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# Class Note for CHEM 6311 with Professor Albright at UH

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Date Created: 02/06/15

1 Bonding and Resonance Elementary Concepts We will cover bonding in much greater depth towards the middle of this course Here however we will review some elementary level material A Lewis Octet Rule amp Formal Charges Please review these for yourself Know how to calcu late formal charges and what this means Let me ask a question Compare BH3 to NH4which central atom is more electron deficient If you answered Nthen you are in real trouble If you answered B make sure you know why B Molecular Orbitals These are the basis for all of spectroscopy Understanding them is increasingly more relevant for mechanistic studies pharmacology dyes etc I Atomic orbitals s p df a Usually f orbitals are not important for bonding b In this course we also will not be concerned with d orbitals cAn atomic orbital is described by a wave function Q i A plot of the magnitude of Q vs rthe distance from the atomic nucleus is characteristic for each type of orbital For all orbitals except 5 this plot also has an angular dependence ii Note that there is a phase associated with Q iii We normally draw these wave functions to have three dimensional character showing boundaries which include 80 of the total value of Q in the following way The maximum value of Q is at the nucleus ie r 0 q S orbital r The minimum value of Q is at the nucleus ie r 0 l P orbital I39 j y 90969688 11 G 35 III III 9 12 iv The perimeter lines correspond in position to the dashed lines in the plots of Q vs r above d lndividual atomic wavefunctions Q do not have any direct physical significance i The phase of Q on an individual atom does not have any meaning ii The sign of one relative to the other atomic orbitals in a molecule does have a physical meaning a It tells us whether the orbital is bonding or antibonding with respect to other atomic orbitals in a molecu lar orbital b It tells in which directions in space the molecular orbital is concentrated eThe value of DZ does have physical reality It is related to the electron density and can be measured directly by xray spectroscopyThe problem here is that most electron density per unit volume is concentrated in the core Hence it is an excellent technique to measure the positions of the nuclei rather than the electron density associated with the valence region 2 Combination ofAtomic Orbitals aThis is the root of aH bonding theory bAny molecular orbital is some sort of combination of constituent atomic orbitals i Example interacting two atomic s orbitals to describe the orbitals in the H2 molecule H H H H potential energy W Here lI CI C2l2 cl amp c2 are called orbital coeffi constructlve Interference bonding same relative Phase ClentsThe larger theIr magnI tudethe more that particular atomic orbital contributes to the molecular orbital Because of symmetry cl Cl in this example This is obviously a bonding molecu lar orbitalThere is no region per pendicular to the internuclear axis where the value of V drops to zero unless the internuclear distance is infinite lz C3l C4l2 4 c3 c4 by symmetry destructive interference D s have opposite phases This is an antibonding molecular orbital node plane where V drops to zero This is half way along the internuclear axis Recall for a p orbital the node is at the nucleus Notice in this case the nodal plane is perpendicular to the internuclear axis iii AE in the interaction diagram shown above is the stabilization energy upon interacting two atomic orbitals We ll talk much more about what factors influence AE later this semester 3Types of molecular orbitals a 039 no nodes along internuclear axis GWWOWC 0 WWW b T one node along the internuclear axis dThe relative stability is determined by overlap i This is a measure of how much constructive interference there is ii In general39 gt T gt 5 related simply by the number of nodes along internuclear axis 4 Hybrid orbitals aThese are formed by mixing atomic orbitals of the same center nucleus b For our purposes we will consider only the mixing of s and p atomic orbitals i If we give them some preassigned mixing coefficients then these hybrid orbitals on different centers can be used to interact with each other ii This is the essence of valence bond VB theory c Alternatively we do not have to require atomic orbitals to mix initially on the same center in some preassigned manner i Hybridization will still occur but the amount of s and p mixing will not relate to our preassumed values ii This is called linear combination of atomic Qrbitals LCAO theory Most of you know something about the former and little about the latter approach This is unfortunate and will be rectified somewhat I hope in the latter stages of this course dWhen atomic orbitals mix with each other on the same center they do so as vectors Z p2 C1py C2pz Orient the shaded lobe along the 1 axis depending on the sign of c c1sczpy y The two hybrids have opposite orienta c1sc2py ampgty tions Z I The hybrid has the same orienta c1s CZPY y tion as the first example but Z an opposite phase 01502Py63Pz ltgy 110 5 Common types of hybrid orbitals Consider an arbitrary hybrid orbitalX which is centered on an atom that has 5 and p atomic orbitals An LCAO expansion in general would have the form X cpl l39 c2lx l39 33 l39 c4lz and since X must be normalized we require that det I c dr c dr c dr c dr notice that there are no crossterms when X is expanded in the integral because clczf lslPd l7 0 Therefore cl2 c cg c3 l a sp3 hybrids let us take methane as an example using the labelling and axis system shown below Since all hydrogen atoms are equivalent are related by symmetry operationsthen it makes sense that the four sp3 hybridsxl X4 should also be equivalent in terms of their composition In other words the ratio of s to p character should be the same in each hybrid and the amount of 5 character should be the same in all four hybrids 10 H2 1094712quot PHI H E HA I I H3 H3 Each hybrid is pointed H2 directly at a hydrogen atom y I x H 111 The general form of the hybrid orbitals are XI cl l l39 c2lx l39 33 l39 c4lz X2 cs l l39 c6lx l39 37 l39 c8lz X3 2 39 l39 c0lx l39 cI I D l39 clz l X4 2 93 cl4lx cl5ly c6lz and c2c c c l etc and c2c c c23l so c2l4 and cl l2 i Let us first consider the hybrid which is pointed at HI I XI clls c2lx 33 34 H but because of the coordinate system X y I l c2c40 and normalizing XI gives 4c32 So c323l4 and c332 The ratio of szp character is given by the square of the mixing coefficients so spl434l3 This is why we call this hybrid an sp3 hybrid The other sp3 hybrids have the same szp ratio and since the s mixing coefficients are all the samethe sum of the p coefficients squared must always equal 34 ii Now let us consider the second hybridX2 11 c I 2 H H L 04 y c7 02887 X c808 I 65 9 180 1094712 705288 and tans c5lc4 so c4tane 5 X2 c5lS c6lX c7ly c8lZ and c6 0 and c5 lZ 4 c c I b sp2 hybrids let us consider borane with the geometry and coordinate system shown below H3 H1 XI clsc2ly H er a r H2 1200 I c 32 23 115 Hz H Therefore szp l2 iesp2 V X 3g 114 c sp hybrids let us consider the hypothetical molecule HzBe 4 H2 9 H1 O H1 180 39 1 2 115 X cllS c2ly where cl c2 leE d generalizations 5P HAH angle 10947 120 180 hybridization Sp3 SF2 5P W 5P2 5 character 25 33 13 50 12 The designation spm hybrid gives only the ratio of s to p atomic character We always use one sAO and three two or one pAO s depending on the number of unhybridized pAO s that we require to form 7 bonds e The hybridization of orbitals involved in equivalent bonds can be determined from the angle 9 between them As a means of simplificationwe choose one bond to be aligned with the y axis 2 cosG i where 9 gt 90 i m I H2 l l m e m XI ml ml 116 A H1y This is spm hybridization s l m x 00 p mlm x l00 Let us consider some examples i For idealized ethylene 9l20O H H cos l200 050 lm cc e ie m 2 gt sp2 H H 33 s 67 p 117 but by crystallography 9l l66O cos 9 0447 lm ie m 223 gt spl23 3 l s 69 p Thus the CC s bond must be 38 s and 62 p The bond should be shorter than an idealized sp2 sp2 and there should be more bond overlap The p bond overlap will also be improved because the internuclear distance is reduced 13 ii In ammonia the H N H bond angle has been mea sured to be l073quot This is less than the idealized l0947quot Does this mean that the orbital containing the nonbonding electron pair will have more or less s character than a standard sp3 hybrid Answer this for yourself before proceeding cos 9 0297 iem 336 sp336 gt 229 s amp 77l p HV HH For the remaining orbitalthere is 3 l3 of the s 13918 e 1o73 so it must have 687 p character and is sp2399 totalp 3 x 77 I 687 300 iii The H O H bond angle in water is l05quot Determine the hybridization of both types of outer shell orbitals bonding and nonbonding on the oxygen atom and the angle between the nonbonding orbitals iv Other structural factors can influence the hybridiza tion of bonding or nonbonding orbitals In the mol ecule shown below bulky R groups will force their A bonding orbitals apart and increase the degree of p character in the C H bonds If the R groups are connected in a small ring as in cyclopropane for examplethe hydrogen atoms will be forced apart See the first problem set f Interacting hybrid orbitals as a bonding model the foundation of valence bond theory This is done just as in the LCAO approach OO v2 C3 c4S For methane then one would expect there to be four 5 equivalent CH 039 bonding orbitals since there are four equivalent sp3 X m hybrids WW1 c1x czs 119 R 14 Actually this is not experimentally true We can measure the potential energy of electrons in their molecular orbitals by photoelectron spectroscopy The experi ment is basically of this form An electron is released with kinetic e39kiquot9 energy energy KE It is collected and analyzed with respect to KE The potential energy Eof the electron in its molecular orbital is therefore photons at constant energy hv WlMIWVW CH4 simply IF hv KE Miami 0 observed P ionization potential 3 4 sp3 CH t2 g 15 g 31 15 I g 0 LP orbital energy i ii ionizmion potebtial ionization potebtial According to this VB model methane should have two peaks in a 4l ratio The actual spectrum has three peakswith two in the ratio 3l for the CH bond e39sThis is one of the major failings ofVB theory we ll see later why this comes about C Resonance I This is another place whereVB theory fails to predict structures aWhere electrons mainly in TE orbitals can migrate from one position to another b In this case one does not have localized 7 bonds rather there are combinations of them 2 Classic examples a Benzene ln aVB calculation one would take six equivalent CH bonds six equivalent CC 039 bonds and a combination of X3 and 90 9 E 2 The doubleheaded arrow is a symbol for resonance not a reaction 15 bThe allyl cation 12 1l2 Note curved arrows show electron flow from one structure to another 3Some rules of resonance aThe positions of nuclei must be the same in each resonance structure if they are not then this is a chemical reaction Resonance structures simply indi cate delocalization of electron pairs bThere must be overlap between orbitals that are involved in resonance GA bg 124 orthogonal orbitals Bredt39s rule cTotal electron spin must be conserved he l 1 fl lt gt1lt x t 50 50 51 125 Note the use of arrows with onehalf ofa head to indicate ow of one electron at a time dThe resonance structures need not all carry the same weight For the benzene and allyl cation problems they obviously do because of symmetry ieall bonds are identical For general situations there are some rules i Structures with more covalent bonds carry more weight l lt gt 1 major minor 3926 ii Increased charge separation decreases resonance structure contribution 0 39wlz lt gt H N H N 39 l H 117 major minor 16 iii Structures with excess electron density on more electronegative atoms carry more weight H H c c H j cHx lt gt H1c cm equal contibutions H H c c H ij 39 39 H1C g 123 quot minor major e Resonance can also occur via the 039 system hyperconjugation H H Delocalization of a pair of elec Hl1 Hz HcCH2 trons from a CH 0 bond overlap l l with an empty Tl orbital 13 H 39 Spin density is transferred to H c CH2 H H IICH2 the H atoms This can be seen H H in esr The odd electrons F F quot coupled to the H s r c39Qh lt gt FCCH2 negative 129 l hyperconjugatlon 4 There are several very real problems with resonance theory a Symmetry is easily violated One might naively think that all three CH or CF bonds in the above example can participate in resonance they cannot bAlthough one particular resonance structure can make a minor contribution a large number of them can add up obscuring what one would like to be a simple pictureThe classic example is again benzene Although the two Kekule structures do contribute the major amount an accurateVB calculation of benzene has shown Op 060 000 14 m 17 Let s digress for a moment and talk quickly about the basic electronic effects in organic chemistry this is a review of your undergraduate organic education I inductive effects this occurs via the 039 system and is attenuated strongly by distance 2 resonance effects this occurs primarily via the 7 system and does not appreciably die off with distance A T acceptor groups all have lowlying 75 orbitals X o o R39 A N CEN gtN c HR o R x OHOR o o o NH2 etc o o O 0 NO W N K 0 0 6 D T donor groups all have highlying lone pair orbitals X D39 quotH1 9 9R FCBrl H1 NH1 NW NH1 C 3 a t 131 sCH2 H can act as either 39Ph Be Sure ThatYou Understand This 1 Stuff c It can sometimes be very hard to tell whether or not resonance occurs eg for a hypothetical case 18 potential energy Is the symmetrical ABC species the ground statewhere its elec tronic structure can be described by two resonance structuresg is it a transition state for the inter conversion of two classical non resonating structures Experimen tal probes have long time periods so only averaged structures are often seen 132 reaction coordinate i Example one cyclobutadiene 179 20 1 rialo 133 We ll talk much more about cyclobutadienes later this semester but the weight of experimental evidence suggests that it has a rectangular shape ii Example two semibullvalenes R1 5 R 3 or A 2 3 Xray structures show that the Cz jAbond length is very long compared to a normal cyclopropane and the C4C6 distance is very short compared to what should be a nonbonded distance d Often times xray structures even when available give ambiguous results because of disorder or because the effects of substituents mask the situation eg tetrakis t butylcyclobutadiene has 19 four nearly equivalent CC bond lengths The activation energy for interconversion between classical structures can be my small a couple of kcalmol or less Normal experimental measurements then cannot tell the differ ence between rapidly alternating classical structures and a static delocalized resonating oneThere is one very useful technique however for investigating this D Saunder s Isotopic Perturbation of Degeneracy Technique Let us study one particularity troublesome example the structure of carbocations l Little direct structural information exists because the molecules are very unstable and in an crystalline environ ment the interaction between the carbonium ion and its counterion may well cause the structure to be distorted from its solution or gas phase structure where this type of interaction probably does not exist 2 Consider the following case where R and R are alkyl groups H I c1gnR cl E C1c2 gt Runquot 1 full R R R R R RfR A C B E H H 135 c c c c I Ruquot 1 2uR Ru 1 gunk R R R R is C is a transition state or the ground state structure a Let us first suppose that this is an equilibrium reac tionwhere C is a transition state i If the rate constant between A and B is small then in the 393C nmr we would see one signal each for CI and C2 in A a Provided that R i R B will give two distinct signalstoo b This would mean a total of four signals com pared to only two for the delocalized reso nance structure 20 ii But the rate constants for going from A to B are normally very large a Therefore 393C nmr sees only an averaged set of resonances b This is not distinguishable from the result for a delocalized molecule where C is the ground state bTo distinguish the two casesfirst consider the spectrum when R R K A B where K equilibrium constant A concofA B concof B i Here A Bso K l ii The two kinds of carbons can be modeled using CH3 ch l3 CHzclk lk CH3 l ch C CHZ CH3 52 32 ppm as a model for CI in A 5 335 ppm as a model for C2 in A iii CI in A is identical to C2 in B and viceversa therefore the averaged value for each is identical signal intensity 51 184 ppm 52 136 chemical shift Note Only signals due to the carbocationic centers are shown or discussed the actual 3CNMR spectrum would contain other signals as well 21 c Now let us consider the case of R i R i In general A i B and K7 l ii 5 in A will be slightly different than 52 in B and SI in B will be slightly different than 52 in A but we can neglect this because it is a minor perturbation iii What is important is that since A i B even by a very small amount we will see two peaks iv Suppose that A gt a The resonance for CI moves slightly upfield since the averaged structure between A and B looks slightly more like A In other words SCI is the average of that for CI in structure A and CI in structure BThis will not be at I84 ppm it will be shifted slightly upfield since A gt b Likewise the resonance for C2 will more slightly downfieldThus there is expected to be two peaks when K i l c The 393C nmr spectrum then looks like this signal intensity 5 184 ppm 5 1 chemical shift 2 137 v One can show that A5 l K l K A note the magnitude of 5 depends on how different K is from unity but A is a constant in the following way 5A52B SCI Allel 62 A 5I B 5C 2 AHB 22 6A6C 6CZW A B dividing the numerator amp denominator by B and K 6A 2 5 62I K Z AI K I K I K a NowAG RT In K i AG free energy difference between A and B a constant ii R gas constant iii T temperature Kelvin b Changing the temperature will cause K to change andthereforeA5 must also change since A is a constant dThe alternative is that C is the ground state structure i If R i R by a minor amount then one will see two different resonances ii Their chemical shift difference will be much smaller and more importantly this difference will be indepen dent of temperature 3 Let us take some real examples aThis is a classical equilibrating situation A5 52 ppm at 70 7 c C A5 32 ppm at 140 1 c H ens Jgt D quotT A5 32 ppm at 70quot c A5 105 ppm at 440a c J12 5 H H36 395 CD 3 D e a c Hat CH5 H H cuJ bThis is an unambiguous case of resonance 4 A5 03 ppm at all measured temperatures 3 H38 CD3 H38 CD 23 c A real system where there has been much contro versy is the 2norbonyl cation i Is it an equilibrating system composed of two classical structures ii Or is it a static structure Cthat one can view as a resonance hybrid ofA and B H H G H H D H H D A C Iquot H H 74 I H 39 D H H D H B iii Experimentally A5 l2 ppm and it is tempera ture independentThis gives strong support for the idea that C is the ground state structure dThere is a series of C6R62 speciesAre they rapidly equilibrating classical structures or a staticvery unusual structure CH3 CH3 cm ch CH3 ch CH3 Hac ten HaC CH3 HaC CH3 ch cu cm CD3 cm etc ch HgC 2 CD 2 i The nmr spectra are averaged down to very low temperature 24 ii This is consistent with either the rapidly set of equilibrating structures or a statictotally delocal ized one iii However A5 04 ppm for the ring carbons and the spectra are temperature independent only the nonclassical structure is consistent with all of these data e There can be problems with this technique Firstly one must make very minor perturbations there fore the value of K is very close to unity For an observable A5 there must be a large intrinsic difference large value of A Secondly this whole technique relies on the fact that AG is tempera ture independent Sometimes if there is a large AS contribution this is not true 25

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