CHM 203 ORGANIC CHEMISTRYII Course Team Prof.FemiPeters,Mr. AdakoleIkpe&Dr. MakanjuolaOki(Course Developers)-NOUN Prof.J. Amupitan(CourseEditor)-NOUN Prof. H.D. Aliyu (Course Reviewer) NOUN NATIONAL OPEN UNIVERSITY OF NIGERIA COURSE GUIDE CHM 203 COURSE GUIDE ii © 2020by NOUN Press National Open University of Nigeria Headquarters University Village 91, Cadastral Zone Nnamdi Azikiwe Expressway Jabi, Abuja Lagos Office 14/16 Ahmadu Bello Way Victoria Island, Lagos e-mail:centralinfo@nou.edu.ng URL: www.nou.edu.ng All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. Printed2009, 2021 Revised in February, 2021 ISBN:978-058-798-5 CHM 203 COURSE GUIDE iii CONTENTS PAGE Introduction……………………………………………...… iv WhatYouWillLearninthisCourse…………………… ...iv Course Aims………………………………………………. iv Course Objectives…………………………………………. v Contents…………………………………………………… v WorkingthroughthisCourse……………………………… v CourseMaterials…………………………………………... v Course Guide…………………………………………… .....vi StudyUnits……………………………………………… …..vi AssignmentFiles……………………………………...… ....vi PresentationSchedule……………………………………… vii iv INTRODUCTION Chemistryisthestudyof matter.Matterisstudied under the three divisions ofchemistry,viz:Physical,InorganicandOrganicchemistry.While inorganicandphysicalchemistryaredetailedelsewhere,organicchemistry whichdealswith hydrocarbons,their numerousderivativesinaddition to their physicalandchemicalpropertieswillbestudiedinthis text.Thiscourse iscodedCHM203.Itformsthesecondpartoforganicchemistrycourses youwillencounter duringthecourseofyour programmeinchemistryand related programmes.CHM203 isatwocredit unitcourse.The course content consists of 3inter-relatedandinteresting modules. WHATYOU WILLLEARNINTHISCOURSE Inthiscourse,youwilllearnaboutthe electronic concepts in organic chemistry where we discussed the various factors affecting the structure and physicalpropertiesof organiccompounds,availability of electrons and stereochemistry,therelationshipbetweenthestructure oforganic compoundsand theirreactivity. Alsothiscourse describesAromatic hydrocarbonsandtheirderivativesinadditionto theirphysicaland chemicalpropertieswhich determine theirusesinthe industry. Youwillfind severalIn-Text Questions(ITQs) andSelf-AssessmentQuestions(SAQs), withanswersprovidedaswellasactivity exerciseineachunit. COURSEAIMS Thecourseaimsatgivingyouanin-depthknowledgeofthephysicaland chemicalpropertiesofselected,importantclassesoforganiccompoundsthus givingyouasolidfoundation inorganiccompoundsof industrialimportance. Theaimofthiscoursecanbesummarizedasfollows: 1.DiscusstheRelationshipbetweenthestructureandreactivityof organiccompounds. 2.Acquaint learners with Aromaticand polynucleararomatic compounds. 3.Tostudythedeterminantsofthepropertiesandidentificationof selectedclassesoforganiccompounds. 4.To determine theimportanceof moleculararchitecture on physical propertiesof organiccompounds. CHM 203 COURSE GUIDE v COURSEOBJECTIVES Eachunithasitsrespectiveobjectivewhichyoushouldalwaysreferto in yourcourse ofstudysothatderailmentfromsettargetwillbeavoided. Always makea listofyourattainmentaftereachunitandcomparethem with theobjectiveslistedbythecoursedevelopers.Thus theoverall objectiveof thecoursecanbesummarizedas: a)Familiarizeourselveswithmolecularstructureoforganic compounds. b)Notethecharacteristics ofselectedgroups of organiccompounds. c)Describehowcompoundsof thesamegroupcanbe identified. d)Describetheimportanceanduses of organiccompounds CONTENTS CHM203OrganicChemistryIIconsistsof three moduleswhich havebeen painstakingly puttogether totakeyou througha unique,structuredlearning experience. WORKINGTHROUGHTHISCOURSE Inordertobeabletosuccessfullycompletethiscourse,youarerequiredto carefullystudyeachunitalongwithrecommendedtextbooksandother materialsthatmaybeprovidedbytheNationalOpenUniversity.Youmay alsoneedtoexploitother e-readingsuchasinternetforfurtheruseful informationonthecourse. EachunitcontainsSAQsand ITQs. Atcertainpointsinthecourseyou would berequiredtosubmitassignmentsforgradingandrecording purposes.You arealsotoparticipateinthefinalexaminationattheendofthecourse.Itis recommendedthatyoudevoteanabundanttimeforreadingand comprehension.Itishighlynecessarythatyouavailyourselvesthe opportunityofattendingthetutorialsessionswhereyouwillbeableto compareyourunderstandingofthecoursecontentswithyourcolleagues. COURSEMATERIALS Thecoursematerialsare made upofthefollowingsections: vi COURSEGUIDE Thisdescribeshow best tostudythis majoraspectoforganicchemistryand italsospellsoutwhatconstitutethecourseitself. STUDY UNITS Eachstudyunit givesanoverview ofthecontent tobecoveredinthiscourse. Thisissub-dividedintosub-headingsas introduction, objectives; thatiswhat tofocuson,thecontent,conclusion,summary,referencesandalistofother materialstobeconsultedinorder toaugmentorfacilitatethestudent’s understandingandfinallythetutor-markedassignments. Module1Electronic Concept and Stereochemistry Unit 1 Factors Affecting Structure andPhysicalPropertiesof OrganicCompounds Unit 2 Factors Affecting Availability of Electrons Unit 3 Stereochemistry Module2Functional Groups and Reactivity in Organic Chemistry Unit 1 Functional Group Chemistry of Main Class Organic Compounds Unit 2 Alkanes, Free Radical Substitution Reactions in Alkanes and The Reactivity-Selectivity Principle Unit 3 Various Organic Reactions Unit 4 Nucleophilic Substitution and Elimination Reactions Module3Aromatic Compounds Unit 1 Benzene and other Aromatic Compounds Unit 2 Reactions in Aromatic Compounds ASSIGNMENTFILES Thefilescontaintutorialquestionsthatcoverthewholecourse.These will enableyoutoassessyourunderstandingof thecoursebythefacilitator.The marksscoredfor the assignmentswill bestoredandwillconstitute30% of thefinalscoreattheendofthesemesterexamination. CHM 203 COURSE GUIDE vii PRESENTATION SCHEDULE Themethod ofoperation,thatis,self-tutoredandfacetofacefacilitation with respecttothecoursewill beavailableattheinformation deskat different studycentresnearesttoyou. Youarewelcome tothestudyofCHM203-OrganicChemistryII,one of thetripodsonwhichthestudyofchemistrystands. CONTENTS PAGE Module1Electronic Concept and Stereochemistry€..1 Unit 1 Factors Affecting Structure andPhysical PropertiesofOrganicCompounds……………. 1 Unit 2 Factors Affecting Availability ofElectrons……21 Unit 3 Stereochemistry……………………………….. 59 Module2Functional Groups and Reactivity in Organic Chemistry€€€€€€€€€€€ 85 Unit 1 Functional Group Chemistry of Main Class OrganicCompounds…………………………... 85 Unit 2 Alkanes, Free Radical Substitution Reactions in Alkanesand The Reactivity-Selectivity Principle……………………………………….. 103 Unit 3 Various Organic Reactions…………………… 125 Unit 4 Nucleophilic Substitution and Elimination Reactions……………………………………… 143 Module3Aromatic Compounds€€€€€€€€€€.. 175 Unit 1 Benzene and other Aromatic Compounds……...175 Unit 2 Reactions in Aromatic Compounds……………. 197 MAIN COURSE CHM 203 MODULE 1 1 䵏䑕䱅‱ 䕌䕃呒低䥃 䍏乃䕐吠 䅎D 協䕒䕏䍈䕍䥓呒Y 䥎呒佄啃呉低 The study of organic chemistry involves the reactions and interactions of molecules. Since molecules are composed ofatoms, it is necessary to study the structure of atoms and how they contribute to the properties of molecules. In this module we will study the factors that affect the structures and physical properties of organic compounds. This knowledge will make learning about organic molecules a little easier. Because organic chemistry is a study of compounds that contain carbon and to have a better understanding of the properties of organic molecules, one has to study their three-dimensional (3D) structure. Why is this important? Our perception of smell and taste depends, in many instances, on the 3D structure of molecules. Enzymes are very selective in the 3D structure of the molecules they interact with. The effectiveness of drugs is highly dependent on their 3D structure. Organic chemists need to be able to determine the 3D structures (stereochemistry) of new and existing molecules to relate 3D structure to reactivity. Hence, the following units will be discussed in this module: Unit 1 Effect of molecular architecture on physical properties of organiccompounds Unit 2 Factors affecting availability of electrons in organic compounds Unit 3 Stereochemistry UN䥔‱ 䕆䙅䍔 但 䵏䱅䍕䱁删 䅒䍈䥔䕃呕剅 低⁐䡙卉䍁L 偒佐䕒呉䕓 但 佒䝁义䌠䍏䵐何乄S 䍏N呅NTS 1.0Introduction 2.0LearningObjectives 3.0MainContent 3.1Molecular ArchitectureandFactors AffectingPhysical Properties 3.1.1IntermolecularForces 3.1.2MeltingPoint 3.1.3BoilingPoint 3.1.4Solubility 4.0Conclusion 5.0Summary CHM 203 ORGANIC CHEMISTRYII 2 6.0Tutor MarkAssignment 7.0References/FurtherReadings ㄮ0INT剏䑕C TI低 Inour previous knowledge in organic chemistry, wehave learntsomeof theimportantaspectsof bondingand thestructuresof organic molecules indetail.Buthaveyou thoughtabouthowweestablishtheidentityand structureof a molecule? Oneanswertothisquestioncouldbecomparing itsphysicalandchemicalproperties with thoseofthe knowncompounds. Earlier methodsof identification involvedthedetermination of physical propertiessuchasmelting point,boiling point,solubilityandrefractive index.Thechemicalmethodsusedfor identificationinvolved,however, either thedegradation of the moleculetosimplecompoundsofknown structure oritssynthesisfromthesimplecompoundsof knownstructure. The structure and properties of organic compounds are considerably influenced by the conditions under which they are formed most especially bond type and temperature.Inthisunit,wewilldiscusstherelationship betweenmolecularstructureand physicalproperties.Thestudyof physicalpropertiesisalso importantinthepurification oforganic compounds. ㈮0OBJECTIVES By the end ofthisunityoushouldbeableto: · identifyorganicmoleculesusingtheirphysicalpropertiessuch as meltingandboiling points,solubilityandrefractiveindex · determinetherelationshipbetweenmolecularstructureand physical properties of organiccompounds. ㌮0M䅉N䍏N呅乔 ㌮1䵯汥捵污r 䅲捨it散瑵牥 慮搠 䙡捴潲猠 䅦晥捴楮g Phys楣慬 Pro灥牴楥s潦⁃潭灯畮d Thebondingandstructuralfeaturesof acompoundaremanifestedinits physicalproperties.Thus,physicalpropertiesofacompoundsuchas meltingpoint,boilingpoint,solubility,etc.,often givevaluableclues aboutitsstructure.Conversely, ifthestructureof acompoundisknown, itsphysicalpropertiescanbe predicted. Thephysicalpropertiesof acompounddependuponthenumberand natureofatomsconstitutingitsstructuralunitsandalsoon thenatureof forcesholdingtheseunitstogether.Youknow thatincaseofionic CHM 203 MODULE 1 3 compounds,thepositiveandnegativeionsareheldtogetherbystrong electrostaticforces.Contrarytothis,incovalentcompounds,the moleculesareheldtogetherbyintermolecularforces.Letusnowstudy brieflywhattheseintermolecularforcesare.Then,youwill learnhow theseintermolecularforcesaffectthephysicalpropertiesofthe compounds. ㌮ㄮ1Intermo汥捵污rF潲捥s Forces between molecules areresponsible for the magnitude of the melting and boiling temperatures and for solubility characteristics of molecules. The greater the attraction between molecules of a speci €c compound, the higher the melting and boiling points are likely to be. Solubility characteristics use the classic saying, like dissolves like. Polar molecules are most soluble in polar solvents and nonpolar molecules are most soluble in nonpolar solvents. Thethree importantintermolecularforcesare:(i) dipole-dipole interactions,(ii)Londonforcesand(iii)hydrogen bonding. Letusnowconsider theseintermolecularforcesonebyone. (i)䑩p潬e-䑩p潬eI湴敲act楯n猺aredefinedas theinteractions betweenthedifferentmoleculesof acompoundhaving permanent dipoles. Dipoles result from unequal sharing of electrons in bonds. If molecules are close to each other, the negative pole of one molecule is attracted to the positive pole of another molecule.Consider theexample ofchloromethane which hasapermanent dipole.Themolecule of chloromethane orientthemselvesinsuchawaythatthepositiveend ofone dipolepointstowards,andisthusattractedby,thenegativeend oftheother dipole.Theseinteractions,calleddipole-dipole interactionare depictedinFig1.1. a) b) Fig.1.1: Int er molec ul ar forces (€ ----- i n dicates interacti on). a ) A 灯l慲桹摲潧敮⁣桬潲楤e mol散ul攠楮瑥牡捴楮朠睩瑨⁡湯瑨敲⁨祤牯来n 捨汯物摥 浯汥捵汥 戩A牲慮来m敮tof捨l潲omet桡湥mol散ul敳 獨owi湧⁰潳iti癥慮d湥条瑩ve灯l敳映潮e浯汥捵汥⁡湤 瑨eDipole- dipoleint敲慣ti潮猠扥twe敮chl潲omet桡湥mol散ul敳. Thedipole-dipoleinteractionsareweakinteractionsandareof theorder of4to12kJmol -1 whereasthebondenergyforanordinarycovalent bondrangesfrom125to420kJ mol -1 . CHM 203 ORGANIC CHEMISTRYII 4 䥮-呥硴⁑略獴楯渠1 When the positive end of a molecule attracts the negative end of another molecule, the electrostatic forces that arise is named ________ A.Electrovalent bond B. Dipole-dipole forces C. Weak forces D. Gaseous forces 䥮-呥硴⁁湳睥爠1 Option B (ii)Lo湤潮Forc敳:Theintermolecularinteractionsexistbetween nonpolar moleculesalso.At any given instant, the electrons surrounding an atom or molecule are not uniformly distributed; that is, one side of the atom may have a greater electron density than the other side. This results in a momentary dipole within the atom. The dipole on oneatom may induce a dipole on another atom. The net result is an attraction between atoms. Consider twononpolar moleculesAand Binwhichthecentreof positive chargecoincideswith thatof the negativecharge. WhenthemoleculesAandBapproacheachother,thereisadistortion inthe distributionofthechargeresultingin asmallandmomentary dipole inonemolecule.Thissmalldipolecanthencreateanother dipole inthe secondmoleculewhichiscalledi湤u捥ddip潬e.Thus, if themomentary dipoleof moleculeA and Bisasshownbelow; Suchadistributionofchargeleadstomutualattractionbetweenthe molecules.Theseinduced dipoleinduceddipole interactionarealso knownasLo湤潮forc敳 ( il lu st ra te d w it h ‚ i n Fi gu re 1. 2) . CHM 203 MODULE 1 5 䙩朮ㄮ㈺ 䱯湤潮 景牣敳 慲攠 楮摵捥d -摩灯汥 ‚楮摵捥d-摩灯汥 楮瑥牡捴楯湳. Londonforcesaretheonlyforcesofattraction possible betweennonpolar molecules.Theseinteractionsareweaker thanthe dipole-dipole interactionsandareoftheorderof 4kJmol -1 .Theseforces vary withthe distancebetweenthemolecules.If €risthedistance betweenthetwo molecules,thentheLondonforcesareproportionalto 1/r 6 .This explains the interaction between helium atoms (Fig. 1.3), that are nonpolar, yet they musthave attraction for each other since they form liquids when cooled suf € c ie n t l y. A l l mo l e c u l e s e x h i b i t d i s p e r s i o n f o rc e s . 䙩朠ㄮ㌺⁌潮摯渠摩獰敲獩潮⁦潲捥猠楮⁴睯⁨敬極洠慴潭s 䥮-呥硴⁑略獴楯渠2 1.Classifythefollowingstatementastrueorf慬獥:Londonforces are the onlyforces operating between polar molecules. 2.At 25 o C, chlorine (Cl2) is a gas whereas bromine (Br2) is a liquid. Why? (iii)Hydr潧enBo湤in机This is a special type of dipole-dipole interaction. It does not refer to an actual bond, but astrong interaction between a covalently bonded hydrogen atom and a molecule containing an atom with nonbonding electrons, such as oxygen, nitrogen, and the halogens. The hydrogen atom undergoing hydrogen bonding must be covalently bonded to an oxygen, a nitrogen, or a ƒuorine ato m, res ul ting in a highly polar covalent bond. This puts a large partial positive charge („ + ) on the covalently bonded hydrogen atom and it seeks an electron pair on another atom. Hydrogen bonds are stronger than most dipole-dipole interactions but weaker than a covalent bond. An example is shown for water in Fig 1.4. 䙩朮‱⸴㨠桹摲潧敮⁢潮摩湧⁩渠睡瑥爠浯汥捵汥s CHM 203 ORGANIC CHEMISTRYII 6 Thestrength ofahydrogenbondrangesfrom10to 40kJmol -1 .Hydrogen bondinghasanimportantinfluenceonphysicalpropertiessuchas melting point,boilingpointandsolubilityofsubstances.Thiswillbe illustrated usingexamplesinthefollowingsubsections. The dipole-dipole,induced dipole-induced dipoleetc.interactionsare collectivelyknownas癡nder坡慬sfo牣es.Someauthorspreferto givethenamevanderWaalsforcesonlyforLondonforces.Having understoodtheintermolecularforces,letusnowstudyhowthevariation inmolecularstructureaffectstheseintermolecularforceswhichinturn isreflectedinthephysicalproperties of themolecules. 䥮-呥硴⁑略獴楯渠3 Which response includes only those compounds that can exhibit hydrogen bonding? (a) AsH3, H2Te (b) AsH3, CH3NH2 (c) CH4, AsH3, H2Te (d) CH3NH2, HF (e) HF, H2Te ㌮ㄮ2䵥lt楮gP潩湴 The melting point (mp) is the temperature at which a solid is converted into a liquid. Purecrystallinesolidshavesharpmelting points.Thus,meltingpointis usedasan importantphysical propertybothfortheidentificationof organiccompoundsandformakingthe generalassessmentofthepurity ofthesecompounds.Purecrystallinesolidshavesharpmelting points andtheymeltoveratemperaturerangeof1 o orless.Incontrasttothis, impurecrystallinesolids meltoverwiderrangesoftemperatures.Ina crystallinesolid,theconstituentionsormoleculesarearranged inan orderlyandrigidfashion.Whensuchassolid is heated,thethermal energyofthe moleculesincreases.Thisfinallyleadsto the disintegration of thecrystalstructureandatthemelting pointa disorderly andrandomarrangement of particles,characteristic ofa liquid,is obtained.Sincetheelectrostaticforcesholdingtheionsare verystrong, theycanbeovercome onlyat hightemperatures.Therefore, theionic compoundsgenerallyhavehighmelting points.Forexample, themelting pointofsodiumchlorideis1074Kandthatofsodiumethanoate is595 K.But,theintermolecularforcesareveryweakascompared to the interionicforcesand hence,thesecan beovercomeat lowertemperatures leading tolowermeltingpointsforcovalentcompounds.Themelting CHM 203 MODULE 1 7 pointofmethane,acovalentcompound,is only90Kandthemelting pointof methanol,anothercovalentcompound, is 179K. Letusnowstudytheeffectofmolecularweightonthemeltingpoint. If you examine the melting points of the alkanes in Table 1.1, you will see that the melting points increase (with a few exceptions) in a homologous series as the molecular weight increases. The increase in melting point is less regular than the increase in boiling point because灡捫楮ginfluences the melting point of a compound. Packing is a property that determines how well the individual molecules in a solid fit together in a crystal lattice. The tighter the fit, the more energy is required to break the lattice and melt the compound. In Figure 1.5, you can see that the melting points of alkanes with even numbers of carbon atoms fall on a smooth curve (the red line). The melting points of alkanes with odd numbers of carbon atoms also fall on a smooth curve (the green line). The two curves do not overlap, however, because alkanes with an odd number of carbon atoms pack less tightly than alkanes with an even number of carbon atoms. Alkanes with an odd number of carbon atoms pack less tightly becausethe methyl groups at the ends of their chains can avoid those of another chain only by increasing the distance between their chains. Consequently, alkane CHM 203 ORGANIC CHEMISTRYII 8 molecules with odd numbers of carbon atoms have lower intermolecular attractions and correspondinglylower melting points. Fi朮1.㔺Pl潴潦m敬瑩湧⁰潩湴猠潦⁳瑲慩杨t -捨慩渠慬歡湥献⁁汫慮敳⁷楴h 敶敮畭扥牳映捡牢潮⁡瑯浳⁦慬氠潮⁡敬瑩湧 -灯楮琠捵牶攠瑨慴⁩s 桩杨敲⁴桡渠瑨攠浥汴楮朠灯楮琠捵牶攠景爠慬歡湥猠睩瑨摤畭扥牳f 捡牢潮⁡瑯浳. Youcanseein thefigure thatthemeltingpointincreaseswith the increaseinthemolecularweight.Thiscanalsobeexplaineddueto increase in theLondonforcesbetween thelargermoleculesofhigher molecular weight.Thus,eachadditionalmethylene(-CH 2)unit contributesto the increase inmeltingpoint. Inahomologousseries,thehigherthemolecular weight,thelargerwill bethemoleculesandthegreaterwillbethe …areaofcontact†between thetwomoleculesandhence thegreaterwill be theLondonforces. YoumusthavenoticedinFig.1.5,thealternatingpatternof melting pointsfor thealkaneshavingoddandevennumber ofcarbonatoms.Itis alsoevidentfromthefigurethatthecompoundshavingeven numberof carbonatomslie onahighercurveascomparedto thecompounds havingoddnumberofcarbonatoms.Thiscanbeexplainedonthebasis thatinsolidstate,theLondonforcesamongthe moleculeshavingodd number ofcarbonatomsareweaker than thosein themoleculeshaving evennumberofcarbonatoms.Thisisbecausethemoleculesofalkanes havingoddnumberofcarbonatomsdonotfitwellinthecrystallattice ascomparetothose of thealkanes havingevennumber ofcarbonatoms. Afterstudyingtheeffectofmolecularweightonmeltingpoint,letus nowsee how theisomericcompoundshavingthesamemolecular weight, showdifferentmelting points.Themeltingpointsofstraightchainand branchedchainisomersof butanearegivenbelow: CHM 203 MODULE 1 9 Thebranchingofthecarbonchaininterfereswiththeregular packingof themoleculesinthecrystal; branchedchainhydrocarbonstendtohave lowermelting pointsthantheirstraightchainisomers. But,incase,thebranched molecule hasasubstantialsymmetry,then its melting point isrelativelyhigh.Thisisclearlyevident whenwecompare the meltingpoints ofisomericpentaneswhichareasgiven below: Thebranchingfrompentaneto 2-methylbutanelowersthemeltingpoint butfurther branchingin2,2-dimethylpropane increasesthe meltingpoint. Thiscanbeexplainedbythefactthatthesymmetricalmoleculesfit together moreeasilyinthecrystallatticeandhencehavehigher melting pointsascomparedto thelesssymmetricalmolecules.Hence, higher meltingpointfor2,2-dimethylpropaneisjustified. Thisisalsoreflectedwhenweanalysethe meltingpoints ofcis-and trans-isomers.Thetrans-isomer beingmoresymmetrical,fitsbetterin thecrystallattice thanthelesssymmetricalcis-isomer.Hence,the trans-isomersgenerallyhave higher meltingpoints. Thenatureofthefunctional groupspresent inamoleculealsoaffectsits physical properties.Forexample,when thefunctional group issuch that itintroducespolarity,and henceleadstoapermanentdipole momentin themolecule;then,duetothedipole-dipoleforcesofattraction between thepolarmolecules,theyshow highermeltingpoints thanthenonpolar moleculesofcomparablemolecular weights.Forexample,themelting pointofpropanone,apolarmoleculehavingmolecularweightof58,is 178K.Youcancompareit withthemeltingpointsofisomersof nonpolar butane(mol.Wt.=58)youhavejuststudiedabove.Thisleads tothe conclusionthatthepolarpropanonehashighermeltingpointthan the nonpolar isomericbutanes. CHM 203 ORGANIC CHEMISTRYII 10 Theeffectof hydrogenbondingonmelting point issmall.But,the hydrogen bonding hassignificanteffect on theboiling point,about which youwillstudyinthenextsubsection. 䥮-呥硴⁑略獴楯渠4 Which compound has the highest melting point? A. decane B. 2,2,3,3- tetramethylbutane C. 2,2,3-trimethylpentane D. 4-methylnonane ㌮ㄮ3B潩汩湧P潩nt Theboilingpoint of asubstanceisthetemperatureatwhichitchanges (vapourizes)fromtheliquid tothegaseousstate.Attheboilingpoint the vapour pressureofaliquid isequaltotheexternalpressure.Thus,the boiling point dependson theexternalpressureand it increaseswith increasein theexternalpressure.Hence,whilereportingtheboilingpoint ofasubstance,externalpressuremustbespecified. Similartothecaseofmeltingpoints,theboiling pointsarealsousedas constantsforidentificationandcharacterizationof liquidsubstances. Theknowledgeof boiling points isalso importantin thepurification of liquids. Letusnowstudysomeof thefactorsaffectingtheboilingpoint. Theboilingpointof asubstancedependsonitsmolecularstructure. In order for a compound to vapourize, the forces that hold the individual molecules close to each other in the liquid must be overcome. This means that the boiling point of a compound depends on the strength of the attractive forces between the individual molecules. If the molecules are heldtogether by strong forces, it will take a lot of energy to pull the molecules away from each other and the compound will have a high boiling point. In contrast, if the molecules are held together by weak forces, only a small amount of energy will be needed to pull the molecules away from each other and the compound will have a low boiling point. For example, relatively weak forces hold alkane molecules together. Alkanes contain only carbon and hydrogen atoms. Because the electronegativities of carbon and hydrogen are similar, the bonds in alkanes are nonpolar. Consequently, there are no significant partial charges on any of the atoms in an alkane. The molecules of an alkane are held together by these induced-dipole induced-dipole interactions, which are known as van der Waals forces. Van der Waals forces are the weakest of all the intermolecular attractions. In order for an alkane to boil, the van der Waals forces must be overcome. The magnitude of the van der Waals forces that hold alkane molecules together depends on the area of contact between the molecules. The CHM 203 MODULE 1 11 greater the area of contact, the stronger are the van der Waals (London) forces and the greater is the amount of energy needed to overcome those forces. If you look at the homologous series of alkanes in Table 1.1, you will see that the boiling points of alkanes increase as their size increases. Generally,thisincreaseinboiling pointamountsto20-30 o forthe addition ofeachcarbonatom inthemolecule. This relationship holds because eachadditional methylene group increases the area of contact between the molecules. The four smallest alkanes have boiling points below room temperature (room temperature is about 25 °C), so they exist as gases at room temperature. Pentane (bp = 36.1 o C) is thesmallest alkane that is a liquid at room temperature. The boiling points of the compounds in any homologous series increase as their molecular weights increase because of the increase in van der Waals forces. So the boiling points of the compounds in a homologous series of ethers, alkyl halides, alcohols, and amines increase with increasing molecular weight. Amongisomericmolecules,sincetheunbranched isomerislinearand henceextended inshape,ithaslargersurfaceareaascomparedto the branched isomers.Therefore,theLondonforcesarestrongerinthe unbranchedisomerleadingtohigherboilingpointforthisisomer. Thus, if two alkanes have the same molecular weight, the more highly branched alkane will have a lower boiling pointThis is illustratedinthe structures belowfor theisomersofpentane. The boiling points of these compounds, however, are also affected by the polar character of the bond (where Z denotes N, O, F, Cl, or Br) because nitrogen, oxygen, and the halogens are more electronegative than the carbon to which they are attached. The magnitude of the charge differential between the two bonded atoms is indicated by the bond dipole moment. The dipole moment of a bond is equal to the magnitude of the charge on one of the bondedatoms times the distance between the bonded atoms. Whenwecomparemoleculeshavingthesameshapeandsize,themore polar moleculehasthehigher boilingpoint.Examplesare: CHM 203 ORGANIC CHEMISTRYII 12 Dipole moment 20.98 x 10 -30 cm -0 Boiling point249.4 K 231 K Molecular weight46 44 Alcoholshavemuch higher boiling points than alkanes or ethers of comparable molecular weight (Table 1.2) because, in addition to London forces and the dipoledipole interactions of the bond, alcohols can form hydrogen bonds. Ahydrogen bond is a special kind of dipoledipole interaction that occurs between a hydrogen that is bonded to an oxygen, a nitrogen, or a halogen and the lone-pair electrons of an oxygen, nitrogen, or halogen in another molecule (seesubsection3.1.1 of this unit).Thus,tovaporizesucha compound,hydrogen bondsbetween the moleculesmust bebroken.Thisrequiresenergy, whichismanifestedas theunusuallyhighboiling pointforsuchcompounds. Table 1.2: Comparative Boiling Points (°C) Alkanes Ethers Alcohols Amines CH3CH2CH3 CH3OCH3 CH3CH2OH CH3CH2NH2 -42.1 -23.7 73 16.6 CH3CH2CH2CH3CH3OCH2CH3CH3CH2CH2OHCH3CH2CH2NH2 -0.5 10.8 97.4 47.8 CH3CH2CH2CH2 CH3 CH3CH2OCH2C H3 CH3CH2CH2CH2 OH CH3CH2CH2CH2 NH236.1 34.5 117.3 77.8 The length of thecovalent bond between oxygen and hydrogen is 0.96 Å. The hydrogen bond between an oxygen of one molecule and a hydrogen of another molecule is almost twice as long (1.691.79 Å), which means that a hydrogen bond is not as strong as an O-H covalent bond. A hydrogen bond, however, is stronger than other dipoledipole interactions. The strongest hydrogen bonds are linear‡the two electronegative atoms and the hydrogen between them lie on a straight line. Although each individual hydrogen bond is weak‡requiringabout 21 kJ/mol (5 kcal/mol) to break‡there are many such bonds holding alcohol molecules together. The extra energy required to break these hydrogen bonds is the reason alcohols have much higher boiling points than either alkanes or ethers with similar molecular weights. CHM 203 MODULE 1 13 The boiling point of water illustrates the dramatic effect hydrogen bonding has on boiling points. Water has a molecular weight of 18 and a boiling point of 100 °C. The alkane nearest in size is methane, with a molecular weight of 16.Methane boils at-167.7 o C. Primary and secondary amines also form hydrogen bonds, so these amines have higher boiling points than alkanes with similar molecular weights. Nitrogen is not as electronegative as oxygen, however, which means that the hydrogenbonds between amine molecules are weaker than the hydrogen bonds between alcohol molecules. An amine, therefore, has a lower boiling point than an alcohol with a similar molecular weight (Table 1.2). Because primary amines have two bonds, hydrogen bondingis more significant in primary amines than in secondary amines. Tertiary amines cannot form hydrogen bonds between their own molecules because they do not have a hydrogen attached to the nitrogen. Consequently, if you compare amines with the same molecular weight and similar structures, you will find that primary amines have higher boiling points than secondary amines and secondary amines have higher boiling points than tertiary amines. Hydrogenbondingisalsoimportantinotherways.Asweshallseein thenextsubsection,hydrogenbondingplaysanimportantroleinthe solubilityof organiccompounds. 䥮-呥硴⁑略獴楯渠5 Which of the following alkanes will have the lowest boiling point? CHM 203 ORGANIC CHEMISTRYII 14 ㌮ㄮ4卯汵扩lity Whenanysubstancedissolvesinasolvent,itsconstituent ionsor moleculesgetseparatedfromeach otherand thespacebetween them is filledbysolvent molecules.Thisisknown asolvationandtheamountof substancedissolvedinacertainamountofsolventisreferredtoasits solubilityinthatsolvent.Solubilitythusdependsontheinteractions betweensolute-solute,solute-solventandsolvent-solvent molecules. The general rule that explains solubility on the basis of the polarity of mo lecules is that ˆlike dissolves like‰ . In other words, polar co mpou nds dissolve in polar solvents, and nonpolar compounds dissolve in nonpolar solvents. This is because a polar solvent such as water has partial charges that can interact with the partial charges on a polar compound. The negative poles of the solvent molecules surround the positive pole of the polar solute, and the positive poles of the solvent molecules surround the negative pole of the polar solute. Clustering of the solvent molecules around the solute molecules separates solute molecules from each other, which is what makes them dissolve.Clearlystrongsolute-solvent molecular interactionsascomparetothose ofsolute-soluteorsolvent solventmoleculeswill leadtodissolution of thesolute. Similar totheprocessesof melting or boiling,dissolution of asubstance alsorequiresthattheinterionicor intermolecularforcesofattraction between theionsormoleculesmust beovercome.Thestrongelectrostatic CHM 203 MODULE 1 15 forcesbetweenthe ionsofan ioniccompoundcanbe overcome bythe solvents whichhavehighdielectricconstant.Thus, water whichhasa highdielectricconstant(Š ) of 80, dissolvesioniccompoundsreadily whereassolventslikecarbontetrachloride(Š=1.2) orether(Š=4.4) areextremelypoorsolventsforsuchcompounds. Hence, ionic compounds have greatersolubilityinpolarsolvents. ThedielectricconstantŠ,of asolventmeasuresitsabilitytoseparatethe ions of thesolute. Theterm polar hasdoubleusagein organicchemistry.When werefer that ithasasignificantdipolemoment, µ.But,whenwetalkabouta polar solvent, weunderstand that ithasahigh dielectricconstant, Š.Thus,the dipole momentisthe propertyofindividualmolecules whereassolvent polarityor dielectricconstantisapropertyof many moleculesacting together. Indeterminingthesolubilityofcovalentcompounds,theruleofthumb islike-dissolves-like.Sincewaterisapolarcompound,itisagoodsolvent for polarcompounds,butisa poorsolventforhydrocarbon whichare nonpolarinnature.Thus,the hydrocarbonsreadily dissolvein other hydrocarbonsor innonpolarsolventssuchasbenzene,ether or tetrahydrofuran. This is because the van der Waals interactions between solvent and solute molecules are about the same as between solvent solvent and solutesolute molecules. Alkanes are nonpolar, which causes them to be soluble in nonpolar solvents and insoluble in polar solvents such as water. The densities of alkanes (Table 1.1) increase with increasing molecular weight, but even a 30-carbon alkane such as triacontane (density at 20 o C = 0.8097 g/mL) is less dense than water (density at 20 o C = 0.9982 g/mL). This means that a mixture of an alkane and water will separate into two distinct layers, with the less dense alkane floating on top. An alcohol has both a nonpolar alkyl group and a polar OH group. So, is an alcohol molecule nonpolar or polar? Is it soluble in a nonpolar solvent, or is it soluble in water? The answer depends on the size of the alkyl group. As the alkyl group increases in size, it becomes a more significant fraction of the alcohol molecule and the compound becomes less and less soluble in water. In other words, the molecule becomes more and more like an alkane. Four carbons tend to be the dividing line at room temperature. Alcohols with fewer than four carbons are soluble in water, but alcohols with more than four carbons are insoluble in water. In other words, an OH group can drag about three or four carbons into solution in water. CHM 203 ORGANIC CHEMISTRYII 16 The four-carbon dividing line is only an approximate guide because the solubility of an alcohol also depends on the structure of the alkyl group. Alcohols with branched alkyl groups are more soluble in water than alcohols with non-branched alkyl groups with the same number of carbons, because branching minimizes the contact surface of the nonpolar portion of the molecule. So tert-butyl alcohol is more soluble thann-butyl alcohol in water. Thesolubilityoforganiccompoundsinwateralsodependsontheextent ofhydrogen bondingpossiblebetween thesoluteand thesolvent(water) molecules.Forexample,thegreatersolubilityofetherinwateras comparedtothatofpentane(inwatercanbeaccountedonthebasisof hydrogenbonding presentintheformercase. Sincetheolefinic,acetylenicor benzenoidcharacter doesnotaffect the polaritymuch,thesolubilityofunsaturatedandaromatichydrocarbons in waterissimilar tothat ofalkanes.Incompoundslikeethers,esters, aldehydes,ketones,alcohols,amides,acidsandamines,solubilityin water dependsonthelengthofthealkylchainandthemembers containing lessthanfivecarbonatomsin themoleculesaresolublein water. Increaseintheintermolecularforcesinasolute,asaresult of increasein the molecularweight,isalsoreflected in the lowsolubilityofcompounds having highmolecularweight.Forexample,glucose issoluble inwater but itspolymer,starch isinsoluble in water. This is because polymers havehighmolecularweight.Thus,inahomologousseries,thesolubility ofthemembersdecreaseswiththeincreaseinmolecular weight. However,branching of thecarbonchainleadsto adecreaseinthe intermolecularforces.Hence,thebranchedchainisomerismoresoluble ascomparedtothestraightchainisomer. Apartfromotherfactorsdiscussedabove,solubilityofacompoundina givensolventgenerallyincreases withtemperature. Sometimeshighsolubilityof acompoundisobservedduetoachemical reaction whichactsasa drivingforce.Onesuchcategoryofreactionsis CHM 203 MODULE 1 17 acid-basereactions.Forexample,thehighersolubilityofaniline in aqueousacidisduetotheformation ofanilinium ion. Although determination of the physical propertiessuchas those discussed abovehelpsintheidentificationof organiccompounds, physicalmethods involvingtheuseofspectroscopyallowdetermination of themolecular structuremuchmorerapidlyandnondestructively usingsmallquantities of material. 䥮-呥硴⁑略獴楯渠6 Which of the following compounds is expected to have the greatest solubility in water? 卅䱆⁁卓䕓卍䕎吠䕘䕒䍉卅 i.Which of the following compounds will form hydrogen bonds between its molecules? ii.Which of the compounds in SAQ 4.1 above will form hydrogen bonds with a solvent such as ethanol? iii.List the following compounds in order of decreasing boiling point: ivRank the following groups of compounds in order of decreasing solubility in water: CHM 203 ORGANIC CHEMISTRYII 18 v.In which of the following solvents would cyclohexane have the lowest solubility: pentanol, diethyl ether, ethanol, or hexane? vi.Rank the following compounds in order of increasing strength of intermolecular forces: CH3CH2CH2CH2CH3(pentane), CH3CH2CH2CH2OH (1-butanol), and CH3CH2CH2CHO (butanal). · Pentane has only nonpolar C-C and C-H bonds, so its molecules are held together by only癡渠摥爠坡慬s forces. · 1-Butanol is a polar bent molecule, so it can have摩灯汥-摩灯汥 interactions in addition to癡渠摥爠坡慬s forces. Because it has an O-H bond, 1-butanol molecules are held together by intermolecular桹摲潧敮⁢潮摳 as well. · Butanal has a trigonal planar carbon with a polar C=O bond, so it exhibits摩灯汥-摩灯汥interactions in addition to癡渠摥爠坡慬s forces. There isnoH atom bonded to O, so two butanal molecules cannothydrogen bond to each other. viiWhich compound in each pair has the higher boilingpoint? viiiWhich compound is water soluble? 楸 Which of the following molecules can hydrogen bond to another molecule like itself? · Which of the following molecules can hydrogen bond with water? CHM 203 MODULE 1 19 㐮0䍏乃L啓I低 Wecansafelyconcludethatagoodknowledgeof thephysicaland chemicalpropertiesof organiccompoundsisparamount inthe identification ofsuchcompounds. 㔮0单䵍 A剙 Duringthecourse ofthis unit, wehave learntabouttherelationship betweenmoleculararchitectureandphysicalpropertiesof organic compounds.Wehavealso learntabout theinterrelationship between intermolecularforces,meltingandboilingpointsandthesolubilityof organiccompounds. The greater the attractive forces between molecules‡London (van der Waals)forces, dipoledipole interactions, hydrogen bonds‡the higher is the boiling point of the compound. A hydrogenbond is an interaction betweena hydrogen bonded to an O, N, or F and the lone pair of an O, N, or F in another molecule. The boiling pointincreases with increasing molecular weight of the homolog. Branching lowers the boiling point. Polar compounds dissolve in polar solvents, and nonpolar compounds dissolve in nonpolar solvents. The interaction between a solvent and a molecule or an ion dissolved in that solvent is called solvation. The oxygen of an alcohol or an ether can drag about three or four carbons into solution in water. 㘮0呕呏删䵁剋⁁卓䥇乍䕎T 㜮0REFERE乃E匯F啒THERRE䅄I乇S Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. Dewick, P. M. (2006).Essentials of organic chemistry: for students of pharmacy, medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R. N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009). Chemistry: the central science. Pearson Education. CHM 203 ORGANIC CHEMISTRYII 20 Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New Age International. Okuyama, T., &Maskill, H. (2013). Organic Chemistry: a mechanistic approach. Oxford University Press. Ghatak, K. L. (2014).A Textbook of Organic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons. CHM 203 MODULE 1 21 UN䥔′ 䙁䍔佒匠 䅆䙅䍔䥎䜠 䅖䅉䱁䉉䱉呙 但⁅䱅䍔剏乓 䍏N呅NTS 1.0Introduction 2.0LearningObjectives 3.0MainContent 3.1Factors Affecting Availability of Electrons in Organic Compounds 3.1.1Inductive Effect 3.1.2Resonance (Mesomeric) Effect 3.1.3Hyperconjugation 3.2Application of Inductive Effect, Hyperconjugation and Mesomeric Effect: Acidity and Basicity 3.2.1Strengths of Acids and Bases 3.3Steric Effect 3.4Tautomerism 4.0Conclusion 5.0Summary 6.0Tutor Mark Assigment 7.0References/FurtherReadings ㄮ0INT剏䑕C TI低 Inthisunit,youwillstudyaboutthe factors that affect the availability of electrons in organic compounds. These factors are known toaffectthe reactivityof organic molecules.Thereactivityof onesubstancetowards anotherismeasuredbytherateatwhichthetwosubstancesreactand theamount of theproductsformed.Theseeffectswhichareassociated withthechangeinmolecularstructurearecalledstruc瑵r慬eff散ts which includes:inductiveeffect,resonanceeffectandstericeffect. Notall moleculesareequallyreactive.But,whatmakesome organic moleculesmorereactivethanothers?Tofindananswertothis question, weshouldhavesome ideaofthe natureofreactionsthatthe organic moleculesundergo.Alargenumberofreactionsthattheorganic moleculesundergocanbereadilyunderstoodassimpleanalogiesof acid-basereactions.Therefore,itisimportantforusto knowthebasic featuresofacid-basereactions.Wewillfamiliarizeourselveswiththe conceptofacid-baseequilibrium in this unit.Here,youwillalsostudy thattheposition of theacid-baseequilibriumisa measureof molecular reactivity;further itisinfluencedbymanyfactors. Although,the functionalgroupspresentinamoleculeareof key importancein determiningthemolecularreactivity,ithasbeenobserved that various compoundscontainingthesamefunctional groupsdifferin their CHM 203 ORGANIC CHEMISTRYII 22 reactivity.Thus,inadditiontothepresenceof thefunctional groups, structuraleffectsvis-a-visthenatureandarrangementofatomsattached tothefunctional groupsalsocontrolthemolecularreactivity.Wewillalso discusssolventeffectsandhydrogenbondingwhicharealsoimportant factorsaffectingtherateandtheextentofsuchreactions.Finally,youwill studyaninterestingequilibriuminvolving aprotonshiftfromoneatom of amoleculetoanother,calledta畴om敲ism. ㈮0OBJECTIVES By the end ofthisunit,youshouldbeableto: · understand some of the factors affectingelectron availability in organic molecules, along with their consequences. · apply these factors on organic substrates to locate electron deficient and electron rich sites. · defineacidsandbases and be able toclassify givencompoundsas acids orbasesaccordingtoBronsted LowryandLewis definitions. · definepKaofanacid and predicttherelativeaciditiesand basicities ofcompounds. · explain theeffectofstructuralchangesontheacidicandbasic behaviourof organicmolecules. · definetautomerismandgiveexamplesofvariouskindsof tautomerism. ㌮0M䅉N䍏N呅乔 ㌮1䙡捴潲猠 䅦晥捴楮朠 䅶慩污扩汩瑹 潦 䕬散瑲潮猠 楮 佲条湩c 䍯浰潵湤s Electronic factors that influence organic reactions include the inductive effect, electromericeffect, resonance (mesomeric) effects, and hyperconjugation. These electronic factors involve organic molecules, most of which are made from a combination of the following six elements: carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur (known collectively as CHNOPS). Yet, the limited number of building blocks does not prevent organic compounds from taking on diverse properties in their physical characteristics and chemical reactivity. The subtle differentiation of various compounds in organic chemistry is essential for the biological functions of the molecules and creates a wide variety of reactions.Letusnowdiscussthesefactorsoneafter the other. CHM 203 MODULE 1 23 ㌮ㄮ1䥮摵捴楶攠䕦晥捴 We have seen that when carbon bonds to an electronegative elementlike O, N, CI, or F, a bond polarization devel ops, makin g the C „+ and the heteroato m or ha logen „ -. The phenomenon of withdrawing electrons through sig ma (‹) bon ds to t he more elec tro negative a to m or gro u p is called an inductive effect. The inductive effect is a permanent state of polarization . The electr on densit y in a ‹ bond between two unli ke ato ms is not uniform and denser toward the more electronegative of the two atoms. The inductive effect is what gives rise to bond polarizations, polarizations within molecules, and bond and molecular dipole moments. Partial charges due to induction The inductive effect is a distance-dependent phenomenon: If the electronegative atomXis connected to a chain of carbon atoms, then the positive charge is relayed to the other carbon atoms.C1, with its positive„charge, exerts a pull on the electrons ofC2, but the pull is weaker than it is betweenXonC1. The effect rapidly dies out and is usually not significant after the2 nd carbon atom, or at most the3 rd . There are two categories of inductive effects: the敬散瑲潮-睩瑨摲慷楮g (-I) effect and the敬散瑲潮-摯湡瑩湧 (+I) effect. In the figure above,Xis electron-withdrawing andYis electron-donating. These relative inductive effects are measured with reference to hydrogen: The-I effect is seen around a more electronegative atom or group, and electron density is higher there than elsewhere in the molecule. Electron- withdrawing groups include halogen, nitro(ŒNO 2),cyano(Œ C N) , carboxyl(ŒC OOH), ester(ŒCOOR), and aryloxy(ŒOAr). On t he ot her hand,the +I effect is observed among the less electronegative atoms of the molecule by electron-releasing (or electron-donating) groups. The alkyl groups are usually considered electron-releasing (or electron- donating) groups. Inductive effect generally influences both physical and chemical properties of organic compounds as seen in the strength of organic acids CHM 203 ORGANIC CHEMISTRYII 24 (to be discussed in later section of this unit), basic strength of amines and amides, ease of substitution reactions of haloalkanes (dipole moment and bond length), the ease of addition reactions of unsaturated hydrocarbons, and so on. 䥮-呥硴⁑略獴楯渠1 a.State whether the following statements or True or False? Explain yourchoice:Inductive effect is the ability of an atom or a group of atoms to cause polarization of electron density along thecovalent bond so that the atom of higher electronegativity becomes electron deficient. b.Explain the comparative stability of primary,secondary and tertiary carbocations using Inductive effect. ㌮ㄮ2剥獯湡湣攠⡍敳潭敲楣⤠䕦晥捴 Wh ilst i nductive ef f ects pull electron s thro ug h the ‹ -bond framework, electrons can also mo ve through the -bon d network. A -bond can stabilize a negative charge, a positive charge, a lone pair of electrons or an adjacent bond by resonance (i.e. delocalisa tion or …spreadi ng out† of the electrons). A resonance effect reflects the ability of an atom or group of atoms to withdra w or donate ele ctrons t hr ough -bonds. This is also sometimes referred to as a mesomeric effect. In a nor mal bo nd, the electrons are localized between the constituent atoms. However, if double and single bonds are present alternately in a molecule, it is called conjugation e.g. in 1,3-butadiene, the double bonds are conjugated. Similarly, if thedouble, single and a lone pair are present, alternatively, it is also called conjugation e.g. vinyl chloride. The presence of conjugation alters the properties of the compound and there is a difference in the actual and expected properties. The theory of resonance explains the anomalous properties of such conjugated compounds. This theory states that when a molecule can be represented by two or more classical structural (or electronic) formula, all of which can explain some but not all the properties, then the molecule has neither of these structures (called contributing or canonical or limiting structures) but is a hybrid of all these contributing structures. For example: CHM 203 MODULE 1 25 Note that the e lectro ns are n ot nece ssaril y pre sent where one would expect them, but are rather delocalised over the entire molecule which gives it extra stability expressed in terms of delocalisation energy or resonance energy. Also, normally, we find chlorine withdrawing electrons towards itself by-I effect but here we find that the same chlorine has got a positive charge and is involved in a double bond. Does this mean that chlorine has lost its-I effect? No, this is not true as-I effect is a permanent effect. In fact, in addition to theI effect, it now also has an electron donating mesomeric or resonance effect (called +M or +R effect). Since the two effects are operating in opposite directions, one of them will overwhelm the other. Remember there was no such possibility if halogen's lone pairwas not conjugated. For example, in the following case, the Cl is not conjugated to the double bond and hence Cl is only exerting itsI effect. The atoms/groups like Cl in which lone pair (or electrons of negative charge) is in conjugation with double or triple bond are electron donating and gain a formal positive charge in the resonating structure in the process and are known to exert +M/+R Effect. Consider the case ofNO2joined to a conjugated system, where the nitro withdraws the conjugated electrons and gives rise to polarization as shown below: The atoms/groups likeNO2which are in conjugation with double or triple bond and are electron withdrawing and gain a formal negative charge in the resonating structure in the process and are known to exert- M/-R Effect. The resonance effect also alters the electron density distribution in the molecule significantly and its direction may be different from the normal inductive effect. In case the two effects are operating in opposite directions, the relative strengths of the two effects will determine which will dominate. Mesomeric/resonance effect introduces total CHM 203 ORGANIC CHEMISTRYII 26 delocalization of charges while inductive effect introduces partial polarization, hence, in general M > I. But there are exceptions to it like when halogens are attached to a conjugated system like benzene,-I > +R (negative inductive effect is greater than resonance effect). In neutral compounds, therewill always be a +M andM group(s): One group donates (+M) the electrons and the other group(s) accepts the electrons (M). All resonance forms are not of the same energy. In phenol, for example, the resonance form with the intact aromatic benzene ring is expected to predominate. As a rule of thumb, themoreresonance structures an anion, cation or neutral -system can have, the more stable it is. 䭥礠灯楮琠慢潵琠牥獯湡湣攺 (i)Resonating/canonical structure are maginary hypothetical, while resonance hybrid is the true strucure. (ii)Resonance involve thedelocalizatio n of lone p air and -electrons. (iii)Resonance is an intramolecular process. (iv)Resonance must follow the Lewis octet rule, i.e. C-atom, N-atom are never pentavalent and O-atom never tetravalent. (v)In the resonating structure arrangement of atoms remainsame, they should differ only with respect to arrangement of electrons. CHM 203 MODULE 1 27 (vi)The energy difference in between resonance hybrid and most stable resonating structure is called resonance energy. (vii)Resonance work only at ortho and para position with equal intensity,it never work at meta position. (viii)Resonance proceeds in the s yste m via -electrons. 䥮摵捴楶攠癥牳畳敳潭敲楣⁥晦散瑳: · Mesomeric effects are generally stronger than inductive effects. A+M group is likely to stabilise ananion more effectively than a +I group. · Mesomeric effects can be effective over much longer distances than inductive effects, provided that conjugation is present (i.e. alternating single and double bonds).Whereas inductive effects are determined by distance, mesomeric effects are determined by the relative positions of +M andM groups in a molecule. 䥮-呥硴⁑略獴楯渠2 a.Mesomeric effect involves delocalisation of __________. b.State whether the following statements or True or False? Explain your choice: TheOH group cannot exhibit Inductive effect. ㌮ㄮ3䡹灥牣潮橵条瑩潮 A ‹-bond can stabilize a neighbouringcarbocation (or positively charged carbon) by donating electrons to the vacant p-orbital. The positive charge is delocalize d or …s prea d out†, and this st abilizi ng ef f ect is known as ˆ no - bond resona nce‰. H yp erconjugatio n helps ex plain the stabi lit y of alk y l radicals. It i nvolves the delocalizati on of ‹ -electrons belonging to the C- H bond of the alkyl group attaching to an atom with an unshared p-orbital. The more the hyperconjugative hydrogen, the more is the stability. Why are more highly substituted alkenes more stable? One explanation involves hyperconjugation: hyper meaning above/beyond and conjugation meaning getting together. Figure 1.6 shows the overlapping of the sp 3 -s orbitals of a C-H bon d with an e mp t y antibon ding orb ital of an adjacent alkenecarbon atom. This overlapping of orbitals and sharing of the C-H bonding electrons, called hyperconjugation, increases the stability of the molecule. CHM 203 ORGANIC CHEMISTRYII 28 䙩朮‱⸶㨠䡹灥牣潮橵条瑩潮⁳瑡扩汩穡瑩潮. Hyperconjugation involvestheconjugationofsigma-electronswith adjacentpielectrons,asshownbelow: Thisinteractionisalsoknownasconjugation.Thistypeofdelocalization leadsto asituation wherethereisno bond betweenthehydrogenandthe carbonatomofthemolecule.Therefore, itisalsoknownashydrogen no-bondresonan捥.Rememberthat the protondoesnotleaveits positionandsincethenucleiortheatomsdo notchangetheirpositions, therefore,thehyperconjugation becomessimilartoresonance. Hyperconjugationalsoresults inthe delocalization ofcharge,asyouwill nowstudyincaseofcarbocations. Hyperconjugationinvolving hydrogensis themostcommon. Thestabilityofcarbocationshasbeenearlierexplainedonthebasisof inductiveeffectofthealkyl groups.Letusconsideragainaprimary carbocation,suchas theoneshownbelowinFig. 1.7 below: CHM 203 MODULE 1 29 䙩gㄮ㜺⁴桥hy灥牣潮橵条t i潮i渠愠捡牢潣慴 i潮 Itisclearfromtheabovestructurethattheelectronsformingthe-C H bondcanoverlap,intotheemptyporbitalof thecarbonatomcarryingthe positivecharge.TheC H bondadjacentto the>C=CN-methylaniline >aniline Asthemethyl groupiselectrondonating,it increasesthe basicity incase ofN-methylanilineascompared toaniline.Thebasicityfurther increases inN,N-dimethylaniline due totheincrease inthenumber ofmethyl groups. ii)Thedecreasingorder ofbasicities isasshownbelow:CH3NH2> NH3>NH2OH Since themethyl grouphas+Ieffect, itincreasesthebasicityofCH3NH2 ascomparedtoNH2.But,thesubstitutionofan OHgroupin NH3 decreasesitsbasicitybecause it has Ieffect. 戩 䵥獯浥物挠敦晥捴猠楮⁰桥湯汳⁡湤⁡特氠⡯爠慲潭慴楣⤠慭楮敳: Mesomeric effects can also stabilise positive and negative charges. In this case,Thenegative charge needs to be on adjacent carbon atom for aM group to stabilise it while the positive charge needs to be on adjacent carbon atom for a +M group to stabilise it. On deprotonation of phenol the CHM 203 MODULE 1 45 phenoxide anion is formed. This stabilised by delocalisation of the negative charge at the 2-, 4-and 6-positions of the benzene ring. 䭥祮潴敳⁩渠浥獯浥物挠敦晥捴s · IfM groups are introduced at 2-, 4-and/or 6-positions, the anion can be further stabilisedby delocalization through the -system, as the negative charge can be spread onto theM group. We can use double-headed curly arrows to show this process. · IfM groups are introduced at the 3-and/or 5-positions, the anion cannot be stabilised by delocalization, as the negative charge cannot be spread onto theM group. There is no way of using curly arrows to delocalize the charge onto theM groups. · If I groups are introduced on the benzene ring, the effect will depend on their distance from the negative charge. The closer the I group is to the negative charge, the greater will be the stabilising effect. The order ofI stabilisation is therefore 2-position > 3- position >4-position. · The M effects are much stronger thanI effects. Example: The NO2group is strongly electron-withdrawing;I andM. CHM 203 ORGANIC CHEMISTRYII 46 The presence of groups such as OH, OMe, or halogen an electron- withdrawing inductive effect, but an electron-donating mesomericeffect when in the o-and p-positions, may, however, cause the p-substituted acids to be weaker than the m-and, on occasion, weaker even than the unsubstituted acid itself, e.g. p-hydroxybenzoic acid: It will be noticed that this compensating effect becomes more pronounced in going ClŽ Br OH, i.e. in increasing order of readiness with which the atom attached to the nucleus will part with its electron pairs. The behaviour of o-substituted acids is, as seen above, often anomalous. Their strength is sometimes found to be considerably greater than expected due to direct interaction between the adjacent groups. Thus intramolecular hydrogen bonding stabilises the anion⡂)from o-hydroxybenzoic (salicyclic) acid⡁)by delocalising its charge, an advantage not shared by its m-and p-isomers, nor by o-methoxy benzoic acid: Intramolecular hydrogen bonding can, of course, operate in the undissociated acid as well as in the anion, but it is likely to be considerably more effective in the latter than in the former-with consequent relative stabilisation-because the negative charge on oxygen in the anion will lead to stronger hydrogen bonding. The effect is even more pronounced where hydrogen bonding can occur with hydroxyl groups in both o-position,and 2, 6-dihydroxybenzoic acid is found to have pKa = 1.30. The below table shows different carboxylic acids and their respective pKa values. CHM 203 MODULE 1 47 Ta扬e1.㘺䑩晦敲敮琠捡牢潸祬楣⁡捩摳⁡湤⁴桥楲⁲敳灥捴楶攠灋愠癡汵敳 䅣楤 灋a 噡汵e 䅣楤 灋愠噡汵e HCO2H 3.77 HO2CCO2H 1.23 CH3CO2H4.76 HO2CCH2CO2H 2.83 CH3CH2CO 2H 4.88 HO2CCH2CH2CO2 H 4.19 C6H5CO2H4.17 HO2CC6H4CO2Ho-2.98;m-3.46;p- 3.51 Similartoacidity, thebasicityofcompoundsisalsoaffectedbythe resonance.Forexample,incaseaminobenzene(aniline), the lone pair of electrons on the nitrogen atom of the aminobenzenecan be stabilized by the delocalization of the electrons onto the 2-, 4-and 6-positions of the benzene ring. Aromatic amines are therefore less basic than aliphatic amines. · IfM groups are introduced at the 2-, 4-and/or 6-positions (but not at the 3-or 5-position), the anion can be further stabilised by delocalization, as the negative charge can be spread onto theM group. This reduces the basicity of the amine. · If I groups are introduced on the benzene ring, the order ofI stabilisationis 2-position > 3-position > 4-position. This reduces the basicity of the amine. · If +M group (e.g. OMe) are introduced at the 2-, 4-or 6-position of aminobenzene, then the basicity is increased. This is because the CHM 203 ORGANIC CHEMISTRYII 48 +M group donates electron density to the carbon atom bearing the amine group. This trend can be illustrated using the scheme below: Resonancestructuresdiscussedinthissectioninvolve -electronsand insomecasesnon-bondedelectrons.Theseresonancestructuresclearly showthatthenonbondingelectrons ofthenitrogenatomaredelocalized overthearomaticring.Thus,theelectrondensityatthenitrogenatom increaseswhichresultsinthe higherbasicityofp-substitutedaniline. Youcantestyourknowledgeofresonancebyanswering thefollowing 䥔Q. 䥮-呥硴⁑略獴楯渠癩i Drawresonancestructuresfor thefollowingspeciestorationalizethe facts givenwith them. a.H2C=:O + Histheconjugateacidofmethanal(formaldehyde) andhas asubstantialpositivecharge oncarbon. b.Inacetonitrileoxide.H3C C=N + O:-,theinnercarboncan actas aLewisacid. So far we have been discussing factors that may influence the relative availability of electrons in bonds, or at particular atoms, in a compound, and hence affect that compound†s reactivity. The operation of these factors CHM 203 MODULE 1 49 may, however, be modified or evenbe nullified by the influence of steric factors; thus effective delocalization via -orbitals can only take place if the p or -orbitals on the atoms involved in the delocalization, can become parallel or fairly nearly so. If this is prevented, significantoverlapping cannot take place and delocalization may be inhibited. Inthenextsection, youwillstudythestericeffectonmolecularreactivity. ㌮3St敲楣E晦散t Theeffectarisingfrom thespatial interactionsbetween the substituent groupsiscalledthest敲楣e晦散t.Resonance ability of an atom is lost if it loses planarity with the other part of the system due to steric crowding by bulky group in adjacent positions. In a way, you havealreadystudied theeffect ofsuch interactionson thestabilityofgeometrical isomers (whereyoustudied thatthetrans-isomer ismorestablethanthecis- isomers)andconformational isomers(whereyoustudied that the staggeredconformationismorestablethantheeclipsedconformation). Astheacid-basebehaviouror themolecularreactivityisrelatedtothe availabilityof theelectrons,stericfactorsmayalsoinfluencethe molecularreactivity.Forexample,theycaninhibitthedelocalizationof charge,asobservedincaseof N,N-dimethyl-o-toluidine.The delocalizationof thenon-bondedelectronpair onnitrogen,asshownin thestructureofN,N-dimethylanilineinFig1.8(a). Fig1.8:(a) Delocalizationofnon-bondedelectronsonnitrogeninto aromaticringin N,N-dimethylaniline(b)Such adelocalizationin not possiblein N,N-dimethyl-o-toluidinerequiresthatthep-orbitalof nitrogenand thoseofthearomaticringshouldbecoplanar.Such coplanarityisinhibited inthecaseN,N-dimethyl-o-toluidine dueto the presenceoftheorthomethylgroup,asshownin Fig.1.8(b).Therefore, inthismolecule theelectronpairisnotdelocalizedbut isavailablefor bondingwith theprotonwhichmakesthismoleculemorebasicthan CHM 203 ORGANIC CHEMISTRYII 50 N,N-dimethylaniline.Thistypeofstericeffectisknownassteric, inhibition ofresonance. Themostcommonstericeffect is,however,thesterichindrancewhere thepresenceofthebulkygroupsmakestheapproachofthereagentto thereactionsitedifficult.Suchsterichindrancecanaccountforthe lower basicityoftertiaryaminesascomparedtosecondaryamines. Rememberthat thesteric hindranceaffectsthe molecularreactivitynot byincreasing ordecreasing theelectronavailabilitybutdue tospatial congestion.Therefore,itis differentfromelectroniceffects. Let†s look a t so me othe r acid and ba se exa mpl es below: In the above compounds, A and B have everything identical except position of the two methyl group. It is expected that Ashould be stronger base than B due to closeness of two electron donating methyl group to NH2. The fact is opposite to this. In compound BNO2is surrounded by two bulky methyl group and they sterically repel theNO2group. In order to minimize thesteric repulsion by the two adjacent methyl group, the nitro group loses planarity with the benzene ring. Therefore,NO2due to lack of planarity-weigh ring is not able to resonate. This is known as steric inhibition of resonance. Thus in B,NO2is not decreasing basic strength by resonance. In ANO2lies in the plane of the ring, it is in resonance CHM 203 MODULE 1 51 with the ring, decreases basic strength ofNH2by resonance, hence weaker base. Using same analogy, we can explain the acidic strength of C and D. C is stronger acid in spite of closeness of two electron donating methyl group to COOH. 䅣瑩癩瑹″⸱: Makemodelsofprimary,secondaryandtertiaryaminesandcompare thesterichindrance observedinthesemolecules. Trimethylamineisthusleaststabilizedbysolvation,leading tothelower basicityof trimethylaminein waterascomparedto dimethylamineand methylamine.However,in thegasphaseor non-aqueousmedia,the electron-donating inductiveeffect ofamethylgroupmakes trimethylamine the mostbasicamong the methylamines. Letusnowstudywhatissolvationand therole ofsolvent on thereactivity of the molecules. Thepresenceof asolventinacid-basereactionsleadstothesolvationof theionizedspecieswhicharetheconjugateacidandtheconjugate base whenwearedealingwithBronstedacidsandbases.S潬癡瑩潮refersto theinteractionof thedissolvedspeciesandsolventmoleculeswherein severalsolventmoleculessurroundthedissolvedspeciesbyforminga s潬癥ntshe汬ors潬癥ntc慧earoundit,asshownbelow: CHM 203 ORGANIC CHEMISTRYII 52 Thegreater thesolvation,thegreater isthedelocalization of thecharge onthespecies.Thus,increasedsolvationincreasesthedissociationofan acidor a basebyincreasingthestability of theions.Theseinteractions areparticularlyimportantwhenwater isusedasasolventwherethe hydrogenbondingplaysanimportantroleinsolvating theanions.The highdielectricconstantofwateralsohelpsinthe dissociationof theacids. Thus,theionizationandtheacidityof asubstanceincreaseswith the increasein the dielectricconstant ofthesolvent.This is illustratedin Table1.7. Tab汥1.㜺E晦ec琠潦s潬ven琠潮p Ka潦e瑨an潩cac楤⁡琠㈹8 K Solvent pKa Benzene 82% Dioxane 18%Water 70% Dioxane 30%Water 45% Dioxane 55%Water 20% Dioxane 80%Water Water Almostunionized 10.14 4.32 6.31 5.29 4.76 Thus,asthepercentageofwaterinthesolventsystemincreases,pKa valueof theaciddecreases.Waterispeculiarsolventasitcanbehaveboth asanacidaswellasa base.Butitsusehasalimitationinthesensethat some organiccompoundsare notsolubleinit. Havingdiscussedthevariousaspectsofacidsandbases,letusnow focus ourattentiononaninternalacid-base processcalledtautomerism. 䥮-呥硴⁑略獴楯渠癩楩 What is a steric effect in organic chemistry? CHM 203 MODULE 1 53 ㌮4T慵tome物sm Thetermtautomerismdesignatesarapidandreversibleinterconversion ofisomerswhicharerelatedtoeach otherwith theactualmovement of electronsaswellasof oneormoreatoms.Suchisomersarecalled ta畴om敲s.Thus,tautomerism isachemicalreactionand istobe differentiatedfromresonancein which thenucleido notmove.Itis, therefore,represented by theequilibriumsign ( )between the tautomers.Tautomerswhichdifferfromeachother onlyinthelocation ofahydrogenatomand adoublebondarecalledprot潮ta畴om敲s. Table1.8showssomeexamples of protontautomers.Incontrast to resonancestructure,tautomersarerealcompoundsandarecapableof independentexistence. 呡扬攠ㄮ㠺⁓ 潭eexa浰汥猠潦⁰r潴潮瑡畴omers Aparticularexampleof tautomerisminvolvingtheketonesascarbonyl compounds iscalledketo-en潬tautomerismandisrepresentedbelow: CHM 203 ORGANIC CHEMISTRYII 54 Theketo-enoltautorismisenormousimportanceasyouwillstudylater inthiscourseandalsoin theOrganicReactionsMechanismcourse.In keto-enoltautomers,theketoformisusuallythemorestableformand, therefore, it predominatesatequilibrium. Themechanism ofenolisationinvolvessolventmediated protontransfer stepsratherthanadirectintramolecularjumpof theprotonfromcarbon tooxygen.Protontautomerisminsomecasesleadstotheformationofa ringinoneofthe tautomers.Such a tautomerism isknownasring-ch慩n ta畴om敲ismandisillustratedbelow for 2-acetylbenzoic acid: Anotherkindoftautomerism,knownas癡汥nceta畴om敲isminvolves ashiftininteratomicdistancewithin a molecule, without theseparation ofanyatomfromtherestofthemolecule,asan intermediatestage.This kindoftautomerismoccursasaresult ofmovement ofvalenceelectrons of themolecule.Anexampleofvalencetautomerismisshownbelow: CHM 203 MODULE 1 55 cyclooctatetraene Thevalencetautomerismmayappearsimilartoresonancebutremember thatthetwoaredifferent.Thedifferenceisthatthevalencetautomerism involvesmakingandbreakingof ‹and electronsorthenonbonding electronsshiftand the ‹ frameworkofthemoleculeisnotdisturbed. Someother differencesbetweentautomerismandresonanceareas follows: ii)Tautomerismmayinvolveachangeinthehybridizationof atoms which mayresultin achangeintheshapeofthemolecule. While inresonancethereis nosuchchangeinthehybridization and geometryofthemolecule. iii)The tautomershave aphysicalrealitywhiletheresonance structuresare imaginery. iv)Tautomerisminvolvesanequilibriumbetweentwoor more tautomers.Ontheotherhand,theresonanceimplies thatthe actualstructureof themoleculein theweightedaveragedof variousresonancecontributorsandnota mixture onthem. 卅䱆⁁卓䕓卍䕎吠䕘䕒䍉卅 i.Explaintheacidicnatureof2,2,2-trifluoroethanolascompared toethanol. ii.ExplainthedifferencebetweenpKa1(4.16)andpKa2(5.61) of butanedioicacid. iii.Drawresonancestructuresfor thefollowing: i)chlorobenzeneii)acetonitrile iii) pyrrole Pyrrole is less resonance stabilized than benzene,as can be seen from the above resonance structure of pyrrole that in four out of the five structures an electronegative nitrogen atom has a positive charge over it. iv.Ethylamine and aniline react with aq. HCl. Write the equation for these reactions. 瘮 Drawall the enol tautomers for each of the following ketones. CHM 203 ORGANIC CHEMISTRYII 56 癩 Which of the following alcoholic proton (OH) is expected to be the most acidic? viiDeduce the correct order of acidity among the following compound I-IV. viiiAre the following pairs of compounds tautomers or resonance forms? CHM 203 MODULE 1 57 㐮0䍏乃L啓I低 Electronic effects are the effects originating or present in the organic molecules due to which the reactivity at one part of a molecule is affected by electron attraction or repulsion originating in another part of a molecule. These electronic effects areof three types mainly, inductive effect, mesomeric (or resonance) effect, and hyperconjugative effect. They are all permanent effects that stability and reactivity of organic molecules and also determines the strength of an organic acid or base. 㔮0单䵍 A剙 Inthisunit,youstudiedthat: · Many reactions of organic compounds can be classified as acid- base reactions. Therefore, the study of acids and bases is important for understanding the organic reactions. · According to Bronsted-Lowry definition, anacid is a proton donor and a base is proton acceptor. · Lewis definition classifies acids as electron pair acceptors and bases as electric pair donors. · The acidities of Bronsted acids can be expressed in terms of their pKa values · A strong acid has a weak conjugate base and a weak acid has a strong conjugate base and vice versa. · Structural changes can bring about marked differences in the acidic and basic behaviour of a molecule which can be explained on the basis of inductive, resonance and steric effects andon the basis of hydrogen bonding. · The inductive effects operate through sigma bonds and decrease rapidly with increase in the distance between the substituent and the reaction site. As a consequence of the fact that inductive effect increases with the number of substituents present, a tertiary carbocation is more stable than a secondary carbocation which is more stable than a primary carbocation. · Resonance stabilization of an anion (or the conjugate base) favours dissociation of the acid. · The steric effectoperates due to the presence of the bulky groups near the reaction site which prevent the approach of the reagent to the reaction site. The steric requirements for Bronsted acids are usually negligible because of the small size of the proton but are important in case of Lewis acids. CHM 203 ORGANIC CHEMISTRYII 58 · In addition to the structural changes mentioned above, the nature of the solvent also plays an important role in the acid-base equilibrium. 㘮0呕呏删䵁剋⁁卓䥇乍䕎T 㜮0REFERE乃E匯F啒THERRE䅄I乇S Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. Dewick, P. M. (2006).Essentials of organic chemistry: for students of pharmacy, medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R. N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009). Chemistry: the central science. Pearson Education. Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New AgeInternational. Okuyama, T., & Maskill, H. (2013). Organic Chemistry: a mechanistic approach. Oxford University Press. Ghatak, K. L. (2014).A Textbook of Organic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons. CHM 203 MODULE 1 59 UN䥔″ 協䕒䕏䍈䕍䥓呒Y 䍏N呅NTS 1.0Introduction 2.0LearningObjectives 3.0MainContent 3.1Definition of Isomers 3.1.1Constitutional Isomers (Structural Isomers) 3.1.2 Stereoisomers (Spatial Isomers) 3.1.3 Optical Isomerism 3.1.3.1Chirality 3.1.3.2Enantiomers 3.1.3.3Optical Activity 3.1.3.4Naming of Enantiomers: The R and S System of Nomenclature 4.0Conclusion 5.0Summary 6.0Tutur Mark Assignment 7.0References/FurtherReadings ㄮ0INT剏䑕C TI低 Compounds that have the same molecular formula but are not identical in structure are called isomers. Isomers fall into two mainclasses: constitutional isomers and stereoisomers. Constitutional isomers differ in the way their atoms are connected (revised your introduction to organic chemistry course). For example, ethanol and dimethyl ether are constitutional isomers because they have the same molecular formula, C2H6O, but the atoms in each compound are connected differently. While the oxygen in ethanol is bonded to a carbon and to a hydrogen, the oxygen in dimethyl ether is bonded to two carbons. Unlike the atoms in constitutional isomers, the atoms in stereoisomers are connected in the same way. Stereoisomers (also called configurational isomers) differ in the way their atoms are arranged in space. Stereoisomers are different compounds that do not readily interconvert. Therefore,they can be separated. There are two kinds of stereoisomers: cistrans isomers and isomers that contain chirality centers. This unit is all about stereochemistry which deals with the arrangement of atoms in space. Here, you will learn about the differentkinds of stereoisomers that are possible for organic compounds. CHM 203 ORGANIC CHEMISTRYII 60 ㈮0佂䩅䍔䥖䕓 At the end of this unit, you should be able to: · Understand the concept of stereochemistry. · differentiate chiral and achiral molecules. · recognize and draw structural isomers (constitutional isomers), stereoisomers including enantiomers and diastereomers and racemic mixture. · identify the stereocenters in a molecule and assign the configuration as R or S. · know the relationship between enantiomers and their specific rotations. ㌮0䵁䥎⁃低呅乔 ㌮1䑥晩湩瑩潮映䥳潭敲s If two or more different compounds have the same molecular formulawe call them isomers. This is the general definition of isomer. Isomers are the compounds with the same composition of elements, therefore their relative molecular weights and general formulas are identical, but their structuresincluding in the 3D arrangement are different. There are two major classes of isomers, and under these major classes there are further classifications of isomers as in Fig 1.9. CHM 203 MODULE 1 61 䙩朠ㄮ㤺⁔祰敳映䥳潭敲s Now let us look individually at the different types of isomer and see some examples for each type. ㌮ㄮ1䍯湳瑩瑵瑩潮慬⁉獯浥牳
却牵捴畲慬⁉獯浥牳) Different compounds that have the same molecular formula are called isomers and when they have different connectivity (i.e. which atom is bonded to which) we call them constitutional isomer or structural isomers. Examples are as shown below: This compound has a molecular formula of C2H6O. Now we can draw two structures1&2for this molecular formula: CHM 203 ORGANIC CHEMISTRYII 62 Dimethyl ether1and Methanol2are constitutional isomers (or structural isomers) because both of them have the same molecular formula C2H6O but they have different structures due to the difference in the bond connectivity of the atoms to each other. In1the bonds are C-O-C and each Catom has 3 H atoms. In structure2the bonds are C-C-O and each carbon has 2 H and there is a one H bonded to O atom. Similarly, the molecular formula of butane is C4H10. Two structures can be drawn for butane in order to obtain a saturated hydrocarbon with molecular formula C4H10. The two structures have the same molecular formula but they have different bond connectivity so they are constitutional isomers. Note that structureㄠ ☠ 2are completely two different compounds, they have different physical (boiling and melting point etc.) and chemical properties. The second example is however chemically active and both isomer㌠☠4can react with carboxylic acid to form ester while 1 will not react. Now if you try to convert1to2or vice versa, you will neverbe able to do so without breaking the bonds and then re-building the molecules. The same thing is applicable in the case of structures㌠☠4. We will never be able to interconvert from one structural isomer to another by rotation about bonds. We only cando that via only breaking bonds. 䥮-呥硴⁑略獴楯渠1 a. Draw three constitutional isomers with molecular formula C3H8O. CHM 203 MODULE 1 63 b. How many constitutional isomers can you draw for C4H10O? ㌮ㄮ2 却敲敯楳潭敲猠⡓灡瑩慬⁉獯浥牳) In stereoisomers (spatial isomers),different compounds that have the same molecular formula and the same bonds connectivity but they have different arranging (orientation) in the space. There are two types of stereoisomers which are: Conformational isomers 1) Configurational isomers Both types are stereoisomers that have the same molecular formula and the same connectivity and both of them have different arrangement in the space. However, in Conformational isomers we can convert from one isomer to another isomer by just rotation about a C-C bond. Ethane is good and simple example on conformational isomers. Let us look at the most two important conformations of ethane staggered and eclipsed conformations. Now apply the definition of stereoisomers on the two isomers and see what you will find out? The two structures have the same molecular formula that means they are isomers, they have the same bond connectivity that mean they are stereoisomers, they have different bonds arrangement in the space and so they may be either conformational isomers or configurational isomers. For sure now we can say they are conformational isomers because we can convert from staggered conformation to the eclipsed one via rotation about the C-C bonds. Let us look at, respectively, the chair conformers and Newman projection representation of the same example. CHM 203 ORGANIC CHEMISTRYII 64 Hence, conformational isomers are stereoisomers that can be converted from one isomer to other isomers by rotation about C-C bond. The conformations of propane, butane and all other conformations have the same property as the ethane so they all are conformational isomers. 䥮-呥硴⁑略獴楯渠2 Using Newman projections, draw the most stable conformer for the following: a. 3-methylpentane, considering rotation about the bond b. 3-methylhexane, consideringrotation about the bond c. 3,3-dimethylhexane, considering rotation about the bond Now let us look at the second type of stereoisomers; Configurational isomers. In the Configurational isomers we cannot convert from one isomer to other isomers by rotationabout bonds. We can only convert from one CHM 203 MODULE 1 65 isomer to other isomers via bond breaking. There are two major types of Configurational isomers which are: 1) Geometrical isomers 2) Optical isomers Geometrical isomers (cis-trans isomers) are isomers that have the same molecular formula, the same bond connectivity but the atoms are in different non-equivalent positions to one another. Geometrical isomers occur as a result of restricted rotation about a carbon-carbon bond. Restricted rotation about C-C bond can arise in two different situations: (a) In a double bond (b) In a cyclic compound That means geometrical isomers can arise only if we have a double bonds and/or cyclic structures. As a result of the restricted rotation about a carboncarbon double bond, an alkene suchas 2-pentene can exist as cis and trans isomers. The cis isomer has the hydrogens on the same side of the double bond, whereas the trans isomer has the hydrogens on opposite sides of the double bond. Cyclic compounds can also have cis and trans isomers. The cis isomer has the hydrogens on the same side of the ring, whereas the trans isomer has the hydrogens on opposite sides of the ring. CHM 203 ORGANIC CHEMISTRYII 66 䥮-呥硴⁑略獴楯渠3 Draw the cis and trans isomers for the following compounds: a. 3-hexene c. 1-bromo-4-chlorocyclohexane b. 3-methyl-2-pentene d. 1-ethyl-3-methylcyclobutane Now we will study the second type of configurational isomers which is the潰瑩捡氠楳潭敲献 ㌮ㄮ3佰瑩捡氠䥳潭敲楳m Optical isomerism is one form of stereoisomerism. In optical isomerism there is presence of a tetrahedral center. By tetrahedral center we mean that a C atom is attached to four other atoms or group with single bonds (sp 3 hybridization). We have the name optical isomers because of their effect on plane polarized light (reacts with light) as we will see later. There are two major types of optical isomers 1) Enantiomers 2) Diastereomers CHM 203 MODULE 1 67 Before we start studying the opticalisomers in details we need to learn some definition of terms related to the stereochemistry. ㌮ㄮ㌮1 䍨楲慬楴y Chirality means "handedness". Chiral objects (molecules) are those objects (molecules) which are not superimposable on (cannot be made to coincide with) their mirror image. In other words, its mirror image is not the same as itself. A hand is chiral because if you look at your left hand in a mirror, you do not see your left hand; you see your right hand (Figure 1.10). In contrast, a chair is not chiral‡it looks the same in the mirror. Objects that are not chiral are said to be achiral. An achiral object has a superimposable mirror image. Some other achiral objects would be a table, a fork, and a glass. 䙩杵牥‱⸱〺⁕獩湧⁡楲牯爠瑯⁴敳琠景爠捨楲慬楴礮 䄠捨楲慬扪散琬⁲楧桴 桡湤Ⱐ 楳 湯琠 瑨攠 獡浥 慳 楴猠 浩牲潲 業慧e -瑨敹 慲攠 湯n - 獵灥物浰潳慢汥⸠䅮⁡捨楲慬扪散琬⁣桡楲Ⱐ楳⁴桥⁳慭攠慳⁩瑳楲牯r 業慧e-瑨敹⁡牥⁳異敲業灯獡扬攮 An object will exhibit handedness if it has no plane of symmetry. Plane ofsymmetry is a position where an object can be cut in half and each half is identical. In other words, a plane of symmetry bisects a molecule into two mirror images halves. CHM 203 ORGANIC CHEMISTRYII 68 Not only objects can be chiral, molecules can be chiral, too. The feature that most often is the cause of chirality in a molecule is an asymmetric carbon. An asymmetric carbon is a carbon atom that is bonded to four different groups. The asymmetric carbon in each of the following compounds is indicated by an asterisk. For example,the starred carbon in 4-octanol is an asymmetric carbon because it is bonded to four different groups (H, OH, CH2CH2CH3and CH2CH2CH2CH3). The starred carbon in 2,4- dimethylhexane is an asymmetric carbon because it is bonded to four different groups‡methyl, ethyl, isobutyl, and hydrogen. An asymmetric carbon is also known as a chirality center. The hybridization on the chiral carbon must be sp 3 . Examples: An sp 3 hybridized carbon with 1, 2, or 3 different atoms or groups attached can be superimposed on its mirror image and is, therefore achiral. None of the following three compounds (a, b, c) is chiral because they do not have 4 different atoms or groups on the sp 3 central C atom. Each of them are superimpose on its mirror image. CHM 203 MODULE 1 69 Now it is important to learn how to draw the mirror image of the structure. Allwhat you need to do is to imagine that you have amirror and draw what you willsee. Let us take 3-methylhexane for example: Another example using 2-chlorobutane can be shown as, 䥮-呥硴⁑略獴楯渠4 a. Name five capital letters that are chiral. b. Name five capital letters that are achiral. ㌮ㄮ㌮㈠䕮慮瑩潭敲s Molecules that are not superimposable on their mirror images are chiral. The existence of chirality is necessary and sufficient condition for the existence of enantiomers i.e. if a compound is chiral; it can exist as enantiomers and if it is achiral it cannot exist as enantiomers. CHM 203 ORGANIC CHEMISTRYII 70 Enantiomers (from the Greekenantion, wh ich means ˆop posite‰) are stereoisomers that are mirror image of each other and they are non- superimposable on their mirror images. In other words, two isomers (the original molecule and its mirror image) have a different spatial arrangement, and so cannot be superimposed on each other. All enantiomers are chiral (we will study them in more details soon). 2-bromobutane can also be presented in a similar manner. CHM 203 MODULE 1 71 The two stereoisomers of 2-bromobutane are enantiomers. A molecule that has a non-superimposable mirror image, like an object that has a non- superimposable mirror image, is chiral. Each of the enantiomers is chiral. A molecule that has a superimposable mirror image, like an object that has a superimposable mirror image, is achiral. To see that the achiral molecule is superimposable on its mirror image (i.e., they are identical molecules), mentally rotate the achiral molecule clockwise. Now let us look at how chemists drawEnantiomers. Chemists draw enantiomers using either perspective formulas or Fischer projections. 健牳灥捴楶攠景牭畬慳 show two of the bonds to the asymmetric carbon in the plane of the paper, one bond as a solid wedge protruding out of the paper, and thefourth bond as a hatched wedge extending behind the paper. You can draw the first enantiomer by putting the four groups bonded to CHM 203 ORGANIC CHEMISTRYII 72 the asymmetric carbon in any order. Draw the second enantiomer by drawing the mirror image of the first enantiomer. A shortcut‡called a䙩獣桥爠 灲潪散瑩潮 ‡for showing the three- dimensional arrangement of groups bonded to an asymmetric carbon was devised in the late 1800s by Emil Fischer. A Fischer projection represents an asymmetric carbon as the point of intersection of twoperpendicular lines; horizontal lines represent the bonds that project out of the plane of the paper toward the viewer, and vertical lines represent the bonds that extend back from the plane of the paper away from the viewer. The carbon chain always is drawn vertically with C-1 at the top of the chain. To draw enantiomers using a Fischer projection, draw the first enantiomer by arranging the four atoms or groups bonded to the asymmetric carbon in any order. Draw the second enantiomer by interchanging two of the atoms or groups. It does not matter which two you interchange. It is best to interchange the groups on the two horizontal bonds because the enantiomers then look like mirror images on paper. Physical properties including melting point, boiling point, colour, hardness, density, etc. All physical properties of pair of enantiomers are the same except for one propriety which is optical activity (we will discuss this in the next subsection). However, enantiomers have identical chemical properties, except toward chiral substances where they will behave differently. That mean a pair of enantiomers react in different way with external chiral molecule. 䥮-呥硴⁑略獴楯渠5 Using Fischer projections, draw enantiomers for each of the following compounds: CHM 203 MODULE 1 73 䥮-呥硴⁁湳睥爠5 ㌮ㄮ㌮㌠佰瑩捡氠䅣瑩癩瑹 Enantiomers share many of the same properties‡they have the same boiling points, the same melting points, and the same solubility. In fact, all the physical properties ofenantiomers are the same except those that stem from how groups bonded to the asymmetric carbon are arranged in space. One of the properties that enantiomers do not share is the way they interact with polarized light. What is polarized light? Normal lightconsists of electromagnetic waves that oscillate in all directions. Plane-polarized light (or simply polarized light), in contrast, oscillates only in a single plane passing through the path of propagation. Polarized light is produced by passing normal light through a polarizer such as a polarized lens or a Nicol prism or other polarizing medium so that all of the vibrations are in the same plane. CHM 203 ORGANIC CHEMISTRYII 74 However, when polarized light passes through a solution of a chiral compound, the light emerges with its plane of polarization changed. Thus, a chiral compound rotates the plane of polarization. A chiral compound will rotate the plane of polarization clockwise or counterclockwise. If one enantiomer rotates the plane of polarization clockwise, its mirror image will rotate the plane of polarization exactly the same amount counterclockwise. Therefore, enantiomers rotate a plane polarized light in different direction. A compound that rotates the plane of polarization is said to be optically active. In other words, chiral compounds are optically active and achiral compounds are optically inactive. If an optically active compound rotates the plane of polarization clockwise, it is called摥硴牯牯瑡瑯特 , indicated by (+). If an optically active compound rotates the plane of polarization counterclockwise, it is called汥癯牯瑡瑯特 , indicated by (-).Dextroand levoare Latin pref ixes f or ˆto the right‰ and ˆto the lef t,‰ respecti v el y. Sometimes lowercasedandlare used instead of (+) and (-). The degree to which an optically active compound rotates the plane of polarization can be measured with an instrument called a灯污物浥瑥r . CHM 203 MODULE 1 75 䙩朠ㄮㄱ㨠卣桥浡瑩挠牥灲敳敮瑡瑩潮映灯污物浥瑥r Now when a polarized light pass through a sample in the polarimeter there are two possibilities: 1)The polarized light will pass straight without any reflection, that means either the substance in the polarimeter is achiral. OR The substance in the polarimeter is an equal mixture of two enantiomers. In this cases we say the substance is optically inactive. Optically inactive substances are those substances which do not rotate the polarized light. An equal mixture of enantiomers is called剡捥浩挠浩硴畲e . This is a mixture containing equal quantities of enantiomers (50:50) (i.e. a chiral molecule and its mirror image). The racemic mixture is optically inactive because one molecule will rotate the polarized light for example by 50 o to the right, the other enantiomers will rotate the polarized light by 50 o to the left so they cancel each other and the resultant rotation is zero. 2)The second possibility is when the polarized light of a polarimeter pass through a sample and there is a rotation of the polarized light either to the left or to the right depending on the nature of the substance. If such a rotation takes place, we say the substance is optically active. For a substance to be optically active it must be chiral and it should be either single enantiomers or unequal mixture of enantiomers (one of them present in excess). The a moun t of rotation is called the observ ed o ptical rotati on, , this v alue is obtained directly from the polarimeter andit depends on: 1. Concentration 2. length of the cell 3. the wavelength 4. solvent 5. temperature To compare samples, a quantity called the獰散楦楣 牯瑡瑩潮 , [ ], is measured. The specific rotation is the number of degrees of rotation caused by a solution of 1.0g of the compound per mL of solution in a sample tube 1.0 dm long at a specified temperature and wavelength. The specific rotation can be calculated from the observed rotation using the following formula: CHM 203 ORGANIC CHEMISTRYII 76 Wh ere, [ ] = specif ic rotation T = temperature in °C ‘ = wavelen gth = observe d rotation l = length of sample container in decimeters c = concentration in g/mL Example: The observed rotation of 2.0 g of a compound in 50 mL of solution in a polarimeter tube 20-cm long is +13.4 o . What is the specific rotation of the compound? lin decimeter = 20/10 = 2 dm(10cm = 1 dm) c in g/mL = 2/ 50 = 0.04 g/mL Theref ore, specif ic rota tion, [ ] T = . . = + 167 . 5 For example, one enantiomer of 2-methyl-1-butanol has been found to have a specific rotation of +5.75 o . Because its mirror image rotates the plane of polarization the same amount but in the opposite direction, the specific rotation ofthe other enantiomer must be-5.75 o . Note: when the sodium D-line is used,is indicated as D. ㌮ㄮ㌮㐠乡浩湧映䕮慮瑩潭敲猺⁔桥 R慮d S卹獴敭f 乯浥湣污瑵牥 We need a way to name the individual stereoisomers of a compound such as 2-bromobutane so that we know which stereoisomer we are talking CHM 203 MODULE 1 77 about. In other words, we need a system of nomenclature that indicates the configuration (arrangement) of the atoms or groups about the asymmetric carbon. Chemists use the letters R and S to indicate the configuration about an asymmetric carbon. For any pair of enantiomers with one asymmetric carbon, one will have theRconfiguration and the other will have theSconfiguration. TheR,Ssystem was devised by Cahn, Ingold, and Prelog. Let us now look at theCahn-Ingold-Prelog (CIP) sequence rules in naming enantiomers. 1.Rank the groups (or atoms) bonded to the asymmetric carbon in order of priority. The atomic numbers of the atoms directly attached to the asymmetric carbon determine the relative priorities.The higher the atomic number, the higher the priority. 2.Orient the molecule so that the group (or atom) with the lowest priority (4) is directed away from you. Then draw an imaginary arrow from the group (or atom) with the highest priority (1) to the group (or atom) with the next highest priority (2). If the arrow points clockwise, the asymmetric carbon has the R configuration (R is forrectus, which is Latin f or ˆright‰). If the arrow points count erclock wise, the as ym metric c arbon has the S configuration (S is forsinister, whic h is Latin f or ˆlef t‰). As an example, we will determine which of the enantiomers of 2- bromobutane has the R configuration and which has the S configuration. CHM 203 ORGANIC CHEMISTRYII 78 From step 1 above, we have: From step 2, if the group (or atom)with the lowest priority is bonded by a hatched wedge, draw an arrow from the group (or atom) with the highest priority (1) to the group (or atom) with the second highest priority (2). If the arrow points clockwise, the compound has the R configuration, and if it points counterclockwise, the compound has the S configuration. If the group with the lowest priority (4) is not bonded by a hatched wedge, then switch two groups so group 4 is bonded by a hatched wedge. Then proceed as in step 2 (above): Draw an arrow from the group (or atom) with the highest priority (1) to the group (or atom) with the second highest priority (2). So if the arrow for the enantiomer with the switched groups points clockwise, the molecule has the R configuration. This means the original molecule before the switch has the S configuration. In contrast, if the arrow points counterclockwise, the enantiomer (with the switched groups) has the S configuration, which means the original molecule has the R configuration. CHM 203 MODULE 1 79 3.In drawing thearrow from group 1 to group 2, you can draw past the group with the lowest priority (4), but never draw past the group with the next lowest priority (3). Now let†s see how to d eter mine the co nf iguration of a compo und dra wn as a Fischer projection. 1.Rankthe groups (or atoms) that are bonded to the asymmetric carbon in order of priority. 2.Draw an arrow from the group (or atom) with the highest priority (1) to the group (or atom) with the next highest priority (2). If the arrow points clockwise, the enantiomer has the R configuration; if it points counterclockwise, the enantiomer has the S configuration, provided that the group with the lowest priority (4) is on a vertical bond. 3.If the group (or atom) with the lowest priority is on a horizontal bond, theanswer you get from the direction of the arrow will be the opposite of the correct answer. For example, if the arrow points clockwise, suggesting that the asymmetric carbon has the R configuration, it actually has the S configuration; if the arrow points counterclockwise, suggesting that the asymmetric carbon has the S configuration, it actually has the R configuration. In the following example, the group with the lowest priority is on a horizontal bond, so clockwise signifies the S configuration, not the Rconfiguration. CHM 203 ORGANIC CHEMISTRYII 80 4.In drawing the arrow from group 1 to group 2, you can draw past the group (or atom) with the lowest priority (4), but never draw past the group (or atom) with the next lowest priority (3). Knowing whether a chiral molecule has the Ror the S configuration does not tell us the direction the compound rotates the plane of polarization, because some compounds with the R configuration rotate the plane to the right (+) and some rotate the plane to the left (-). We can tell by looking at thestructure of a compound whether it has the R or the S configuration, but the only way we can tell whether a compound is dextrorotatory (+) or levorotatory (-) is to put the compound in a polarimeter. For example, (S)- lactic acid and (S)-sodium lactate have the same configuration, but (S)- lactic acid is dextrorotatory whereas (S)-sodium lactate is levorotatory. When we know the direction an optically active compound rotates the plane of polarization, we can incorporate (+) or (-) into its name. 乯瑥:When comparing two Fischer projections to see if they are the same or different, never rotate one 90° or turn one over, because this is a quick way to get a wrong answer. A Fischer projection can be rotated 180° in the plane of the paper, but this is the onlyway to move it without risking an incorrect answer. Now answer the following ITQ to check your understanding. CHM 203 MODULE 1 81 䥮-呥硴⁑略獴楯渠6 Indicate whether each of the following structures has the R or the S configuration: 卅䱆⁁卓䕓卍䕎T 䕘䕒䍉卅 i.20 mgmandelic acid was dissolved in 1 cm 3 of ethanol and the solution placed in a 10 cm long polarimeter cell. An optical rotation of4.35ºC was measured (that is, 4.35º to the left) at 20ºC with light of wavelength 589 nm. What is the specific rotation of theacid? ii.Neglecting stereoisomers, give the structures of all compounds with molecular formula C5H10. Which ones can exist as cis- trans isomers? iii.The relationship between the following two structures is: (A)enantiomers (B) diastereomers (C) structural isomers (D) identical ivThe specific rotation of pure (R)-2-butanol is-13.5°. What % of a mixture of the two enantiomeric forms is (S)-2-butanol if the specific rotation of this mixture is-5.4°? (A) 40%(B) 30%(C) 60%(D) 70%(E) None of the above vWhich of the following objects are chiral? a.A mug with DAD written on one side of the handle. b.A mug with MOM written on one side of the handle. c.A mug with DAD written opposite the handle. d.A mug with MOM written opposite the handle. e.A wheelbarrow. f.A remote control device. g.A nail. h.A screw. CHM 203 ORGANIC CHEMISTRYII 82 viDo the following structures represent identical molecules or a pair of enantiomers? 十儠㐮7 Name each of the following compounds using R,S and E,Z designations where necessary: 㐮0C低CLUS䥏N The excitement that chemists feel for the area of stereochemistry has hopefully rubbed off during your reading of this unit. From simple structural isomers to geometric isomers and then optical isomers where we mentioned enantiomers and diastereomers, stereochemistry continues to challenge organic chemists to create molecules of increasing complexity, which inevitably leads to molecules with intriguing properties and simple aesthetic beauty. Furthermore, stereochemical concepts shed important light on the study of reaction mechanisms. It is this topic that we still need to develop further. In our analyses of reaction mechanisms, we will rely heavily upon the concepts and terminology introduced in this unit. All introductory organic chemistry courses teach the fundamentals of stereoisomerism, and we have only brie ƒ y re view that information here. CHM 203 MODULE 1 83 㔮0S啍䵁 RY · Stereochemistry is the field of chemistry that deals with the structures of molecules in three dimensions. · Compounds that have the same molecularformula but are not identical are called isomers; they fall into two classes: constitutional isomers and stereoisomers. Constitutional isomers differ in the way their atoms are connected. Stereoisomers differ in the way their atoms are arranged in space.There are two kinds of stereoisomers: cistrans isomers and isomers that contain chirality centers. · A chiral molecule has a nonsuperimposable mirror image. An achiral molecule has a superimposable mirror image. The feature that is most often the cause of chirality is an asymmetric carbon. · Nonsuperimposable mirror-image molecules are called enantiomers. Diastereomers are stereoisomers that are not enantiomers. · The letters R and S indicate the configuration about an asymmetric carbon. If one molecule has theR and the other has the S configuration, they are enantiomers; if they both have the R or both have the S configuration, they are identical. · Chiral compounds are optically active-they rotate the plane of polarized light; achiral compounds are optically inactive. If one enantiomer rotates the plane of polarization clockwise (+), its mirror image will rotate the plane of polarization the same amount counterclockwise (-). · A racemic mixture (recemate) is optically inactive. 㘮0呕呏删䵁剋⁁卓䥇䵅乔 7.0剅FERENCES/FUR呈䕒 R䕁䑉NGS Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. Dewick, P. M. (2006).Essentials of organic chemistry: for students of pharmacy, medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R. N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009). Chemistry: the central science. Pearson Education. Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New Age International. CHM 203 ORGANIC CHEMISTRYII 84 Okuyama, T., & Maskill, H. (2013). Organic Chemistry: a mechanistic approach. Oxford University Press. Ghatak, K. L. (2014).A Textbook of Organic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons. CHM 203 MODULE 2 85 䵏䑕䱅′ 䙕乃呉低䅌 䝒何偓 䅎D 剅䅃呉噉呙⁉丠佒䝁义䌠䍈䕍䥓呒Y 䥎呒佄啃呉低 The main classes of compounds you will study in organic chemistry are alkanes, alkenes, alkynes, alkyl halides, ethers, alcohols, and amines. As you learn various propertiesand reactivity of compounds, you will need to be able to refer to them by name. So you will begin your study of this module by learning how to identify and name some of these main classes of organic compounds. First you will learn how to name alkanes and the way they react because they form the basis for the names of almost all organic compounds. Subsequently you will study the various organic reactions. It is important that you are conversant with the names and formula of at least the first 10 members of each functional group in organic chemistry. The following units will be discussed in this module: Unit 1 Functional group chemistry of main class organic compounds Unit 2 Alkanes, free radical substitution reactions in alkanes and the reactivity-selectivity principle Unit 3 Electrophilic and nucleophilic substitution reaction Unit 4 Various organic reactions e.g. addition free radicals, elimination reaction etc. UN䥔‱ 䙕乃呉低䅌 䝒何倠䍈䕍䥓呒夠但⁍䅉N 䍌䅓匠佒䝁义C 䍏䵐何乄S 䍏乔䕎呓 1.0Introduction 2.0Learning Objectives 3.0Main Content 3.1Functional Groups in Organic Chemistry 3.2An Overview of Functional Groups 3.2.1Hydrocarbons 3.2.2Compounds Containing C-Z € Bonds 3.2.3Compounds Containing C=O Group 3.2.4Alcohols 3.2.5Amines 3.2.6Thiol 3.3Functional Groups and Reactivity 4.0Conclusion 5.0Summary 6.0Tutor Mark Assignment CHM 203 ORGANIC CHEMISTRYII 86 7.0References/FurtherReadings ㄮ0乔剏䑕䍔䥏N Chemists have learned through years of experience that organic compounds can be classified into families according to their structural features and that the members of a given family often have similar chemical behaviour. Instead of 40 million compounds with random reactivity, there are a few dozen families of organic compounds whose chemistry is reasonably predictable. Well study the chemistry of the specific families throughout much of this unit, beginning in this module with a look at the simplest family, thehydrocarbons. The structural features that make it possible to classify compounds into families are calledfunctional groups.A functional group is a group of atoms within a molecule that has a characteristic chemical behaviour. Chemically, a givenfunctional group behaves in nearly the same way in every molecule it is a part of. For example, compare ethylene, a plant hormone that causes fruit to ripen, with menthene, a much more complicated molecule found in peppermint oil. Both substances containa carbon‚carbon double-bond functional group, and both therefore react with Br2in the same way to give a product in which a Br atom has added to each of the double-bond carbons. This example is typical:the chemistry of every organic molecule, regardlessof size and complexity, is determined by the functional groups it contains. ㈮0佂䩅䍔䥖䕓 When you have studied this unit, you should be able to: · know the major classes of organic compounds and identify important functional groups. · understand thefactors that determine the properties of organic compounds. · have a better understanding of functional groups and reactivity of organic compounds. · describe the importance and purpose of functional groups in organic reactions. ㌮0䵁䥎⁃低呅乔 ㌮1䙵湣瑩潮慬⁇牯異猠楮⁏牧慮楣⁃桥浩獴特 What are the characteristic features of an organic compound? Most organic molecules have C-C and C-H € bonds. These bonds are strong, CHM 203 MODULE 2 87 nonpolar, and are not readily broken. Organic molecules may have the following structural features as well: ƒ 䡥瑥牯慴潭s -atoms other than carbon or hydrogen.Common heteroatoms are nitrogen, oxygen, sulfur, phosphorus, and the halogens. ƒ €䉯湤献The most common „ bonds occur in C-C and C-O double bonds. These structural features distinguish one organic molecule from another. They determine a molecules geometry, physical properties, and reactivity, and comprise what is called a晵湣瑩潮慬 杲潵瀮 ƒ A functio nal g roup 楳 慮 慴潭 潲 愠 杲潵瀠 潦 慴潭猠 睩瑨 捨慲慣瑥物獴楣 捨敭楣慬 慮搠 灨祳楣 慬 灲潰敲瑩敳⸠ 䥴 楳 瑨e reactive par t 潦⁴桥潬散畬攮 Why do heteroatoms and „ bonds confer reactivity on a particular molecule? ƒ Heteroatoms have lone pairs and create electron-deficient sites on carbon. ƒ „Bonds are easily broken in chemical reactions. A„bond makes a molecule a base and a nucleophile. Dont think, though, that the C-C and C-H € bonds are unimportant. They form the捡牢潮⁢慣止潮e or獫敬整潮to which the functional groups arebonded. A functional group usuallybehaves the same whether it is bonded to a carbon skeleton having as few as two or as many as20 carbons. For this reason, we often abbreviate the carbon and hydrogen portion of the moleculeby a capital letter刬and draw theRbonded to a particular functional group. 䕴桡湥,for example, has only C-C and C-H € bonds, so it hasno functional group. Ethane has no polar bonds, no lone pairs, and no „ bonds, so it has湯⁲敡捴楶攠獩瑥献 Because of this, ethane and molecules like it are very unreactive. CHM 203 ORGANIC CHEMISTRYII 88 䕴桡湯氬on the other hand, has two carbons and five hydrogens in its carbon backbone, as well as an OH group, a functional group called a 桹摲潸祬 group. Ethanol has lone pairs and polar bonds that makes it reactive with a variety of reagents, including the acids and bases. The hydroxyl group makes the properties of ethanol very different from the properties of ethane. Moreover, any organic molecule containing a hydroxyl group has properties similar to ethanol. Ethane All C-C and C-H € bonds Polar C-O and O-H bonds No functional group Two lone pairs Most organic compounds can be grouped into a relatively small number of categories, based on the structure of their functional group. Ethane, for example, is an慬歡湥,whereas ethanol is a simple慬捯桯氮 ㌮2䅮⁏癥牶楥眠潦⁆畮捴楯湡氠䝲潵灳 We can subdivide the most common functional groups into several types. ƒ Hydrocarbons ƒ Compounds containing a C-Z € bondwhere Z = an electronegative element ƒ Compounds containing a C=O group ƒ Others. 䥮-呥硴⁑略獴楯渠1 What is a functional group? Give at least two examples of functional groups. ㌮㈮1䡹摲潣慲扯湳 䡹摲潣慲扯湳 are compounds made up of only the elements carbon and hydrogen.They may bealiphaticoraromatic. ㄮ 䅬楰桡瑩挠桹摲潣慲扯湳 can be divided into three subgroups. ƒ 䅬歡湥shave only C-C € bonds and no functional group. Ethane, CH3CH3,is a simple alkane. CHM 203 MODULE 2 89 ƒ 䅬步湥sare hydrocarbons that contain one or more double bonds between neighboring carbon atoms. Theyhave a C-C double bond as a functional group. Ethylene,CH2=CH2,is asimple alkene. ƒ 䅬歹湥scontain one or more triple bonds between neighboring carbon atoms. Theyhave a C-C triple bond as a functional group. Acetylene,HC…CH,is a simplealkyne. ㈮ 䅲潭慴楣 桹摲潣慲扯湳. The additional functional group that contains only carbon and hydrogen is an aromatic ring which is a six- carbon ring with alternating double bonds. The aromatic ring can also be shown as a ring with a circle in the middle representing the double bonds. Aromatic rings are found in many compounds including steroids and medications. This class of hydrocarbons was so named because many of the earliest known aromatic compounds had strong, characteristic odours. The simplest aromatic hydrocarbon is扥湺敮攮 Thesix-membered ring and three „ bonds of benzene comprise asinglefunctional group. When a benzene ring is bonded to another group, it is called a灨敮祬 杲潵瀮In phenylcyclohexane, for example, a phenyl group is bonded to the six-membered cyclohexane ring. Alkanes, which have no functional groups, are notoriously unreactive except under very drastic conditions. For example,灯汹整桹汥湥 is a synthetic plastic and high molecular weight alkane, consisting of chains of‚CH2‚groups bonded together, hundreds or even thousands of atoms long. Because it is an alkane with no reactive sites, it is a very stable CHM 203 ORGANIC CHEMISTRYII 90 compound that does not readily degrade and thus persists for years in landfills. 䥮-呥硴⁑略獴楯渠2 a.Identify the type of hydrocarbon in each structure. b.Give the systematic name of the following compound: ㌮㈮2䍯浰潵湤猠䍯湴慩湩湧⁃ -Z Bo nds The electronegative heteroatom Z createsa polar bond, making carbon electron deficient. The lone pairs on Z are available for reaction with protons and other electrophiles, especially when Z = N or O. Several simple compounds in this category are widely used. ㄮ 䅬歹氠桡汩摥猺 The haloalkanes, also known as alkyl halides, are a group of chemical compounds that comprised of an alkane with one or more hydrogens replaced by a halogen atom (Group 17 atom). There is a fairly large distinction between the structural and physical properties of haloalkanes and the structural and physical properties of alkanes. As an example, chloroethane (CH3CH2Cl, commonly called ethyl chloride) is an alkyl halide used as a local anesthetic. Chloroethane CHM 203 MODULE 2 91 quickly evaporates when sprayed on a wound, causing a cooling sensation that numbs the site of an injury. Haloalkanes are found in fire extinguishers, refrigerants, propellants, solvents, and medications. They are also a significant source of pollution and their use has been reduced or eliminated in some products. Chlorofluorocarbons (CFCs) were used as refrigerants in air-conditioners but were found to be a major cause of the depletion of the ozone layer. Research and development of alternatives began in the 1970s. Hydrochlorofluorocarbons (HCFCs) have been usedfor many years since they cause less damage to the ozone layer, but many countries agreed to eliminate HCFCs by the year 2020. ㈮ 䕴桥牳:Theetherfunctional group consists of an oxygen atom that forms single bonds with two carbon atoms. Molecules containing these functional groups may be simple or very complex. Diethyl ether, the first common general anesthetic, is a simple ether because it contains a single O atom, depicted in red, bonded to two C atoms. Hemibrevetoxin B, on the other hand, contains four ether groups, in addition to other functional groups. Diethyl ether Although ethers themselves are relatively unreactive, they can be converted to peroxides after prolonged exposure to oxygen. Peroxides are very reactive and are often explosive at elevated temperatures. Many commercially available ethers come with a small amount of a peroxide scavenger dissolved in them to help prevent thistype of safety hazard. 䥮-呥硴⁑略獴楯渠3 Determine the molecular formula of a 5-carbon hydrocarbon with one bond and one ring CHM 203 ORGANIC CHEMISTRYII 92 ㌮㈮3䍯浰潵湤猠䍯湴慩湩湧⁃㵏⁇牯異 Many different types of functional groups possess a C-O double bond (a 捡牢潮祬⁧牯異 ).The polar C-O bond makes the carbonyl carbon an electrophile, while the lone pairs on O allow it to react as a nucleophile and base. The carbonyl group also contains a „ bond that is more easily broken than a C-O € bond. 剥慣瑩癥⁦敡瑵牥猠潦⁡⁣慲扯湹l 杲潵p ㄮ 䅬摥桹摥: A very common structural component of organic structures is the捡牢潮祬, which is simply a carbon atom and an oxygen atom connected by a double bond. The reactivity of carbonyls is primarily dictated by the polarization of the C=O bond, but the surrounding atoms also play a role in its specific reaction pathways. The other group attached to the carbonyl may be an R-group or a hydrogen atom. Because the hydrogen atom is so small, the partial positive charge on the carbonyl carbon is very easy for other molecules to approach, making aldehydes a particularly reactive type ofcarbonyl. Aldehydes are versatile reactants for a wide variety of organic syntheses. Many aldehydes also have distinctive flavours and aromas. For example, the flavour of cinnamon is primarily due to the molecule cinnamaldehyde, and vanillin is the Aldehyde most responsible for the smell and taste of vanilla extract. A special aldehyde is the molecule in which the carbonyl is bonded to two hydrogen atoms. This molecule, called formaldehyde, has a wide variety of uses. By itself, it can be used as a tissue preservative or as a very harsh disinfectant. It is also used as a precursor to various materials, including plastics, resins, and other polymers. ㈮ 䭥瑯湥:A步瑯湥involves a carbonyl in which the carbon atom makes single bonds with two R-groups. Ketones undergo most of the same reactions as aldehydes, but they tend to be slightly less reactive. The CHM 203 MODULE 2 93 simplest ketone is acetone, in which the carbonyl carbon is bonded to two CH2groups. This ketone is commonly used to remove fingernail polish and serves asan industrial solvent. Methyl ethyl ketone is used as a paint stripper and a solvent. Ketones are also used in the production of various polymers, either as a building block or as a solvent. The R-group in a ketone can be the same or different as seen in the example. ㌮ 䍡牢潸祬楣 慣楤猺 are another carbonyl-containing functional group, in which the carbon atom is bonded to a hydroxyl group on one side and either a carbon or hydrogen atom on the other. As the name implies, carboxylic acids are weakacids. An OH group that is directly connected to a carbonyl will ionize to a small extent when dissolved in water. The reason for this is the relative stability of the resulting anion. A carboxylate ion, in which the negative charge is spread over two different oxygen atoms through resonance structures, is more stable than an isolated oxygen-centered anion. The carboxylic acid and carboxylate ion are interchangeable. Carboxylate ions are often present in amino acids. Carboxylate ion 㐮 䕳瑥爺Anesteris similar to a carboxylic acid, in that it contains a carbonyl where the carbon is bonded to one additional oxygen atom and one carbon or hydrogen atom. However, the second oxygen atom is bonded to another carbon instead of to an acidic hydrogen atom. Structurally, carboxylic acids and esters are related to one another in the same way as alcohols and ethers. CHM 203 ORGANIC CHEMISTRYII 94 Esters can be formed by heating carboxylic acids and alcohols in the presence of an acid catalyst. This process is reversible, and the starting materials can be regenerated by reacting an ester with water in the presence of a weak base. Some esters have very pleasant odours, so they are used in the manufacture of many perfumes. Propyl acetate contributes to the odour of pears, while isoamyl acetate gives bananas their smell. This ester also serves as an alarm signal for honeybees. Esters are employed in the manufacture of fabrics (polyesters) and Plexiglass. Anesthetics such as procaine and benzocaine also contain esters. 㔮 䅭楤敳:Anamideis a carbonylin which the carbon is attached to one nitrogen atom and one carbon or hydrogen atom. Alternatively, we could define an amide as an慭楮ein which one of the carbon atoms attached to the nitrogen is part of a carbonyl. An amide can be formed bycombining a carboxylic acid and an amine. Only primary and secondary amines can be sued to form amides, since they have a hydrogen that can be replaced with the carbonyl carbon; tertiary amines will not form amides. Amides are used as colouring agents incrayons, pencils, and ink. They are employed in the paper, plastic, and rubber industries. Polyacrylamide is a very widely used amide; it is involved in the treatment of drinking water and sewage, and in plastics manufacture. The amide Kevlar is widely employed for the production of body armor, and nylon is another type of amide-based polymer. Atenolol and donepezil are examples of useful drugs that contain a variety of functional groups.Atenolol is a ‚ bloc ke r, a group of drugs used to treat hypertension. Donepezil, sold under the trade name Aricept, is used to treat mild to moderate dementia associated with Alzheimer's disease. CHM 203 MODULE 2 95 ㌮㈮4䅬捯桯汳 The慬捯桯lfunctional group involves an oxygen atom that is bonded to one hydrogen atom and one carbonatom. The carbon atom will be part of a larger organic structure. One way to indicate a generic alcohol would be with the formula R-OH. R represents any organic fragment in which a carbon atom is directly bonded to the explicitly indicated functional group (in this case, OH). The R group is typically a chain of carbon atoms. Alcohols can be classified as primary, secondary, or tertiary based on the characteristics of the carbon to which it is attached. In a primary alcohol, the carbon bonded directly tothe oxygen atom is also bonded to exactly one carbon atom, with the other bonds generally going to hydrogen atoms. In a secondary alcohol, the carbon is attached to two other carbon atoms, and in a tertiary alcohol, the carbon is bonded to three other carbon atoms. The type of alcohol being used will determine the product of certain reactions. We are already familiar with several common alcohols. For example, ethanol is the alcohol present in alcoholic beverages. It is also widely used in the industrial manufacture of other chemicals. Methanol is used as a gasoline additive or alternative. Additionally, methanol can be used to manufacture formaldehyde, which is employed in the production of CHM 203 ORGANIC CHEMISTRYII 96 plastics, paints, and other useful substances. Isopropanol is commonly known as rubbing alcohol. In addition to its industrial uses, isopropanol is used to clean various surfaces, including computer monitors, whiteboards, and even skin (e.g., before getting blood drawn). ㌮㈮5䅭楮敳 An慭楮econsists of a nitrogen atom bonded to some combination of carbons and hydrogens. Like alcohols, amines can be classified as primary, secondary, or tertiary. However, the rules for assigning these categories are slightly different. In an alcohol, the oxygen atom is always bondedto exactly one carbon atom, so we look at the branching on the adjacent carbon, not the oxygen atom itself. In a neutral amine, the nitrogen can be bonded to one, two, or three carbon atoms, and this is how we decide whether it is called a primary, secondary, or tertiary amine. Neutral amines are weak bases, because the lone pair on nitrogen can act as a proton acceptor. Many smaller amines have very strong and offensive odors. For example, the aptly-named compounds cadaverine and putrescine are foul-smelling amines, formed as a part of the decay process after death. Amines serve a wide variety of uses. Diphenylamine acts as a stabilizer for certain types of explosives. Amines are found as components in some lubricating materials, in developers, and are apart of waterproofing textiles. Some amines, such as Novocain, are used as anesthetics. Many pharmaceutical compounds contain amines, including 8 of the 10 most prescribed medications in 2012. 䥮-呥硴⁑略獴楯渠4 1.The general formula for amines is__________. A. R-CH2B. R2CH C. R-NH2D. R-COOH 2.Glutamic acid is the parent compound of monosodium glutamate (known as MSG), which is used as a flavor enhancer. Glutamic acid has the following structure: CHM 203 MODULE 2 97 Name the functional groups you recognize in this molecule. Do you think there are other groups of atoms in this molecule that might qualify as functional groups? ㌮㈮6周楯l The瑨楯lfunctional group contains a sulphuratom bonded to a hydrogen atom. It is very similar to an alcohol functional group with the sulphur replacing the O. Thiols are also called mercaptans which are derived from the Latin phrase for "capturing mercury" because of the strong bonds it forms with mercury-containing compounds. Some thiol compounds have a distinctive smell similar to rotten eggs. They are often added to natural gas, which itself has no odor, as a way to detect leaks since its odor can be detected by humans in very small amounts.A thiol group is also present in the amino acid cysteine. ㌮3䙵湣瑩潮慬⁇牯異猠慮搠剥慣瑩癩瑹 A functional group also determines reactivity. What type of reaction does a particular kind of organic compound undergo? Begin by recalling two fundamental concepts. · Functional groups create reactive sites in molecules. · Electron-rich sites react with electron-poor sites. All fu nctiona l gr oups contai n a hete roato m, a € bond , or bot h, and 瑨敳攠 晥慴畲敳 浡步 敬散瑲潮 -摥晩捩敮琠 ⡯爠 敬散瑲潰桩汩挩 獩瑥猠 慮d 敬散瑲潮-物捨
潲畣汥潰桩汩挩⁳楴敳⁩渠愠浯汥捵汥. Molecules react at these sites. To predict reactivity, first locatethe functional group andthen determine the resulting electron-rich or electron-deficient sites it creates. Keep three guidelinesin mind. CHM 203 ORGANIC CHEMISTRYII 98 ƒ An electronegative heteroatom like N, O, or halogens makes a carbon atomelectrophilic. ƒ A lone pair on a heteroatom makes it basic and nucleophilic. ƒ „ Bonds createnucleophilicsites and are more easily broken than r bonds. By identifying the nucleophilic and electrophilic sites in a compound you can begin to understand how it will react. In general, electron-richsites react with electron-deficient sites: ƒ An electron-deficient carbon atom reacts with a nucleophile, symbolized as :Nu ‚ . ƒ An electron-rich carbon reacts with an electrophile, symbolized as E + . At this point we dont know enough organic chemistry to draw the products of many reactions with confidence. We do know enough, however, to begin to predict if two compounds might react together based solely on electron density arguments, and at what atoms that reaction is most likely to occur. For example, alkenes contain an electron-rich C‚C double bond and so they react with electrophiles, E + . On the other hand, alkyl halides possess an electrophilic carbon atom, so they react with electron-rich nucleophiles. CHM 203 MODULE 2 99 䥮-呥硴⁑略獴楯渠5 1.Why are alkanes sometimes called paraffins? 2.What is the principal difference in properties between alkenes and alkanes? How are they alike? 㐮0䍏乃䱕卉低 Propose structures for simple molecules that contain the following functional groups: ⡡)Alcohol ⡢)Aromatic ring ⡣)Carboxylic acid ⡤)Amine ⡥)Both ketone and amine ⡦)Two double bonds 十儠㐮2 Identify the functional groups in each of the following molecules: CHM 203 ORGANIC CHEMISTRYII 100 十儠㐮3 Propose structures for two isomers with the formula C2H7N. There are two isomeric structures. One has the connection C-C-N, and the other has the connection C-N-C. 十儠㐮4 The correct IUPAC name for the following structure is. (A) 5-hexen-3-ol (B) 1-hexen-4-ol (C) 3-hydroxy-5-hexene (D)Isohexen-3-ol (E) 4-hydroxy-1-hexene 十儠㐮5 Write a structure for each of the following compounds: a. isopropyl alcohold. neopentyl chloride b. isopentyl fluoride e. tert-butylamine c. sec-butyl iodide f. n-octyl bromide 㐮0䍏乃䱕卉低 In this unit, we have been able to classify the numerous organic compounds into several major categories based on the functional groups CH3CH2CHCH2CHCH2 OH CHM 203 MODULE 2 101 they contain. We also explained how functional groups present in organic molecules determine the chemical reactivity ofthose molecules. 㔮0单䵍䅒Y With over twenty million known organic compounds in existence, it would be very challenging to memorize chemical reactions for each one. Fortunately, molecules with similar functional groups tend to undergo similar reactions. A晵湣瑩潮慬 杲潵pis defined as an atom or group of atoms within a molecule that has similar chemical properties whenever it appears in various compounds. Even if other parts of the molecule are quite different, certain functional groups tend to reactin certain ways. Organic molecules vary greatly in size and when focusing on functional groups, we direct our attention to the atoms involved in the functional group. As a result, the abbreviation R is used in some examples. The letter R is used in molecular structures to represent the †Rest of the molecule‡. It consists of a group of carbon and hydrogen atoms of any size. It is used as an abbreviation since a group of carbon and hydrogen atoms does not affect the functionality of the compound. In some molecules, you will see R, R, or R which indicates that the R groups in the molecule can be different from one another. For example, R might be‚CH2CH3while R is‚CH2CH2CH2CH3. 㘮0呕呏删䵁剋⁁卓䥇䵅乔 7.0剅FERENCES/FUR呈䕒 R䕁䑉NGS Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. Dewick, P. M. (2006).Essentials of organic chemistry: for students of pharmacy, medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R. N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009). Chemistry: the central science. Pearson Education. Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New Age International. Okuyama, T., & Maskill, H. (2013). Organic Chemistry: a mechanistic approach. Oxford University Press. CHM 203 ORGANIC CHEMISTRYII 102 Ghatak, K. L. (2014).A Textbook of Organic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons. CHM 203 MODULE 2 103 UN䥔′ 䅌䭁久匬 䙒䕅 剁䑉䍁䰠 单䉓呉呕呉低 剅䅃呉低匠䥎 䅌䭁久匠 䅎䐠 呈E 剅䅃呉噉呙 -卅䱅䍔䥖䥔夠偒䥎䍉偌E 䍏N呅NTS 1.0Introduction 2.0LearningObjectives 3.0MainContent 3.1Physical Properties of Alkanes 3.2Chemical Reactions of Alkanes 3.2.1Combustion of Alkanes 3.2.2Halogenation of Alkanes (Free Radical Substitution Reaction) 3.2.2.1 Factors that Determine Product Distribution 3.2.2.2 The Reactivity‚Selectivity Principle 4.0Self-Assessment Questions (SAQs) 5.0Conclusion 6.0Summary 7.0References/FurtherReadings ㄮ0INT剏䑕C TI低 It will be well to reiterate what an alkane is, lest you be confused as to the difference between alkanes and alkenes. Alkanes are compounds of carbon and hydrogen only, without double bonds, triple bonds, or rings. They all conform to the general formula CnH2n+2and sometimes are called paraffin hydrocarbons, open-chain saturated hydrocarbons, or acyclic hydrocarbons. The nomenclature of alkanes has been discussed in the previous units, and you may find it well to review Module 1-unit 1 and Module 2-unit 1 before proceeding. This unit is concerned with the chemistry of only one class of compounds, saturated hydrocarbons or alkanes, several fundamental reaction principles are developed that we shall discussed extensively in later units. The study of some of these principles such as activation energy of a reaction has been associated traditionally more with physical chemistry than with organic chemistry. We include them here because they provide a sound basis for understanding the key questions concerning the practical use of organic reactions. Is the equilibrium point of a given reaction far enough toward the desired products to be useful? Can conditions be found in which the reaction will take place at a practical rate? How can unwanted side reactions be suppressed? Initially, we will be concerned with the physical properties of alkanes and how these properties can be correlated by the important concept of CHM 203 ORGANIC CHEMISTRYII 104 homology. This will be followed by a brief survey of the occurrence and uses of hydrocarbons, with special reference to the petroleum industry. Chemical reactions of alkanes then will be discussed, with special emphasis on free radical substitution reaction. These reactions are employed to illustrate how we can predict and use energy changes- particularlyˆH, the heat evolved or absorbed by a reacting system, which often can be estimated from bond energies. Then we consider some of the problems involved in predicting reaction rates in the context of a specific reaction, the chlorination of methane. The example is complex, but it has the virtue that we are able to break the overall reaction into quite simple steps. ㈮0䱅䅒义乇⁏䉊䕃呉噅S By the end of this unit, you should be able to: · understand the chemistry of alkanes. · know the characteristic physical properties of alkanes. · describe the chemical reactions alkanes undergo. · understand the concept of free radical substitution reactions in alkanes. · use the reactivity-selectivity principle and activation energy to predict expected products yield in a radical substitution reaction. ㌮0䵁䥎⁃低呅乔 ㌮1偨祳楣慬⁐牯灥牴楥猠潦⁁汫慮敳 The series of straight-chain alkanes, in whichnis the number of carbons in the chain, shows a remarkably smooth gradation of physical properties (see Table 2.1). Asnincreases, each additional CH2group contributes a fairly constant increment to the boiling point and density, and to a lesser extent to the melting point. This makes it possible to estimate the properties of an unknown member of the series from those of its neighbors. For example, the boiling points of hexane and heptane are 69 o and 98 o , respectively. Thus a difference in structure of one CH2group for these compounds makes a difference in boiling point of 29 o ; we would predict the boiling point of the next higher member, octane, to be 98 o + 29 o = 127 o , which is close to the actual boiling point of 126 o . CHM 203 MODULE 2 105 呡扬攠㈮ㄺ⁐桹獩捡氠偲潰敲瑩敳映䅬歡湥猬⁃H 3⡃H2)n-1H a At the boiling point. b Under pressure. c For the supercooled liquid. Members of a group of compounds, such as the alkanes, that have similar chemical structures and graded physical properties, and which differ from one another by the number of atoms in the structural backbone, are said to constitute a桯浯汯杯畳⁳敲楥s . When used to forecast the properties of unknown members of the series, the concept of homology works most satisfactorily for the higher-molecular-weight members because the introduction of additional CH2groups makes a smaller relative change in the overall composition of such molecules. Branched-chain alkanes do not exhibit the same smooth gradation of physical properties as do the continuous-chain alkanes. Usually there is too great a variation in molecular structure for regularities to be apparent. Nevertheless, in any one set of isomeric hydrocarbons, volatility increases with increased branching. This can be seen from the data in Table 2.2 and Figure 2.1, which lists the physical properties of the five hexane isomers. The most striking feature of the datais the 19 o difference between the boiling points of hexane and 2,2-dimethylbutane. CHM 203 ORGANIC CHEMISTRYII 106 呡扬攠㈮㈺⁐桹獩捡氠偲潰敲瑩敳映䡥硡湥⁉獯浥牳 䅣瑩癩瑹″⸱ Use the data of Tables 2.1 and 2.2 to estimate the boiling points of tetradecane, heptadecane, 2-methylhexane, and 2,2-dimethylpentane. CHM 203 MODULE 2 107 䙩朠㈮ㄺ⁇牡灨映䉯楬楮朠偯楮瑳映䅬歡湥s Straight chain alkanes have a higher boiling point than branched alkanes due to the greater surface area in contact and therefore greater van der Waals interactions. Since alkanes are hydrocarbons and there is no significant difference in electronegativity between carbon and hydrogen, they are nonpolar. In being nonpolar, alkanes dissolve in nonpolar solvents because the van der Waals interactions between nonpolar alkane molecules and nonpolar solvent molecules are similar to the alkane-alkane and nonpolar solvent- nonpolar solvent molecule interactions. Alkanes are thus soluble in nonpolar solvents (i.e. other alkanes) and insoluble in polar solvents (such as water). The liquid alkanes are therefore good solvents for nonpolar covalent compounds. 䥮-呥硴⁑略獴楯渠1 For each of the following pairs of compounds, select the substance you expect to have the higher boiling point. 1.octane and nonane. 2.octane and 2,2,3,3‑tetramethylbutane. Note: The actual boiling points are nonane = 150.8°C, octane = 125.7°C, 2,2,3,3‑tetramethylbutane = 106.5°C CHM 203 ORGANIC CHEMISTRYII 108 ㌮3䍨敭楣慬⁒敡捴楯湳映䅬歡湥s Alkanes have only strong € bonds. Because the carbon and hydrogen atoms of an alkane have approximately the same electronegativity, the electrons in the C-H and C-C € bonds are shared equally by the bonding atoms. Consequently, none of the atoms in an alkane have any significant charge. This means that neither nucleophiles (electron pair donors) nor electrophiles (electron pair acceptors) are attracted to them. Alkanes, therefore, are relatively unreactive compounds. Their failure to undergo reactions prompted early organic chemists to call them paraffins, from the Latinparum affinis, which means †little affinity‡ (for other compounds). Thus none of the C-H or C-C bonds in a typical saturated hydrocarbon, for example ethane, are attacked at ordinary temperatures by a strong acid, such as sulfuric acid (H2SO4), or by an oxidizing agent, such as bromine (in the dark), oxygen, or potassium permanganate (KMnO4). Under ordinary conditions, ethane is similarly stable to reducing agents such as hydrogen, even in the presence of catalysts such as platinum, palladium, or nickel. However, all saturated hydrocarbons are attacked by oxygen at elevated temperatures and, if oxygen is in excess, complete combustion to carbon dioxide and water occurs. Vast quantities of hydrocarbons from petroleum are utilized as fuels for the production of heat and power by combustion. Natural gas consists of about 75% methane. The remaining 25% is composed of small alkanes such as ethane, propane, and butane. Petroleum is a complex mixture of alkanes and cycloalkanes that can be separatedinto fractions by distillation. The fractionthat boils at the lowest temperature (hydrocarbons containing three and four carbons) is natural gas that can be liquefied under pressure. This gas is used as a fuel for cigarette lighters, camp stoves, and barbecues. The fraction that boils at somewhat higher temperatures (hydrocarbons containing 5 to 11 carbons) is gasoline; the next fraction (9 to 16 carbons) includes kerosene and jet fuel. The fraction with 15 to 25 carbons is used for heating oil and diesel oil, and the highest-boiling fraction is used for lubricants and greases. The nonpolar nature of these compounds is what gives them their oily feel. After distillation, a nonvolatile residue called asphalt or tar is left behind. CHM 203 MODULE 2 109 䙩朠㈮㈺†䑩晦敲敮琠晲慣瑩潮猠潦⁰整牯汥畭 Alkanes (the most basic of all organic compounds) undergo very few reactions. The two reactions of more importance are combustion and halogenation (i.e., substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane). The halogen reaction is very important in organic chemistry because it opens a gateway to further chemical reactions. 䥮-呥硴⁑略獴楯渠2 1.Liquefied petroleum gas is mainly composed of _____________ a) Methane and ethaneb) Ethane and propanec) Propane and butane d) Butane and hexane 2.If you drop a lighted match onto some petrol (gasoline), it ignites almost explosively. If you drop a lighted match onto some tar on the road, almost certainly nothing will happen. Both contain alkanes, Explain the difference. ㌮㌮6䍯m扵獴楯渠潦⁁汫慮敳 In general, the formula for complete combustion of alkanes is: When there is suf‰cient oxygen, alkanes will burn and form water and carbon dioxide as products: CHM 203 ORGANIC CHEMISTRYII 110 It isquite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Some are easier than others. For example, with alkanes, the ones with an even number of carbon atomsare marginally harder than those with an odd number. For example, with propane (C3H8), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be: C3H8+ O2Š 3CO2+ 4H2O Counting the oxygensleads directly to the final version: With butane (C4H10), you can again balance the carbons and hydrogens as you write the equation down. C4H10+ O2Š 4CO2+ 5H2O Counting the oxygens leads to a slight problem-with 13 on the right- hand side. The simple trick is to allow yourself to have "six-and-a-half" O molecules on the left. If you are not comfortable with that, double everything: The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules do not vaporize so easily-the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas. Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame. Incomplete combustion (where there is not enough oxygen present) can lead to the formation of carbon or carbon monoxide. As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over. The presenceof glowing carbon particles in a flame turns it yellow, and black carbon is CHM 203 MODULE 2 111 often visible in the smoke. Carbon monoxide is produced as a colourless poisonous gas. For example, incomplete combustion of octane can be represented as: 䥮-呥硴⁑略獴楯渠3 Write equations for the complete combustion of a) pentane, C5H12b) ethane, C2H6 c) decane, C10H22d) cyclohexane, C6H12 ㌮㈮㈠†⁈慬潧敮慴楯渠潦⁁汫慮敳
䙲敥⁒慤楣慬⁓畢獴楴畴楯渠剥慣瑩潮) Alkanes do react with chlorine (Cl2) or bromine (Br2) to form alkyl chlorides or alkyl bromides. These halogenation reactions take place only at high temperatures or in the presence of light (symbolized by hv). They are the only reactions that alkanes undergo‹with the exception of combustion (burning), a reaction with oxygen. Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple獵扳瑩瑵瑩潮⁲敡捴楯n in which a C-H bond is broken and a new C-X bond is formed. When a bond breaks so that both of its electrons stay with one of the atoms, the process is called桥瑥牯汹瑩挠扯湤⁣汥慶慧e or桥瑥牯汹獩s. When a bond breaks so that each of the atoms retains one of the bonding electrons, the process is called桯浯汹瑩挠扯湤⁣汥慶慧e or桯浯汹獩s. The mechanism for the halogenation of an alkane is well understood. The high temperature (or light) supplies the energy required to break the or bond homolytically. Homolytic bond cleavage is the楮楴楡瑩潮⁳瑥p of the 乯瑥:Whyis carbon monoxide said to bepoisonous? Oxygen is carried around the blood by haemoglobin. Carbon monoxide unfortunately binds to exactly the same site on the haemoglobinthat oxygen does. The difference is that carbon monoxide binds irreversibly (or very strongly)-making that particular molecule of haemoglobin useless for carrying oxygen. Hence, if you breath in enough carbon monoxide you will die from a sort of internal suffocation. CHM 203 ORGANIC CHEMISTRYII 112 reaction because it creates theradical that is used in the first propagation step. A radical (often called a free radical) is a species containing an atom with an unpaired electron. A radical is highly reactive because it wants to acquire an electron to complete its octet. As an example, in the mechanism for the monochlorination of methane, the chlorine radical formed in the initiation step abstracts a hydrogen atom from methane, forming HCl and a methyl radical. The methyl radical abstracts a chlorine atom from Cl2, forming methyl chloride and another chlorine radical, which can abstract a hydrogen atom from another molecule of methane. These two steps are called灲潰慧慴楯渠 獴数s because the radical created in the first propagation step reacts in the second propagation step to produce a radical that can repeat the first propagation step. Thus, the two propagation steps are repeated over and over. The first propagation step is the rate- determining step of the overall reaction. Because the reaction has radical intermediates and repeating propagation steps, it is called a牡摩捡氠捨慩n 牥慣瑩潮. Any two radicals in the reaction mixture can combine to form a molecule in which all the electrons are paired. The combination of two radicals is called a瑥牭楮慴楯渠獴数 because it helps bring the reaction to an end by CHM 203 MODULE 2 113 decreasing the number of radicals available to propagate the reaction. Any two radicals present in the reaction mixture can combine in a termination step, so radical reactions produce a mixture of products. The radical chlorination of alkanes other than methane follows the same mechanism. The reaction of an alkane with chlorine or bromine to form an alkyl halide is called a牡摩捡氠獵扳瑩瑵瑩潮⁲敡捴楯n because radicals are involved as intermediates and theend result is the substitution of a halogen atom for one of the hydrogen atoms of the alkane. In order to maximize the amount of monohalogenated product obtained, a radical substitution reaction should be carried out in the presence of excess alkane. Excess alkane in the reaction mixture increases the probability that the halogen radical will collide with a molecule of alkane rather than with a molecule of alkyl halide‹even toward the end of the reaction, by which time a considerable amount of alkyl halide will have been formed. If the halogen radical abstracts a hydrogen from a molecule of alkyl halide rather than from a molecule of alkane, a dihalogenated product will be obtained. Bromination of alkanes follows the same mechanism as chlorination. ㌮㈮㈮ㄠ䙡捴潲猠瑨慴⁄整敲浩湥⁐牯摵捴⁄楳瑲楢畴楯n CHM 203 ORGANIC CHEMISTRYII 114 The distribution of products depends on probability and reactivity. Two different alkyl halides are obtained from the monochlorination of butane. Substitution of a hydrogen bonded to one of the terminalcarbons produces 1-chlorobutane, whereas substitution of a hydrogen bonded to one of the internal carbons forms 2-chlorobutane. The expected (statistical) distribution of products is 60% 1-chlorobutane and 40% 2-chlorobutane because six of butanes 10hydrogens can be substituted to form 1-chlorobutane, whereas only four can be substituted to form 2-chlorobutane. This assumes, however, that all of the bonds in butane are equally easy to break. Then, the relative amounts of the two products would dependonly on the probability of a chlorine radical colliding with a primary hydrogen, compared with its colliding with a secondary hydrogen. When we carry out the reaction in the laboratory and analyze the product, however, we find that it is 29% 1-chlorobutaneand 71% 2-chlorobutane. Therefore, probability alone does not explain the regioselectivity of the reaction. Because more 2-chlorobutane is obtained than expected and the rate-determining step of the overall reaction is hydrogen atom abstraction, we conclude that it must be easier to abstract a hydrogen atom from a secondary carbon than from a primary carbon. Alkyl radicals have different stabilities, and the more stable the radical, the more easily it is formed because the stability of the radical is reflected in the stability of the transition state leading to its formation. Consequently, it is easier to remove a hydrogen atom from a secondary carbon to form a secondary radical than it is to remove a hydrogen atom from a primary carbon to form a primaryradical. When a chlorine radical reacts with butane, it can abstract a hydrogen atom from an internal carbon, thereby forming a secondary alkyl radical, or it can abstract a hydrogen atom from a terminal carbon, thereby forming a primary alkyl radical.Because it is easier to form the more CHM 203 MODULE 2 115 stable secondary alkyl radical, 2-chlorobutane is formed faster than 1- chlorobutane. After experimentally determining the amount of each chlorination product obtained from various hydrocarbons, chemists were able to conclude that at room temperature it is 5.0 times easier for a chlorine radical to abstract a hydrogen atom from a tertiary carbon than from a primary carbon, and it is 3.8 times easier to abstract a hydrogen atom from a secondary carbon than from a primary carbon. The precise ratios differ at different temperatures. To determine the relative amounts of products obtained from radical chlorination of an alkane, both probability (the number of hydrogens that can be abstracted that willlead to the formation of the particular product) and reactivity (the relative rate at which a particular hydrogen is abstracted) must be taken into account. When both factors are considered, the calculated amounts of 1-chlorobutane and 2-chlorobutane agree with the amountsobtained experimentally. The percent yield of each alkyl halide is calculated by dividing the relative amount of the particular product by the sum of the relative amounts of all the alkyl halide products (16 + 15 = 21). CHM 203 ORGANIC CHEMISTRYII 116 Radical monochlorinationof 2,2,5-trimethylhexane results in the formation of five monochlorination products. Because the relative amounts of the five alkyl halides total 35 (19.0 + 7.6 + 7.6 + 5.0 + 6.0 = 35), the percent yield of each product can be calculated as follows: Because radical chlorination of an alkane can yield several different monosubstitution products as well as products that contain more than one chlorine atom, it is not the best method for synthesizing an alkyl halide. Addition of a hydrogen halide to an alkene or conversion of an alcohol to an alkyl halide is a much better way to make an alkyl halide. Radical halogenation of an alkane is nevertheless still a useful reaction because it is the only way to convert an unreactive alkane into a reactive compound. Once the halogen is introduced into the alkane, it can be replaced by a variety of other substituents. 䥮-呥硴⁑略獴楯渠4 How many alkyl halides can be obtained from monochlorination of the following alkanes? Neglect stereoisomers. CHM 203 MODULE 2 117 The relative rates ofradical formation when a bromine radical abstracts a hydrogen atom are different from the relative rates of radical formation when a chlorine radical abstracts a hydrogen atom. At 125 °C, a bromine radical abstracts a hydrogen atom from a tertiary carbon1600 times faster than from a primary carbon and abstracts a hydrogen atom from a secondary carbon 82 times faster than from a primary carbon. When a bromine radical is the hydrogen-abstracting agent, the differences in reactivity are so great that the reactivity factor is vastly more important than the probability factor. For example, radical bromination of butane gives a 98% yield of 2-bromobutane, compared with the 71% yield of 2- chlorobutane obtained when butane is chlorinated. A bromine radical is less reactive and more selective than a chlorine radical. In other words, bromination is more highly regioselective than chlorination. Similarly, bromination of 2,2,5-trimethylhexane gives an 82% yield of the product in which bromine replaces the tertiaryhydrogen. Chlorination of the same alkane results in a 14% yield of the tertiary alkyl chloride. CHM 203 ORGANIC CHEMISTRYII 118 䥮-呥硴⁑略獴楯渠5 Calculate the percent yield of each product obtained in Problems 1a and b above if chlorination is carried out in the presence of light at room temperature. ㌮㈮㈮2周攠剥慣瑩癩瑹 ƒ卥汥捴楶楴礠偲楮捩灬e Why are the relative rates of radical formation so different when a bromine radical rather than a chlorine radical is used as the hydrogen- abstracting reagent? To answer this question, we must compare theˆH o values for the formation of primary, secondary, and tertiary radicals when a chlorine radical is used, as opposed to when a bromine radical is used. TheseˆH o values can be calculated using the bond dissociation energies in Table 2.3 below. Remember thatˆH o is equal to the energy of the bond being broken minus the energy of the bond being formed. CHM 203 MODULE 2 119 呡扬攠㈮㌺⁈潭潬祴楣⁂潮搠䑩獳潣楡瑩潮⁅湥牧楥猠Y -Z „ Y … ⬠Z … We must also be aware that bromination is a much slower reaction than chlorination. The activation energy for abstraction of a hydrogen atom by a bromine radical has been found experimentally to be about 4.5 times greater than that for abstraction ofa hydrogen atom by a chlorine radical. CHM 203 ORGANIC CHEMISTRYII 120 Using the calculated values and the experimental activation energies, we can draw reaction coordinate diagrams for the formation of primary, secondary, and tertiary radicals by chlorine radical abstraction (Figure 2.3a) and by bromine radical abstraction (Figure 2.3b). 䙩杵牥′⸳㨠⡡⤠剥慣瑩潮⁣潯牤楮慴攠摩慧牡浳⁦潲⁴桥⁦潲浡瑩潮f 灲業慲礬 獥捯湤慲礬 慮搠 瑥牴楡特 慬歹氠 牡摩捡汳 慳 愠 牥獵汴 潦 慢獴牡捴楯渠潦⁡⁨祤牯来渠慴潭⁢礠愠捨汯物湥⁲慤楣慬⸠周攠瑲慮獩瑩潮 獴慴敳⁨慶攠牥污瑩癥汹楴瑬攠牡摩捡氠捨慲慣瑥爠扥捡畳攠瑨敹⁲敳敭扬e 瑨攠牥慣瑡湴献
戩⁒敡捴楯渠捯潲摩湡瑥⁤楡杲慭猠景爠瑨攠景牭慴楯渠潦 灲業慲礬 獥捯湤慲礬 慮搠 瑥牴楡特 慬歹氠 牡摩捡汳 慳 愠 牥獵汴 潦 慢獴牡捴楯渠潦⁡⁨祤牯来渠慴潭⁢礠愠扲潭楮攠牡摩捡氮⁔桥 瑲慮獩瑩潮 獴慴敳⁨慶攠愠牥污瑩癥汹⁨楧栠摥杲敥映牡摩捡氠捨慲慣瑥爠扥捡畳攠瑨敹 牥獥浢汥⁴桥⁰牯摵捴献 Because the reaction of a chlorine radical with an alkane to form a primary, secondary, or tertiary radical is exothermic, the transition states resemble the reactants more than they resemble the products. The reactants all have approximately the same energy, so there will be only a small difference in the activation energies for removal of a hydrogen atom from a primary, secondary, or tertiary carbon. In contrast, the reaction of a bromine radical with an alkane is endothermic, so the transition states resemble the products more than they resemble the reactants. Because there is a significant difference in the energies of the product radicals‹ depending on whether they are primary, secondary, or tertiary‹there is a significant difference in the activation energies. Therefore, a chlorine radical makes primary, secondary, and tertiary radicals with almost equal ease, whereas a bromine radical has a clearpreference for formation of the easiest-to-form tertiary radical (Figure 2.3). In other words, because a bromine radical is relatively unreactive, it is highly selective about which hydrogen atom it abstracts. In contrast, the much more reactive chlorine radical is considerably less selective. These observations illustrate the 牥慣瑩癩瑹 ƒ獥汥捴楶楴礠 灲楮捩灬e , which states that the greater the reactivity of a species, the less selective it will be. CHM 203 MODULE 2 121 Because chlorination is relatively nonselective, it is a useful reaction only when there is just one kind of hydrogen in the molecule. By comparing the values for the sum of the two propagating steps for the monohalogenation of methane, we can understand why alkanes undergo chlorination and bromination but notiodination and why fluorination is too violent a reaction to be useful. The fluorine radical is the most reactive of the halogen radicals, and it reacts violently with alkanes (ˆH o =-31 kcal/mol). In contrast, the iodine radical is the least reactive of the halogen radicals. In fact, it is so unreactive (ˆH o = 34 kcal/mol) that it is unable to abstract a hydrogen atom from an alkane. Consequently, it reacts with another iodine radical and reforms I2. 䥮-呥硴⁑略獴楯渠6 Carry out the calculations that predict that a. 2-bromobutane will be obtained in 98% yield. b. 2-bromo-2,5,5,-trimethylhexane will be obtained in 82% yield. 卅䱆-䅓卅卓䵅乔 䕘䕒䍉卅 i.Explain why chlorination or bromination of methylcyclohexane will produce a greater yield of 1-halo-1-methylcyclohexane. a.Would chlorination or bromination produce a greater yield of 1- halo-2,3 dimethylbutane? b.Would chlorination or bromination produce a greater yield of 2- halo-2,3-dimethylbutane? CHM 203 ORGANIC CHEMISTRYII 122 c.Would chlorination or bromination be a better way to make 1-halo- 2,2-dimethylpropane? iiIodine does not react with ethane even though I2is more easily cleaved homolytically than the other halogens. Explain. iiiWhen 2-methylpropane is monochlorinated in the presence of light at room temperature, 36% of the product is 2-chloro-2-methylpropane and 64% is 1-chloro-2-methylpropane. From these data, calculate how much easier it is to abstract a hydrogen atom from a tertiary carbon than from a primary carbon under these conditions. Let total amount of product = 100 The relative amount of 1-chloro-2-methylpropane = number of primary hydrogen x ease of the primary hydrogen removal (X) i.e. 64 = 9 x X X = 7.11 Similarly, The relative amount of 2-chloro-2-methylpropane =number of tertiary hydrogen x ease of the tertiary hydrogen removal (Y) i.e. 36 = 1 x Y Y = 36 Relative ease of removal of hydrogen atom from a tertiary carbon than from a primary carbon are Y : X i.e. 36 : 7.11 = 5 : 1 Hence, it is five times easier toabstract a hydrogen atom from a tertiary carbon than from a primary carbon under the given conditions. ivIf 2-methylpropane is brominated at 125 °C in the presence of light, what percent of the product will be 2-bromo-2-methylpropane? Compare your answerwith the percent given in Problem 2 above for chlorination. v.a)Why does a bunsen burner with the air-hole closed produce a yellow flame? b)Explain why carbon monoxide, formed by incomplete combustion of hydrocarbons, is poisonous. 4.0䍏乃䱕卉低 Alkanes (the most basic of all organic compounds) undergo very few reactions. The two reactions of more importance are combustion and halogenation. Halogenation is the replacement of one or more hydrogen atoms in an organic compound by a halogen (fluorine,chlorine, bromine or iodine). Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple substitution reaction in which a C-H bond is broken and a new C-X bond is formed. CHM 203 MODULE 2 123 㔮0单䵍䅒Y · Alkanes are called saturated hydrocarbons because they do not contain any double or triple bonds. Since they also have only strong € bonds and atoms with no partial charges, alkanes are very unreactive. · Alkanes do undergo radical substitution reactions with chlorine (Cl2) or bromine (Br2) at high temperatures or in the presence of light, to form alkyl chlorides or alkyl bromides. The substitution reaction is a radical chain reaction with initiation, propagation, and termination steps. · Unwanted radical reactions are prevented byradical inhibitors‹ compounds that destroy reactive radicals by creating unreactive radicals or compounds with only paired electrons. · The rate-determining step of the radical substitution reaction is hydrogen atom abstraction to form a radical. · The relative rates of radical formation are 3 o > 2 o > 1 o > methyl. · To determine the relative amounts of products obtained from the radical halogenation of an alkane, both probability and the relative rate at which a particular hydrogen is abstracted must be taken into account. · The reactivity‚selectivity principle states that the more reactive a species is, the less selective it will be. · A bromine radical is less reactive than a chlorine radical, so a bromine radical is more selective about which hydrogen atom it abstracts. · Fluorine radicals (F Œ ) are the most reactive of all of the halogen atoms and iodine radicals (I Œ ) are the least reactive. I Œ is so unreactive that it does not abstract H from most alkanes. The overall reactivity trend for halogen atoms is F > Cl> Br > I and this is the same order as they appear in the next to the last column of a periodic table. 㘮0呕呏删䵁剋⁁卓䥇䵅乔 7.0剅FERENCES/FUR呈䕒 R䕁䑉NGS Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. CHM 203 ORGANIC CHEMISTRYII 124 Dewick, P. M. (2006).Essentials of organic chemistry: for students of pharmacy, medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R. N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009). Chemistry: the central science. Pearson Education. Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New Age International. Okuyama, T., & Maskill, H. (2013). Organic Chemistry: amechanistic approach. Oxford University Press. Ghatak, K. L. (2014).A Textbook of Organic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons. CHM 203 MODULE 2 125 UN䥔″ 噁剉何匠佒䝁N 䥃⁒䕁䍔䥏乓 䍏乔䕎呓 1.0Introduction 2.0Learning Objectives 3.0Main Content 3.1Writing Equations for Organic Reactions 3.2Reaction Mechanism 3.2.1Bond Cleavage 3.2.2Radicals, Carbocations, and Carbonions 3.2.3Bond Formation 3.2.4Kindsof Arrows 3.3Various Organic Reactions 3.3.1Substitution Reactions 3.3.2Elimination Reactions 3.3.3Addition Reactions 3.3.4Redox Reactions 3.3.5Rearrangement Reactions 3.3.6Pericyclic Reactions 3.3.7Free Radical Reactions 3.4Typesof Reagents 4.0Conclusion 5.0Summary 6.0Tutor Mark Assignment 7.0References/FurtherReadings ㄮ0䥎呒佄啃呉低 An understanding of chemical processes has made possible the conversion of natural substances into new compounds withdifferent, and sometimes superior, properties. Aspirin, ibuprofen, nylon, and polyethylene are all products of chemical reactions between substances derived from petroleum. Reactions are difficult to learn when each reaction is considered a unique and isolated event. Virtually all chemical reactions are woven together by a few basic themes. In this unit, we shall be looking at the equations for organic reactions, reaction mechanisms, various organic reactions, how a reaction occurs, and types of reagents. Understanding the details of an organic reaction allows us to determine when it might be used in preparing interesting and useful organic compounds. A chemical reaction is the transformation of one chemical or collection of chemicals into another chemical or collection of chemicals. This involves the making of new chemical bonds and the breaking of old chemical bonds. Organic reactions are the chemical reactions that are undergone by CHM 203 ORGANIC CHEMISTRYII 126 organic compounds. Reactions of organic compounds can be organized broadlyby the types of reactions and how these reactions occur. The types of reactions organic compounds undergo are divided into substitution reactions, elimination reactions, addition reactions, redox reactions, rearrangement reactions, pericyclic reactions andfree radical reactions. ㈮0佂䩅䍔䥖䕓 When you have studied this unit, you should be able to: · understand how equations for organic reactions are written. · have a better understanding of reaction mechanisms. · become familiar with the various classes oforganic reactions. ㌮0䵁䥎⁃低呅乔 ㌮1坲楴楮朠䕱畡瑩潮猠景爠佲条湩挠剥慣瑩潮s Like other reactions, equations for organic reactions are usually drawn with a single reaction arrow (Š) between the starting material and product, but other conventions make these equations look different from those encountered in general chemistry. The牥慧敮琬the chemical substance with which an organic compound reacts, is sometimes drawn on the left side of the equation with the other reactants. At other times, the reagent is drawn above the reaction arrow itself, to focus attention on the organic starting material by itself on the left side. The solvent and temperature of a reaction may be added above or below the arrow.Th e sy mb ols † h v ‡ a nd † ˆ ‡ are used for re acti ons 瑨慴⁲敱畩牥 light 潲 heat, 牥獰散瑩癥汹. Figure 3.1 presents an organic reaction in different ways. When two sequential reactions are carried out without drawing any intermediate compound, the steps are usually numbered above or below the reaction arrow. This convention signifies that the first step occurs beforethe second, and the reagents areaddedin sequence,not at the same time. CHM 203 MODULE 2 127 In this equation only the organic product is drawn on the right side of the arrow. Although the reagent CH3MgBr contains both Mg and Br, these elements do not appear in the organic product, and they are oftenomitted on the product side of the equation. These elements have not disappeared. They are part of an inorganic by-product (HOMgBr in this case), and are often of little interest to an organic chemist. 䙩杵牥′⸴㨠䑩晦敲敮琠睡祳映睲楴楮g 潲条湩挠牥慣瑩潮s ㌮2剥慣瑩潮⁍散桡湩獭s Having now learned how to write and identify some common kinds of organic reactions, we can turn to a discussion of牥慣瑩潮散桡湩獭. Due to the unique properties of carbon, namely, its ability to catenate, form chains and rings, and form multiple bonds with itself or other atoms, an enormous number of organic compounds are possible. Moreover, an organic compound can undergo many different reactions. For example, ethanol may react to form acetaldehyde, acetic acid, ethylene, formic acid, carbon dioxide, and so on. So, it may appear that studying organic reactions should be a very difficult and almost futile exercise. But, CHM 203 ORGANIC CHEMISTRYII 128 fortunately, closer studies have revealed that the reactions of enormously large number oforganic compounds actually take place in such ways that these may be classified into a few groups of reactions. In most cases, the organic reactions are brought about with inorganic compounds like acids, bases, oxidizing agents, reducing agents and so on.The inorganic compounds that bring about the organic reactions are termed asreagents, and the organic compounds undergoing the chemical transformations are termed assubstrates. A reactio n m echanis m 楳 愠 摥瑡楬敤 摥獣物灴楯渠 潦 桯眠 扯湤猠 慲e 扲潫敮⁡湤 景牭敤⁡猠愠獴慲瑩湧慴敲楡氠楳⁣潮癥牴敤⁴漠愠灲潤畣琮 A reaction mechanism describes the relative order and rate of bond cleavage and formation. It explains all the known facts about a reaction and accounts for all products formed, and it is subject tomodification or refinement as new details are discovered. A reaction can occur either in one step or in a series of steps. 䄠潮e-獴数⁲敡捴楯渠楳⁣慬汥搠a co ncerted reac tion. No matter how many bonds are broken or formed, a starting material is converteddirectlyto a product. A „ B 䄠獴数睩獥⁲敡捴楯n involves more than one step. A starting material is first converted to an unstable intermediate, called areactive intermediate, which then goes on to form the product. A „ reacti ve inter me diate „ B ㌮㈮1䉯湤⁃汥慶慧e Bonds are broken and formed in all chemical reactions. No matter how many steps there are in the reaction, however, there are only two ways to break (cleave) a bond: the electrons in the bond can be dividedequallyor unequallybetween thetwo atoms of the bond. ƒ Breaking a bond byequally dividing the electronsbetween the two atoms in the bond is calledhomolysisorhomolytic cleavage. Homolysis or homolytic cleavage ƒ Breaking a bond byunequally dividing the electronsbetween the two atoms in the bond is calledheterolysisorheterolytic cleavage. CHM 203 MODULE 2 129 Heterolysis of a bond betweenAandBcan give eitherAorBthe two electrons in the bond. WhenAandBhave different electronegativities, theelectrons normally end up on the more electronegative atom. Heterolysis or heterolytic cleavage Homolysis and heterolysis require energy.Both processes generate reactive intermediates, but the products are different in each case. · Homolysis generates uncharged reactive intermediates with unpairedelectrons. · Heterolysis generateschargedintermediates. Each of these reactive intermediates has a very short lifetime, and each reacts quickly to form a stable organic product. ㌮㈮2剡摩捡汳Ⱐ䍡牢潣慴楯湳Ⱐ慮搠䍡牢慮楯湳 The curved arrow notation works fine for heterolytic bond cleavage because it illustrates the movement of anelectron pair.For homolytic cleavage, however, one electron moves to one atom in the bond and one electronmoves to the other, so a different kind of curved arrow is needed. ƒ To illustrate the movement of a single electron, use a half-headed curved arrow, sometimes called afishhook. Homolysis Two桡汦-桥慤敤curved arrows are needed for瑷漠獩湧汥 敬散瑲潮s. Heterolysis One晵汬-桥慤敤curved arrow is needed for潮攠敬散瑲潮⁰慩r . 䄠晵汬-桥慤敤⁣畲癥搠慲牯w shows the movement of an electronpair.A 桡汦⁨敡摥搠捵牶敤⁡牲潷 shows the movement of asingleelectron. Homolysisof the C‚Z bondgenerates two uncharged products with unpaired electrons. CHM 203 ORGANIC CHEMISTRYII 130 ƒ A reactive intermediate with a single unpaired electron is called a radical. Most radicals are highly unstable because they contain an atom that does not have an octet of electrons. Radicals typically haveno charge. They are intermediates in a group of reactions calledradical reactions. Heterolysisof the C‚Z bond can generate acarbocationor acarbanion. · Giving two electrons to Z and none to carbon generates a positively charged carbon intermediate called acarbocation. · Giving two electrons to C and none to Z generates a negatively charged carbon species called acarbanion. Both carbocations and carbanions are unstable reactive intermediates: A carbocation contains a carbon atom surrounded by only six electrons. A carbanion has a negative charge on carbon, which is not a very electronegative atom.Carbocations (electrophiles)andcarbanions (nucleophiles)can be intermediates inpola r reactions -reactions in which a nucleophile reacts with an electrophile. 䙩杵牥 ㈮㔺 周牥攠 牥慣瑩癥 楮瑥牭敤楡瑥猠 牥獵汴楮朠 晲潭 桯浯汹獩s 慮搠桥瑥牯汹獩猠潦⁡⁃ ƒ娠扯湤 Thus, homolysis and heterolysis generate radicals, carbocations, and carbanions, the three most commonreactive intermediates in organic chemistry. CHM 203 MODULE 2 131 ƒ Radicals and carbocations are electrophiles because they contain an electron-deficient carbon. ƒ Carbanions are nucleophiles because they contain a carbon with a lone pair. 䥮-呥硴⁑略獴楯渠1 1.Select the correct statement on carbanion from the following option. a)Carbanion is an intermediate compound b)In carbanion, central carbon atom carries negative charge c) It possess an unshared pair of electron d)All of the mentioned 2.Carbonium ionsare the intermediates in which the positive charge is carried by the carbon atom with ___________ electrons in the valence shell. a) 6b) 5c) 4d) 3 ㌮㈮3䉯湤⁆潲浡瑩潮 Like bond cleavage, bond formation occurs in two different ways. Two radicals can each donateone electronto form a two-electron bond. Alternatively, two ions with unlike charges can come together, with the negatively charged ion donatingboth electronsto form the resulting two electron bond.Bond formation always releases energy. CHM 203 ORGANIC CHEMISTRYII 132 ㌮㈮4䭩湤猠潦⁁牲潷s Table 3.1 below summarizes the many kinds of arrows used in describing organic reactions. Curved arrows are especially important because they explicitly show what electrons are involved in a reaction, how these electrons move informing and breaking bonds, and if a reaction proceeds via a radical or polar pathway. 呡扬攠㈮㐺⁁⁓畭浡特映䅲牯眠呹灥猠楮⁃桥浩捡氠剥慣瑩潮s 䅲牯w 乡浥 啳e Reactive arrowDrawn between the starting materials and products in an equation Double reaction arrows (equilibrium arrows) Drawn between the starting materials and products in an equilibrium equation Double-headed arrow Drawn between resonance structures Full-headed curved arrow Shows movement of an electron pair Half-headed curved arrow (fishhook) Shows movement of a single electron 䥮-呥硴⁑略獴楯渠2 Use full-headed or half-headed curved arrows to show the movement of electrons in the following equations. 䥮-呥硴⁁湳睥爠2 a.In this reaction, the C‚O bond is broken heterolytically. Because only one electron pair is involved,one full-headed curved arrowis needed. CHM 203 MODULE 2 133 b.This reaction involves radicals, so half-headed curved arrows are needed to show the movement of single electrons. One new two-electron bond is formed between H and Cl, and an unpaired electron is left on C. Because a total of three electrons are involved,three half headed curved arrows are needed. ㌮3噡物潵猠佲条湩挠剥慣瑩潮s Like other compounds, organic molecules undergo the following reactions: ㌮㌮1卵扳瑩瑵瑩潮⁒敡捴楯湳 A substitution reactionis that in which an atom or a group of atoms is replacedby another atom or group of atoms. A general substitution reaction: (Z = H or heteroatom). In a general substitution reaction, Y牥灬慣敳 Z on a carbon atom. Substitu tion reactio n s involve bonds: one bond breaks and 慮潴桥爠景牭⁡琠瑨攠獡浥⁣ 慲扯渠慴潭. The most common examples of substitution occur when Z is hydrogen or a heteroatom that is more electronegative than carbon. Examples: CHM 203 ORGANIC CHEMISTRYII 134 ㌮㌮2䕬業楮慴楯渠剥慣瑩潮s Anelimination reactionis one in which elements of the starting material are †lost‡ and a „ bond is formed. A general elimination reaction can be represented as below: In an elimination reaction, two groups X and Y are removed from a starting material.Tw o bo nds are broken, and a € b ond is f or me d 扥瑷敥渠慤橡捥湴⁡瑯浳. The most common examples of elimination occur when X=H and Y is a heteroatom more electronegative than carbon. Examples: ㌮㌮3䅤摩瑩潮⁒敡捴楯湳 Addition reactions occur whenone or two molecules of the reagent are added to a multiple bond or a small ring of the substrate. The two starting materials add together to form only one product with no atoms left over. A + BŠ A-B In an addition all parts of the adding reagent appear in the product; two molecules become one. When you take an alkene (or alkyne) and add certain types of reagents to them, you get results like this. In this reaction,− − are broken and − − bonds are formed. CHM 203 MODULE 2 135 A typical addition reaction: All addition reactions follow the same pattern (break a „bond, form two single bonds). However, these do not always proceed through the same mechanism.One of the challenges with learning addition reactions is in keepingtrack of which kinds of reagents lead to †regioisomers‡ (i.e. constitutional isomers) and which lead to †stereoisomers‡ as some even lead to both! ㌮㌮4剥摯砠剥慣瑩潮s In these reactions, the reagents bring about a change in the oxidation number of theC-atom of the substrate molecule. A reduction reaction is that in which the oxidation number of the C-atom in the molecule decreases to form the product, while in an oxidation reaction, oxidation number increases. For example, It is a reduction reaction as the oxidation number of the C1-atom of the substrate decreases from 0 to-2. It is an oxidation reaction as the oxidation number of the C1-atom of the substrate increases from-2 to +2. ㌮㌮5剥慲牡湧敭敮琠剥慣瑩潮s In these reactions, the bond connectivity within the molecule changes to give a different compound having same molecular formula. Such a change in the bond connectivity takes place due to the migration of an atom or a group from one location to a new location within the molecule. Such a migration of an atom or a group may take place within the same molecule in Intramolecular processes or may involve two different molecules of the same species in intermolecular processes. For example, CHM 203 ORGANIC CHEMISTRYII 136 䥮-呥硴⁑略獴楯渠3 Classify each reaction as addition, elimination, substitution, or rearrangement. (a)CH3CH2Br + LiOHŠ CH3CH2OH+ LiBr (b)CH3CH2CH2Cl Š CH3CH=CH2+HCl (c)CH3CH=CH2+ H2Š CH3CH2CH3 ㌮㌮6健物捹捬楣⁒敡捴楯湳 In these reactions, the electron reorganization of the substrate molecule takes place through a cyclic transition state in a single-step reaction, giving stereo specific products under theinfluence of either heat or light. For example, CHM 203 MODULE 2 137 In these reactions, the products that are obtained by heating the substrate cannot be formed by irradiation with light and vice versa. The mechanism of a reaction of any one of the above types consists of everything that happens as the starting materials are converted to the products. The reactions may take place in a single step or in a multiple step. In a multiple-step reaction, the product of a step that takes part in the successive step is considered as an intermediate in the overall reaction. For example, B and C are intermediates forthe overall reaction: The reactions take place due to electron reorganization in the reactant molecule. ㌮㌮7䙲敥⁒慤楣慬⁒敡捴楯湳 As discussed earlier in unit 2 of this module, free radicals are chemical species having one or more unpaired electrons in their valence shells. Homolytic bond fission leads to the formation of free radicals. Thus, free radicals are odd electron molecules,e.g. ƒCH3, ƒC2H5, ƒC6H5, (C6H5)3ƒC, etc., and are highly reactive. Free radicals are paramagnetic, that is, they possess a small permanent magnetic moment, due to the presence of unpaired electron(s). This property is used for the detection of the presence of free radicals. 䙯牭慴楯渠潦⁆牥攠剡摩捡汳 Free radicals are formed by homolytic fission of a covalent bond, when a molecule is supplied with sufficient energy‚thermal or photochemical. (i)Thermal Cleavage‚Most of the covalent bonds are thermally stable up to a temperature of 200℃. However, there are a few group of compounds (peroxy and azo) which undergo homolytic cleavage at temperatures below200℃. These are designated as initiators. CHM 203 ORGANIC CHEMISTRYII 138 (ii)Photochemical Cleavage‚A second general method of obtaining radicals is through irradiation with either UV or visible light. The energy transferred to the molecule by the interaction must be of the order of bond dissociation energy or grater to produce homolysis. Alkanes undergo substitution reaction with halogens. The hydrogen atom of the alkane is substituted with a halogen atom to give an alkyl halide in the presence of light or heating at250℃-400℃. This free radical reaction is less vigorous in bromine and more discriminatory. Bromine radical will prefer secondary carbon to the primary. This reaction proceeds via a free radical mechanism. 䵥捨慮楳洠潦⁆牥攠剡摩捡氠剥慣瑩潮 (i) Chain I nitiati on St ep: This is the step in which energy is absorbed and reactive particles are generated. Here the chlorine molecule is broken into chlorine radicals. This cleavage is homolytic. (ii) Chain Propa gation St ep: Here, a reactive particle (free radical) is consumed and another reactive particle is generated. CHM 203 MODULE 2 139 (iii)Chain Termination Step: Here, the reactive particles are consumed but no new particles are generated. 䥮-呥硴⁑略獴楯渠4 Radical chlorination of alkanes is not generally useful because mixtures of products often result when more than one kind of C-H bond is present in the substrate. Draw and name all mono-chloro substitution products C6H13Cl you might obtain by reaction of 2-methylpentane with Cl2. ㌮4呹灥猠潦⁒敡来湴s Many reagents that bring about numerous organic reactions can be classified into two main groups. This classificationis done on the basis of the ability of the reagents to accept or donate electrons to the substrate molecule. The reagent that accept electrons are termed asElectrop hiles (philic meaning loving, that is, electron loving) and often abbreviated by the symbol E "• .The reagents that donate electrons to the electron-deficient species are calledN uc leophiles (nucleus or positive centre loving)and often abbreviated by the symbolN u "– . Electrophi les are electron-deficient species. All positively charged species are electrophilic in nature, for example,H "• , C H "• , " C l "• , N O "• ,and so on. Neutral molecules or species where the central atom is unsaturated with respect to electrons also behave as electrophiles. For example,AlC l , BF , a nd C ̈ H ,where the central atom contains the sextet of electrons. Other neutral molecules where the centralatom can expand its octet to decet or dodecet also behave as electrophiles. For instance,S Fcan form species likeS F , andS iC l and formS iC l "– ,so,S Fand S iC l behave as electrophiles. Transition metal compounds that act as Lewis acids due to the ability of the metal ions to accommodate electrons in thed-AOs also behave as electrophile, for example,ZnC l , S nC l , Fe C l ,and so on. Odd electron species like free radicals where the unpaired electron is accommodated in a low energy. Singly OccupiedMolecular Orbital (SOMO) also act as electrophiles, for example,C H O . Nucleophi les, on the other hand, are electron-rich species. CHM 203 ORGANIC CHEMISTRYII 140 All negatively charged species by nature are Nucleophilic, for example, H "– , C H "– , C l "– , HO "– , HS "– ,and so on. Neutral species that have loosely held electrons as in „-bonds, act as nucleophiles. For example, alkenes, alkynes, and so on, are Nucleophilic in nature. Neutral molecules that have one or more lone pairs on an atom, usually a heteroatom, behave as the nucleophiles. For example,H O ̈ : , N ̈ H ,and so on. Odd electron species where the unpaired electron is accommodated in a high-energy SOMO also behave as nucleophiles, for example,C ̇ H . 䥮-呥硴⁑略獴楯渠5 1.Which of the following is not an electrophi le? a) (CH3)4N + b) Cl2 c) HBr d) Br2 2.Which reagent is a good nucleophile? a) NH3 b) BH3 c) Br2 d) HBr 卅䱆-䅓卅卓䵅乔 䕘䕒䍉卅 i.Classify each of the following reactions as an addition, elimination, substitution, or rearrangement: (a)CH3Br+KOHŠ CH3OH+KBr (b)CH3CH2BrŠ H2C=CH2+HBr (c)H2C=CH2+H2Š CH3CH3 ii Which of the following species are likely to be nucleophiles and which electrophiles? Which might be both? iiiWhat product would you expect from reaction of cyclohexene with HBr? With HCl? ivAdd curved arrows to the following polar reaction to show the flow of electrons: CHM 203 MODULE 2 141 瘮 Add curved arrows to the following polar reactions to indicate the flow ofelectrons in each: 㐮0䍏乃䱕卉低 We have been able to explain how equations for organic reaction are written; the reaction mechanism involving bond cleavage, radicals, carbocation and carbonions, bond formation and kind of arrows. Also the varioustypes of organic reactions with examples and lastly the types of reagents used in organic reactions. 㔮0单䵍䅒Y There are common patterns to how organic reactions occur. In a substitution reaction, one atom or a group of atoms in a substance is replacedby another atom or group of atoms from another substance. Bulky groups that prevent attack cause the reaction to be sterically hindered. In an elimination reaction, adjacent atoms are removed with subsequent formation of a multiple bond and a small molecule. An addition reaction is the reverse of an elimination reaction. Radical reactions are not very selective and occur in three stages: initiation, propagation, and termination. Oxidation-reduction reactions in organic chemistry are identified by the change in the number of oxygen in the hydrocarbon skeleton or the number of bonds between carbon and oxygen or carbon and nitrogen. 㘮0呕呏删䵁剋⁁卓䥇乍䕎T 7.0剅FERENCES/FUR呈䕒 R䕁䑉NGS Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. CHM 203 ORGANIC CHEMISTRYII 142 Dewick, P. M. (2006).Essentials of organic chemistry: for students of pharmacy, medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R. N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009). Chemistry: the central science. Pearson Education. Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New Age International. Okuyama, T., &Maskill, H. (2013). Organic Chemistry: a mechanistic approach. Oxford University Press. Ghatak, K. L. (2014).A Textbook of Organic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons. CHM 203 MODULE 2 143 啎䥔‴ 乕䍌䕏偈䥌䥃⁓啂協䥔啔䥏丠䅎D 䕌䥍䥎䅔䥏丠剅䅃呉低S 䍏乔䕎呓 1.0Introduction 2.0Learning Objectives 3.0Main Content 3.1The Discovery of Nucleophilic Substitution Reactions 3.2The SN2Reaction 3.3Characteristicsof the SN2Reaction 3.3.1The Substrate: Steric Effects in the SN2 Reaction 3.3.2The Nucleophile 3.3.3The Leaving Group 3.3.4The Solvent 3.4The SN1Reaction 3.5Characteristics of SN1 Reaction 3.5.1The Substrate 3.5.2The Leaving Group 3.5.3The Nucleophile 3.5.4The Solvent 3.6Biological Substitution Reactions 3.7Elimination Reactions 3.8The E2Reaction and the Deuterium Isotope Effect 3.9The E2Reaction and Cyclohexane Conformation 3.10The E1and E1CB Reactions 3.10.1The E1 Reaction 3.10.2The E1CB Reactions 3.11Biological Elimination Reactions 4.0Self-Assessment Questions (SAQs) 5.0Conclusion 6.0Summary 7.0References/FurtherReadings ㄮ0䥎呒佄啃呉低 The carbon-halogen bond in an alkyl halide is polar and the carbon atom is electron-poor. Thus, alkyl halides are electrophiles, and much of their chemistry involves polar reactions with nucleophiles and bases. Alkyl halides do one of two things when theyreact with a nucleophile/base, such as hydroxide ion: either they undergosubstitutionof the X group by the nucleophile, or they undergoeliminationof HX to yield an alkene. CHM 203 ORGANIC CHEMISTRYII 144 Substitution Elimination Nucleophilic substitution and base-inducedelimination are two of the most widely occurring and versatile reaction types in organic chemistry, both in the laboratory and in biological pathways. Well look at them closely in this unit to see how they occur, what their characteristics are, and how they can be used. Well begin with nucleophilic substitution reactions. ㈮0佂䩅䍔䥖䕓 By the end ofthis session, you should be able to: · have a better understanding of nucleophilic substitution and elimination reactions. · have a better understanding ofthe types of nucleophilic substitution and elimination reactions. · understand the characteristics of nucleophilic substitution and elimination reactions. · become familiar with biological nucleophilic substitution and eliminationreactions ㌮0䵁䥎⁃低呅乔 ㌮1周攠䑩獣潶敲礠潦⁎畣汥潰桩汩挠卵扳瑩瑵瑩潮⁒敡捴楯湳 The discovery of the nucleophilic substitution reaction of alkyl halides dates back to work carried out in 1896 by the German chemist Paul Walden. Walden found that the pure enantiomeric (+)-and (-)-malic acids could be interconverted through a series of simple substitution reactions. When Walden treated (-)-malic acid with PCl5, he isolated (+)- chlorosuccinic acid. This, on treatment with wet Ag2O, gave (+)-malic acid. Similarly, reaction of (+)-malic acid with PCl5gave (-)- chlorosuccinic acid, which was converted into (-)-malic acid when treated with wet Ag2O. The full cycle of reactions is shown in䙩杵牥′⸶ below. CHM 203 MODULE 2 145 䙩杵牥′⸶: Wa lden‰s cy cle of reaction s inte rconvertin g ( +)-慮搠(-)- 浡汩c慣楤献 At the time, the results were astonishing. The eminent chemist Emil Fischer called Waldens discovery †the most remarkable observation made in the field of optical activity since the fundamental observations of Pasteur.‡ Because (-)-malic acid wasconverted into (+)-malic acid,some reactions in the cyclemust have occurred with a change, or inversion, in configuration at the chirality center. Today, we refer to the transformations taking place in Waldens cycle as nucleophilic substitution reactions because each step involves the substitution of one nucleophile (chloride ion, Cl - , or hydroxide ion, HO - ) by another. Nucleophilic substitution reactions are one of the most common and versatile reaction types in organic chemistry. Following the work of Walden, further investigations were undertaken during the 1920s and 1930s to clarify the mechanism of nucleophilic substitution reactions and to find out how inversions of configuration occur. Among the first series studied was one that interconverted the two enantiomers of 1-phenyl-2-propanol⡆楧畲攠 ㈮㜩. Although this particular series of reactions involves nucleophilic substitution of an alkyl p-toluenesulfonate (called atosylate) rather than an alkyl halide, exactly the same typeof reaction is involved as that studied by Walden. For all practical purposes, the entire tosylate group acts as if it were simply a halogen substituent. (In fact, when you see a tosylate substituent in a molecule, do a mental substitution and tell yourself that youre dealing with an alkyl halide.) CHM 203 ORGANIC CHEMISTRYII 146 䙩杵牥′⸷: 䄠坡汤敮⁣祣汥⁩湴敲捯湶敲瑩湧
 +⤠慮搠(-⤠敮慮瑩潭敲s 潦‱-灨敮祬-2-灲潰慮潬. In the three-step reaction sequence shown in䙩杵牥′⸷, (+)-1-phenyl-2- propanol is interconverted with its (-) enantiomer, so at least one of the three steps must involve an inversion of configuration at the chirality center. Step 1, formation of a tosylate, occurs by breaking the O-H bond of the alcohol rather than the C-O bond to the chiral carbon, so the configuration around carbon is unchanged. Similarly, step 3, hydroxide- ion cleavage of the acetate, takes place without breaking the C-O bond at the chirality center.The inversionof stereo chemical configuration must CHM 203 MODULE 2 147 therefore take place in step 2, the nucleophilicsubstitution of tosylate ion by acetate ion. From this and nearly a dozen other series of similar reactions, workers concluded that the nucleophilic substitution reaction of a primary or secondary alkyl halide or tosylate always proceeds with inversion of configuration. 䥮-呥硴⁑略獴楯渠1 What product would you expect from a nucleophilic substitution reaction of (R)-1-bromo-1-phenylethane with cyanide ion, − "a, as nucleophile? Show the stereochemistry of both reactant and product, assumingthat inversion of configuration occurs. ㌮2周攠SN2剥慣瑩潮 In every chemical reaction, there is a direct relationship between the rate at which the reaction occurs and the concentrations of the reactants. When wemeasure this relationship, we measure the kinetics of the reaction. For example, lets look at the kinetics of a simple nucleophilic substitution‹ the reaction of CH3Br with OH - to yield CH3OH plus Br - . At a given temperature, solvent, and concentrationof reactants, the substitution occurs at a certain rate. If we double the concentration of OH - , the frequency of encounter between the reaction partners doubles and we find that the reaction rate also doubles. Similarly, if we double the concentration of CH3Br, the reaction rate again doubles. We call such a reaction, in which the rate is linearly dependent on the concentrations of two species, a second-order reaction. Mathematically, we can express this second-order dependence of the nucleophilic substitution reaction by setting up arate equation.As either [RX] or [ - OH] changes, the rate of the reaction changes proportionately. Reaction rate=Rate of disappearance of reactant "a ? CHM 203 ORGANIC CHEMISTRYII 148 =kx [RX] x [ - OH] Where [RX] = CH3Br concentration in molarity [ - OH] = - OH concentration in molarity k= A Constant value (the rate constant). The essential feature of the SN2 mechanism is that it takes place in a single step without intermediates when the incoming nucleophile reacts with the alkyl halide or tosylate (thesubstrate) from a direction opposite the group that is displaced (theleaving group). As the nucleophile comes in on one side of the substrate and bonds to the carbon, the halide or tosylate departs from the other side, thereby inverting the stereo chemical configuration. The process is shown in䙩杵牥′⸸ for the reaction of (S)-2-bromobutane with HO - to give (R)-2-butanol. 䵥捨慮楳m The mechanism of the SN2 reaction takes place in a single step when the incoming nucleophile approaches from a direction 180° away from the leaving halide ion, thereby inverting the stereochemistry at carbon. Step 1:The nucleophile‚OH uses its lone-pair electrons to attack the alkyl halide carbon 180° away from the departing halogen. This leads to a transition state with a partially formed C‚OH bond and a partially broken C‚Br bond. Step 2:The stereochemistry at carbon is inverted as the C‚OH bond forms fully and the bromide ion departs with the electron pair from the former C‚Br bond. 䙩杵牥′⸸㨠䵥捨慮楳洠潦⁓ N㈠牥慣瑩潮 As shown in the mechanism above, the SN2 reaction occurs when an electron pair on the nucleophile Nu: - forces out the group X: - , which takes with it the electron pair from the former C-X bond. This occurs through a transition state in which the new Nu-C bond is partially forming at the same time that the old C-X bond is partially breaking and in which the negative charge is sharedby both the incoming nucleophile and the outgoing halide ion. The transition state for this inversion has the remaining three bonds to carbon in a planar arrangement⡆楧畲攠㈮㤩 . CHM 203 MODULE 2 149 䙩杵牥 ㈮㤺 周攠 瑲慮獩瑩潮 獴慴攠 潦 慮 S N㈠ 牥慣瑩潮 桡猠 愠 灬慮慲 慲牡湧e浥湴映瑨攠捡牢潮⁡瑯洠慮搠瑨攠牥浡楮楮朠瑨牥攠杲潵灳. 䕬散瑲潳瑡瑩挠灯瑥湴楡氠浡灳⁳桯眠瑨慴 湥条瑩癥⁣桡牧e 楳⁤敬潣慬楺敤 楮⁴桥⁴牡湳楴楯渠獴慴攮 The mechanism proposed by Hughes and Ingold is fully consistent with experimental results, explaining both stereo chemical and kinetic data. Thus, the requirement for backside approach of the entering nucleophile from a direction 180° away from the departing X group causes the stereochemistry of the substrate to invert, much like an umbrella turning inside out in the wind. The Hughes‚Ingold mechanism also explains why second-order kinetics are found: the SN2 reaction occurs in a single step that involves both alkyl halide and nucleophile. Two molecules are involved in the step whose rate is measured. ㌮3䍨慲慣瑥物獴楣猠潦⁴桥⁓ N㈠剥慣瑩潮 Now that we know how SN2 reactions occur, we need to see how they can be used and what variables affect them. Some SN2 reactions are fast, and some are slow; some take place in high yield and others in low yield. Understanding the factors involved can be of tremendous value. Lets begin by recalling a few things about reaction rates in general. The rate of a chemical reaction is determined by the activation energy ˆG, the energy difference between reactant ground stateand transition state. A change in reaction conditions can affectˆG either by changing the reactant energy level or by changing the transition-state energy level. Lowering the reactant energy or raising the transition-state energy increasesˆG and decreases the reaction rate; raising the reactant energy or decreasing the transition-state energy decreasesˆG and increases the reaction rate⡆楧畲攠㈮㄰) . Well see examples of all these effects as we look at SN2 reaction variables. CHM 203 ORGANIC CHEMISTRYII 150 䙩杵牥′⸱〺 周攠敦晥捴猠潦⁣桡湧敳⁩渠牥慣瑡湴⁡湤⁴牡湳楴楯n -獴慴e 敮敲杹敶敬猠潮⁲敡捴楯渠牡瑥. ⡡)䄠桩杨敲⁲敡捴慮琠敮敲杹敶敬
牥d 捵牶攩⁣潲牥獰潮摳⁴漠愠晡獴敲⁲敡捴楯渠⡳浡汬敲 Š G ‹ ).⡢)䄠桩杨敲 瑲慮獩瑩潮 獴慴攠 敮敲杹 汥癥氠 ⡲敤 捵牶攩 捯牲敳灯湤猠 瑯 愠 s 汯睥r 牥慣瑩潮
污牧敲 Š G ‹ ). ㌮㌮1周攠卵扳瑲慴攺⁓瑥物挠䕦晥捴猠楮⁴桥⁓ N㈠剥慣瑩潮 The first SN2 reaction variable to look at is the structure of the substrate. Because the SN2 transition state involves partial bond formation between the incoming nucleophile and the alkyl halide carbon atom, it seems reasonable that a hindered, bulky substrate should prevent easy approach of the nucleophile, making bond formation difficult. Inother words, the transition state for reaction ofa sterically hindered substrate, whose carbon atom is †shielded‡ from approachof the incoming nucleophile, is higher in energy and forms more slowly thanthe corresponding transition state for a less hindered substrate⡆楧畲攠㈮ㄱ) . 䙩杵牥 ㈮ㄱ: 却敲楣 桩湤牡湣攠 瑯 瑨攠 S N㈠ 牥慣瑩潮⸠ 䅳 瑨攠 浯摥汳 楮摩捡瑥Ⱐ瑨攠捡牢潮⁡瑯洠楮 ⡡)扲潭潭整桡湥⁩猠牥慤楬礠慣捥獳楢汥, 牥獵汴楮朠楮⁡⁦慳琠S N㈠牥慣瑩潮⸠周攠捡牢潮⁡瑯浳⁩n ⡢)扲潭潥瑨慮e ⡰物浡特⤬ ⡣)2-扲潭潰牯灡湥 ⡳散潮摡特⤬ 慮d ⡤)2-扲潭o-2- 浥瑨祬灲潰慮攠⡴敲瑩慲礩⁡牥⁳畣捥獳楶敬礠浯牥⁨楮摥牥搬⁲敳畬瑩湧⁩n 獵捣敳獩癥汹⁳汯睥爠S N㈠牥慣瑩潮献 As䙩杵牥′⸱1 shows, the difficulty of nucleophile approach increases as the three substituents bonded to the halo-substituted carbon atom increase in size. Methyl halides are by far the most reactive substrates in SN2 CHM 203 MODULE 2 151 reactions, followed by primary alkyl halides such as ethyl and propyl. Alkyl branching at the reacting center, as in isopropyl halides (2°), slows the reaction greatly, and further branching, as intert-butyl halides (3°), effectively halts the reaction. Even branching one carbon removed from the reacting center, as in 2,2-dimethylpropyl(neopentyl)halides, greatly slows nucleophilic displacement. As a result, SN2 reactions occur only at relatively unhindered sites and are normally useful only with methyl halides, primary halides, and a few simple secondary halides. Relative reactivities for some different substrates are as follows: Vinylic halides (R2C=CRX) and aryl halides are not shown on this reactivity list because they are unreactive toward SN2 displacement. This lack of reactivity is due to steric factors: the incoming nucleophile would have to approach in the plane of the carbon‚carbon double bond and burrow through part of the molecule to carry out a backside displacement. CHM 203 ORGANIC CHEMISTRYII 152 ㌮㌮2周攠乵捬敯灨楬e Another variable that has a major effect on the SN2 reaction is the nature of the nucleophile. Any species,either neutral or negatively charged, can act as a nucleophile as long as it has an unshared pair of electrons; that is, as long as it is a Lewis base. If the nucleophile is negatively charged, the product is neutral; if the nucleophile is neutral, the product is positively charged. A wide array of substances can be prepared using nucleophilic substitution reactions. Table 2.5 lists some nucleophiles in the order of their reactivity, shows the products of their reactions with bromomethane, andgives the relative rates of their reactions. Clearly, there are large differences in the rates at which various nucleophiles react. What are the reasons for the reactivity differences observed in呡扬攠㈮5? Why do some reactants appear to be much more †nucleophilic‡ than others? The answers to these questions arent straightforward. Part of the problem is that the termnucleophilicityis imprecise. The term is usually taken to be a measure of the affinity of a nucleophile for a carbon atom in the SN2 reaction, but the reactivity of a given nucleophile can change from one reaction to the next. The exact nucleophilicity of a species in a given reaction depends on the substrate, the solvent, and even the reactant concentrations. Detailed explanations for the observed nucleophilicities arent always simple, but some trends can be detected in the data of呡扬e ㈮5. CHM 203 MODULE 2 153 呡扬攠㈮㔺⁓潭攠S N㈠剥慣瑩潮猠睩瑨⁂牯浯浥瑨慮e ƒ Nucleophilicity roughly parallels basicity whencomparing nucleophiles that have the same reacting atom. Thus, OH - is both more basic and more nucleophilic than acetate ion, CH3CO2 - , which in turn is more basic and more nucleophilic than H2O. Since †nucleophilicity‡ is usually taken as the affinity ofa Lewis base for a carbon atom in the SN2 reaction and †basicity‡ is the affinity of a base for a proton, its easy to see why there might be a correlation between the two kinds of behavior. ƒ Nucleophilicity usually increases going down a column of the periodic table. Thus, HS - is more nucleophilic than HO - , and the halide reactivity order is I - .Br - .Cl - . Going down the periodic table, elements have their valence electrons in successively larger shells where they are successively farther from the nucleus, less tightly held, and consequently more reactive. The matter is complex, though, and the nucleophilicity order can change depending on the solvent. ƒ Negatively charged nucleophiles are usually more reactive than neutral ones. As a result, SN2 reactions areoften carried out under basic conditions rather than neutral or acidic conditions. ㌮㌮3周攠䱥慶楮朠䝲潵p Still another variable that can affect the SN2 reaction is the nature of the group displaced by the incoming nucleophile. Because the leaving group is expelled with a negative charge in most SN2 reactions, the best leaving groups are those that best stabilize the negative charge in the transition state. The greater the extent of charge stabilization by the leaving group, the lower the energy of the transition state and the more rapid the reaction. CHM 203 ORGANIC CHEMISTRYII 154 Thus, weak bases such as Cl - , Br - , and tosylate ion make good leaving groups, while strong bases such as OH - and NH2 - make poor leaving groups. Its just as important to know which are poor leaving groupsas to know which are good, and the preceding data clearly indicate that F - , HO - , RO - , and H2N - are not displaced by nucleophiles. In other words, alkyl fluorides, alcohols, ethers, and amines do not typically undergo SN2 reactions. To carry out an SN2 reaction with an alcohol, its necessary to convert the - OH into a better leaving group. This, in fact, is just what happens when a primary or secondary alcohol is converted into either an alkyl chloride by reaction with SOCl2or an alkyl bromide by reaction with PBr3. Alternatively, an alcohol can be made more reactive toward nucleophilic substitution by treating it withpara-toluenesulfonyl chloride to form a tosylate. As noted previously, tosylatesare even more reactive than halides in nucleophilic substitutions. Note that tosylate formation does not change the configuration of the oxygen-bearing carbon because the C-O bond is not broken. CHM 203 MODULE 2 155 The one general exception to the rule that ethers donttypically undergo SN2 reactions occurs with epoxides and the three-membered cyclic ethers. Epoxides, because of the angle strain in the three membered rings, are much more reactive than other ethers. They react with aqueous acid to give 1,2-diols, and theyreact readily with many other nucleophiles as well. Propene oxide, for instance, reacts with HCl to give 1-chloro-2- propanol by SN2 backside attack on the less hindered primary carbon atom. ㌮㌮4周攠卯汶敮t The rates of SN2 reactions are strongly affected by the solvent. Protic solvents‹those that contain an-OH or-NH group-are generally the worst for SN2 reactions, while polar aprotic solvents, which are polar but dont have an‚OH or-NH group, are the best. Protic solvents, such as methanol and ethanol, slow down SN2 reactions by solvation of the reactant nucleophile. The solvent molecules hydrogen bond to the nucleophile and form a cage around it, thereby lowering its energy and reactivity. A solvated anion (reducednucleophilicity due to enhanced ground-state stability). In contrast with protic solvents, which decrease the rates of SN2 reactions by lowering the ground-state energy of the nucleophile, polar aprotic solvents increase the rates of SN2 reactions by raising the ground-state energy of the nucleophile. Acetonitrile (CH3CN), dimethylformamide [(CH3)2NCHO, abbreviated DMF], dimethyl sulfoxide [(CH3)2SO, abbreviated DMSO], and hexamethylphosphoramide {[(CH3)2N]3PO, abbreviated HMPA} are particularly useful. These solvents can dissolve many salts because of their high polarity, but they tend to solvate metal cations rather than nucleophilic anions. As a result, the bare unsolvated anions have a greater nucleophilicity and SN2 reactions take place at CHM 203 ORGANIC CHEMISTRYII 156 correspondingly faster rates. For instance, a rate increase of 200,000 has been observed on changing from methanol to HMPA for the reaction of azide ion with 1-bromobutane. ㌮4周攠SNㄠ剥慣瑩潮 Most nucleophilic substitutions take place by the SN2 pathway just discussed. The reaction is favoured when carried out with an unhindered substrate and a negatively charged nucleophile in a polar aprotic solvent, but is disfavoured when carried out with a hindered substrate and a neutral nucleophile in a protic solvent. You might therefore expect the reaction of a tertiary substrate (hindered) with water (neutral, protic) to be among the slowest of substitution reactions. Remarkably, however, the opposite is true. The reaction of the tertiary halide (CH3)3CBr with H2O to give the alcohol 2-methyl-2-propanol is more than1 million timesas fast as the corresponding reaction of CH3Br to give methanol. A nucleophilic substitution reaction is occurring-a halogen is replacing a hydroxyl group-yet the reactivity order seems backward. These reactions cant be taking place by the SN2 mechanism weve been discussing, and we must therefore conclude that they are occurring by an alternative substitution mechanism. This alternative mechanism is called the SN1 reaction, forsubstitution, nucleophilic, unimolecular. In contrast to the SN2 reaction of CH3Br with OH - , the SN1 reaction of (CH3)3CBr with H2O has a rate that depends only on the alkyl halide concentration and is independent of the H2O concentration. In other CHM 203 MODULE 2 157 words, the reaction is a first-order process; the concentration of the nucleophile does not appear in the rate equation. Reaction rate=Rate of disappearance of alkyl halide =kx [RX] Many organic reactions occur in several steps, one of which usually has a higher-energy transition state than the others and is therefore slower. We call this step with the highest transition-state energy therate-limiting step, orrate determiningstep.No reaction can proceed faster than its rate- limiting step, which acts as a kind of traffic jam, or bottleneck. In the SN1 reaction of (CH3)3CBr with H2O, the fact that the nucleophile concentration does not appear in the first-order rate equation means that it is not involved in the rate limiting step and must therefore be involved in some other, non‚rate-limiting step. 䵥捨慮楳m Step 1:Spontaneous dissociation of the alkyl bromide occurs in a slow, rate-limiting step to generate a carbocation intermediate plus bromide ion. Step 2: The carbocation intermediate reacts with water as nucleophile in a fast step to yield protonated alcohol as product. Step 3: Loss of a proton from the protonated alcohol intermediate then gives the neutral alcohol product. CHM 203 ORGANIC CHEMISTRYII 158 The mechanism of the SN1 reaction of 2-bromo-2-methylpropane with H2O involves three steps. Step-the spontaneous, unimolecular dissociation of the alkyl bromide to yield a carbocation-is rate-limiting. Unlike what happens in an SN2 reaction, where the leaving group is displaced at the same time the incoming nucleophile approaches, an SN1 reaction takes place by loss of the leaving groupbeforethe nucleophile approaches. 2-Bromo-2-methylpropane spontaneously dissociates to the tert-butyl carbocationplus Br - in a slow rate-limiting step, and the intermediate carbocation is then immediately trapped by the nucleophile water in a faster second step. Water is not a reactant in the step whose rate is measured. The energy diagram is shown in䙩杵牥′⸱2. 䙩杵牥 ㈮ㄲ: 䅮 敮敲杹 摩慧牡洠 景爠 慮 S Nㄠ 牥慣瑩潮⸠ 周攠 牡瑥 - 汩浩瑩湧⁳瑥瀠楳⁴桥⁳灯湴慮敯畳⁤楳獯捩慴楯渠潦⁴桥⁡汫祬⁨慬楤攠瑯⁧楶e 愠 捡牢潣慴楯渠 楮瑥牭敤楡瑥⸠ 剥慣瑩潮 潦 瑨攠 捡牢潣慴楯渠 睩瑨 a 湵捬敯灨楬攠瑨敮捣畲猠楮⁡⁳散潮搬⁦慳瑥爠獴数 Becausean SN1 reaction occurs through a carbocation intermediate, its stereo chemical outcome is different from that of an SN2 reaction. Carbocations, as weve seen, are planar,sp2-hybridized, and achiral. Thus, if we carry out an SN1 reaction on one enantiomerof a chiral reactant and go through an achiral carbocation intermediate, the product must lose its optically activity. That is, the symmetrical intermediate carbocation can react with a nucleophile equally well from either side, leading to a racemic, 50;50mixture of enantiomers. CHM 203 MODULE 2 159 The conclusion that SN1 reactions on enantiomerically pure substrates should give racemic products is nearly, but not exactly, what is found. In fact, few SN1 displacements occur with complete racemization. Most give a minor (0‚20%) excess of inversion. The reaction of (R)-6-chloro-2,6- dimethyloctane with H2O, for example, leads to an alcohol product that is approximately 80% racemized and 20% inverted (80%R,S+20%Sis equivalent to 40%R+60% S). This lack of completeracemization in SN1 reactions is due to the fact that ion pairsare involved. If a certain amount of substitution occurs before the two ions fully diffuse apart, then a net inversion of configuration will be observed. ㌮5䍨慲慣瑥物獴楣猠潦⁴桥⁓ Nㄠ剥慣t楯n Just as the SN2 reaction is strongly influenced by the structure of the substrate, the leaving group, the nucleophile, and the solvent, the SN1 reaction is similarly influenced. Factors that lowerˆG, either by lowering the energy level of thetransition state or by raising the energy level of the ground state, favor faster SN1 reactions. Conversely, factors that raiseˆG, either by raising the energy level of the transition state or by lowering the energy level of the reactant, slow down the SN1 reaction. ㌮㔮1周攠卵扳瑲慴e According to the Hammond postulate, any factor that stabilizes a high- energy intermediate also stabilizes the transition state leading to that intermediate. Since the rate-limiting step in an SN1 reaction is the spontaneous, unimolecular dissociation of the substrate to yield a carbocation, the reaction is favored whenever a stabilized carbocation intermediate is formed. The more stable the carbocation intermediate, the faster the SN1 reaction. CHM 203 ORGANIC CHEMISTRYII 160 The stability order of alkyl carbocations is 3°>2°>1° >-CH3. To this list we must also add the resonance-stabilized allyl and benzyl cations. Just as allylic radicals are unusually stable because the unpaired electron can be delocalized over an extendedporbital system, so allylic and benzylic carbocations are unusually stable. An allylic cation has two resonance forms. In one form the double bond is on the †left‡; in the other form its on the †right.‡ A benzylic cation has five resonance forms, all of which contribute to the overall resonance hybrid. 䙩杵牥′⸱㌺ 剥獯湡湣攠景牭猠潦⁡汬祬楣⁡湤⁢敮穹汩挠捡牢潣慴楯湳. 周攠灯獩瑩癥⁣桡牧攠楳⁤敬潣慬楺敤癥爠瑨e p獹獴敭⁩渠扯瑨 Because of resonance stabilization, a primary allylic or benzylic carbocation is about as stable as a secondary alkyl carbocation and a secondary allylic or benzylic carbocation is about as stable as a tertiary alkyl carbocation. This stability order of carbocations is the same as the order of SN1 reactivity for alkyl halides and tosylates. ㌮㔮2周攠䱥慶楮朠䝲潵p We said during the discussion of SN2 reactivity that the best leaving groups are those that are most stable; that is, those that are the conjugate bases of strong acids. An identical reactivity order is found for the SN1 reaction because the leaving group is directly involved in the rate-limiting step. Thus, the SN1 reactivity order is Note that in the SN1 reaction, which is often carried out under acidic conditions, neutral water is sometimes the leaving group. This occurs, for example, when an alkyl halide is prepared from a tertiary alcohol by reaction with HBr or HCl. The alcohol is first protonated and then spontaneously loses H2O to generate a carbocation, which reacts with halide ion to give the alkyl halide. Knowing that an SN1 reaction is involved in the conversion of alcohols to alkyl halides explains why the CHM 203 MODULE 2 161 reaction works well only for tertiaryalcohols. Tertiary alcohols react fastest because they give the most stable carbocation intermediates. 䵥捨慮楳m 却数‱:The‚OH group is first protonated by HBr. 却数′:Spontaneous dissociation of the protonated alcohol occurs in a slow, rate limiting step to yield a carbocation intermediate plus water. 却数″:The carbocation intermediate reacts with bromide ion in a fast step to yield the neutral substitution product. 䙩杵牥′⸱㐺 周攠浥捨慮楳洠潦⁴桥⁓ Nㄠ牥慣瑩潮映愠瑥牴楡特⁡汣潨潬 睩瑨⁈䉲⁴漠祩敬搠慮⁡汫祬⁨慬楤攮⁎敵瑲慬⁷慴敲⁩猠瑨攠汥慶楮朠杲潵p ⡳瑥瀩. CHM 203 ORGANIC CHEMISTRYII 162 ㌮㔮3周攠乵捬敯灨楬e The nature of the nucleophile plays a major role in the SN2 reaction but does not affect an SN1 reaction. Because the SN1 reaction occurs through a rate-limiting step in which the added nucleophile has no part, the nucleophile cant affect the reaction rate. The reaction of 2-methyl-2- propanol with HX, for instance, occurs at the same rate regardless of whether X is Cl, Br,or I. Furthermore, neutral nucleophiles are just as effective as negatively charged ones, so SN1 reactions frequently occur under neutral or acidic conditions. ㌮㔮4周攠卯汶敮t What about the solvent? Do solvents have the same effect in SN1reactions that they have in SN2 reactions? The answer is both yes and no. Yes, solvents have alarge effect on SN1 reactions, but no, the reasons for the effects on SN1 andSN2 reactions are not the same. Solvent effects in the SN2 reaction are due largelyto stabilization or destabilization of the nucleophilereactant,while solventeffects in the SN1 reaction are due largely to stabilization or destabilization ofthetransition state. The properties of a solvent that contribute to its ability to stabilize ions by solvation are related to the solvents polarity. SN1 reactions take place much more rapidly in strongly polar solvents, such as water and methanol, than in less polar solvents, such as ether and chloroform. In the reaction of 2-chloro-2-methylpropane, for example, a rate increase of 100,000 is observed on going from ethanol (less polar) to water (more polar). The rate increases on going from a hydrocarbon solvent to water are so large they cant be measured accurately. CHM 203 MODULE 2 163 Both the SN1 and the SN2reaction show solvent effects, but they do so for different reasons. SN2 reactions aredisfavouredin protic solvents because theground-state energyof the nucleophile is lowered by solvation. SN1 reactions arefavouredin protic solvents because thetransition-state energyleading to carbocation intermediate is lowered by solvation. 䥮-呥硴⁑略獴楯渠2 Predict whether each of the following substitution reactions is likely to be SN1 or SN2: ㌮6䉩潬潧楣慬⁓畢獴楴畴楯渠剥慣瑩潮s Both SN1 and SN2 reactions are well-known in biological chemistry, particularly in the pathways for biosynthesis of the many thousands of plant-derived substances calledterpenoids. Unlike what typically happens in the laboratory, however, the substrate in a biological substitution reaction is usually an organodiphosphate rather than an alkyl halide. Thus, the leaving group is the diphosphate ion, abbreviated PPi, rather than a halide ion. In fact, its useful to think of the diphosphate group as the †biological equivalent‡ of a halogen. The dissociation of an organodiphosphate in a biological reaction is typically assisted by complexation to a divalent metal cation such as Mg 2+ to help neutralize charge and make the diphosphate a better leaving group. CHM 203 ORGANIC CHEMISTRYII 164 ㌮7 El i mi na tio n Reactions : Za i ts ev ‰s Rul e We said at the beginning of this unit that two kinds of reactions can take place when a nucleophile/Lewis base reacts with an alkyl halide. The nucleophile can either substitute for the halide by reaction at carbon or can cause elimination of HX by reaction at neighbouring hydrogen. Elimination reactions are more complex than substitution reactions for several reasons. One is the problem of regiochemistry. What products result by loss of HX from an unsymmetrical halide? In fact, elimination reactions almost always give mixtures of alkene products, and the best we can usually do is to predict which will be the major product. According to Zaitsevs rule, formulated in 1875 by the Russian chemist Alexander Zaitsev, base-induced elimination reactions generally (although not always) give the more stable alkene product-that is, the alkene with more alkyl substituents on the double-bond carbons. In the following two cases, for example, the more highly substituted alkene product predominates. Za itsev‰s Rule In the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates. Another factor that complicates a study of elimination reactions is that they can take place by different mechanisms, just as substitutions can. Well consider three of the most common mechanisms-the E1, E2, and CHM 203 MODULE 2 165 E1cB reactions-which differ in the timing of C-H and C-X bond- breaking. In the E1 reaction, the C-X bond breaks first to give a carbocation intermediate that undergoes subsequent base abstraction of H + to yield the alkene. In the E2 reaction, base-induced C-H bond cleavage is simultaneous withC-X bond cleavage, giving the alkene in a single step. In the E1cB reaction (cB for †conjugate base‡), base abstraction of the proton occurs first, giving a carbanion (R: - ) intermediate. This anion, the conjugate base of the reactant †acid,‡ then undergoesloss of X - in a subsequent step to give the alkene. All three mechanisms occur frequently in the laboratory, but the E1cB mechanism predominates in biological pathways. 䔱⁒敡捴楯渺 C‚X bond breaks first to give a carbocation intermediate, followed by base removal of a proton to yield the alkene. 䔲 剥慣瑩潮: C‚H and C‚X bonds break simultaneously, giving the alkene in a single step without intermediates. 䔱捂⁒敡捴楯渺 C‚H bond breaks first, giving a carbanion intermediate that loses X‚to formthe alkene. 䥮-呥硴⁑略獴楯渠3 What product would you expect from reaction of 1-chloro-1- methylcyclohexane with KOH in ethanol? CHM 203 ORGANIC CHEMISTRYII 166 ㌮8周攠䔲⁒敡捴楯渠慮搠瑨攠䑥畴敲極洠䥳潴潰攠䕦晥捴 The E2 reaction (forelimination, bimolecular) occurs when an alkyl halide is treated with a strong base, such as hydroxide ion or alkoxide ion (RO - ). It is the most commonly occurring pathway for elimination. 䵥捨慮楳m Step 1:Base (B:) attacks a neighboring hydrogen and begins to remove the H at the same time as the alkene double bond starts to form and the X group starts to leave. Step 2: Neutral alkene is produced when the C‚H bond is fully broken and the X group has departed with the C‚X bond electron pair. 䙩杵牥′⸱㔺 䵥捨慮楳洠潦⁴桥⁅2 牥慣瑩潮映慮⁡汫祬⁨慬楤攮⁔桥 牥慣瑩潮 瑡步猠 灬慣攠 楮 愠 獩湧汥 獴数 瑨牯畧栠 愠 瑲慮獩瑩潮 獴慴攠 楮 睨楣栠瑨攠摯畢汥⁢潮搠扥杩湳⁴漠景牭⁡琠瑨攠獡浥⁴業攠瑨攠䠠慮搠X 杲潵灳⁡牥敡癩湧. Like the SN2 reaction, the E2 reaction takes place in one step without intermediates. As the base begins to abstract H + from a carbon next to the leaving group, the C-H bond begins to break, a C=C bond begins to form, and the leaving group begins to depart, taking with it the electron pair from the C-X bond. Among the pieces of evidence supporting this mechanism is that E2 reactions show second-order kinetics and follow the CHM 203 MODULE 2 167 rate law: rate=kx[RX]x[Base]. That is, both base and alkyl halide takes part in the rate-limiting step. A second piece of evidence in support ofthe E2 mechanism is provided by a phenomenon known as the deuterium isotope effect. For reasons that we wont go into, a carbon‚hydrogen bond is weaker by about 5 kJ/mol (1.2 kcal/mol) than the corresponding carbon‚deuterium bond. Thus, a C- H bond is moreeasily broken than an equivalent C-D bond, and the rate of C-H bond cleavage is faster. For instance, the base-induced elimination of HBr from 1-bromo-2-phenylethane proceeds 7.11 times as fast as the corresponding elimination of DBr from 1-bromo-2,2-dideuterio-2- phenylethane. This result tells us that the C-H (or C-D) bond is broken in the rate-limiting step, consistent with our picture of the E2 reaction as a one-step process. If it were otherwise, we couldnt measure a rate difference. Yet a third piece of mechanistic evidence involves the stereochemistry of E2 eliminations. As shown by a large number of experiments, E2 reactions occur withperiplanargeometry, meaning that all four reacting atoms‹the hydrogen, the two carbons, and the leaving group‹lie in the same plane. Two such geometries are possible: syn periplanar geometry, in which the H and the X are on the same side of the molecule, and anti periplanar geometry, in which the H and the X are on opposite sides of the molecule. Of the two, anti periplanar geometry is energetically preferred because it allows the substituents on the two carbons to adopt a staggered relationship, whereas syn geometry requires that the substituents be eclipsed. 䥮-呥硴⁑略獴楯渠4 What stereochemistry do you expectfor the alkene obtained by E2 elimination of (1S,2S)-1,2-dibromo-1,2-diphenylethane? ㌮9周攠䔲⁒敡捴楯渠慮搠䍹捬潨數慮攠䍯湦潲浡瑩潮 Anti periplanargeometry for E2 reactions is particularly important in cyclohexane rings, where chair geometry forces a rigid relationship between the substituents on neighboring carbon atoms. The anti CHM 203 ORGANIC CHEMISTRYII 168 periplanar requirement for E2 reactions overrides Zaitsevs rule andcan be met in cyclohexanes only if the hydrogen and the leaving group are trans diaxial. If either the leaving group or the hydrogen is equatorial, E2 elimination cant occur. The difference in reactivity between the isomeric methyl chlorides is due to the difference in their conformations. Neomethyl chloride has a conformation with the methyl and isopropyl groups equatorial and the chlorine axial-a perfect geometry for E2 elimination. Loss of the hydrogen atom at C4 occurs easily to yield the more substituted alkene product, 3-menthene, as predicted by Zaitsevs rule. Methyl chloride, by contrast, has a conformation in which all three substituents are equatorial. To achieve the necessary geometry for elimination, methyl chloride must first ring-flip toa higher-energy chair conformation, in which all three substituents are axial. E2 elimination then occurs with loss of the only trans-diaxial hydrogen available, leading to the non-Zaitsev product 2-methene. The net effect of the simple change in chlorinestereochemistry is a 200-fold change in reaction rate and a complete change of product. The chemistry of the molecule is controlled by its conformation. ㌮㄰周攠䔱⁡湤⁅ㅣ䈠剥慣瑩潮s ㌮㄰⸱⁔桥⁅ㄠ剥慣瑩潮 Just as the E2 reaction is analogous to the SN2 reaction, the SN1 reaction has a close analog called the E1 reaction (forelimination, unimolecular). The E1 reaction can be formulatedfor the elimination of HCl from 2- chloro-2-methylpropane. 䵥捨慮楳m Step 1: Spontaneous dissociation of the tertiaryalkyl chloride yields an intermediate carbocation in a slow, rate-limiting step. CHM 203 MODULE 2 169 Step 2: Loss of a neighboring H + in a fast step yields the neutral alkene product. The electron pair from the C‚H bond goes to form the alkene€ bond. 䙩杵牥′⸱㘺 䵥捨慮楳洠潦⁴桥⁅ㄠ牥慣瑩潮⸠呷漠獴数猠慲攠楮癯汶敤, 瑨攠晩牳琠潦⁷桩捨⁩猠牡瑥 -汩浩瑩湧Ⱐ慮搠愠捡牢潣慴楯渠楮瑥牭敤楡瑥⁩s 灲敳敮琮 E1 eliminations begin with the same unimolecular dissociation to give a carbocation that we saw in the SN1 reaction, butthe dissociation is followed by loss of H + from the adjacent carbon rather than by substitution. In fact, the E1 and SN1 reactions normally occur together whenever an alkyl halide is treated in a protic solvent with a nonbasic nucleophile. Thus, the best E1 substrates are also the best SN1 substrates, and mixtures of substitution and elimination products are usually obtained. For example, when 2-chloro-2-methylpropane is warmed to 65°C in 80% aqueous ethanol, a 64;36 mixture of 2-methyl-2-propanol (SN1) and2-methylpropene (E1) results. Much evidence has been obtained in support of the E1 mechanism. For example, E1 reactions show first-order kinetics, consistent with a rate- CHM 203 ORGANIC CHEMISTRYII 170 limiting, unimoleculardissociation process. Furthermore, E1 reactions show no deuterium isotope effect because rupture of the C-H (or C-D) bond occurs after the rate-limiting step rather than during it. Thus, we cant measure a rate difference between a deuterated and nondeuterated substrate. A final piece of evidence involves the stereochemistry of elimination. Unlike the E2 reaction, where anti periplanar geometry is required, there is no geometric requirement on the E1 reaction because the halide and the hydrogen are lost in separate steps. We might therefore expect to obtain the more stable (Zaitsevs rule) product from E1 reaction, which is just what we find. To return to a familiar example, methyl chloride loses HCl under E1 conditions in a polar solvent to give a mixtureof alkenes in which the Zaitsev product, 3-menthene, predominates. ㌮㄰⸲周攠䔱捂⁒敡捴楯n In contrast to the E1 reaction, which involves a carbocation intermediate, the E1cB reaction takes place through a carbanion intermediate. Base- induced abstraction of a proton in a slow, rate-limiting step gives an anion, which expels a leaving group on the adjacent carbon. The reaction is particularly common in substrates that have a poor leaving group, such as -OH, two carbons removed from a carbonyl group, HO-C-CH-C=O. The poor leaving group disfavors the alternative E1 and E2 possibilities, and the carbonyl group makes the adjacent hydrogen unusually acidic by resonance stabilization ofthe anion intermediate. ㌮ㄱ䉩潬潧楣慬⁅汩浩湡瑩潮⁒敡捴楯湳 Allthree elimination reactions-E2, E1, and E1cB-occur in biological pathways, but the E1cB mechanism is particularly common. The substrate is usually an alcohol rather than an alkyl halide, and the H atom removed is usually adjacent to a carbonyl group, just as in laboratory reactions. Thus, 3-hydroxy carbonyl compounds are frequently converted to unsaturated carbonyl compounds by elimination reactions. A typical example occurs during the biosynthesis of fats when a 3-hydroxybutyryl thioester is dehydratedto the corresponding unsaturated (crotonyl) thioester. The base in this reaction is a histidine amino acid in the enzyme, and loss of the-OH group is assisted by simultaneous protonation. CHM 203 MODULE 2 171 䥮-呥硴⁑略獴楯渠5 Tell whether each of the following reactions is likely to be SN1, SN2, E1, E1cB, or E2, and predict the product of each: 卅䱆-䅓卅卓䵅乔 䕘䕒䍉卅 i.What product would you expect to obtain from a nucleophilic substitution reaction of (S)-2-bromohexane with acetate ion, CH3CO2 - ? ii.What productwould you expect to obtain from SN2 reaction of OH - with (R)-2-bromobutane? Show the stereochemistry of both reactant and product. iiiWhat product would you expect from SN2 reaction of 1- bromobutane with each of the following? ⡡)NaI ⡢)KOH ⡣) H − C "a C − Li⡤)NH3 CHM 203 ORGANIC CHEMISTRYII 172 ivWhich substance in each of the following pairs is more reactive as a nucleophile? Explain. ⡡)(CH3)2N - or (CH3)2NH⡢)(CH3)3B or (CH3)3N⡣)H2O or H2S 瘮 Rank the following compounds in order of their expected reactivity toward SN2reaction: CH3Br, CH3OTos, (CH3)3CCl, (CH3)2CHCl viWhat product(s) would you expect from reaction of (S)-3-chloro- 3-methyloctane with acetic acid? Show the stereochemistry of both reactant and product. ViiRank the following substances in order of theirexpected SN1 reactivity: 㐮0䍏乃䱕卉低 In this unit, we have succeeded in explaining that nucleophilic substitution and base-induced elimination are two of the most widely occurring and versatile reaction types in organic chemistry, both in the laboratory and in biological pathways. 㔮0单䵍䅒Y The effects on SN1 reactions of the four variables-substrate, leaving group, nucleophile, and solvent-are summarized in the following statements: Substrate:The best substrates yield the most stable carbocations. As a result, SN1 reactions are best for tertiary, allylic, and benzylic halides. Leaving group: Good leaving groups increase the reaction rate by lowering the energy level of the transition state for carbocation formation. Nucleophile: The nucleophile must be nonbasic to prevent a competitive elimination of HX, but otherwise does not affect the reaction rate. Neutral nucleophiles work well. Solvent: Polar solvents stabilize the carbocation intermediate by solvation, thereby increasing the reaction rate. SN1, SN2, E1, E1cB, E2‹how can you keep it all straight and predict what will happen in any given case? Will substitution or elimination occur? Will the reaction be bimolecular or unimolecular? There are no CHM 203 MODULE 2 173 rigid answers to these questions, but its possible to recognize some trends and make some generalizations. ƒ Primary alkyl halides: SN2 substitution occurs if a good nucleophile is used, E2 elimination occurs if a strong, sterically hindered baseis used, and E1cB elimination occurs if the leaving group is two carbons away from a carbonyl group. ƒ Secondary alkyl halides: SN2 substitution occurs if a weakly basic nucleophile is used in a polar aprotic solvent, E2 elimination predominates if a strongbase is used, and E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group. Secondary allylic and benzylic alkyl halides can also undergo SN1 and E1 reactions if a weakly basic nucleophile is used in a protic solvent. ƒ Tertiary alkyl halides: E2 elimination occurs when a base is used, but SN1 substitution and E1 elimination occur together under neutral conditions, such as in pure ethanol or water. E1cB elimination takes place if the leaving group is two carbons away froma carbonyl group. 㘮0呕呏删䵁剋⁁卓䥇乍䕎T 7.0剅FERENCES/FUR呈䕒 R䕁䑉NGS Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. Dewick, P. M. (2006).Essentials of organic chemistry: for students of pharmacy,medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R. N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009). Chemistry: the central science. Pearson Education. Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New Age International. Okuyama, T., & Maskill, H. (2013). Organic Chemistry: a mechanistic approach. Oxford University Press. Ghatak, K. L. (2014).A Textbook ofOrganic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. CHM 203 ORGANIC CHEMISTRYII 174 Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons. CHM 203 ORGANIC CHEMISTRYII 175 䵏䑕䱅″ 䅒位䅔䥃⁃位偏啎䑓 䥎呒佄啃呉低 Compounds which have relatively few hydrogens in relation to the number of carbons, such as benzene, are typically found in oils produced by trees and other plants. Early chemists called such compounds慲潭慴楣 compounds because of their pleasing fragrances. In this way, they were distinguished from aliphatic compounds, with higher hydrogen-to-carbon ratios, that were obtained from the chemical degradation of fats. The chemical meaning of the word €aromatic nowsignifies certain kinds of chemical structures. This module will lead us to the discussion of two units where we will examine the criteria that a compound must satisfy to be classified as aromatic and their reaction types. These units are: Unit 1 Benzene and other Aromatic Compounds Unit 2 Reactions in Aromatic Compounds UN䥔‱ 䉅乚䕎䔠 䅎䐠 佔䡅删 䅒位䅔䥃 䍏䵐何乄S 䍏乔䕎呓 1.0Introduction 2.0Learning Objectives 3.0Main Content 3.1Isolation of Benzene 3.2The Structure of Benzene 3.2.1Resonance 3.2.2Hybridization and Orbitals 3.3Nomenclature of Benzene Derivatives 3.3.1Monosubstituted Benzenes 3.3.2Disubstituted Benzenes 3.3.3Polysubstituted Benzenes 3.3.4Naming Aromatic Rings as Substituents 3.4Interesting Aromatic Compounds 3.5Stability of Benzene 3.6The Criteria for Aromaticity-H‚ckelƒs Rule 3.7Examples of Aromatic Compounds 3.7.1Aromatic Compounds with a Single Ring 3.7.2Aromatic Compounds with More Than One Ring 3.7.3Aromatic Heterocycles CHM 203 MODULE 3 176 3.7.4Charged Aromatic Compounds 3.8Basis of H‚ckelƒs Rule 3.8.1Bonding and Antibonding Orbitals 3.8.2Molecular Orbitals Formed When More Than Two p Orbitals Combine 4.0Conclusion 5.0Summary 6.0Tutor Mark Assignemt 7.0References/FurtherReadings ㄮ0䥎呒佄啃呉低 Benzene (C6H6) is the simplest aromatic hydrocarbon (or arene).Since its isolation by Michael Faraday from the oily residue remaining in the illuminating gas lines in London in 1825, it has been recognized as an unusual compound. Benzene has four degrees of unsaturation, making it a highly unsaturated hydrocarbon.But, whereas unsaturated hydrocarbons such as alkenes, alkynes, and dienes readilyundergo addition reactions, benzene does not.For example, bromine adds to ethylene to forma dibromide, but benzene is inert under similar conditions. Ethylene addition product Benzene Benzenedoesreact with bromine, but only in the presence of FeBr3(a Lewis acid), and the reaction is a獵扳瑩瑵瑩潮, notan addition. Substitution: Br replaces H Thus, any structure proposed for benzene must account for its high degree of unsaturation and its lack of reactivity towards electrophilic addition. CHM 203 ORGANIC CHEMISTRYII 177 For 6Cƒs, the maximum number of Hƒs = 2n+ 2 = 2(6) + 2 = 14.Because benzene contains only 6Hƒs, it has 14„6 = 8 Hƒs fewer than the maximum number. This corresponds to ’ ’ for each degree of unsaturation =景畲 摥杲敥猠潦⁵湳慴畲慴楯n 楮⁢敮穥湥. In the last half of the nineteenth century August Kekuléproposed structures that were close to the modern description of benzene. In the Kekulé model, benzene was thought to be a rapidly equilibrating mixture of two compounds, each containing a six-membered ring with three alternating … bonds. These structuresare now called䭥歵泩⁳瑲畣瑵牥献 In the Kekulé description, the bond between any two carbon atoms is sometimes a single bond and sometimes a double bond. Kekulé description: An equilibrium Although benzene is still drawn as a six-membered ring with three alternating … bonds, in reality瑨敲攠楳漠敱畩汩扲極洠扥瑷敥渠瑷漠摩晦敲敮琠歩湤猠潦 扥湺敮攠浯汥捵汥献 Instead, current descriptions of benzene are based on resonance and electron delocalization due to orbital overlap. In the nineteenth century, many other compounds having properties similar to those of benzene were isolated from natural sources. Because these compounds possessed strong and characteristic odours, they were called arom atic com pounds. It is their chemical properties, though, not their odour that make these compounds special. † Aromatic compounds resemble benzene-they are unsaturated compounds that do not undergo the addition reactions characteristic of alkenes. ㈮0佂䩅䍔䥖䕓 When you have studied this session, you should be able to: · understand the processes of benzene isolation. · get acquainted with the nomenclature of benzene and benzene derivatives. · understand the stability of benzene and the criteria forAromaticity. CHM 203 MODULE 3 178 ㌮0䵁䥎⁃低呅乔 ㌮1䥳潬慴楯渠潦⁂敮穥湥 Benzene and other arenes can be obtained by distilling coal. This is a particular messy process in the laboratory, and requires a lengthy business of separating the products from one another. However, in industry it is an economic way of isolating benzene. There is strong demand for coke, which is produced by heating coal in the absence of air. For every tone of coal turned into coal, about 70 dm 3 of coal tar is made. This is an oily liquid, which contains a variety of products. If the coal tar is separated by fractional distillation, around 30 dm 3 of benzene can be collected. Methylbenzene, naphthalene and anthracene are also obtained in smaller quantities. In the laboratory a quicker way to make benzene is to heat the calcium salt of benzoic acid, (C6H5COOH)2Ca, with soda lime (soda lime contains calcium hydroxide together with sodium hydroxide): (C6H5COO)2Ca(s) + Ca(OH)2(s)‡ 2C6H6(l) + 2CaCO3(s) ㌮2周攠却牵捴畲攠潦⁂敮穥湥 Any structure for benzene must account for the following: † It contains a six-membered ring and three additional degrees of unsaturation. † It is planar. † All C-C bond lengths are equal. Although the Kekuléstructures satisfy the first two criteria, they break down with the third, because having three alternating … bonds means that benzene should have three short double bonds alternating with three longer single bonds. ㌮㈮1剥獯湡湣e Benzene is conjugated, so we must use resonance and orbitals to describe its structure. The resonance description of benzene consists of two equivalent CHM 203 ORGANIC CHEMISTRYII 179 Lewis structures, each with three double bonds that alternate with three single bonds. The resonance description of benzene matches the Kekulé description with one important exception.The two Kekulé representations arenotin equilibrium with each other.Instead, the true structure of benzene is a resonancehybridof the two Lewis structures, with thedashed lines of the hybrid indicating the position of the … bonds. The resonance hybrid of benzene explains why all C„C bond lengths are the same. Each C„C bond is single in one resonance structure and double in the other, so the actual bond length (139 pm) is intermediate between a carbon„carbon single bond (153 pm) and a carbon„carbon double bond (134 pm). The C„C bonds in benzene are equal and intermediate in length. ㌮㈮2䡹扲楤楺慴楯渠慮搠佲扩瑡汳 Each carbon atom in a benzene ring is surrounded by three atoms and no lone pairs of electrons, making itsp2 hybridized and trigonal planar with all bond angles 120°.Each carbon also has aporbital with one electron that extends above and below the plane of the molecule. The six adjacentporbitals overlap, delocalizing the six electrons over the six atoms of the ring and making benzene a conjugated molecule. Because each porbital has two lobes, one above and one below the plane of the benzene ring, the overlap of theporbitals creates two €doughnuts of electron density. CHM 203 MODULE 3 180 † Benzene€s six elect ro ns make it ele ctron rich and s o it readily reacts with electrophiles. 䥮-呥硴⁑略獴楯渠1 In principle, which of the following is true regarding benzene and1,3,5- cyclohexatriene? a.Theoretically they are the same molecules b.Both have same length of all their C-C bonds c.Both have same enthalpy of hydrogenation d.Cyclohexatriene has three different C-C bond lengths while benzene has only onetype of C-C bond length whose value is between those of cyclohexatriene ㌮3乯浥湣污瑵牥映䉥湺敮攠䑥物癡瑩癥s Many organic molecules contain a benzene ring with one or more substituents, so we must learn how to name them. Many common names are recognized by the IUPAC system, however, so this complicates the nomenclature of benzene derivatives somewhat. ㌮㌮1䵯湯獵扳瑩瑵瑥搠䉥湺敮敳 To name a benzene ring with one substituent,name the substituent and add the wordbenzene.Carbon substituents are named as alkyl groups. Many monosubstituted benzenes, such as those with methyl (CH3„), hydroxy („OH), and amino („NH2) groups, have common names. 䥮-呥硴⁑略獴楯渠2 CHM 203 ORGANIC CHEMISTRYII 181 Draw the structure of each of the following compounds: a. 2-phenylhexane b. benzyl alcohol c.3-benzylpentane d. bromomethylbenzene ㌮㌮2䑩獵扳瑩瑵瑥搠䉥湺敮敳 There are three different ways that two groups can be attached to a benzene ring, so a prefix-潲瑨漬整愬 or灡牡-can be used to designate the relative position of the two substituents. Ortho, meta, and para are also abbreviated as o, m , and p,respectively. If the two groups on the benzene ring are different,alphabetize the names of the substituentsprecedingthe wordbenzene.If one of the substituents is part of acommon root,name themolecule as a derivative of that monosubstituted benzene. Alphabetize two different substituent names: Use a common root name: CHM 203 MODULE 3 182 ㌮㌮3偯汹獵扳瑩瑵瑥搠䉥湺敮敳 Forthree or more substituents on a benzene ring: · Number to give the lowest possible numbers around the ring. · Alphabetize the substituent names. · When substituents are part of common roots, name the molecule as a derivative of that monosubstitutedbenzene. The substituent that comprises the common root is located at C1. Examples of naming polysubstituted benzenes 4-chloro-1-ethyl-2-propylbenzene · Assign the lowest set of numbers. · Alphabetize the names of all the substituents. 2,5-dichloroaniline · Name the molecule as a derivative of the common roota niline. · Designate the position of the NH2group as €1 and then assign the lowest possible set of numbers to the other substituents. 䥮-呥硴⁑略獴楯渠3 Tell whether the followingcompounds are ortho-, meta-, or para- disubstituted: CHM 203 ORGANIC CHEMISTRYII 183 ㌮㌮4乡浩湧⁁牯浡瑩挠剩湧猠慳⁓畢獴楴略湴s A benzene substituent (C6H5„) is called aphenyl group,and it can be abbreviated in a structure asPh„. † A phenyl group (C6H5 ‚) is formed by removingone hydrogen from benzene (C6H6). Benzene, therefore, can be represented as PhH, and phenol would be PhOH. Thebenzylgroup, another common substituent that contains a benzene ring, differs from a phenyl group. Finally, substituents derived from other substituted aromatic rings are collectively called慲祬⁧牯異献 Examples of aryl groups: CHM 203 MODULE 3 184 ㌮4䥮瑥牥獴楮朠䅲潭慴楣⁃潭灯畮摳 Benzeneandtoluene,the simplest aromatic hydrocarbons obtained from petroleum refining, are useful starting materials for synthetic polymers. They are two components of thearomatic compoundmixture added to gasoline to boost octane ratings. 偯汹捹捬楣 慲潭 慴楣 桹摲潣慲扯湳 ⡐䅈猩. Naphthalene, the simplest PAH, is the active ingredient in mothballs. 䉥湺潛 a嵰祲敮攬a more complicated PAH is formed by the incomplete combustion of organic materials. It is found in cigarette smoke, automobile exhaust, and thefumes from charcoal grills. When ingested or inhaled, benzo[a]pyrene and other similar PAHs are oxidized to carcinogenic products. 䡥汩捥湥 and瑷楳瑯晬數are two synthetic PAHs. Heliceneconsists of six benzene rings. Because the rings at both ends are not bonded to each other, all of the rings twist slightly, creating a rigid helical shape that prevents the hydrogen atoms on both ends from crashing into each other. Similarly, to CHM 203 ORGANIC CHEMISTRYII 185 reduce steric hindrance between the hydrogen atoms on nearby benzene rings, twistoflex is also nonplanar. † Benzo[a]pyrene, produced by the incomplete oxidation of organic compounds in tobacco, is found in cigarette smoke. ㌮5却慢楬楴礠潦⁂敮穥湥 Consideringbenzene as the hybrid of two resonance structures adequately explains its equal C-C bond lengths, but does not account for its unusual stability and lack of reactivity towards addition. Heats of hydrogenation, used to show that conjugated dienes are morestable than isolated dienes, can also be used to estimate the stability of benzene. Equations (1)-(3) compare the heats of hydrogenation of cyclohexene, 1,3- cyclohexadiene, and benzene, all of which give cyclohexane when treated with excess hydrogen in the presence of a metal catalyst. The addition of one mole of H2to cyclohexene releases-120 kJ/mol of energy. If each double bond is worth-120 kJ/mol of energy, then the addition of two moles of H2to 1,3-cyclohexadiene should release 2 ×-120 kJ/mol =- 240 kJ/mol of energy. The observed value, however, is-232 kJ/mol. This is slightly smaller than expected because 1,3-cyclohexadiene is a conjugated CHM 203 MODULE 3 186 diene, and conjugated dienes are more stable thantwo isolated carbon-carbon double bonds. The hydrogenations of cyclohexene and 1,3-cyclohexadiene occur readily at room temperature, but benzene can be hydrogenated only under forcing conditions, and even then the reaction is extremely slow. If each double bond is worth-120 kJ/mol of energy, then the addition of three moles of H2to benzene should release 3 ×-120 kJ/mol =-360 kJ/mol of energy. In fact, the observed heat of hydrogenation is only-208 kJ/mol, which is 152 kJ/mol less than predicted and even lower than the observed value for 1,3- cyclohexadiene. The huge difference between the hypothetical and observed heats of hydrogenation for benzene cannot be explained solely on the basis of resonance and conjugation. † The low heat of hydrogenation of benzene means that benzene is especially stable, even more so than the conjugated. This unusual stability is characteristic of aromatic compounds. Benzeneƒs unusual behaviour in chemical reactions is not limited to hydrogenation. Benzene does not undergo addition reactions typical of other highly unsaturatedcompounds, including conjugated dienes.Benzene does not react with Br2to yield an addition product. Instead, in the presence of a Lewis acid, brominesubstitutesfor a hydrogen atom, thus yielding a product that retains the benzene ring. Addition does not occur.An addition product would no longer contain a benzene ring. Addition occurs.A substitution product still contains a benzene ring This behavior is characteristic of aromaticcompounds. CHM 203 ORGANIC CHEMISTRYII 187 ㌮6周攠䍲楴敲楡⁦潲⁁牯浡瑩捩瑹 - H€ckel s Rul e Four structural criteria must be satisfied for a compound to be aromatic: 䄠浯汥捵汥畳琠扥⁣祣汩挮 † To be aromatic, eachporbital must overlap withporbitals on two adjacent atoms. Theporbitals on all six carbons of benzene continuously overlap, so benzene is aromatic. 1,3,5-Hexatriene has sixporbitals, too, but the two on the terminal carbons cannot overlap with each other, soㄬ㌬5- 桥硡瑲楥湥⁩猠湯琠慲潭慴楣. 䄠浯汥捵汥畳琠扥⁰污湡 爮 † All adjacentporbitals must be aligned so that the o electron density can be delocalized. For example, cyclooctatetraene resembles benzene in that it is a cyclic molecule with alternating double and single bonds. Cyclooctatetraene is tub shaped, however,湯琠 灬慮慲, so overlap between adjacent … bonds is impossible.䍹捬潯捴慴整牡敮攬⁴桥牥景牥Ⱐ楳 not慲潭慴楣,so it undergoes addition reactions like those of other alkenes. 䄠浯汥捵汥畳琠扥⁣潭灬整敬礠捯湪畧慴敤. Both 1,3-cyclohexadiene and 1,3,5-cycloheptatriene contain at least one carbon atom that does not have aporbital, and so they are not completely conjugated and thereforenot慲潭慴楣. CHM 203 MODULE 3 188 A molecule mu st satis fy H€ck els rule, and contain a part icular n umbe r 潦…敬散瑲潮献 Some compounds satisfy the first three criteria for aromaticity, but still they show none of the stability typical of aromatic compounds. For example, 捹捬潢畴慤楥湥 is so highly reactive that it can only be prepared at extremely low temperatures. † Aromatic compounds must have aporbital on every atom. It turns out that in addition to being cyclic, planar, and completely conjugated; a compound needs a particular number of … electrons to be aromatic. Erich Hückel first recognized in 1931that the following criterion, expressed in two parts and now known asH€c k els rule, had to be satisfied, as well: † An aromatic compound must contain 4n+ 2…electrons (n= 0, 1, 2, and so forth). † Cyclic, planar, and completely conjugated compounds thatcontain 4n …electrons are especially unstable, and are said to beantiaromatic. Thus, compounds that contain 2, 6, 10, 14, 18, and so forth … electrons are aromatic.Benzene is aromatic and especially stable because it contains 6 … electrons.Cyclobutadiene is antiaromatic and especially unstable because it contains 4…electrons. CHM 203 ORGANIC CHEMISTRYII 189 呡扬攠㌮ㄺ⁔桥⁎畭扥爠潦 … Electro ns Tha t Satisf y H€ck els Rule Considering aromaticity, all compounds can be classified in one of three ways: 䅲潭慴楣 -A cyclic,planar, completely conjugated compound with 4n+ 2 … electrons. 䅮瑩慲潭慴楣 -A cyclic, planar, completely conjugated compound with 4n … electrons. 乯琠 慲潭慴楣 -A compound that lacks one (or more) of the four requirements (or nonaromatic) to be aromaticor antiaromatic. 乯瑥: † An aromatic compound ismorestable than a similar acyclic compound having the same number of…electrons. Benzene is more stable than 1,3,5-hexatriene. † An antiaromatic compound islessstable than an acyclic compound having the same number of…electrons. Cyclobutadiene is less stable than 1,3-butadiene. † A compound that is not aromatic issimilarin stability to an acyclic compound having the same number of …electrons. 1,3- Cyclohexadiene is similar in stability tocis, cis-2,4-hexadiene, so it is not aromatic. 䥮-呥硴⁑略獴楯渠4 a.What is the value of n in H‚ckelƒs rule when a compound has nine pairs of …-electrons? b.Is such a compound aromatic? ㌮7䕸慭灬敳映䅲潭慴楣⁃潭灯畮摳 ㌮㜮1A牯浡瑩挠䍯浰潵湤猠睩瑨⁡⁓楮杬攠剩湧 Benzene is the most common aromatic compound having a single ring. Completely conjugated rings larger than benzene are also aromatic if they are planar and have 4n+ 2 … electrons. CHM 203 MODULE 3 190 † Hydrocarbons containing a single ring with alternating double and single bonds are calledannulenes. To name an annulene, indicate the number of atoms in the ring in brackets and add the wordannulene.Thus, benzene is [6]-annulene. Both嬱㑝- 慮湵汥湥 and嬱㡝-慮湵汥湥 are cyclic, planar, completely conjugated molecules that follow H‚ckelƒs rule, and so they are aromatic. 嬱そ-䅮湵汥湥 has 10 … electrons, which satisfies H‚ckelƒs rule, but a planar molecule would place the two H atoms inside the ring tooclose to each other, so the ring puckers to relieve this strain. Because嬱そ-慮湵汥湥 楳 湯t 灬慮慲,the 10 … electrons canƒt delocalize over the entire ring and it is湯t 慲潭慴楣. ㌮㜮2䅲潭慴楣⁃潭灯畮摳⁷楴栠䵯牥⁔桡渠佮攠剩湧 H‚ckelƒsrule for determining aromaticity can be applied only to monocyclic systems, but many aromatic compounds containing several benzene rings joined together are also known. Two or more six-membered rings with alternating double and single bonds can be fused together to form灯汹捹捬楣 慲潭慴楣⁨祤牯捡牢潮猠⡐䅈猩. Joining two benzene rings together forms 湡灨瑨慬敮攮 There are two different ways to join three rings together; forming慮瑨牡捥湥 and灨敮慮瑨牥湥, and many more complex hydrocarbons are known. As the number of fused benzene rings increases, the number of resonance structures increases as well. Although two resonance structures can be drawn for benzene, naphthalene is a hybrid of three resonance structures. CHM 203 ORGANIC CHEMISTRYII 191 Three resonance structures for naphthalene ㌮㜮3䅲潭慴楣⁈整敲潣祣汥s Heterocycles containing oxygen, nitrogen, or sulphur-atoms that also have at least one lone pair of electrons-can also be aromatic. With heteroatoms, we must always determine whether the lone pair is localized on the heteroatom or part of the delocalized … system. Two examples,灹物摩湥and 灹牲潬攬illustrate these different possibilities. ㌮㜮4䍨慲来搠䅲潭慴楣⁃潭灯畮摳 Both negatively and positively charged ions can also be aromatic if they possess all the necessary elements. These charged aromatic compounds include: cyclopentadienyl anion and tropylium cation. 䥮-呥硴⁑略獴楯渠5 Which of the following compounds are aromatic? ㌮8 Ba si s of H€ckel s Rul e? Why does th e nu mbe r of ‚ electrons deter mine w hether a co mp ound i s 慲潭慴楣? Cyclobutadiene is cyclic, planar, and completely conjugated, just like benzene, but why is benzene aromatic and cyclobutadiene antiaromatic? A complete explanation is beyond thescope of an introductory organic chemistry text, but nevertheless, you can better understand the basis of aromaticity by learning more about orbitals and bonding. CHM 203 MODULE 3 192 ㌮㠮1䉯湤楮朠慮搠䅮瑩扯湤楮朠佲扩瑡汳 So far we have used the following basic conceptsto describe how bonds are formed: † Hydrogen uses its 1sorbital to form ˆ bonds with other elements. † Second-row elements use hybrid orbitals (sp, sp 2 , orsp 3 ) to form ˆ bonds. † Second-row elements useporbitals to form … bonds. This description of bonding is called癡汥湣攠扯湤⁴桥潲礮 In valence bond theory, a covalent bond is formed by the overlap of two atomic orbitals, and the electron pair in the resulting bond is shared by both atoms. Thus, a carbon„carbon double bond consists of a ˆ bond, formed by overlap of two sp 2 hybrid orbitals, each containing one electron, and a … bond, formed by overlap of twoporbitals, each containing one electron. This description of bonding works well for most of the organic molecules we have encountered thus far. Unfortunately, it is inadequate for describing systems with many adjacentporbitals that overlap, as there are in aromatic compounds. To more fully explain the bonding in these systems, we must utilize浯汥捵污爠潲扩瑡氠⡍伩⁴桥潲礮 MO theory describes bonds as the mathematical combinationof atomic orbitals that form a new set of orbitals called浯汥捵污爠潲扩瑡汳
䵏猩. A molecular orbital occupies a region of spacein a moleculewhere electrons are likely to be found. When forming molecular orbitals from atomic orbitals, keep in mind: † A set ofnatomic orbitals formsnmolecular orbitals. † When twoporbitals of similar phase overlap side-by-side, a … bonding molecular orbital results. † When twoporbitals of opposite phase overlap side-by-side, a …* antibonding molecular orbital results. A … bonding MO is lower in energy than the two atomicporbitals from which it is formed because a stable bonding interaction results when orbitals of similar phase combine. A bonding interaction holds nuclei together. Similarly, a …* antibonding MO ishigher in energy because a destabilizing node results when orbitals of opposite phase combine. A destabilizing interaction pushes nuclei apart. CHM 203 ORGANIC CHEMISTRYII 193 If two atomicporbitals each have one electron and then combine to form MOs, the two electrons will occupy thelower energy … bonding MO. ㌮㠮2䵯汥捵污爠 佲扩瑡汳 䙯牭敤 坨敮 䵯牥 周慮 呷o p 佲扩瑡汳⁃潭扩湥 The molecular orbital description of benzene is much more complex than the two MOs formed. Because each of the six carbon atoms of benzene has ap orbital,six atomicporbitals combine to form six … molecular orbitals. A description of the exact appearance and energies of these six MOs requires more sophisticated mathematics and understanding of MO theory than is presented in this unit. Nevertheless, note that the six MOs are labeledy1„ y6, with ‰1 being the lowest in energy and ‰6 the highest. The most important features of the six benzene MOs are as follows: † The larger the number of bonding interactions, the lower in energy the MO.The lowest energy molecular orbital (‰1) has all bonding interactions between theporbitals. † The larger the number of nodes, the higher in energy the MO.The highest energy MO (‰6*) has all nodes between theporbitals. † Three MOs are lower in energy than the startingporbitals, making them bonding MOs (‰1, ‰2, ‰3), whereas three MOs are higher in energy than the startingporbitals, making them antibonding MOs (‰4*, ‰5*, ‰6*). † The two pairs of MOs (‰2 and ‰3; ‰4* and ‰5*) with the same energy are calleddegenerate orbitals. † The highest energy orbital that contains electrons is called thehighest occupied molecular orbital(HOMO).For benzene, the degenerate orbitals ‰2 and ‰3 are the HOMOs. † The lowest energy orbital that doesnotcontain electrons is called the lowest unoccupied molecular orbital(LUMO).For benzene, the degenerate orbitals ‰4* and ‰5* are theLUMOs. To fill the MOs, the six electrons are added, two to an orbital, beginning with the lowest energy orbital. As a result, the six electrons completely fill the bonding MOs, leaving the antibonding MOs empty. This is what gives benzene and other aromatic compounds their special stability and this is why six … electrons satisfy H‚ckelƒs 4n+ 2 rule. † All bonding MOs (and HOMOs) are completely filled in aromatic compounds. No … electrons occupy antibonding䵏献 CHM 203 MODULE 3 194 † Depicted in this diagram are the interactions of the six atomicp orbitals of benzene, which form six molecular orbitals. When orbitals of like phase combine, a bonding interaction results. When orbitals of opposite phase combine, a destabilizing node results. 卅䱆-䅓卅卓䵅乔 䕘䕒䍉卅 i.Give IUPAC names for the following compounds: ii.Azulene, a beautiful blue hydrocarbon, is an isomer of naphthalene. Is azulenearomatic? Draw a second resonance form of azulene in addition to that shown. iii.How many electrons does each of the four nitrogen atoms in purine contribute to the aromaticpsystem? ivThe [10]-and [12]-annuleneshave been synthesized, and neither has been found tobe aromatic. Explain. CHM 203 ORGANIC CHEMISTRYII 195 v.How many bonding, nonbonding, and antibonding molecular orbitals doescyclobutadiene have? In which molecular orbitals are the electrons? 4.0䍏乃䱕卉低 In this unit, we have been able to explain the concept of aromaticity, isolation of benzene, structure of benzene, nomenclature of benzene and spectroscopic properties of aromatic compounds. Also, we were able to give examples of aromatic compound, explainwhy benzene is exceptionally stable, and criteria and basis for H‚ckelƒs rule. 㔮0单䵍䅒Y Aromatic rings are a common part of many biological structures and are particularly important in nucleic acid chemistry and in the chemistry of several amino acids. In this unit, weƒve seen how and why aromatic compounds are different from such apparently related compounds as cycloalkenes. The word aromatic is used for historical reasons to refer to the class of compounds related structurally to benzene. Aromaticcompounds are systematically named according to IUPAC rules, but many common names are also used. Disubstituted benzenes are named as ortho (1,2 disubstituted), meta (1,3 disubstituted), or para (1,4 disubstituted) derivatives. The C6H5 - unit itself isreferred to as a phenyl group, and the C6H5CH2 - unit is a benzyl group. Benzene is described by valence-bond theory as a resonance hybrid of two equivalent structures and is described by molecular orbital theory as a planar, cyclic, conjugated molecule with sixelectrons. According to the Hückel rule, a molecule must have 4n+2 electrons, where n=0, 1, 2, 3, and so on, to be aromatic. Planar, cyclic, conjugated molecules with other numbers of electrons are antiaromatic. Other kinds of substances besides benzene-like compounds can also be aromatic. The cyclopentadienyl anion and the cycloheptatrienyl cation, for instance, are aromatic ions. Pyridine and pyrimidine are six-membered, nitrogen-containing, aromatic heterocycles. Pyrrole and imidazole are five- membered, nitrogen-containing heterocycles. Naphthalene, quinoline, indole, and many others are polycyclic aromatic compounds. 䅲潭慴楣⁣潭灯畮摳⁨慶攠瑨攠景汬潷楮朠捨慲慣瑥物獴楣s : ƒ Aromatic compounds are cyclic, planar, and conjugated. CHM 203 MODULE 3 196 ƒ Aromatic compounds are unusually stable. Benzene, for instance, has a heat of hydrogenation 150 kJ/mol less than we might expect for a cyclic triene. ƒ Aromatic compounds react with electrophiles to give substitution products, in which cyclic conjugation is retained, rather than addition products, in which conjugation is destroyed. ƒ Aromatic compounds have 4n+2 electrons, which are delocalized over the ring. 㘮0呕呏删䵁剋⁁卓䥇乍䕎T 7.0剅FERENCES/FUR呈䕒 R䕁䑉NGS Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. Dewick, P. M. (2006).Essentials of organic chemistry: for students of pharmacy, medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R.N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009). Chemistry: the central science. Pearson Education. Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New Age International. Okuyama, T., & Maskill, H. (2013). Organic Chemistry: a mechanistic approach. Oxford University Press. Ghatak, K. L. (2014).A Textbook of Organic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons. CHM 203 ORGANIC CHEMISTRYII 197 啎䥔′ 剅䅃呉低匠但⁁剏䵁呉䌠䍏䵐何乄S 䍏乔䕎呓 1.0Introduction 2.0Learning Objectives 3.0Main Content 3.1Electrophilic Aromatic Substitution 3.1.1The General Mechanism 3.2Halogenation 3.3Nitration and Sulphonation 3.4Friedel-Crafts Alkylation and Friedel-Crafts Acylation 3.4.1General Features 3.4.2Other Facts about Friedel-Crafts Alkylation 3.4.3Intramolecular Friedel-Crafts Reaction 3.5Substituted Benzenes 3.5.1Inductive Effects 3.5.2Resonance Effects 3.6Electrophilic Aromatic Substitution of Substituted Benzenes 3.7Activation and Deactivation of Benzene Ring 3.8Orientation Effects in Substituted Benzenes 3.8.1The CH3Group„An ortho, para Director 3.8.2The NH2Group„An ortho, para Director 3.8.3The NO2Group„A meta Director 4.0Conclusion 5.0Summary 6.0Tutor Mark Assignment 7.0References/FurtherReadings ㄮ0䥎呒佄啃呉低 In this unit, weshall look at the chemical reactionsof benzene and other aromatic compounds. Although aromatic rings are unusually stable, benzene acts as a nucleophile with certain electrophiles, yielding substitution products with an intact aromatic ring. We begin thisunit with the basic features and mechanism of electrophilic aromatic substitution, the basic reaction of benzene. Next, we will discuss the electrophilic aromatic substitution of substituted benzenes, and conclude with other useful reactions of benzene derivatives. The ability to interconvert resonance structures and evaluate their relative stabilities is crucial to understanding this unit. CHM 203 MODULE 3 198 ㈮0佂䩅䍔䥖䕓 By the end ofthis session, you should be able to: · understand the concept of electrophilic aromaticsubstitution reaction. · understand the general mechanism of electrophilic aromatic substitution reaction. · get acquainted with the chemistry of substituted benzenes and their electrophilicsubstitution reaction. · understand how benzene can be activated anddeactivated. · understand the orientation effects in substituted benzene. ㌮0䵁䥎⁃低呅乔 ㌮1䕬散瑲潰桩汩挠䅲潭慴楣⁓畢獴楴畴楯n Based on its structure and properties, what kinds of reactions should benzene undergo? Are any of its bonds particularlyweak? Does it have electron-rich or electron-deficient atoms? Benzene has six o electrons delocalized in sixporbitals that overlap above and below the plane of the ring. These loosely held … electrons make the benzene ring electron rich, and so it reactswith electrophiles. Because benzeneƒs six … electrons satisfy H‚ckelƒs rule, benzene is especially stable. Reactions that keep the aromatic ring intact are therefore favored. As a result,the characteristic reaction of benzene iselectrophilic aromatic substitution-a hydrogen atom is replaced by an electrophile. Benzene doesnotundergo addition reactions like other unsaturated hydrocarbons, because addition would yield a product that is not aromatic. Substitution of hydrogen, on the other hand,keeps the aromatic ring intact. CHM 203 ORGANIC CHEMISTRYII 199 Addition (The product is not aromatic) Substitution (The product is aromatic) ㌮ㄮ1周攠䝥湥牡氠䵥捨慮楳m No matter what electrophile is used, all electrophilic aromatic substitution reactions occur via a瑷o-獴数散桡湩獭: addition of the electrophile E + to form a resonance-stabilized carbocation, followed by deprotonation with base, as shown in the mechanism below. 1.Halogenation-Replacement of H by X (Cl or Br). 2.Nitration-Replacement of H by NO2. 3.Sulfonation-Replacement of H by SO3H 4.Friedel-Crafts alkylation-Replacement of H by R CHM 203 MODULE 3 200 5.Friedel-Crafts acylation-Replacement of H by RCO 䵥捨慮楳m General Mechanism of Electrophilic Aromatic Substitution 却数‱:Addition of the electrophile (E + ) to form a carbocation. Resonance-stabilized carbocation † Addition of the electrophile (E + ) forms a new C-E bond using two … electrons from the benzene ring, and generating a carbocation. This carbocation intermediate is not aromatic, but it is resonance stabilized -threeresonance structures can be drawn. † Step 1 is rate-determining becausethe aromaticity of the benzene ring is lost. 却数′:Loss of a proton to re-form the aromatic ring. † In Step 2, a base (B:) removes the proton from the carbon bearing the electrophile, thus re-forming the aromatic ring. This step is fast because the aromaticity of the benzene ring is restored. † Any of the three resonance structures of the carbocation intermediate can be used to draw the product. The choice of resonance structure affects how curved arrows are drawn, but not the identity of the product. The first step in electrophilic aromatic substitution forms a carbocation, for which three resonance structures can be drawn. CHM 203 ORGANIC CHEMISTRYII 201 This two-step mechanism for electrophilic aromatic substitution applies to all electrophiles.The net result of addition of anelectrophile (E + ) followed by elimination of aproton (H + ) is substitution of E for H. ㌮2䡡汯来湡瑩潮 The general mechanism outlined above can now be applied to each of the five specific examples of electrophilic aromatic substitution. For each mechanism we must learn how to generate a specific electrophile. This step isdifferentwith each electrophile. Then, the electrophile reacts with benzene by the two-step process of the mechanism outlined above. These two steps are thesamefor all five reactions. 䙩杵牥″⸱:䕮敲杹⁤楡杲慭⁦潲 敬散瑲潰桩汩挠慲潭慴楣 獵扳瑩瑵瑩潮: 偨䠠⬠E + „ PhE + H + CHM 203 MODULE 3 202 † The mechanism has two steps, so there are two energy barriers. † Step 1 is rate-determining; its transition state is at higher energy. In桡汯来湡瑩潮, benzenereacts with Cl2or Br2in the presence of a Lewis acid catalyst, such as FeCl3or FeBr3, to give thearyl halideschlorobenzene or bromobenzene, respectively. Analogous reactions with I2and F2are not synthetically useful because I2is too unreactive andF2reacts too violently. Chlorination Bromination In bromination, the Lewis acid FeBr3reacts with Br2to form a Lewis acid- base complex that weakens and polarizes the Br„Br bond, making it more electrophilic. This reaction is Step 1 of the mechanism for the bromination of benzene. The remaining two steps follow directly from the general mechanism for electrophilic aromatic substitution: addition of the electrophile (Br + in this case) forms a resonance-stabilized carbocation, and loss of a protonregenerates the aromatic ring. 䵥捨慮楳洠景爠瑨e 䉲潭楮慴楯渠潦⁂敮穥湥 却数‱:Generation of the electrophiles † Lewis acid„base reaction of Br2with FeBr3forms a species with a weakened and polarized Br„Br bond. This adduct serves as a source of Br + in the next step. CHM 203 ORGANIC CHEMISTRYII 203 却数′:Addition of the electrophile to form a carbocation † Addition of the electrophile forms a new C-Br bond and generates a carbocation. This carbocation intermediate is resonance stabilized- three resonance structurescan bedrawn. † The FeBr4„also formed in this reaction is the base used in Step 3. 却数″:Loss of a proton to re-form the aromatic ring. † FeBr4„removes the proton from the carbon bearing the Br, thus re- forming the aromatic ring. † FeBr3, a catalyst, is also regenerated for another reaction cycle. Chlorination proceeds by a similar mechanism. Reactions that introduce a halogen substituent on a benzene ring are widely used, and many halogenated aromatic compounds with a range of biological activity have been synthesized, as shown in Figure 3.2. CHM 203 MODULE 3 204 䙩杵牥″⸲:䕸慭灬敳映扩潬潧楣慬汹⁡捴楶e 慲祬⁣桬潲楤敳 ㌮3乩瑲慴楯渠慮搠卵汰桯湡瑩潮 Nitrationandsulphonationof benzene introduce two different functional groups on an aromatic ring. Nitration is anespecially useful reaction because a nitro group can then be reduced to an NH2group, a common benzene substituent. Generation of the electrophile in both nitration and sulphonation requires strong acid. In湩瑲慴楯測 the electrophile is + NO2(the湩瑲潮極洠 楯n), formed by protonation of HNO3followed by lossof water. 䵥捨慮楳洠景r 䙯牭慴楯渠潦⁴桥⁎楴牯湩畭⁉潮
 + 乏2⤠景爠乩瑲慴楯n . In獵汰桯湡瑩潮, protonation of sulfur trioxide, SO3, forms a positively charged sulfur species ( + SO3H) that acts asan electrophile. CHM 203 ORGANIC CHEMISTRYII 205 These steps illustrate how to generate the electrophile E + for nitration and sulphonation, the process that begins any mechanism for electrophilic aromatic substitution. To complete either of these mechanisms, you must replace the electrophile E + by either + NO2or + SO3H in the general mechanism. Thus,瑨攠瑷o-獴数⁳敱略湣攠瑨慴⁲数污捥猠䠠批⁅⁩猠瑨攠獡浥 牥条牤汥獳f E + . 䥮-呥硴⁑略獴楯渠1 Draw a stepwise mechanism for the nitration of a benzene ring. ㌮4䙲楥摥l-䍲慦瑳⁁汫祬慴楯渠慮搠䙲楥摥l -䍲慦瑳⁁捹污瑩潮 Friedel-Crafts alkylationandFriedel-Crafts acylationform new carbon- carbon bonds. ㌮㐮1䝥湥牡氠䙥慴畲敳 InFriedel-Crafts alkylation,treatment of benzene with an alkyl halide and a Lewis acid (AlCl3) forms an alkyl benzene. This reaction is analkylation because it results in transfer of an alkyl group from one atom to another (from Cl to benzene). Friedel-Crafts alkylation„General reaction CHM 203 MODULE 3 206 Examples of Friedel-Crafts alkylation InFriedel-Crafts acylation,a benzene ring is treated with anacid chloride (RCOCl) and AlCl3to form a ketone. Because the new group bonded to the benzene ring is called anacyl group,the transfer of an acyl group from one atom to another is anacylation. Acid chlorides are also called慣祬⁣桬潲楤敳. 䵥捨慮楳m The mechanisms of alkylation and acylation proceed in a manner analogous to those for halogenation, nitration, and sulphonation. The unique feature in each reaction is how the electrophile isgenerated. InFriedel„Crafts alkylation,the Lewis acid AlCl3reacts with the alkyl chloride to form a Lewis acid„base complex, illustrated with CH3CH2Cl and (CH3)3CCl as alkyl chlorides. The identity of the alkyl chloride determines the exact course of the reaction. CHM 203 ORGANIC CHEMISTRYII 207 † For CH3Cl and 1° RCl, the Lewis acid„base complex itself serves as the electrophile for electrophilic aromatic substitution. † With 2° and 3° RCl, the Lewis acid„base complex reacts further to give a 2° or 3° carbocation, which serves as the electrophile. Carbocation formation occurs only with 2° and 3° alkyl chlorides, because they afford more stable carbocations. In either case, the electrophile goes on to react with benzene in the two-step mechanism characteristic of electrophilic aromatic substitution, illustrated in Mechanism 18.6 using the 3° carbocation, (CH3)3C + . Mechanism ofFriedel-Crafts Alkylation Using a 3° Carbocation. † Addition of the electrophile (a 3° carbocation) forms a new carbon„ carbon bond in Step (1). † AlCl4 „ removes a proton on the carbon bearing the new substituent, thus re-forming the aromatic ring in Step (2). InFriedel-Crafts acylation,the Lewis acid AlCl3ionizes the carbon- halogen bond of the acid chloride, thus forming a positively charged carbon electrophile called anacylium ion,which is resonance stabilized. The positively charged carbon atom of the acylium ion then goes on to react with benzene in the two-step mechanism of Electrophilic aromatic substitution. CHM 203 MODULE 3 208 To complete the mechanism foracylation, insert the electrophile into the general mechanism and draw the last two steps, as illustrated in the ITQ 2 below. 䥮-呥硴⁑略獴楯渠2 Draw a stepwise mechanism for the following Friedel„Crafts acylation. ㌮㐮2佴桥爠䙡捴猠慢潵琠䙲楥摥l -䍲a晴猠䅬歹污瑩潮 Three additional facts about Friedel-Crafts alkylations must be kept in mind. ㄮ 噩湹氠 桡汩摥猠 慮搠 慲祬 桡汩摥猠 摯 n ot牥慣琠 楮 䙲楥摥l …䍲慦瑳 慬歹污瑩潮 Most Friedel-Crafts reactions involve carbocation electrophiles. Because thecarbocations derived from vinyl halides and aryl halides are highly unstable and do not readily form, these organic halides donotundergo Friedel„Crafts alkylation. ㈮ 剥慲牡湧敭敮瑳⁣慮捣畲 The Friedel-Crafts reaction can yield products having rearranged carbon skeletons when 1° and 2° alkyl halides are used as starting materials, as shown in Equations (1) and (2). In both reactions, the carbon atom bonded to the halogen in the starting material (labeled in red) is not bonded to the benzene ring in the product, thus indicating that a rearrangement has occurred. CHM 203 ORGANIC CHEMISTRYII 209 The result in Equation (1) is explained by a carbocation rearrangement involving a 1,2-hydride shift:瑨攠汥獳⁳瑡扬攠㊰⁣慲扯捡瑩潮
景牭e 搠晲潭 瑨攠㊰⁨慬楤攩⁲敡牲慮来猠瑯⁡潲攠獴慢汥 ㎰⁣慲扯捡瑩潮, as illustrated below. 却数猠⠱⤠慮搠⠲) Formation of a 2° carbocation † Reaction of the alkyl chloride with AlCl3forms a complex that decomposes in Step(2) to form a㊰⁣慲扯捡瑩潮. 却数⁛㍝Carbocation rearrangement † ㄬ2-䡹摲楤攠獨楦t converts the less stable 2° carbocation to a more stable 3°carbocation. 却数猠⠴⤠慮搠⠵) Addition of the carbocation and loss of a proton CHM 203 MODULE 3 210 † Friedel„Crafts alkylation occurs by the usual two-step process: 慤摩瑩潮映瑨e 捡牢潣慴楯n followed by汯獳映a灲潴潮to form the alkylated product. A Rearrangement Reaction Beginning with a 1° Alkyl Chloride 佴桥爠 晵湣瑩潮慬 杲潵灳 瑨慴 景牭 捡牢潣慴 楯湳 捡渠 慬獯 扥 畳敤 慳 獴慲瑩湧慴敲楡汳 Although Friedel-Crafts alkylation works well with alkyl halides, any compound that readily forms a carbocation can be used instead. The two most common alternatives are alkenes and alcohols, both of which afford carbocations in the presence of strong acid. † Protonation of an alkene forms a carbocation, which can then act as an electrophilein a Friedel-Crafts alkylation. † Protonation of an alcohol, followed by loss of water, likewise forms a carbocation. CHM 203 ORGANIC CHEMISTRYII 211 Each carbocation can then go on to react with benzene to form a product of electrophilic aromatic substitution. For example: ㌮㐮3䥮瑲慭潬散畬慲⁆物敤敬 …䍲慦瑳⁒敡捴楯湳 All of the Friedel„Crafts reactions discussed thus far have resulted from intermolecular reaction of a benzene ring with an electrophile. Starting materials that contain both units are capable of楮瑲慭潬散畬慲⁲敡捴楯測 and this forms a new ring. For example, treatment of compound䄬which contains both a benzene ring and an acidchloride, with AlCl3, forms Š- tetralone by an intramolecular Friedel-Crafts acylation reaction. ㌮5卵扳瑩瑵瑥搠䉥湺敮敳 Many substituted benzene rings undergo electrophilic aromatic substitution. Common substituents include halogens, OH, NH2,alkyl, and many functional groups that contain a carbonyl. Each substituent either increases or decreases the electron density in the benzene ring, and this affects the course of electrophilic aromatic substitution. CHM 203 MODULE 3 212 What makes a substituent on a benzene ring electron donating or electron withdrawing? The answer is楮摵捴楶攠敦晥捴s and牥獯湡湣攠敦晥捴猬 both of which can add or remove electron density. ㌮㔮1䥮摵捴楶攠䕦晥捴s Inductive effects stem from the敬散瑲潮敧慴楶楴y of the atoms in the substituent and the灯污物穡扩汩瑹 of the substituent group. † Atoms more electronegative than carbon-including N, O, and X- pull electron density away from carbon and thus exhibit an electron- withdrawing inductive effect. † Polarizablealkyl groups donate electron density, and thus exhibit an electron-donating inductive effect. Considering inductive effectsonly,an NH2group withdraws electron density and CH3donates electron density. Electron-withdrawing inductive effect · N is moreelectronegative than C. · N inductively withdraws electron density. Electron-donating inductive effect · Alkyl groups are Polarizable, making them electron-donating groups. ㌮㔮2剥獯湡湣攠䕦晥捴s Resonance effects can either donate orwithdraw electron density, depending on whether they place a positive or negative charge on the benzene ring. † A resonance effect is electron donating when resonance structures place a negative charge on carbons of the benzene ring. † A resonance effect is electron withdrawing when resonance structures place a positive charge on carbons of the benzene ring. CHM 203 ORGANIC CHEMISTRYII 213 An electron-donating resonance effect is observed whenever an atom Z having a lone pair of electrons is directly bonded to a benzene ring(general structure-C6H5 „ Z:). Common examplesof Z include N, O, and halogen. For example, five resonance structures can be drawn for aniline (C6H5NH2). Because three of them place a negative charge on a carbon atom of the benzene ring, anNH2group donates electron density to a benzene ring by a resonance effect. Three resonance structures A, B, and C place a (-) charge on atoms in the ring. In contrast,an electron-withdrawing resonance effect is observed in substituted benzenes having the general structure C6H5 „ Y=Z,where Z is more electronegative than Y. 䥮-呥硴⁑略獴楯渠3 Classify each substituent as electron donating or electron withdrawing. ㌮6䕬散瑲潰桩汩挠䅲潭慴楣⁓畢獴楴畴楯渠潦⁓畢獴楴畴敤⁂敮穥湥s Electrophilic aromatic substitution is a general reaction ofallaromatic compounds, including polycyclic aromatic hydrocarbons, heterocycles, and substituted benzene derivatives. A substituent affects two aspects of electrophilic aromatic substitution: † T桥⁲慴攠潦⁲敡捴楯渺 A substituted benzene reacts faster or slower than benzene itself. † 周攠潲楥湴慴楯渺 The new group is located either ortho, meta, or para to the existing substituent. The identity of the first substituent determines the position of thesecond substituent. CHM 203 MODULE 3 214 Toluene (C6H5CH3) and nitrobenzene (C6H5NO2) illustrate two possible outcomes. ㄮ 呯汵敮e Toluene reacts晡獴敲than benzene in all substitution reactions. Thus, its electron-donating CH3group activates the benzene ringtoelectrophilic attack. Although three products are possible,compounds with the new group ortho or para to the CH3group predominate. The CH3group istherefore called anortho, para director. ㈮⁎楴牯扥湺敮e Nitrobenzene reacts浯牥⁳汯睬y than benzene in all substitution reactions. Thus, itselectron withdrawing NO2group deactivates the benzene ringto electrophilic attack. Although threeproducts are possible, the compound with the new group meta to the NO2group predominates.The NO2group is called ameta director. Substituents either activate or deactivate a benzene ring towards electrophiles, and direct selective substitution at specific sites on the ring.All substituents can be divided into three generaltypes. 佲瑨漬⁰慲愠摩 牥捴潲猠慮搠慣瑩癡瑯牳 † Substituents thatactivatea benzene ring and direct substitution ortho and para. CHM 203 ORGANIC CHEMISTRYII 215 General structure 佲瑨漬⁰慲愠摥慣瑩癡瑯牳 † Substituents thatdeactivatea benzene ring and direct substitution ortho and para. 䵥瑡⁤楲散瑯牳 † Substituents that direct substitution meta. † All meta directorsdeactivatethe ring. General structure„Y(‹ + or +) 䥮-呥硴⁑略獴楯渠4 Draw the products of each reaction and state whether the reaction is faster or slower than a similar reaction withbenzene. ㌮7䅣瑩癡瑩潮⁡湤⁄敡捴楶慴楯渠潦⁂敮穥湥⁒楮g † Why do substituents activate or deactivate a benzene ring? CHM 203 MODULE 3 216 † Why are particular orientation effects observed?Why are some groups ortho, para directors and some groups meta directors? To understand why some substituents make a benzene ring react faster than benzene itself (activators), whereas others make it react slower (deactivators), we must evaluate the rate-determining step (the first step) of the mechanism. Recall the first step in Electrophilic aromatic substitution is the addition of an electrophile (E + ) to form a resonance-stabilized carbocation. The Hammond postulate makes it possible to predict the relative rate of the reaction by looking at the stability of the carbocation intermediate. † The more stable the carbocation, the lower in energy the transition state that forms it, and the faster the reaction. The principles of inductive effects and resonance effects can now be used to predict carbocation stability. † Electron-donating groups stabilize the carbocation and activate a benzene ring towards electrophilic attack. † Electron-withdrawing groups destabilize the carbocation and deactivate a benzene ring towards electrophilic attack. The energy diagrams in Figure 3.3 illustrate the effect of electron-donating and electron withdrawing groups on the energy of the transition state of the rate-determining step in Electrophilic aromatic substitution. † All activators are either R groups orthey have an N or O atom with a lone pair bonded directly to the benzene ring. CHM 203 ORGANIC CHEMISTRYII 217 䙩杵牥猠 ㌮㌺ 䕮敲杹 摩慧牡浳 捯浰慲楮朠 瑨攠 牡瑥 潦 敬散瑲潰桩汩c 慲潭慴楣⁳畢獴楴畴楯渠潦⁳畢獴楴畴敤⁢敮穥湥s † Electron-donor groupsDstabilize the carbocationintermediate, lower the energy of the transition state, and increase the rate of reaction. CHM 203 MODULE 3 218 † Electron-withdrawing groupsWdestabilize the carbocation intermediate, raise the energy of the transition state, and decrease the rate of reaction. † All deactivators are either halogens or they have an atom with a partial or full positive charge bonded directly to the benzene ring. ㌮8佲楥湴慴楯渠䕦晥捴猠楮⁓畢獴楴畴敤⁂敮穥湥s To understand why particular orientation effects arise, you must keep in mind the general structures for ortho, para directors and for meta directors. There are two general types of ortho, para directors and one general type of meta director: † All ortho, para directors are R groups or have a nonbonded electron pair on the atom bonded tothe benzene ring. † All meta directors have a full or partial positive charge on the atom bonded to the benzene ring. To evaluate the directing effects of a given substituent, we can follow a stepwise procedure. 䡯眠瑯䑥瑥牭楮攠瑨攠䑩牥捴楮朠䕦晥捴猠潦⁡⁐慲瑩捵污爠卵扳瑩瑵敮t 却数‱:Draw all resonance structures for the carbocation formed from attack of an electrophile E + at the ortho, meta, and para positions of a substituted benzene (C6H5-A). † There are at leastthree resonance structures for each site of reaction. CHM 203 ORGANIC CHEMISTRYII 219 † Each resonance structure places a positive chargeorthoorparato the new C„E bond. 却数′:Evaluate the stability of the intermediate resonance structures. The electrophile attacks at those positions that give the most stable carbocation. ㌮㠮1周攠䍈 3䝲潵p-䅮 o rt ho , pa ra 䑩牥捴潲 To determine why aCH3group directs electrophilic aromatic substitution to the ortho and para positions,first draw all resonance structures that result fromelectrophilic attack at theortho, meta, and para positions to the CH3 group. To evaluate the stability of the resonance structures, determine whether any are especially stable or unstable. In this example,attack ortho or para to CH3 generates a resonance structurethat places a positive charge on a carbon atom with the CH3group.The electron-donating CH3groupstabilizesthe adjacent positive charge. In contrast, attack meta to the CH3group doesnotgenerate any resonance structurestabilized by electron donation. Other alkyl groups are ortho, para directors for the same reason. The CH3group directs electrophilic attack ortho and para to itself because an electron-donating inductive effect stabilizes the carbocation intermediate. CHM 203 MODULE 3 220 ㌮㠮2周攠么 2䝲潵p-䅮牴桯Ⱐ灡牡⁄楲散瑯r To determine why anamino group (NH2) directs electrophilic aromatic substitution to the ortho and para positions,follow the same procedure. Attack at the meta position generates the usual three resonance structures. Because of the lone pair on the N atom, attack at the ortho and para positions generates a fourth resonance structure, which is stabilized becauseevery atom has an octet of electrons. This additional resonance structure can be drawnfor all substituents that have an N, O, or halogen atom bonded directly to the benzene ring. The NH2group directs electrophilic attack ortho and para to itself because the carbocation intermediate has additional resonance stabilization. ㌮㠮3周攠乏 2䝲潵p-䄠浥瑡⁄楲散瑯r To determine why anitro group (NO2) directs electrophilic aromatic substitution to the meta position,follow the same procedure. CHM 203 ORGANIC CHEMISTRYII 221 Attack at each position generates three resonance structures. One resonance structure resulting from attack at the ortho and para positions is especially destabilized,because it contains a positive charge on two adjacent atoms. Attack at the meta positiondoes not generate any particularly unstable resonance structures. With the NO2group (and all meta directors), meta attack occurs because attack at the ortho or para position gives a destabilized carbocation intermediate. 䥮-呥硴⁑略獴楯渠5 The Friedel-Crafts reaction of benzene with 2-chloro-3-methylbutane in the presence of AlCl3occurs with a carbocation rearrangement. What is the structure of the product? 卅䱆-䅓卅卓䵅乔 䕘䕒䍉卅 ) i.Predict the major product of the sulphonation of toluene. iiRankthe compounds in each of the following groups in order of their reactivity to electrophilic substitution: ⡡)Nitrobenzene, phenol, toluene, benzene ⡢)Phenol, benzene, chlorobenzene, benzoic acid ⡣)Benzene, bromobenzene, benzaldehyde, aniline iiiPredict the major products of the following reactions: ⡡)Nitration of bromobenzene ⡢)Bromination of nitrobenzene CHM 203 MODULE 3 222 ⡣)Chlorination of phenol ⡤)Bromination of aniline ivDraw resonance structures for the intermediates from reaction of an electrophile at the ortho, meta, and para positions of nitrobenzene. Which intermediates are most stable? 瘮 What product would you expect from bromination of p- methylbenzoic acid? viAt what position would you expect electrophilic substitution to occur in each of the following substances? 㐮0䍏乃䱕卉低 In this unit, we looked at some of the unique reactions that aromatic moleculesundergo and their mechanisms. These reactions include halogenation, nitration andsulphonation, Friedel-Crafts Alkylation and Friedel-Crafts Acylation. We also lookedatthe chemistry of substituted benzene, activation and deactivation of benzene andlastly orientation effects in substituted benzenes. 㔮0单䵍䅒Y In the preceding chapter, we looked ataromaticity-the stability associated with benzene and related compounds that contain a cyclic conjugated system of 4n+2 electrons. The mostcommon reaction of aromatic compounds is electrophilic aromatic substitution, in which an electrophile (E + ) reacts with an aromatic ring and substitutes for one of the hydrogens. The reaction is characteristic of all aromatic rings, not just benzene and substituted benzenes. In fact, the ability CHM 203 ORGANIC CHEMISTRYII 223 of a compound to undergo electrophilic substitution is a good test of aromaticity. Many different substituents can be introduced onto an aromatic ring through electrophilic substitution reactions. To list some possibilities, an aromatic ring can be substituted by a halogen (-Cl,-Br, I), a nitro group (-NO2), a sulphonic acid group (-SO3H), a hydroxyl group (-OH), an alkyl group (-R), or an acyl group (-COR). Starting from only a few simple materials, itƒs possible to prepare many thousands of substituted aromatic compounds. 㘮0呕呏删䵁剋⁁卓䥇乍䕎T 7.0剅FERENCES/FUR呈䕒 R䕁䑉NGS Bruice, P. Y. (2004).Organic Chemistry, 7 th Edition. Pearson Education: London. Dewick, P. M. (2006).Essentials oforganic chemistry: for students of pharmacy, medicinal chemistry and biological chemistry. John Wiley & Sons. Morrison, R. T., & Boyd, R. N. (2007).Organic Chemistry text book, 6 th editions.Prentice-Hall ofIndiaPvt.Ltd. Brown, T. L. (2009).Chemistry: the central science. Pearson Education. Mukherji, S. M., Singh, S. P., Kapoor, R. P., & Dass, R. (2010).Organic Chemistry, vol. I.New Age International. Okuyama, T., & Maskill, H. (2013). Organic Chemistry: a mechanistic approach. OxfordUniversity Press. Ghatak, K. L. (2014).A Textbook of Organic Chemistry and Problem Analysis. PHI Learning Pvt. Ltd. CHM 203 MODULE 3 224 Brown, W. H., & Poon, T. (2016).Introduction to organic chemistry. John Wiley & Sons.