S3-S14.(b)Plaitakis,A.;Fesdjian,C.O.;Shashidharan,S Drugs1996,5,437-456.(8)Frantz,A.;Smith,A.Nat.Re V.Drug Dico V ery2003,2,9.Published on Web12/30/200310.1021/ja035671m CCC:$27.50©2004American Chemical Society J.AM.CHEM.SOC.2004,126,823-8339823There are many types of receptors able to interact with carboxylic acids and amino acids in organic solvents,10-13yielding selective complexation in some instances.However,the number of reported receptors of glutamate in aqueous solution is very scarce.In this sense,one of the few reports concerns an optical sensor based on a Zn(II)complex of a 2,2′:6′,2′′-terpyridine derivative in which L -aspartate and L -glutamate were efficiently bound as axial ligands (K s )104-105M -1)in 50/50water/methanol mixtures.14Among the receptors employed for carboxylic acid recogni-tion,the polyamine macrocycles I -IV in Chart 1are of particular relevance to this work.In a seminal paper,Lehn et al.15showed that saturated polyamines I and II could exert chain-length discrimination between different R ,ω-dicarboxylic acids as a function of the number of methylene groups between the two triamine units of the receptor.Such compounds were also able to interact with a glutamic acid derivative which has the ammonium group protected with an acyl moiety.15,16Compounds III and IV reported by Gotor and Lehn interact in their protonated forms in aqueous solution with protected N -acetyl-L -glutamate and N -acetyl-D -glutamate,showing a higher stability for the interaction with the D -isomer.17In both reports,the interaction with protected N -acetyl-L -glutamate at physiological pH yields constants of ca.3logarithmic units.Recently,we have shown that 1H -pyrazole-containing mac-rocycles present desirable properties for the binding of dopam-ine.18These polyaza macrocycles,apart from having a highpositive charge at neutral pH values,can form hydrogen bonds not only through the ammonium or amine groups but also through the pyrazole nitrogens that can behave as hydrogen bond donors or acceptors.In fact,Elguero et al.19have recently shown the ability of the pyrazole rings to form hydrogen bonds with carboxylic and carboxylate functions.These features can be used to recognize the functionalities of glutamic acid,the carboxylic and/or carboxylate functions and the ammonium group.Apart from this,the introduction of aromatic donor groups appropriately arranged within the macrocyclic framework or appended to it through arms of adequate length may contribute to the recognition event through π-cation interactions with the ammonium group of L -glutamate.π-Cation interactions are a key feature in many enzymatic centers,a classical example being acetylcholine esterase.20The role of such an interaction in abiotic systems was very well illustrated several years ago in a seminal work carried out by Dougherty and Stauffer.21Since then,many other examples have been reported both in biotic and in abiotic systems.22Taking into account all of these considerations,here we report on the ability of receptors 1[L 1]-6[L 6](Chart 2)to interact with L -glutamic acid.These receptors display structures which differ from one another in only one feature,which helps to obtain clear-cut relations between structure and interaction(9)Rebek,J.,Jr.;Askew,B.;Nemeth,D.;Parris,K.J.Am.Chem.Soc.1987,109,2432-2434.(10)Seel,C.;de Mendoza,J.In Comprehensi V e Supramolecular Chemistry ;Vogtle,F.,Ed.;Elsevier Science:New York,1996;Vol.2,p 519.(11)(a)Sessler,J.L.;Sanson,P.I.;Andrievesky,A.;Kral,V.In SupramolecularChemistry of Anions ;Bianchi,A.,Bowman-James,K.,Garcı´a-Espan ˜a,E.,Eds.;John Wiley &Sons:New York,1997;Chapter 10,pp 369-375.(b)Sessler,J.L.;Andrievsky,A.;Kra ´l,V.;Lynch,V.J.Am.Chem.Soc.1997,119,9385-9392.(12)Fitzmaurice,R.J.;Kyne,G.M.;Douheret,D.;Kilburn,J.D.J.Chem.Soc.,Perkin Trans.12002,7,841-864and references therein.(13)Rossi,S.;Kyne,G.M.;Turner,D.L.;Wells,N.J.;Kilburn,J.D.Angew.Chem.,Int.Ed.2002,41,4233-4236.(14)Aı¨t-Haddou,H.;Wiskur,S.L.;Lynch,V.M.;Anslyn,E.V.J.Am.Chem.Soc.2001,123,11296-11297.(15)Hosseini,M.W.;Lehn,J.-M.J.Am.Chem.Soc.1982,104,3525-3527.(16)(a)Hosseini,M.W.;Lehn,J.-M.Hel V .Chim.Acta 1986,69,587-603.(b)Heyer,D.;Lehn,J.-M.Tetrahedron Lett.1986,27,5869-5872.(17)(a)Alfonso,I.;Dietrich,B.;Rebolledo,F.;Gotor,V.;Lehn,J.-M.Hel V .Chim.Acta 2001,84,280-295.(b)Alfonso,I.;Rebolledo,F.;Gotor,V.Chem.-Eur.J.2000,6,3331-3338.(18)Lamarque,L.;Navarro,P.;Miranda,C.;Ara ´n,V.J.;Ochoa,C.;Escartı´,F.;Garcı´a-Espan ˜a,E.;Latorre,J.;Luis,S.V.;Miravet,J.F.J.Am.Chem.Soc .2001,123,10560-10570.(19)Foces-Foces,C.;Echevarria,A.;Jagerovic,N.;Alkorta,I.;Elguero,J.;Langer,U.;Klein,O.;Minguet-Bonvehı´,H.-H.J.Am.Chem.Soc.2001,123,7898-7906.(20)Sussman,J.L.;Harel,M.;Frolow,F.;Oefner,C.;Goldman,A.;Toker,L.;Silman,I.Science 1991,253,872-879.(21)Dougherty,D.A.;Stauffer,D.A.Science 1990,250,1558-1560.(22)(a)Sutcliffe,M.J.;Smeeton,A.H.;Wo,Z.G.;Oswald,R.E.FaradayDiscuss.1998,111,259-272.(b)Kearney,P.C.;Mizoue,L.S.;Kumpf,R.A.;Forman,J.E.;McCurdy,A.;Dougherty,D.A.J.Am.Chem.Soc.1993,115,9907-9919.(c)Bra ¨uner-Osborne,H.;Egebjerg,J.;Nielsen,E.;Madsen,U.;Krogsgaard-Larsen,P.J.Med.Chem.2000,43,2609-2645.(d)Zacharias,N.;Dougherty,D.A.Trends Pharmacol.Sci.2002,23,281-287.(e)Hu,J.;Barbour,L.J.;Gokel,G.W.J.Am.Chem.Soc.2002,124,10940-10941.Chart 1.Some Receptors Employed for Dicarboxylic Acid and N -AcetylglutamateRecognitionChart 2.New 1H -Pyrazole-Containing Polyamine Receptors Able To Complex L -Glutamate inWaterA R T I C L E SMiranda et al.824J.AM.CHEM.SOC.9VOL.126,NO.3,2004strengths.1[L1]and2[L2]differ in the N-benzylation of the pyrazole moiety,and1[L1]and3[L3]differ in the presence in the center of the polyamine side chains of an amino group or of a methylene group.The receptors4[L4]and5[L5]present the central nitrogens of the chain N-functionalized with benzyl or phenethyl groups,and6[L6]has large hydrophobic octyl groups.Results and DiscussionSynthesis of3-6.Macrocycles3-6have been obtained following the procedure previously reported for the preparation of1and2.23The method includes a first dipodal(2+2) condensation of the1H-pyrazol-3,5-dicarbaldehyde7with the corresponding R,ω-diamine,followed by hydrogenation of the resulting Schiff base imine bonds.In the case of receptor3,the Schiff base formed by condensation with1,5-pentanediamine is a stable solid(8,mp208-210°C)which precipitated in68% yield from the reaction mixture.Further reduction with NaBH4 in absolute ethanol gave the expected tetraazamacrocycle3, which after crystallization from toluene was isolated as a pure compound(mp184-186°C).In the cases of receptors4-6, the precursor R,ω-diamines(11a-11c)(Scheme1B)were obtained,by using a procedure previously described for11a.24 This procedure is based on the previous protection of the primary amino groups of1,5-diamino-3-azapentane by treatment with phthalic anhydride,followed by alkylation of the secondary amino group of1,5-diphthalimido-3-azapentane9with benzyl, phenethyl,or octyl bromide.Finally,the phthalimido groups of the N-alkyl substituted intermediates10a-10c were removed by treatment with hydrazine to afford the desired amines11a-11c,which were obtained in moderate yield(54-63%).In contrast with the behavior previously observed in the synthesis of3,in the(2+2)dipodal condensations of7with 3-benzyl-,3-phenethyl-,and3-octyl-substituted3-aza-1,5-pentanediamine11a,11b,and11c,respectively,there was not precipitation of the expected Schiff bases(Scheme1A). Consequently,the reaction mixtures were directly reduced in situ with NaBH4to obtain the desired hexaamines4-6,which after being carefully purified by chromatography afforded purecolorless oils in51%,63%,and31%yield,respectively.The structures of all of these new cyclic polyamines have been established from the analytical and spectroscopic data(MS(ES+), 1H and13C NMR)of both the free ligands3-6and their corresponding hydrochloride salts[3‚4HCl,4‚6HCl,5‚6HCl, and6‚6HCl],which were obtained as stable solids following the same procedure previously reported18for1‚6HCl and2‚6HCl.As usually occurs for3,5-disubstituted1H-pyrazole deriva-tives,either the free ligands3-6or their hydrochlorides show very simple1H and13C NMR spectra,in which signals indicate that,because of the prototropic equilibrium of the pyrazole ring, all of these compounds present average4-fold symmetry on the NMR scale.The quaternary C3and C5carbons appear together,and the pairs of methylene carbons C6,C7,and C8are magnetically equivalent(see Experimental Section).In the13C NMR spectra registered in CDCl3solution, significant differences can be observed between ligand3,without an amino group in the center of the side chain,and the N-substituted ligands4-6.In3,the C3,5signal appears as a broad singlet.However,in4-6,it almost disappears within the baseline of the spectra,and the methylene carbon atoms C6and C8experience a significant broadening.Additionally,a remark-able line-broadening is also observed in the C1′carbon signals belonging to the phenethyl and octyl groups of L5and L6, respectively.All of these data suggest that as the N-substituents located in the middle of the side chains of4-6are larger,the dynamic exchange rate of the pyrazole prototropic equilibrium is gradually lower,probably due to a relation between proto-tropic and conformational equilibria.Acid-Base Behavior.To follow the complexation of L-glutamate(hereafter abbreviated as Glu2-)and its protonated forms(HGlu-,H2Glu,and H3Glu+)by the receptors L1-L6, the acid-base behavior of L-glutamate has to be revisited under the experimental conditions of this work,298K and0.15mol dm-3.The protonation constants obtained,included in the first column of Table1,agree with the literature25and show that the zwitterionic HGlu-species is the only species present in aqueous solution at physiological pH values(Scheme2and Figure S1of Supporting Information).Therefore,receptors for(23)Ara´n,V.J.;Kumar,M.;Molina,J.;Lamarque,L.;Navarro,P.;Garcı´a-Espan˜a,E.;Ramı´rez,J.A.;Luis,S.V.;Escuder,.Chem.1999, 64,6137-6146.(24)(a)Yuen Ng,C.;Motekaitis,R.J.;Martell,A.E.Inorg.Chem.1979,18,2982-2986.(b)Anelli,P.L.;Lunazzi,L.;Montanari,F.;Quici,.Chem.1984,49,4197-4203.Scheme1.Synthesis of the Pyrazole-Containing MacrocyclicReceptorsNew1H-Pyrazole-Containing Polyamine Receptors A R T I C L E SJ.AM.CHEM.SOC.9VOL.126,NO.3,2004825glutamate recognition able to address both the negative charges of the carboxylate groups and the positive charge of ammonium are highly relevant.The protonation constants of L 3-L 6are included in Table 1,together with those we have previously reported for receptors L 1and L 2.23A comparison of the constants of L 4-L 6with those of the nonfunctionalized receptor L 1shows a reduced basicity of the receptors L 4-L 6with tertiary nitrogens at the middle of the polyamine bridges.Such a reduction in basicity prevented the potentiometric detection of the last protonation for these ligands in aqueous solution.A similar reduction in basicity was previously reported for the macrocycle with the N -benzylated pyrazole spacers (L 2).23These diminished basicities are related to the lower probability of the tertiary nitrogens for stabilizing the positive charges through hydrogen bond formation either with adjacent nonprotonated amino groups of the molecule or with water molecules.Also,the increase in the hydrophobicity of these molecules will contribute to their lower basicity.The stepwise basicity constants are relatively high for the first four protonation steps,which is attributable to the fact that these protons can bind to the nitrogen atoms adjacent to the pyrazole groups leaving the central nitrogen free,the electrostatic repulsions between them being therefore of little significance.The remaining protonation steps will occur in the central nitrogen atom,which will produce an important increase in the electrostatic repulsion in the molecule and therefore a reduction in basicity.As stated above,the tertiary nitrogen atoms present in L 4-L 6will also contribute to this diminished basicity.To analyze the interaction with glutamic acid,it is important to know the protonation degree of the ligands at physiological pH values.In Table 2,we have calculated the percentages ofthe different protonated species existing in solution at pH 7.4for receptors L 1-L 6.As can be seen,except for the receptor with the pentamethylenic chains L 3in which the tetraprotonated species prevails,all of the other systems show that the di-and triprotonated species prevail,although to different extents.Interaction with Glutamate.The stepwise constants for the interaction of the receptors L 1-L 6with glutamate are shown in Table 3,and selected distribution diagrams are plotted in Figure 1A -C.All of the studied receptors interact with glutamate forming adduct species with protonation degrees (j )which vary between 8and 0depending on the system (see Table 3).The stepwise constants have been derived from the overall association constants (L +Glu 2-+j H +)H j LGlu (j -2)+,log j )provided by the fitting of the pH-metric titration curves.This takes into account the basicities of the receptors and glutamate (vide supra)and the pH range in which a given species prevails in solution.In this respect,except below pH ca.4and above pH 9,HGlu -can be chosen as the protonated form of glutamate involved in the formation of the different adducts.Below pH 4,the participation of H 2Glu in the equilibria has also to be considered (entries 9and 10in Table 3).For instance,the formation of the H 6LGlu 4+species can proceed through the equilibria HGlu -+H 5L 5+)H 6LGlu 4+(entry 8,Table 3),and H 2Glu +H 4L 4+)H 6LGlu 4(entry 9Table 3),with percentages of participation that depend on pH.One of the effects of the interaction is to render somewhat more basic the receptor,and somewhat more acidic glutamic acid,facilitating the attraction between op-positely charged partners.A first inspection of Table 3and of the diagrams A,B,and C in Figure 1shows that the interaction strengths differ markedly from one system to another depending on the structural features of the receptors involved.L 4is the receptor that presents the highest capacity for interacting with glutamate throughout all of the pH range explored.It must also be remarked that there are not clear-cut trends in the values of the stepwise constants as a function of the protonation degree of the receptors.This suggests that charge -charge attractions do not play the most(25)(a)Martell,E.;Smith,R.M.Critical Stability Constants ;Plenum:NewYork,1975.(b)Motekaitis,R.J.NIST Critically Selected Stability Constants of Metal Complexes Database ;NIST Standard Reference Database,version 4,1997.Table 1.Protonation Constants of Glutamic Acid and Receptors L 1-L 6Determined in NaCl 0.15mol dm -3at 298.1KreactionGluL 1aL 2aL 3bL 4L 5L 6L +H )L H c 9.574(2)d 9.74(2)8.90(3)9.56(1)9.25(3)9.49(4)9.34(5)L H +H )L H 2 4.165(3)8.86(2)8.27(2)8.939(7)8.38(3)8.11(5)8.13(5)L H 2+H )L H 3 2.18(2)7.96(2) 6.62(3)8.02(1) 6.89(5)7.17(6)7.46(7)L H 3+H )L H 4 6.83(2) 5.85(4)7.63(1) 6.32(5) 6.35(6) 5.97(8)L H 4+H )L H 5 4.57(3) 3.37(4) 2.72(8) 2.84(9) 3.23(9)L H 5+H )L H 6 3.18(3) 2.27(6)∑log K H n L41.135.334.233.634.034.1aTaken from ref 23.b These data were previously cited in a short communication (ref 26).c Charges omitted for clarity.d Values in parentheses are the standard deviations in the last significant figure.Scheme 2.L -Glutamate Acid -BaseBehaviorTable 2.Percentages of the Different Protonated Species at pH 7.4H 1L aH 2LH 3LH 4LL 11186417L 21077130L 3083458L 4083458L 51154323L 6842482aCharges omitted for clarity.A R T I C L E SMiranda et al.826J.AM.CHEM.SOC.9VOL.126,NO.3,2004outstanding role and that other forces contribute very importantly to these processes.26However,in systems such as these,which present overlapping equilibria,it is convenient to use conditional constants because they provide a clearer picture of the selectivity trends.27These constants are defined as the quotient between the overall amounts of complexed species and those of free receptor and substrate at a given pH[eq1].In Figure2are presented the logarithms of the effective constants versus pH for all of the studied systems.Receptors L1and L2with a nonfunctionalized secondary amino group in the side chains display opposite trend from all other receptors. While the stability of the L1and L2adducts tends to increase with pH,the other ligands show a decreasing interaction. Additionally,L1and L2present a close interaction over the entire pH range under study.The tetraaminic macrocycle L3is a better(26)Escartı´,F.;Miranda,C.;Lamarque,L.;Latorre,J.;Garcı´a-Espan˜a,E.;Kumar,M.;Ara´n,V.J.;Navarro,mun.2002,9,936-937.(27)(a)Bianchi,A.;Garcı´a-Espan˜a,c.1999,12,1725-1732.(b)Aguilar,J.A.;Celda,B.;Garcı´a-Espan˜a,E.;Luis,S.V.;Martı´nez,M.;Ramı´rez,J.A.;Soriano,C.;Tejero,B.J.Chem.Soc.,Perkin Trans.22000, 7,1323-1328.Table3.Stability Constants for the Interaction of L1-L6with the Different Protonated Forms of Glutamate(Glu) entry reaction a L1L2L3L4L5L6 1Glu+L)Glu L 3.30(2)b 4.11(1)2HGlu+L)HGlu L 3.65(2) 4.11(1) 3.68(2) 3.38(4) 3Glu+H L)HGlu L 3.89(2) 4.48(1) 3.96(2) 3.57(4) 4HGlu+H L)H2Glu L 3.49(2) 3.89(1) 2.37(4) 3.71(2)5HGlu+H2L)H3Glu L 3.44(2) 3.73(1) 2.34(3) 4.14(2) 2.46(4) 2.61(7) 6HGlu+H3L)H4Glu L 3.33(2) 3.56(2) 2.66(3) 4.65(2) 2.74(3) 2.55(7) 7HGlu+H4L)H5Glu L 3.02(2) 3.26(2) 2.58(3) 4.77(2) 2.87(3) 2.91(5) 8HGlu+H5L)H6Glu L 3.11(3) 3.54(2) 6.76(3) 4.96(3) 4.47(3) 9H2Glu+H4L)H6Glu L 2.54(3) 3.05(2) 3.88(2) 5.35(3) 3.66(4) 3.56(3) 10H2Glu+H5L)H7Glu L 2.61(6) 2.73(4) 5.51(3) 3.57(4) 3.22(8) 11H3Glu+H4L)H7Glu L 4.82(2) 4.12(9)a Charges omitted for clarity.b Values in parentheses are standard deviations in the last significantfigure.Figure1.Distribution diagrams for the systems(A)L1-glutamic acid, (B)L4-glutamic acid,and(C)L5-glutamicacid.Figure2.Representation of the variation of K cond(M-1)for the interaction of glutamic acid with(A)L1and L3,(B)L2,L4,L5,and L6.Initial concentrations of glutamate and receptors are10-3mol dm-3.Kcond)∑[(H i L)‚(H j Glu)]/{∑[H i L]∑[H j Glu]}(1)New1H-Pyrazole-Containing Polyamine Receptors A R T I C L E SJ.AM.CHEM.SOC.9VOL.126,NO.3,2004827receptor at acidic pH,but its interaction markedly decreases on raising the pH.These results strongly suggest the implication of the central nitrogens of the lateral polyamine chains in the stabilization of the adducts.Among the N-functionalized receptors,L4presents the largest interaction with glutamate.Interestingly enough,L5,which differs from L4only in having a phenethyl group instead of a benzyl one,presents much lower stability of its adducts.Since the basicity and thereby the protonation states that L4and L5 present with pH are very close,the reason for the larger stability of the L4adducts could reside on a better spatial disposition for formingπ-cation interactions with the ammonium group of the amino acid.In addition,as already pointed out,L4presents the highest affinity for glutamic acid in a wide pH range,being overcome only by L1and L2at pH values over9.This observation again supports the contribution ofπ-cation inter-actions in the system L4-glutamic because at these pH values the ammonium functionality will start to deprotonate(see Scheme2and Figure1B).Table4gathers the percentages of the species existing in equilibria at pH7.4together with the values of the conditional constant at this pH.In correspondence with Figure1A,1C and Figure S2(Supporting Information),it can be seen that for L1, L2,L5,and L6the prevailing species are[H2L‚HGlu]+and[H3L‚HGlu]2+(protonation degrees3and4,respectively),while for L3the main species are[H3L‚HGlu]+and[H4L‚HGlu]2+ (protonation degrees4and5,respectively).The most effective receptor at this pH would be L4which joins hydrogen bonding, charge-charge,andπ-cation contributions for the stabilization of the adducts.To check the selectivity of this receptor,we have also studied its interaction with L-aspartate,which is a competitor of L-glutamate in the biologic receptors.The conditional constant at pH7.4has a value of3.1logarithmic units for the system Asp-L4.Therefore,the selectivity of L4 for glutamate over aspartate(K cond(L4-glu)/K cond(L4-asp))will be of ca.15.It is interesting to remark that the affinity of L4 for zwiterionic L-glutamate at pH7.4is even larger than that displayed by receptors III and IV(Chart1)with the protected dianion N-acetyl-L-glutamate lacking the zwitterionic charac-teristics.Applying eq1and the stability constants reported in ref17,conditional constants at pH7.4of 3.24and 2.96 logarithmic units can be derived for the systems III-L-Glu and IV-L-Glu,respectively.Molecular Modeling Studies.Molecular mechanics-based methods involving docking studies have been used to study the binding orientations and affinities for the complexation of glutamate by L1-L6receptors.The quality of a computer simulation depends on two factors:accuracy of the force field that describes intra-and intermolecular interactions,and an adequate sampling of the conformational and configuration space of the system.28The additive AMBER force field is appropriate for describing the complexation processes of our compounds,as it is one of the best methods29in reproducing H-bonding and stacking stabiliza-tion energies.The experimental data show that at pH7.4,L1-L6exist in different protonation states.So,a theoretical study of the protonation of these ligands was done,including all of the species shown in5%or more abundance in the potentiometric measurements(Table4).In each case,the more favored positions of protons were calculated for mono-,di-,tri-,and tetraprotonated species.Molecular dynamics studies were performed to find the minimum energy conformations with simulated solvent effects.Molecular modeling studies were carried out using the AMBER30method implemented in the Hyperchem6.0pack-age,31modified by the inclusion of appropriate parameters. Where available,the parameters came from analogous ones used in the literature.32All others were developed following Koll-man33and Hopfinger34procedures.The equilibrium bond length and angle values came from experimental values of reasonable reference compounds.All of the compounds were constructed using standard geometry and standard bond lengths.To develop suitable parameters for NH‚‚‚N hydrogen bonding,ab initio calculations at the STO-3G level35were used to calculate atomic charges compatible with the AMBER force field charges,as they gave excellent results,and,at the same time,this method allows the study of aryl-amine interactions.In all cases,full geometry optimizations with the Polak-Ribiere algorithm were carried out,with no restraints.Ions are separated far away and well solvated in water due to the fact that water has a high dielectric constant and hydrogen bond network.Consequently,there is no need to use counteri-ons36in the modelization studies.In the absence of explicit solvent molecules,a distance-dependent dielectric factor quali-tatively simulates the presence of water,as it takes into account the fact that the intermolecular electrostatic interactions should vanish more rapidly with distance than in the gas phase.The same results can be obtained using a constant dielectric factor greater than1.We have chosen to use a distance-dependent dielectric constant( )4R ij)as this was the method used by Weiner et al.37to develop the AMBER force field.Table8 shows the theoretical differences in protonation energy(∆E p) of mono-,bi-,and triprotonated hexaamine ligands,for the (28)Urban,J.J.;Cronin,C.W.;Roberts,R.R.;Famini,G.R.J.Am.Chem.Soc.1997,119,12292-12299.(29)Hobza,P.;Kabelac,M.;Sponer,J.;Mejzlik,P.;Vondrasek,put.Chem.1997,18,1136-1150.(30)Cornell,W.D.;Cieplak,P.;Bayly,C.I.;Gould,I.R.;Merz,K.M.,Jr.;Ferguson,D.M.;Spelmeyer,D.C.;Fox,T.;Caldwell,J.W.;Kollman,P.A.J.Am.Chem.Soc.1995,117,5179-5197.(31)Hyperchem6.0(Hypercube Inc.).(32)(a)Fox,T.;Scanlan,T.S.;Kollman,P.A.J.Am.Chem.Soc.1997,119,11571-11577.(b)Grootenhuis,P.D.;Kollman,P.A.J.Am.Chem.Soc.1989,111,2152-2158.(c)Moyna,G.;Hernandez,G.;Williams,H.J.;Nachman,R.J.;Scott,put.Sci.1997,37,951-956.(d)Boden,C.D.J.;Patenden,put.-Aided Mol.Des.1999, 13,153-166.(33)/amber.(34)Hopfinger,A.J.;Pearlstein,put.Chem.1984,5,486-499.(35)Glennon,T.M.;Zheng,Y.-J.;Le Grand,S.M.;Shutzberg,B.A.;Merz,K.M.,put.Chem.1994,15,1019-1040.(36)Wang,J.;Kollman,P.A.J.Am.Chem.Soc.1998,120,11106-11114.Table4.Percentages of the Different Protonated Adducts[HGlu‚H j L](j-1)+,Overall Percentages of Complexation,andConditional Constants(K Cond)at pH7.4for the Interaction ofGlutamate(HGlu-)with Receptors L1-L6at Physiological pH[H n L‚HGlu]an)1n)2n)3n)4∑{[H n L‚HGlu]}K cond(M-1)L13272353 2.44×103L2947763 4.12×103L31101324 3.99×102L423737581 2.04×104L51010222 3.51×102L6121224 3.64×102a Charges omitted for clarity.A R T I C L E S Miranda et al. 828J.AM.CHEM.SOC.9VOL.126,NO.3,2004。
chemical-reaction-engineering-3ed-edition作者-octave-Levenspiel-课后习题答案
chemical-reaction-engineeri ng-3ed-edition作者-octave-Levenspiel-课后习题答案Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING (1)CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS (3)CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA (7)CHAPTER 4 INTRODUCTION TO REACTOR DESIGN (20)CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR (23)CHAPTER 6 DESIGN FOR SINGLE REACTIONS (27)CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR (34)CHAPTER 11 BASICS OF NON-IDEAL FLOW (36)CHAPTER 18 SOLID CATALYZED REACTIONS (45)Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for a small community (Fig.P1.1). Waste water, 32000 m 3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in the tank attack and break down the organic material (organic waste) +O 2 −−−→−microbes CO 2 + H 2OA typical entering feed has a BOD (biological oxygen demand) of 200 mg O 2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.Figure P1.1Solution:)/(1017.2)/(75.183132/100010001)0200()(313200031320001343333s m mol day m mol day molgm L mg g L mg day day m dayday m VdtdN r A A ⋅⨯=⋅=-⨯⨯⨯-⨯-=-=--1.2 Coal burning electrical power station. Large central power stations (about 1000 MW electrical) using fluiding bed combustors may be built some day (see Fig.P1.2). These giants would be fed 240 tons of coal/hr (90% C, 10%H 2), 50% of which would burn within the battery of primary fluidized beds, the other 50% elsewhere in the system. One suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and containing solids to a depth of 1 m. Find the rate of reaction within theWaste Waste Clean200 mgMean residenZerobeds, based on the oxygen used.Solution:380010)1420(m V =⨯⨯⨯=)/(9000101089.05.01024033hr bed molc hrkgckgcoal kgc hr coal t N c ⋅-=⨯-=⨯⨯⨯-=∆∆ )/(25.111900080011322hr m kmolO t N V r r c c O ⋅=-⨯-=∆∆-=-=)/(12000412000190002hr bed mol dt dO ⋅=+⨯= )/(17.4800)/(105.113422s m mol hr bed mol dt dO V r O ⋅=⋅⨯==-Chapter 2 Kinetics of Homogeneous Reactions2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction?Solution: Because we don’t know whether it is an elementary reaction or not, we can’t tell the index of the reaction.2.2 Given the reaction 2NO 2 + 1/2 O 2 = N 2O 5 , what is the relation between the ratesof formation and disappearance of the three reaction components? Solution: 522224O N O NO r r r =-=-2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rateexpression-r A = 2 C 0.5 A C BWhat is the rate expression for this reaction if the stoichiometric equation is written asA + 2B = 2R + SSolution: No change. The stoichiometric equation can’t effect the rate equation, so it doesn’t change.2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance ofsubstrate is given by-r A =A06]][[1760C E A + , mol/m 3·sWhat are the units of the two constants? Solution: ][]6[]][][[][03A A C E A k s m mol r +=⋅=- 3/][]6[m mol C A ==∴sm mol m mol m mol s m mol k 1)/)(/(/][3333=⋅⋅=2.5 For the complex reaction with stoichiometry A + 3B → 2R + S and withsecond-order rate expression-r A = k 1[A][B]are the reaction rates related as follows: r A = r B = r R ? If the rates are not so related, then how are they related? Please account for the sings , + or - .Solution: R B A r r r 2131=-=-2.6 A certain reaction has a rate given by-r A = 0.005 C 2 A , mol/cm 3·min If the concentration is to be expressed in mol/liter and time in hours, what wouldbe the value and units of the rate constant?Solution:min)()(3'⋅⨯-=⋅⨯-cm molr hr L mol r A A 22443'300005.0106610)(minAA A A A C C r r cm mol mol hr L r =⨯⨯=⋅⨯=-⋅⋅⋅=-∴ AA A A A C C cmmol mol L C cmmolC L mol C 33'3'10)()(=⋅⋅=∴⨯=⨯2'42'32'103)10(300300)(AA A A C C C r --⨯=⨯==-∴ 4'103-⨯=∴k2.7 For a gas reaction at 400 K the rate is reported as -dtdp A= 3.66 p 2 A , atm/hr (a) What are the units of the rate constant?(b) What is the value of the rate constant for this reaction if the rate equation isexpressed as-r A = - dtdN V A1 = k C2 A , mol/m 3·s Solution:(a) The unit of the rate constant is ]/1[hr atm ⋅ (b) dtdN V r AA 1-=-Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to22)(66.366.3)(1RT C RTP RT dt dP RT dt dP VRT V r A A A A A ==-=-=-22)66.3(A A kC C RT ==So we can get that the value of1.12040008205.066.366.3=⨯⨯==RT k2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.How much faster the decomposition at 650℃ than at 500℃?Solution:586.7)92311731()10/(314.8/300)11(3211212=-⋅⋅=-==KK K mol kJ mol kJ T T R E k k Ln r r Ln7.197012=∴r r2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ants, I findRunning speed, m/hr150160230295370Temperatu re, ℃13 16 22 24 28 What activation energy represents this change in bustliness? Solution:RTE RTE RTE ek eak t cons ion concentrat f let ion concentrat f ek r ---=⋅⋅=⋅='00tan )()(RET Lnk Lnr A 1'-=∴ Suppose Tx Lnr y A 1,==, so ,REslope -= intercept 'Lnk =)/(1-⋅h m r A 150 160 230 295 370 A Lnr-3.1780 -3.1135 -2.7506 -2.5017 -2.2752CT o / 13 16 22 24 28 3101-⨯T3.4947 3.4584 3.3881 3.3653 3.3206-y = 5417.9x - 15.686R2 = 0.9712340.00330.003350.00340.003450.00351/T-L n r-y = -5147.9 x + 15.686Also K REslope 9.5147-=-=, intercept 'Lnk == 15.686 , mol kJ K mol J K E /80.42)/(3145.89.5147=⋅⨯-=Chapter 3 Interpretation of Batch Reactor Data3.1 If -r A = - (dC A /dt) =0.2 mol/liter·sec when C A = 1 mol/liter, what is the rate ofreaction when C A = 10 mol/liter? Note: the order of reaction is not known.Solution: Information is not enough, so we can’t answer this kind of question.3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A isconverted in a 5-minute run. How much longer would it take to reach 75% conversion?Solution: Because the decomposition of A is a 1st -order reaction, so we can express the rate equation as:A A kC r =-We know that for 1st -order reaction, kt C C LnAAo=, 11kt C C LnA Ao =, 22kt C CLn A Ao = Ao A C C 5.01=, Ao A C C 25.02=So 21)24(1)(11212Ln kLn Ln k C C Ln C C Ln k t t A Ao A Ao =-=-=- equ(1) min 521)(111===Ln kC C Ln k t A Ao equ(2) So m in 5112==-t t t3.3 Repeat the previous problem for second-order kinetics. Solution: We know that for 2nd -order reaction, kt C C A A =-011, So we have two equations as follow:min 511211101k kt C C C C C AoAo Ao A A ===-=-, equ(1)2123)1(31411kt kt C C C C C AoAo Ao Ao A ===-=-, equ(2) So m in 15312==t t , m in 1012=-t t3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product by a 21-order rate. What would be the fraction converted in a half-hour run?Solution: In a-21order reaction: 5.0AA A kC dt dC r =-=-, After integration, we can get:5.015.02A Ao C C kt -=, So we have two equations as follow:min)10(5.0)41(15.05.05.05.015.0k kt C C C C C Ao Ao AoA Ao ===-=-, equ(1) min)30(25.025.0k kt C C A Ao ==-, equ(2)Combining these two equations, we can get:25.05.1kt C Ao =, but this means 05.02<A C , whichis impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted 1=A X .3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?Solution: The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,monomer monomer kC r =-And k C C Lnoomin)34(8.0= 1min 00657.0-=kmonomer monomer C r )min 00657.0(1-=-3.6 After 8 minutes in a batch reactor, reactant (C A0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction. Solution:In 1st order reaction, 43.1511111111212==--=Ln Ln X Lnk X Ln k t t A A , dissatisfied.In 2nd order reaction, 49/4/912.0111.01)11(1)11(11212==--=--=Ao Ao Ao Ao Ao Ao Ao A Ao A C C C C C C C C k C C k t t, satisfied.According to the information, the reaction is a 2nd -order reaction.3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike —into the joint with his week’s salary of ＄180, steady gambling at “2-up” for two hours, then home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable —at a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with ＄135. How much was his raise? Solution:180=Ao n , 13=A n , h t 2=,135'=A n , h t 3;=, A A kn r α-So we obtain kt n n LnAAo=, ''')()(tn n Ln t n n Ln AAo A Ao= 3135213180'Ao n Ln Ln =, 28'=An3.9 The first-order reversible liquid reactionA ↔ R , C A0 = 0.5 mol/liter, C R0=0takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium conversion is 66.7%. Find the equation for the this reaction. Solution: Liquid reaction, which belongs to constant volume system,1st order reversible reaction, according to page56 eq. 53b, we obtain121112102110)(1)(-+-+=+-==⎰⎰AX A A tX k k k k Lnk k X k k k dX dt t Amin 8sec 480==t , 33.0=A X , so we obtain eq(1)33.0)(1min8sec 480211121k k k k Ln k k +-+= eq(1) Ae AeAe c X X M C C k k K -+===1Re 21, 0==AoRo C C M , so we obtain eq(2) 232132121=-=-==AeAe c X X k k K ,212k k =∴ eq(2)Combining eq(1) and eq(2), we obtain1412sec 108.4m in 02888.0---⨯==k 14121sec 1063.9m in 05776.02---⨯===k kSo the rate equation is )(21A Ao A AA C C k C k dtdC r --=-=- )(sec 1063.9sec 108.401414A A A C C C -⨯-⨯=----3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor itsconcentration drops from C A0 = 2.03 mol/liter to C Af = 1.97 mol/liter. Find the rate equation from the reaction if the kinetics are second-order with respect to A.Solution: It’s a irreversible second -order reaction system, according to page44 eq 12, we obtainmin 103.2197.111⋅=-k , so min015.01⋅=mol Lkso the rate equation is 21)min 015.0(A A C r -=-3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzymesucrase as follows:Aucrose −−→−sucraseproductsStarting with a sucrose concentration C A0 = 1.0 millimol/liter and an enzyme concentrationC E0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or-r A =MA E A C C C C k +03 where C M = Michaelis constantIf the fit is reasonable, evaluate the constants k 3 and C M . Solve by the integral method.Solution: Solve the question by the integral method:AA M A A Eo A A C k Ck C C C C k dt dC r 5431+=+=-=-, M Eo C C k k 34=, MC k 15= AAo A Ao A Ao C C C C Lnk k k C C t -⋅+=-4451hrt ,AC ,mmol /L A Ao AAo C C C C Ln-AAo C C t -1 0.84 1.0897 6.25 20.681.20526.25C A , millimol /liter0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 0.0025t,hr 1 2 3 4 5 6 7 8 9 10 113 0.53 1.3508 6.38304 0.38 1.5606 6.45165 0.27 1.7936 6.8493 6 0.16 2.1816 7.14287 0.09 2.6461 7.69238 0.04 3.3530 8.33339 0.018 4.0910 9.1650 10 0.006 5.1469 10.0604 110.00256.006511.0276Suppose y=A Ao C C t-, x=AAo A Ao C C C C Ln-, thus we obtain such straight line graphy = 0.9879x + 5.0497R 2 = 0.99802468101201234567Ln(Cao/Ca)/(Cao-Ca)t /(C a o -C a )9879.0134===Eo M C k C k Slope , intercept=0497.545=k k So )/(1956.00497.59879.015L mmol k C M ===, 14380.1901.09879.01956.0-=⨯==hr C C k k Eo M3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A −−→−enzyme R, -r A =min22000⋅+liter molC C C A E AIf we introduce enzyme (C E0 = 0.001 mol/liter) and reactant (C A0 = 10mol/liter) into a batch rector and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.. Solution:510001.020021+=⨯+=-=-AA A A A C C C dC dt r Rearranging and integrating, we obtain:10025.0025.0100)(510)510(⎥⎦⎤⎢⎣⎡-+=+-==⎰⎰A Ao A Ao A A tC C C C Ln dC C dt t min 79.109)(5025.01010=-+=A Ao C C Ln3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at22.9℃:H 2SO 4 + (C 2H 5)2SO 4 → 2C 2H 5SO 4HInitial concentrations of H 2SO 4 and (C 2H 5)2SO 4 are each 5.5 mol/liter. Find a rate equation for this reaction.Table P3.20 t, minC 2H 5SO 4H , mol/li ter t, minC 2H 5SO 4H , mol/li ter1804.1141 1.18 194 4.31 48 1.38 212 4.45 55 1.63 267 4.86 75 2.24 318 5.15 96 2.75 368 5.32 127 3.31 379 5.35 146 3.76 410 5.42 1623.81∞(5.80)Solution: It’s a constant -volume system, so we can use X A solving the problem: i) We postulate it is a 2nd order reversible reaction system R B A 2⇔+ The rate equation is: 221R B A A A C k C C k dtdC r -=-=- L mol C C Bo Ao /5.5==, )1(A Ao A X C C -=, A A Ao Bo B C X C C C =-=, A Ao R X C C 2=When ∞=t , L mol X C C Ae Ao /8.52Re == So 5273.05.528.5=⨯=Ae X , L mol X C C C Ae Ao Be Ae /6.2)5273.01(5.5)1(=-⨯=-== After integrating, we obtaint C X k X X X X X LnAo AeA Ae A Ae Ae )11(2)12(1-=--- eq (1)The calculating result is presented in following Table.t,mi nLmol C R /,Lmol C A /,AXAAe AAe Ae X X X X X Ln---)12()1(AeAX X Ln -0 0 5.5 0 0 041 1.18 4.91 0.10730.2163 -0.227548 1.38 4.81 0.12540.2587 -0.271755 1.63 4.685 0.14820.3145 -0.329975 2.24 4.38 0.20360.4668 -0.488196 2.75 4.125 0.25 0.6165 -0.642712 7 3.31 3.8450.30090.8140 -0.845614 6 3.76 3.620.34181.0089 -1.044916 2 3.81 3.5950.34641.0332 -1.069718 0 4.11 3.4450.37361.1937 -1.233119 4 4.31 3.3450.39181.3177 -1.359121 2 4.45 3.2750.40451.4150 -1.4578267 4.86 3.07 0.4418 1.7730 -1.8197 318 5.15 2.925 0.4682 2.1390 -2.1886 368 5.32 2.84 0.4836 2.4405 -2.4918 379 5.35 2.825 0.4864 2.5047 -2.5564 4105.42 2.79 0.4927 2.6731 -2.7254 ∞5.82.60.5273——Draw AAe AAe Ae X X X X X Ln---)12(~ t plot, we obtain a straight line:y = 0.0067x - 0.0276R 2= 0.998800.511.522.530100200300400500tL n0067.0)11(21=-=Ao AeC X k Slope ,min)/(10794.65.5)15273.01(20067.041⋅⨯=⨯-=∴-mol L kWhen approach to equilibrium, BeAe c C C C k k K 2Re 21==, so min)/(10364.18.56.210794.642242Re 12⋅⨯=⨯⨯==--mol L C C C k k Be Ae So the rate equation ism in)/()10364.110794.6(244⋅⨯-⨯=---L mol C C C r R B A Aii) We postulate it is a 1st order reversible reaction system, so the rate equation isR A AA C k C k dtdC r 21-=-=- After rearranging and integrating, we obtaint k X X X Ln AeAe A '11)1(=-eq (2) Draw )1(AeAX X Ln -~ t plot, we obtain another straight line: -y = 0.0068x - 0.0156R 2 = 0.998600.511.522.530100200300400500x-L n0068.0'1-==AeX k Slope ,So 13'1m in 10586.35273.00068.0--⨯-=⨯-=k133Re '1'2min 10607.18.56.210586.3---⨯-=⨯⨯-==C C k k AeSo the rate equation ism in)/()10607.110586.3(33⋅⨯+⨯-=---L mol C C r R A AWe find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(1) and eq.(2), especially when X Ae =0.5 , the two equations are identical. This means these two equations would have almost the same fitness of data when the experiment data of the reaction show that X Ae =0.5.(The data that we use just have X Ae =0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and C A alone determines this rate:C A ,mol/liter1 2 4 6 7 9 12-r A , mol/liter·hr0.06 0.1 0.25 1.0 2.0 1.0 0.5We plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from C A0 = 10 mol/liter to C Af = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.04812162002 4 68 10 12 14Ca-1/Ra3.31 The thermal decomposition of hydrogen iodide 2HI → H 2 + I 2is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:T,℃ 508427 393 356 283k,cm 3/mol·s0.10590.003100.00058880.9×10-60.942×10-6Find the complete rate equation for this reaction. Use units of joules, moles, cm 3,and seconds.According to Arrhenius’ Law,k = k 0e -E/R Ttransform it,- In(k) = E/R·(1/T) －In(k 0)Drawing the figure of the relationship between k and T as follows:y = 7319.1x - 11.567R 2= 0.987904812160.0010.0020.0030.0041/T-L n (k )From the figure, we getslope = E/R = 7319.1 intercept = - In(k 0) = -11.567E = 60851 J/mol k 0 = 105556 cm 3/mol·sFrom the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate expression is- r A = 105556e -60851/R T ·C A 2Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, C A0 = 100, C B0 = 200, A +B→ R + S, X A = 0.8. Find X B ,C A ,C B . Solution: Given a gaseous feed, 100=Ao C , 200=Bo C , S R B A +→+0=A X , find B X , A C , B C0==B A εε, 202.0100)1(=⨯=-=A Ao A X C C4.02008.01001=⨯⨯==Bo A Ao B C X bC X 1206.0200)1(=⨯=-=B Bo B X C C4.2 Given a dilute aqueous feed, C A0 = C B0 =100, A +2B→ R + S, C A = 20. Find X A , X B , C B .Solution: Given a dilute aqueous feed, 100==Bo Ao C C ,S R B A +→+2, 20=A C , find A X , B X , B CAqueous reaction system, so 0==B A εε When 0=A X , 200=V When 1=A X , 100=VSo 21-=A ε, 41-==Ao Bo A B bC C εε8.01002011=-=-=Ao A A C C X , 16.11008.010012>=⨯⨯=⋅=Bo A Ao B C X C a b X , which is impossible. So 1=B X , 100==Bo B C C4.3 Given a gaseous feed, C A0 =200, C B0 =100, A +B→ R, C A = 50. Find X A , X B , C B . Solution: Given a gaseous feed, 200=Ao C , 100=Bo C ,R B A →+, 50=A C .find A X , B X , B C75.02005011=-=-=Ao A A C C X , 15.1>==BoAAo B C X bC X , which is impossible. So 100==Bo B C C4.4 Given a gaseous feed, C A0 = C B0 =100, A +2B→ R, C B = 20. Find X A , X B , C A . Solution: Given a gaseous feed, 100=+Bo Ao C C ，R B A →+2, 20=Bo C , Find A X , B X , A C0=B X , 200100100=+=B A V ,1=B X 15010050=+=R A V25.0200200150-=-=B ε, 5.01002110025.0-=⨯⨯-=-A ε842.02025.010020100=⨯--=B X , 421.0100842.010021=⨯⨯=A X34.73421.05.01421.0110011=⨯--⨯=+-=A A A AoA X X C C ε4.6 Given a gaseous feed, T 0 =1000 K, π0＝5atm, C A0=100, C B0=200, A +B→5R,T =400K, π=4atm, C A =20. Find X A , X B , C B .Solution: Given a gaseous feed, K T o 1000=, atm 50=π, 100=Ao C , 200=Bo CR B A 5→+, K T 400=, atm 4=π, 20=A C , find A X , B X , B C .1300300600=-=A ε, 2==Ao Bo AB bC C a εε,5.0410********=⨯⨯=ππT T According to eq page 87,818.05.010020115.0100201110000=⨯⨯+⨯-=+-=ππεππT T C C T T C C X Ao A AAo A A409.0200818.0100=⨯==Bo A Ao B aC X bC X130818.011200)818.0100200(1)(0=⨯+⨯-=+-=A A Ao A Ao Bo B X C T T X a b C C C εππ4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-literpopcorn to be operatedin steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/minof mixed exit stream. Independent tests show that when raw corn pops its volumegoes from 1 to 31.With this information determine what fraction of raw corn is popped in the unit.Solution: 301131=-=A ε, ..1u a C Ao =, ..281281u a C C Ao A ==%5.462813012811=⨯+-=+-=∴AA Ao A Ao A C C C C X εChapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A → R + 2S with unknown kinetics. If a spacevelocity of 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time and mean residence time or holding time of fluid in the plug flow reactor.Solution: min 11==sτ,Varying volume system, so t can’t be found.5.2 In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min.What space-time and space-velocity are needed to effect this conversion in a plug flow reactor and in a mixed flow reactor? Solution: Liquid reaction system, so 0=A ε According to eq.4 on page 92, min 130=-=⎰AX AAAo r dC C t Eq.13, AAAo A A Ao R F M r X C r C C -=--=..τ, R F M ..τ can’t be cert ain. Eq.17, ⎰-=AX AAAo R F P r dX C 0..τ, so m in 13...==R B R F P t τ5.4 We plan to replace our present mixed flow reactor with one having double thebolume. For the same aqueous feed (10 mol A/liter) and the same feed rate find the new conversion. The reaction are represented byA → R, -r A = kC 1.5 ASolution: Liquid reaction system, so 0=A εA A Ao Ao r X C F V -==τ, 5.1)]1([)(A Ao A A Ao A Ao X C k X r C C C -=-- Now we know: V V 2=', Ao Ao F F =', Ao Ao C C =', 7.0=A X So we obtain5.15.15.15.1)1()2)1(2A Ao A A Ao A Ao Ao X kC X X kC X F VF V -='-'==''52.8)7.01(7.02)1(5.15.1=-⨯='-'∴A AX X794.0='A X5.5 An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) isto be converted to product in a plug flow reactor. The kinetics of the reaction is represented byA +B→ R, -r A = 200C A C Bmin⋅liter molFind the volume of reactor needed for 99.9% conversion of A to product.Solution: Aqueous reaction system, so 0=A εAccording to page 102 eq.19,⎰⎰-=-==Af AfX AA X A A AoAo Ao r dX r dC C C t F V 001⎰-==AfX AAAo or dX C Vντ, m in /400liter o =ν, L r dX r dX C V AAX A A o Ao Af3.1244001.0999.000=-⨯=-=∴⎰⎰ν5.9 A specific enzyme acts as catalyst in the fermentation of reactant A. At a givenenzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (C A0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given byA −−→−enzymeR , -r A = litermolC C A A ⋅+min 5.011.0Solution: P.F.R, according to page 102 eq.18, aqueous reaction, 0=ε⎰-=A X AA Ao r dX F V 0 )11(21251.05.010A AX A A A Ao X X Ln dX C C F V A+-⨯=+=∴⎰\L Ln4.986)95.005.01(125=+=5.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R.Find the size of mixed flow reactor needed for 95% conversion of reactant in a feed stream (25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation at this enzyme concentration are given byA −−→−enzyme R , -r A =litermolC C A A ⋅+min 5.011.0Solution: min /25L o =ν, L mol C Ao /2=, m in /50mol F Ao =, 95.0=A X Constant volume system, M.F.R., so we obtainmin 5.199205.05.01205.01.095.02=⨯⨯+⨯⨯⨯=-==AAAo or X C Vντ,39875.4min /25min 5.199m L V o =⨯==τν5.14 A stream of pure gaseous reactant A (C A0 = 660 mmol/liter) enters a plug flowreactor at a flow rate of F A0 = 540 mmol/min and polymerizes the as follows3A → R, -r A = 54min⋅liter mmolHow large a reactor is needed to lower the concentration of A in the exitstream to C Af = 330 mmol/liter?Solution: 321131-=-=A ε, 75.0660330321660330111=⨯--=+-=Ao A A Ao A A C C C C X ε 0-order homogeneous reaction, according to page 103 eq.20A Ao AoAooX C F VC kVkk ===ντ So we obtainL X k C C F V A Ao Ao Ao 5.75475.05401=⨯==5.16 Gaseous reactant A decomposes as follows:A → 3 R, -r A = (0.6min -1)C AFind the conversion of A in a 50% A – 50% inert feed (υ0 = 180 liter/min, C A0 =300 mmol/liter) to a 1 m 3 mixed flow reactor.Solution: 31m V =, M.F.R. 1224=-=A εAccording to page 91 eq.11, AAAoAAo AAAo oX X C X C r X C V+-=-==116.0ντmin/1801000)1(6.0)1(L LX X X A A A =-+=So we obtain 667.0=A XChapter 6 Design for Single Reactions6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in aseries. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the concentration in the exit stream of the second reactor. The reaction is second-order with respect to A and V 2/V 1 =2.Solution:V 2/V 1 = 2, τ1 =011υV =A A A r C C --10 , 2τ = 022υV = 221A A A r C C --C A0=1mol/l , C A1=0.5mol/l , 0201υυ=, -r A1=kC 2 A1 ,-r A2=kC 2 A2 (2nd-order) , 2×2110A A A kC C C -=2221A A A kC C C - So we obtain 2×(1-0.5)/(k0.52)=(0.5-C A2)/(kC A22)C A2= 0.25 mol/l6.2 Water containing a short-lived radioactive species flows continuously through awell-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but we’d like to lower it still more.One of our office secretaries suggests that we insert a baffle down the middle ofthe tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so calculate the expected activity of the exit stream compared to the entering stream.Solution: 1st-order reaction, constant volume system. From the information offeredabout the first reaction,we obtain1τ=01100117171A A A A A A C k C C kC C C V ⋅-=-=υ If a baffle is added,022220212122212υυτττV V +=+==011υV =2222221210A A A A A A kC C C kC C C -+-=007176A A kC C =6/k …… ①。
欧洲药典7.5版
INDEX
To aid users the index includes a reference to the supplement in which the latest version of a text can be found. For example : Amikacin sulfate...............................................7.5-4579 means the monograph Amikacin sulfate can be found on page 4579 of Supplement 7.5. Note that where no reference to a supplement is made, the text can be found in the principal volume.
English index ........................................................................ 4707
Latin index ................................................................................. 4739
EUROPEAN PHARMACOPபைடு நூலகம்EIA 7.5
Index
Numerics 1. General notices ................................................................... 7.5-4453 2.1.1. Droppers...................
coumaric acid 3-hydroxylase -回复
coumaric acid 3-hydroxylase -回复Coumaric Acid 3-Hydroxylase: Unveiling the MysteryIntroduction:Enzymes play a crucial role in various biochemical reactions within living organisms. One such enzyme of interest is Coumaric Acid3-Hydroxylase (CA3H). CA3H is an essential enzyme involved in the phenylpropanoid pathway, responsible for the synthesis of a wide range of phenolic compounds, including lignin, flavonoids, and phytoalexins. In this article, we will delve into the intricate world of CA3H, exploring its structure, function, regulation, and potential applications.Structure:CA3H belongs to the family of cytochrome P450 monooxygenases, specifically the CYP98A subfamily. These enzymes contain a heme prosthetic group, which is involved in redox reactions. CA3H is a membrane-bound protein typically found in the endoplasmic reticulum of plants. The enzyme consists of a single polypeptide chain with approximately 500-600 amino acids. The precise structure of CA3H has been elucidated through X-ray crystallography, providing valuable insights into its catalyticmechanism.Function:CA3H catalyzes the hydroxylation of coumaroyl-CoA, an intermediate of the phenylpropanoid pathway. This hydroxylation reaction leads to the formation of caffeoyl-CoA, a key precursor for various phenolic compounds. The reaction is fueled by the transfer of an electron from NAD(P)H to the heme group within CA3H, which initiates a series of redox reactions resulting in the conversion of coumaroyl-CoA into caffeoyl-CoA. The enzyme’s active site contains specific amino acid residues that facilitate the binding and activation of the substrate, ensuring high specificity and efficiency of the reaction.Regulation:The regulation of CA3H expression is tightly controlled to maintain the balance between phenolic compound synthesis and other metabolic processes. Several factors can influence the expression of CA3H, including light, hormones, developmental stage, and environmental stressors. Light has been found to be a key regulator, with certain wavelengths promoting CA3H expression. Hormones, such as auxins and gibberellins, have also been shown to enhanceCA3H activity. Moreover, the expression of CA3H is often upregulated in response to pathogen attack, suggesting its involvement in plant defense mechanisms.Applications:The phenolic compounds synthesized through the action of CA3H have numerous applications in various industries. Lignin, for instance, is a key component of plant cell walls and is widely used in the production of paper, textiles, and biofuels. Flavonoids, another class of phenolic compounds, are known for their antioxidant properties and are utilized in the food and cosmetic industries. Phytoalexins, which are produced in response to pathogen attack, have potential applications as biopesticides. Therefore, understanding CA3H and manipulating its activity could have significant implications for biotechnological processes, such as plant breeding, lignocellulosic biomass utilization, and pharmaceutical development.Conclusion:Coumaric Acid 3-Hydroxylase is a remarkable enzyme that plays a critical role in the synthesis of various phenolic compounds. Itsstructure, function, and regulation have been extensively studied, providing valuable insights into its catalytic mechanism and potential applications. As our understanding of CA3H continues to grow, we are hopeful that it will pave the way for innovative biotechnological advancements, benefiting various industries and contributing to the sustainable utilization of plant resources.。
生物酶工程公开课一等奖优质课大赛微课获奖课件
Ch6 生物酶工程
生物酶工程（Biological Enzyme Engineering）： 是指在基因水平上，对酶蛋白分子进行修饰、改造， 改进酶蛋白催化特性或酶蛋白蛋白质特性等。 本章主要简介 核酶、进化酶、杂合酶和抗体酶相关 基本概念和基本知识。
第1页
Ch6-1 核酶(Ribozyme) 关关剪脱于于到核接氧核到酶型核当酶应当核酶用用前前酶:于前为为:景基这止止因类，治,核已除疗酶经研了主发究要觉我存催天们在化然主传m核要R酶统N问作A概题前用念体底修物酶饰都—加是—工核蛋，酸， 关处①②③④ACG按剪相当mo切子C除究前久1割e于于ln9体切本毒eu开e其切称前e白质这区核研r9除和4发述就RmciSc41内割身性hbN催型于发酶究发ah年3内组觉核发in质里分—ae研A运位稳和研阶a碱化核核觉UC公，含Ⅱ，酶觉rDnys—究究段送点定免外 核 开ai基反酶酸剪NvB子 其 化 了内n司e。,核发a与 特 性 疫Arr主i应 内 切:m酶 来,s，学切含Inei分这n觉i还t细异问原要atdy酸可切型rt并本割子k要,r子类o目四胞性题性eo分酶核发 前n完质。Rr。的n，核，膜吸选问:和N利两作酶毕是本是B-觉 者命酶并虫淋A取择题细用类用包治核将片R身和名主进核另 是及问:巴其N体。括:疗胞段含D酸为要行糖A残题应外瘤三一 酶对N拼有白，脱催用片体酶留:A选种象接催类 化在血于氧D化段R问择:（；化N基N发核本①病拼含 学题技AA因R觉N主酶身接核:前锤治术N有本uR要临临一（或功A酶疗c体头，N催质l床床、前包D异e,效。，型A初总研前前二aN体括核体，化是s含核体次A期期期究e剪组酶R进上z有酶临）发活核阶yN辑Ⅰ讲后一m床本觉A段概性酸R,功e内，步切当身）了. 效含不念前物；研割切切。， Sym后on者s研是究指一些催植化物人核病免毒酸疫R水缺NA点解本酶身剪临, 其床切前功化期效学后本，提质出 In了no锤是vir头蛋状白二级质结；构模病型毒；Ⅰ② 发卡型核临酶床前发期卡型核酶二 OR级碱isbao结基k1Rzay9NUm构共8neAi1象同vP能e年hr一构.sIint够Cyc个 成. e发一在ch卡个没等装发人病有发，卡免毒由结酶觉疫Ⅰ构5蛋四0缺个，白膜点碱辨基认存虫构顺在核临一临成序床、床情糖，是前二前它N期期期况体G临和下前U床底C,体，物本在14身N个TGMok之1y催oR间UNn化切iAve两割切rsi个t。y除部③内分蛋构涵白乙丙成质子肝肝，－,其病病完R中N毒毒成A蛋，复白加合质工酶临分过床子由前量程蛋期D白；a，质发R和第2N觉页 A部
三羟甲基氨基甲烷-盐酸分子式
三羟甲基氨基甲烷-盐酸分子式
三羟甲基氨基甲烷-盐酸的分子式是C4H13NO3Cl。
这个化合物
是一种有机化合物，也被称为盐酸甲羟胺。
它的分子式表示了化合
物中每种元素的原子数量，C代表碳，H代表氢，N代表氮，O代表氧，Cl代表氯。
根据这个分子式，我们可以得知这种化合物由4个
碳原子、13个氢原子、1个氮原子、3个氧原子和1个氯原子组成。
这种化合物在化学和生物学领域中具有重要的应用，例如作为药物
中间体或生物化学实验中的试剂。
这种化合物的分子式为
C4H13NO3Cl，这是根据化学元素的原子量和化合物中原子的数量推
导出的。
希望这个回答能够满足你的需求。
桦褐孔菌酚类化合物激活过氧化氢酶的作用及机制
山西农业科学 2023，51（9）：1050-1059Journal of Shanxi Agricultural Sciences桦褐孔菌酚类化合物激活过氧化氢酶的作用及机制郝瑞林 1，单树花 1，胡晓燕 2，张志怡 2，李荣山 3，李卓玉1（1.山西大学 生物技术研究所，山西 太原 030006；2.忻州师范学院 生物系，山西 忻州 034000；3.山西省人民医院，山西 太原 030006）摘要：利用制备型高效液相色谱仪从桦褐孔菌中分离并鉴定出12种酚类化合物，通过钼酸铵显色法分析酚类化合物对过氧化氢酶的影响。
结果表明，原儿茶醛、Phelligridin E 和Phelligridin I 显示出增强过氧化氢酶活性的能力，特别是Phelligridin E ；经终浓度分别为0.5、1.0、2.0、3.0、4.0、5.0 μmol/L 的Phelligridin E 处理后，过氧化氢酶的活性分别增加了41.90%、44.68%、52.66%、58.10%、62.78%和68.10%。
通过分子对接和分子动力学模拟分析Phelligridin E 激活过氧化氢酶的机制，结果显示，Phelligridin E 与过氧化氢酶的ASN369、HIS372、ILE373形成较强的疏水作用力，与过氧化氢酶的ASP25、ASN369、HIS372、PRO391和CYS393形成氢键。
应用gmx_MMPBSA 进行自由能分解，结果显示，过氧化氢酶的GLU67、ILE373和PRO391对Phelligridin E 与过氧化氢酶的结合贡献较大，其中，ILE373与Phelligridin E 形成距离为3.63 Å的疏水作用力；PRO391与Phelligridin E 形成距离为2.54 Å的氢键；过氧化氢酶的GLU67未与Phelligridin E 形成4 Å范围内的作用力，但是GLU67在空间上位于ASP25附近，ASP25与Phelligridin E 之间的氢键可能导致GLU67的能量分布发生改变。
用于寡核苷酸合成的硫试剂
用于寡核苷酸合成的硫试剂
常用于寡核苷酸合成的硫试剂包括：
1. 二乙基五硫化二磷（DEP-PTS）： DEP-PTS 是一种优秀的
硫试剂，能够有效地合成短链核酸。
它能够在DNA或RNA
合成过程中起到保护羟基的作用，并且能够促进亲核反应的进行。
2. 二乙基五硫化二磷酸三酯（DEP-PTS-Triester）： DEP-PTS-Triester 是 DEP-PTS 的三酯形式，具有更高的反应活性。
它可
以在较短的反应时间内完成核苷酸的合成，适用于高效快速的寡核苷酸合成反应。
3. 三硫丙胺（Tris(2-mercaptoethyl)amine）：三硫丙胺是一种
常用的还原剂，也可以作为硫试剂用于核酸合成反应中。
它能够还原硫化保护基，从而使羟基产生反应，实现核苷酸的连接。
以上是常用的硫试剂，可以用于寡核苷酸的合成。
不同的硫试剂在具体的反应条件和反应体系中可能会有不同的适用性，具体选择硫试剂要根据实验需求和条件来决定。
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• Assymetrical response • Assymetrical response • Static power consumption • Static power consumption • tpL= 0.69 RLCL • tpL= 0.69 RLCL
VSS VSS
EE141 Integrated © Digital
VOUT Vin VTN V :SS In saturation region
©
VSS pseudo-NMOS Combinational Circuits
12
depletion EE141 Integrated Digital Circuits2ndload
VTC Analysis of Pseudo-NMOS Inverter 1
When the input is low, the PDN is cut-off, however the PUN is on and then pulls the output VOH to VDD.
Vin VOL VTN NMOS devices in cut-off region
Vout VTP PMOS device in linear region
NMOS Pseudo-NMOS Switch Logic Dynamic Logic
EE141 Integrated © Digital
Circuits2nd
2
Combinational Circuits
§6.2.2 Ratioed Logic
EE141 Integrated © Digital
EE141 Integrated © Digital
Circuits2nd
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Combinational Circuits
Active Loads Ratioed Logic
VDD Depletion Load PMOS Load VSS F In1 In2 In3 PDN In1 In2 In3 PDN F VDD VT < 0
+ Fewer transistors -> smaller gates, esp. for large number of inputs + less capacitative load on gates that drive inputs – larger power consumption – less noise margin (VOL > 0) – additional design considerations due to ratioed logic
EE141 Integrated Circuits2nd © Digital
In2 PMOS In3
ቤተ መጻሕፍቲ ባይዱ
PDN
Combinational Circuits
Pseudo-NMOS Logic Gates
VDD PMOS Load VSS F In1 In In2 In3
Pulldown Network
Out Idn
Pullup Device: depletion Mode n-transistor (Vt < 0) OUT
Tradeoffs:
+ + Simpler processing Smaller gates higher power! Additional design considerations for ratioed logic
I DD K P [(VG VS VTP )2 (VG VD VTP )2 ] = K P [(VDD VTP ) 2 (Vout VTP ) 2 ] 0
VOH VDD
EE141 Integrated © Digital
Circuits2nd
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Combinational Circuits
VSS depletion load NMOS
VSS pseudo-NMOS
EE141 Integrated © Digital
Circuits2nd
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Combinational Circuits
Pseudo-NMOS Logic
Basic StructureVDD
PMOS Load VSS
VDD
Pulldown Network
EE141 Integrated © Digital
Circuits2nd
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Combinational Circuits
Pseudo-NMOS Logic
Same idea, as NMOS, but use p-transistor for pull-up "ratioed logic" required for proper design (more about this next) Tradeoffs:
EE141 Integrated © Digital
Circuits2nd
1
Combinational Circuits
Alternatives to Static CMOS
There existed some alternative logic families that either are simpler or faster.
region
NMOS： F Vin VGS VTN ,VOUT VDS VGS VTN In1 In1 In2 In linear region In2 VOUT Vin VTNPDN
In3
Out In
PDN
VGS VTN ,VDS VGS VTN
NMOS
In3
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1 BasicInOperation
VTC Analysis of Pseudo-NMOS Inverter 2
When the input is high, the PDN is on, and then pulls the output VOL to the ground.
Vin VDD: (Vout Vin VTN ) NMOS devices in linear
Circuits2nd
3
Combinational Circuits
Major Ratioed Logic Structure
(a) Resistive Load NMOS (b) Depletion Load NMOS
(c) Psuedo-NMOS
GOAL: To reduce the number of devices over complementary CMOS.
2
KN (VDD VTP )2 , Kr 2 Kr (VDD VTN ) KP
14
Circuits2nd
Combinational Circuits
Pseudo-NMOS Inverter VTC
3.0
2.5
Circuits2nd
5
Combinational Circuits
Depletion Load Ratioed Logic
Used before CMOS was widely available Uses only n transistors
Normal n transistors in pull-down network depletion-mode n transistor (Vt < 0) used for pull-up "ratioed logic" required
4
EE141 Integrated © Digital
Circuits2nd
Combinational Circuits
Resistive Load Ratioed Logic
VDD VDD Resistive Resistive Load Load • N transistors + Load • N transistors + Load • VOH = V DD • VOH = V DD • VOL = • VOL = RPN RPN RPN + RL RPN + RL RL RL F F In1 In1 In2 In2 In3 In3 PDN PDN
Pulldown Network
Pullup Device: P-Transistor
OUT
EE141 Integrated © Digital
Circuits2nd
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Combinational Circuits
Basic Structure of Ratioed Logic
Instead of a combination of active pull-down and pullup networks, such a gate consists of an NMOS pulldown network that realizes the logic function, and a simple load device. The entire PUN is replaced with a single unconditional load device that pulls up the output for a high output.