### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Mechanisms CHEM 6311

UH

GPA 3.71

### View Full Document

## 101

## 0

## Popular in Course

## Popular in Chemistry

This 114 page Class Notes was uploaded by Dominic Kling on Saturday September 19, 2015. The Class Notes belongs to CHEM 6311 at University of Houston taught by Thomas Albright in Fall. Since its upload, it has received 101 views. For similar materials see /class/208157/chem-6311-university-of-houston in Chemistry at University of Houston.

## Reviews for Mechanisms

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/19/15

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 lt gt r The minimum value of Q is at the nucleus ie r 0 l p orbltal i 60960688 x III III lll III III III III III 003 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 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 N H H H H potential energy W Here lI CI C2l2 constructive interference bonding cl amp c2 are called orbltal coef Same relative Phase CientsThe larger their magni tudethe more that particular atomic orbital contributes to This is obviously a bonding molecu the meleCUlar orbltal Be ause lar orbitalThere is no region per Of Symmetry c39 Cl m thls pendicular to the internuclear axis example where the value of W drops to zero unless the internuclear distance is infinite lz C3 I 39 C4 2 c3 c4 by symmetry LQ destructive interference D s have opposite phases This is an antibonding molecular orbital node 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 6 WC 6 O m W W b T one node along the internuclear axis 1 two nodes one along and one perpendicular to W theInternuclear aXIs 177 V c 5 two nodes along the internuclear axis 4 I r 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 C2 C1py C2pz 39C1py C2pz Orient the shaded lobe along the 1 axis depending on the sign of c Z V 01502Py Z The two hybrids have Z 3 y Z opposite orienta y c1sczpy y tIons Z Z The hybrid has the same orienta y 815 82py y tIon as the first example but 2 Z an opposite phase 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 110 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 H2 1094712quot xc HI H H H1 3 H3 Each hybrid is pointed Z directly at a hydrogen atom H2 y I I 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 Z XI cls c2lx 33 34 H but because of the coordinate system c2c40 and normalizing XI gives ll4c32 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 c I 2 H H L 04 y c7 02887 quot 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 c 3c223 1 Hz 1200 1 115 I H H 2 Therefore szp l2 iesp2 V 3g 114 c sp hybrids let us consider the hypothetical molecule HzBe H2 931 ltgtH1 I z H2 O cllS c2ly where cl c2 lxE d generalizations 5quot3 HAH angle 10947 120 180 I sp2 sp hybridization sp3 W 5P2 5 character 25 115a 0 SP 33 13 50 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 where 9 gt 90 i m I H2 l l m e m XI ml ml 116 A 1y 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 c 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 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 R 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 H 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 Wz C396 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 Actually this is m 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 photons I d h KE constant energy ana yze wnt respect to m CH4 The potential energy Eof the electron in its molecular orbital is therefore simply IF hv KE expected 0 observed P ionization potential 0 0 4 sp3 CH 2 u 539 15 g 31 1s 3 8 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 0 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 lt gt xa xb 2 The doubleheaded arrow is a symbol for resonance not a reaction bThe allyl cation J gtV Note u m w 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 cpnserved r 1 y lt gt lt X gt x 1 t s o s o 5 1 1 25 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 lt gt l 126 major mlnor ii lncreased charge separation decreases resonance structure contribution O 0 39 wk lt gt H N H N I I 117 H H major minor iii Structures with excess electron density on more electronegative atoms carry more weight H H c c H j cHx lt gt H1c CH1 equal contibutions H H c c H1C E3 39 39 H1C o 123 391 quot minor major e Resonance can also occur via the 039 system hyperconjugation H Delocalization of a pair of elec trons from a CH 0 bond overlap H H 1cm 4 H CH2 l with an empty Tl orbital H H H H 413 Spin denSIty IS transferred to H IZ CH2 H H IICH2 the H atoms This can be seen H H in esr The odd electrons coupled to the H s if F Fc 4 F CCH2 negative l hyperconjugatlon 129 F F 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 0p mom 000 14 m 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 N c H39R39 R OHOR x 0 o NH2etc o o o 0 NO W N K lt gt 0 H D T donor groups all have highlying lone pair orbitals D NH1 396H 3R 3EX Km NH1 NH1 N C 1st lt4quot o Q sCH2 H Ph 131 can act as either 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 potential energy A Bc Is the symmetrical ABC species v 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 PHEEQ 0 d 1 LU l E l a E 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 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 I H H C ElquotMR gt flp gt qu Cz 1 z R RI 1 ruR R R R RI R R R A C B or H H 135 Rvc1 c H c c RV 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 I ch C CHZ cm I Hp 52 32 ppm as a model for CI in A CH3 ch C CHZCHg 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 5 b 82523984 PM E A A51 52303 ppm I Bl chemical shift184 ppm 52 136 Note Only signals due to the carbocationic centers are shown or discussed the actual 3CNMR spectrum would contain other signals as well 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 2 E E A n 396 JUL I 5 184 m 5 Ichemical shift pp 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 5152BlAl mac 5c2 AB dividing the numerator amp denominator by B and K 6A 2 6 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 3 7 A5 52 ppm at 70quot C H c A5 32 ppm at 440quot c H36 395 CD 3 H CD D D D A 7 D A5 32 ppm at 70quot c H36 339 A6 105 ppm at 440a c H M H CH 9 bThis is an unambiguous case of resonance 4 A5 03 ppm at all measured res t mp t CD3 3 3 up 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 x H H G H H D H H D A C III 34 A 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 3cm ch crla HaC CH3 HaC cu3 ch cu cm CD3 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 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 Cgt QCgt 3 m nodes ltD lt3 m 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 177 178 2 Sample calculation ethylene approximations For details see Chapter Appendix l of Lowry and Richardson Go over this material by yourself aLOB 96 These are only used for the 71 system and involve many B H39Lickel calculations 95 W V 39i 39i 39 S 39i 397 39i 39 39y L 39i 39974 39 V rv 39 0 a4 yxxlO 39 17 s 000002 Vamp Vamp 39 7 VA39 VA39 V i 39A V 717 V 39f 39 39 V V 39 39J 74 39 VI 39 39 s 39 V 39I VJq 39 V v hexatriene i i i AV w 7A39 39fxn VIq 39 39 Vil 39A i 39 74 39z 3 gThe 71 orbitals of butadiene through heptatrienyl are shown below Learn those for ethylene through 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 07l a 06i83 a L4I4B W oso o37 cmoso alGIBB 05 om 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 odd O N C Ci 03 L 99 88 02 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 911 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 e en odch b PuttIng e ectrons Into t IS 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 I gt 2n1 orbitals H H 4n2 1t electrons 916 H J c An antiaromatic compound has 2 fewer electrons it obeys the 4n rule therefore its orbital structure must have the following situation 183 l l 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 lt3 1e39 2e39 3e39 lt 4e39 4quot39 quot39 lt1 5e39 39 6e39 2 7e quot 3 8e39 I x 5e l The pattern of these orbital combinations is important butadiene2 has the same E HOMOLUMO symmetry K 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 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 m W A A W Vquot s s Vquot LUMO LUMO LUMO LUM W g 5 5 W W g A A W HOMO HOMO HOMO HOMO net destabilization 923 g E E m 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 X substitution GLf f x gt X x H H 9 25 carbonium ion intermediate xxx 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 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 gt CO 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 a 28 10 9 B 7 6 5 4 5 m bThis deshielding is aftributed 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 H39 Ho 1 p c Examples H H 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 931 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 A 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 H LUMO l 933 HOMO H So IEO E l IEO E2letcis large b However for a 4n polyene H LUMO l HOMO 39H39 l 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 934 Stru ctu re 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 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 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 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 X x ix x X l 5 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 Versus 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 a 85 ppm 0 66 PMquot 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 6 4 CH3C02H 942 T50 rate is H H much faster 9 4 CH3C02H H 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 e 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 ltgt ltgt ltgt 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 2e 1 a2 799 eV 32 a2 gg i 1b1 922 b2 l l39lF39 H b2 19 106 947 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 to Ionization Potential eV 0 949 Ionization Potential eV 1o 950 3 199 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 8 E 9 2 cc 395 1O c 132 a 5 D 11 s 2 5 12 5 s 2 13 8 9101112131415 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 II 39 0 01 I O Ionization Potential eV U l 201 l O COUNTS SEC 7A 1 I l 7 8 9 10 11 12 13 14 15 16 17 18 19 2O 21 COUNTS SEC 7t2 n 2 1 H20 3 E 1 1 391 11 M L393 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 COUNTS SEC IV 1112 1 H N 111 1 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 C OUNTS I SEC IP eV 1 1 1 J N n3 N n n 3 1 1 1 111 J i 39 1 7 8 9 1O 11 12 13 14 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 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 9 9 8 5 go amp gin 12 0 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 A 6 gt 3 3g 752 quotE 3 a 4 O 0 965 n 310 a N E n O quot12 O 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 mi mm W r 0 01 I 100 L 0 01 I L L O WEEK ENSEVZVA V 7 7 W 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 0 396 g x 6 A6 4 390 1 639 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 2a1g b1u b2u 2e1u a2u 2929 0 4 C 3 o L I I I I I I 18 16 14 12 1o Ionization Potential eV C60 gghg hu Counts f 15 14 13 12 11 1o 9 8 7 962 Ionization Potential eV 207 VII Acids and Bases A Equilibrium measurements I Ideal behavior a For the dilute aqueous system k1 A39 H30 71 HA H20 acid base k4 conjugate conjugate base acid k A H30 K e bTherefore kil a HA 2 ln concentrated solutions one cannot assume ideal behavior aThe equilibrium constant takes a modified more precise form mew aHA Ka where aHA etc are activities and K 2 A 1H301YArllYH3ol a HA YHA Here the Y s are activity coefficientsThese correct for deviations from ideal behavior eg anioncation aggragation etc 3 Measurements of extremely strong or weak acid strength aThe directly measurable range of pKa in H20 is 3 lt plgI lt l l bSuppose the pK1 of an acid HA lies in this range but that of HB does not i One can in principle establish an equilibrium between HB and the conjugate base of HA ii Provided the two conjugate bases are close enough to one another in strengththe equilibrium can be measured This is after allwhat pH is about using H20 as the conjugate base of H3O 122 c Given the system KAB HB A39 then the equation B HA X A H30 B H30 H I gtI V HA HB allows calculation of the new K from the equilibrium constant B39 HA 72 KAB X KAHA KAHB Examples of this idea are the Hammet acidity function and the related acidity functions discussed in your book d Consider a series of acids HAI HA2 HA3 each of increasing acidity Futhermore lets say that HAI only is sufficiently weak that its degree of ionization can be measured in H20 i The following equations describe the system HA1 H20 Ai H30 72a K aHSOA1vA HHo1A1 ii To measure the K HA2 we need to increase the amount of H3O present a HA2 is a stronger acid than HAI b Without adding an external strong acid the equilibrium listed above will be shifted completely to the right iii Now YAZ and YHAZ are mt unity and the full equilibrium expression must be written K AmAE aHA1 aHZOHA2YHAZ and no further simplification can be made because we are now away from ideal behavior 123 iv Using the previous equilibrium idea one can easily see that KaHAo AfllHAzl X YAIYHAl KaHA1 AHAI YAE YHA notice that aH3O and aHZO have been fac tored out v We can establish the concentrations experimen tally to give a suitable equilibrium among HAI HA2 AI39 and A2 vi KaHA has been measured previously thus KaHA2 can be found a Provided that the activity coefficients are known b Normally this is not the case but one assume YA YA YHA YHAZ c Therefore the ratio in the expression above iv is unity d This will be reasonably correct if two conditions are met i The structure of HA and HA2 are close ii The same solvent is used to measure Ka HA and the equilibrium between K HA and KHA2 vii The same method then can be employed for HA3 using HA2 to set the equilibrium etc 4Acidity functions aWe can use similar reasoning to establish a new acidity scale i Rewriting our original expression from which HA2 HA3 are indirectly found HA aHS YA Ka HA A 1 am 124 ii Now let ho K HA HAA39 iii and define H0 log ho ivThen Ho 2 logKaHA PKaHA log IFS A log aHS log M A bThe H0 scale was established by Hammett i The scale spans a wide range of sulfuric acid water solutions ii It was built using a series of substituted anilines whose relative pKa s had been established through the method of direct comparison iii Once the solvent system had been character ized in this way it was possible to measure the strength pKa of a new acid directly without concurrent comparison of a second acid by mea suring its ability to ionize in a solvent mixture of known H0 iv Note that H0 is used here in exactly the way pH is used in dilute aqueous solutions c A method for measuring the basicities of very strong bases has been set up along similar lines and an analogous scales H39 has been derived i Solvents used to study strong bases must have a very low dielectric constant ii There is then the additional complication of ion pairing and cluster formationWith very strong acids solvents of a high dielectric constant are used and so ion pairing is not such an important issue Richardson and Lowry go over several other scales that have been developed Please cover these yourself e It appears that there will be no universal acidity scale i This is due to our inability to understand solva D 125 tion the nonideal behavior of concentrated sol vents and there is no universal solvent to carry out the entire spectrum of measurements ii As we shall see shortly the acidity of a com pound varies tremendously from one solvent to another B Bronsted Catalysis Law estimation of pK 1 by kineticsThe idea behind this can be expressed by considering the following freeenergy diagram lThe Bronsted Catalysis Law k G Ka or logk0LlogKa C a CC constants b OL a constant for a group of structurally related HA acids c 0c is really given by AGE AGzi 0L AGI AGZ d Consider an acid catalyzed reaction kobs S gt P 74 where the reaction is catalyzed by H3O HAI HA2 Here rate kabsS and kmS k0kH3Ok2HAk3HA2 e One evaluates km for each HAn separately and the slope of the line 0L slope 41 log kobs PKa HA 126 Normally for aqueous systems the concentration of H3O is held constant from one experiment to the next by using a buffer ie by varying HAnAn39 2This is of course another linear freeenergy rela tionship the plot of log k vs log K is a straight line of slope 0L 3 Notice that 0L bears some relationship to the mechanism and is directly related to x r in Marcus theory aWhen 0L gt lthen the TS is structurally very close to the products iethe HA bond is nearly completely broken bWhen 0L gt 0 then the TS is structurally close to the reactants iethe HA bond is broken to a small extent 4 It was believed that 0L could be used quite simply to measure the degree of proton transfer in the transi tion stateThen negative values of 0c and values of 0c greater than I were found in certain reactions Obvi ously there is more going on here but 0c is still useful as an approximate indicator of the position of the transition state 5 Likewise for base catalyzed reactions one can define logkalS BkB C etcWhere again 5 relates to the extent of proton transfer in this base catalyzed reaction C pH Rate Profiles l Many reactions are catalyzed by acids or bases aThe most common catalysts are simply H and 39OH b Nearly all enzymatic reactions involve acid or base catalysis as well 2 One way to investigate whether a reaction is acid or base catalyzed is by a pH rate profile 3 Each type of curve can be interpreted mechanistically a Curve a acid base catalyzed with an appreciable rate for an uncatalyzed reaction b Curve b acid base catalyzed no rate for an uncatalyzed reaction c Curve c acid catalyzed some rate for an 127 log kobs pH uncatalyzed reaction d Curve dthe reaction is only acid catalyzed e Other results are also possible D Specific and General Acid Catalysis I In considering acid catalysiswe will use the following idealized sequencewkhere HA is an acid 1 s H30 SH H20 k 1 k2 S HA SH A39 77 k z SH gt P H k3 NOTE Both H3O and HA protonate S and catalyze the transformation to product Without any approximations we can write down for the rate d S kliHo1 kHA1s kHo kA 1Hs1 Now assuming the steadystate approximation for HS leads to kliHo1 kHA1s kIHZO k2A k3 so therefore st 128 dS k3kIHzo1 kziHA1s dt kHZO k2A k3 2 Specific acid catalxsis proton transfer occurs prior to the rate determining step a Suppose proton transfer from H3O and HA is fast compared to the last step In this event k3 ltlt k H20 and k z A dS k3kIHgo1k2HA1s dt kIHZO k2A Since HA and H3O are linked by an equilibrium reaction HA Hzolii and so kl km dS K dt WWW Furthermore using the equilibrium conditions for the protonation reaction leads to k2 5HA an E H305 kiz HAS kl 5HH20 kzkrl A H30 K then a kizkl HAH20 H20 or klk 2 Ka krHZO Substituting this back into the rate equation gives know ds 39 Limo H mm dt kHZOk2A 3 3 and rearranging we have 129 dSHkIHZOk72A k H Um dt kHZOk2A kHZO 3 3 klk3 kilHZO H30 S b Specific acid catalysis then shows only a dependance on H30 and M on HA cThere will be no primary isotope effect for substitut ing DA for HA dThis corresponds to the situation where OL l 3 General acid catalysiszThe proton is transferred in the rate determining step aThe original rate expression with no assumptions except the steady state approximation was dS Z k3kIHo1kHA1s dt kIHZO k2A k3 In this case we have that k3gtgtkHZO amp k2A39 so dS E z kH30 k2HAS If there were several acids then KB3930 2kiHAiS bThe rate therefore depends on both H3O m HAi cThere will be a primary isotope effect for the substi tution of HA with BA d 0L will be in general less than l00 eA slighly different form of general acid catalysis can be encountered An illustration is given below 3 HA SHA fast 3HA gt P A39 slow 73 W A P HA fast i Assuming steady state for S HA and that k gtgt k2 130 dsdt wag mots HA ii m in this form of general acid catalysis the proton maybe more associated with S thanA in the hydrogen bonded complex but it is not until the two break apart in the rdsthat this makes a kinetic difference 4 Differentiation between specific and general acid catalysis can be made by plotting the rate at constant pH as a function of the concentration of acid other than H3O general acid catalysis rate hulfer E HAIA39 Note that by varying the concentration of the buffer the pH and ionic strength of the solution remains the same 5 Specific and general base catalysis are analogous and are treated in the same way E Factors influencing acidity and basicity recall the more acidic an acid is the more stable weaker will be the conju gate base I Please go over HardSoftAcidBase Theory by yourself for now we ll come back to it later 2 Electronic factors a Resonance Acid Ph3CH PhZCHz PhCH3 CH4 pKa 3 l 5 33 409 48 This resonance stabilization is modeled in the following way Q CH3 B BH O flz i 710 GCHZH CgtCH2lt gt CH2 131 Of coursethe more phenyl groups there are the more resonance structures can be drawn and the more stable will be the carbanion But notice what happens to the pKa differences in this series Why is the difference between Ph3CH and PhZCHz much less than that between PhZCHz and PhCH3 which in turn is less than that between PhCH3 and CH4 b Aromaticity i 4n 2 TE e s aromatic increased stability ii 4n 7 es antiaromatic decreased stability H H H g i H Ph Ph Ph pKa 18 41 50 III III 711 Ph lt Ph Ph The higher acidity of cycloheptadiene compared to cyclopropenyl denvative is attributed to the cycloheptadienyl anion s ability to become non planar c Percent of s character i The greater the s character in the lone pair the more stable the anion is ii Examples HC E CH HZC CH2 CH4 plgI 24 44 48 d Electronegativity i The more electronegative the atom is that contains the lone pair the more stable the anion is ii Examples HF H20 NH3 CH4 pK 1 32 I6 33 48 2 Solvation a very important factor 132 aAqueous solvation i Acidities of hydrogen halides in H20 Hl HBr HCI HF plgI 9 8 7 32 a Based on electronegativity considerationsthe trend in these values should be completly the opposite b This is al mt the case in the gas phase The reason behind this can be understood by considering the following energy cycle So the pH of HA can be expressed in terms of the electron affinity ofA and the homolytic bond dissociation energy of HA We have the following for this series in the gas phase in kcalmol F 78 36 Cl 83 03 Br 78 88 70 7 Thereforethe BDE decreases much more rapidly than EA increasesThis is a general phenomena H20 H25 HZSe plgI l57 70 39 c In an aqueous solutionAS is also much more negative for the solvation of F39 which is a very small anion than it is for l39 a large diffuse anion ii Another example that is not so clear solution pr MezNHz lt MeNHz lt Me3N lt NH3 gas phase pr Me3N lt MezNH lt MeNHz lt NH3 The latter set is consistent with a methyl group being slightly e donating bThe pKa s can also change dramatically on going from one solvent to another 133 Eia AID K3 2Q m CH3COZH 47 L6 69 CH3NOZ 02 59 57 i DMSO CH32SO does not hydrogen bond to A39 ii Other polar aprotic nonhydrogen bonding sol vents DMF CH32NCHO HMPA CH32N3PO F Solvation effects on a reaction I General A39B gtAE B5 r gt AB Strongly favored by nonpolar solvent AB gt AE B5 r gtA B39 Strongly favored by polar solvent A B gt AB r gtAB lnsensitive to solvent polarity AB gt AB r gtA B Slightly favored by polar solvent charge concentrated more in one region in TS A B gt AB r gtAB Slightly favored by nonpolar solvent charge delocalized over greater region in TS 2 Quantitative can we quantify the polarity of a solvent and its effects on a reaction aThe most general approach is the ET scale i log k r ET This is a linear free energy scale r sensitivity parameter to solvation effects for a given reaction ET constant for a solvent measures relative polarity ii ET is based on the UV spectrum of R R I Q R 4 R m 50 E4 713 E4 0 AEE E kcalmol olar nonpolar solvent solvent 134 iii A polar solvent stabilizes E0 more than EL while a nonpolar solvent stabilizes El more than E0 T Solvent 63 2 536 80 EtOH20 H20 5 L9 EtOH 450 DMSO 4 l l CHZC l2 374 THF 345 C6H6 325 CCI4 bThis is an overly simplified view of solvent effects on reactions however i Suppose one was interested in the attack of nucleophilesA39 on some sort of electrophiles a The reactivity in DMSO is much much higher than in other more polar solvents egHZO EtOH b This is due to hydrogen bonding in the latter HOHA39 HOH effectively reducing the nu cleophilicity of the ion ii We shall have more to say about this later 135 4 Thermochemistry A Useful Definitions The key equation to all thermochemistry is AG AH TAS AH enthalpy change AS entropy change 2 For a given reaction A B gt C D AH AH C AHfD AHA AHB aAHf heat of formation AHf is relative to the elements in their standard state bThis is the enthalpy change going from the molecule to the elements 3AS SC SD SA SBwhere S is the entropy of each component 4 SW Benson has developed a method for determining AH and S for any organic compound in the gas phase aThis is presented systematically in the text 23 b Please go through the method and problems by yourself and know how to do these calculations B A Related Example HowTo MakeA Lot Of Money Calculation of AH and AS are very important in determin ing whether or not a reaction is energetically favorable or not Remember that a very endothermic reaction AG lt 0 will not be viable since all reactions are potentially reversible Let us consider a trivial example of an industrially impor tant process the bromination of methane initiation Br Br gt 2Br AH 46 kcalmol propagation CH4 Br gt CH3 HBr HBr 87 CH3H 04 AH l7 kcalmol CH3 Br2 gt CH3Br Br BrCH3 70 BrBr AH 24 kcalmol 49 termination Br Br gt Brz CH3 Br gt CHzBr CH3 CH3 gt CHzCH3 Since the chain propagation steps occur with greatest fre quency chain length is the ratio of chain propagation steps compared to chain terminationthen AH l7 24 7 kcalmol This is a nicer exothermic process however the fact that the first step in the chain propagation sequence is so endothermicthe chain length is small The Dow Chemical way around this is to mix in a little bit of Cl2 with the Brz Now the chain propagation events may be modified with CH4 Cl gt CH3 HCI HCl 03 HCH3 m AH l kcalmol Br Cl2 gt Cl Br Cl ClCl 52 Br Cl AH 6 kcalmol 6 Kinetics Kinetics is probably the most basic tool for determining reaction mechanisms An important point is that reaction mechanisms cannot be directly proven We have only indirect ways to measure how the atoms were arranged at the transi tion state Therefore a reaction mechanism can only be inferred often from many varied pieces of experimental data A Determining Reaction Mechanisms lProduct studies indicate how the ratios of products vary with respect to changes in the reagents 2 Stereochemistry we saw this with methylene addition to an alkene 3 Isotopic labelling following a label or labels from reactants to products can tell us unambiguously what bonds are broken 4Trapping or observation of intermediates relies on the formation of an intermediate with a definite lifetime how long depends on the trapping reagent or spectroscopic method 5 MO calculations can be used to model reactions 6 Linear free energy relationships we have seen their use in the Hammett equationWe will see several more in this course Basically they are attempts to relate what happens at the transition state to the ground state prop erties 7 Kinetics is probably the most important technique 65 BThere are several rules that we should briefly discuss on how chemists propose a reaction mechanism When propos ing a reasonable mechanism I Use Occam s Razor William of Occam was born l 280 in the town of Occam near London He studied theology at Oxford University l 3 l9 His philosophical views were controversial with Pope John 22nd and as a result he fled to Germany where he died of the Black Plague in I349 He said Plurality is not to be assumed without necessity and What can be done with fewer assumptions is done in vain with more 2 Each individual step in a reaction should be either unimolecular or bimolecular aThe molecularity is the kinetic order for a single reaction step i A rate law can be written rate An Bm ii The kinetic order n m o bA twobody collision is l000 times more probable than a 3body collision solely on statistical grounds 3 Each step should be energetically and chemically feasible this requires experience C Rate laws of typical reactions I General cases A B gt C2D Rxn rate dAdt dBdt dCdt l2dDdt The elementary reaction steps of a mechanistic hypothesis allows a rate equation to be written This would be compared experimen tally with the kinetics of the reaction of interest a Unimolecular A gt B i dBdt kA dAdt or kdt dA A where k the rate constant for this A gtB reaction ii ln integrated form kt ln A0 Awhere A0 initial concentra tion ofA iii Plotted 66 In E An slope k time gt halflives at tIz 2AA0 then kn2ltm this is a crude and easy way to get an estimate of the rate constant k by measuring tIz b Bimolecular 2A gt B i dBdt kA2 or in integrated form 2kt A A0 ii Plottec L i A A0 slope 2k time gt cBimoecuar A B gt C i dC k AB or kt I InAoB BOJ Ao 801w slope kAo 30 In E A time gt ii In many cases we will have much more complex rate equationsThere are general solutions for all of them which we shall not cover here 67 Concentration dThis is the most simple situation for a twostep reaction A gt B gt C 2 It can be shown by some tedious math that A Aide k t k A0 rk t 7th B e e c1 A0 A B A01I 39 kze k kle kl kk2 k1 30k2 k1 k2 A C 80 5 A1 so C E 40 39 B 64 u 39 E B U 20 65 0 Time gt Time gt k1 033k2 80 A C 2 5 so E E 40 o C U 20 B as 0 Time gt 2 Simplifying approximations a Rate determining step i Let s now take a look at a specific multistep reaction to see how the prior rapid equilibrium and 68 the rate determining step aDD 39 39 are used ii Reaction k1 M R A B k C gt D gt E F 391 67 iii Analysis a step I dCdt k AB kC b step 2 dDdt k2C c step 3 dEdt dFdt k3 D iv Lets suppose that k2 ltlt k4 kl k3 a Then step 2 is the rate determining step i the overall rate depends only on this step ii It depends indirectly on any one which proceeds step 2 b Thereforethe reaction rate k2 C v But suppose C is an intermediate whose con centration cannot be measured a C is connected toA and B by means of a reversible step b At the middle of reaction tIme not at the beginning nor at the end we can assume thatA and B and C are connected by a fast faster than k2 equilibrium step ie i Kc1A1B kiki ii c1 k1ki AB iii The overall rate k2 kll k AB k AB vi An example is the acid catalyzed nucleophilic substitution of alcohols ROH H ROH 3F L RBr H20 68 a Assuming the lst step is a rapid equilibrium dRBrdt k ROHf Br39 ROHE kl b N w ROHH k4 69 i Thus ROHf kk ROHH ii and rate k2k4 kROHHBr39 Vii In this case we could evaluate kllk in a sepa rate experiment by using a nonnucleophilic coun teranion however we shall see a much more elegant direct method later b Pseudo first order Another simplification can come into play when one reagent is present in large excess over the others AB gtC rate k A B i If B gtgtAthen rate kals Awhere kals k B and B rs constant ii The above is said to be a psuedo fi order rather than bimolecular reaction iii Normally this will only apply if a B is the solvent or b there is a buffered solution so that if B is an acid or basethere will be a constant c B gt gt A usually 2 l00 fold cThe steady state approximation i This is a more general and more rigorousway to deal with intermediates ii It assumes that one has an intermediate whose concentration is small and approximately constant over a large portion of the reaction time iii As an example consider the following k1 k l Step1 AB C 69 Step2 CD LEF a The rate of product formation dEdt k2 CD b If we make the steady state approximation dCdt 0 i Then using this we have that m AB k C1 k2 c D 70 dt k4 k2D This same ex ression can be found for d A dt tr it P Y c We can further simplify this by assigning a rate determining step i Suppose step 2 is rate determining Then a dCdt from step 2 lt dCdt from step lor b k2 CD lt kn C c So k2 D lt klandtherefore dE klkz E A B D dt k4 1 ii suppose step I is rate determining a so dCdt from step 2 gt dCdt from step I b then k2 CD gt kl C or k2 D gt k c and the rate 5 k4 d Steady state kinetics is also used in the analysis of enzyme catalyzed reactions i In the following reaction an enzyme E catalyzes the conversion of a substrate S to a product P by first forming an enzymesubstrate complex E S which rapidly attains a steady state concentration 5 Substrate Enzyme ES Enzymesubstrate complex P E K ys k2 Es 71 Product k1 ES BS 1 k E39s gt2 EP 639quot a At steady state defined as dE Sdt 0 i the rate k2 ES dPdt and ii m Ens k ES k2 Es W k k2 The problem here is that at any given time the E will be essentially impossible to measure We can get around this in the following way b Knowing that E0 the concentration of enzyme initially added ES we can substitute into the steady state equation 0 m EoSki E SS k k M ii Eons E SS kzlws BS 2 Eons This can be rearranged to kl Therefore V rate k4 k2 K let k m lt2 the Michaelis constant not I an equilibrium constant The substituting back into the initial rate expression leads to v Km S iii This is the MichaelisMenten equation lt ex presses the rate of the reaction in terms of several constants and only one quantity which varies during the reaction iv Under special circumstances the MichaelisMenten equation reduces to simple 72 expressions a If S ltlt K then V kabsS a pseudo first order process b If S gtgt Km the rate k2E0 i This is the maximum possible rate or Vmax ii Athax the reaction is zero order in substrate concentration and the enzyme is considered to be saturated hence saturation kinetics iii In other words atVrmX there is essentially no free enzyme E39s E0 c If S Km thenV i This gives a practical way of determining K it is simply equal to the concentration of the substrate when the velocity of the reaction is halfmaximal ii Km is easily determined graphically Note Km has the dimensions of concentration Does this I kzlEo Evmax make sense d Cases a b and c above are indicated on the graph ofV vs Vmax 39 39 kzlEoIIS 39V V k E I Km s max 2l 0 I 1 I I I 112wax 39 V 612 m s gt e Finally if k and kl gtgt k2 then E5 k4kl i This is equivalent to the case of rapid prior equilibrium discussed previously ii In this case only Km can be treated as equivalent to an equilibrium constant that is Km k k e Oscillatory reactions i In these interesting multistep reactions the concentrations of intermediates and sometimes products rise and fall periodically ii Here is a very simple hypothetical case a The first step is autocatalytic in X I step I A X gt 2X b The second step is autocatalytic inY and de stroys X k2 step 2 X Y gt 2Y thereforedXdt k AX k2 Equation l c The third step involves the destruction on 3 step 3 Y gt P thereforedYdt k2 k3 Equation 2 and dPdt k3 Y iiiThe course of the reaction a Assume that A is very large ie its concentra tion is constant for the entire reaction period b lnitially ie at t to Yis small so i dXdt m AX ii X increases rapidly c But the IS term in Eq2 is also first order in X i Initially dYdt k2X Y ii As X becomes large starts to increase rapidly iii As becomes larger the 2nd term in Eql increases d At some point t n X reaches its maximum 74 Concentrations value i continues to increase but ii X decreases as the rate of decomposition of X increases until iii at t t2 dXdt k2 X Y iv X continues to decrease to a small value e At t t3 is maximized i At large and very small X dYdt k3 ii Thus must start to decrease f Eventually we have returned back to the initial state where X and are both small so the cycle starts over k First step A X lgt 2X Semndsfep XY 1 2v Third step y L p t3 t2 t1 m 613 to X iv A plot of X and and as a function of time is shown below 75 v It can be shown that the solution of the three rate equations is k2X Y k3 In X1 lg A In Y c where C is a constant try it vi In the real world this can be a model of an ecological system A grass X sheep Y lions P humans kill lions AX gt 2X XY gt 2Y Y gt P a In the fist stage of the reaction sheep eat the grass and become very numerous b In the secondthe sheep are eaten by lions which then reproduce until c their population becomes an annoyance to the human population who d in turn kill off a large portion of the lion popula tion and probably increase themselves vii An example of an oscillatory reaction which takes place in a cell is the hydrolysis of esters by the enzyme papain a The general mechanism is thought to be 0 o RlOR39 H20 Papalquot RJKO R39OH Hquot 5 P 615 b The environment can be depicted where E amp E are ionized inactive forms of enzyme cell wall 76 c Analysis i kP ks kH are diffusion constants through the cell wall ii Initially there is no S inside the cell and most of the enzyme is in an ionized form ie E ltlt E F iii As S migrates into the cell its concentration builds up a S migrates into cell rather slowlya realistic situation would be k gtgt ks gt kP kH b Initially however there is little chance for S to encounter E iv As molecules of S do encounter E there is a generation of H v The H does not diffuse out of the cell but rather converts E to E and E to E through their respective equilibra a The rate of consumption of S increases b S decreases until there is virtually no more vi With no more S availablethe enzymatic gen eration of H ceases a H and P diffuse from the cell b The majority of enzyme returns to the E or E forms c The initial state is once again reached d Graphically this behavior looks much like a heartbeat i Notice that there is again an autocatalytic cycle ii The cell wall acts as an inhibitor in a feedback loop DAbsolute Rate Theory There are three ways in general to display a potential energy surface which is a plot of the potential energy for the molecules as a function of geometrical variables that change during the course of a chemical reactionThese are shown below for an arbitrary reaction where a reactant R goes first to an intermediate l and then to a product P by means of two transition statesTS 77 39 a sum O 390 e 390 Transition SIoIb s s Producis o quot9390 Q 0 Q zzt OI O O O O m Q Q 3 QM V Reactor o N 0 0 0 Q 0 011 3309357 Nit 00 391 Q 0 III quot5 9quot unqz AH 2e s 39gt i ix S S b d D lts6 b 17 6173 The potential energy surfaces shown above show what happens when thiophene is oxidized I Let us consider the most simple reaction aThe reaction potential energy surface is g I 0 E R p gt Reaction Coordinate H HH a HH H e 180 H H H A r1 r2 4 I I Energies in eV 1eV 23061 kcallmol 3 I I L20 2 618 r1r2186 5 1 20 40 I I I I gt 1 2 3 4 distances in au 1 au 053 r1 1875 H 2500 1250 1250 1375 2500 r2 An expanded view 78 The reaction coordinate is indicated by the dashed line bActually our potential energy surface for H Hz was idealized i H really could collide with H2 at any angle ii The potential surface must be recomputed for each HHH angle iii For an angle of 60 for example the surface is a Note that the Ea is higher in this case b Actually the activation energy is the lowest for 9 80 n I pig so I392 cWe can compute the rate constant for this reaction using tra39ectory calculations i The H H2 system is assigned a kinetic energy a This also implies a momentum b Naturally the kinetic energy is related to the temperature ii The process is repeated many times iii One trajectory with enough kinetic energy to overcome the barrier and with the two particles moving in the right direction is shown on the next page a A portion of the starting translational energy is turned into vibrational energy after crossing the barrier b A portion of the starting vibrational energy is 79 turned into translational energy after crossing the barrier I I 39 39 h fInlsh II HI I I r fInls I g 39 I I 0 I I I l 39 I I I III I I 39 39 39 l U 2 I I m 39 l I 1 II I I j 39 39 39 I D g I I I j I Lo I I 39 1 I I Il I O 1 I I II III 39 39 I l I I11 I I 39 39 I z I I I 39 39 I J 39I i t 39 I j 39 I I ll l l 39 In an I l I I 39I 39 39 I 39 I I j z I I If g In 3 g D I I I I 39 39 I l l l l l I I I l I I 2 I I I I I I I 39 i 39 g i u l Z 39 2 I Z quot I 39 l o 3 I 39 l in I 39 n 39 39 i39 39 j 39 I t 39 III I I Q 391 I I l39 39 39 Ln 3 I 39 39 I I I I I I It I I I I l39 3 quot52 v39II x 39 I qw vl 39 39 I all l I 39 I I I C I I I I I I I 395 39 I I 0 I o I I 5 I Q I I M II I I 39 Ln 39 U l l HI I I I a 393quot I I I 39 c I I L I quot39 Q 39 Q I I I I I i 39 39 39x I In I I 39 I I 39u u I I u U i I O u c a y u u u o u u g I o u c o u c v b a n 1 l I 6 0 A I x I b O O o a a o a a o c o u a a o u o o u c a u c o u p 4 c 4 I 39 39 39 I l X quot I o 39 39 39 39 O o o o o o a n o o o s c n u c o a o OI hy v quotI 39 39 I I I l l I I I n I I I I lvI Imuii2333Ziliiiliii27271iiilii1235liiiliiiiillliilllll 1 I I I I I I t I I I I I I I I I I 39 5 39 I V 39 39 39 39 I I v quotI 39 quotn 39 l I 39 39 t u v I c I v N n I I I q quotmunumvuNuuu I I I vvvv quotrmv 39 39 I 39 39 5 pun I o u n a n c I u a 5 I a u u o c u a n I u u I a n s n n I u 39 I 39 o 39 1 I u o u I II a u c c u c o I 5 I u I u o o v I U 2 LIIIIIIIIII I 39 39 IIIIIIIIIIIIIIIIIIIIImmmu Immm mmm mmm finish 35 45 CCIo DISTANCE 5 5 6 5 start 15 35 45 C Clo DISTANCE 55 65 start o o o I I u 9 n I n I I I v o t 55 I finish Oommaww and 4 on i u 65 2 I 1 I 55 9 Q o LLJ 8 s g Q Q J Z Z 39 V 39 ft In j 10 33 a O o s f v 339 73 p quot u F 39 I 5 I 39 39 f 39 I Q I I I I I I I I I 3 I l I I In I x I j I I I l I I u m a C u j x mu I I Iquot I I I j I i V xi V u e I I 0 2 O quot 0 g a x I nu o 39I quotIquot g I I 39 g 39 39 o u v I h I H quota I 5 t 39 39 o u 3901 a 39 Q i u U u 39 U I u a m H b w n Q 39 39 39 U O o o o p u I u o s o u u o u o o o 01 n 39 39 U 39 39 I I t 39 c h 39 39 39 o u c a n a o u o u u u c u s o u c u e o o u u I u 39 39 o o a u c o o a c o a o u a o o o I04 39 I I u NI quotnI Hnuhn v39 39 39 39 o o n u u o u n a a a c u u u o u o o c o o c o a u o c 1 I I s 0 u o I u 39 I I u a n n 233quotquot393312E39 311 quotquotquotquot 39 v f 39 39 I n n 5 v u u g c n o a o n o e n I I o s I I I H I nu I39 3 7 x I u x m m 39 m3 39 I I v I I n a u t I u 1qu Iquot r was I In 3939Ii3939 125 I x I Q 2 quotMRI I IIIII o guooaaooooo ca oo onnocoo aaoo4 o accoaou39 ao o guyonaooq ouununonaauoo o 4 gonen cm o oq u 0 39 39 u u HONOHMMON mm 0 O HO 55 IIIIIII IIIIII Mm 35 45 CClo DISTANCE rx 53mm Issag vlimwssmi 55 55 start 25 65 start 15 35 45 000 DISTANCE 620 c The trajectories shown above are calculations from the Cla39 CH3Clb ClaCH3 Clb39 reaction The functional form of the reaction is identical to the H H2 reaction we have just considered For each of the trajectoriesthe particles start at the lower right portion of the potential energy surface and finish on the upper left side Notice that in trajectory a there is just enough potential energy to cross the barrier ln trajectory b notice the loop that the particles make in the vicinity of the transi tion state an event we would never consider 8O Trajectory c shows what happens when there is not enough kinetic energy to get over the barrier Finally trajectory d shows what happens when there is more than enough kinetic energy The excess translational kinetic energy is released as vibrational energy in the product d In general we will need to evaluate 3N6 vibrational coordinates where N number of atoms to find the path of least energy and the transition state i Clearly this is an extremely difficult task and we cannot hope to display the results graphically ii What is most commonly done is to simply plot the path of least energy in dimension t gt 9 transition state ABc at quot1 m n energy E mamaquot path activation at o n ABC J energy of reaction k minima 63921 Reactionpoordinate transition state energy rises in all directions except along the reaction coordinatewhere it decreases see H2 H surface e Characteristics of the reaction energy diagram i The reaction coordinate a This is a collection of vibrational coordinates b It corresponds to the dashed line in the H2 H example iiThe energy minima a These are defined by their energy gradients the derivatives of the energy with respect to all 81 coordinates b The energy gradients are uniformly positive iii The transition m a Uniquely defined by the energy gradients b The energy gradients are positive in every set of directions except one c The energy gradients are negative along one line the reaction path in mdirections iv Both the transition state and the energy minima are called stationary states v The reaction path a This interconnects the stationary states b It is much less well defined in generalfor a polyatomic system c We won t go through all of the technical prob lems associated with this 2The BUrgiDunniu approach aThe BLirgiDunniu approach is an experimental way to get a rough idea about what happens along the reaction path i This uses crystal structures to map that path ii For any reaction the least energy path must be unique a In moving away from the minimumthe energy rises more steeply in any direction other than the reaction path b An external or internal force will cause a mol ecule to readjust its geometry in the energetically least costly direction ie along the reaction path b Example ring whizzing in cyclopropeniumML2 complexes Consider the following molecules and the reaction R R TV RV R gt R 5Aquot 622 R P 393 3 PR3 M Ni Pd pr R3P PR3 X Cl 4l PFs 82 i The full structures of three complexes are shown below a Note that the relative orientations of the phenyl rings on the triphenylphosphine rings change b This causes differences in the intramolecular contactswhich cause the basic structure to change ii The structures below show just the three cyclopropenium ring carbons the metal and the two phosphines from the top We see a gradual progression from a ground statewhere the metal is bonded to two carbons to just about the transition state where the metal is bonded to all three car bons quot2332 nevi3v quot 7 quotquotquot quotquot3939 r39 quot39r7 iii These results agree nicely with the computations of the potential energy surface for the reaction next page left and the projection of the PZM plane on the cyclopropenium ring as the species travels along the reaction coordinate next page right iv While this experimental method offers direct evidence for what structural changes occur it does not of coursetell us anything about the energies or rate constants associated with the reaction 83 l s ll ElilllllLllllllllllllllllllllllllll 625 cAs another example consider the nucleophilic addition to carbonyl c a A100 100 626a 626c ompounds This shows the relative positions of the atom from the nucleOphile which attacks the carbonyl carbon substitu ents connected to the carbonyl carbon and the position of the carbonyl oxygen atom for structures abeedA L The calculated surface is shown below 3The Arrhenius equation a For a one step reaction kobs A e39EaRT bThis is called the Arrhenius equation i A preexponential factor related to ASiE ii E3 activation energy iii R gas aw constant iv T temperature in Kelvin cPlotting ln kobs vs lT i The slope EaR ii The intercept dArrhenius estimates lnA of halflife for the reaction84 A gtB i UsingA e W assumes AS 0 see below an ii defining tIz time when A B and tI99 time when 99A T500C T IOOOC tIz t199 tIz tH99 l2 sec 80 sec l7 x lo 393 sec 0l sec 49l min 542 hrs l39 sec l54 min 8 l 4 days l46 yrs 326 hrs 26 hrs 533 yrs 3534 yrs I I4 days 208 yrs l27 XI 06 yrs 844 xl06 yrs 264 yrs l748 yrs 4Transition state theory a Calculations such as those above give a good idea of the relative rates of reactions as a function of the activation energy i But the rates are not particularly accurate and ii the theoretical basis for the relationship is also not clear bA more rigorous way to relate rate constants to activation energies can be given if we assume an equilibrium between reactants and the transition state i For the reaction A gt B where Al is the transi tion state a Ki A A and h K KkT K h k Boltzmann s const h Planck s const T temp K K the transmission factorcommonly assumed to be l00 ratio of molecules that will cross over when they get to Al ln extremely large molecules ie enzymes for example the top of the transition state may be illdefined so that K lt l 85 Potential Energy Reaction Coordinate c Go over this in Appendix I Chapter 2 in your book ii Recalling that AGi RTln Ki a where AGi AHi TASi b then k KTkTerAGWRT KkTerAHRTeASR c Where i AG is the free energy of activation ii AHi is the enthalpy of activation the potential energy difference between the reactants and the transition state iii ASi is the entropy of activation related to the rigidity and structure of the transition state versus the reactants c Comparing this to the Arrhenius Equation for the unimolecular reaction 0 A ii E1 AHt RT iii AGi AHi amp ASi can be computed by plotting lnkT vs lT A5 Inamp R h eKkT 6A5 R h intercept observable range In L T 627E 86 1 T D a slope gt AHJ small error b intercept gt ASi large error extrapolation goes far outside the experimental region of tempera ture c AH and AS errors tend to compensate giving AGi a smaller error than ASK iv m the value of AG depends on the tem peraturewhereas AH and AS are temperature independent quantities Rate dependence on AH and AS i Lets say we have two competing reaction paths A e C gt B and that AS is the same for both reactions ii Then for the following rate ratios kak7 AH AHbi 2 04l kcalmole l0 l37 eVery large rate differences for IO 549 very small changes in AHi a Variations of AHi by themselves do not imply much about the structure of the transition state b Perhaps one can get clues as to what is happen ing electronically when close comparisons are made iii On the other hand if AHi is held constant then ka kb ASJ AS 2 l4 eu calmole OK l0 l45 IO l83 iv ASi does provide clues about the structre of the transition state a Exp l a typical SNZ reactionAS 42 eu T Ph Ph CHzBr gt HzO 3k Br gt Ph CHZOH2 Br H H 528 87 i AS is negative iethe TS is highly ordered ii There is a lower degree of vibrational and rotational freedom than in the reactants b Exp 2 a typical bond breaking reactionAS l 0 eu T RO OR gt R0 OR gt 2R0 RfBu 629 CH3CH3 gt 2CH3 AS 7eu i The positive value of ASi indicates a less ordered TS ii Herethe more positive this number is the more closely the TS resembles the products One particle becomes twoThe number of rotational and vibrational degrees of freedom increases c Example threetwo special cases where ASi has atypical values We talked before about whether cyclobutadiene was a square structure where there are two different resonance structures or is it a rectangle which undergoes a conversion from one rectangular form to another Berry Carpenter has devised a very clever experiment to probe this issue A1 A2 lt R K21 EgtltE 88 Potential Energy A compound R was decomposed by a reaction that is very wellknown to produce cyclobutadiene with shape A if cyclobutadiene is a rectangle and A then can rearrange with a rate constant kl to isomer ABoth A and A1 are intercepted by an olefin which undergoes another very well defined reaction to produce three isomers BC and D The only difference between these products is the placement of the deuterium labels Now if cyclo butadiene was a squarethen A and A1 are really only resonance structures and B CD On the other hand if cyclobutadiene was a rectangle and provided that k22kthen B lt CD In fact this was found to be the case so cyclobutadi ene is indeed rectangular in shape It can easily be seen that kl x B C D Carpenter repeated these reactions at different temperatures and as shown below there is curva ture instead of being a straight line At lower temperatures the value of lnkT is larger than that QUANTUM MECHANICAL TUNNELING 631 Reaction Coordinate 89 extrapolated by the dashed line at higher tempera ture Furthermore a very good quantum mechani cal calculation puts the activation energy for the conversion from A to A1 to be l08 kcalmol However Carpenter s estimate is that AHJ 46 kcalmol and ASi l5 euThis is a very big differ ence between the experimental and calculated value andfurthermore the large negative value of ASi is difficult to explain One idea that has come forward and has been reproduced by calculations is that this compound undergoes quantum me chanical tunnelling A fraction of the molecules cross the classical barrier and a fraction tunnels from one side to the other The probability to cross the classical barrier is very strongly related to the temperature but quantum mechanical tunnel ling does not Consequently the apparent rate constant becomes too large at lower temperatures as predicted by the classical activation energies Another way to put this is recall that k erAGfRT h so one could express the amount of tunnelling by how much K was greater than I Using this it was found that K l00 at l0 OC and 800 at 50 0C A second example is given by a simple bondbreaking reaction 39 H 4 fast 1 H3C 632 AS 2eu a The activation energy for this reaction is small because the compound is strainedThe important point is that AS is small in magnitude unlike the 90 potential energy CH3CH3 gt 2CH3example that was presented previously Here the two radicals are held to gether However the reaction shown below has been studied and AS was found to be very differ ent An explanation that has been advanced for the difference can be outlined as follows a 939 a slow 539 AS 16eu H 81 triplet HJC I 0 0 h H l singlet C HJC 634 reaction coordinate rcc The key to understanding this problem is that the diradical that is formed has two electronic states singlet and tripletAs shown abovethe first mol ecule starts out with the two electrons paired in the reactant a singlet state and ends up with the singlet diradical At some energy above this is the triplet state where the two electrons are parallel Why the energy ordering is this way is beyond what we need to know however the important point is that in the second example this energy 91 ordering is reversedThereforethe reaction path needs to undergo spinforbidden crossing STATE CROSSING l 1 lt singlet potential energy g l triplet 635 reaction coordinate rcc The molecule in this case is a substituted trimethylenemethane As shown below the T orbitals are very straightforwardWith a total of four 7 electrons two go into the lowest level and that leaves two for the degenerate pair In this case it is energetically much more favorable l0 kcal mol for trimethylenemethane itself for the triplet state where the two electrons are unpaired i 333 5 In order for the molecule to undergo a state crossing there must be a coupling of vibrational and electronic states i The number of possible nuclear configurations in which this is allowed is quite small ii Thus ASi becomes much more negative iii Since ASi becomes more negativeAG be comes larger that is the apparent bond energy is 92 apparently so large that the bond energy is nega tive ie AHi is negative but so too is ASi so that AGi becomes positive e Diffusion controlled reactions i These are bimolecular reactions where every encounter is successful a Ea lt kinetic energy of molecules in their first vibrational state 0 to 3 kcalmol b The rate constant is the collisional rate con stant ii In solutionthe rate constant depends only on the viscosity of the solvent AiB A 34gtA B aAB 3n6 a w T viscosity of solvent dAB collision distance at the activated complex 039 mean effective radius related to diffusion iii For typical solvents at 25 C k l0I3 if Ea 0 k IOquot l0390 if Ea 23 kcalmol iv Examples CH3 CH3 gt CH3 CH3 gas k2x l0390 CH3 CH3 gt CH3 CH3 H20 k32 x IOquot Ea 3 kcalmol H 39OH gt H20 H20 solvent k 4 x loquot l l gt l2 H20 solvent k 82 x IOquot k 93 f Volume of activation i This is another noteworthy experimental rate based tool to investigate the structure of the transition state a For the reaction A B gtiA Bi c 638 V1 V2 V3 V4 nk 1 slope i Define P AW V3 VI V2 AV V4 VI V2 ii The closer AW is to AV the more the transition state resembles the product C iii The closer AW is to zero the more the transi tion state resembles the reactants A B b AW can be measured by measuring reaction rates at different pressures keeping the tempera ture constant at each pressure i dln k dPT AWRT where P pressure atmospheres ii Plotting ln k versus P gives a straight line with slope of AW RT iii A negative value of AW implies a tight structur ally rigid transition state where the molar volume of the transition state is less than that of the reac tants I v lt gt gt 63 94 AW 33 cm3mol AV 37 cm3mol iv A positive value of DW implies a loose structur ally flexible transition state R NN R gt lRNEN R gt 2R NEN 639b AW l5 cm3mol AV 60 cm3mol v The interpretation of the magnitudes of AW are not always so straightforward For example in the reaction below CH CH lt l H E39 AW 447 cm3mol AV 333 cm3mol It is not possible that the transition state occupies less volume than the product Obviously the way that the solvent solvates the transition state rela tive to the product is of importance here A more straightforward example of this is given in the following hypothetical reaction AB gtA5 B5 gtA B39 640 Since two molecules are formed one might think that AV and AW should be positive However in a polar solvent the formation of the ions causes the solvent shell to contractTherefore AV and AW is expected to be much smaller This is called electrostriction and it is a big effect For the reac tion shown above AW might be close to zero and might even be negative vi The drawbacks in measuring AW are basically twofold a The temperature must be kept very constant throughout the pressure range studied b It is a small effect Suppose AW l 5 cm3mol 95 then going from I to l000 atms at 25 C increases the rate constant only l8 times E Experimental Techniques for Measuring Rates of Reactions method time scale Ea jkcalmoll classical isolation l hr weeks 25 45 Stopflow technique l0quot sec min l0 20 NMR l0396 00 sec 5 25 ESR l0quot 0395 sec 2 l0 Flash photolysis picosecondl0quot2 l sec l l5 femtosecond l0quot5 sec molecular vibrations Relaxation techniques temperature jump l0398 l sec 3 l5 pressure jump l0396 l sec 5 IS I For relaxation techniques consider a typical reaction sequence 1 AE quotk 3 fast 1 B C T D slow 6A1 Often times the first step is a proton transfer Using the steady state approximation for B we have klk A1E1c1 k4 k2C Then using the fact that the first step is fast kl gt k2C rate and rate2AE1c1 a Now let K kjlk BAEand b dlnkdTp AHRTZ where AH enthalpy change for the lst step c Suddenly changingT when AH 0 causes Ink to change dThis in turn causes the relative concentrations ofA E and B to change 2What is done in a temperature jump experiment is to discharge a capacitor in a small reaction cell aThis occurs roughly in l08 sec and the temperature 96 changes by l to IOOC b Concentration changes can be measured by normal optical methods ie UV vis spectroscopy etc 3 Suppose AH is negative for the above reaction aA temperature increase will shift the equilibrium in the first step to the left b Suppose we can measure A relatedto k1 related to A I l 39 x A l l j I I I I 39 l l I I I l l l I I I I j j j j temperature temperature temperature temperature jump jump jump jump time 4 Likewise a sudden pressure change can perturb the equilibrium dan dPT AV RT F Rough Descriptions ofTransition State Reaction Path Changes l Principle of Least Motion a Stated first by Muller and Peytral in I924 and then later more precisely by Rice and Teller b Basically this says an activation energy will be lowest when the atoms change their positions as little as possible or given two possible reaction pathsthe one requiring the least nuclear motion will be energetically favored i The basic idea is quite simple the form of any reaction path is approximately given by two parabolas ii The closer they are to each other ie smaller span that the reaction coordinate must takethe lower the value will be of their intersection point the activation energy iii This is illustrated below 97 Potential Energy Reaction Coordinate Clearly Ea forA gt B is less than Ea forA gt B cThis is obvioust a very crude idea but it works quite well for most cases dThere is a lot of common sense in this idea i It costs energy to stretch bend and rotate around bonds from their equilibrium ground m geometries to approach the transition state ii The fewer changes in these coordinates that need to be madethe smaller will be the energetic cost to deform them to a productive ie reacting geometry e If one really examines what lies in back of the prin ciple of least motion it implies that the transition state structure can be viewed as a resonance combination of the electronic stucture ofA and B in the previous case i The closer thatA structurally resembles B the more resonance should occur ii Consequently the transition state becomes stabilized more fThere are instances when the principle of least motion breaks downthe most obvious is in a reaction of an atom with a diatomic moleculezABC gtAB C From the potential energy surface on the next page it would appear that the BC bond should stretch before A comes close to atom B Clearly this is in error Actually the principle of least motion is only a qualita tive indicator of what the reaction path should look like 98 ABCgtABC r BC The prInCIple of least motIon correctly predicts that a motion like 1 A B I I 4 4 A B 645 C C will not be favorable and it is not g Spectacular failures in qualitative detail of this principle are very interesting i They imply that something extremely unusual is occuring to the electronic structure along the least motion path ii Two examples that we shall examine in detail at H I l gt or H the end of the semester are gt D 646 H n Both reactions take place by paths which lie my far from the least motion paths 2The Hammond Postulateza highly exothermic reaction will have aTS geometrically close to the reactants aThis can be illustrated as follows H D Potential Energy 99 A B Reaction Coordinate Pnlenlizl Energy TS2 is geometrically closer to the reactants b By extension the more endothermic a reaction is the more the geometry of the TS resembles the products cThree general solutions are pictured below exothermic thermoneutral endothermic Pnlenlizl Energy Pnlenlizl Energy A B Reaction Cnnrdinzle l A B Reaction Cnnrdinzle RelAzclinn CnnrdinBzIe 3Thornton s rules a more detailed analysis of Hammond s concepts For a review and more examples than are covered here or in your book seeW PJencks Chem Rev 855l l I985 a Energy changes along the reacton path i Suppose we model the region in the immediate vicinity of the transition state by a parabolaThe diagram for theA gt B reaction in the thermoneutral example above is shown on the left side of the drawings below ii Now consider a perturbation in which the energy of B is raised relative to A by an amount SAEO a fraction of AEO aThe change will be transmitted in a linear way along the reaction coordinate b This can be expressed as m X 5AE where m is the ratio of the energy Increase SAEO to the distance betweenA and B along the reaction coordinateThe transition state is initially defined as being at X0 for the thermoneutral case c This can be illustrated as indicated in the draw ing on the right side m slope Potential Energy Potential Energy toB toA to A I I 00 00 toB X 100 Reaction Coordinate X Reaction Coordinate i The line of slope m represents the incremental increase in energy due to SAEO as the reaction moves from A to B Here the reaction is made endothermic so m is a positive number ii We raised the energy of B relative toA ESAE0 is positive so m is positive and X as defined in the diagram is positiveThe arrows indicate the value of SAE0 when X is a negative number then 5AE is negative ie stabilizing and when X is a positive number then SAEO is positive ie destabilizing Notice that the transition state has moved in the positive X direction towards B which is in agree ment with the Hammond principle iii The same analysis holds if the energy ofA is raised relative to B but then m is a negative num ber and the TS is shifted towardsA i eX is nega tive b Energy changes normal to the reaction path or dimensions other than the reaction coordinate i Consider the simple case of an atom reacting with a diatom a The enegy surface may be represented as ABC gtABC 101 Potential Energy gt lt 4 to A BC to AB C In other wordsthe dashed line perpendicular to the reaction path at the transition state in this case corresponds to the symmetric vibrational mode b The crosssection of the energy surface along the dashed line along the one dimensional vibra tional coordinateZ again defined at 20 for the thermoneutral case at the transition state also gives a parabolic shape gt lt 4 to AB C to AB C Potential Energy 00 Vibration Coordinate Z 00 Vibration Coordinate Z ii Now apply a perturbation SAEO which makes the vibration more difficult in the stretching direc tion a SAEO m Z where m a positive number b We can again apply a line of slope m to the transition state to see what happens i Raising the energy in the direction towards the right side of the vibration coordinate moves the transition state to the left ii The values of rAB and ch will be smaller in the transition state iii Decide for yourself the effect of raising the energy to the left cThe net result is that a perturbation which lowers the energy in a direction along the reaction coordi nate shifts the transition state structure My from the energy lowering but for directions perpendicular to the reaction coordinate the transition state will shift towards the energy lowering 102 re Ifficult h her energy A gt r d Let us consider theBe zlmination of HX from an alkyl halide as an example that illustrates the utility of this way of looking at energy changes and their conse quences on reaction paths Below is an idealized repre sentation of the potential energy surface for a base Br reacting with an alkyl halide RX to form the proto nated base BH an olefin and XThis is a twodimen sional contour surface where the two major variables are plottedzthe distance between the base and the proton being pulled off from the alkyl halide rBH and the CX bond distance rcX There are therefore two minima corresponding to the reactants and products on the upper left and lower right of the diagram BH A g3 xx x 6 J x ml B rBH X The structure on the lower left corresponds to com plete CX bond breaking to form a carbonium ion and that on the upper right side has complete CH bond breaking to form a carbanion Now depending upon R B and X a reaction path that passes through either a carbonium ion carbanion etc could be present how ever for our reference reaction the reaction path follows a diagonal from upper left to lower right and there is only one transition state with no intermedi ate located symmetrically between the reactant and product This is called the E2 mechanism i Suppose then one changes the R group so that the carbonium ion R on the lower left side of the idealized potential energy surface now becomes stabilized look back at Chapter I where we talked about carbonium ions being stabilized by reso nance The result of this perturbation is shown in the PE surface below on the left side i rEH stabilize X39 x hllze x39 carbonium ion x The net consequence is to stabilize the lower left corner and since this is perpendicular to the reac tion path directionthe transition state will move towards the direction of energy lowering the arrow in the drawing and the new reaction path will be curvedThe new transition state will occur at larger values of rcX the CX bond will be more broken and larger values of rBH the CH bond is less brokenThe formation of a stable carbonium ion intermediate is called the El mechanism ii Suppose that we make X become a better 104 leaving group That means that X is stabilizedThe resultant surface is displayed on the right side of the drawing above Now X39 occurs two places on the twodimensional potential energy surface It appears as the product lower right side and since this is along the reaction path the transition state moves away from the energy lowering X also occurs on the lower left diagonal which is perpend icular to the reaction path andthereforethe transition state is moved towards it The result as shown is the addition of two vectors which again curves the reaction path Here however rcX does not change while rBH occurs at a longer distance as before 4The ProssShaik model of reactivity This was developed byAddy Pross and Shason Shaik to view many reactions in terms of ground state changesThe idea here can be demonstrated by the first step in the SNI reaction RX gt R X Reaction Path a On the left side are the energies of the ground and first excited state of RX In the ground state the two 105 electrons between C and X are shared covalently However the first excited state is one where the two electrons are both associated with X Now the first step in the SNI reaction is one where the CX bond is broken and so the covalent ground state evolves in the R X diradical state The RX excited ionic state evolves into R X39 ionic state The problem here is that at the product side the ionic solution is more stable than the diradical state But the two states given by the lines do not cross they both have the same symmetry in terms of their wavefunctions Instead the two states mix with each other so that the lower state becomes stabilized and the upper state becomes destabilizedThey undergo an avoided cross ing shown by the dashed lineThat amount of inter mixing between the two wavefunctions can be viewed as resonancewhere the amount of stabilization and destabilization is represented by b Suppose that R is stabilized RX39 Potential Energy Reaction Path 106 The first excited state of the reactant and the ground state of the product is stabilized If we make the assumption that the configuration interactionreso nance term 5 is not changed then the resultant transition state T52 is moved closer to the reactant and occurs with a lower activation energy It is clear that the stabilization of either the product ground state or the reactant excited state will be linearly related to a stabilization of the transition state c Suppose we move now to a more complicated reactionthe 5N2 processThis is modelled bsz RX gt YR 39XThe first excited state is one where an electron has been transferred from the nucleophileY to the RX antibonding orbital y Rx Y39R X gt E 0 C AElY39ARx E E E 9 o n Y RX YR X Reaction Path 657 That energy difference between the ground and ex cited state is a function of the ionization potential on ly and the electron affinity of RX ARX Raising the electron affinity of RX and RY by modifying R will as shown lower the activation energy Making only the 107

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.