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by: Michelle Gulgowski DVM


Marketplace > University of Kentucky > Chemistry > CHE 538 > PRIN OF ORGANIC CHE
Michelle Gulgowski DVM
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This 50 page Class Notes was uploaded by Michelle Gulgowski DVM on Friday October 23, 2015. The Class Notes belongs to CHE 538 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/228302/che-538-university-of-kentucky in Chemistry at University of Kentucky.




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Date Created: 10/23/15
CHE538 Lecture 17 the Basis for LCAO and Hijckel MO Theory The Operator Postulate Physical observables have mathematical operators which are used in conjunction with the wavefunction Wavefunctions are mathematical expressions that completely describe the system under study Everything chemical needs to be describable via quantum mechanics Quantum Mechanical Operators don t commit these to memory they all have classical counterparts for example fix Fla E KE lingr functinn of pnsition such as x or potential War 3 component nf momentum y and 2 same form Hamiltnnian time indemn nt Hamiltonian time dependent Kinetic energy 1 component of angular rnn mentum fix it El rquot 333 p x 2m Eiqan Values Eigan values QLPn ann The operator Q acts on the wave function LP and gives back an eigan values q and wave function For example there is second derivative in the time independent Hamiltonian Once this acts on the wavefunction q in the equation above is an energy eigan value Expectation Values The expectation value postulate states that observable values can be had by the following Ob j L110 p dedydz iwzdxdydz When Op is the energy operator the expectation value is the energy of the system The Expectation Value ob L140 pLPalxdydz E j LPH dedydz When normalized orbitals are used the equations above are valid We are summing energetic contributions of the electron in really small volumes over all space to get the total energy At right above the Schrodinger equation Applying MOs to the preceding notice not applying the preceding to MOs The Schrodinger equation has two major approximations One approximation it does not account for relativistic effects The theory of relativity states in part that momentum near light speed is not a simple linear function of velocity P i mv for v z c Another problem there is no explicit solution for more than two bodies The argument is a bit circular sorrv In other words we have to begin by assuming LP C1Z1 C2952 C3Z3 is a decent way to handle the electrons in allyl We assume that the ADS E Wdedydz Expanding the Expression EC12 IggfdvC22 j 12de 032 154 2C1C2 131126quot 02C1C3 2519535111 2 C12C3 X2953 dv39 239 If ou can t see where C1C2J 11H12dv C1C2I12H11dv y the cross terms are coming from do this XaXb X2 aX bX ab X2 abX ab C1C3 9511 1953615 C1C3 9531 1951625 C2C3 9521 1953615 C2C3 9531 1952615 Notation to Embellish an Uqu Expression HZ dv the coulomb integral This is the energy of an isolated atomic orbital Hrs Ierzsdv the resonance integral This is the stabilization achieved when an electron in xr is also allowed in adjacent xs SW Izrzdv the normalization integral 1 if normalized orbitals are used the overlap integral The extent to which cr and cs overlap in space dv Rewriting the Ugly EC12Su 022822 032833 20105 2010320203 034 0221ng 032Jr 20102H12 ZCIC3H13ZCZC3H23 OI39 EC12811 3225224r C32333 20102 20193 F29293 The values of the integrals are fixed determined by molecular geometry The values of the coefficients will determine the energy of the molecular orbitals Differentiation of the expression above with respect to the coefficients result in energy minima for the equation 11 Differentiation and Groupinq Terms Further Simplification dEdc1 H11 Es11c1 le E812c2 H13 ES13c3 0 dEdc2 Hm E82001 Ha E82902 H23 ESZ3C3 O dEdc3 Hm E83001 H32 ES3202 H33 ES33C3 0 We are looking for the coef cients and the energies In the case Where C1 C2 and C3 are not all zero We can solve for these coef cients using Cramer s rule Hopefully the form of the secular determinate is obvious A Determinant to Solve for CE x1 x2 x3 X1 HM Esll Hm ESQ HE 13813 X2 Hm 13821 Ha 13822 HE ESE 0 X3 Hm E831 Ha 13832 H33ES33 coulomb inteqral 0L resonance inteqral 8 no overlap normalization inteqral 1 x1 x2 x3 x1 oc E B O x2 B cc E B 0 x3 0 B oc E Divide Throuqh bv the Value for Resonance x1 x2 x3 x1 Cc EV B x2 1 Cc EV B x3 0 Cc EV B De ning X ocEB x1 x2 x3 x1 X l 0 x2 1 X l 0 x3 0 l X When X O we have taken E to equal alpha The energy doesn t differ from the atomic value When X gt O E is negative and thus stabilizes the MO system of AOs Principles of Organic Chemistry lecture 19 page 1 LCAO APPROXIMATIONS OF ORGANIC Pi MO SYSTEMS o BENZENE MOS 0 Collecting all the molecular orbitals for benzene and arranging them according to energy and putting in the electrons gives the following energy diagram o What is the resonance energy of the benzene ring 0 Imagine putting three walls between the double bonds to isolate them M I What would be the pienergy of this version of benzene o E 76ml 0 Where does this result come from I The pienergy of benzene is E 2 ZI I 4rl l SIHI o The difference and therefore the resonance energy of benzene is 72 o Aromaticity o 4N2 0 lled atomic orbitals ipaired Principles of Organic Chemistry lecture 19 page 2 I lead to stability 0 Note the electronic con guration of C5H5 o It is like electronic configuration of the monoatomic noble gasses 0 lled molecular orbitals o aromatic species I need to be cyclic and fully conjugated H H I 135hexatriene above is fully conjugated but acyclic and 135 cycloheptatriene is cyclic but not totally conjugated 0 There is a methylene unit insulating the p orbitals at 1 and 6 from parallel interaction 0 Frontier Molecular Orbital Theory 0 reactivity and coef cients I these are important for anions and cations o kinetic results versus thermodynamic results I How does this relate to M0 calculations I Electronic changes are instantaneous compared to changes in atomic displacement 0 Let s talk about the protonation of 13butadiene 8 8 1618 3 0618 0618 1 Q L r so 59 1618 W I electron density functions as c2 so 602372 26 0 7 E e Principles of Organic Chemistry lecture 19 page 3 o The electron density is heavily weighted toward the termini Where is the molecule going to like protonate in the rst elemental event What should we consider if we are going to talk about the stability of species on the reaction pathway in the protonation of 13butadiere o Frost Hiickel diagrams 0 Only valid for fully conjugated systems 0 2Bcosm21tn Em n atoms m integer 0 1 S n2 39 for cyclic conjugated molecules 0 for benzene n6 l for linear conjugated molecules 0 for allyl o for butadiene Principles of Organic Chemistry lecture 19 page 4 O o The frost method gives the energies straight away 0 2Bcosm1In1 Em n atoms m integer 1 n I IN BOTH fully conjugated cyclic systems and linear systems we are looking form the projection of the radius of the circle onto the energy axis I Don t commit these equations to memory Establish an operational definition and work them out every time 0 Can you envision how to use these with the secular determinants to solve more complex systems I You can get the energies with these Frost diagrams and then use the method of expansion by cofactors to get the MO coef cients 0 The cofactor process is simpli ed by the fact that you are not looking for roots of polynomials 0 You are looking for ratios of expressions as a function of energies o The energies are calculated directly above for many interesting systems 0 OddEven Alternate and non alternate hydrocarbons o For the operational de nition in conjugated molecules mark the conjugated atoms so that the number of marked atoms is maximized I An alternate fully conjugated molecule is one in which alternate atoms can be marked with no marks adjacent to each other 0 Examples Principles of Organic Chemistry lecture 19 page 5 I The fully conjugated molecule is non alternate if two unmarked atoms or two marked atoms are adjacent after the making operation 0 so azulene at right is nonalternate 0 some more examples of nonalternate conjugated hydrocarbons H 39c quot399 39 Q g9 b o All linear and branched structures are by de nition alternate HH M MA I Even alternate conjugated molecules have the same number of marked versus unmarked atoms 39 c39 E CE H H IC H 39 Odd alternate conjugated molecules have an unequal number of marked versus unmarked atoms HgEKH H H and see the benzyl cation above 0 example 0 Now here is the utility in all this Large structures lead to difficult secular determinants However the following rules allow some prediction regarding the characteristics of the HOMOs of these orbitals 0 Alternate conjugated molecules dispose orbital energies symmetrically about zero Principles of Organic Chemistry lecture 19 page 6 I The nonbranched acyclic even alternate conjugated molecules dispose orbital energies above and below zero but not at zero I The odd alternate conjugated molecules dispose orbital energies such that at least one orbital is nonbonding and has value 0 5 0 We would have had a good head start on allyl with this knowledge 0 The nonalternate simple cyclic systems like cyclopropenyl cyclopentenyl cycloheptenyl dispose orbital energy dissymmetrically about zero I Nonalternate bicyclic systems and nonalternate cyclic branched systems loose this property Symmetrical systems of this variety can dispose orbital energies symmetrically about the zero line 0 What can we say about trimethylenemethane o Is it alternate If it is alternate is it odd or even 0 We want to maximize the number of marks on the structure I When we maximize the number of marks the nodes in the NBOs non bonding orbitals occur at the unmarked atoms The sum of the coef cients of the atomic orbitals on the starred atoms directly linked to ONE unmarked atom is zero I Example 4 H 1 H 2 HJIKH 02c3c40 H H 3 I How should the orbitals of trimethylenemethane disposed in energy 0 How many molecular orbitals are there four It is an odd alternate conjugated system After you put one molecular orbital at the zero line you have three left 0 There is no way to symmetrically dispose these remaining three orbitals about zero without putting another at GE If you carefully workout the secular determinant you will nd roots 0 0 SQRT3 SQRT3 O I Can you do this Principles of Organic Chemistry lecture 19 page 7 I There are two unique ways to symmetrically satisfy C2 C3 C4 0 0 C2 C3 and C2 C3 I Think about how this relates to our solution of the benzene molecule when we used mirror planes to simplify the secular determinant I These two solutions lead to C4 0 and C4 ZCZ After you normalize you get I T lSQRT6X2 lSQRT6X3 ZSQRT6x4 I T3 lSQRT2XZ lSQRT2X3 4 v m gig lMDeJ me mde M o This structure may look so esoteric that you think it is not important to consider it I If we add to electrons to trimethylenemethane and change the C atoms to N atoms we get guanidinium cation I lfwe change the C in the middle to a boron atom and the atoms connected to oxygen atoms we can think about an inorganic structure er H H H N N39 02502 039 H T H H g H HJY H HquotquotH H H p H I the pi systems in these structure are isoelectronic 0 Their MO s will be similar with perturbations by the electronegativity of the N B and O atoms 0 Your author does the benzyl example for you I You should read it page 200204 I I don t agree with the analysis that resonance structures predict 25 on C1 I You need atomic coefficients to predict charge Principles of Organic Chemistry lecture 19 page 8 I don t agree that the benzyl system was the only system for which we get different results when we do Huckel MO theory versus resonance theory Remember we found bond order differences in simpler allyl system Principles of Organic Chemistry Prologue 0 lecture 1 page 1 My email address is acammers ukvedu Feel free to email me about anything You will find a syllabus on the website The rules and regulations tell me that I do not have to print one if I make it freely available on the web httpwww chem iikv GR 7006 FALL httpwwwchemukyeducoursesche5382006FALLsyllabus5382006html or point and click your way to the 230001 website from the UK chemistry website If you are having a problem getting the syllabus from the web please see me I Lecture notes will be posted at the class website 0 I will email you when I post new notes Textbook Perspectives on Structure andMechanism in Organic Chemistry by Felix Carroll I We will also use other texts and the chemical literature You should acquire the text Other sources will be made available I I will consider this class a success ifI manage to get you to begin believing that you can perhaps contribute to the field of organic chemistry or if you come to believe you can apply some of the things we discuss to your research There are undergrads and graduate students in the course The undergrads will be evaluated based on one 1hour exam and the 2hour final only I The grads will have an extra hour burden Outside reading will be assigned and pertinent problem sets will be contribute to the final grade See the syllabus I Describing Chemistry 0 0 Think about describing a person to someone else You have various choices between a full paragraph and autobiography or a purely physical description The following would likely not be sufficient for most purposes Yet much of organic chemistry is described at a similar level compared to the overall complexity of the system We call this description arrow pushing or arrow formalism I Arrow Formalism O O Describes chemical processes as a chronological series of elemental steps showing electronic change from one chemical intermediate to another I This formalism takes us from one Lewis structure to another Instructor draws an energy diagram explaining this Principles of Organic Chemistry lecture 1 page 2 0 These stable structures share features in common I l Octets and duets Maximize bonding I 2 i Charges on electronegative atoms charges on the electropositive atoms I Examples 0 Transition structures break these rules I Examples I A truism all chemical change is electronic I Most of chemistry can be described by the coulombic electrostatic force I This is amazing There are very few statements that can be condensed so tightly about a subject so complex as chemistry I Strong nuclear force holds protons and neutrons together mediator particle gluon strength l I Weak nuclear force prevents beta decay in this process a neutron in a nucleus converts to a proton with the concomitant ejection of an electron and an antineutrino mediator particle intermediate vector boson strength lxlO s o httphyperphysicsphyastrgsueduhbasenuclearbetahtmlc2 I Gravity mass to mass attraction graviton strength 1x107 I Electromagnetic charge to charge photon strength 1137 I Electrons are smaller N2000x and faster moving than nuclei I Chemists have a variety of models to depict and conceptualize electrons in molecules 0 We will discuss these further I We worry about nuclei to a lesser extent 0 For example we do calculations to determine how stable the electrons are in any given system 0 Differences in electronic energy levels determines the amount to photonic energy that a molecules can absorb I With these concepts in mind what is wrong with the following example 0 0 H 39 I N H F D N G F 9 1 O o F o o F F F o This example might insult your intelligence but I see it every year both at the undergraduate and the graduate level 0 I When you are describing chemical mechanisms ask yourself if you are moving electrons I Balance charge in every elemental step I Balance spin in every elemental step Principles of Organic Chemistry lecture 1 page 3 Balance mass 0 Formalism Course content Two headed arrow Single headed arrow Equilibrium arrow Resonance arrow Unidirectional arrow 0 We are going to include arrow formalism in our descriptions but we are going to try to dig deeper for more complete descriptions of structure and function 0 To appreciate the deeper aspects of chemical structure we are going to have to discuss symmetry 0 Organic molecules are certainly characterized by dissymm etry so why is symmetry so darn important to understand molecular structure 0 Instructor explains Concepts of local symmetry Symmetry in Lewis structures translating to racemic lowenergy conformations I Example cis l 2dim ethylhexane 4 amp I we draw the molecule as a symmetrical Lewis structure I it has a symmetrical high energy conformation o the two halves of this conformation are enantiomers Enantiomers have identical chemical and physical propertiesianother hard truism 0 Thus these two halves must have access to both sides of mirror image conform ational space when the molecule rests in the asymmetric lowenergy conformations 0 Thus we can call cis l2dimethylhexane achiral though the conformations are chiral I This situation is different from I which is simply achiral Principles of Organic Chemistry lecture 21 page 1 DEMYSTIFYING THE REACTION ENERGY DIAGRAM The Relationship between Kinetics and Thermodynamics 1 N Equot zquot The difference 11 chemical Kinetics is a focus on rates ofprocesses 12 chemical Thermodynamics is a focus on the stabilities of constituents of a dynamic system at equilibrium 121 Dynamic is a key word in the last sentence 122 Without exchange between the elements making up the system we can not be certain that the system has minimized its chemical potential energy 13 the similarities 131 Both chemical thermodynamics and kinetics depend on the stabilities of species Sources 21 Keith Laidler Chemical Kinetics 3rd ed 1987 Harper and Row 22 Steinfeld J 1 Francisco J S Hase W L Chemical Kinetics and Dynamics 2quotd ed 1999 Prentice Hall Constants 31 h 6626 075 5 X 10 34 Js Planck s constant 32 k 1380 658 X 10 23 JK Boltzmann s constant RNA 321 Don t confuse this k with the rate constant for reactions kR or Kc an equibrium constant I will reserve just plain k for Boltzmann 33 R 8314 510 JmolK molar gas constant 34 one calorie cal is 4184 J 35 NA 6022 136 7 X 10A23 molA39l Avogadro s number Reactions occur along Vibrational modes 41 Think about a Vibration in a diatomic molecule AWB AMB AWB A 411 t 0 1 2 3 4 5 4111 We are marking time on the t aXis above 412 What is happening on the Energy aXis Principles of Organic Chemistry lecture 21 page 2 4121 5 Energv 4122 4123 When we depict bond breakage we use the same graph only we need a Morse type potential at right instead of a quadratic function 4124 There is some distance that atoms A and B can obtain at which there is no longer a bond 42 To apply the concept to transition states and kinetic theory we need to think more deeply about motions in molecules Molecular motion is strongly related to the concept of degrees of freedom 421 This concept is very neatly described by P W Atkins in Physical Chemistry 3rd ed p 455 If you look this up just hit the index for degrees of freedom in whatever edition you can get your hands on It should also be adequately covered in other physical chemistry text books 422 We begin by calculating the number of Vibrational modes of an N atom molecule 423 The total number of coordinates needed to specify the locations of every atom is 3N 4231 X y and z for each atom Principles of Organic Chemistry lecture 21 page 3 424 Each atom may change its location by varying one of its coordinates so there are 3N degrees of freedom in the molecule 425 It is pro table to group the degrees of freedom 4251 Three coordinates are needed to specify the location of the centre of mass ofthe molecule and so three ofthese degrees of freedom correspond to the translational motion of the molecule 42511 All other atoms must follow the center of mass so they are dependent on this locus 42512 When the center of mass changes coordinates they must change coordinates 42513 For example fix the coordinates of the C atom in methane to 000 42514 The remaining 3N 3 degrees of freedom are non translational modes of the molecule 425141It is like we are following the gas phase molecule as it bounces around so with respect to our frame of references it is not moving from a to b 42515 In the case ofa nonlinear molecule like methane three coordinates are needed to specify its orientation in space 42516 Let s wipe these degrees offreedom out 425161You can visualize this by xing a HC bond of methane on the x axis 425162The C atom is at 0 0 0 425163To fix the H1 atom on the xaxis we limit its y and 2 values to 0 4251631 This leaves us with 3N 32 degrees of freedom 4251632 But we have left methane still able to tumble about the x axis 4251633 Let s nail this down by limiting the z coordinate of H2 to 0 4251634 We have now frozen methane in space It can t translate and it can t tumble only vibrations are left 425164This leaves 3N 7 6 degrees of freedom for relative displacements of the atoms in the molecule 4251641 For methane 3x4 6 6 vibrational modes 5quot Principles of Organic Chemistry lecture 21 page 4 Drift quotquotquotquot 39 39 a H 425165 42517 For a linear molecule HCl we only needed to limit C1 to 000 and H to X00 425171Tllis leaves 3N 7 5 degrees of freedom for relative vibrational displacements of the atoms in the molecule 4252 H20 is a triatomic N 3 nonlinear molecule and has three modes of vibration and three modes of rotation 3N 6 vibrational modes 3x576 9 4253 C02 is triatomic N 3 and linear and has four modes of vibration and only two modes of rotation 3x375 4 vibrational modes 42531 If you have Chemscape Chime by MDL Information Systems on the computer you are at you can see these modes at the following site httpwwwchempurdueedugchelpvibscthtml Statistical Mechanics 51 Inareaction aAbB quotquotyYZZ eq 01 eq 02 75 RT 511 The equilibrium constant KC quotnggqqu 6 5111 E0 the total energy change at absolute zero 51111 This is only a value we can approximate It is not rigorously measurable since 0 K is not attainable 5112 q partition function 5113 q E Zglql eq 03 51131 g is the degeneracy of the ith state 511311F0r the ground state of Butane at a certain temperature 511312The anti conformer has g 1 511313The gauche conformer has g 2 P Principles of Organic Chemistry lecture 21 page 5 kT E e 5114 q Zg eq04 l 51141 Small energies count more at low temperature 51142 q is an expression of stability Ask yourself what happens to the value of q as the bond strength decreases in a diatomic molecule 511421 q decreases 5115 qm qeqvqrqt 51151 The partition function can be broken down into contributions from 511511electronic vibrational rotational and translational motions 5116 eq 02 states that the equilibrium hence the free energy for any reaction can be calculated by knowing the partition functions for all the modes of molecular motion starting from the zero point energy Wow 512 Let s focus on the Vibrational components of the partition functions because this is going to be the most important for the plain vanilla organic reaction Thermodynamic Formulation of Conventional Transition State Theory 61 The rst thing we are going to do is insult your intelligence by proposing that a stationary state is in equilibrium with a species that does not exist the transition state 611 This is okay for two reasons 6111 1 On the other side of the transition state there is a product to catch the material We can write an equilibrium involving the reactant and the tstate and the product and the tstate in which the tstate cancels out 6112 2 We can assume that the tstate is a very unstable intermediate possessing a lifetime far shorter than our ability to detect it This way we can do the math without having issues with division by zero This approximation puts a very low concentration of the transition state in solution This approach is really the same thing we are doing in item 1 above 612 In a unimolecular chemical reaction R gt P with X the transition state KC i 613 R X eq 05 Principles of Organic Chemistry lecture 21 page 6 614 Kffq qRe39E RT eq06 615 If R is a nonlinear Natom molecule then is has 3N 6 vibrations 6151 One of these vibrations is different than the rest since it corresponds to the reaction coordinate 61511 The transition state has 3N 7 normal vibrations 61512 The vibrational partition function has to re ect this fact 61513 The partition function for a vibration is 1 7 eihvm 71 61514 v frequency of the vibration 61515 The limit of this function has to be calculated as v approaches 0 Why 615151 As the force holding the two atoms together increases the v increases Likewise as the force decreases v decreases 61516 kThv eq 07 1040 O llelx l x 61517 The functions le39lx391 and UK above are compared for various values for X starting at the left with the smallest values 61518 You can see that with smaller arguments the functions look very much alike Now we can write a new expression for equation 06 that accounts for the reaction coordinate in the transition state and the stationary state portion of the transition state energy maximum 711 We will denote the partition function for the stationary state portion of the transition as q Thus 7111 q kThv0q eq 08 Principles of Organic Chemistry lecture 21 page 7 7112 and eq 06 becomes K314 kTh Vq qR am eq 09 7113 VX kThRq qR e E RT eq 10 71131 v is the Vibrational frequency corresponding to the elongating bond along the reaction coordinate 71132 Here is the part that will mess with your mind 711321 The left side of eq 10 is the rate ofthe reaction 711322 It is a frequency with no return on the reaction coordinate and it is multiplied by the concentration of the transition state The units certainly are right Msec 711323 Accepting that this entity is indeed a rate of conversion dR 7114 kThR61 6111 e kRR eq11 kR 7115 R gt P eq 12 712 A portion of eq 11 is stationary q qR eiEO RT Ke q eiAG RT eq 13 7121 This is the stationary part of the rate expression X int R p 7122 Reaction Coordinate 1 7123 By putting an imaginary barrier at X we can posit that X is in equilibrium with R 713 The bit in eq 10 and eq 11 that turns the argument into a kinetic one is the 7131 kTVh 7132 This is the contribution of the reaction coordinate 7133 Mass is leaking over the barrier Via bond breakage 7134 This is not really equilibrium 8 Principles of Organic Chemistry lecture 21 page 8 7135 All the mass that becomes X can not return to R 714 Combining eq 1 and eq 13 gives 7141 kThe AG RTR kRR eq 14 7142 kR kThe AG RT 0r kR kThe AH RTAs R eq 15 7143 kR kThe39AH RTeAs R eq 16 Law of exponents 7144 lnkR lnkTh AS R AH RT eq 17 A nice equation to work with experimentally RTlnkThkR AG eq 18 Principles of Organic Chemistry lecture 24 page 1 CHEMICAL MECHANISMS chapter 6 For many reasons chemists and biochemists are interested in describing chemical change in material in mechanistic terms 11 For the Development of synthetic reactions 12 solvent solute interaction 13 surface chemistry 14 to describe life processes at the molecular level 141 to describe protein protein interactions 142 drug protein interaction A mechanism is a stepwise description of chemical change at the molecular level 15 An argument is made for the logical succession of elemental steps from reactants to intermediates through transition states to products 16 An element step takes material from one stationary state to another Methods Tool Kit Challenges in the description of mechanisms 1 Trapping 11 Radical Cationic molecular clocks are traps 111 We can set up the structure so the intermediate is unstable and reacts irreversibly 112 We have discussed this previously 113 Let s say we were curious about whether the following reaction precedes by a one or a twoelectron reduction M1 9 He 5 Principles of Organic Chemistry lecture 24 page 2 115 The cyclopropyl moiety traps radical intermediate C in the ringopen forms D and E These anions can be protonated by water in the usual way 1151 Smlz is a oneelectron reducing agent Crossover experiments 21 The transformation below one can question whether the allyl on the oxygen atom in the reactant is the same allyl moiety on the aromatic carbon atom in the product 0 OH on co 0H GOA C 9 A gtlt OH OxK O HS 3 0 m D 22 H H 221 If the anionic bits and the cationic bits can diffuse away faster than or on a competitive timescale they can recombine in the solvent cage we would expect to see crossover products C and D when we run a mixture of A and B in the same reaction vessel 2211 Isotopic Labeling 31 Furthermore when the carbon atoms indicated are labeled with 13C the label is attached directly to the aromatic ring in the product C and D are not observed OOWEIH OH 4 C 311 H H 32 At this point we can postulate that the reaction belongs to the 33sigmatropic rearrangement variety I A A o 743 o OH O gt gt 321 3211 The labeling experiment eliminates a solventcaged charged dissociation recombination mechanism postulated above because the cation is Cw symmetric 3212 In that mechanism the labeled atom should only attach itself to the aromatic carbon atom 50 of the time Determination of AH and AS 41 AH will be ambiguous 411 42 However AS should lt 0 for the associative mechanism in which a cycle is formed in the transition state It is most useful when one tries to understand the effect of substituents Product reactant stationary state characterization 511 Based on spectroscopy 512 NOT Trivial 513 Sometimes an intermediate is stable enough to get spectrometric information about it by isolation and characterization 5131 In this case you want to subject the isolated material to the reaction conditions Principles of Organic Chemistry lecture 24 page 3 5132 If you can get the same product ratio by doing this while avoiding the original reactant you can establish that this intermediate is on the reaction pathway 5133 If you get an equilibrium mixture of the original reactant and products the intermediate may be on the reaction pathway graph 1 or may be a participant in a nonproductive equilibrium graph 2 5134 This approach is less useful if you have only one product P2 1 Graph 1 Graph 2 5135 Synthesize the intermediate under the reaction conditions 0 6 1 Check the product ratio it should be the same as that produced by the reaction 62 Hypothesis carbene intermediates E and C are present in the rearrangement benzocyclobutene to styrene below H H H H H H H H A H OH gt C H OH gt E II H H I H H H e N We e 621 C E 63 Isotopic labeling synthesis of the hypothetical intermediate and isolation of the intermediates were employed in the study of the rearrangement of benzocyclobutane A 64 Isotopic Labeling H H HH H H 11 H H H H H HH A B H H H i loop H H H D D39 c E H H a H H H direct H 2 i H H H H 65 C F F39 G 65 1 By the scheme above starting from A the 13C label should be terminal and ortho in the styrene product 90 I w Principles of Organic Chemistry lecture 24 page 4 CH3 4 4 9 gt gt gt Co I H 652 multiple highenergy intermediates I benzocyclobutane r39 Syrene 653 Reaction Coordinate 6531 Putting any of these highenergy intermediates into the reaction conditions should produce the same ratio of styrene benzocyclobutane 654 There is now quite a bit of evidence for the l246cyclohepatetraene intermediates 6541 They have been isolated in solid Ar and studied via IR spectrometry 654ll Chapman 0 L39 Johnson J W McMahon R 1 West P R J Am Chem Soc 1988 110 501509 httn39 nub an 39 i biquot 39 hi mi iacsatl988l 10 13902 ndfia NOaO W ndf ht Wwwor anicdivisionor essa s 2004munsch df 6542 They have been isolated at room temperature in cyclic ether cages 6543 They are chiral due to the symmetry of the allene functionality H H CH3 I C E C H H H J 6544 6545 The barrier to racemization has been determined to be l94i03 kcalmol 65451 Warmuth RJ Am Chem Soc 2001 123 6955 65452 httn39 nub an r Um39n m39 1 mi iar at mm 17 13978 ndfia015743epdf 7 Principles of Organic Chemistry lecture 24 page 5 6546 This rearrangement should occur through the at singlet shown above 65461 WHY MUST THTS TRANSITION STATE BE A STNGLET 65462 By the way Where should the electrons be in the structure Should they be in a hybridized orbital 66 Carbocation intermediates 161 Nonclassical carbocations George Olah 1611 See the discussion of Norbornyl cation on p 296 of Carroll k racemizationeq kracemizationax kequat kaxial ACOK ACOK A P ACOH AcOH O Q OAc IO S Br 0 kracemization q kequat B 162 kracemizationax kaxial r 1621 What does this rate study indicate ACOK H ACOH 14C gt 14C H CA I OBS OAc C H H H 739 9quot 9 163 H 164 In another study Resolved 13 C NMR spectra of carbonium ions at cryogenic temperatures The norbomyl cation at 5 K Costantino S Yannoni Volker Macho Philip C MyhreJ Am Chem Soc 1982 10425 73807381 1641 If classical cations then the barrier to exchange must be less then 02 kcalmol Pnncxples of Orgamc Chemxstry lecture 24 page 6 L25 200K 152K V2 6 129K 132 NMR above 170 K C3V5ymmemc H 1561 ppm J 39 1625ppm J537 W 1018ppm J533 6 6 H H H c c 1 H H 2 H H1 2H Be oW 130 K Cssymmetnc Principles of Organic Chemistry lecture 36 page 1 E2 E1 and the stuff in between 1111 Do you have any thoughts about the likely product of this reaction 11111 Is there anything fundamentally different between developing a carbanion by proton abstraction or by reduction 112 Saytzeff s Elimination 1121 The thermodynamically favored product is obtained Br EtOH I EtOK gt 1122 71 2 113 Hofmann Elimination 9 1131 The thermodynamically least favored product is obtained s EtOH Jl f EtOK gt 1132 26 74 114 The best explanation I have seen the discrepancy between Hofmann and Saytzeff elimination is offered by Hughes and lngold and focuses on the increased acidity of the 3 proton 1141 The competition between the two pathways depends on which Bproton is removed by the base 1142 The Bproton on the less substituted carbon atom is more affected by the charged highly electronegative leaving group because the Bproton on the more substituted carbon atom is stabilized by the electron donating effect of the substituents 9 H S H s KD s lt gt H lt gt L H63 H 4 H H H 99H H H H H 1143 H H 11431 The acidity of the less substituted carbon atom is more enhanced by the presence of the leaving group than the acidity of the more substituted carbon atom 11432 The less substituted carbon atom lack the hyperconjugation provided by the methyl in the example above 11432 1 Substitution destabilizes anions 11433 Other Hofm ann substituents are trialkylphosphonium and trialkylammonium salts 1144 Another explanation that you will see focuses on the hypothesis that the Hofmann reactants pass through an earlier transition state Principles of Organic Chemistry lecture 36 page 2 11441 By the Hammond postulate this means that the TS should resemble energy and structure the product more 11442 However such an explanation begs for the question why do Hofmann reactants pass though early TSs 11443 The explanation above gives a reason for the early TS 11443 1 The TS must be more like the Elcb mechanism developing negative charge at carbon 114432 LFER show that the Hofmann elimination develops more charge at C than the Saytzeff reaction 115 The E2 reaction is a subset of reactions that depend heavily on the antiperiplanar relationship 11511 At least it is mentally economical to think of it this way OMS tBuOk 40 C b gt 1152 100 1153 This is not a new concept for you See the anomeric effect in lecture 13 p 2 11531 In that example the lone pair needed to be antiperiplanar to the electronegative group 1 1 6 Grob fragm entations G OMes 11621 From which conformation of carvone left does fragmentation occur 116211 You need to be able to see this 117 Syn elimination 1171 Pyrolytic eliminations occur via a syn pathway 1172 often these do not have the ApKa pushing the reaction forward 5 s s 7 Y H 84 o H O H quotH 5 SH 1173 11731 This is an example of the Chugaev Reaction demonstrated by Cram to occur with a syn elimination mechanism Principles of Organic Chemistry lecture 36 page 3 11732 Consider the similarity between this reaction and the ene reaction A H gt H 0 O 11733 7 7lt 11734 If you want to conceptualize this reaction in terms of MOs you want to start with the fivecenter six electron system 117341 The energy distribution of the molecular orbitals will look similar to the way the cyclopentadienyl anion distributes MOs 1174 Syn elimination is the only way to rationalize the following product distribution from this single diastereomer Q YO 0 Ph H Ph H H I I Hm gt D D Ph Ph H 11741 Principles of Organic Chemistry lecture 15 page 1 LCAO APPROXIMATIONS OF ORGANIC Pi MO SYSTEMS I The allyl system cation anion 01 radical o 1 Draw molecule and set up determinant 1 2 3 H H 1 X 1 0 lo 3 2 1 x 1 0 H ICZ H O H 3 o 1 x x 1 I 1 1 0 1 x O X 1 x 1 0 x 0 1 o XX21 7X 0 X3 2X XX22 X 0 X SQRTQ X SQRT2 0 These are the energies of the orbitals in terms of beta bonding SQRT2B non bonding 039 antibonding SQRT23 I discuss the cation anion and radical with regard to the solutions 1 el 1 I CC C CCC CC C Pi MOS 0f Allyl anion pi MOS ofAIIyI radical pi MOS of Allyl cation 0 The bonding orbital contributes ZSQRT2B to the energy of the molecule I The nonbonding orbital contributes nothing to the energy of the molecule I It should be relatively easy to put electrons in and take electrons out of the allyl I In general we observe anionic radical and cation chem istries for allyl I This is similar to the simplest nonbonding orbital of atomic hydrogen 0 This why the chemistry of H is neutral and i I Proton hydrogen atom radical H dot and hydride I Does this mean that the allyl anion and cation are really isoenergetic I No there are ee repulsion terms that are not considered in the Huckel formulation I There are other effects that Huckel does not include I Huckel is approximating what the orbital spacing is 0 When we put electrons into the system the Huckel levels don t change but the overall energy of the molecule certainly does Principles of Organic Chemistry lecture 15 page 2 o Huckel LCAO is only a first approximation I You will notice that all the atomic orbitals that make up the MOs are not the same size 0 O O This is the stuff of Cum the weighting constants in the MOs wave function or the AO coefficients These AO coefficients can be obtained by expansion of minors of the secular determinant into cofactors I For example 2 3 X 1 X 1 A11I 1 X 3 1 X I A11X21 likewise I A12X 39 A131 We can determine the relative ratios of the coefficients c1c3 A11 A13 XZl I CzCg A12 A13 X I as before Lets arbitrary set 03 1 0 FROM All I For wl X SQRTQ so cu l I For w2X0 so c12l I For W3 X SQRT2 so c13 l 0 FROM A1 I For wl X SQRTQ so cm SQRT2 I For w2X0 so c220 I For W3 X SQRT2 so c23 SQRT2 0 FROM A13 I For wl X SQRTQ so c13 l I For w2X0 so c23l I For W3 X SQRT2 so c33 l X All Xi 1 A12X A13 1 SQRT2 l SQRT2 l 0 l 0 l Principles of Organic Chemistry lecture 15 page 3 SQRT2 1 SQRT2 1 0 Now we can write out the MOs W1 13611 SQRT2X12 13613 W2 391X21 1X23 W3 1X31 SQRT2X32 13633 0 These are the coefficients Normalize and you are done To do this we need to find the normalization factor to apply to the MO above for WI and W3 SQRT12 SQRT2212 2 for w SQRT1212 SQRT2 Thus 2 and l SQRT2 are the normalization factors for wl W3 and 12 respectively W1 147611 1SQRT2X12 143613 W2 1SQRT2X21 1SQRT2Xz1 W3 147631 1SQRT2X32 1433 Bond order and Atomistic Electron Density can be approximated by the LCAO method 0 The concept of bond order 9 SSH 39o39 R3C CR3 has CC bond order of one R3CCR3 has CC bond order of two R3CO has CC bond order of two R3COiH has CC bond order lt two 9 0 H H H of o How might we try to gather experimental evidence for the above shift in bond order I The resonance second from the right is more important when the O atom is protonated Likewise one might wonder about the CC bond order in allyl cation 0 Bond order PNij SumNcic N is the number of electrons in the MO cicj is the product of the coefficients for the bound atoms Principles of Organic Chemistry lecture 15 page 4 0 We can get these parameters from calculations 0 The better the calculations the more agreement there is with experimental results 0 Why should this calculation give the bond order I Think about interaction between AOs in the molecule 0 It is a zero sum game due to normalization o If an orbital grows another has to shrink elsewhere 0 The extent of interaction bonding between the p orbital above is a function of how much volume is gained for the electron occupying both versus only one orbital o The interaction bonding scales linearly with the coefficient ofthe A0 in the MO Pi Bond Order in ALLYL o Pccallyl cation 2121SQRT2 1SQRT2 071 I There are 2 electrons I Coefficient 1 12 I Coefficient 2 1SQRT2 I This is the pi bond order There is also a sigma bond so the actual bond order is 171 o The number is the same for the radical and anion by this method because these added electrons are in the NBO o This is not always the case When the next electron up is but in a bonding orbital we get more bond order from more occupation Molecular Orbitals of pi systems and their application to organic chemistry I Instructor compares the C12 bond order in allyl cation with two hypothetical electronically excited states of allyl cation 0 Instructor talks about the differences in dynamics that one would expect between ground and excited states 0 If asked could you predict which electronic state of allyl cation would be more conforrnationally stable 0 Could you do the same thing for ally radical 0 WHAT IF THE COEFFICIENTS JUST DISTRIBUTED e density uniformly 39 W1 1SQRT3X11 1SQRT3X12 1SQRT3X13 Principles of Organic Chemistry lecture 15 page 5 Pccallyl cation 21SQRT3 olSQRT3 23 066 I The real distribution leads to more bonding in the lowest energy orbital than 066 I Notice that the lowenergy orbital is also more spherical o What is the pi bond order from resonance I Instructor draws resonance structures and talks about weighting HH HH 39IC 39 H HHHamp H 1He 1H3 2 2 0 Thus the Hiickel LCAO solutions ive us a clear icture wh con39u ation stabilizes molecular systems I Instructor explains this effect is not obvious from resonance structures without a lot of hand waving o The double bond in ethene has a 66 kcalm ol barrier to rotation o The bond is order is 2 in ethene Therefore the rotation about allyl cation according to Huckel should be 071 66 kcalmol 47 kcalmol of the naked gas phase species This is really slow on the molecular time scale Remember that you need a barrier of about 24 kcalmol between molecules 1 and 2 to isolate molecule 1 or molecule 2 at room temperature 0 Checking the number against literature values give some discrepancy the literature values are interesting I Experimental value for the allyl radical is N15 kcalmol I Korth GG39 Trill H39 Sustmann R lZHAllyl Radical Barrier to Rotation and Allyl Delocalization Energy J Am Chem Soc 1981 103 4483 High level calculation for the allyl radical gives N13 kcalmol I Gobbi A Frenking G quotResonance Stabilization in Allyl Cation Radical and Anionquot J Am Chem Soc 1994 116 92759286 note the year this stuff is just being figured out currently It makes you wonder how mature Organic Chemistry is if the fundamental behavior of such simple molecules is unknown High level calculation for the allyl cation gives N39 kcalmol I High level calculation for the allyl anion gives N23 kcalmol One might hypothesize that the massive difference in electronic distribution between ground and transition states in the anion and the cation versus the radical could account for the difference in barrier to rotation Principles of Organic Chemistry lecture 15 page 6 I Demonstrate this on the board


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