How to Determine Aromaticity: The Lone Pair Predicament · 2012-04-28 · • This means that the...
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How to Determine Aromaticity: The Lone Pair Predicament Still haven’t unwrapped your thinkbook? Don’t worry! Here’s a brief overview of aromaticity: DEFINITION Aromaticity a special kind of stability a molecule possesses when it meets these three requirements: 1) There is a Closed loop of p‐orbitals somewhere in the molecule • This means that the atoms in the ring must be sp 2 or sp (rarely) hybridized Image 1 • Remember: sp 3 hybridized atoms do not contribute any p‐orbitals. Images 2, 3, 4 and 5 clockwise from top left All the carbons in this benzene molecule are sp 2 hybridized.
Transcript of How to Determine Aromaticity: The Lone Pair Predicament · 2012-04-28 · • This means that the...
HowtoDetermineAromaticity:TheLonePairPredicamentStillhaven’tunwrappedyourthinkbook?Don’tworry!Here’sabriefoverviewofaromaticity:DEFINITIONAromaticityaspecialkindofstabilityamoleculepossesseswhenitmeetsthesethreerequirements:1)ThereisaClosedloopofp‐orbitalssomewhereinthemolecule
• Thismeansthattheatomsintheringmustbesp2orsp(rarely)hybridized Image1
• Remember:sp3hybridizedatomsdonotcontributeanyp‐orbitals.
Images2,3,4and5clockwisefromtopleft
Allthecarbonsinthisbenzenemoleculearesp2hybridized.
Images6and7,fromlefttoright
• Thep‐orbitalsdonotnecessarilyhavetobefilledwithelectrons
Image8
2)Thesep‐orbitalsmustoverlap.Inotherwords,themoleculemustbeplanarImage9
• Note:Ifmoleculehasalargesubstituent(s),torsionalstraincanforceitoutofplanarity
o Thiscasehappenswhentorsionalstrain>aromaticstability HowcanItell?Belowareafewcasesyoushouldremember Note:Awronganswerwiththerightassumption(assumingthattorsional
strainwillovercomethearomaticstability)willearncreditifcorrectreasoningisshown
Thepositively‐chargedcarbonhasasp2hybridizationandstillhasap‐orbitalthatcontributes
totheclosedringeventhoughelectronsdonotoccupyit.
Theelectronsofthep‐orbitals
aredelocalizedanddistributedintotwoofitsownplanes,oneaboveandbelowtheplanar
molecule.Allaromaticmoleculesareconjugated.
o Ex)Annulene
o Whyisn’tthisaromatic?Hint:don’tforgetto
drawinthehydrogens!Image10
Hydrogenatomsthatarenotdrawnoncarbons2and7trytooccupythesamespaceThisleadstotorsionalstrainthatoutweighsthestabilitythemoleculewouldgainfrom
aromaticityThemoleculeassumesanon‐planarconformationandisnotaromatic
o Ex)Benzenecanholdupto5‐tertbutylgroupsuntiltorsionalstrain>aromaticstability
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3)Huckel’srule:4n+2pielectrons;nisanyinteger(0,1,2…)
• Trickyrule:takeresonancestructuresintoaccount• Pielectrons=pibond(s)withintheclosedloop(1pibond=2electrons)+lonepairs
participatinginresonancestructures’pibondso Whichlonepairscount?Wellthat’stheLonePairPredicament!
THELONEPAIRPREDICAMENTRememberbeforewhenwesaidthatsp3hybridizedatomscouldnotparticipateinaromaticity?Wellthat’strue,butit’snotaseasyasaruleasitsounds.Sometimes,anatomcanlooklikeitissp3hybridized,butitisactuallybesp2hybridized.
Howcanthishappen?Hasyourlifecomecrashingdownattheverythoughtofsp2atomsappearingtobesp3?Ustoo.Hopefully,thisexplanationwillhelpyougettosleeptonight:Anatomthatlookslikeitissp3canbeclassifiedassp2onlywhenitwouldbenefitthemoleculeforthistobeso.Whatexactlydoesthatmean?Wellconsiderthiscircumstance:Sayyou’redraggingyourselfthroughtheChemistry14Cmidtermandyoucomeacrossthearomaticityproblem.ItkindlyasksyoutodetermineifmoleculeXisaromatic.Therearenoformalcharges.Itlookssomethinglikethis:
Image12You’reasmartfellasoyourememberthatsp2hybridized(containsp‐orbitals)atomsinaclosedringcanproduceanaromaticmolecule(assumingitfollowsHuckel’srule).
4carbonatomsaresp2hybridizedⅩ Nitrogenatomissp3hybridizedbecauseithasthreesigmabondsandonelonepair.→ Done.Notaromatic.Itdoesn’tfollowHuckel’sRule(4pielectrons)ANDnotallatomsin
theringhavep‐orbitals.WRONG!!!!!Theresonancestructuresofthismoleculeare:Image13
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Combinedresonancestructure:
Thismoleculeisnotaromaticbecauseit
disobeysHuckel’srule.Thenitrogenscontributeits’lonepairstotheresonancestructures(thereforearesp2hybridized).Thisisalsothe
casewithSulfur,butSulfurcanonlycontributeoneofitslonepairs.Becauseitonlyusesonelonepair,thetotalnumberofpielectronsis8,
whichdoesnotfitwith4n+2(n=1.5–notaninteger).Remember:anatomwillneverusebothlonepairstocontributetoaromaticity!
Oneofthelonepairsofthenitrogenatomcontributetotheseresonancestructures,creatingpartialpibondsandcausingthenitrogentohaveapartialp‐orbital.Thereforethelonepaircontributestothearomaticity.ThestructurefollowsHuckel’sRule(6electrons=4(1)+2)ANDallatomsintheringhavep‐orbitals.Thep‐orbitalsoverlapmakingthemoleculeplanar.Themoleculeisaromatic.See!You,myfriendjustfellvictimtolonepairploy–acheaptrickusedbytheseconartiststhatwillcostyoupointsonexamsandadmissionintomedicalschool.Howdoyouknowifaseeminglysp3hybridizedatomisactuallysp2hybridized?Here’showwefixthis:ASKYOURSELFTHESEQUESTIONS,INORDER:1.Doesthisatompossessloneelectronsthatcontributetoresonancestructures?2.Doesthisseeminglysp3atomisruinthearomaticity?*Remember,atomsreallywanttobearomatic.Itdecreasestheirenergyandmakesthemmorestable,sodon’tbefooledbysp3camouflage.3.Wouldtheatomthatisruiningthearomaticityactuallycompleteaclosedloopofoverlappingporbitalsifithadaporbital?Ifallaboveistrue,feelfreetoassignthemoleculeansp2hybridization.
→ Note:Whenthishappens,theselonepairscancountaspielectrons.PRACTICEPROBLEMSAreyoustillconfused?Weunderstandifyouare.Thistopicisprettyconfusingandgoesagainstsomanytopicswecoveredin14Aand14B.Fearnot!HerearesomeSUPERFUNexamplestogetyoupumpedforhybridizingeventhemostcomplexmolecules:Image15
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Thismoleculeisnotaromaticbecausethecarbonatthetopoftheclosed‐ring(wherethe
ethylgroupisattached)hasnoporbital.Itissp3hybridized.Don’tbefooled,eventhoughthemoleculehasthecorrectnumberofpielectrons
tofollowHuckel’srule!
Thismoleculeisaromatic.Theoxygen’s
electronsdonotcontributetoresonancestructure,andthereforedonotlieinthep‐orbital.Nevertheless,oxygenisstillsp2
hybridizedANDitfollowsHuckel’sRulewith6pielectrons,soitallworksout!
Thismoleculeisaromatic.Thenitrogenonthetopdoesnotcontributeit’slonepairsinresonancestructures(thereforedonotcountas
pielectrons).Thisisshownintheresonancecontributorsbelow.Thenitrogenonthebottomcontributesit’slonepairsinresonance
structures,thereforetheydocountaspielectrons.Thetotalpielectroncount=6(followsHuckel’sRule).
Thismoleculeisnotaromatic.Thelonepaircannotmakeapibondwiththeothercarbonsbecauseitwillcreateapentacarbon(unstable).
Therefore,thatcarbonissp3moleculesanddonothaveap‐orbital.Also,thetotalnumberofpielectrons=4(n=1.5–notaninteger).
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CitedWorksImages1and9:Wikipedia.WikimediaFoundation.Web.27Apr.2012.<http://en.wikipedia.org/wiki/File:Benzene_Orbitals.svg>.Images2,5,6:http://www.meta‐synthesis.com/webbook/45_vsepr/VSEPR.htmlImages3,4,7:http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/Valence_Bond_Theory/HybridizationImage8:http://www.nku.edu/~russellk/tutorial/reson/C7H7+.GIFImage10:http://science.uvu.edu/ochem/wp‐content/images/A/aromatic12.pngImage11:CreatedonChemDrawImage12:http://en.wikipedia.org/wiki/File:Pyrrole_structure.pngImage13:http://en.wikipedia.org/wiki/File:Pyrrole_Resonance.pngImage14:http://en.wikipedia.org/wiki/File:Pyrrole‐2D‐full.svgImage15:CreatedonChemDrawImages16,17,20:http://chemwiki.ucdavis.edu/Organic_Chemistry/Aromatics/Hückel's_RuleImage18:http://upload.wikimedia.org/wikipedia/commons/3/31/Imidazole_structure.pngImage19:http://en.wikipedia.org/wiki/File:Resonance‐imidazole.png
