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Organc Chem I

by: Jesus Koepp

Organc Chem I CHM 2210

Jesus Koepp
GPA 3.66

Stanislaw Wnuk

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Stanislaw Wnuk
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This 20 page Class Notes was uploaded by Jesus Koepp on Monday October 12, 2015. The Class Notes belongs to CHM 2210 at Florida International University taught by Stanislaw Wnuk in Fall. Since its upload, it has received 14 views. For similar materials see /class/221829/chm-2210-florida-international-university in Chemistry at Florida International University.


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Date Created: 10/12/15
Chapter 4 Alkanes Nomenclature Conformational Analysis and an Introduction to Synthesis Simplest alkane methane CH We can build additional alkanes by adding CHZ units CnH an saturated no double bonds no rings Straightchainquot alkanes have a zigzag orientation when they are in their most straight orientation W J n mm N V a J a of J l J a a J an i J J J J J At least one carbon attached to more than two others J mm 1 J m J gquot min A J J Same molecular formula but different connectivity of atoms Constitutional isomers have different physical properties melting point boiling point densities etc Structural mp hi1 0 F mula Cl l1 3h piI imam 95 ea 7 CH ithHpHDH A lh i an n CH hexanes CHLli rlHCllpl39lj ill 63 1 CaHM CH CH cwcucn 423 5 58 Possible Number of Molecular Formula Constitutional lsomers CAHin 2 CBH 3 CBHH 5 cHG 9 cHm 18 CyH20 35 CmH22 75 0 an 4347 Omit2 366319 CJDHB 4111846763 04an 62481801147341 All hydrocarbons are combustible Combine with oxygen and release energy Heat of combustion is very high CsHia 725402 lacng 139s goa A emh 5452kJmnl AnnaQ UPAC Nomenclature Early chemicals were given common or trivial names based on the source ofthe compound or a physical property The International Union of Pure and Applied Chemistry IUPAC started devising a systematic approach to nomenclature in 1392 The fundamental principle in devising the system was that each different compound should have a unique unambiguous name Q The basis for all IUPAC nomenclature is the set of rules used for 39 es naming alkan lNomenclature ofUnbranched Alkanes Leam the rst ten Dena uruccar Dedecane rVCHs on amen c cwctl incur an an LL CHJCH MCH CHjCHg cu mmclature or Unbranched Alkyl groups The unbranched alkyl groups are obtained by removing one hydrogenf o the alkane a d n med b 39 y replacing the ane of the corresponding alkane with y ALKAVE CH ill bratimes Mrtlunr Menu batlimo ixurmee LK L GROUP alsltkavln39l ION lin Nommcla lre ofBranchefLChain Alkanes IUI AC Locate the longest continuous chain orcarbons this is the parent chai name n and determines the parent nu up u llrl mn l I 7 carbuns aheptane u z Numberthe longest chain beginning with the end orthe chain nearer the substituent 3 Designate the location orthe substituent ll llri llrt lu IH39IL llr ll JHumlllrll hAI I e39 1 a When two or more substituents are present give each substituent a number corresponding to its location on the longest chain Suhstituents are listed alphahetic 4ethyl3methylheptane 5 wn di tri tetra e c en two or more substituents are identical use the prefixes al Commas are used to separate numhers horn each other by Repeatthe numherinwosuhstituents on same car o cl The pre xes are NOT used in alphahetical prioritization 4ethyl23SMmethylheptane wn n two chains of equal length compete to be parent choose the chain with the greatest number of substituents III tll II 4 l II IL n t n tu t u end ofthe parent chain choose the namethat gives the low number atthe first point of differenc t39ll t iisiil tllst 7 When branching rst occurs at an equal distance from eithe er mendature of Branched Alkyl Ch 5 Two alkyl groups can be derived from propane in II t V irrolnl uruup Hui 39 Li Prupum ril i iii cl l ikll nlvll grim Example Arisnprnpylheptane How many butyl groups are possible 39ll 7 lll l1 nmp um HL u IL lllllmw gt39Il l rn mlnllyl gmllp ll m39llri n a l quotl lfll iwlmm group H II 1 mm LII 7 Cll Irnhulgl group What type of hydrogens are in 22 dimethylpropane 22764melhylpmpane Classi cation of Hydrogen Atoms Hydrogens take theirclassi cation from the carbon they are attache t llulrugruulunh 39 7 397 11 L Il ll 39ll L 39L 3 Ilylrngrnulmn llnlugndlnm an the parent chain 0 Halo and alkyl substituents are considered to be of equal ranking quotH 1 1quot Cll Ll llCllCllfll Cll ll39ll llfll Cl CI Aflllmenwvll plmum LCllemyLm mlpwllmw as alkyl halides o and still used l ll rum u rm wm Mu m rm m um umquot um llr l ll 1 rum my wmm mlmm a u mm mm lmwuhl amn mum Nomenclature of Alcohols IUPAC Nomenclatu re of Alcohols In the l39LlPAC naming systemtl1eremayheas Witkl any rm quotquotmp m m M Mknmmm Select the longest chain containing the hydroxyl and tn the name 1mm whatlayman Infnme endan on t W mW mm quotunme r 339 number Exam The other substituents take their locations accordingly 0H CHscrHCHchon Cl lzCHICHCHlCHIClel CchHz HCHg CH l 3 CH f parenl zebutanul rmeLh lrlrbutanul V L y Armethylrlrhexannl 1quot lncant Pm x locant 39x 0 Examples le ClCHz Hz HZOH cH3cll ElQc ll3 Cl l 3aman mp anal t rdimedlylrlpenlanul 0 Common Names of simple alcohols are still often used and are approved by IUPAC CILCHJT 0H CllfolLCH H H llx HCH 39 39 39 l l39mlul lluliwl MN nlmllnl mllnlll nlculml ll cell rnr39on mama cu mum alumni l olmul nlluhol o Ring compounds are very common Named as cyco parent namefor number of C s inthe ring BOO cyclobutane cyclopentane cyclooctane Nomeqpla ulj 0 Vubstitu ted ybitlziikahes o If only one substituent is present it is assumed to be at position one and is not numbered 9 When two or more alkyl substituents are present the one with alphabetical priority is given positionl m erngcon nuesrdm u other subsm en ercloc lse orcounterclockwlse to glveme owest numbers If one of the subsumens ls afunctlonal group leg hydroxylj lt ls given snlon 1 CI 39H y tirwmm mm mmquot cu 3H3 IrE lyllmmelrerIQm o If otherparts orthe molecule are dominant or ira long chain is attached to a ring with fewer carbons the cycloalkane is c nsid red the substituent Cl lscyclnblltylpenmne Lchhmrlrcyclnpenlylrlmelhyh hm chem has mure carbuns than nng M l sdicyclnprnpylprupzne CH INCH 2 H T A V CHCH 39Iulxullm mrllnkvdlrh39 noun It dum lsupnnn It BUT keep numbers as low as possible Il hctgcl Cl L39hl l lk1439lh39lrl mm chllyn a eiml min is clulwmmx m ihlaclnneane nllcl ane to ene lowest nunber l cllI lnun In summer n mm llnxturl O The double bond 1 and 2 cu of a cylcoalkene must be in pos ons s AJt m m nl wquotum anrmnm lgyrhhnnm39 ulmlurlowmrnn lml lbdlmtm3lurllmnmegt y r r VIalkendlsg Vinxlandjallyigrpup The vinyl and allyl groups are common names that The hydroxyl is the group with higher priority and need m be readin recagnlzed must be given the lowest possible number CHVCH CHCHClI3 W The illyl gmup The any group an if on H siVlclluiJelmudn nl 2MulilylZrlullccnl rl Hy 1 n n 0 dnmlnIpcmacnzol or Zmt39illjIQCIKIIICVlwIOi RE c H n ctm awn u W now 1 manhunt Note thatpositions oihoth the cc and the 0H mustbe indicated MW inquot o lftwo groups occur on the same side ofthe double bond the compound is cis o lfthey are on opposite sides the compound is trans 1 Choose the longest continnons chain oi carhons containing the triple hond H 2 The name oi the parent alkane ismodi ed by dropping quotanequot and adding quotynequot H H H U 3 Number Lheparem chain from the end their Diihiurnrihcm munl v hithlomrdnnr closest m the triple bond 0 Several alkenes have common names which are 5 4 3 z 1 recognized by IUPAC CHSCHZCHZCECH CH3CECCH3 H i 17 on e Lbut e Cli CH CHL39IKCHI Cl C L p tyn y 39 V 3 Iprnpcnc sohnh han wPAc thrnc Propom unumon I39IIIIcn 39 Pmpyll39nr 4 Positions oi substituonts are determined by the usual rules 4 3 7 1 5 4 3 2 l CHSC CCHZCI CHsCH2HC CH Cll3 lachloraZVbutyne Samethyla 17p entyne 5 In an allrynol the alcohol has priority in numbering 4 3 2 1 HcCCHZCHZOH Sabutynalaol laalkynes are also called terminal allrynes acidicH quot p 125 Sigma Bonds and Bond Rotation far apart from each other as possible The drawlng to the right ls called a Newman projectlon II H f 4 J J J xx t The eclipsed conformation has all CH bonds on adjacent carbons directly on top of each other 2 g 1 I 2 M Th dtlredral angle Id oerwccn gens ll llyllm is I How to draw a Newman projection I H H f39AH or t v gt HEB H H H HH as Sight down the cc bond sigma bond is symmetric Groups on front carbon intersect in center Groups on back carbon end at circle The potential energy diagram or L 39 I ethane The staggered conformation is more stable than eclipsed by kJ mol391 n Rotation rate Ionsec n unrwl Energy barrier called torsional strain Bond Romnons 1n Propane One can my shuw me or hand ztztime Cnnsider as ethanewith 1 H replaced by a CH Bamerl4kTmol on around 243 of butane gives three energy mi 39 e two gauche eonromau39ons Rota Th nua are less stable than the am conformation by 33 kJ mol1 because of repulsive Van derWaals force n roups s bmee the two meahyl g Conforms onalAnaly oI Butane The ngered an ismuxe slahle dlznllle m equivzknl ngered gzudle cun lmzl39nns n the amicanmman39nn the m CH gmups are m m 5 mm dsmeyznd m ummmmsm ngyisinnud u Conformal nal Analys s of Butane 3 minima and energy barriers between each 3 maxima u n o Heats ofcombustion perCH2 unit reveal cyclohexane has no ring strain and othercycloalkanes have some ring strain Heal ol Heat 0 combustion Rlng Combusllon par cmcmup sualn cydoalkane our n kJ molquot kJ molquot kl moiquot y wallal a 20m 697 0 H5 Cyl lulwl l iv A 2744 885 O 9 Cytlopelllmt 5 3320 66L 0 27 Cyclolrcxaoe 5 3952 658 l o Cytiniteptant 7 4637 662 4 27 Cyl louci B ECHO 663 B J Cylilolldllr no a 7951 664 6 all Uyrlrt r ne to am am a By mnlatlecawe is was 559 n u Urltlmnclwd nlkam 65H 6 9 Angle strain is caused by bond angles different from 1095 c c angle also means drnltal overlap 5 reduced 9 Torsional strain is caused by eclipsing CH bonds on adjacent C s cycldprdpane has both high angle and torsional strain O Cyclobutane has considerable angle strain but not as much as in cyclopropane O Cyclobutane bends to relieve some torsional strain even though bending increases angle strain 90 becomes 88 u are 103quot closeto the idealized tetrahedral n bond angles Thus a planar cyclopentane 103 would have very little angle strain But a planar geometry would have very severe torsional strain 10 eclipsed H Consequently the geometry or cyclopentane is bent The tnrsiomlstrainis reduced in thabentsmrrtine Four H arestiil erlipsadbntd arestaggen e Cyclohexane is not planar since the angle of a simple hexagon is 120 e Cyclohexane adopts a CHAIR CONFORMATIDN that has no ring strain and no angle strainl e All bond angles are1095 and all CH bonds are perfectly staggered ib in n Conformations of Cyclohexane Chair conformation Conformations of Cyclohexan Chair conformation Equatorial hydrogen Conformations of Cyclohexane Chair conformation up and down on39al close m plane of ring Conformations of Cyclohexane Boat conformation Conformations of Cyclohexane Boat conformation Steric interactions Conformations of Cyclohexane Boat conformation tibn 397 Ii 39 Rf Chair ips to another chair going through a boat Every axial position becomes equatorial and vice versa Ring ipping occurs rapidly at room temperature ll 3 llll iiii V AL 7 ll ll i H i i ll 11 ampUllpQ ii 11 CH ll Energy Diagram of Energy Barrier H39eii their Flips about 100 K per sec 999 of molecules in chair l orm iWisrimar i No 0 Chair Note kcalmol x 4 Z kJmol To draw chair cyclohexanes follow these steps 1 Draw the carbon chair 2 Add the axial hydrogens mun i rm 3 Draw the c1 and c4 equatorial hydrogens 4 Draw the remaining equamriai hydrogens rn nud imiin minim min i39 In Inlte cyclohexane rlngs Conformations ofSubstituted Cyclohexanes O Monosubstituted Cyclohexanes o Methylcyclohexane m CH3 H3 Conformations of Substituted Cyclohexanes O Monosubstituted Cyclohexanes o MethyIcyclohexane m CH3 Eqmtun39almethyl group Axialmethyl group CH3 Conformations ofSubstituted Cyclohexanes O Monosubstituted Cyclohexanes MethYICYCIOhexane Steric interactions CH3 Hm M AH Conformations of Substituted Cyclohexanes O Monosubstituted Cyclohexanes o Methylcyclohexane H CH3 a mm steric interactions L 4 g L a it it Conformations of Substituted Cyclohexanes O Monosubstituted Cyclohexanes o Methylcyclohexane Hm H3 5 sten39c interactions L L L a more stable Subs uwd Cyclohexanes i quot Mine mine by 75 kJmnl f 95 equatorial at room temp Axial Minn cumnwhml Zwlwalvmemvi unlumuhnn axial ups due to In axiai methylcyclohexane there are two 13ediaxiai interactions zm39alrmethylcyclnhmne E zch 137diaxizl interaction introduces 375 kJmoi oi39 smic strain The larger the group the greater the equatorial preference trbutylcyclohexane is gt99 9 equatorial A lltili an i l Conformations ofSubstituted Cyclohexanes O Monosubstituted Cyclohexanes o tenButyicyciohexane tertrbutyl substituent Disubstituted Cycloalkanes With two substituents on the ring in different positions con gurational isomers arise sis gimp nn same side nf ring trans gimp nn nppnsite side nf ring g lerichlorocyclobumne mm ismner For 3 4 or 5 membered rings can treat as if planar 13dimethylcyclnp entanes F 39 Isomers in 39 39 Cyclohexanes Three options for pairs of cistrans cyclohexanes II I Mei in idinntiw s u f m H i N LIL u mi mi N Kn ii 1 7 tn us en in H Menu iniqeinttw quotweMum indium Preferred conforms ns trans 1A Dimetnyicyionexane prefers trans diequatoriai conformation cHZ H C H CH3 CH3 H CH W my 1 bath methyls equatnna Nate methyl nn uppnsite sides nfnng ci 1443irnetnyicycionexane exists in an axialequatorial conformation Note that both are on the same side or ring top side in this drawing Con rming whether sis or tram Identify each subsdment as on the upper or lawu39 side or ring H Hume upperhund sil owermi CH3 lnwerh and 39 H ls 39 I always trans Direguatorial isomer 14 u 39 trans H in H 13 ms CH3 39 u H 12 CH3 trans Synthesis ofAlkanes and C Pbly y liciAlkahb s Hydrogenation of Alkenes and Alkynes d b39a zikbb n39 7 mum kmznrm 7 n I KR r gt gt I Iknln39 Hu 39x Cholesterol 25mm mmww Synthesls of Alkanes and Cycloalkanes Reduction of Alkyl Halides R X Zn IIX R H ZnX 7 m em R H Specific Example Ilkr 2 crLCIIZC HCH T 2 CII Cll lIClL ZnHr m u xeollulyl hrumidr man 2 mmuhulnnn Synthes s of Alkanes and Cycloalkanes Alkylation of Terminal Alkynes o Alkynes can be subsequently hydrogenated to alkanes a mu m ur a mm mm c me u m iH quotNquot CH CHCHICH H Synthe of Alkanes and Cycloalkanes Example synthesize 2methylpentane starting from 3methyl1butynequot CH CH c x u H4 um c M 4CHCIICEC7CH cum km LI quotNquot CHK HCHJCHICH


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