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by: Michelle Gulgowski DVM


Marketplace > University of Kentucky > Chemistry > CHE 232 > ORGANIC CHEMISTRY II
Michelle Gulgowski DVM
GPA 3.91


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This 36 page Class Notes was uploaded by Michelle Gulgowski DVM on Friday October 23, 2015. The Class Notes belongs to CHE 232 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/228303/che-232-university-of-kentucky in Chemistry at University of Kentucky.




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Date Created: 10/23/15
Carbonyl Condensation Reactions Carbonyl Compounds Earlier we talked about how the carbonyl carbon was such a great electrophile Later we talked about how we can make the carbons alphato the carbonyl group into great nucleophiles enolates Does anyone see the problem here A very common problem with aldehyde enolates in particular is that they tend to react with themselves to dimerize This can be a good thing This self reaction process is called an Aldol Reaction Why Aldol Because one aldehyde reacts with one enol The first step in this reaction can be summed up as follows 0 O Qk KOH H 2 H Ethanol H O KOH 0 H20 H Remember of course that this reaction is reversible a retroaldol o 09 H eVH AH p H0 Q ON The Forward reaction ie formation of the aldol product is initially only favorable for non sterically hindered aldehydes Ketones undergo the retro reaction quite easily You should also note that any carbonyl compound with alpha hydrogens can undergo the aldol reaction ie esters acids etc As a note to make your life easier every time you see an O C C C O arrangement in a molecule you should be thinking aldol you ll understand this better when I explain it in class The next question to ask is how do I tell whether a molecule is going to undergo an aldol like reaction or an enolate like reaction There are a couple of clues 1 STRONG base If bases like LDA are being used and a decent electrophile is present then you re likely looking at an enolate type reaction 2 GOOD electrophiles If there s another strong electrophile around A halogen Br2 for example or something like iodomethane CH3I you re looking at an enolate type of reaction 3 Catalytic amounts of a weak base If there are no other electrophiles added then the use of a weaker base such as sodium ethoxide NaOEt or hydroxide H0 is usually a clue that an aldoltype reaction will be taking place The Aldol Condensation This is what I would call a money reaction If you get the details of this one down you re golden First off what do you think of when you hear the word condensation Yup water So it makes sense that the aldol condensation is an aldol reaction that gives off water And that s exactly what it does the initial aldol product is dehydrated to give a conjugated carbonyl lt0 6 9 o H H HOpH V I H HO HO As you see the ene al is simply the aldol product minus the elements of water a condensation compound reaction This dehydration reaction is your fourth method for forming double bonds although it only works under very specific conditions ie it only makes conjugated carbonyl compounds although this is still VERY usefull So one more time the aldol condensation takes a carbonyl compound usually an aldehyde ketone or ester and joins two of them together to form a conjugated carbonyl compound Here s another example with the carbons from one of the pair marked in bold to show where everything is going 1 2 Generally with aldol condensations you are combining two of the same molecule to form a dimer However in a few cases it is possible to combine two different molecules to form the aldol product There are two MAJOR restrictions 1 One of the pair must not have any enolizable ie alpha protons for example aryl aldehydes amp ketones like benzaldehyde or benzophenone OR 2 One of the pair must form an enol MUCH more easily than the other ie a 13 dicarbonyl compound versus a normal ketone Some examples H H OEt O NaOEt O gt LOEt EtOH O Gaaa It s that amp Ethyl Cinnamate again There must be a million ways to make N0 EnOlizable PFOtOIlS hereu this stuffand you39ll see 39em all 0 O o O O t B OK M W u gt tBuOH L Protons here much more easily enolized But what about the enolates of other carbonyl compounds such as esters Do they undergo these self condensation reactions as well Of course they do However the thing to note is that esters have essentially a leaving group attached to the carbonyl carbon Thus the products of the condensation reaction are slightly different In essence the product of an aldol type reaction between two esters this reaction is called a Claisen condensation is almost always a 13 dicarbonyl compound remember these In its simplest form O NaOEt O 0 e H3o LOEt MOE i EtOe o 3 30 0 lt WQXH 03 39 M03 C EtOe note why do we use NaOEt and not NaOH Again this is a fairly general reaction just about any ester can undergo this type of self condensation The driving force of this reaction is not dehydration however it is the formation of the stable enolate anion ie the one between the two carbonyl groups that makes the final step irreversible Is it possible to do mixed Claisen reactions Of course provided the same rules are followed as for the mixed aldol One component must either have no enolizable protons or must be enolized VERY rapidly Here s an example in addition to what s in your book Ethyl Picolinate O 2 2 NaH N 0 Et 39 Acetone N IN I ll i Hint ammonium acetate is basically a mixture of ammonia NH3 and acetic acid CH3COOH The nature of the electrophile is also worth considering What if we have a conjugated carbonyl compound Recall that we then add a stabilized nucleophile to the alkene portion a 14 addition Malonates and keto esters are particularly good at this type of reaction called a Michael addition O O O NaOEt gt H2CCOOEt2 NaOEt Oe COOEt2 gt CHCOOEti CHCOOEt If a large excess of the conjugated compound is used it is of course possible to get a double Michael dialkylated product 0 OEt O O O O M excess BK NaOEt OEt I The Stork enamine reaction is just another way of performing Michael type additions with less activated ketones and aldehydes You should familiarize yourself with this reaction from the text but don t put too much emphasis on it You now know how aldehydes ketones and esters can react to give addition or condensation products both in conjugated and non conjugated systems as well as a very limited way of doing mixed condensations Another way to get a reaction between two different kinds of carbonyl groups is if they are both part of the same molecule This is called intramolecular condensation and is an excellent way to make 5 and 6 membered rings 0 W NaOH go H gt EtOH H H lNaOH 1 O HmOH gt gt H 4 H H C Ql I H C H 9 06 O As stated above this reaction generally only works to make 5 and 6 membered rings you generally won t form 347 or 8 membered rings in this way Also note that you could have nearly any substituent hanging off either of the carbonyl groups or off of the carbons in between the carbonyl groups this reaction is VERY general Also note you can do this reaction with estersas well in which case it is called the Dieckmann reaction Take a look at the synthesis below you need to be able to recognize what types of reactions are taking place at each step here 0 O COOMe MeO lt OMe M 1 NaH M6000 COOMe CC MeO lt gt 0Me Meo OMe DEED MeOOC COOMe o o OOMe COOMe M 00c H HzPd e H COOMe 3NaH 3BrCH2COOMe MeOOC H COOMe 00Me COOH COOMe Iko HOOC COOH Hgy MeOOC COOMe gt gt Han H00 COOH NEOH MeOOC COOMe HOOC MeOOC MeOOC O O 1Naowm O COOMe ng O 2 H3O Heat MeOOC o 0 We can now put all of the knowledge we ve gained in the last chapter together to look at the world famous Robinson Annulation This is a classic organic reaction which works splendidly to bring two different molecules together to form a ring or to form several rings from a single molecule It involves both Michael additions and Aldol condensations Let s watch one in action H 0 CH3 CH3 CH3 d CQ O I O O H O 0 Okay that s the overall picture Let s look at a second example step by step complete with mechanistic commentary OCH 3 O OCH 3 KOH o x f OCH3 0 OH i O OCH3 0 o voo N d M o The first step is the enolization of one of the ketones The trick is which ketone Well you have to do a little thinking here Which enolate anion will be the most stable In this case the protons on the carbon next to the aromatic ring are the most acidic the resulting double bond is in conjugation with the benzene ring which helps We then do a Michael addition to the conjugated ketone and the resulting enolate tautaumerizes to the ketone What happens next An aldol condensation OCH3 omg 0mg 0 1533 OH 9 0 OCH3 H 34 OCH3 g 41 Fairly standard aldol condensation chemistry The dangling ketone is enolized and the resulting enolate attacks the ring ketone to form the initial aldol product Following a proton transfer to protonate the alcohol and deprotonate to form a new enolate hydroxide is lost to form the new conjugated ketone which in the case of the Robinson is in a ring That about wraps it up for this chapter As you have seen it is all basically one reaction just applied to a variety of molecules Rules to remember look for the most easily enolized carbonyl group attack the most electrophilic center then spit out the smallest molecule possible That s it Aldehydes and Ketones Nomenclature form text The best way to think of an aldehyde or ketone or just about any carbonyl compound is with a slight positive charge on carbon and a slight negative charge on oxygen 9 RaO Just about all of the chemistry of carbonyl compounds is explained by the oxygen being slightly nucleophilic thus easily protonated and the carbon being strongly electrophilic Remember this Preparation Aldehydes 1 Oxidation of a primary alcohol with PCC 2 Ozonolysis of an alkene REVIEW IT 3 Reduction of an Acyl Halide Acyl halides can be reduced with a special reagent lithium trit butoxyaluminum hydride LiAlOt Bu3H O LiA1o H 0 RJKCI 43 RJLH Your text states that aldehydes can be easily prepared from esters with DIBAH diisobutylaluminum hydride Typically it is easier to reduce all the way to a primary alcohol you need 2 equivalents of DIBAH for this then re oxidize JOLOR JOH FCC r j gt Diisobutyl aluminum hydride R or DIBAH or DIBAL H R Ketones 1 Oxidation of secondary alcohols usually be the Swern oxidation or with PCC 2 Ozonolysis of an alkene 3 Friedel Crafts Acylation Below is the preparation of a ketone sequentially from a primary alcohol through an intermediate aldehyde OH L o WOH FCC W0 lgtj PCC j Some ketones can also be prepared from acyl halides and organo copper reagents called lithium dialkylcuprates as shown below 0 O CuBr Li R CI M VLI 39 VCu gt CuLI gt R LiBr 2 Lithium dialkylcuprate Further oxidation of aldehydes and ketones As you might imagine most ketones are inert to all but the harshest oxidative conditions and thus there is no synthetic utility in trying to oxidize them However aldehydes can generally be oxidized to carboxylic acids under relatively mild conditions Jok AgZO H20 JL R H NH4OH EtOH R OH Reactivity of Aldehydes and Ketones These carbonyl compounds generally have two reaction pathways they react with strong nucleophiles generally strong nucleophiles have a formal negative charge under neutral generally anhydrous conditions or with weak nucleophiles those with lone pairs but no charge under mild acid catalysis If you take a good look at the nucleophile and reaction conditions you ll be able to figure out which way it will go Reactivity aldehydes are much more reactive than ketones nuff said Addition of water or alcohols to from a hydrate or alcoholate ketal Ketones and aldehydes in aqueous or alcoholic media frequently react with the medium to form hydrates or alcoholates The extent to which this occurs correlates to many things including the electrophilicity of the carbonyl carbon While acetophenone exists mostly as the ketone trichloroacetaldehyde chloral exists almost entirely as the hydrate if exposed to water 0 0 HO OH H20 gt Very little Cl H i CI H Very little Cl CI Cl CI H I I Similar things happen in neutral alcoholic medium take chloroacetaldehyde in methanol for example 0 HO OMe cLH MeOH Cgtlt H Why do these hydrates form better w e withdrawing substituents Remember the first figure shown in these notes The mechanism for these additions is relatively straightforward HO OM C9CH MeOH CgtltH e 39 Proton M e H transfer 39i39 c3589 H This is the mechanism for the reaction in neutral media The mechanism in basic media is left as an exercise for the reader ie you In acidic alcoholic media the reaction behaves like the EnergizerTM Bunny it just keeps going and going until a completely new product is formed It is called an acetal 0 HO OMe MeO OMe The mechanism is quite straightforward H 6 e I H H V44 H L H HOH OJ MeOH 055839 MeO OMe THN 39 H W H gt C39gtltH Med OMe T CgtltH H 0 T 0 quot J H H oMe Me CI H H lt6 CI 9 gt k gtCIlt Me XH gt CIQLH quotoquot H L Me H Basically a series of protonation nucleophile attack deprotonation steps Big Note Acetal formation CANNOT occur under basic catalysis Convince yourself that this is trueif you can t come see me Remember that these reactions are all in equilibrium it can be forced to the acetal by doing it under anhydrous conditions or by distilling off the water or forced back to the ketonealdehyde by the addition of excess water making it the perfect protecting group O EtO OEt EtO OEt EtO OEt EtO OEt o H H M H 3 Br H K05I 1 BuLi HzOH I Benzene I I I 2 A Ey atron I I I I Br etc H R H In general simple alcohols like methanol and ethanol are not used in the formation of acetals particularly from less reactive ketones The main reason is entropy you ve got to get three molecules together to form one that s not so good The very common way around this is to use a glycol ethylene or propylene glycol to form a cyclic acetal H 0 OH 0 o gt H b TMS TMS enzene TMS TMS As you would expect for ethers acetals are stable to base and most nucleophiles such as Grignard reagents and alkyllithiums They revert back to the carbonyl compound on exposure to aqueous acid Enamines Just as with alcohols amines can add to ketones and aldehydes Primary amines add to give imines while secondary amines give enamines pronounced ene amines The reactions are generally catalyzed by a small amount of acid and need to be buffered to a pH of 45 An Imine in this case the amine is hydrazine and the product is called a hydrazone O N NHZ gt trace H N EtOH The mechanism shown here for cyclobutanone is pretty straightforward H8 H I 2 N N H2 H e o e I lh9rNH2 proton H NNH2 H2 WNHZ transfer b H gt H gt H gt H lle le NNH2 5 Io In the case of secondary amines we have a lack of protons that can easily be removed from the amine the mechanism thus requires that the offending positive charge be neutralized by removing a proton from the alkyl group O b 6 Overall H on e a 52er 0quot C 2 N39 6 No proton on amine to remove gt remove proton from former ketone DeOxygenation reactions There are two general reactions for the complete de oxygenation of ketones and aldehydes The general scheme for de oxygenation is R can 2 H JOL De Oxygenate R R39 RA R39 The two methods are the Wolff Kishner runs under basic conditions and the Clemmensen under acidic conditions Below find an example for each one WolffKishner HZNNHZ KOH fo HEAT CgtNNH2 gt Clemmensen ZnHg HCl Heat The Greatest Double Bond Forming Reaction Ever Invented The Wittig Reaction This reaction is what I would call cute chemistry One of the cool thing about Wittig reactions is that they just about always work The most general scheme an alkyl halide usually the bromide and an aldehyde or ketone are taken to an alkene If an aldehyde is used with a primary alkyl halide generally the trans product results Either R39 or Rquot MUST be hydrogen JOL Br H R Rquot R R39 R39 R39quot gt J R In a more detailed picture of this reaction the alkyl bromide is allowed to react with triphenyl phosphine to form an alkylphosphonium salt Bre Ph H Br PPh XH 3 ph G Ez llt R R Ilh These salts are generally quite stable and can be stored over long periods of time ie many of these salts are even sold commercially The phosphonium salt is then deprotonated usually with sodium hydride or butyllithium to form the ylide An ylide is simply a charge separated species as shown below R RI 36 Ph H R Ph R PhQRJlt Ph G K Rh R39 I5h R39 quotylidequot This ylide is then allowed to react with a ketone or aldehyde to form a betaine intermediate This intermediate cyclizes to form another intermediate an oxaphosphetane O R39JLH m go R O Rlll R R ph gR M135 RI 39 Ph3 gt R R betaine oxaphosphetane The oxaphosphetane decomposes rapidly to form the alkene and triphenylphosphine oxide R39quot H R39 Ph 0 R g Ph I5 39 I Ph 0 3 RI R R39 I5h The main driving force for this reaction is the formation of the phosphorous oxygen double bond this is one of the strongest bonds known and its formation pulls the reaction to completion Below are a few examples of the Wittig reaction at work OTMS o OTMS 0H A 1 PPh3 A OTMS H 0 CI gt V 2 BuLi O PPhg gt O OTMS 0H PPh B 0H PBr3 Br PPh 3 r OH Br PPh gBI39 PCC NaH PPh3 o CC 0 PPh3 A simple modification of the Witti g reaction leads to a dibromo olefin which is an excellent precursor to alkynes TMSO TMSO TM 80 PPh3 CBr4 BuLi H l gt 0 Br TMS TMSO B TMSO Conjugate Addition Okay I m afraid things are going to get a bit ugly again We now focus on conjugated carbonyl compounds If you look at resonance structures for these particularly in comparison with the figure on the top of page 1 O 339 0a Unconju gated 6 Conjugated You see that in conjugated systems there are TWO sites for nucleophilic attack Let s do a little nomenclature before we continue when we look at carbonyl compounds the carbonyl carbon is the real center of attraction Any carbons directly attached to the carbonyl are called the alpha 0L carbons in the case above the one with the slight negative charge As you ll see in upcoming chapters the alpha carbon is often quite nucleophilic Unsurprisingly the carbon attached to the alpha carbon is called the beta 5 carbon And as you can see in the case of conjugated carbonyl compounds which are often called alphabeta unsaturated compounds the beta carbon is electrophilic This leaves us with a bit of a puzzle If we add a nucleophile to this conjugated compound will it add to the carbonyl or to the double bond eNu 0 HO NU gt gt gt or 0 OH 0 x XL NU Nu Addition to the carbonyl you ve seen before Addition to the double bond to form an enolate which tautaumerizes to the ketone is new This type of addition is called conjugate addition or more frequently Michael addition How can we tell if we ll be getting carbonyl addition or conjugate addition Fortunately it is quite easy Non carbon nucleophiles we frequently call these soft nucleophiles like nitrogen amines and oxygen alcohols will always add in a conjugate fashion particularly if the reaction is carried out under neutral conditions Some examples 0 H o gt EtOH NO ACID O 0 o MeOH M NO ACID OMe A way to force only conjugate addition with a hard nucleophile ie RM is to form the copper salt in this case a dialkylcuprate You ve already seen these used to prepare ketones from acid chlorides formation of copper salts is a great way to soften a hard nucleophile The dialkyl cuprates are usually formed from alkyllithiums hence lithium dialkylcuprates Your text gives a number of good examples but I thought I should throw in my two cents on the subject as well 0 Br Li 0 39 CI Cl 5 L1 5 05 CuBr Cuu 2 Think carefully about what s happening here gt 0 r 39 Mo As you might expect conjugate addition to an alkynyl ketone leads to a vinyl ketone Here is an example using a different kind of soft nucleophile we ll discuss this type in a few weeks o o NaH 0 0 0e 0 0 06 M03 gt M03 M03 M03 A highly stabilized and thus quotsoftquot nucleophile C o o o o o o o H Dom e H H 03 EtO c j lt gtEto 9 gt EtO H OEt H o o EtO EtO Of course as we have already seen organolithium e g butyllithium or magnesium Grignard reagents always add to the carbonyl Carbonyl Condensation Reactions Conjugate Addition If we look at resonance structures for conjugated carbonyl compounds often called a5unsaturated compounds we ll see that there are TWO sites for nucleophilic attack Thus we ll find that the alpha carbon is often quite nucleophilic and the 5 carbon is electrophilic O 039 O 5quot RJ W R C7 6 L39nconjugated a Conjuglied Addition of a nucleophile to the double bond to form an enolate which tautomerizes to the ketone is called conjugate addition When carbonyl addition or conjugate addition will occur In other words when will a nucleophile add to the carbonyl or the double bond of a conjugated carbonyl compound We have seen that hard nucleophiles such as present in RMgBr and RLi reagents add to the carbonyl A way to force only conjugate addition with an organometallic reagent is to form the copper salt in this case a dialkylcuprate We ve already seen these are used to prepare ketones from acid chlorides formation of copper salts is a great way to soften a hard nucleophile Thus soft nucleophiles favor conjugate addition to a5unsaturated compounds Hence noncarbon nucleophiles like nitrogen amines and oxygen alcohols will always add in a conjugate fashion particularly if the reaction is carried out under neutral conditions Some examples EKOH 39O ACID 3 0 O O niEOH Mao Olle Enolates are also soft nucleophiles and hence add to conjugated carbonyls in conjugate fashion 0 Two molecules of an aldehyde or ketone react with each other in the presence of base to form a 5hydroxy carbonyl compound O HO HQO H HO 2J H HO 0 So O J QkH H H V n thrs reactron an enorate rs the nucrophne one ardehyde morecure becomes the enorate whne the other morecure serves as the erectrophne Wrth ardehydes the eourhonum usuaHy tavors the ardor products But wrth stamng rnat ourse there are it E 3 e u e 1 K J lt o a Q i an 2 we teH whether a morecure rs gong to undergo an ardohhke reactron or an eno ateahke reactron7 e a coupe or dues 1 STRONG oase f oases rke LDA are bemg used and a d present then you re hkery ookmg at an enolate type rea OOD erectroohnes f there s another strong erectroohne around such as a ecent erectrophne rs CUOH 2 G harogen rnor ure Brz torexarnpre or somethmg hke rodornethane Chm you re 0 r g at an enolate type of reactrorr 3 Cata ytrc amoun s o a weak base If there are no other electrophiles ded th th s eakerquot base such as sodrum ethoxrde NaOEt or ad hydroxrde HOa rs usuaHy a due that an aldoItype reactron M be takmg prace paHydroxy carbonyr compounds torrned m the ardor reactron dehydrate rnore readny than other arcohors n tact under oasrc condmons the shydroxy carbony compound 5 Often not so ated rt Oses e ements of H20 from the a and p carbons to torrn an copaunsaturated cornpound For ex rnpre 0 9 aapH g JH r HO HO As you see the eneal is simply the aldol product minus the elements of water a condensation reaction In fact elimination of water is spontaneous and the 3hydroxy carbonyl compound is not isolated when the oc3 unsaturated product is further conjugated with a CC or a benzene ring O O O HO H2O HO i Ph gt Ph 2 Ph CH3 H3C H20 HO Ph H3C Ph NOT ISOLATED So one more time the aldol condensation takes a carbonyl compound usually an aldehyde ketone or ester and joins two of them together to form a conjugated carbonyl compound Generally with aldol condensations you are combining two of the same molecule to form a dimer However in a few cases it is possible to combine two different molecules to form the aldol product There are two MAJOR restrictions 1 One of the pair must not have any enolizable ie alpha protons for example aryl aldehydes amp ketones like benzaldehyde or benzophenone H O H OEt NaOEt 0 AOEt EtOH 8N0 Enolizable Protons here 0 OR 2 One of the pair must form an enol MUCH more easily than the other O G u ti 3 M t EHDK l Eu l I Pinto115 here much more easily ampli ed Claisen Condensation But what about the enolates of other carbonyl compounds such as esters Do they undergo these selfcondensation reactions as well Of course they do However the thing to note is that esters have essentially a leaving group attached to the carbonyl carbon Thus the products of the condensation reaction are slightly different In essence the product of an aldoltype reaction between two esters is almost always a 13dicarbonyl compound In its simplest form 0 NaDEI 0 6 H30 0 0 A03 MOB MOB Etcve o o a 09 60 0 AWOH MOB OH 6 l Etca note why do we use NaOEt and not NaOH Again this is a fairly general reaction rjust ab out any ester can undergo th39 type of selfcondensation The driving forcequot of this reaction is not dehyd is ration ii i s the formation of the stable enolate anion ie the o between the two carbonyl groups that es the nal step irreversible is it possible to 39 followed as do mixed Claisen reactions Of course provided the same rules ar for the mixed ado One component must either have no enolizable protons or must be enolized VERY rapidly Ethyl Picolmate 0 O O O 3 2 NaH N N N OEt AFCIOIIE Mechanism Practice Draw the mechanism for the reaction above H E The nature of the electrophile is also worth considering What if w conjugated carbonyl compound Recall that we then add a stabilized nucleopliile o be w t39 l 4 addition M I 4 t r good at this type of reaction called a Michael addition 0 0 Jr H3CC00Et3 N 0E CHCO0EQ B I 09 c0050l cmcoom w a arge excess of the conjugated compoundws used n s of course posser to get a doudechhae da Ky ated prodJct on o 0 0 OEt O Conjugation Conjugation relies on the partial overlap of p orbitals on adjacent double or triple bonds One of the simplest 39 J is 13 bntm quot J Conjugation comes in three flavors the simplest of which is the normal straight through linear conjugation seen in many biomolecules such as Vitamin A OH However it is possible for two systems to be in cross conjugation with each other as in the example below the two benzene rings are cross conjugated NOT conjugatedl O D Conjugation is broken completely by the introduction of saturated sp3 carbon MAM There are a lot of double bonds but there is NO conjugation in this molecule For linearly conjugated systems it is quite straightforward to look at the molecular orbital picture of the various energy levels in the molecule Here are a list of guidelines for the preparation of such a picture 1 For every p orbital there is a n molecular orbital 2 Each molecular orbital M O has its own unique energy associated with it 3 For molecules with an even number of J39lZ bOHdS half of the MO s are higher in energy than the starting p orbitals and half are lower in energy For molecules with an odd number n of carbons in the conjugated framework ie allyl radical n 12 M O s are higher in energy n 12 are lower in energy and there is ONE nonbonding MO with the same energy as the p orbitals 4 The lowest energy MO has 0 nodes The highest energy MO has n l nodes where n is the total number of MO s 5 The number of nodes increases by one for each higher energy level a MO s with an odd number of nodes always have a node in the middle b MO s with an even number of nodes NEVER have a node in the middle 6 Each M 0 will hold 2 electrons The MO diagram for butadiene is shown below Things you should note 1 the progression of nodes from 0 to 1 to 2 to 3 2 A filled set of bonding orbitals 3 An EMPTY set of anti bonding orbitals 4 Electrons with PAIRED SPINS Below the MO diagram I have put calculated Electron Density Plots for each MO M Energy Level I I I I 8r r s I I I Anti Bonding Orbitals I I a 2 Lowest Unccupied MO LUMO I I I I 4 1 HighestOccupiedMO HOMO I Bonding Orbitals i 39 0 A I I I Level 0 lowest energy Level 1 HOMO Level 2 LUMO Level 3 Highest Energy These plots show that the lowest energy level has electron density spread over the entire conjugated backbone The HOMO looks more like two double bonds and is the best representation of the way we write the structure namely N The LUMO has its double bond character in the center of the molecule while the Highest anti bonding orbital shows NO interaction between any of the p orbitals What if we put the molecule into its first excited state The orbital diagram then looks like the diagram below The actual structure of the molecule is best represented by the electron density shown in the LUMO diagram above yielding the diradical structure shown on the right You should be able to derive structures such as this from the orbital pictures of the HOMO and LUMO of any MO diagram will a 4 m 8 2 Lowest Unccupied MO LUMO I 1 1 HighestOccupiedMO HOMO 4 W 0 Preparation of Conjugated Systems There are a number of methods for the preparation of conjugated systems One possibility is by allylic bromination followed by either normal 2 or conjugate 1 elimination 1 A gt WE gid M Conjugate Elimination O d OBr gt OH U HM Br M NBS hv t Butoxide 2 W i gtDMSO N Reactions of Conjugated Systems As your text states conjugated systems do not give simple products on addition of HX or X2 Simply put the charge generated from the initial electrophilic addition to one of the double bonds is delocalized over the entire conjugated system leading to multiple products W Br ea 39 r r A Br WBr 0 M W 49 gt r viAe WVBF The ratio of products varies with the reaction temperature and depends both on which material is formed faster and which product is most stable See Ch 146 UVVis Spectroscopy UV Vis spectroscopy is based on exciting the electronic levels in conjugated molecules What occurs is simply a promotion of one electron from the molecule s HOMO into its LUMO The molecule generally takes on the electronic character of the LUMO in this instance generally having a diradical character The greater the degree of conjugation in the molecule ie the more levels in the MO picture the easier it will be to excite an electron into the LUMO At sufficiently low energy level differences the energy required to promote an electron to the LUMO can be provided by visible light yielding a colored compound Another way to say this is that if a compund is colored an easy route must exist for the promotion of an electron Along with increasing the degree of conjugation there are other ways to facilitate the excitation of an electron One method is to facilitate what is called intramolecular charge transfer This is usually accomplished by the preparation of a push pull system as shown below 0 O I o O 39 39 The ease with which this charge transfer reaction takes place depends on the strength of the push component in this case the dimethylamine group and the strength of the pull component the nitro group Thus by varying the push and pull moieties and by changing the length of the conjugated bridge separating them we can control the color of the molecule Some examples of colored compounds are shown below be sure you understand why they have different colors Azure A Auramine 0 Strong quotPushquot end Strong quotPullquot H Weak quotPullquot N KI 39 39 w s o o N N Strong quotPushquot The Diels Alder Reaction One of the most spectacular reactions in organic chemistry In linking two carbon molecules together it forms two single bonds and one double bond all in one step At its simplest run a o W A diene reacts with an alkene called a dienophile in a concerted reaction to give a cyclohexene No intermediates of an ionic or radical nature have been detected for this reaction It goes in one step There are some qualifications however The reaction depicted above requires extreme temperatures and pressures in order to go For a Diels Alder reaction to work at room temperature a number of criterial must be met 1 An electron poor dienophile is required Some examples mm Hoocm 0mm 0 Q 2 An electron rich diene is helpful OCH3 TMSO 3 The diene MUST be in an S cis configuration I 1 CG GOOd Excellent Bad can TOtate to 8015 Locked into S cis Locked S trans Booo Note also that in some cases the Diels Alder reaction is reversible ie a cyclohexadiene can revert back to a diene and a dienophile Occasionally the retro Diels Alder is more favorable than the forward reaction In these cases there are some special tricks that can be used to force the reaction along One such case is the use of an orthoquinodimethane I ti a lt33 o quinodimethane The quinodimethane regains aromatic characted after the Diels Alder reaction which acts as a great impetus to drive the reaction Your text has an excellent description of the endo preference of the Diels Alder reaction However I would like to make the point here that configuration is retained after the Diels Alder reaction If your dienophile is cis the product MUST be syn and if the dienophile is trans the product will be anti I will close this chapter with a brief look at a couple of fragments involving Diels Alder Chemistry of E Corey s synthesis of Gibberellic Acid Remember a diene can be formed from AHA Here we see a cyclohexene ring elimination of an allyl halide C1 or Br These can be easily formed by a Diels Alder reaction OH An excellent example of a TETHERED Diels Alder system these reactions work QUITE well H O o Igt OMe OMe O HO HO HO OH A Diels Alder reaction forms the A Precursor to the stuff shown above bicyclic system contains a cyclohexene gt Diels Alder Aromaticity The remarkable stability and inertness of benzene is partly derived from the circulation of JI electrons around the perimeter This circulation leads to a number of measurable properties eg the C C bond lengths in benzene are all 139A In drawing MO diagrams for benzene the first most obvious factor is that while it is possible to draw a diagram where there are no nodes However you will note that it is not possible to draw a diagram where there is only ONE node This restriction against having an odd number of nodes is due to the cyclic nature of benzene and the high degree of symmetry involved To make up for this there are two representatives of the 2 node system Although it is possible to draw an M0 level for benzene with 3 nodes we cannot use it due to symmetry considerations We thus jump to looking at levels with 4 nodes of which again there are two quot39 R v I a a a quot A i f quotv a a o39 And finally there is but one level with 6 nodes What does this all mean when we re drawing the energy levels The energy level diagram is shown below Note that the HOMO is comprised of two orbitals of equal energy these are called degenerate orbitals There are also two degenerate but unfilled orbitals in the LUMO Anti Bondin g Orbitals Lowest Unccupied MO LUMO Highest Occupied MO HOMO X Bonding Orbitals There is a simple trick to quickly writing the orbital level diagrams 1 For conjugated cycles with an EVEN number of atoms there will be a single high energy level and a single low energy level For conjugated cycles with an ODD number of atoms there will be a single low energy level and a degenerate ie double high energy level 2 There are as many levels as there are atoms and the rest come as degenerate pairs 3 The nonbonding energy level runs through the middle of the diagram So as an example the levels for cyclooctatetraene and cyclopentadienyl radical are shown below Anti Bonding Orbitals Lowest Unccupied MO LUMO Nonbonding level 1 1 Highest Occupied MO HOMO Bonding Orbitals iv 1 Odd of carbons gt a single low energy level a degenerate high energy level 2 5 carbons minus the two high energy levels and the one low energy levels leaved TWO Which must come as a degenerate pair 3 The nonbonding level runs through the middle of the diagram 4 Add the electrons five electrons leave the HOMO only 34 full a bad situation Anti Bonding Orbitals Lowest Unccupied MO LUMO 1 1 Nonbonding Level 1 1quot Highest Occupied MO HOMO Bonding Orbitals 1 Even of carbons gt a single lowest energy level a single highest energy level 2 That leaves 6 more orbitals arranged as degenrate pairs 3 The nonbonding level runs through the middle through the HOMOh 4 Fill it from the bottom up With the 8 electrons leaving the HOMO half filled We know that aromaticity provides the benefits of high stability etc But what is required for a molecule to be aromatic There are some simple rules one of which you should be able to derive from the previous diagrams 1 The molecule must be cyclic 2 This cycle must be fully conjugated 3 The cycle must be planar 4 The electrons must be able to circulate 5 The conjugated cycle must contain 4n2 electrons n 01234 If the conjugated cycle has only 4n electrons it is anti aromatic and will either be highly reactive or will distort in order to violate one of the other rules 1 4 Some examples C 139 4r on 1 J preparation rapidly dimerizes to givea U gt IEEIII nonconjugated molecule what sort of reaction is this When some sharp chemist made a cyclobutadiene derivative that was too sterically hindered to react it was discovered that the cycle had distorted in this instance by strongly localizing the bonds and violating rule 4 the Distorted bond localized What does this distortion do to help out with the half filled shell If we look at the orbital energy levels we see that the distortion removes some of the orbital degeneracy by lowering the molecule s symmetry Both electrons populate the lower energy level f f HHlt 1 HHlt it Please notice that this leaves a HOMO and LUMO that are VERY close in energy highly conjugated antiaromatic cycles which distort in this way are often highly colored because of the closeness of these two energy levels See Ch 14 UV Vis spectroscopy Cyclooctatetraene This molecule contains 8 4n where n22 electrons and is antiaromatic In order to avoid this nasty situation it distorts by bending or folding into a tub shape I gt M We re now violating rule 3 and 4 and thus avoiding the aromaticity question altogether Is there a way to make anti aromatic molecules aromatic Of course simply by adding or removing electrons For example if cyclooctatetraene is reduced electrochemically or with a metal like K it easily picks up two electrons The resulting dianion is planar and aromatic 10 JI electrons Similarly cyclopentadiene non aromatic is very easily deprotonated to form cyclopentadienyl anion 6 JI electrons Polycyclic aromatic hydrocarbons which result from fusing benzene rings together are also aromatic Below are a few examples 980 Naphthalene Anthracene Pyrene It is also possible to fuse rings of different sizes together you should think about whether these would be aromatic or not Annulenes Aromatic systems that are not comprised of benzene rings often called non benzenoid aromatics are a fascinating area of chemistry These large ring compounds are called annulenes One of the most significant members of this family is 18annulene the number in brackets denotes the number of carbons in the ring and usually the number of JI electrons 18annulene hexadehydro18annulene 18annulene is a planar fully aromatic compound The ring is just large enough that the hydrogens sticking into the middle of the ring do not interact to twist the molecule out of planarity In order to avoid these types of interactions we also study dehydro annulenes like hexadehydro18annulene above These molecules have all internal hydrogens removed and in this case also form aromatic systems Anti aromatic annulenes also exist such as the dehydro12annulene shown below These compounds are usually easily reduced by two electrons to form aromatic ions dehydro12annulene


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