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# Class Note for CHEM 6311 with Professor Albright at UH 2

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Date Created: 02/06/15

IX Aromaticity Before we actually discuss what aromaticity is and the ways that people have tried to measure it we need to know how p orbitals interact with each other A Linear acyclic 7t systems Use the LCAOMO approach aWe can proceed exactly as before except this time using only those p orbitals on adjacent atoms which overlap in a it fashion b In essencewe are once again ignoring all of the filled level interactions and taking the 5 bond frame as given c Example constructing the three 7t orbital allyl system combining it amp p g 8 8 8738 8 age i These orbitals are drawn to indicate relative signs only Their sizes do not reflect their relative coeffi cients ii They look just like those for H2 and H3 a Look at the p orbitals from the top of the chain b For the middle level of the allyl system Q l HEM 175 dThe TE orbitals of butadiene can be constructed from two ethylenesThis is shown in detail in Lowry and Richardson please read it carefully eThere is a set of rules associated with the shapes of the orbitals see Lowry and Richardson p78 iThe orbitals alternate in symmetry with respect to a mirror plane passing through the middle of the T systemThe lowest energy orbital is always S iiThe lowest energy orbital has no nodes perpendicu lar to the chain parallel to the mirror plane iiiThe number of nodes increases by one going from one orbital to the next highest ivThe highest MO has nodes between each adjacent atom vTo preserve the symmetry of the systemthe nodes must be symmetrically placed with respect to the central mirror plane vi ln chains with an oddnumber of carbonsthe A levels must always have a nodal plane at the central carbon atom which lies in the mirror plane viiThere are the same number of MO s as there are atoms in the chain provided that each atom has one and only one p orbital that it contributes to the p system viiiThe number of electrons are equal to the number of p orbitals minus the charge taken algebra ically of the system if all atoms in the chain are carbons f Example the pentadienyl system 176 94 m of Cgt QCgt m nodes 92 i The 71 orbitals we have drawn out are identical for quot Et Me I m 93 H H there are 5 I 4equot 11 amp 12 are filled M there are 5 6equot 11 II3 are filled there are 5 3 8 equot nil IVA are filled 177 y s I 130 A 74 39wxJ m 39zza V 9 yx 39 m 1 0 02 39 0 O m m V a 39 39 V 0 mm 0 0 9 7 5 0 mm 1 7 7 a 74 0 mm 39Wx 7 39wzj yxzs I 7 wm 4 0 quot4 7 39zxa Vxxz mm 39 yxz 7 39 39rx 0 mm yxxx V 39 39 axz 4 mm 39 39zxx 39zxa xx4 74 IxxO m m 7 74 0 1 1 rm S A S A 8 EL y m d m V nmm V m w a yxx T m a A s A gSh B H39Lickel calculations These are only used for the 71 system and involve many approximations For details see Chapter Appendix l of Lowry and Richardson Go over this material by yourself 2 Sample calculation ethylene aLOB 0c TB 96 178 a 06 energy of a p orbital on carbon before any interaction occurs bB interaction energy defined as a negative num ber typically l8 kcalmol 3 For any acyclic polyene in units of B Ej2cosj7tn l crj 2n I 2 sin rj7tn l ETQT 2 nj Ej where Ej orbital energy of molecular orbital j n of p orbitals j orbital index l 2 n r atom index l 2 n nj number of e39s in orbitan crj coefficient on atom r in molecular orbital j 4Two more examples a b a LeiaB 037 oso 060 o37 05 3907l 05 a 44 B 2 a 06 l 8 B 060 03 03 060 a W 07 00 O7 al4l4B W 05 07 05 2 9 7 037 060 060 037 C Cyclic 7 systems These are pictured below Please learn all of them Again there are a number of patterns that emerge 179 aThe number of nodes increases with each increase in orbital energy bThe lowest orbital contains no nodes cThe lowest orbital is nondegenerateThe next orbitals on up come in degenerate pairs except in systems with an even number of atomswhere the highest orbital is nondegenerate ie gig bag a2 i i 82 239 lg I Il 88 02 even odd dA linear combination of two degenerate orbitals gives two orbitals which are equivalent in every re spect eg 180 dThe number of MOS is equal to the number of p orbitals in the cyclic polyeneThe number of electrons is equal to the number of atoms minus the algebraic charge for cases with all carbon atoms eThe MO s of benzene below are a good example 2There are two simple ways to get the orbital energies within the H39Lickel approximation a Frost s circle construct a circle of diameter 4B and inscribe the cyclic polyene so that one of the verticies of the polygon is exactly at the bottom of the circle The points at which the verticies of the polygon touch the circle are the orbital energies eg 181 9 36005 720 92 900720 I80 sinl80 03I 93 72 8 54 sin54 08I bAnalytical method E 2 cos ZjTE In Cr ne21tiri ln n of carbon atoms j 0ili2 in l2 when n is odd or i n2 when n is even notei l l The coefficients are given in their complex form D HLickels rule the first real triumph of theoretical chemis try The rule For planar or systems that are totally conjugated an unbroken chain of p orbitals in a planar molecule number ofTEe s 4n 2 n 0 l2 gt aromatic 4n nl23 gt antiaromatic 2 Definitions a An aromatic compound is one where delocalization of the p electron creates stability eg bezene where there are 67Ee s 4n2 nl so benzene is more stable than l35 cyclohexatriene benzene cyclohexatriene 913 bAn antiaromatic compound is one where delocaliza tion of TE electrons creates instability 182 0 Notice the alternating CC bond lengths and39the nonplanar ring create little overlap between the 7 bonds cThe problem with these definitions as we shall see later is that both l35cyclohexatriene and flat antiaromatic cyclooctatetraene are hypothetical mol ecules Chemists cannot make them andtherefore the stability difference cannot be measured in a direct fashion E Derivation of Hiickel s rule Going through this two ways gives us an idea about what aromaticity and antiaromaticity are really all about Orbital topology a Recall that the 7 orbitals of cyclic polyenes were as shown below In each casethere are 2n degenerate levels 915 b Putting glggtrons intgdtlais pattern in such a way as to completely fill all degenerate energy levels there are therefore 2n I filled orbitals 4n 2 electrons delocalized in a cyclic manner39 aromatic H I 2n1 orbitals H H 4n2 1t electrons 916 H c An antiaromatic compound has 2 fewer electrons it obeys the 4n rule therefore its orbital structure must have the following situation 183 quot quot 2n1 orbitals 39 4n n electrons n n H dThe key feature of an aromatic compound then is that it is closed shell i All degenerate orbitals are doubly occupied ii There is a probably large HOMO LUMO gap It should be very stable e An antiaromatic structure has two electrons in a degenerate set i Therefore the HOMOLU MO gap is zero ii This means that there must be a very high lying HOMO and a lowlying LUMO iii In other words antiaromatic compounds are ex pected to be extraodinarily reactive 2 Hoffmann Goldstein Approach aThis uses the interaction of two acyclic polyene chains ribbons to form a cyclic system ie is this interaction stabilizing or destabilizing in terms of HOMOLUMO interactions 7 b Symmetry properties i Any polyene ribbon will have the following symmetry properties m m A Wd 5 lp LUMO LUMO m S H WP A H Wu HOMO HOMO 919 HoffmannGoldstein nomenclature 184 ii For any polyene ribbon with an even or odd number of n electrons there will be four possibilities LUMO Wu Wu wp Wp HOMO wp W H w w H mode 1 2 3 04 i 1639 ze39 1 e39 2e39 1639 1 X quot lt 5e39 39 6e39 2 7e quot 3 8e39 g 5e39 6e39 7e39 2 8e39 a The pattern of these orbital combinations is important butadiene2 has the same E l HOMOLUMO symmetry W properties as hexatriene iii Modes a We can symbolize each ribbon by a shorthand notation nZ ie where n of carbons p orbitals z molecular charge 922 185 b The mode of each ribbon is n z modulo 4 i Modulo 4 is the operation in which the quantity n z is divided by 4 and the remainder is the value used ii Example For the ribbons shown on the previ ous page 42 mode 4 2 modulo 4 2 modulo 4 2 6 mode 6 0 modulo 4 2 5392 mode 5 2 modulo 4 3 c Interacting ribbons i There are three ways that mode 0 and mode 2 ribbons can interact 33 made 22 mode 00 m m m in Va A A W Vquot s s VP LUMO LUMO LUMO LUMO in g s 53 Vp Va g A A w HOMO Homo HOMO HOMO 923 net destabilization in mode 20 net destabilization m m w A 5 Vi LUMO LUMO Wp lt l 5 Agamp V d HOMO HOMO stabilization a 2 Therefore with these two types only mode 20 or 02 interactions are stabilizing 186 Work out all of the other pairwise combinations for yourself ii Let s see what are the total number of TE electrons associated with these interactions a For any ribbon the number of TE electrons n z b Mode 0 ribbons contain 4 8 l2 4n electrons mode 2 ribbons contain 2 6 l0 4n 2 electrons c A mode 0 0 interaction contains 4n 4n 4n total 7 e39s a mode 2 2 interaction contains 4n 2 4n 2 4n total 7 e39s a mode 2 0 interaction contains 4n 2 4n 4n 2 total 7 electrons iii Thus HLickel s rule automatically falls out from the GoldsteinHoffmann analysis While this is rather difficult compared to the other method of proving HLickel s rule it is easily applied to other kinds of interactions between polyene ribbons as we shall see later E Criteria for aromaticity and antiaromaticity How do we tell whether a compound is aromatic vs nonaromatic l Ultimately what we are looking for are the effects of cyclic conjugation a Of interacting polyene TE orbitals versus separated noninteracting pairs of TE electrons to what degree does this effect differ from normal conjugation b Putting this in another context is there any special 7 stabilization or destabilization associated with cyclic TE conjugation interaction 2 Reactivity this is not a particularin good one a Special reactivity effects rely on there being some connection between ground state stability and what happens at the transition state bThe most common example of this argument is found in the attack by electrophiles on cyclic polyenes specifically that aromatic compounds undergo substi tution reactions whereas nonaromatic compounds undergo addition reactions 187 H lt1 H H x X1 substitution V gt X X H H 9 25 mamaquot x H addition cThe key feature here is what happens to the carbocation intermediate does it get attacked by the nucleophileX39or is a proton pulled off i The primary thesis in this argument is that there is an intrinsic bias for nucleophilic attack but that since aromatic compounds are extra stablethe path leading to an aromatic product is more exothermic and is favored ii This may be partly true However it is certainly not the whole story a For one thing this is a very exothermic second step so the transition state must lie close to the intermediate not the products b Secondly it does not always work out this way egthe compound below has never been isolated It is said that the reason for this is the extra stability in the product of the decomposition reaction benzene COgt 9 26 i Prismane has 3 fused cyclobutane rings and 2 cyclopropane rings about I30 kcalmol strain energy ii Combine this with more than 30 kcalmol of aromatic stabilization in benzene and this reaction must be exothermic by at least I60 kcalmol iii Yet prismane can be isolated and upon heating it does not directly rearrange to benzene 188 gt 927 3 Diamagnetic anisotropy This is nmr information usually 39H nmr Energetic issues are avoided aAromatic protons resonate at lower fields than nonaromatic protons aromatic ole nic protons protons 9 28 10 9 a 7 6 5 4 5 ppm bThis deshielding is attributed to diamagnetic anisot r39OP i The general model that is used is shown below Orienting an aromatic molecule perpendicular to the applied magnetic field He causes the T electrons to circulate like the free electrons in a circular copper wire ii The circulating electrons create their own magnetic field H which is much much smaller than H a Within the polyene ring where the e39s are circulatingthe direction of the field is opposite to that at Ho and it extends outward as shown b Ha a proton outside of the polyene ring experi ences an increased field He H and is expected to resonate at a higher applied frequency c This model also predicts that a proton inside of the polyene ring such as H will feel a diminished field He H Thus it should resonate at a lower frequencyThis has been experimentally tested and found to be true H Ho I 189 c Examples H H 930 l8 annulene prepared by Frans Sondheimer l8 7 electrons 4n2 when n4 Accordingly the l2 outside protons resonate at l025wherasthe six inside protons are found at 305 Compare this with H H 5 56 ppm 5 575 ppm However thIs IS stIll not the whole story for example 932 For I 6 annulene I67E electrons 4n with n4 Fur thermore the l2 outside protons resonate at 5l 5 The 4 inside protons resonate at l03 5 dAromatic vs antiaromatic i Do the aromatic electron circulate in the presence of an applied magnetic field opposite to antiaromatc electronsThis is clearly nonsense Electrons are electrons ii While this classical model of a current in the wire works beautifully for polyene rings with 4n 2 T 190 electrons it breaks down for other electron counts iii We will not cover this in detail but diamagnetic anisotropy really depends on the mixing of excited states into the ground state under the influence of a magnetic field a For a 4n 2 polyene LUMO l 39l39l39 933 HOMO H So IEO E l IEO E2letcis large b However for a 4n polyene LUMO H 934 HOMO 39H39 l 39 lEo E l IE0 E2l etc is small because the HOMOLUMO gap if any is smallThere is therefore a large amount of mixing of excited states into the ground state in this case 4 Structure aThe idea is that a molecule will always distort itself to the most stable situation i Thus if conjugation is stabilizing then the structure should reflect this fact ii On the other hand if conjugation is destabilizing then its structure should be distorted in a way that reflects this iii In terms of aromaticity aromatic compounds should show no alternation of CC single and double bonds whereas nonaromatic compounds will ie 935a 191 tBu tBu j E ABA LB 148A tBu R iv Both compounds are perfectly stable They can be stored indefinitely in a bottle b Is tetrakis t butylcyclobutadiene then antiaromanc Lets look at why one should expect a rectangular structure Dam ggfgg 3ng singlet IAE1 lt IAEzl 936 triplet i The singlet structure of any antiaromatic compound should be distorted because there will be at least one kind of distortion which splits the degeneracy hence stabilizes one member for any cyclic polyene a This is the Jahn Teller theorem b In cyclooctetraene the distortion is not only changing bond lengths it also involves puckering the ring ii Returning to tetrakis t butylcyclobutadiene does the square structure mean that it is a triplet in the solid state a Probably not lts solution properties suggest 192 that it is a singlet b Where then is the driving force associated with the destabilization antiaromaticity in it i It turns out that cyclobutadiene itself is definiter rectangular Recall the Carpenter work described in Chapter 8 ii We ll talk later about the squarerectangular energy difference 5 Stability aThis should be the most direct method But square cyclobutadiene or nonaromatic cyclohexatriene are hypothetical molecules so how can we compare them to rectangular cyclobutadiene or benzene bThe most common argument takes the following form H2 AH 498 kcalmol 937 2 AH 286 kcalmul x 3 758 kcalmol By this calculation benzene has 36 kcalmol stabiliza tion cThe problem with this measurement is that is does not take into account any special stabilization that occurs when CC bonds are adjacent to each other i For example at the HLickel level 0 193 butadiene has 048B more 7 stabilization than two totally noninteracting ethylenesTaking B l8 kcal molthis would amount to 86 kcalmol ii For cyclohexatriene we would expect three such interactionsfor a total of 26 kcalmol This seems to explain the bulk of the stabilization dA different argument claims that there should be additional stabilization in cyclohexatriene because the CC single bonds are made up of sp2 sp2 hybrids i These are stronger that sp3 sp3 hybrids ii This suggests that the aromaticity that special stability due to conjugation may in fact be an illusion iii Indeed some people have argued that there is m stability associated with the regular structure in ben zene X xXx x X L XQX x X X X CH a 383 N Li H For X N and Hthere is very good evidence that the cyclic aromatic structure on the left is not stable When X Li it turns out that it is Much work has been undertaken in the last few years for the XCH case There are a couple of very good pieces of evi dence which show that in fact if it were not for the C C 039 system benzene would be a distorted l35 cyclohexatriene eWhat then about antiaromatic compounds is there destabilization in them i We talked before about why they should have a distorted structure the Jahn Teller theorem a Destroying TE conjugation in this case does lead to stabilization b Does this imply a destabilization associated with a planar flat structure That is is there a T destabi lization Probably not ii Let s first take a real system 194 2 AH 98 kcalmul 939 H2 AH 23 kcalmol X 4 92 kcalmol a Cyclooctateraene shows 6 kcalmol of destabilization Is this from antiaromaticity b The CCC angles in tub conformation of cyclooctatetraene are l265quotThis molecule cer tainly has ring strain cyclooctene probably has much less strain c One can argue that cyclooctatetraene is nonaromatic there is effectively no 7 overlap between CC bonds so it should not show any antiaromaticity iii Now let s take a look at cyclobutadiene a At the HLickel level O versus 940 ll 000 H H Loop 44 antiaromatic aromatic Where is the stability difference a One can argue reasonably in fact that the cyclobutadiene dianion should be more destabilized by resonance than is cyclobutadiene b There are two more electrons hence there should be more e39 e39 repulsion c A recent very goodVB calculation has found that square singlet cyclobutadiene is stabilized by resonance to the tune of 30 kcalmol fThe moral of this story is that there probably is not195 any special stabilization associated with aromaticity i T conjugation in any form is stabilizing in an ener getic sense ii All cyclic conjugated polyenes are stablized in this way iii For the hypothetical antiaromatic ones however in the latter we can expect distortions to a yet more stable form or a very high reactivity F Homoaromaticity l Homoaromaticity involves conjugation interrupted by one or more sp3 insulating carbon atoms a It follows HUCkel s rule b Often one observes a special reactivity c Example one protonation on cyclooctatetraene a homotropylium ion 9 85 ppm 0 66 ppm i The endo proton on C8 is strongly shielded 5 06 ii The ring protons appear at 5 85 H23456 or 5 66 H I 7 iii The C8 exo proton is at 5 52 iv There is a barrier to ring flipping of 225 kcalmol v All of this is taken as evidence of homoaromaticity d Example two decomposition of anti7norbornenylptoluenesulfonate 196 Tso 4 CH3C02H 942 T50 rate is H much faster 9 4 CH3C02H The stability associated with homoaromaticity makes the formation of this intermediate more facile e Example threethe bicyclo32loctadienyl anion tBuOK lt gt DMSO very acidic E 943 proton 3 e lt gt 2 Homoaromatic stabilization is really important only for anions or cations a For neutral molecules nothing in particular is gained b For example l35 cycloheptatriene and l47 cyclononatriene show no evidence of stabilizing TE overlap across insulating units G Spiroaromaticity a new twist on homoaromaticity lThis is a special case related to homoaromaticity a HLickel s rule is obeyed b However the orientation of the 7 system changes across the insulating carbon 2 Examples gtltl E W 944 o c 197 a In these systems only wd can interact with 1Id ie 3 ltgt D l d d WdWp wpwp bThis results in the following analysis mode 00 mode 22 mode 20 946 le l p Il ol Altgt Wd l d l p Wd l d Wp le Wd destabilization no effect stabilization c A specific example is shown below The numbers in parenthesis are experimental ionization potentials This molecule is very reactivefor example it under goes hydrogenation much more readily than expected 1 a2 32 a2 1 b1 i I K I I 3 I 2e 1 a2 799 eV 32 a2 gg i 1b1 922 b2 l l39lF39 H b2 19 106 H Bicycloaromaticity there are two types 198 948 longicyclic laticyclic l Each has its own intrinsic stability patterns see Goldstein and Hoffmann jAm Chem Soc 936l9397 2A couple of examples which show experimental ioniza tion potentials are shown below D Ionization Potential eV 0 949 Ionization Potential eV 1o 950 a 199 3Work through each interaction with care I Heterocycles and Through Bond Conjugation I Using the perturbation theory ideas from the previous chapter let us work through what happens to the 7 system on going from benzene to pyridine aThe degenerate p sets of benzene are split one where the AO coef cient on the carbon atom that is becoming nitrogen is stabilizedThe other is left un changed bAs mentioned in the previous chapter electron density is increased on the more electronegative atom for the bonding MOs and increased on the more electropositive atom for the antibonding MOs This is reflected in the drawings aboveThe shapes are drawn for one value of I Therefore more electron density is associated with contours that are closer to a nucleus cThe amount of stabilization is proportional to the square of the AO coefficient and how much more elec tronegative the atom has become relative to carbon This is demonstrated in the plot below The ionization potential ofX the atomic rst lP for the 4832 gt 3P0 state of X measures the relative electronegativity of X 200 O 11 12 Ionization Potential 9V w 00 9101112 131415 Ionization Potential X eV 2The slope of the a2 MO is zero There is a node on the X atom for this MO so it should not be perturbedThe coefficient at X for the lbl MO according to H39L39Ickel theory is 6 while that for 2b2 is 3 The slopes of the two lines are O89 and O373for the lb and 2b MOs respectively These are very close to the values of the H39L39Ickel coefficients squared 2The PE spectra of the three disubstituted azines along with pyridine are shown on the next page aThe assignments of the peaks is a lengthy process and one not without difficulties Let us accept how ever these results and look at a correlation of the IPS for the two 71 MOs derived from elg in benzene g3 Ionization Potential eV 120 COUNTS SEC COUNTS SEC COUNTS SEC C OUNTS I SEC 33n 1 18 19 2O 21 2 WWW IP M I I Y O N N i w 14 15 16 17 18 19 20 21 11P eVl 3 1 IV H20 IL M 10 11 12 13 14 15 16 17 18 19 20 21 IPeV I 1 I g I w 0 N 7t2 1 1 lt5 A r M 1 y w J 1M 8 9 1O 11 15 16 17 18 19 20 21 I 01 202 bThere is a very nice relationship here between the size of the AO coefficient and the stabilization of the MO c Replacing a CH group for a nitrogen atom also replaces one CH 039 bond with a lone pair These are also at low ionization potentials and have been marked on the spectra dWhen there are two nitrogen atoms there are then two lone pair MOs They are the plus and minus combinations and are thus listed as n and n respec tively A correlation of these MOs is shown below 1 9 s Egg wag e For the l2diazine there is 7 type of overlap be tween the adjacent lone pairs So one expects and one sees that there is substantial throughspace over lap between the lone pairs which leads to a sgnificant nn splitting f But the nn splitting for the l3diazine is almost as large and what is worse is that the nn pattern for the l4diazine is reversed from what it should be This is called throughbond conjugationTo look at this we shall first see what happens in the 222 diazabicyclooctane molecule Lo JV 0 o o Ionization Potential eV 203 3Throughbond conjugation One expects the lone pairs in 222 diazabicyclooctane to be that shown below W n N N 0 lgtltgt l 0 M 956 n aThe distance between the lone pairs is enormous so the splitting between the lone pairs is expected to be very small bThis however neglects the CC o and 5 orbitlas which have the same symmetry and therefore can mix with the lone pair combinations 957 cActually the mixing with 5 is quite small since 5 lies at too high of an energy However the mixing between o and n is quite strong The net conse quence is that n is pushed to high energy relative to n Therefore the PE spectrum for 222 diazabicyclooctane is that shown on the next page 204 cps Ionization Potential eV 0 N I 958 dThe extension to the 4diazine case is straightfor ward For 3diazinethere are complications and this analysis needs to take into account not only the CH bond but also the bond path in the other direction The actual wavefunctions for the n and n combina tions are shown below quot O lt gt Aaltnm llVIIVAIIMN Wmll llll lam mum manna t 7 0 01 I L 0 o I L 0 01 I L O MN 85252 V 7 V7 7 Ionization Potential eV 5 120 eThroughspace overlap dies out exponentially how ever throughbond conjugation does not The trans mission of electrons in proteins is one example where throughbond conjugation is expected to play a pri mary role fAnother example is provided by the stability and singlettriplet splitting in the three isomeric benzynes 205 AH theo AH exp ST kcallmol kcallmol I 109 106i3 36 V 123 116i3 17 960 0 O 134 138i1 2 f meta and para benzyne are only about IO and 25 kcalmol respectively less stable than ortho benzyne Clearly there is a 7 bond between the two hybrids in ortho benzyne For the other two compounds it is through bond congugation that makes them so stable a typical CC 7 bond is worth about 60 kcalmol and makes the singlet state more stable than the triplet one J Buckyballs what review of 7 systems would be without them Definitions and some shapes a o c Q s a ef g39 wow o 0 3 6 5 6 O a V 0 e 3 6 961 C73 nanotube 206 2There is a definite pattern of six and five membered rings A six membered ring causes the structure to be flat graphite is the ultimate example of flat sheets It is the five membered rings that give curvature 3The PE spectra for bezene and C60 are compared below The a2u and ezg ionizations correspond to the 71 levels in benzene Because of the very high symmetry in C60 there are much larger degeneracies Note that the 71 ionizations in C60 are at much lower energies The electron affinity the position of the LUMOs in C60 are also much smaller lt s electron affinity is 266 eV ie 38l eV lower than the e2u set in benzene C60 is then a good electron donor as well as an excellent electron acceptor e19 C6H6 25 19 b1u b2u 2e1u a2u 2929 0 4 C 3 o O I I I I I I 18 16 14 12 1o Ionization Potential eV C60 Counts f 15 14 13 12 11 1o 9 8 7 962 Ionization Potential eV 207

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