Organic Reactions and Pharmaceuticals
Organic Reactions and Pharmaceuticals CHEM 14D
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This 101 page Class Notes was uploaded by Michael Reilly on Friday September 4, 2015. The Class Notes belongs to CHEM 14D at University of California - Los Angeles taught by S. Hardinger in Fall. Since its upload, it has received 11 views. For similar materials see /class/177969/chem-14d-university-of-california-los-angeles in Chemistry at University of California - Los Angeles.
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Date Created: 09/04/15
RADICALS 0 Reactions with Brz Addition to an alkene with Brz gt No Reaction in the dark Br2 H Br hv in the light During a demonstration by Dr H the reactants changed colors when held against light This is a substitution reaction but what kind 0 8N2 No There s no leaving group 0 SM No No leaving group o EAS No The reactant is not aromatic o I v Mechanism This symbol is a shhook which has half of an arrow head This means that only one electron is moving instead of a pair of electrons h hv or heat lt Unpaired electron free radical and open octet BrCEr gt Br The Br Br bond weakens with heat or light it takes 46 kcalmol to break it Each electron in the bond goes to a different Br I 39239 Three Fates of Radicals a Addition to pi bond p 4 Br H Br r gt C The bromine radical and one electron from a pi bond form a new bond on the least substituted end of the alkene The other electron in the pi bond is transferred to the more stable carbon atom making a 2 radical b Atom or group transfer QH gt H Br The bromine radical and an electron from the 0 H bond form a new bond forming a new molecule H Br The remaining electron in the 0 H bond transfers onto to the attached carbon making a carbon radical Hydrogen transfer is often preferred to transfer of a methyl group even ifthe product is less stable The removal of hydrogen from the molecule of CHZCHCHs is called hydrogen abstraction c Radical combination BrnKBJr39 gt Br Br This step is rare because it is a termination step see notes below radicals are eliminated Radicals are found in low concentrations and many times react with other molecules before radical combination can occur I 0 Radical Structure and Stability m H Br 00 Q gt H Br Process of removing hydrogen from a molecule and forming a new radical is called hydrogen abstraction Is the given product 3 carbon radical the best position for the radical Structure The electron density from the radical is distributed throughout the empty orbital sp2 geometry o Substituents Carbocation stability Radical stability CH3 lt CH2R lt HCR2 lt CR3 CH3 lt CH2R lt CHR2 lt CR3 Methyl 1 2 3 Methyl 1 2 3 Same trend for radical stability as for carbocation stability more substituted carbon makes the more stable radical but making a medical radical isn t as39 quot 39 as making a methyl 39 quot l Methyl and 1 radicals exist 0 Carbon radicals aren t as electrondeficient as carbocations so it is possible to form a methyl radical o Resonance r Carbocation resonance with pi bonds H2C H2C The positive charge is delocalized by resonance and the H20 CH2 pair of electrons forming the pi bond shift to the adjacent bond Hzpzv gt H20 Radical resonance with pi bonds HzC39 CHz An electron from the alkene forms a new bond with the carbon radical leaving an electron to the carbon at the other end of the alkene Carbocation resonance with lone pairs Transfer of cation from carbon to oxygen Very good resonance contributor because the structure has full octets H2CTJOH gt H20OH Give a carbon attached to nine electrons Radical resonance with lone pairs o Too many electrons NOT POSSIBLE This mechanism would H2O 3H OH which violates the octet rule j Radical resonance 9 radical on ox en OH gt H2COH BADII Will not happen this way H20 I 0 Brz Radical Reaction Mechanism Br Br2 gt H Br 0 hv or heat 6 Mechanism eh Br gt 2Br39 Initiation hv Starts without radicals and produces radicals H f x 00 H Br Br Radical intermediate is formed now what Three ways to get to the desired product Br 09 Termination Radicals are consumed but not produced going from radicals to no radicals 2 0 r H T 3 039s 8 B rCr Propagation A radical produces a new radical Looks good but there s a kinetics problem the reaction would be too slow As soon as the radicals are produced Bro they re consumed So the probability of one radical finding another is very slim Not the best choice Unlikely Br H bond is strong and harder to break Br Br is a weaker bond so this is more probable New radical Bro is formed which will react with hydrogen on cyclohexane again creating a new 3 radical This is called a chain reaction one or more mechanism steps repeated in a cycle 39239 Other Radical Halides CI H C12 Others hv CI2 is not selective at all H I2 gt No Reaction hv 2 bonds are weak so atoms separate and form radicals easily I is not reactiveis happy as a radical so it doesn t have energy to react H F2 gt It dMM hv F2 is extremely reactive It will react with anything hence the kaboom The more reactive the less selective F2 is not selective at all hence it is extremely reactive I 39239 H Br and Alkenes HBr Br gt Br Pure gt The product follows Markovnikov s Rule The hydrogen was added to the least substituted end of Br H the alkene in pure HBr HBr Br Br gt The product Is antiMarkovnikov H202 Peroxide impurity The hydrogen was added to the most H Br substituted end of the alkene when the HBr solution was impure 39239 H Br Addition Mechanism in the Presence of Impurities Which radical mechanism leads to the antiMarkovnikov product Br H Br 0 Br Br Br 37C gt 39 H 2 C H Br Primary radical Secondary radical Mechanism HQ gt 2Ho39 Ho39 Ht gt H20 Br V F Br Br Br Br Ht J C H Br H Br Therefore the secondary radical intermediate forms the antiMarkovnikov product in the presence of peroxide This effect peroxide impurities leading to antiMarkovnikov addition of a bromine only applies to brominell HCI or HF add to an alkene according to Markovnikov s rule See Practice Probems Radicals in the Thinkbook problems 1720 pg186 for more on this topic 39239 Combustion CH4 02 gt 002 H20 energy Structure of 02 CLO A diradical NOT OX0 doublebonded oxygen atoms Mechanism H m I o39 o M H c H H C O39OH any many more IV 39 3 mechanism steps H 39239 Biological Oxidation o39 o o o Product when an electron is added to an oxygen diradical is Ozo39 superoxide Humans make about 10kg per year ofthis potentially cancerous molecule O O DNA H gt 39DNA HO O Fragment of DNA where hydrogen was abstracted can lead to cancer When superoxide in the body reacts with DNA it may pull of a hydrogen abstract a hydrogen from DNA forming OOH and a DNA radical la Subs 11 WmmZEIEIS and Uvgzmc Cmvmmy byPun aYurkxmsEmlce 439ch What dues 1 nahde anra snbsmmwn chm L Ifarnuclenphmc mzd39he my ummalecula meamng that the rate 5 dependent um ahe Iactm SM 15 a mechamsmpalhway wrysxmxlar m 5 First zers hue1y review 5N2 shz is subsmunan nucleaphxhc bxm 712ch ar In suz nee general mechanism is The nucleaphxle attacks the electmphle from the backsxde expeumg a 1eavmg gmup place at the same me Now let 5 by malanglhe reaclmn occur In two slaps n wauld lack hke ths nucquot stands far nucleaphxle ms wauld absalutely hm accu n wauld create a pentavalent carbon W mm m second step the nucleaphde anackedv n wauld hmkhke ths n a M 4 a ewe Ric Lu w a vOu I e Here the leavmg gmupleaves creating apasmve charge unthe carban The nucleaphxle TM m h m m mulated Ittums cutthat ths xnter medmte structu wnh apasmve charge an carban is crucial w 5N1 mechanism a earboeataons read m away that wrll complete the open etet Let39s address the lssue of earboeatrons before proceedlng further wrth the snl meeh ls Cmtzmzz n General serueture Carboeatrons are sp2 hybndrzed whreh means that they are m a planar arrangement u The pz orbrtal ls empty so you ran thrnk of the posrtaye eharge as loeated here Stability The stabrlrty o a earboeauon depends on two factors r rutmherotsuhshments leswanr Flrsl we 11 luck at the number afsubstxtuents oetet on the earbou make Carbocanoxls emem er the posrtrye ehatgr and open lnstable Let39s look at an example seeondary teruary v Methyl earboeata on Alkyl groups are weakly eleetron donatrng 50 m the ab Xample the more CHs groups there are the more eleetron densrty ean be donatedto earbon Thls added eleetron reasonrng methyl earborattons are so unstable that we usually wlU not Emounle hem a s s ch Methyl pnmary seeondary tertaary Nov121 s consider resonance In thls wnl and adjacent c Thls ereates aresonanee eontnbutor The rwo resonance eonmbutors Thls sta brllzes the earboeataon oeaaon benzyl e oeaaon r allyl anolbenz arboeaaons have alot of stab ty The allyl earboeatron has resonanee truetures beeause ofthe pl bo d anolthebenz l arboeaaon ls even more stable beeause ofresonance lnvolvlng the attaeheol benzene group Allyl earb arb yl e lll s D l The lone parrs move from the attaeheol group x to the slngle bonol between x and c ereaang a double bonol Thls ereates an eleetron de clency on x movlng the posrtrve charge to x Agaan the earboeatron 1 up N For example The most eleetronegaave atom oxygen has aposrave charge m the seeonol shouldbe on the least eleetronegatlve atom However all the atoms have full oetets so resonanee does stablllze the moleeule H s o r r For example a pnmary earboeaaon wth resonanee ls more stable than a seeonolary earboeaaon wrthoutresonanee e UK pnmary wth resonanee W Jquot 39 a seeonolary wrthoutresonanee Let s by an Example Choose the most stable carbacmmn A ll n gym Nr V 49 V pnmary tertaarywres ternary seeonolary T m r The Three Fates at z Carhncztinn open n r H tllustxate thern A mum at nucleapr s L s r L 3 Hr m wu nht W W lls the open oetet on e b H wever now oxygen has three bonds so rt ame on the more eteetronegattve atom t Thts ar on o the postttve charge am the postttve charge rs now but all atorns have afull oetet so thts rs apreferable arrangem 2 use a prom afar7111 band Ag ct Q rtetwat Here the electrons on the o ofHOCH form abond wrth aproton on the h w u w rh h quotA been transfenedto CH30H2 Notr that carbocanons are so reatnvr they are desperatequot thstts Lhey Lan be depmtenated by many rnoteeutes thnt are nntparneutnrty good nusteophttrs sueh as water 2 r ea ang rrzm we L amp G 0 Whenever you see a earboeanon always eheekto see 1frean39angement rs possrbte to te and the Now we are ready to present the general rneehantsrn for snl Mechan m arS 1 Le 1 or m 39T L t the earboeatron and forms abond Here39s a speer e example new t 3 1 I V rs tep e e 7 Lhks e r J gr an a LLLH3c131 if a 7 lg 239 gt r t th t w r no t c1 as rt1eaves Water attaeks the earboeatron thatwas formed Agam the transrtron state shows the water attackmg m a d partaal bon wrth c onee water attaehes the Oh a posmve eharge s aterrnustprotonate rt Th Hr t t n 1ost Nucleophxhc attaekrs energetreany produetave beeause xtgmns abon T e alook at the energy dAagram The larger V rs the larger the energy of aetrvatron Thrs slows down the reaetaon Here rt39s so slow that rt deterrnrnes the rate of the reaetron so rtrs eane t e rare deternnnrng step All subsequent steps happen rnueh rnore quxckly m e m anson 1 quot l AT K For Sm rate rs proportrona1 to R3C rLG N there are other steps that oeeur as well Erawpze Problemi 1 D aw therneehanrsrn occurs Again the transition state shows apaitial bond between the carbon and incoming nucleophile Lastly the nucleophile attaches 1 us g39llll H the wrist e quot rn 394 o 2 l lllt SH quotw a o tcnzmlllmlmc is 2 Draw the mechanism that will produce both products A For the first product I leaves the carbocation fonns and Cl attacks aeyxas 4 r 4 Na 7 V For the secondproduct the caroocation re ges rst A c H group shi s This A H occurs as normal Vumblesthatu ett le Rate at an Sag Rea n Stereochemistry Remember that inversion of stereochemistry is associated with SNZ What about SNI InSNl 39 quot 39 A quot39 39 39 of stereochemistry We may say that we get aracemic mixture in SNl which means that there are equal zmuunts quhetvm enznuumers Huwever Lhsxs aslmph ed Way uf luukmg at n In the 51 reamun the leevmg guup actually gets m the Way quhe nucleuphde xfmeuueleupbue attacks un resultmgmmvemun Hen e er u u the umErs s usually nutpurely equal su we Banjust say that we get 2 mnzmre e m annnmm m Sm Nurleuphle In SNZ a beast nucleuphde meant that the rate uf reamun WEIde mares Because the rate Expressmn fur 52 15 rate 15 pmpumunal cu se Whyv nue R3C 7 LG meznmg n hm rwu quotbu mereases the rate ufnudeuphdm attack whmh ultimately mereases the rate u reamun In 5111 huweva the ts law Expressmn 15 le 15 pmpumunal m Rgc 7 LG meanmg n therate ufreammn Leaving sump In 5112 abetta39 leemug guup unease the rate ufreammn 1n sex 2 better Raving emu als umsesubeme ufxeamun Thereamunrate depmds sulely upuu huw g the leevmg guup leaves Subshtunun atthe leaving gunp FurSuZ chrLG gt RC 27LG gt chHeLG gtgtgt Rch Ths 15 because mure substituted muleeules mhxbxt nucleuphdm attack mmugb ancs Fursul chrLG lt RcHrLG lt chHeLG ltltlt ch Ths 15 because mure substituted carbucanuns are mure stable Salve vb 39b my b1 Vvaww is 215 calledmm ap thaluss ufaleavmg guup zanun The mefm39sd suhrmt s pals A solycnt wltln ahlgh dleleetxle constant ls polar Thls mcreases the rate of rcacuon A hlgh dleleetne constant rncans the solycntnas ahlgh ablllty to scparatc opposlte charges Thls stablllzes the charges ln the transltlon statc whleh lowcrs the energy of actwatlon and speeds up the rate of the rcacuon Methanol and Water are verypala solvents W dr w h wdw w h lrnportant here 1 s a 1 warmth A lt p r c In thls example the carbocauon ls shown as rcactcol wltln L Howcycr there ls TV a greater Usually the solycntwoulol act as the nuclcoplulc Example Problem thch reaction proceeds slnww Nuclcoplulc solycnt andleavlng group are all the sarnc for these two reactlons The only a stable so ltwlll rcactrnorc slowly A Rearanalllz S 1 Remain sq Lw LAT Rcrncrnbcr the rate olctcrrnlnlng stcp ls the L of There are three factors that make a reasonable reaetlon l arnoolcratc or a n roup 2 a stable carbocatlon a lncrc a pnrnary Wlthout rcsonancc probably would not work but a pnrnary Wlth rcsonancc woulol lvent 3 apolarso Example Is an Sul rncchanlsrn forthls rcactlon arcasonablc one7 h k rClaseeond n l r solycntDMso Thls seemsreasonable The factors fora sn2 nowcyc so seem reasonable rCl ls agoodleavmg group rBrls arnoolcratc nuclcoplulc and DMSOls a be amlxture ofthe two For thls class nowcycr wc lcarn that M ls usually favored oycr snl wnyt Cummings g ands z Carbonlehemlstry Fundamentals lmages and Informanon from Emloe P Drganu Chzmlsny Pearsons Prenuce Hall 2004 llarohngen s Che urry 14D Tilmkbook 2006 Lecture Supplement Carbonyl Che urry Fundamznmu Carbonyl group 7 a carbon double bonded to an oxygen 0 xquot Acyl group 7 carbonyl group attached to an alkyl or aryl group 0 o R C C Carbonyl compounds compounds contammg carbonyl groups Carbonyl compounds can be dlvlded mto two classes 7 Has a group attached to the acyl group that can functlon as a leamng goup Carboxylzte Acyl chlonoe A nhydnde Thloester o acyl haholes zcld anhydndes esters throesters carbonylates and zmldes are carboxyllc OH group Functlonal g up 7 re e the termtnal e of the alkane name wlth olc acldquot o mes group mrboxyl ro plac r 0 mon na d for carboxyllc aclds wth slx or fewer carbons r lacuc acld m muscles clmc acld m clmc trul s etc 7 Carboxyllc aclds wth amlno goup on the arczrbon amlno aclds llnlted together by amlde bonds to torm pepudes and protelns r Acyl Halids o OH group of the carboxyllc acld replaced by halog 7 Replace lc acldquot wth yl ch 7 Add Anhydrldes o Loss of mterfrom two molecules of a carboxyllc acld results m an acld anhydnde r Symmetrlcal anhydnde lf two carboxyllc add molecules are the same Rep ace l quot en londequot or l bromldequot acldquot wlth anhydr de 7 Mlxed anhydnde o carboxyllc acld molecules are dlfferent State names of both aclds m alphabeucal order followed by anhydnde Esters o OH group of the carboxyllc acld replaced by JR goup 7 Name the group 1139 attached to the carboxyl oxygen rst followed by the name of the add wth lc addquot replaced by ate 7 Lac ones cycllc e ers r commonly cause the tragances of trults and flowers 7 Amides o OH goup of the carboxyllc acld replaced by NHZ NHR or NR2 goup r Named by uslng the acld name replaclng olc acldquot or lc acldquot wlth acld r Aclds endlng wth Garbo llc l th r ctams cycllc amtdes Named 2rzzzcyclozlkznonesquot aza deslgnates the N atom zcld quot yllc acldquot ls rep aced wl 2m1dequot Suuauls of Carboxylic Acids and Cmboxylic Add Daivalives 7 three sptz orbltals to torm obonds to the mrbonyl oxygen the cedarbon and a subsutuent o tn nal planar o 8 5 llquot grees ndlzed sptz orbltal torms a obond wlth the carbonyl carbon 7 eac of the other two sptz orbltals contatns a lone palr o 7 bond forms between p orbltal of carbonyl oxygen and p orbltal of carbonyl carbon Compound Structure Aldehyde if R H o Ketone ll c R R H and alkyl ax aryl a an gully at aldzhydzs ana Imam an ma has a he nplacedhy a nucleaph e a many man s hamzd m a H and m an alkyl ax aryl yvnp a emeponn fm39maldzhydz z hamzd m Wm H nplace 11mm bquotfmm an nann at n parem hydracarhan mm 31quot have pnng n dms sex Van hn many man a hamzd a Ma alkyl ax aryl mcamn and mm WW nplace wquot Hann n nann at the pann ham 5 nnmheled m an mndmn ma gves an many man an number x Spmnm have x pmgesumm am msmsmmm Phme hwpuus mCulrmyl Cmnpmmds Relatwe Bu mg Pmnts e bagth mml estervacylcl39dnnde raldhydw hymns am Baum Hun Huanameaxamaamnaam annaxacmaaaayaaanzs mm am 4 mmemmsnmnammmaeamananaaacaaneagaa Carhaxyhc and farm mmrmalzculax hymngen hands 7 hgh hmhng palm Amldzs have an hghzs hmhng palms because an lawman munmum wnhsepanmd charges canmhmes slgm can y a an Wen campmmd swam mm amps am am an leavmg yvnp an male nacnve mm 7 an nnan lawman gains Hann leavmg an nnan as elechuns ale pus an an male stable hecamzsr gams lesamnc gmd LG 2 Atmnlc mm mg m hmlse mw elecuuns mm 12 3 human mac 7 male ekchvn wnhdlawmg mlghhms X Furmal c e r nnam Hann 9 nanua aldzhydzs am Imam na have Iza ung gully m cannm mdzrga snhstmmnn nammns m an many campmmds at Class RelauveRucuvmes afcarban lCam aundsTawardNuclea n 5 cyl gt and gt aldehyde gt man estuN carbuxyhcgt anndegt cubaxylale a a an huh annyan cxd mastxeactwe leastxucuve Aldzhydz lass acme than nacnvethar a p n r resides m the pnlanty ur the carbnnyl gmup 0 Oxygen is mare electmnegauve than Canaan nuclaopmle 2370 lane p an H p bonds or eleclmph e high e density R Y mm 1 Anne Nucleu hue atCarhnn lCarbnn a the ghly EN nygen acnep39s an electmn pair a carhnnyl adm39mn and manuan mechamsms can39am Lhs imp 1 0 L TE hednlAdth P r pmdn rdbynlwhuphlhc anuk M w 7 OH OH spquot2 0 9 spquot3Cte nthnl 2 Accept Electmphue Usually mum at oxygen a lane pans and p1 hand acceptan electmphue b Need smug Kean veamd Ex H30 HZSOA ROHzem 7 Wm alarm 15 mt stmng emugh m pmmnahe the narbunyl nygen w w 3 Ennlate Pumaan a dnven by cumugahe base resnnamers39abmzaunn M Gemnhzedcal na a cmm mum pnxrmmb w 1quot Hun m Inn p l n mmnMe mum mm m m K WWW x mm 10mm K nun Ho x addmnn mm revers nlz Wm ml mm a w present mm pmlzm Suhg m nn occurs r ln cuss n Whether x7 m 27 us expelled deperds on their relauve bsmmes r diveahxditase Lbbet znussam nz ismuch wash has than Y z v lhe ude 0 9 05 lg W i c g Z R t R W kquot 5 HR hm ankummm humming 1 nuclzuph e avach Lb carbonyl carbun 2 nmleophlle leaves and lefmms the mums T E H E T Edge nulls marsh 4 4 ng ss nulls readmn ngass nulls readmn Y so glen when mun moms WP ml mm ml gong 39 sucsunm mm mm opnun txlnmm tsnbsunmpmlm Relauvmes of Carboxyhc Aclds and Carboxyhc Acid Derlvanves Relahve haslslhcles of the L65 0 u weakest n 7 mm est base Ll VOCR 70R0H 7 M12 bag Relzuve reacuvmes of mrboxyllc zcld dmvzuves 9 O O 9 must l lslst c gt C c gt c gt c lolsllvu R U R O R R OF R OH R m mum Mylch undc lsld ester anhydnde 0 l c nulhnxy nd If uucleophlllc attack dlsrupts malor resonance then the molecule ls less rem Ex The reso lee nance between the mrbonyl and the arrllne group ls rrlore lrrlporlant than the resonance between the mr onyl and an oxygen bemuse N ls less EN so shares lts electrons rrlore radlly addlng rrlore slahlllty A r Ml n lr the newly added hase than the ongmal group ln the reactant Cll ll c c 4 c Cl P Cl R CL R O 1 l 9 C Cl t annumun l R o p Gelersl Mechanism for NudeophilicAcyl Substitution Reactions If the nudeoghlle ls lug 39 e1 charged mzulmly clmzti elmum39mr uf Eh View mdnphu ucrs m If the nudeoghxle rs neutral Innl Nude Z 95 O c H20 7 3 v y R r L OH R 0H mum nuclearhdn 3mm C0 Substimtion Rac ons Example from Thmkbook Lecture Supplement pg 28 7 Why rs proteru hydrolysrs much slower than phosgene hydrolysxs7 9 a O O H O u o RKN a k HINR mm X a k 21130 9 a OH c1 or 0H 0H Phasgmz hydlulysls fusteneactmn meem hydmlysls slnwznzuctmn med enzym cutalys What rs the mterdetermmmg step 7 mu RDS 1 meeuarusm step uudeopurue mack at 00 mrbon rrude era Chlonde Lezvrug Group LG emu rutmuou L 017 EN 3 0 EN 3 0 use LG does Irretemt bemuse LG does not leave m RDS Condusrou Irrelemt been notlave m RDS R unspeafwd R uuspeerned Stene Effects Condusrou 5 00 Nitrogai EN o cniorine EN o Magnifies carbonyl 5 Magnifies carbonyl 5 Conclusion 7 Resonance 8 NR KR 1K Kc Conclusion More resonance stabilization Slower Resonance stunninth magnum 39 0 Gum L H OFirupunum J ininu 393 ii 1 Km Second Mechanim Step LG leaves mide Cid Chloride LG LG NR2 a niLmnion L r r Conclusion Slower Faster Resonance Gain Gain resonance With oxygai Conclusion Same Same Conclusion amide siowerbecause slower LG amino acids HOW ALDEHYDES AND KETONES REACT CLASS in A substitution reaction does not occur because the carbonyl group of an aidenyde or a ketone is auached to a group that cannot leave 7 undergo addition reaction If tne nuclethile 2 adds to an aldehyde or keton o form io e s alkoxide n that can be protonated either by the solvent or added acid m HE rl R Ir z inu nuuiunii Rd Rel a or Clas ll Carbonyl Communds g on Carbonyl Carbon 7 a ck pamal posrhye charge on the mrbonyl carbon alzbles rllo acl as an electrophlle and be lla ed by a nudeophlle An aldehyde h e a kel o r llc as a grealer parual posrhye charge on us carbonyl carbon than a ketone one has electron donahng alkyl groups C n hm n akelone t 9 9 3 ms 0 gt c gt c lust mauve H H R H R R mauve fammldehyde aldehyde room Steric Fact rs Aldehyde carbonyl carbons are more accessrble srnce the H groups are smaller lhan the alkyl groups of kelone e kel ones wlth srnall alkyl groups are more ranclrye lhan kelones wth large alkyl groups 0 nr Wu C Co Addition Ramon s Example from Thlnkbook Leclure Supplernenl pg 29730 Whlch ls aslerv O H OH OCH3 O u OH OCIl3 JK NlUEI39D gtlt AUDIE 7 vexN5 cll3 Hm CH H CH3 smug gtltCH llZ CH1 3 Whal ls the mterdetermlmng step 7 ocll3 Rds Nudeophlllc allack on carbonyl mlbon I Kelone I ldehyde lenc Effecls cll larger ll smaller Conduslon slower Fasler 5e 00 cll ls electron onor ll ls eleclroueuual decreasmg uucleophlllclly Conduslon slower because smaller 5e Fasler Aldehde gt kelones Problem from Thlnkbook Leclure Supplemeul pg 31 NamX I K CH3 OCH CH3 0 a wme llle mechanlsm 01 0H 10 4 new 4 a 5 1 7 CHZKOCHK CPIsow cm 0 Heocm memyl never Telmlreaml leaves 7 eleel m 1mm faster ereeepl H mler ls Islm ybc oe ls very small can t mkz mw eleelmrs om rrslrlmlue lxnnsmnnshle thzncm leme l hzlp dzlncahze chmgz allaws ORm lelm 2 lore pms xzpel weakznbands b Rewnle the reacllou so ll ls obvlously faster 5 e m 01 OH HO 7K aczkoi 7 3 CH3 Ir ls a better LG 1 ls Keq lt l esler favored or Keg gt l carboxylzte f2vored7 4 04 quot all 0 Addition to CarbonCarbon Pi Bonds H20 gt H3C catalyst OH H2CCH2 Need a catalyst This is one way ethanol is manufactured How does this reaction proceed 39239 Pi Bonds H H Alkene Alkyne Two election clouds Many mechanisms that apply alkenes also apply to alkynes o v Generic Addition Mechanism R R elec electrophile nucleophile Glee R nuc gt R C R gt R R RDS R J Rgt R ltR Carbocation Intermediate Which fate Rearrangement nothing better than tertiary Lose proton no protons to lose Capture nucleophile 0 Addition of H X to a Pi Bond 0 X Cl Br I H C H30 0 H30 H 3 H Br gt26gt gt Cd or gt CH2 H30 H30 H30 Tertiary Primary less stable carbocation 3 is more stable carbocation Which fate Rearrangement won t stabilize charge Lose proton reaction will go backwards Capture nucleophile Br39 H C Major Product H3C H gr 3 Formed from tertiary carbocation intermediate C gt CH3 H30 H30 Br H3C Br H3C Minor Product CH4 gt CH2 l gt Formed from primary carbocation Intermediate H30 H30 Br This noted addition of Br to the more substituted end of the alkene was discovered by Markovnikov Markovnikov s Rule When a hydrogen halide adds to an alkene the hydrogen is added to the carbon bearing the most hydrogen substituents least substituted carbon and the halide is added to the end of the alkene with the least number of hydrogens H r f H x H goes to the least substituted carbon CH2 gt X goes to the more substituted carbon 39239 Addition of H Br to a Triple Bond H Br H HTCH3 gt C H2CC CH3 CH3 1quot Minor Product 2 Maior Product Major Product H c c 2 CH3 Carbocation fate Lose a proton 9 unproductive would move reaction backwards Rearrangement 9 Vinyl carbocations CANNOT rearrange H H2cc H2clt EH C H H Capture a nucleophile K H Br Br39 mH Br H20lt C 1quot without resonance CH3 H 00 gt 2 CH3 CH3 H Br Br Br Br Br H3clt 4 H3C C a H3C CH3 CH3 CH3 2 with resonance FINAL PRODUCT One contributor with complete octets 39239 Addition of H20 H2CCH2 H OH gt H30 OH Ethylene Example No H OH gt C H OH is not a leaving group unless you have a very good acid Mechanism A H OH2 gt a C H2O Carbocation Intermediate Rearrangement is unproductive Lose proton 9 reaction goes backwards Capture nucleophile H20 H3O This reaction is reversible It is in EQUILIBRIUM HO gt H2O Favors alkene just a little How do you control the direction of equilibrium Think of la Chatelier s Principle increase the concentration on one side reactants or products to shift the equilibrium to the other side 39239 Addition of H20 0 H2504 611 gt H30 H Ph Phenyl CeH5 A m H OH H20 H ph gt Ph gt C gt gt lt Ph 0 Ph Carbocation intermediate Resonance but no extra stability Rearrange to 3 with resonance pH Ax 0 39239 Alkynes and H30 H H0 H H30 Ph Ph Alkene and H30 CH2 CH3 OH H 0 Ph 3 Ph gt Alkyne and H30 CH CH2 Enol Mechanism H h KHIOH bow 03H 0H2 H quotOH Ph Ph C gt Ph gt Ph lt gt OH CH2 CH2 CH2 OH OH CH3 H 2 with resonance 1quot without resonance O H OH2 O Ph 2 Ph H30 CH3 CH3 Resonance contributor Major Product with complete octets The intermediate labeled 1quot without resonance can proceed to deprotonation and formation of an enol The keto product is favored greatly OH 0 Ph M14 H CH3 H Enol Keto These products are tautomersI which are created through tautornerization a process where a hydrogen atom migrates to another carbon and a double bond shifts to an adjacent bond Why not OH OH H20 OH Ph C V 2 Ph This product will go backwards to the happiest stage keto form CH3 CH3 o 0 Miscellaneous Additions a Catalytic Hydrogenation H2 gt Pt Synaddition new things are added on the same face Markovnikov s rule does not apply No nucleophile or electrophile Catalyst is a transition metal Think of it as removing the pi bond C39 v Addition of Brz Antiaddition Markovnikov s rule does not apply Electrophile Brz induced 5 539 when Brz approaches the pi bond Nucleophile alkene c HydroborationOxidation 39 1 BH3 gt 2 H202 HO39 quot OH Synaddition AntiMarkovnikov Electrophile BH3 Borane because ofthe open octet Nucleophile alkene Addition of water on the alkene converts it into an alcohol SN2 Ionic Substitution Reactions em 14D 8N2 Ionic Substitution Reactions Substitution can occur in organic compounds that have an electronegative atom or group bonded to an sp3 hybridized carbon S stands for Substitution General Mechanism RLG Nuc 9 R Nuc LG Nucleophile enters as leaving group leaves Nuc Nucleophile LG Leaving Group N stands for Nucleophilic Nuc Elec HOQHggl 9 HOCH3 Cl 2 stands for Bimolecular Rate kalkyl halidenucleophile Rate kCH3ClHO for the reaction above 2nd order reaction Concerted reaction because nucleophile attacks and leaving group leaves simultaneously No intermediates are formed Gedunker experiment Kinetics factors that effect reaction rate HO CH3C 9 reaction collision More often molecules collide the faster the reaction Example 1 Cars on a freeway The more cars the more accidents 2 Recall throwing molecule balls in class Chance of collision increases as concentration increases 2 x HO 2 x rate CH30 10 rate 10 Rate HOCH30 Rate k HOCH30 Caution No Trimolecular collisions A B C 9 D linear relationship e 9 Cadillac Audi More likely Moment of Collision HOCH3 Cl a collision Measured rate is related to the mechanism Kinetics can disprove a mechanism or AudiBenzCadillac collide altogetr 9 HOCH3 CL Lifetime s 1015s femto second doesn t last very long Transition state studied by Zewail H Getting closer Partial bond lone pair beginning to become OC bond Bond getting weaker further Cl starting to leave Halogens are more electronegative than carbon so they have a larger share of the electrons This polar C halogen bond causes alkyl halides to undergo substitution and elimination reactions Trigonal bipyramidal is the best way to arrange 5 atoms around a central atom Uh ohpentavaent carbon No central C still has 8 e so the rule is not violated ln Backside Attack the nucleophile attacks from the backside of the carbonleaving group bond due to o Electrostatics Negative charges on nucleophile and leaving group repel o Sterics steric hindrance Crowding leaving group blocks approach of nucleophile to the front 0 Hughes lngold noticed that the stereocenter will change won t happen in front side attack front Retention of Q N H n n c i N Stereochemistry HHIC SCH3Z CH3 A Inversion of CH5 bac Q Stereochemistry N CW 99 89R or Res but not always true 0H2 See virtually 100 Inversion of stereochemistry also called a Walden inversion this is good and bad 0 When a chiral alkyl halide undergoes an 8N2 reaction only one substitution product is formed I quotfquot f x f l r1 I I 4 Ho c em 1 R HE R R RJ3t Iqon H H Bruice p 366 0 Good says backside attack is an accurate model 0 Bad questions the 2 reasons above There must be another factor Real reason for backside attack Oquot nuczlt 1gt c i410 W Greatest stabilization occurs when orbitals overlap end to end The overlap between the orbital containing the pair of donated e by the nucleophile and the carbonleaving group antibonding orbital is maximized a Backside attack b Frontside attack empty of antibonding MO 5 c m emP Y Q I Bl Qty antibonding MO 4 an invphase Z on Ing 39 f b d i an out of pha V h se a anln p ase linterac non l antibonding bonding interaction NU interaction fl 9 c grill is d quot3quot u quote quot Lian 59 bonding MO O C i quot 15quot G bonding M0 Bruice p 364 Practice Problem from Thinkbook 4 Provide the organic products of this reaction If more than one product is formed indicate which product if any is the major one If no reaction occurs write NR 3 NaI acetone quot Ci Answer CH3 CH3 NaI gt u n acetone 8N2 With 1nver310n quot CI 1 8N2 Energetics o G HTS I Gibbs free energy G energy of whole system Enthalpy H differences energy due to bond changes usually tens of kcamo Entropy S freedom of the system molecules like to be floppy or more molecules entropy is a small factor perhaps only one calmol and even when multiplied by T temperature in Kelvin it is still smaller than H 0 At reasonable temperatures G z H Products and reactants aren t the only thing that matter the transition state has its own energy really important 0 Energy Profile TS can go forward or backward matters because need to get over this hill for reaction to occun Example breakfast Bombshelter If you have enough energy you will leave the classroom and eat breakfast CH30 HO39 Energy CH30H G is negative so it wants to Cl39 spontaneously go to products Reaction Coordinate Transition state T836 the highest energy point in the energy reaction profile due to partial bonds Energy of Activation G Energy needed to reach transition state controls the rate of a reaction 2 partial bonds don t make up for 1 full bond 2 partial bonds in 36 is energetically expensive Breakfast metaphor Rate of room emptying is a function of 1 Energy of the students 2 Height of the hill of stairs leading up to the doors G influences the position of the equilibrium but not the rate G influences rate of reaction but not equilibrium Example Cells 9 amino acids Can t wait for years for a reaction to complete so enzymes make the TS more stable This 3 G so reactions go faster A0 constant as fast as reaction can possibly go G energy of activation height of k A e G RT hill 0 t R gas constant ra e T temperature in Kelvin Arrhenius Eguation figured out before 73 Rate and G are inverses so UG rate Exponential fac or small c anges in G can lead to large rate changes Temperature temp rate Spontaneity If G lt 0 spontaneous reaction 8 02 9 802 H O 0 72 kcalmol Very spontaneous thermodynamically but in terms of rate kinetics it s extremely slow Block of sulfur isn t going to 802 If you heat block of sulfur it will convert to 802 Like a match heating initiates a rxn There are lots of hills in the rxn that convert sulfur G of one of the hills is probably very large So if G lt 0 then thermodynamically spontaneous If G lt 25kcalmol then kinetically spontaneous Practice Problem from Thinkbook 5 Consider this reaction a Write the rate expression for this reaction Cl 91 b Write a curved arrow mechanism for this reaction DMSO Br c Draw the transition state Answer a rate k R Cl Br1 or rate 0t RCl Br b Igiridgt ali gtgt7l3r1 lt gt i1 i C 5 B r ca C16 SN2 Variables A v Nuc R3CLG 9 nus 0R3 LG Nucleophile how does the nucleophile influence the rate TS IR 7 Nuc C B LG R R Role of nucleophile in partial bonds Share electrons it s making the partial bond The more complete the bond is the more stable the TS The single most important factor which controls the nucleophilicity or basicity of any molecule or ion is the ability desire or driving force to share an e pair Nucleophilicity ability to share e pairs with electrophile Basicity ability to share e pair with H Stronger bases are better nucleophiles Vi XY When comparing molecules wrth attacking atoms that vary greatly In size the polarizability ability to skew the electron cloud of the atom and the reaction conditions determine whether the greater polarizability of the larger atoms makes up for their decreased basicity The relationship between basicity and nucleophilicity becomes inverted when the reaction is carried out in a protic solvent the solvent molecules have a hydrogen bonded to a nitrogen or oxygen This will be explained in the solvents section 4 Factors influencing Nucleophilicity similar to basicity factors because both nucleophiles and bases share e 1 Resonance can increase or decrease e density at the atom that shares e with the electrophile but generally decreases nucleophilicity O 03 Q5 emo kw to H x Methoxide vs Acetate hybrid is reality Stronger nuc bc neg charge Weaker nuc bc neg charge spread out more concentrated over 2 oxygens a No res 9 no res CHgO CHg39Cl 9 CH30CH3 39 0 H we 9 0 Loses res when reacts so more hesitant to do so KOQ 3 A 06 H Doesn t want to lose resonance it s stabilizing like 2 Atomic Size refers to size of atomic radius of atom doing e sharing not molecule as a whole just the business end the end that forms a new bond with electrophiles CH3039 VS CH3S39 O is smaller so charge is more concentrated greater drive to share e better nuc O and 8 both have the same of valence e and formal charge of 1 but 0 has a smaller atomic radius Smaller atoms are better nucleophiles because they have a more concentrated e density and thus a stronger driving force to share e 3 Electronegativity measure of e greediness think of it as antinucleophilicity Higher electroneqativity means lower nucleophMy because the role of a nucleophile is to share e If the atom is more electronegative it is less willing to share its e and wants to hold onto them F39 vs HO39 EN 40 35 Poorer nucleophile better nucleophile 4 Inductive Effect the electronic effect of atoms other than the atom that is sharing e density can increase or decrease e density on the atom sharing e with the carbon Electron withdrawing groups decrease nucleophilicity Electron donating groups increase nucleophilicity CH3CH2039 VS CF3CH2039 Since F is very EN it draws e density away from 0 making for a poorer nucleophile XFactor doesn t consistently fit into this sequence of decreasing importance Formal Charge If you have more e density better nucleophile HOquot VS H2O More e density formal neg charge neutral Better nucleophile F I vs NH3 I A T033 Up Here s a case where EN More e densrty stronger nUcleophrle overrides formal charge But high EN Practice Problem from Thinkbook 8 Rank the nucleophilicity in aprotic solvent and brie y outline your reasoning e S 0 A56 use Answer consider 4 factors 1 Resonance No res No res Has res Least nucleophilic least willing to share 2 Atomic Size 0 is smaller S is larger better nuc concentrated charge So overall answer Best nuc Medium nuc Poorest nuc 11 Select the poorest nucleophile HO CH3CO2 CH30 hydroxide acetate methoxide Answer No res Has res No res Poorest nucleophile because res reduces e density Leaving Group portion of molecule that leaves with the pair of e that was the bond between the leaving group and some other atom role of leaving group is to accept e and leave the more easily the leaving group can accept and spread out e density the better it is R Metaphor Your significant other packs your bags Nuc p I G weakens your bond encouraging you to leave 39 Want to encourage LG to accept e and leave R R Relatively good leaving groups make alkyl halides convenient to study for substitution reactions Cells of plants and animals exist in mostly aqueous environments Since alkyl halides are insoluble in water biological systems use compounds in which the group that is replaced is more polar than a halogen and thus more soluble in water Consider the same 4 nucleophile factors 39ust reverse your logic The weaker the basicity of a group the better is its leaving ability because weak bases readily bear the electrons they formerly shared with a proton Better bases are better nucleophiles which are poorer leaving groups I B1 391quot 1 lquot 39 E s asef l I wake ham l least stablehirsg most etabiggie I Brurce p 367 1 l R Br tquot r lquot f worst leaving best leaving 33quotquot i gmup group Since basicity and leaving group ability are inversely related weaker conjugate bases 2 better heaving groups alkyl fluorides are least reactive while alkyl iodides are the most reactive of the alkyl halides quot 39 L3 V 1 x mil s Ri 1 Ia ggc g VE EU a hijl L Bruice p 367 1 Resonance acetate is a better LG bc it accommodates e density better due to res RQLW a o RQ JCH3 gt CH3C9 0 KOQ vs acetate 2 Atomic Size the larger the size of the business atom the better the LG because its e shell is less concentrated and can accept more e more readily and leave Alkyl iodides are least basic while halide fluorides are most basic because larger atoms are better able to stabilize their negative charge 3 Electronegativity the higher the EN the better the LG because it has a higher affinity for e 4 Inductive Effect can be e withdrawing usually or e donating so can work either way Withdraws e better LG Donates e worse LG X Factor Formal Charge RLG 9 LG Best leaving group charge being quenched RLG 9 LG39 Bad leaving group Neutral 9 charged unfavorable RLG39 9 LGZ39 Horrible very unlikely Almost never see 2 charge except in metals Charged 9 more charged Good Leaving Groups lodides and sulfonates are the best leaving groups lodides l39 good because of large size 0 O n equot Sulfonates dood LGs due to resonance ROSR 9 98R 2 o Diazonium good because it s a gas so once you make it it leaves Also good because 9 neutral Metaphor It s hard to get a dozen cats out the door because R191 E N 9 N E N they come back in but a gas only leaves Moderate Leavin Grou s Br Cl Like l butsmaller Like sulfonates but less resonance 0 H20 ROH O can accept e but it s small f R H2 Bad Leaving Groups not leaving groups unless special circumstances F Too Small HO39 CH3039 039 not stabilized by resonance Never Leaving Groups H Poor EN very small H3C carbon anion Negative charge on C not stabilized by resonance Practice Problems from Thinkbook 17 Label these leaving groups as best middle or poorest CF3803 CH3803 CH3C02 Answer 3 res contributors 3 res contributors 2 res contributors Inductive effect e donation by CH3 e donation by CH3 Effective e dispersion Medium e dispersion Worst e dispersion Best Middle Poorest Steric Effects results from repulsion by two or more atoms or atom groups decreases reactivity when groups are in the way at the reaction site Note Steric effects affect nucleophilicity but not basicity Strength of a base only depends on its willingness to share its electrons A bulky nucleophile cannot approach the back side of a carbon as easily as a less sterically hindered nucleophile R3CI HQ H3CCI9HOCH3 V Ho a H3C a H3C l A CH2 CHQ Cl HO Siv2 Rate 8N2 reactions are sensitive to increasing steric hindrance at the electrophilic carbon H3CLG gt RCHZ39LG gt RQCH39LG gtgtgt R3CLG Methyl 1 2 3 no reaction Fastest slowest Tertiary alkyl halides cannot undergo 8N2 reactions because the three alkyl groups make it impossible for the nucleophile to come within bonding distance of the tertiary carbon The larger the group attached to the C attached to the LG the more hindrance harder for nucleophile to reach electrophilic carbon the slower the reaction 39SNZ 1225mm 12125 decvezse 25mg rumba m 522 m submems uses manual a t o gmo 7 aneur39 r R cu y no h n 7 k 77 7 mngmnuu nmmun mgmmruw numn Eiiemn m 2 aenczHy hmdeved secundzvy mm mm W hydmwde mn Emma p 365 me Supplemem Hzr ingnr s Yhinkhnnk p s E m IndranceinlhchRL mm m um um mu m m ml acll I uWynnewer mm In m 52 wadlnu m mm mmmr Mn N Hm I x lpluh llanldppmuhlhuulvbm t m groupImmummum mlIhc5N2mmllnummwhcn themde wu mx mnulvhmvquot u In mm M mc mm uulym Hm quotMum 1m mudle wuqu A rrznnnnhh qmwu m gmummmm vmu quotI ma mu mm Hm mulcu culru mml nhmg Ilvc KI mm mhlhclmlm pm m Anny I39hnvslluszlmcwrwmIhcvulclemvlulclhuMu c5 mm H ruamgnmm rcplxlccnl by mmle Humopium Bullnnd id Mud sum mu MmM anorech umeansnc mom accurate demcnan medel evely calhsmn 3 Mcmyl mm L 391 1 ms mm W V rm39rlwl WWW s M M m 113 we s ac p quotWe cannec ed m another c b tutore methyl groups I mlrwlJr39mprrrrr more bfuclraga l lx rrpyrip lHLIhILI Mobil L D 0 quotlm rrate a Hui L L u39ru mrr rrr39t39 r39 1quot i Less collisions are productive Harhm is completely blockedquot 1 I will mull Jrtrain k irer lairljpl rnrlrrlcr r39ll39ri l L r39e39l39llrrr v39 I39139 39r Practice Problems from Thinkbook 19 Select the slowest reaction Briefly explain your choice 69 e 69 6 W Na SCH Olt O Na SCH3 Br 7 SCH Br CH OH 3 a 3 scrr3 OR b CH3OH Answer Reaction a is more sterically hindered at the electrophilic carbon so it is slower than reaction b Solvent Effects What is the role of a solvent in a reaction Serve as buffer absorb heat in reaction so temp doesn t rise too quickly Dissolve stuff in solution so particles can react with each other Can very significantly influence the rate of an SN2 reaction H20 KF gt Kaq F39aq Solid H K has charge wants to be neutral H0I H a4 0 F39 has charge wants to be neutral 2 H quotI H 0 r39 quot Quanti vs Qualit I 39 V I 0quot I o E n H 0 Lots of FH bonds but quality not that good n H 7 However quantity makes it strong H H 0 gt4 0 H F is a nucleophile H bonding decreases ability of F to share e Z4060 H20 mOIeCUIeS around K solvent occupies some of its e density How does the solvent effect the reaction rate The energy of activation determines the rate of an 8N2 reaction If solvent stabilizes reactants more than the transition state reaction is slower because East energy of activation increases If solvent stabilizes transition state more than reactants reaction is faster because Em decreases 8N2 reactions need polar solvents to 1 generate nucleophile dissociate salts to give cations and anions 2 help leaving group leave and stabilize the LG 3 stabilize transition state partial charges H Bonding In general H bonding decreases nucleophilicity because it ties up e density Protic capable of donating H for H bonding attach to O N F XH OH bond most common Separate cation from anion enhances nucleophilicity Example a referee in a boxing match separates boxers then they re more exposed If you increase increase separation increase nucleophilicity Dielectric constant measure of ability of substance to separate ions Nucleophilicity is highest when the solvent is polar but cannot H bond with the nucleophile polar aprotic solvents are the best for 8N2 reactions The relationship between basicity and nucleophilicity becomes inverted when the reaction is carried out in a protic solvent the solvent molecules have a hydrogen bonded to a nitrogen or oxygen Definitions from Hardinger s Thinkbook and added details from Bruice Dielectric constant measure of ability of substance to separate ions Polar solvent has high dielectric constant 520 stabilizes reactants or transition states with large charges better than those with small charges do Nonpolar solvent has low dielectric constant s20 stabilizes species with smaller or no charge In most nonpolar solvents ionic compounds are insoluble but they can dissolve in aprotic polar solvents such as DMF dimethylformamide or DMSO dimethylsulfoxide Protic solvent capable of H bonding usually needs an NH or O H bond solvent molecules arrange themselves so their partially positively charged hydrogens point toward the negatively charged species causing an ion dipole interaction The solvent shields the nucleophile so at least one of these interactions must be broken before the nucleophile can undergo an 8N2 reaction It is easier to break the iondipole interaction between an iodide ion a weak base and the solvent than between a fluoride ion a stronger base and the solvent because weak bases interact weakly with protic solvents while strong bases interact more strongly Aprotic solvent does not have an H atom that can participate in H bonding No NH or O H bonds Aprotic polar solvents Do not have an H attached to an O or N so there are no positively charged H s to form iondipole interactions Have partial negative charge on their surface that can solvate cations but partial positive charge is on inside of molecule making it less accessible Thus a naked anion like uoride is a better nucleophile in DMSO than it is in water Lemar Supplulml Hamiwur39s mama p I I Dun39l memorize Dapmhlems mmquot m camman so venls Allmllerm mm 1 numnwmm mm wnussunmmmmn mu er 3 A mammnm m wwwm um m u y 1 mm MW L39Jm39 mmquot Sulunu Hm N m mmquot kwmdmm Imxxmm mm rmquot Snz veaexions lever polar apmlic solver m mm Organic Suhmb used in Sn2 rxns H J n uman mm Humuquot M v m u m Mquotva mmm 1 mm y y an m u m H mm muin mm m m mm 5 1m 1 mm u I Wm m way m w m I um mm mm pawn u Wu pmm me Snpplmzm Hzrdinnu s YMnkhnnk p 7 Sulwul mm nu n Rum n a T1h152 mm g s m H mm Ma nlytampamgHbonmng nudeapmhc y n u 1mm umnqummzm mlcuph v mw W nmcnmlmm MW 9 m 1 v m um Hum m m 7 5pm Y MunWm m u W 1m mun a m w w Mnm 4 LINuh M NH mm me wwm w nmw thw rwerss Relauunshm Evenone lakes a decrease m nuc eopmhcny mix bunakedm ammmts L 3x rams 01F ha more amused er Lame atoms don t agtlte as much a a loss Small alums lake more m a 055 rm smuwwmmm c Numqnxmyavauwams 2 Hugglossmuumaonhmony Infancumumlmlutusnl n in Immune cw Fsuaerbaa pro t3pmquot palarnonpo ar Var Snz Practice Problem from Thinkbook 13 Select the slowest reaction Briefly explain your choice 6 6 6301 Na OCH3 Br quotgt SCH3 Bf M OCH3 CH3OH OR CH3OH Answer CH30H is a protic solvent Because of hydrogen bonds protic solvents decrease nucleophilicity Therefore CH38 is a better nucleophile than CH30 because smaller atoms have more concentrated electron density so they hydrogen bond better As a result the reaction involving CH30 is slower 30 Consider this reaction Iquot iBr e 9 I Na CH3OH CH3 a Draw the product of the reaction b Write the curved arrow mechanism for this reaction including the transition state for each step c Changing only the electrophile write a complete reaction that is clearly slower than the reaction given above Briefly explain why your new reaction is slower d Changing only the nucleophile write a complete reaction that is clearly faster than the reaction given above Briefly explain why your new reaction is faster Answer a Note inversion of stereochemistry Isak CH3 b e 1135 s ng3 CH3S H H H Qt Mpm 39 z ZCH3 CH3 x Br 6 5617716 as same as BT CSCH3 c CH3 CH3 CH3 CH3 mu CHasNagt quot SCH3 CH3OH CH3 CH3 F is a poorer leaving group than Br so this reaction is slower You can also increase steric hindrance at the carbon undergoing substitution to make the reaction slower d 1 Br quot NaI gt CH3OH 3 CH3 CH Carbonyl Chemistry Survey of Reactions and Mechanisms Course Notes Chemistry 14D Images and sample reactions taken from the Chemistry 14D Thinkbook for Fall 2004 and Organic Chemistry by Paula Yurkanis Bruice 4th edition Carbonyl Chemistry Survey of Reactions and Mechanisms When dealing with Carbonyls we consider two general mechanism types 3a IfX is a leaving group then you kick out the leaving group and the result is a SUBSTITUTION REACTION XLC A Nucleoph ilx c gt carbonyl 1 3 N uc substitution 3 x 3 N no X a LG H k2 Nuc l we start with the carbonyl which is attacked by the nucleophile X V A ddin39on g N uc 3b IfX is NOT a leaving group the O39 accepts a H and the result is an ADDITION REACTION 2 we then get a tetrahedral intermediate whose fate is determined by the presence of a leaving group Carbonyls Have X as a Leaving Group Don t have X as a Leaving Group Carboxylic Acid RCOzH Aldehyde RCHO U 0 H C m H313 ll Acyl Halide ROX Ketone RCOR 0 lt1 Hf 1 H r CH Acid Anhydride RCOzCOR H I ll3L39 L le Ester RCOzR 0 H10 OCH3 Amide RCONR z l H NH Nitrile RCN doesn t look like a Carbonyl because it doesn t have a CO but it reacts very similarly H3C4CEN C 0 Survey of Reactions 1 HEMIACETALACETAL and HEMIKETALKETAL FORMATION Aldehyde 9 Hemiacetal 9 Acetal o no R39 A r e ou 399 do quoti j oR39 R H RH R H Ketone 9 Hemiketal 9 Ketal quot Ketonealdehyde alcohol hemiacetalhemiketal after one equivalent of alcohol acetalketal after two equivalents of alcohol hemiacetal hemiketal and half all start with the letter h this is a reminder that when you produce the hemiacetal or hemiketal you are halfway to the acetalketal A hemiacetahemiketal has a carbon attached to an ether on one end and an alcohol on the other 0 ie RO C OH quot An acetalketal has acarbon that is attached to two ethers if you take ahemiacetal and replace the OH with an OR group then you get an acetal o ie RO C OR First let us look at a generic mechanism for the formation of an acetal or ketal Here we start with the ketone so we will be forming a hemiketal and then aketal 39 in step 1 the oxygen is protonated This makes the carbonyl carbon more electrophilic by giving it a greater partial positive charge 39 Then the rst equivalent of alcohol attacks the electrophile in step 2 39 Then the tetrahedral intermediate is formed when this is deprotonated by the B the hemiketal is formed as seen inside the orange box The mechanism doesn t stop here the remaining alcohol group on the molecule is protonated by H B to form H2O as aleaving group with this a double bond is formed Then the second equivalent of alcohol attacks the electrophilic carbonyl carbon quot Another tetrahedral intermediate is formed o The uocutis deprotonated by the B and ultimately the ketal is formed as shown in the pink box Now let s look at an example of an intramolecular formation of hemiacetal as seen in the assembly of cyclic glucopyranose from acyclic glucose IIH Hm HH Hm H l UH t l ufl39 HM n HU on HI I H1 IN I L i glmulnrllmwu I uly39mu 39Llrkglucopmnch laCCILl We start with this acyclic glucose We get these two diastereomers as products through identical mechanisms This is an addition reaction where a nucleophilic alcohol attacks the most reactive part of the molecule which is the aldehyde Since we are focusing on the aldehyde let the rest of the glucose loop structure be denoted byC Mechanism remember that this whole process is reversible 58 aquot r quot iron m Thurnigh 7 quot A 39 H 8H E 3 quotV l quot 20 a H x 5 09 39quot39C OH quot C 0H t I I l O H H H 1 2 3 4 0 In step 1 the alcohol attacks the carbonyl carbon forming the tetrahedral intermediate found in step 2 The H20 here serves as a catalyst The OH is underlined in bold yellow it serves as a proton bus in the reaction by shuttling the proton around For instance in the second step it protonates the negatively charged oxygen in the second step In the third step it deprotonates the alcohol Why should we protonate the 0 before deprotonating the OH In this case it doesn t really matter what the sequence of the proton transfer is The nal product is formed and water is regenerated in step 4 Is the following a valid mechanism step NO This is not valid because 4 member ring 90 0 intermediate has ring strain 39Cl T Fl 0 There is a 13 H shift C 00 c OH 0 This does not happen because of the nature of 391 l 3 l the orbitals H 0 This process forbidden H How to speed up the reaction Why are hemiacetalacetal and hemiketalketal formation faster in the presence of an acid 0 Because alcohol is a poor nucleophile it usually helps to have an acid catalyst Protonating the alkoxide R0 of the tetrahedral intermediate shifts the equilibrium towards the product side Additionally acid can protonate the carbonyl oxygen this increases the electrophilicity of the carbonyl group Why are hemiacetalacetal and hemiketalketal formation faster in the presence of a base 0 Deprotonating the tetrahedral intermediate s oxonium ion R20H makes the deprotonated atom a poorer leaving group The deprotonation also converts the weakly nucleophilic alcohol into a stronger alkoxide nucleophile How would this reaction occur in a living cell 0 Inside a liVing cell the proton shuttle is more likely to be an enzyme containing a suf ciently acidic proton and the lone pair would function as the base 11 IMINE FORMATION AldehydeKetone Primary Amine Imine u R N O 11 u R L1H RAH low Ki NHz KUH RAR39 Imine is a compound with a CN bond 0 Imine formation requires small amount of catalytic acid 0 An example of this is the reaction of retinal amp opsin to form rhodopsin l39 NH opsin R I csl u prulein R C H39 N in rcmml rhodopsin In the mechanism we will use RNH2 as the opsin and O as the retinal ret mechanism for imine formation First the amine attacks the carbonyl carbon as seen in st Then the alkoxide 0gams aproton 2a andthe Nitrogen 1025 a proton 2b to form a neutral tetrahedral mtzrmzdmtz O W nitrogen is more basic than oxygen it can be forcedtowards the imine by removing water Precipitation ofimine produced want water to leave 39 39 39 39 39 v p hence we need to deprotonate the positively charged nitrogen first in 5 Now in 4 we protonate the a1coho r then deprotonated to yield the imine This proceeds in an E1 like fashion ea A 9 R H K I l H Ha A H H e iLmij 4 7 les 1 e 0 lt gt HN t H 0H gtH N U 4 RNHa 3 1 r 3 H rul H rcl H C 0H1 R as W N a gt N 0 2 gt i 4 k gt i H30 Km H 9 m H m H rei 39ihereacu39on 39 w 7 a a am am 39 NC bond forming the imine R N H k at a The mechanism step shown below CANNOT REALLY HAPPEN in a lab It can only really happen in an enzyme Where the substrate is held tightly Why can t it happen Because it is a TRIMOLECULAR COLLISION Which is unlikely low do imines HOE OH 1 compare with carbnnv 7 let R e a I H9H2 1 N Car on ls Imines Ncarbonyl can react because N carbon ofan imine Nucleophilic of5 on carbonyl carbon has smaller 5 because nitrogen attack at the oxygen can readily accept pair is less electronegatiVe carbon of electrons on C70 pi on can not accept electrons in the CiN pi bond as readily as 0 en of a carbon 1 H20 0 9 Nimine is less readily attacked by OH2 nucleophiles than a carbonyl CH3 39NCH3 H gt N a X OH Electrophilic Nimine nitrogen has a lone pair and is attack at the Ncarbonyl can react because of two less electronegatiVe than ox en Oxygen in lone pairs on oxygen Nimines will react more readily with the case of electrophiles than a carbonyl carbonyls H or HinH O 9 0 CH3 Nitrogen in I H the case of H20 HinH N 9 imines T CH H20 Nsame number ofresonance forms as N same number ofresonance forms as Enolate the imine the carbonyl formation Noxygen heteroatom nitro gen heteroatom 39 more electronegative 39 less electronegative hence it 39 39 can t stabilize negative charge I gtamp RC as well A CH2 Hz Nimines won t form enolates as easily 6 as carbonyls HOJ need a strong base like LDA III FISCHER ESTERIFICIATION Carboxylic Acid Alcohol Strong Acid 9 Ester Suppose you want to make an Ester what would you use If we react a carboxylic acid with an alcohol there is NO REACTION You might be thinking that perhaps carboxylic acid is not a good enough electrophile Ph Olt it s lone pairs give it resonance therefore it is resistant to nucleophilic H You might want to deprotonate CH30H to make it CH3039 but this doesn t work because it would rather grab the acidic proton from the carboxylic acid instead of attacking the carbonyl carbon In other words instead of following the black arrow it would rather follow the pink arrow Ph Ph CH30 9 Olt NO REACTION OH OCHs In that case you might want to change the leaving group from an OH to a halogen such as Cl ie react with an acyl halide instead of a carboxylic acid however since halogens like Chlorine are reactive we realize that we want to start with a carboxylic acid 0lt 39 The reason that there is some delta on the carbonyl carbon is because the more electronegative oxygen is taking some electron density away from the carbon in a carboxylic acid 0 5 9 this 5 is not very big because of electron donation from resonance OH So how do we make the carbonyl carbon more electrophilic We can protonate the carbonyl oxygen this would make the oxygen more electron de cient and make the 5 bigger Ph Ph O vs HOi protonating the carbonyl oxygen OH O H makes it more electrophilic What can we use to protonate the carbonyl oxygen We can use a strong acid like H2804 which does nothing but protonate When do we protonate and when do we not protonate Nprotonation increases electrophilicity of carbonyl carbon making it more susceptible to nucleophilic attack Protonate Doesn t need to be Protonated Less reactive carbonyls such as esters reactive carbonyls like aldehydes and esters and amides ketones only the strongest nucleophiles like when the nucleophile is very weak such LiALH4 and Grignard reagents as H20 So what actually happens carboxylic acid alcohol strong base an ester Ph CH30H Ph 0 5 4 OH H2804 OCH This reaction is called a Fischer esterification where we convert the carboxylic acid into an ester using a small alcohol and a strong aci Fischer Esteri cation Mechanism CH30H HTSSOgH 9 CH30H2 N rst we protonate the alcohol using strong acid Now one of two possibilities can happen either A the proton goes on the carbonyl oxygen This is better The oxygen atoms O are more stable because of resonance 0 KXHQDCII L 39Ph OH 39 gtPh 6 B The proton goes on the alcohol of the carboxylic acid Doesnyt increase resonance but i ngc 0 due to subtle inductive effects a 1 alcohol is more basic PhAOH A quot5 Ph kom lt gt Ph 0 After protonating the carbonyl oxygen we move on to the step with a yellow star near it This is the attack of the alcohol on the carbonyl carbon Then we see in step 3 that the alcohol grabs the hydrogen from the positively charge oxygen in the HOCH3 group in the tetrahedral intermediate Then in step 4 one of the alcohol groups gets protonated so that it can leave as water in step 5 The carbocation intermediate in step 6 has a resonance form where oxygen takes the positive charge that was formerly on the carbon when the oxygen of this resonance structure is deprotonated by the alcohol in step 7 the end product is an ester and a protonated alcohol H II 191 l H n H Q 3 llQWH t UH m 1H HH I H p gt gt X X 1h UH Hi In 4 M u m u H CHM 1ampsz 4 5 H bl lmV39H1 H 0 r 9 l H Hl 6 gt A 7 gt U gin 7 xvi9 urn 1m mum l h or 1 quot Ph W Dellydralion may also occur by an l39iZlilxc pallm ny rn fmt an EB k HA 1 H H Ph Ht K39Hi K 11 Tetrahedral intermediates that are the same are underlined in light blue to show that the different mechanisms arrive at the same products For the E2 mechanism a bimolecular interaction occurs where dehydration occurs simultaneously with the oxygen reforming a double bond with the carbon while an alcohol deprotonates it In the above reaction the carbonyl is protonated after methanol because methanol is a stronger base In real life There isn t much stability difference between a carboxylic Acid and an ester To drive the reaction towards the ester we can use Le Chatelier s Principle 0 To go towards the ester use a large amount of alcohol or remove the water as it is formed 0 To go towards the alcohol add a lot of water Note that adding a lot of acid does not in uence the position of the equilibrium you just have to make sure that the acid is strong sufficiently acidic to protonate the carbonyl or alcohol In the reaction where H49 PhCOzH CH3OH 4 PhCOzCH3 HzO In the case of an aldehyde or a ketone it s easier for the alcohol to attack the carbonyl carbon however the alcohol has more dif culty attacking the ester because of the resonance involved OH OR I OR R OH l 0 R O RiCiR in the case of an ester 9 07 4 RiCiOR R 4 R ketone OH OH IV ESTER HYDROLYSIS WITH A BASE OCH3 NH3 NH3 Olt gt In this reaction a new CiN bond is formed First the nitrogen of the NH3 attacks the carbonyl carbon forming a tetrahedral intermediate Afterwards the OCH3 leaves and the new CiN bond is formed V FORMING NEW C C BONDS carbonyl Grignard reagentalkyne 9 a new C C bond GRIGNARD REACTIONS carbonyl Grignard reagent 9 alcohol and a new C C bond In order for carbon C to be nucleophilic enough to attack the electrophilic carbonyl carbon it must have a sufficiently negative charge ie R3C39 functions as the nucleophile H3C39 this is the carbon anion or carbanion In these reactions we want to form a new CiC bond Can we use methy H3CiH e H3C39 this doesn t happen because pka N 50 for the hydrogens attached to the carbon in the methyl therefore it is hard to take away the proton But in reality we don t need the full negative charge in the Carbon for it to be sufficiently nucleophilic To give Carbon a 6 then it has to be attached to something that is less electronegative than it 39 Carbon is more electronegative than any metal 39 Metals generally have low electronegativity 6 6 H3C Mgt electronegativity less than 25 make a C Metal bond Electronegativity 25 Usually M a metal This metal is usually Magnesium Mg Hence Grignard reactants are organometallic this means that they have a CarboniMetal bond Why Magnesium Mg prefers to be in a 2 oxidation state base How do you prepare the Grignard reactant H3C Br Mg I H3CngBI the solvent has to be an ether because it doesn t have any Cl ether lone pairs to donate I How to use the Grignard reactant 6 6 0 OH h H LNNQBr rh0 Phi 05 R H ether I CH3 CH3 Overall Grignard Reaction 1 Mg ether OH Ph 2 carbonyl CH3Br 3 H30 Ph CH Nthis makes a new CNC bond N very versatile reaction N can use a wide variety of reactants the only real limitataion is that because Grignards are such strong reagents you can t make them in a reasonable acidic solution N can make a lot of alcohols can use alcohols in the reaction N de nitely cannot stick this in water or any acid for that matter until the Grignard is completely done reacting Grignard Reactions AldehydesKetones vs Esters Ketone or aldehyde butwe39re usmg a ketone here 1 HC7 Br 0 v i Mggt Phi R 2H30 OH H30 H 39 PhifiH CH CH We have seen that aldehydes and ketones undergo addmon 7 or 5 or 0cm 1 gt7 Br H the reaetron does not end here It keeps gorngt o Phici CHchchg R Ph 2 H30 No leavrng group to There rs rnore Gngnard m the s eject so addmon oeeurs o1 tron so the new Ejectleavmg carbonyl reacts wrth rnore Nueleophueattaeks group Gngnardreaetant e bonvl carbon 39 y 06 x e M u 9wquot am 7quot 00 Fh M5 The ester proeess and eonverts rnto the alcohol An attaek on the esterarmde rs SLOW39ER than the attaek on aldehydeketone YN39E ANION REA TIONS reaetrons were for one or more carbons R7 ice Na g e 1270 Thrs H has pKa of 25 Hybnduan on of carbon Meehanrsrn CH3 H30 R7CC39 olt 4 Rig 0 4 R CH3 the alcohol stru has the c cbond N biggest drawback if you wanted to onl one add carbon it is di cult because it s hard to separate the CfC bond so that only one carbon attaches V METAL HYDRIDE REACTIONS Metal Carbonyl alcohol With the Grignard reactants we have created new CiC bonds what happens when you want to make a new CiH ond We use metals with a hydrogen attached to it Why We want a hydride ion H that has a negative charge large enough to attack the carbonyl carbon 0 We need to use a metal that is less electronegative than hydrogen 0 2 are really useful for this process I NaBH4 a borohydride H I it isn t very polar but it has a negative charge that it would like to get rid of the blue bond or any bond for that matter can be seen essentially H H 39 LiAlH4 Mechanism for reactions with NaBH4 or LiAlH Net addition 039 O H o occurs because 8 NXBHLI L Doug none Iofthe ph case 3 ph Oquot Ph quot3 substrtuents are a 8 H H 3 leaving group we could have used NaBH4 as a proton source but it s concentration is low compared to the solvent so we use CH30H instead Although NaBH4 is a good source of creating a new CiH bond it does not react with anything that is less reactive than an aldehyde or a ketone O H NaBHg alcohol OCH OCH NaBH4 is chemoselective in the above reaction this means that it operates on one mctional group and leaves the other alone If we want to react with less reactive carbonyls we have to ne something stronger ianNaBPh HBH3 gt BH3 in order for it to be more reactive you have to make B less willing to have a a negative charge Nlooking down at the same row as Boron we find that we can use Aluminum Al which is further down than Boron This is less electronegative so it is more inclined to dump the hydrogen More reactive source of hydride Good enough to attack esters anhydrides amides etc HOWEVER YOU CANNOT USE A PROTIC SOLVENT due to it s high reactivity So our new reactant is Li H i AlH3 LiAlH4 reacm similarly to NaBH4 except that it is stronger For aldehydes and ketones reaction is simply addition because there s no leaving group For other carbonyls with a leaving group you get two new CiH bonds when reacting with LiAlH4 Look at the following mechanism 0 6 g DH 0 z akomc gt BriaIr PhGDR p H PM a Pk H P AlHg Al Ha lt reacm just like NaBH4 except that since there s a leaving group substitution occurs before net addition is achieved LiAlH4 can also react with amides to form terminal amines by getting rid of the carbonyl jk 1 LiAlH4 RgCHzN39Rz R NR 2 H30 LiAlH4 is so strong that it can even react with a carboxylate ion X le H OH R o 2 H30 I mgtlt H Overall things to keep in mind when doing Carbonyl Chemistry In terms of rotonat39 the Carbon 1 with an acid Must take into account Strength of the acid 0 Protonation of a carbonyl requires the strength to be HgO or stronger Basicity of the nucleophile Protonation may not always be required Some nucleophiles like NaBH4 LiAlH4 alkyne anions organolithiums and Grignard reactanm are strong bases and will not do well in the presence of a strong acid therefore in the presence of these reactanm do not protonate the carbonyl with a strong acid What happens if a Catalyst is Present See Thinkbook Lecture Supplement Carbonyl Reaction Catalysis General Base Catal sis catalyst base N use a base to remove a hydrogen from the nucleophile N Nuc is a stronger nucleophile than NuciH NuuH Bust H 6 V uXL 3 Example Peptide Hydrolysis H HI 9 1 fl D l I the pink line indicates where the bond is broken when the base attacks the carbonyl carbon Hydrolysis is faster with OH than without it 0 Without OH the nucleophile would be HZO General Acid Catalysis catalyst acid Acid speeds up the reaction 0 By enhancing carbonyl electrophilicity o It shi s the equilibrium towards the tetrahedral intermediate a neutral tetrahedral intermediate is less likely to kick out the leaving group Strong acids provide H some examples are 4 o H30 o ROHZ Example Peptide hydrolysis the reaction is slower without H30 because the protonation of the carbonyl oxygen increases the electrophilicity of the carbonyl carbon making it easier for the water to attack it Emmatic Catalysis catalyst enzyme Enzymes catalyze by 0 Providing proper orientation 0 Stabilizing the transition state with factors such as hydrogen bonding and other stabilizing features Example Peptide Hydrolysis with Chymotrypsin o Chymotrypsin is a serine protease it effects the selective hydrolysis of the peptide at the carboxyl end of the phenylalanine Phe N0 Chymotrypsin VERY slow reaction What does Chymotrypsin do 39 Mechanism occurs at the active site First the imidazole deprotonates the OH The imidazole deprotonates the OH Start the nucleophilic attack on the carbonyl dmpnnhin pm Tetrahedral intermediate fragm ents W nlevwdmlk 3le M W VJW ma Hydrogen bonds stabilize the transition state mnsl prntnnate the bad leaving grnnp Both acid catalysis and ase catalysis occur in areas near the yellow circles JQ kl i u H ELIMINATION REACTIONS E2 and E1 Chem 14D Winter 2006 Credit to Professor Steven Hardinger s Chemistry 14D Thinkbook Mnter 2006 blue version and Paula Bruice s Organic Chemistm 4th edition from which information diagrams and examples for this project were used In an Elimination Reaction in particular a elimination an alumni gmup ar atoms lemma Admith Ilnnd conEEquEmlv arms group is remved from a naruun between the two carbons 0 while a pmtun BH l5 rennveilfmman a Team Eamon Thus in an elimination reaction a CC pi bond is formed Le 00 sp3 bond alkane a CC sp2alkene a 050 sp alkyne E2 Mechanism General 9 v 4A E L l m wn 1 Base takes awayH39 lrom ie deprotonates the carbon that is adjacent to the carbon attached to the leaving group 2 The pair olelectrons from the CH bond move to occupy the p orbital between the HCCLGA pi bond is formed 3 Leaving group leaves Kinetics Notice Steps 1 3 occur in one step The E2 mechanism is concerted No intermediates are orme Also lormation olthe product with a CC pi bond depends on Mthe concentration olboth reactants R3CLG and base Thus the rate law for E2 can be written Rate kR3CLGbase The rate law depends on the rst orderconcentration oltwo reactants making it a 2quotd order bimolecular elimination reaction and giving us the 2 in Example from Thinkbook Lecture Supplement pg 8 H 39 0041 1 li F G lt9 U 2 s Naocna i i 3 x 5 3 f Du r Step 1 DEPROTONAHON OF HYD ROGEN 39OCH3 is a strong base It deprotonates by taking the H 39om Carbon 2 Step 2 FORMATION OF CC PI BOND The electrons 39om the 3H bond form a pi bond between C2 the carbon and C3 carbon Step 3 LEAVING GROUP LEAVES CI39 the leaving group is ejected 39om C3 Note H could have also been taken 39om C4 However in this case the cyclohexane is mmetrical so the product would have been identical to the one provided In other cases when carbon is asymmetrical elimination ofhydrogens 39om different adjacent carbons adjacent to the C LG will result in the formation of multiple products In such instances we must consider hctors ofst ility Using Example from Lecture Supplement pg 8 Hquot Ti KOH A w IFHiIx removed 9 Br iFHnixmmovea First let us consider Torsional Strain of the Transition State TS We can use Newman projections to determine which transition state ofthese three products is most stable and therefore which subsequent product is most likely to form 5 7i same as HX same as x Hat H C H same as V B l 5 l Transition States A C are listed in order of increasin torsional strain due to van der Waals interactions The groups of atoms highlighted in red show us potential energetically expensive overlapping positions From this demonstration we are able to visualize an important requirement for the E2 mechanism Observation 1 E2 products favor antiperiplanar arrangement of HC CLG bonds Anti to reduce torsional strain minimize van der Waals interactions Periplanar so that p orbitals can overlap to form the pi bond Second because transition state exhibits qualities of both the reactants and products we must also consider Torsional Strain of the Products in particular the alkene For this we can use space filling models to show the differences in torsional strain of substituents in both cis and trans conformation Lecture Supplement pg 10 Alkeue Slruclure Space Filling Model Strain Less torsional s lxai n E V LransZbutene More torsional strain Cis and trans apply only when there are two carbon substituents on opposite ends of an alkene Cis means the two substituents are on the same side of the chain Trans means the substituents are on opposite sides of the chain From this diagram it is evident that when substituents are arranged in a cis conformation torsional strain increases and stability decreases Since E2 reactions generally favor the most stable product the cisalkene is typically not the favored product This brings us to our second conclusion about product stability Observation 2 Trans alkene is more stable than cis alkene Third In our previous analysis we looked at alkene stability based on the positions of carbon groups attached to the alkene Now we will take a look at the position of the alkene itself on the carbon chain Lecture supplement pg 10 A A terminal alkene A An internal alkene By empirical observation we find that Observation 3 Internal alkene is generally more stable than terminal alkene Thinkbook Practice Problem 15 For the reaction shown below a Select the major product b Write the mechanism for the major product of this reaction c Very briefly explain your choice for the reaction mechanism gtlt Y CHSOH Solution a The first alkene is more substituted and is therefore the major product b 60 r L f j iH Hzo Ere SB 0 The internal alkene is favored because it is the most stable product Fourth we want to see how the number of substituents degree ofsubstitution on the alkene affects its overall stability We know that sp2 bonds are stronger and therefore more stable then sp3 bonds Thus it follows that the more sp2 bonds a molecule has the more likely it is to be the major product A has only 1 sp2 alkyl substituent so in this A A 0 analysis it is the least stable B has 2 sp2 alkyl substituents so it is more d V stable C also has 2 sp2 alkyl substituents even though its stability is decreased by torsional t 7 strain MK 80 we have ourfourth conclusion about alkene stability The sp2 bonds are circled in blue Observation 4 Alkene isomer with more s92 jCCj bonds is more stable From Thinkbook Practice Problem 10 Pickthe most stable product Explain Ax Ax 1 HJPOtr K x 1 L 39 J lt0H gt T Solution OH L This product has 4 alkyl groups attached to either side of the the alkene Things to remember When strong acid is in aqueous solution in water consider it as H30 in the mechanism Don tforget about rearrangement We have now covered the major factors of transition state and alkene stability pertinent to elimination reactions However we have one more factorto consider If you recall when we consider stability of carbocations resonance and degree of substitution are competing factors Similarly alkenes have two major competing factors of stability number of alkyl substituents and torsional strain In general we can observe that Degree of substitution outweighs torsional strain unless torsional strain is severe A good way to summarize the product predictions made bythe above factors is Zaitsev s Rule which states Major product of elimination is the more substituted alkene As stated before when we do reactions we want to obtain the most stable product By favoring alkenes with electron donating carbon group substituents Zaitsev s rue helps us predict the most stable ie major produc However when the base is large or LG is large NR3 SR2 or F39 the product is an exception to Zaitsev s Rule That is it follows Hofmann Orientation which states The less substituted alkene is the major product Or in other words the major product is notthe more substituted alkene Thinkbook Practice Problem 17 Forthe reaction shown below a Provide a curved arrow mechanism including all transition states showing how the major product is formed KO H CHSOH Solution When stereochemistry is involved use chair conformation for cyclohexanes Remember draw dashes down CH3 KoH CHSOH 3 CI same a same a 5 0H 1 CH3 CH3 i i i Cl 5 In this cyclohexane the reactant there are two hydrogen options 1 the H that shares the C attached to CH3 2 the two H s attached to the second which would produce carbon which gives us This product corresponds to Zaitsev s rule This Pmd Ct does quot0 correspond with beCause the most substituted akene is ZaItsev s rule It s not Hofmann s onentatIon produced However to form his cyclohexaney either because neither the base nor the LG is the hydrogen used was ot diaxial to the large However the less substituted LG therefore is not the major product Praduct Is maiar because a diaxial H was available to form a pi bond lfthe leaving group is not in an axial position to begin with chair flipThen look for axial This will guide you to diaxial leaving group and hydrogens ifany are present By looking for L d H in diaxial positions you will achieve a HCCLG that is antiperiplanar lt E 0 Lo We are now ready to summarize the E2 Requirements 1 Strong base 2 Moderate or better leaving group 3 Hydrogen HCCLG in antiperiplanar arrangement Requirements 1 and 2 base and leaving group interact with one another which means that if you have a really stron base ou can get awa with a moderately weaker leaving group and vice versa The third requirement for the sake of eiminations is nonnegotiable E1 Mechanism Looking at the requirements for the E2 mechanism the base and leaving group requirements are somewhat exible However ifthe leaving group is poor E2 probably won t occur no matter how mechanism is no longer concerted As long as a hydrogen is present you can still have an Elimination Reaction only we will call this mechanism E1 l E1 stands for I Implication l Comparable to Elimination l hydrogen will be eliminated and a pi E2 Product bond will form between two carbons pi bond formation Unimolecular Rate is determined by the leaving group 8N1 Requirements leaving and a carbocation forming carbocation intermediate Example Mechanism from Thinkbook Practice Problem 20c Draw the curved arrow mechanism and major products Solution urxme product Reason for multiple products Carbocation Mantra Whenever we see a carbocation in a mechanism step regardless of where it comes from we rst consider resonancequot Resonance helps us determine how stable a carbocation is and thus how likely it is to form The carbocation above is tertiary and is located next to an adjacent pi bond from the benzene ring so we can say it is tertiary with resonance making it a relatively very stable carbocation Next we consider the 3 fates of a carbocation 1 Rearrangement is possible if it leads to more stable carbocation 2 If the carbocation captures a nucleophile the reaction will proceed by the 8N1 mechanism 3 The carbocation can also lose a proton to form a pi bond proceeding by the E1 mechanism Note each of the three fates is equally weighed by a carbocation and because a carbocation is desperate but not fussyquot it will o en under go more than one of these fates in any given reaction Also note because SN1and E1 mechanisms both concern the formation ofa carbocation intermediate a reaction that involves a carbocation will o en undergo both SN1and E1 simultaneously Therefore the alkene product of the previous problem is the result of an elimination reaction E1 while the alcohol is produced from substitution 8N1 Kinetics The formation ofa carbocation involves losing a bond RSOLG without gaining any bonds so we consider it the rate determining step Therefore the rate law of E1 Is wri en as39 Rate kmcLG E1 Requirements 1 Stable carbocation 1 w resonance or better 2 Moderate or better leaving group 3 Polar solvent 4 Hydrogen HCCLG Finally between Elimination and Substitution mechanisms how can we predict which is most likely Order of preference Requirements I E2 1 Strong base 2 Moderate or better leaving group 3 Antiperiplanar HCCLG I 5N2 1 Not 3 RECLG 2 Moderate or better leaving group 3 Moderate or better nucleophile 4 Polar aprotic solvent I E1SN1 1 Stable carbocation 1 w res or better 2 Moderate or better leaving group 3 Polar protic solvent 4 hydrogen for E1 Strategy When approaching a reaction in which the mechanisms is not speci ed it maybe helpful to think of the mechanism requirements and order ofpreference as a checklist If a requirement of a higher priority mechanism example E2 is not met cross it offthe list and move on to the next mechanism SN2 Again if a requirement for this mechanism is not met we are only le with the two equally lower ranking mechanisms E1 and SN1 Because both E1 and SN1 share the same rate determining step it is generally safe to presume that both mechanisms occur simultaneously Last Example Thinkbook Practice Problem 18 For the reaction shown below a Write all the products of this reaction b Provide a mechanism that clearly shows how all of your products are formed c For each reaction mechanism listed give a single brief reason why it was not chosen for this reaction SN1 E1 SN2 and E2 Br CHSCHIOH Electrophilic Aromatic Substitution Reactions Course Notes Archive 1 Electrophilic Aromatic Substitution Reactions An organic reaction in which an electrophile substitutes a hydrogen atom in an aromatic compound Always think about 0 Do I have resonance 0 How can I regain resonance 0 How does resonance in uence the rate of this reaction Aromaticity is VERY important and nice to have Benzene rings are very stable aromatic molecules Therefore the electrophile must be VERY strong in order to disrupt aromaticity When doing EAS reactions consider a few important things 0 How can I make my electrophile stronger 0 Strong electrophiles REALLY need electrons therefore electrophiles are stronger when they are MORE POSITIVE 7 Use a STRONG ACID to protonate a weak electrophile 0 How can I make my nucleophile benzene ring stronger 0 Strong nucleophiles REALLY want to donate electrons therefore nucleophiles are stronger when they are MORE NEGATIVE 7 lone pair electrons pi bond Does your benzene ring have electron donating substituents with resonance 0 How can I make my reaction FASTER 0 Reaction rate depends on the rate determining step slowest step 0 The rate determining step is the step which leads to the formation of the arenium ion carbocation 0 Therefore you can make your reaction faster by increasing the stability of the arenium ion I resonance electron delocalization I andor better electron donating groups substituents on the benzene Two step mechanism of all EAS reactions Aluminum 1 z 1 FSII39 g a 1 with adjustments T hinkbook P 164 CFQ7 ReacLiun Cmniinai Disclaimer All images are borrowed from Bmice P Organic Chemistry Pearsons Prentice Hall 2004 Hardinger S Chemistry 14D Thinkbook Steven A Hardinger 2005 c m c unlin quotarr ElectrophilicAromolic Substitution Reactions Course NotesArchive 2 1 ELECTROPHILIC ATTACK 39 Electrophile E attacks the aromatic ring I THE rote determining step Requires a high energy of activation because aromaticity is lost Leads to the formation of a resonance stabilized carbocation known as an arenium ion 0776577E Hardinger Thinkbook P 161 CFQ2 An Arenium Ion IS a carbocation It has resonance Think of the 3 Fates Hardinger Textbook Chapter on EAS P 4 1 Capture a nucleophile 7 addition reaction won t restore aromaticity up Al o o H D L4 D10 4 V 2 Lose a proton gain a pibond 7 restores aromaticity 6 This occurs r 9 3 7 H OD 3 Rearrangement 7 won t accomplish much 2 DEPROTONATION 3 B E GE 7 O m e Hardinger Thinkbook P 161 CFQ2 0 The arenium ion is deprotonated by a weak base B I Requires a lower energy of activation because aromaticity is regained k Two ways in which substituents donate or withdraw electrons What makes the substituent UNIQUE compared to the other hydrogen atoms attached to the benzene ring resonance higher electronegativity Disclaimer All images are borrowed from Bruice P Organw chem Pearsons Prmtice H211 2004 card r 5 Chemistry 14D Thmkbook Steven A Hardinger 2005 q we Electrophilic Aromatic Substitution Reactions Course NotesArchive 3 vs Inductive Effect Electron Donating If electrons on the sigmabond of substituent move more readily towards the benzene ring the substituent has a greater electron donating ability 0 FASTER REACTION Electron Withdrawing If the substituent is more electronegative than the other hydrogen atoms attached to the benzene ring the substituent has a greater electron withdrawing ability 0 SLOWER REACTION Resonance Effect delocalized lone pairs on the substituent can donate electron density to the benzene ring at the same time the substituent is more electronegative than hydrogen and may inductively withdraw electrons but has a lesser effect Example Hardinger Thinkbook Practice Problem P 178 15a OH H H OH OH g x e I KOH 1 Hhon l H H H n a 3 4 1 2 esonance contributors important resonance contributor full octet for all atoms C ofEl A Sn lone pairs except alkyl aryl and CHCHR groups resonance delocalizes positive charge of a carbocation which increases carbocation stability w ofE u r a 4394 i positive charge or partial positive charge high electronegativity electron greedy Substituent Directing Effect The electrophile attacks the nucleophi c pibond on the benzene ring but the substituent 1i on the benzene influences WHERE THE ELECTROPHILE ATTACIIES in order to have the FASTEST RATE OrthoPara and Meta Disclaimer All images are borrowed from Bmice P Organic Chemistry Pearsons Prentice Hall 2004 Hardinger s Chemistry 14D Thinkbuar Steven A Hardinger 2005 Haw n m s Electrophilic Aromatic Substitution Reactions Course Notes Archive 4 X X x X E E T A E E zlzrrrnphil E par mm mm Important Qualities of Substituent in order of importance 1 MECHANISM How do resonance and the degree of substitution on the arenium ion in uence the reaction rate RESONANCE Increases the stability of the carbocation 0 There are always at least 3 resonance contributors for ortho para and meta 0 Think of the resonance contributor Where the positive formal charged carbon is directly attached to the substituentdirector dire quotor How do 3 s the subsntumt in uence the carbonation elec H o If the substituent is an electron donating substituent it now can donate electrons to stabilize the electrondeficient carbocation Resonance Contributors Hardinger Thinkbook Lecture Supplement P 17 DEGREE OF SUBSTITUTION increases the stability of the carbocation Tertiary gt Secondary gt Primary Hardinger Thinkbook P 162CFQ 3 Ortho para gt meta 2 Probability 7 2 ortho 1 para and 2 meta positions Disclaimer All images are borrowed from Bmice P Organic Chemistry Pearsons Prentice Hail 2004 Hardinger s Chemistry 14D Thinkbuok Steven A Hardinger 2005 u A s Wuquot quotH q z Electrophilic Aromatic Substitution Reactions Course Notes Archive 5 meta ortho gt para 3 Steric E ects Electrophile attaches where there is the least amount of steric hindrance Para meta gt ortho Activators Substituents that make benzene MORE REACTIVE donate electrons into the benzene ring 0 Increase benzene nucleophilicity more reactive towards electrophiles stabilizes carbocation intermediate stabilizes transition state leading to carbocation formation faster reaction rate for electrophilic attack Deactivators Substituents that make benzene LESS REACTIVE withdraw electrons from the benzene ring 0 decrease benzene nucleophilicity less reactive towards electrophiles destabilize carbocation intermediate increases the net positive charge on the carbocation destabilize transition state leading to carbocation formation 0 slower rate of electrophilic attack All orthopara directors are activators except for halides All meta directors are deactivators no exceptions P 632 Bruice Organic Chemistry Contains a table of many more Orthopara and Meta directors OItho Para Directors all substituents that donate by resonance andor inductive effects are orthopara directors Common OrthoPara Directors 0 alkyl groups electron donors 0 Ph adjacent pi bond resonance 0 X provides more resonance Disclaimer All images are borrowed from Bruice P Organic Chemistry Pearsons Prentice Hall 2004 Hardinger S Chemistry 14D Thinkbook Steven A Hardinger 2005 Human q m c Electrophilic Aromatic Substitution Reactions Course Notes Archive 6 H 39 39 0 amine 0 C 39 0 ester 39 OH 39 OR ether 39 NR2 39 Halides Cl Br I remember they re also deactivators In uence on Rate Which reaction is faster OCH3 O lt j 39 Has 4 resonance contributors The most important resonance contributor has a full octet on all atoms OCH 5 H 39 Increased stability 39 Decreased activation energy 39 Faster rate Meta Directors all substituents that withdraw electrons by resonance and or inductive effects EXCEPT halogens which donate electrons by resonance but withdraw electrons inductively Common Meta Directors N02 CF3 NR3 RC O 39 l C7N Why are electron withdrawing groups meta directors If you have a substituent which removes electron density from the benzene ring you want to make the most stable carbocation possible Compare the relative stability of resonance contributors for carbocations with a strongly electron withdrawing substituent like nitro N02 Disclaimer All images are borrowed from Bruice P Organic Chemistry Pearsons Prentice Hall 2004 Hardinger S Chemistry 14D T hinkbook StevenA Hardinger 2005 C 1710 arlinnor Electrophilic Aromatic Substitution Reactions Course Notes Archive 7 N02 No2 02 NoZ gt let an 2 Ortho Meta Para Ortho and Para will form one resonance contributor that looks like this gt 539 a This resonance contributor does not make a signi cant contribution to the stability of the arenium ion In fact the adjacent positively charged N destabilizes the carbon s open octet 560 Therefore Ortho and Para positions have TWO resonance contributors while the Meta position has THREE resonance contributors none of which orms this destabilizing resonance contributor Thus you can conclude that meta directors are orthopara inhibitors In uence on Rate Which reaction is faster gt Nq Fewer resonance contributors none of which has a full octet on all atoms 0 Less stable 0 Increase activation energy 0 Slower Rate Regioselectivity 0r Major Product Rule the most stable mechanism intermediate will have the lowest energy of activation and the fastest rate and therefore be the major product Why a substituent on the benzene directs an electrophile to a certain position depends on the stability of the carbocation intermediate Some Conditions Required in Order to Carry out a Reaction 0 FriedelCra s reactions won t happen with meta directors because the rings are too unreactive The order of a reaction is important in order to achieve the desired product Bruice Organic Chemistry P 643 0 A STRONG ELECTROPHILE is necessary in order to overcome benzene s aromaticity Important EAS reactions Disclaimer All images are borrowed from Bruice P Organic Chemist Pearsons Prentice Hall 2004 Hardinger 5 Chemistry 14D Thinbook Steven A Hardinger 2005 mam q me Carbonyl Fundamentals A1 dehy dc A11 Informalth and examples burrowedram Chem 14D Thlnldmakfar Fall 20 Dr Hardmger s Lecture Retardmgs an Paula Ymkams Bruce s Orgam Chemlsvy 4 Edman unless aLherwlse mdlmted Ketone Carbonyl the functional group is comprised of a carbon atom doublebonded to an oxygen atom The Various types of Cafb xyh carbonyl containing functional groups are listed in the table on the quot1 right gt Carbonyls can function as both a nucleophile and an electrophile Ester The 39ty ofthe carbon oxygen bond causes there to be a artial positive charge e carbon atom and a partial Amide negative charge on oxygen atom This electron deficient pdmaryy c n atom acts as an electrophile Secondary The carbonyl gro p can also as anucle phile The xyge Tern y of this functional group is electr n rich because of its lone shown ne pairs th same result because the products are resonance structures Enone and thus identical Three co Fates Acer Chlon39de gt Accept nucleophile at carbon A ti 0ccurs in ALL Carbonyl addition mechamsmsL L Aziyf de arbon goes from sp2 to sp3 hybn39dizatm carbonyl c 39 n Depending on ifthere is a leaving group attached to the carbonyl carbon there are two possibilities for the mum error product of a nucleophilic attack on the carbonyl carbon uergo the three electmphlllc attach an Tab le n39n rates nueelophilie attac u enolate faxmatmn ms www wlhpedu ma 0 Addition no leaving gong LG present The rst atom to create the tetrahedral adduct resulting in step involves nucleophilic attack on the partially pos an oxygen With a itive carb on 39 fo rmal positive charge The oxygenis then protonated to create an alcohol Dugnm Taken me http www chemhelper cammludd html o Nucleophilic Carbonyl Substitution LG is present Occurs with ester amide anhydride and acid chloride functional groups I e t39 39 addition Cirl elimlnatlon I e H C X iI39 H Il I H Ll 3 Nu Nue N leavlng nucleophile gm up l Diagram Taken From httpchemistry2csudhedurpendarViscarboxderhtml gt Accept electrophile usually H at oxygen 39 There are two possible sites for nucleophilic attack of the carbonyl on an electrophile the lone pairs and the pi bond 39 Is there a preference for the electrophile usually H to be accepted by the lone pair or the pi bond No Both sites are nucleophilic and both lead to the same product through resonance gt Form Enolate 39 Assisted by resonance stabilization of conjugate base 39 Initiated by a base deprotonating the carbon adjacent to the CO carbon 0 x Up k re ux 39 I39Lm L a 39 quot W i H2 H EIgH 393 Rq FFI I rquot any Th l Diagram From httpwwwchemleedsaculdPeopleCMlUlabstufflitreViewscarbonylchemistrydoc 2 How do we determine the rate of a CO reaction 0 The rate determining step rds is the slowest step of the reaction the step with the highest energy of activation This is the step that controls the rate of the reaction 0 In order to determine the rate determining step of the CO we must use its mechanism 0 Proton transfers are usually not the ratedetermining step so we can ignore them for our analysis 0 Nucleophilic attack on the CO is usually the RDS 2 What factors control the rate of nucleophilic attack on CO in order of importance 1 Magnitude of 6 on the carbonyl carbon A larger 6 makes the carbonyl group more electrophilic nucelophilic attack will occur faster as a result 2 Resonance Resonance can increase or decrease the 6 on the carbonyl carbon Resonance may also be lost when the tetrahedral adduct is formed Resonance adds extra stability to the molecule As Dr H put it Resonance is like money Molecules L V 4 V that have it want to keep it and molecules that don t have it want to get it Therefore if resonance is lost the energy of the transition state is higher since the molecule is less stable This will make the molecule more resistant to nucleophilic attack As a general rule resonance between atoms in the same row of the periodic table is stronger than resonance between atoms that are in different rows Leaving group The presence of a good leaving group controls whether addition or substitution occurs Study the mechanisms for addition and substitution to better Visualize this concept Steric effects If there are large groups attached to the carbonyl carbon the nucleophile may be sterically hindered from attacking the carbonyl carbon and thus nucleophilic attack will be slowed because the carbon is blocked 39239 What to think about when asked to identify the fastest reaction relative to other reactions 0 0 Look for differences between the reactions considering similarities will not help in your analysis Analyze each of the four factors Examining nucleophilic addition and nucleophilic substitution in the various carbonyl conta I ining functional groups As mentioned before the presence of a good leaving group determines if nucleophilic addition or nucleophilic substitution will occur With that in mind when observing the various carbonyl containing functional groups aldehydes and ketones will undergo nucleophilic addition since alkenes and hydrogen are not good leaving groups By the same logic thiolester RCOSR ester amides and carboxylate ion RCOO39 can undergo nucleophilic substitution since they contain good leaving groups Nucelophilic addition reactivity for aldehydes vs ketones v Considering the four factors mentioned in the previous section 0 We will consider the differences between aldehydes and ketones Note that neither have a leaving group and that resonance is the same Therefore these two factors do not affect the relative reactivity of the functional group in this case 0 6 CO The 6 on the aldehyde carbonyl carbon is greater than that of the ketone carbonyl carbon because the ketone contains two alkyl groups that donate electron density to the carbonyl carbon whereas the aldehyde only has one alkyl group The more the magnitude of the 6 is reduced the more resistant the carbonyl carbon is to nucleophilic attack If the 6 is reduced the molecule will be more stable initially and there will be less of a need to get rid of the partial positive charge Diagram From httpwwwchemleeds acukPeopleCMRla bstufflitrevieWscarbo nylchemistrydoc o Sterics The ketone has two large alkyl groups attached to the carbonyl carbon whereas the aldehyde only has one large alkyl group attached Therefere the ketone has more steric hinderance which inhibits the nucleophile from attacking the carbonyl carbon more than in the aldehyde case As a result ketones are less reactive than aldehydes 39239 Conclusion The rate of addition of aldehydes is faster than that of ketones Nucleophilic substitution in thiolesters esters amides and carboxylate ion Adapted from Chem 14D Thinkbook CFQ 9 Carbonyl ChemistryFundamentals v Considering the four factors that affect reaction rate 0 Carbonyl carbon 6 and Resonance controlled by the magnitude of resonance between the carbonyl carbon and its attachments We must consider the resonance lost as a result of substitution in addition to the a ects of resonance on the magnitude of the 6 on the carbonyl carbon described above The magnitude of resonance lost will affect the rate nucleophilic attack on the carbonyl carbon Let s consider the resonance of each function group in question individually 39 Thiolester Resonance is between carbon and sulfur Resonance is best between atoms in the same row of the periodic table This is not the case here as carbon is in the second row and sulfur is in the third row Ester Resonance between carbon and oxygen Stronger resonance because carbon and oxygen are within the same row of the periodic table Amide Better resonance than ester case because C and N are on the same row of the periodic table and N is less electronegative than 0 meaning it is more willing to share its lone pair electrons Carboxylate ion Resonance with a negative oxygen in the carboxylate case results in much more significant resonance contributors than with neutral nitrogen in the amide case 39 Thus we conclude based on carbonyl carbon 6 and resonance that the order of increasing resonance stabilization is thiolester lt ester lt amide lt carboxylate ion 0 Steric effects Difficult to determine SR OR and NR2 are larger than 039 however it is dif cult to distinguish between the others Overall resonance plays a more crucial role in determining the rate of nucleophilic attack In this analysis steric effects can be ignored because its effect on the reaction rate is minimal 0 Leaving group 39 Thiolester 78R is a good leaving group because of its high polarizability 39 Ester OR is a poorer leaving group than 78R because it is less polarizable than S however its higher electronegativity makes it a better leaving group than 7NH2 39 Amide NH2 a poorer leaving group than 70R because N is less electronegative and less willing to accept electrons 39 Carboylate 02 is a horrible leaving group It has a formal charge of 2 and low polarizability o 0 Conclusion In order of nucelophilic carbonyl substitution reactivity thiolester gt ester gt amide gt carboxylate ion gt Recognizing Mechanism Patterns The owchart below can help you recognize the mechanistic patterns present in carbonyl chemistry anal help you logically reason out what comes next when thinking about mechanisms l CO I I I 2 Nucleophilic 3 Protonate carbonyl 4 Enolate formation attack at carbonyl oxygen carbon 5 Tetrahedral 6 Form Enol 7 i carbon acts as Intermediate nucleophile 8 Eject a Leaving 9 0 acts as a Group nucleophile This owchart was derived from OWLSSurvey of Carbonyl Reactions and Mechanisms Problem 1 Note At every tetrahedral intermediate step 5 you should ask yourself Is there a leaving group in order to know if nucleophilic carbonyl substitution or nucleophilic carbonyl addition will occur gt Examples of the Various Mechanism Steps The numbers at the various mechanism steps correspond to the numbers in the owchart Look at the owchart while following the mechanisms to reinforce the concepts learned and help guide you through the possibilities of what can come next Mechanism 1 e Oi 9 O H 2 it H OH I I W a u 6 O H a OH v H5Cl C Hf 39Hb 0 H P 1 Start with C0 3Tetrahedral 9 0 acts as a mallet 2 Nucleophilic attack intermediate no nucleophile LG present Denrotonate H20 Mechanism 2 u y N O 0 vi 2 015 O H 4 iow ii c r k 3amp1 I Product 1 Start With C0 3Tetrahedral 8 Elect a Leav1ng 2 Nucleophilic attack intermediate LG Group present 1 Reform C0 Deprotonate Mechanism 3 0 6 hEH 4 EH as e 39 6 9 tea 7 1 Start with 30 Draw Resonance 7 carbon acts structure as nucleophile Product 4 Enolate Formation Mechanism 4 0 0 ll H CH2 2 TbC tz r 9x Se orderquot 1 l 8 d r We Llb a H 53 1 Start with C0 6 Form Enol Product 4 Enolate Formation Mechanism 5 Fischer Esterification Even though we have not explicitly learned this mechanism yet using the owchart of possible mechanism steps as well as our knowledge of carbonyl chemistry we can predict the mechanism We will learn the intricacies of this reaction in our Carbonyl Survey of Reactions in the next topic we will cover But for now let s analyze the mechanism steps with respect to our flowchart Some Notes about Fischer Esterification o In Fischer Esterification we begin with a carboxylic acid in the presence of strong acid and a large excess of alcohol The reaction results in the formation of an ester 0 The carboxylic acid carbonyl must be protonated first because the carbonyl carbon is a poor electrophile due to significant resonance with the hydroxyl group Also methanol is a poor nucleophile due to the high electronegatiVity of oxygen Putting a formal negative charge on the methanol by deprotonating it in the presence of strong acid makes the methanol a better nucleophile With this information we can now analyze the mechanism see next page Carbonyl Chemistry Fundamentals What is a carbonyl group and what are its propem39es CarbonyIs can behaye as a nocIeophIIe OR an eIectrophI e A carbonyI Is any cornpoonothat has a carbon doubIeebonded to an oxygen I I ophIIIc oxygen eIectronerIch pI bond Ione bans on oxygen or charge on oxygen gt EIectrophIIIc carbon 6 charge on carbon A EN2 5 Geometry trIgonaI pIanarI sp2 hybrIdIzed Co R R eenerIc carbonyI compound Three Fundamental Carbonyl Mechanism Steps using Acetone 1 Nucleophilic mack at carbon J M ooerate nocI end at sarne tI 2 YIeIos a shah c ang TIP thIs ge 1 3 vIeIos newpru Does NOT oxygen attacks rnoIecoIe reso a k e A who TIP 1 ProtonatIon In TIP 2 Iftnere Is st 3 Enolate Fonnan39On protonate the carb rnore eIectrone es hybrIoIzatIon no nerIc step occurs In aImostALL carbonyI reactIons rnatternhetherthe pI bon presence of STRONG acIo It IS somewhat erroneous El IaIm that these e and 5 charges anse trorn contrIbotor shown here that It Is not consIoereo eophIIe H20 attacxsthe eIectrophIIIc carbon rne the pI bond eIectrons are shIrteo to the gatIye oxygen edra addud Interrn rn s 2 eolIatenot product that SF noct ynth torrnaI ea charge on oxygen 1 or Ione paIron proton or Hao because resoIts In sarne nance contnbotors or each other or oxygen and or carbonyI can ONLY occur I e Hasoc Rot I2 H o rong ach present In a maximum osoaIIy I nrst ony 1 Strong base deprotonates R group of carbonyl 2 Welds enolate with new pi bond formed and negative charge TIP Although incorrect in terms of reasoning this step can be better remembered by seeing that it is very similar to the carbocation fate step lose a proton form a pi on quot Depending on the nature of the leaving group nucleophilic attack at carbon can lead to two results 1 Nucleophilic carbonyl substitution If there is a leaving group present kick out the leaving group and form new pi bond substitution product 2 Addition If there is no leaving group present in the tetrahedral adduct addition product intramolecular step and make new carbonoxygen pl on If group X is NOT a leaving group ie CH3 H F39 and no other mechanism steps can occur protonate the oxygen to make it neutral intermolecular step If group X is a leaving group ie Cl 39 Br39 I39 OHZ then pack its bags and kick it out b d QUESTION Why not have the nucleophile attack the carbonyl and the leaving group leave simultaneousyin a concerted action as in SN2 reactions NSWER Remember SN2 reactions ONLY occur at sp3 hybridized carbons carbonyl compounds are sp2 hybridized What F Group our Factors to Consider in Finding Relative Reactivity of Carbonylecontaining Functional s Remeniuei 39 i quot 39 39 reactions o r r o i i i V H R R H 7 R 9 3 5 Esiei R Aldeliy ile Czuboxylic acid Carboy m Tliiriester p lc A1de I Aunde 3quot Amide Auliydurle Acid chloride Acyl clllmidet Whe asked to find the fastest reaction comparing compounds with different carbonyle containing functional groups COMPARE A Resonance IZI OOD resonance 9 slower reaction IZI NOPOOR resonance 9 faste I Carbonyl will be less Ii r reaction kely to react due to resonance stabilization that reduces the partial 57539 charg Loss ofresonance will spike up the transition state energy for nucleophilic attack step same row and less electronegative 51 VS ch c1 H3C NH Acid chloride iunciionai group Primary Amide functional group Weaker re o ance SIGNIFICANT res c is 2 row element but CI is 339 nd N are 0th 2 0W row element elements B Leaving Gro 1 Better leaving g Wh roup of tetrahedral adduct 9 faster reaction tconstitutes a GOOD leaving group eview potential gain of energyrich resonance ie ROSOgH N electron density ATOMIC RADIUS larger radius can better make room for newlygained 3 ELECTRONEGATIVITY more electronegative the atom the more greedy it is about electron density the more it can keep to itself when it leaves 4 INDUCTIVE EFFECT greater the inductive power ofleaving group the better the ability to accommodate extra electron density 5 FORMAL CHARGE if charge changes from positivenegative to neutral the more stable the product the better the leaving group C Magnitude of 6 charge on carbon Larger 6 charge on carbon will make carbon Inore electrophilic and more easily attacked by nucleophiles I Carbons with large 6 charge can be attacked by weaker nucleophiles Neighboring groups with high electronegativity suck awa d ct carbon of carbonyl ren en39ng it more ele Y rophilic or electron electron density from poor I 6 charge on carbon can be reduced by neighboring electrondonating groups such as alkyl groups CHg D Steric Effects Bulky substituents hinder nucleophilic attack 9 slower reaction In general acidchloride anhydride thioester gt ester gt arnide i ll l r 9 R r n R39 s R 5 gtgt carb u oxyl ate Factors in Acid Anhydride Thioester Ester Amide Carboxylate finding reaction chloride rates A Resonance Poor Good Poor Good Great Superior resonance resonance resonance resonance resonance resonance C and Cl with two C and S with with N with oxygen NOT same oxygen NOT same oxygen C and N atom row atoms row atom same row C and 0 same elements C and N elements C and 0 elements row elements same row same row and N is elements elements les greater electroneg delocalization ative than of charge 0 better resonance B Leaving group OCOR good SR good OR H2 CI good leaving group leaving moderate moderate VERY poor leaving due to gained group due leaving bad leaving group group to larger group due leaving due to stabilization atomic to highly group due formal charge and size electroneg to less electronegati ative O electroneg ve O ative N and small radius C Magnitude of 6 Larger Larger delta Moderate Larger Moderate Largermodera charge on carbon delta due due to delta due delta due small delta te delta due to electronegati to less 0 due to to highly electronegat ve O electroneg electroneg less electronegativ ive Cl atom ative S ative O electroneg e 0 but good ative N resonance and good resonance D Steric effects Steric effects hard to compare among these functional groups resonance trumps or group can be clearly defined as the rate determining step Since there is no clearcut explanation why one step should be slower than the other the ratedetermining step happens to depend Conclusion Although we can determine the faster reaction by comparing only the 1magnitude of 6 charge on carbon and 2 steric effects for carbonyl addition reactions we need to consider all 4 factors for nucleophilic carbonyl substitution To protonate or not protonate that is the question Hardinger When do we know when or when not to protonate 39 Nucleophile the better the nucleophile the less necessary it is to protonate in the 15 step 39 Electrophile the better the electrophile the less necessary it is to protonate in the 15 step 39 lf strong acid is present protonate NHiOHz ea C e aq H2 SO4 k D14 0 f T T 1 Strong acid aqH2804 is present protonate 4 6charge on carbonyl carbon larger due to using H30 even less electron density on carbonyl oxygen better electrophile 2 Moderate nucleophile H2O presence implied by aqueous 5 Now moderate nucleophile H2O can attack carbonyl carbon more easily even with 3 Resonance stabilization makes carbonyl resonance present carbonyl C turned into carbon less likely to undergo nucleophilic better electrophile attack When can OH or OR ever be a leaving group I thought they can never leave A In the tetrahedral adduct when there is a formal negative charge on the oxygen quotOH or quotOR can leave due to unfavorable electron repulsion B If conjugation can be gained quotOH or quotOR can be leaving groups enolate mechanisms CH3 0 Electron Repulsion VOCHg Ph OCHg Ph CH3 Tetrahedral adduct CH3 CH3 0 CH3 conjugation H X gt M OH chocic OH HO CH3 H HO39 Tetrahedral adduct Conjugation acquired Some images and reactions taken from Chemistry 14D Thinkbook 2006 but otherwise created by Microsoft Word Structure Reactivit Geueml formation Eno late 0 L l e Enamine lCHQQ H g 4 HSCL Enol 0 H30 nl OH k x as MCH 3951 A Order ofDecreasin Nucleo hllle reactivi LATE gt x NlCHJI x Enolates Enols and Enamines Ennlate negative charge on oxygen with adjacent CC double bond Enamine NR grou adjacent CC double bond Reacts in similar Way as enolates Ennl hydroxyl group with adjacent CC double bond Alkene alcohol enol ENOL 1 source of ln all Lhre Ennlzte ls nucleophlllc due El Oxygen s small alomle l El Formal negatlve charge Dellaewls lo nucleophlllelty xygen ls more electronegatlve than N Enamine ls modaately nucleophlllc due lo El N EN3 ls less electronegatlve than oxygen EN 3 5 Enul ls modaale nucleophlle due to El Addltlonal electron dmslty from oxygen of hydroxyl goup electronerlch pl bond ls nucleophlllc ch e a aracler When looking at which side is favored products or reactants IZl Equilibrium favors side wi R acidbase pair I Weakest acid and weakest base should be on SAME side IZI W en comparing p a w ues juxapose just the acids or just the bases ide with larger pka Value weaker acid favored in equili rium Remember that differences in p a Values reflect a difference quantity of 10quot Example compound HrA pKa3 base 900mpound A pKa1o Hrbase equilibnumlies to the RIGHT by factor of 10 Some Useful Ka Values from more acidic 9 basic H2S04pk 20 k H30 pka 1 cHscocHs pka 19 B diketonepka 9 CHzCO CH3 pKa25 HCN pk LDAHpka36 a 91 CHZOH pka 155 How do we know which roton to de mwnam 8 o gnaw 1 me all resonance contributors of formed enolam for each use 2 Pick hydrogen that results in enolate with the most number of HvJPORTANT resonance contributors IZI Example 1 3 possible choices to ake away a hydrogen 0 from terminal methyl group on the left 2 important resonance contributors O H from CH2 inbetween 2 carbonyl i portant resonance contributors MOST ACIDIC HYDROGEN e H from methyl attached to oxygen 2 important resonance contributors enzmine ennl fnrm39 39 r ii ii i i r gained by 39 electmn density IZI Enamina sabihzed by similar conjugation Enols a1though eto automermoie sab1e than enol automer eno1s may be y stabilized by Hrbonding conjugation or gained ammahci 39 39 39 a whydon39tth 39 N2 39 Aliza are usually good nucleophiles m i l m romanon of 39 carbon BUT major productis enolate formation due to l STERIC effects If base could easily snatch a hydrogen from one of the neighboring less stencally hindered atorns it will and it won39t work hard to 2 Prouun uun m thdnlia in t be ALL enolate fonnan39on lly hindered base such as LDA lithium Is there a wa for there to Instead of just any old base use stenca diisopropyl arnide cc I I T I 39 rnlln ii rznnni undergo SNZrtype addition and can ONLY deprotonate carbonyl39s less stencally hindered neighboring hydrogen LDA is a superior base so itrea s din lakes an H atom until no more left Enolates are Reacu39ve Nucleophiles 0 13991 Although the major enolate contributor shows concentration of electron density on the electronegatiye oxygen 01 When it reacts with an electrophile it behaves like the electron density is A CH concenirated on C like so Enolates React with Alkyl Halides AldehydesKetones Esters Enolates react Example Mechanism unpouam uung to b Wlt remem er 1 Alkyl halides In step 2 this nucleophi Ic attack of enolate s pi bond is much like S 2 SN2E2 mechanism pathways Must follow the rules of SN2 reactions 1 Ele rophile cannot be tertia O 2 Good leaving group O 3 Good 1 OCH3 base deprotonates most acidic H inbetween 2 nucleophile carbonyls gives most no ofsigni cant res contributors most 9 4 Polar solvent stable product electrons of CH bond made into pi bond and aprotic preferred electrons of original p39 bond move to oxygen 2 Nucleophilic pi bond attac s carbon of Cl bond in CH3CH2Cl mhese rues do not alkyl halide and superior leaving group I le ves apply no SN2 rxn 3 In same step electrons on oxygen move back to remake pi Q feectrop1e 5 3 ban 2 reaction occurs 4 Result is new CC bond inbetween 2 carbonyls enol te is moderate base 2 Aldehyde ketones Aldol reaction a o laxmeK b we r H OE 39CH MO H om F 3h 1 OH base deprotonates most acidic H H of methyl group directly attached to carbon of carbonyl ofketone electron of CH bond m 39 39 ade In bond move to oxygen 2 Nucleophilic pi bond attacks carbon of s to pi bond and electrons oforiglnal pi the other carbonyl molecule aldehyde and pi electrons of electrophile move to oxygen 3 V th presence ofalcohol CH30H base H30 protonates n with 4 Leaving group OH neg charge make int leaves due to ette deprotonating adjacent H atom with base 0 b r leaving gro new conjugation gained by up Notice that OR can leave if 9 In tetrahedral adduct there is another oxygen with a negative charge 9 The leaving ofthis ecule can gain re readily OH an the mol the mo attached to carbonyl Originally named aldol reactionquot because this rxn originally thought to only occur between aldehydes and alcohol 9 BUT it turns out that this reaction also occurs with ketones 3 Esters Claisen and 39 nn Condensation Reactions ace ow ove to 0 en 9 9 YES OCHa will leave due 4 As OCHa leaves electrons on oxy product formed is Bketoester 5 ea a SEC electrons of CH bond moving next to gen bond and electrons of original pi bond moving u 6 Pi bond attacks H of H30 and forms B ketoester again c3 0 X it 1 OCH3 base deprotonates most acidic H H of methyl group directly attached to carbon ofcarbonyl elect bond m de into pi bond and electrons oforiginal pi bond rons of CH xygen 2 Nucleophilic pi bond attacks carbon of original carbonyl molecule ester and pi electrons of electrophile move to 3 Tetrahedral adduct formed and negative charge on oxygen Is there a leaving group pres en electron repulsion m ve to make pi bond Deprotonation ofH by base MUST occur R ct OND time with base OCH3 deprotonating H 39 to pi carbonyl to form 39 p to oxygen quot Bketoester product Acid Hao of aqueous HCI drives equilibrium to 39ght Yields desired B 4 Claisen condensatio Condensations of2 esters to form desired Bketoester intermolecular reaction Enols can be A01 catal zed or Baseecatal zed OH 7 x H09 Base 0 i EEQH e D HOAH Acid L of HioiL 3014 OH kart OH H10 Notice that the 70H is regenerated at the aid catalyst motiee that the 70H is regemated at the aid catalyst
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