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Class Note for CHEM 6311 with Professor Albright at UH


Class Note for CHEM 6311 with Professor Albright at UH

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Date Created: 02/06/15
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 4TraDDing 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 kln2ltm 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 l A lAo ii Plotted L i A A0 slope 2k time gt 61 62 cBimoecuar A B gt C i dC k AB or kt I InAoB BOJ Ao 801w Slope kIIo 30 63 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 80 60 40 20 dThis is the most simple situation for a twostep reaction A gt B gt C kl k2 It can be shown by some tedious math that A Aide k t k A0 rk t 7th B k k e e 2 C A0 A B Ao I kze k t kle k1 kk2 k1 30k2 k1 k2 A C 80 5 A1 60 C E 40 39 B 64 u E B U 20 65 g 0 Time Time gt k1 033k2 80 A I 2 5 an E g 40 I O U 20 B 66 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 approximations are used ii Reaction k1 M A B 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 1 68 a Assuming the lst step is a rapid equilibrium dRBrdt k ROHf Br39 ROHI i b Now ROHH kil 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 Step1 AB C 4 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 KWM MWP F memm dt k7 k2D This same expression can be found for dAdt try it 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 hMMlthM c So k2 D lt klandtherefore C rat z Mmmm 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 Substrate Enzyme Enzymesubstrate complex P E K ys k2 Es 71 p 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 am k E39s ii Eons E SS kzlws 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 k2EoS a If S ltlt Km then v K km S a m 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 Es E0 lvmax 2S 2 2 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 make sense d Cases a b and c above are indicated on the graph ofV vs c If S Km then V vmax k E HS L vmax kzlEo 12Vmax S 612 73 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 kl 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 k3 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 Y 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 k1 First step A X gt 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 Rl R 20 papain RJKO R39OH H 5 E P 615 b The environment can be depicted where E amp E are ionized inactive forms of enzyme cell wall 616 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 Roocioms I 3 zzquot 4 1 I I O 0 Transition 00352345quot 039 I J 313 a i 3533933903 o 3 25quot 3 9 0 039003905 lutermodioto 395 6 3 t g o o o o h Winch w s 222240 9 o 9 o 9 9 9 0251 quoto zogozozi o q o oolgl a ozz Transition sm o II M F z 5 quot i 82quot Q a39 e aw 32 X 9 391 l 7 I If ghtgi gt 3027 ggv f Mk gg 2quotquot a I r Producis o g X 82 We Energy R 6173 The potential energy surfaces shown above show what happens when thiophene is oxidized Reac onCoordinata I Let us consider the most simple reaction H HH HH H aThe reaction potential energy surface is 9 180 H H H H Al I391 r2 39 4 2500 W I l Energies in eV 1eV 23061 kcallmol distances in au 1 au 053 r1 1875 1250 618 1250 1375 2500 r2 I An expanded view 1 2 3 4 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 r1 I e 60 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 lf fInISh 1 g I I 39 MEI l l W 32v m 0 39 22 HH 39 39 i 0 a Iusi o 390 I v ii39 I a I b 39 1 I inti 39 j 0 ll IIH i j 39 l w 1139 i s 39 I l 1 I l l l I j I E a i f I m my 3 tr 39 MIN 3 39 a39 If I HI In I I r I a II I a t 2 I 39 g jv 3 I I i quotI Z W Z 22m 1 I EH1 q lt 391 39 quot I I W ssg I39 Hquot Isquot39a 3 It I v 3 It I 3 I ll l 39 4 o I l C3 C3 I v Q g I I w D D i l s a W 39 Ln 5 II I 0 n n g A39 a a l j m I 25 l In I 12u my I quot39l 3quot In Jquot 5r39iil 3quot 24312211223 quot2 39 393939 393 quotlimiiiihoimmmmmu mumu momu mmm 39 mm 3mm quotquotquot quotquotquotquot39 mm m u m 55 45 55 55 CClo DISTANCE c N 39 39 u i o I 39 h o n H u v 39 39 I n n 39 39 r n u I 39 i c o 39 n s v t i I I p o a y u q 2 y n y a u s u o g I l I I V I h I u p j 39 I u 1 39 39 I I I u v a n o A u n n u I n u a n n n u c l a a v u a 5 n v u u u n 4 l n 1 n u y a y I I I 39 39t 39 I l u 39 I 39 I b I I O I v v n o o a t v u n u u e c 1 e n v a u q p u v q g r v n qu 15 25 35 45 55 65 C Clo DISTANCE start finish m 3 i tn 3 co 55 45 CCIb DISTANCE 45 35 u CClb DISTANCE quotquotquotquot 3 3939 39 KMquotE KT o 1 1 o a 1 sc v 10 00 00 351 to a so no on at II I I in a o Q i n n u 39 39 v o u u u quotsu uuuuuuuuuuuu v 39 0 p nnnnnn quotn39uuz u 00 I ll II 0 oiacuvg ue ntv u n a u n a n c n o a o n o o u o n n 39 39 39 39 39 quot M a w nh in C ziif39miil39iilli39 ihiumfll mlhl Illll IIIIIIIIIII 15 25 35 45 55 CClo DISTANCE 65 start 15 25 35 45 55 65 000 DISTANCE 6 20 Start 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 u u gt 9 transition state ABc a LI 5 u c 9 AB c O n J energy of reaction K minima 621 p 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 V V R gt R I llquot 622 R P 393 3 PR3 M Ni Pd pr R3P PR3 X Cl 4l PFs R Ph 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 5132 Effff Eiryn r Vquotquotquot rgtr rr55 r NV nur mquot V I 2quot in 7 7 1 Hquot w 7e777 eVVn EL 25quot 7 n I A quot r quot33i gt K 739 quotJack Hi 739quot 2 r Li 12139 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 A gtB i UsingA e W assumes AS 0 see below and ii defining tIz time when A B and tI99 time when 99A E1 T500C T IOOOC tIz t199 tIz tH99 20 l2 sec 80 sec l7 x lo 393 sec 0l sec 25 49l min 542 hrs l39 sec l54 min 30 8 l 4 days l46 yrs 326 hrs 26 hrs 35 533 yrs 3534 yrs I I4 days 208 yrs 40 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 KkT Adm KkT rAHWRT AsR bthenk h e h e e 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 eKkT AStR A I h e ii E AH RT iii AGi AHi amp ASi can be computed by plotting lnkT vs lT intercept In observable range k AH In slope T T 6 27b 86 ue 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 d Rate dependence on AH and AS i Lets say we have two competing reaction paths A e C gt B ka kb 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 20quot Br gt Ph CH20H2 Br H H 628 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 R0 on gt 2R0 R fBu 629 CH3CH3 gt 2CH3 ASi 7 eu 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 D H H D H H EC02Me Kzl Kzl E E E E k gtk D D 630 EHE HEHT HEHE B c D 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 631 TUNNELING gt 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 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 Q 39 H l singlet C HJC 634 p 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 t singlet singlet g l triplet 635 p potential energy 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 ME H 636 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 ll 92 In fact in the example shown above the strain is 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 k CH3 CH3 gt CH3 CH3 gas k2x 039 CH3 CH3 gt CH3 CH3 H20 k32 x IOquot Ea 3 kcalmol H OH gt H20 H20 solvent k 4 x loquot gt l2 H20 solvent k 82 x IOquot 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 Ink AV slope W 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 r r lt 39K 0 63 94 AW 33 cm3mol AV 37 cm3mol iv A positive value of DW implies a loose structur ally flexible transition state 1 R NN R gt RuNEN R 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 k1 A E 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 k4 k2C Then using the fact that the first step is fast kl gt k2C 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 M L related to k1 and k2 A I l I I A I I 642 l l I I 39 I I l I I I I I I I I I I x I I I I 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 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 M L related to k1 and k2 A I l I I A I I 642 l l I I 39 I I l I I I I I I I I I I x I I I I 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 D l gt or D ABCgtABC r BC The prInCIple of least motIon correctly predicts that a motion like B A l 1 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 the end of the semester are gt 646 Both reactions take place by paths which lie very 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 Potential Energy 99 B Reaction Coordinate Pnlenlizl Energy T b S2 is geometrically closer to the reactants 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 A Reaction Cnnrdinzle 3Th thermoneutral endothermic Pnlenlizl Energy Pnlenlizl Energy 648 B A B A Reaction Cnnrdinzle Rez innCnnrdinme ornton 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 855I I I985 a Potential Energy 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 in slope 649 Potential Energy toA toB 00 Reaction Coordinate 00 Reaction Coordinate X X 100 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 2 650 101 Potential Energy Vibration Coordinate Z 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 I 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 away from the energy lowering but for directions perpendicular to the reaction coordinate the transition state will shift towards the energy lowering 102 ABC gtABC more difficult higher energy ABC gt quot 652 new Tsl p133 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 I X 653 stahl In x 39 carbonium ion 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 rEH stabilize X39 x 39 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 RX39 3 a R x C m E E R X39 9 o n 655 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 656 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 YquotR X39 gt 9 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 electron affinity of RX to change by modifying X will do the same however the transition state will shift to the left It can also be seen that in any case AEi flYARX b 5 Marcus theoryThis was originally developed in the l960 s by Rudy Marcus to explain electron transfer It has been more recently used to investigated proton transfer atom transfer reactions and 5N2 reactions a Let us consider a proton transfer acidbase reac tion which is exothermic AH B39 A39 HB etAGlt0 658 bTwo identity reactions are used The Identity Reactions Reaction 1 AH A39 A39 HA AG 0 Reaction 2 BH BN7 B HB AG0 659 cThe primary idea behind Marcus theory is that the activation energy and the position of the transition state is a function of the intrinsic reaction barrier which is an average of the two identity reactions and the overall thermodynamics of the reactionAG d For this case we construct two sets of parabolas for the identity reactions as follows 660 00 05 10 00 05 10 x AG gt AGE x In each case the reaction coordinates have minima at X 0 and X l and the transition state occurs at x 05 Note that AG for reaction I is greater than that for reaction 2Why7 108 eThe same two parabolas are then drawn for the averaged case where the intrisnic reaction barrier is defined by Ac 2 A In AG 1 t AGin AG X I 661 I 39 gt 00 C05 10 X The product parabola is lowered in energy by an amount AG given by the dashed ineThe activation energy is expressed as AG and the position of the transition state by Xi f One can show see problem l7 p247 in your text for this treatment that 2 AGWAGI AG 4A Gt m 2 normally w AG E AG this last term 2 l6AGl is small where w the work term the standard free energy required to bring the reactants together in the right configuration for the reaction Most importantly this includes changes in solvation and Lg 2 8AG 109 For the present reaction k AHB39 k1 A39HB AGlto 662 1 it is easy to see that X lt 05 and how AG is affected by AG gWe shall see in the next chapter that there is an other linear free energy relationship the Bronsted catalysis law which states for the reaction above that lnkl uaneq C where C a constant OL the extent of proton transfer at the transition state and this runs from 0 to l0 It then suggests that somehow 0c is related to X in Marcus theory We will see that this is true later h Let us return to the Hammett equation logkLJO39p which leads to o ESAGt SAGp or SAG SAG p using the Marcus equation aAG aw I AG p BAG BAG 2 4Aan and neglecting changes in the work term H l p 2 8Aan We used a very similar treatment before when we looked at the physical basis for the Hammett equation S r p spsr where sr slope of the reactant parabola at the transition state 110 and sP the slope of the product parabola at the transition state Both approaches use equivalent approximations and starting points for their derivation Both are useful ways to view linear free energy relationships For example suppose that the stretching force constant in HA for the proton transfer problem we just finished looking at then since sr is larger and srlsPsr is larger so p becomes larger 00 05 10 Notice that the activation energy becomes larger and since p is larger the reaction becomes more sensitive to changes in substituents Suppose we make AG larger It is clear then that p should become larger 00 05 10 X In this situation notice that the transition state moves in the opposite direction as in the previous case but again the activation energy becomes largerThis points out a general concept called the reactivity selectivity principleThe less reactive a substance is the more selective it is and vice versa 11 AG kcallmol h One interesting concept from Marcus theory is that when reactions become very exothermic AG be comes smaller there reaches a point when AG is zero ie when AG 4AG in then AGi 00 39inverted region AG 665 0 40 AG kcallmol i As shown in the graph on the left when this condition is met not only does AGi become zero but also when the reaction becomes even more exothermic the value of AG rises This is called the inverted region One can graphically show this by the plot on the right side Three values of AG are shownThe lowest parabola on the reactant side shows a very small value of AG The middle parabola is one where AGi cs 00 Finally destabiliz ing the reactant parabola even more causes AG to rise again Now the intersection of the parabola occurs on the left side ii This does in fact occurThe experiment shown below relates to electron transfer in a series of organic moleculesThe biphenyl group on one side of this steroid is selectively reduced The electron is then transferred to the acceptor group by a slow enough process that the overall rate of the reaction can experimentally be measured Once the AG for the reaction is l5 eV there is a decided peak in the observed rate constant 5 l 09 which recall from the discussion on diffusion controlled reactions corresponds to Ea 00 kcalmol 112 lntramolecular rate constant 590391 0 39 0 Cl f0 0 A O 106k I I 0 1 2 Change in free energy eV from J Phys Chem 90 3675 1986 4 Kinetic versus equilibrium control a Consider the following set of generalized reactions k2 k C2 4 AIB lt C1 where C1 FCZ 667 k2 39I I bThe normal situation for this system is A l AH AE 2 l A B c1 2 l AH2 quotquotquot quotlquot39A39H4 quotquotquotquot quot 668 gt Reaction Coordinate i Here AHI lt AH2 and AE gt AEZ ii so kl lt k2 and CI lt C2 under all condition 113 c However sometimes one can have the following situation AH Reaction Coordinate i Here AHI lt AH2 as before ii but now AEI lt AEZ so k gt k2 and kgtgtk2 iii There will be a difference in the product distri bution depending on whether the reaction is carried out under kinetically controlled or eguilib m controlled conditions dTo exert kinetic control over a reaction supply just enough energy forA B to get over the barrier in the forward direction i ie use the lowest possible temperature a short reaction time mild reagents a noninteracting nonpolar solvent etc ii In the second example above this will favor CI vs C2 e Equilibrium control of this same reaction will favor C2 vs CI i This is because k2 is very small compared to kl ii That is by using long reaction times high tem peratures very reactive reagents polar solvents etc one can cause C2 to build up at the expense of CI G lsotope Effects l lsotope effect arise from the differences in the masses of atomic isotopes aThey are most evident in the replacement of protium 114 H by deuterium D because D has a mass twice that of H b For the most part we will discuss only such replace ments 2 Primary isotope effects a Recall the energy curve for a CH stretch 67 Potential Energy T zero point energy correctlon i For earcthibrational levelthe energy 8quot n l2h1where a h Planck s constant b n 0l23are vibrational levels and c 1 frequency of CH stretch ii The frequency of any bond may be expressed as k V where 27 u a k stretching force constantand b u reduced mass mlmz m m2 b Comparing a C H vs C D bond being broken i MH 2 3 ii Mo 24l4 iii DD lt DH since up gt LLH and as shown below R H 9 Potential Energy 115 iv 80 D lt 80H v It is also clear that AGH lt AGD a kH gt kDor h K I k0 gt I NoteThis is a purely vibrational effect The elec tronic situation ie the bonding is identical with the same force constant for H and D cA maximum value for kH k0 is approximately kH kD ehAvoZRT ADO is related to reduced mass difference i Considering various bond types XH makakDatZSOC CH 70 OH 110 NH 60 FH 149 MH 2742 where M a transition metal These values are related to the value of the vibra tional frequency itself if 1 is small then ADO will also be small etc ii The maximum value is reduced somewhat under normal circumstances see Appendix 2 in your book In the general situation the HA bond is not broken by itself but rather abstracted by another reagent ie A H B gt AHB gt A H B 672 If the hydrogen atom is transferred from A to B early of late along the reaction path ieX is greater or less than 05 using the Marcus approachthen the kinetic isotope effect is reduced Likewise if the transition state has a nonlinear hydrogen transfer then it is also reduced 116 A H B gt N39 3 gt A H B 390r E73 A H B gt AHB gt A H B iii Examples CHc Br gt CH2 HBr kHkp46 674a MeiHC C CHMei OH gt Me1CZ CHMe1 kHKD639I Potential Energy C02Me 047 0 he 0 iv In rare instances kHkD can be less than l0eg kHkD R R 140 Phgc I PthH 03992 C H ocvll 674b oc co Here the WH stretching frequency is only l845 cmquot whereasthe CH stretching frequency is 2900 cmquotThereforethe W H C stretching frequency at the transition state which becomes later as kHkD becomes smaller may well be greater than that forWH in the reactant Using the equation previously given the max kHkD 35 In the reaction CpWCO3H CpWCO339 gt CpWCO339 CpWCO3H the kHkD 37 v Also in extremely rare instances tunnelling may increase kHkD greatly since the probability of tunnelling increses with mass in an exponential manner 2 H H H H H 6 730 H3 i Bu CH3 1quot CH3 CH3 classical 1 4 trajectory 730 730 barrierwidth at 77 0K 674 quantum anlcal tunnelling Reaction Coordinate 117 a Here protons tunnel but D s do not b Values of kH kD as high as 50 at room tempera ture have been observed c Tunnelling will occur only if there is a narrow barrier not much configurational space must be transversed 3 Secondary kinetic isotope effects aThese may occur when the bond involvmg the substitution eg D for H is not broken i l0 lt kH kD lt l5 gt normal secondary isotope effect ii 07 lt kH kD lt l gt inverse secondary isotope effect b Consider a reaction which does M involve any change in the zero poi t energy on going from reac tants to TS l M gt on k m C m E B 75 C m El H 8o quot 800 Reaction Coordinate i If ADO ADJ then AGHi AGE and kH k0 00 ii However this is usually not the case for reac tions which occur at the carbon upon which the isotopic replacement is made reaction occurs CH at this center D 6 76 c A secondary isotope effect results when ADO and A001 are not equal because of changes on going from ground state to transition state 118 NYC vcH bend 1340 cm1 Potential Energy i Not only do ADO and A001 depend on the reduced mass differencethey also depend on the frequency of vibration itself Recall that 8quot n 2hv ii As the frequency decreases so do ADO and A001 a Consider a change in hybridization quotC H quotc 677 gt vc44 bend 1350 cm1 vcH bend 800 cm1 b On the reaction coordinate this can be repre sented by 8003 Reaction Coordinate c Since A001 lt ADO AGHi lt AGDJ and kH kD gt 00 d Just how much larger kH k0 is than I depends on the degree of hybridization change in reaching the TS d General rules for secondary kinetic isotope effects i If the s character increases on going from reactants to TS a then kH kD gt 00 b This is a normal secondary isotope effect ii If the s character decreases a then kH kD lt 00 1 19 b This is an inverse secondary isotope effect e Examples 1 Ph Ph 5 I 5 Ph quot2 H quotc gtC39 2 39C quotquotC39 gt Hzo CVH Cl 39 kHkD13O DH H mm HD 5 1 679 NC N Ph39u C CN 5 C quot gt CO gt c ou u D H Ph Ph kHkp 073 DH DH 4ACS I2 I 3933 I999 a nice example of IO and 20 isotope effects The calculated reaction of CCI2 and propene is shown below 0972 0913 H CH3 679b 1027 1005 LozelC C 1004 H 0986 theory 03932 0339 experiment The transition state is not well defined 2766 AGT l Ol kcalmol however ASi36 eu so 2190 39 99ml AHT 5 kcalmol A negatIve actIvatIon o r24A r22A As can be seen there is pretty good agreement between experiment and theory the two CC bonds do not form at the same rate Why does the one bond form before the other Notice that the carbene lies down on its side when it approaches the olefin We will discuss this feature in Chapter 8 5 Other isotope effects a For example H AT Now reduced mass is three times as large HoweverT is radioactive so this reac tion is not carried out on a molar basisThe techniques are different Radioactivityscintillation counters are used primarily for biochemical reactions where the concentrations are very low b 393C a 39ZC 34S a 32 etcThe reduced mass ratio is 120 now much closer to unity However one can use them for primary isotope effects which are easier to analyse and understand since the bond being broken be comes the dominate portion of the reaction path HAn Example of aVery Small Barrier jAm Chem Soc I20 0423 I998 The IR spectrum of norbornadieneFeCO3 in the CO stretching region shows a very interesting temperature behaviorAt low temerature there are three peaks which correspond to that expected for a molecule with CS symmetry 2At higher temperatures there are two peaks in the ratio of 2 This is consistent with an FeCO3 group having C3v symmetry 3A mechanism for this behavior is shown below on the right side quot7 2050 2000 1950 39 VCan g quotmmmm Fe mm 13 a C o quot39quotm m I 680 O 4The IR spectrum as a function of temperature is shown below on the left sideThis can be simulated just like NMR spectra so that there is a rate constant for the rotation process illustrated at each tempera tureThis is shown below on the leftThe rate con stants then can be used to make an Eyring plot on the right sideThe AH of 07 kcalmol is extraordinarily small to have been measured 226 AHr 07 kcalmol ASI 03 calmoldeg O 121 2050 2020 1990 1900 1930 1900 682 I 39 I 39 l f Y 0003 0004 0005 0006 0007 0008 1T Ir 3 CmI


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