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# Adv Physical Chemistry CHEM 6320

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This 34 page Class Notes was uploaded by Mr. Clementine Gottlieb on Wednesday October 21, 2015. The Class Notes belongs to CHEM 6320 at Tennessee Tech University taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/225694/chem-6320-tennessee-tech-university in Chemistry at Tennessee Tech University.

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CHEM 6320 Fall 2005 Study and Description of Reaction Mechanisms A Chemical Thermodynamics 1 Review a Thermodynamics tells us the relative energy levels and thus the feasibility of a particular process but nothing about the reaction path and thus no insight into the mechanism or the rate b Relation between the Gibbs free energy enthalpy and entropy AG 2 AH T AS c Spontaneous processes In general a system evolves spontaneously in a direction that lowers its energy or enthalpy and that increases its disorder ie entropy For an isolated system no matter or energy exchange with the surrounding dS gt 0 spontaneous process dS 0 reversible process The Helmholtz energy A U T S The Gibbs energy GH TSA PV 2 Enthalpy of reaction ArH a Usually computed from data tables representing gasphase reactions b For a general case aA bB gt cC dD can be calculated as 39 f 39 f ArH Z boi irfiii n Z boi irfg iid ArH Z npmdAfH prod Z nreactAfH react where AfH is the enthalpy of formation ArH Z nreactAcH react Z npmdAcH prod where ACH is the enthalpy of combustion c AH predictions in solution require estimates of solvation effects 105 CHEM 6320 Fall 2005 3 Chemical Equilibrium 3 Relation between the Gibbs free energy and the equilibrium constant VAA VBB gt VYY VZZ aVYaVZ K Y Z AG RT1nK aVAaVB A B eq Q AGT ArG T RTan lennE aVY aVZ where Q g is the reaction quotient a a IfQltKgtAGlt0gt IfQgtKgtAGgt0gt b Le Chatelier s Principle If a chemical reaction at equilibrium is subjected to a change in conditions that displaces it from equilibrium then the reaction adjust toward a new equilibrium state The reaction proceeds in the direction that partially offsets the change in conditions c GibbsHelmholtz equation aAG W AH 6T P T2 aanT danT AH l 6T 12 dT RT2 106 CHEM 6320 Fall 2005 B Chemical Kinetics l Reaction Rate 3 Consider a general reaction VAA VBB gt VYY VZZ b The number of moles of reactants and products AI MUDmega 17130 B0VB I quotY I quotY 0 Vy l an z0 Vz gU Where 5 is the extent of reaction c The change in time of the number of moles of reactants and products dnAz dnBz V w dz dz dz B dz dny r 2 WY 0152 dnzm W dz dz dz dz d Dividing by the volume V one obtain the change in time of the concentrations 1 ma dw VA 0152 1 0an dB VB 0152 V dz dz V dz V dz dz V dz idnyltrgt vY idnzltrgt gw V dz dz V dz V dz dz V dz e Define the rate of reaction vz Rate Laws 3 The relationship between vz and the concentrations is called the rate law The rate laws describe the experimentally determined dependence of the rate of the reaction with the reactant and products concentrations b Rate law expression c In general the rate law cannot be deduced from the chemical equation 107 CHEM 6320 Fall 2005 d The overall order of the reaction in equal to the sum of the order of the reaction with respect to each species appearing in the rate law 2 e The reaction rates can be followed by spectral changes conductance changes for reactions which create or consume ions pH changes for reactions which create or consume H or HOT polarimetry etc f Examples of rate laws 2Noltg 02ltg 2N02g v klNOlzlozl H2g12g gt2Hlg VZlezlllzl CH3CHOg gt CH4g COg v kCH3CHO32 C12 g COg gt C12COg v kC1232CO k39H2Br212 H2 g Brz g gt 2HBrg 1 1 k HBr Br2 g Integrated rate law eXpression Give the timedependence of the concentrations Firstorder reaction consider A B gt products reaction dA v z k A gt dl l The halflife of the reaction z12 is time required for half of ln 2 0693 the reactant to disappear z12 7 T Secondorder reaction consider A B gt products reaction dlAl 2 v z k A gt dl l The halflife z12 kAlo norder reaction consider A gt P reaction dz dz dz 1 l 1 l gt kz j where x P nllino xquot 1 Am 1 108 CHEM 6320 Fall 2005 3 Temperature Dependence of the Rate Constants a The rates of chemical reactions strongly depend on the temperature and almost always they increase with the temperature The dependence of the rate with the temperature is due to the dependence of the rate constant with the temperature Commonly the dependence of the rate constant with the temperature can be described by the empirical equation called Arrhenius equation dlnk Ea lnk dT RT2 lnczlnA Ea RT k Ae Ea RT 1000 T where Ea is the activation energy and A is the preeXponential factor The activation energy can be determined from the slope of In k versus 1 T representation or from a twopoint fit knowing k at two different temperatures T1T2 kT1 T1 T 2 k T 2 A simplistic but not completely true interpretation of the activation energy is as the energy that the reactants should have in order for the reaction to proceed from the reactants to products along a reaction coordinate The In k versus lT representation may not be linear so the dependence of k to T can be written as kzc mk EUW whmen11 gem gt Ea E39mRT and AzaTmem ln EazR 109 CHEM 6320 Fall 2005 4 Understanding the Phenomenological Rate Laws a b One should consider how reactants transform into products and how this path determines the kinetics of the chemical reaction Many reactions involve chemical intermediates and therefore can be written as Reactants gt Intermediates gt Products Example Nucleophilic substitution for tertiary alkyl halides CH33C C1TgtCH33 C LOU CH33C OH Intermediates are bound chemical compounds relative unstable and therefore reactive located at a minimum energy along the reaction coordinate not to be confused with transition state that is located at or in the close proximity of a maximum in energy along the reaction coordinate Chemical reactions can be divided into Elementary reactions reactions that occur in a single step and that do not involve any reaction intermediate Complex reactions reactions that does not occur in a single step and that do involve reaction intermediates Chemical kinetics deals with determining the sequence of elementary reactions by which a complex reaction occurs ie with determining the mechanism of the complex reactions 5 Kinetics of Elementary Reactions a Elementary reactions or processes are reactions that occur in a single step and that do not involve any reaction intermediate are processes that involve a limited a maximum of 34 number of chemical bonds being broken andor formed are denoted in some Physical Chemistry textbooks by gt for a forward reaction and lt for a reverse reaction These notations distinguish the elementary reactions or elementary steps from the other reactions 110 CHEM 6320 Fall 2005 b The rate law of an elementary reaction step or process can be deduced from the balanced chemical equation itself c The number of reactant molecules involved in an elementary chemical reaction is called molecularity The elementary reactions are termed unimolecular bimolecular and termolecular if they have one reactant two reactants or three reactants Unimolecular reactions A gt Products Bimolecular reactions A B gt Products Termolecular reactions A B C gt Products 6 Kinetics of CompleX Reactions 3 Kinetics of some special chemical reactions Reversible reactions reactions that occur in both directions k1 k l Assuming that both reactions are firstorder or are elementary with respect to A and assuming B0 0 B k1A k1B B A10 A k1 k1A k 1Alo 3 A 2 k1 k1 or A Alo Aleq gte k1k t Weq Alo Alo lll CHEM 6320 Fall 2005 Parallel reactions reactions in which one reactant is transformed in two or more products k1 B A k2 C Assuming that both reactions are firstorder or are elementary with respect to A AAOe k1k2tv1k1 V2 Cl k2 0693 The halflife for A I12 k1 k2 The activation energy Ea for the disappearance of A klEal k2Ea2 a k1 k2 where Em is the activation energy for the first reaction A gt B and Ea is the activation energy for the second reaction A gt C Consecutive reactions a series of reactions in which the product of one reaction is the reactant in another one ALBLC Assuming that both reactions are firstorder or are elementary and assuming B0 C0 0 k1A 2 A Aioe kl all k1A k2B k1 Aioe kl k2B k1 kt kt gt B 2 1 2 k2k1e e Alo dC kt kt k2B gt C1Ml dz k2 k1 lwo 112 CHEM 6320 Fall 2005 ClAla AlA137 BlA127 ClAn b The ratedetermining step If one step of the reaction mechanism is much slower than any other steps that step controls the overall reaction rate and is called ratedetermining step The presence of a ratedetermining step in the mechanism of a compleX reaction is not required though Working out the kinetics of a compleX reaction the final result is the rate law which is an algebraic eXpression containing one or more rate constants and concentrations of all reactant species involved in or prior to the rate determining step Kinetic data provide information only about the rate determining step and steps preceding it Example N02 g COg A NOg C02 g Twostep mechanism k1 N02 g N02 2 gt NOg N03 g k2 N03 g COg gt N02 g C02 g The rst step is much slower than the second step v1 ltlt v2 The rate of the overall reaction is given by the rate of the ratedetermining step v1 k1 N022 113 CHEM 6320 Fall 2005 k1 k2 k3 A B C gt D gt E F Example 391 dlcl 7 dDl Tr If k2 is slow relative to k3 then D will go on to E F as soon as it is formed so that the rate becomes rate If the first step is a rapid but unfavorable equilibrium 161 gtgt k1 then one can write 2 2 C k l AB and the rate becomes rate c The steadystate approximation If the concentration of a reaction intermediate is approximately constant over a period of time one can approximate that dIdt 0 which means that the rate of formation is equal with the rate of consumption for that intermediate This happens when the elementary steps in which the intermediate is consumed are faster and very little intermediate can build up This approximation dIdl 0 is called the steadystate approximation and is used often to simplify the mathematics of a kinetic model The steadystate assumption corresponds to a case in which the intermediate is so reactive that I E 0 The steadystate approximation is a poor approximation when the rate constants of the two steps are comparable ll4 CHEM 6320 Fall 2005 In A B Example 391 k C C D 2gt Products Assume C in reactive intermediate and apply steadstate approximation for its concentration dz gtC gt rate If the first step is ratelimiting ltgt k2D gtgt k4 gt rate If the second step is ratelimiting lt3 k2D ltlt 161 gt rate The first step is a preequilibrium d The principle of microscopic reversibility The same pathway that is traveled in the forward direction of a reaction will be traveled in the reverse direction since it affords the lowest energy barrier for either process Each reaction coordinate defines two reactions the forward one and the reverse one e The principle of detailed balance It states that when a compleX reaction is at equilibrium the rate of the forward process is equal to the rate of the reverse process for each and every step of the reaction mechanism f More comments If a proposed mechanism supports an experimentally determine rate law it doesn t mean that the reaction follow that mechanism Kinetics itself cannot prove a mechanism but can rule out possible mechanisms that are inconsistent with kinetics 115 CHEM 6320 Fall 2005 Kinetics indistinguishability refers to the case of two possible mechanisms providing the same rate law Extrakinetic evidence like spectroscopic evidence of one intermediate that appear in one mechanism but not in the other is needed to distinguish between the two mechanisms g Guidelines for deducing mechanisms from rate laws The following guidelines are helpful in deducing mechanistic information from a known rate law The more compleX the rate law the more information can be deduced All species in the numerator of a rate law must appear in the mechanism as reactants They will appear as many times in the mechanism before or during the ratedetermining step as the order to which they appear in the rate law Species missing from the numerator of the rate law occur in the mechanism after the ratedetermining step Fractional orders 12 32 etc indicate the dissociation of the molecule in question A simple rate law with a species having a negative order 1 2 etc implies a rapid equilibrium step involving that species before the ratedetermining step A compleX rate law having two or more terms in the denominator means that there is at least one intermediate chemical species and suggests the use of the steady state approximation The terms in the denominator of a compleX rate law give specific information about what the intermediates react with how they disappear The species in the numerator of a compleX rate law give information about how an intermediate is formed If the numerator of the rate law has a high power of any species that species must enter into the mechanism several times sequentially 116 CHEM 6320 Fall 2005 7 Catalysis a Catalysis is another way to increase the reaction rate other than increasing the temperature by making the reaction occur through a different mechanism that has lower activation energy lower rate constant A catalyst is a substance that participates in a chemical reaction but is not consumed in the process The catalyst provides a new mechanism by which reaction can occur When a catalyst is present the heat of reaction exothermicity or endothermicity of the reaction does not change gt The position of the chemical equilibrium does not change with the presence of the catalyst A catalyst lowers the activation energy for both the forward and reverse reaction Cgt it increases the reaction rate for both the forward and reverse reaction The reaction coordinate for an uncatalyzed chemical reaction is different than the reaction coordinate of a catalyzed chemical reaction Depending on the phase in which the catalyst is compared to the reactant and product one finds homogeneous catalysis same phase and heterogeneous catalysis different phases Enzymatic reactions are catalyzed biological processes that involve enzymes Enzymes are protein molecules that catalyze specific biochemical reactions The reactant molecule that the enzyme acts upon is called substrate The region of the enzyme where the substrate reacts is called active site The specificity of an enzyme depends on the geometry of the active site and the spatial constrains imposed by the rest of the enzyme ll7 CHEM 6320 Fall 2005 8 Transition State Theory a b Transition state theory also called activatedcomplex theory is a theory of the rate of elementary reactions The theory focuses on transient species called activated complex or transition state locate din the vicinity of the top of the barrier height activation energy of a reaction Deducing the rate law Consider an elementary reaction A B gt Products The rate law is given by v 2 2 kA B According to the activated compleX theory the model of the elementary reaction is a twostep process process here does not mean reaction but rather halfprocess A B AB gt P Where AB1 is the activated compleX The transition state activated compleX quantities are denoted by a double dagger sign I not 7 Another important assumption is that the reactants and the activated compleX are in equilibrium with each other and the equilibrium constant for this equilibrium is ABic ABic Ac B c A B Where 0quot is the standardstate concentration this is often 100 moldm3 The activated complexes are assumed to be stable Within a small region of Width 6 center at the barrier top According to the transitionstate theory the rate of the reaction is given by the rate of a unimolecular process with a rate constant kit that transform the activated compleX into products 039 1D I Kg ki ABi d l 118 CHEM 6320 Fall 2005 The rate constant kit is proportional to the frequency with which the activated complexes cross over the barrier top V0 Where the proportionality constant K is called transmission coefficient and is assumed to be 1 if no other information is available ki 10 2 VC Because ABi Ki A Bc gt VCK ABc Compare to v 2 2 kA B V i Ifc AB C i gt k VCKC k has units of concils 1 o The equilibrium constant written in terms of partition functions i ABic qi Vc 6 MB 61A VqB V Where 61A 13 and 611 are the partition functions of A B and ABi Assume that the motion of the reacting system over the barrier top is a onedimensional vibrational motion The vibrational kBT W C partition function qvib for this degree of freedom is qvib Write the partition function of the activated compleX as a product of the partition function of the motion of passing over the energy barrier and a partition function for the rest of degrees of freedom of the activated complex 1 o q qv1bqim c th 61A VqB V 1ch kBT qilht Vc kBT i c hc 61A VX6113 V hc Where Kit is the equilibrium constant for the formation of the transition state from the reactants but with one degree of freedom gtk ll9 CHEM 6970 Fall 2004 BriefReview onuantum Mechanics Origins of Quantum Mechanics 7 The classical physics predicts the precise trajectory of a particle and allows the translational rotational and vibrational modes of motion to be excited to any energy by controlling the applied force These conclusions agree to everyday experience but do not extend to individual atoms Classical mechanics fails when applies to transfers of very small quantities of energy and to objects of very small mass Some early experimental evidence showedthat several concepts of classical mechanics are invalid 7 In the beginning of the 1920 s it was well established that the light behaves as a wave in some experiments and as a stream of photons in others This was known as the Wavepar cle duality of the light In 192324 Louis de Broglie proposed the idea that the matter might also display wavelike properties under certain conditions Under those conditions similar equations should hold for the matter A particle of mass m moving with the speed v will exhibit a de Broglie wavelength given by A g g 7 Schrodinger equation is the equation for nding the wave function of a particle and come up based on idea that if the matter possesses wavelike properties there must be a wave equation that governs them VWMW A Waufundinn 7 Wave nature ofthe particles 7 The wavefunction or state function of a system evolves in time according to the time a PUJ 71Enlh at 39 dependent Schrodinger equation Iflk xj ih where Pxt Vxe 7 The solutions of timeindependent Schrodinger equation are called stationary state wave functions i Woo 72m 1X2 Vxtx7Etx 7 Rewrite the equation and generalize to three dimensions hi 62 62 62 j 2 2 2 wxyzVxyzvxyzEvXJJ 239quot 8x 8y 82 CHEM 6970 Fall 2004 7 Rewrite including the notation for the Laplacian operator 62 62 a 2 2 2 V2 Laplacian operator 6x By 62 h2 2 3 V Vxyz lIEl 2m 7 Rewrite including the notation for the Hamiltonian operator 2 2 V2 Vx y z H Hamiltonian operator m 3 ill1 E 1 Simple form of the Schrodinger equation 7 By solving the Schrodinger equation for the hydrogen atom one obtains the allowed energies eigenvalues and the wave function of the electron eigenfunctions which will supply the atomic orbitals 7 Schrodinger equation in terms of the spherical coordinates 2 2 h L2326 h 2 1 15in96i 2me r Br 6r 8 2me r 51116 r h2 l 6211 82 2m 47rgrlEW e r s1n m9 0 7 Solutions are given as zr t9 RrYt9 and Yt9 6 12 l 1 39 32 Rn r 7 2 J rle7rna0 131431 lZnKn l3l quot00 na 0 Rquot r depends on two quantum numbers 2r Lilfl1 are associated Laguerre polynom1als nao 12 m 211llmlgt lml W Y 6 4 HM P cost9e lml I I P x are called associated dx m Pquotquot39x1 x27 Legendre polynomials CHEM 6970 Fall 2004 The Bomroppenhelmer approxlm at on 7 allow separatlon of eleetronle and nuelearmotlon Wmdleenlnr vR Weleetmnre n R lmlclzuR rHamlltoman for H2 moleeule t 2 2 2 2 v v i vfvg EE g 2M 2m2 475mm 47an LeL L i 39M r 3 l K 475D 47an 47an AnsDR H 24 7 Due to larger mass of the nuelel eompared wth the eleetrons thls appromma o onslders the nuelel flxedln posltlons relatlye to the mohon of the eleetrons negleetlng the nuelearmotlon Thls results ln droppmg two an lunehe energy terms ln the expresslon for the Hamlltonl 2 2 A i y y E 2 we 475mm 47an e The Hamlltonlan beeomes 2 2 2 2 2 eieiii 4mm Answzg 47an WEEK 7 The Hamlltonlan ln atomle unlts H 1VfV 1 i 2 R rThe ease of H a one eleetron system 7 Avolds the lntereleetronle repulslon term that makes the equahon not to be solved exaetly eThe Schrodlnger equatlon ean be solved exactly for H Wthln the Bom Oppenhelmer BO apprommatlon H looks more llke hydrogen atom oner W V m CHEM 6970 Fall 2004 Potential energy surfaces 7 An important property in discussing beam results and calculating reaction cross sections 7 According to the BomOppenheimer approximation the wavefunction of a system composed of N nuclei and n electrons is written as a product of nuclear wavefunction depending on the positions of the nuclei and an electronic wavefunction depending on the positions of the electrons within a xed nuclear con guration This allows solving the Schrodinger equation for the wavefunction of the electrons alone at a speci c nuclear con guration Modifying the nuclear con guration and solving again the Schrodinger equation one get a different energy Energy 7 The case of diatomic molecules representing the electronic energy versus the interatomic distance one obtains a potential energy curve 7 The case of triatomic molecules there are three geometric parameters that de ne the molecular geometry and representing all three of them plus the energy requires a 4dimensional representation 7 Example the geometry of water is de ned by bond lengths and one angle or A 3 bond lengths or 1 bond length and 2 bond angles or 3 bond angles HA 0 HB 7 The minimum number of geometric parameters necessary to de ne the geometry of a system ofN atoms N Z 3 is 3N 6 7 When the potential energy depends on more than one single geometric parameter then use the phrase potential energy surface 7 The entire potential energy function cannot be plotted because the plotting is limited to 3 dimensions To overcome this one usually represents the energy function of only 2 geometric parameters keeping the otherothers at a xed value Such a plot is a cross sectional cut of the full potential energy surface 7 Example water 7 a 3dimensional plot VrOHA rOHB a constant versus rOHA and rOHB gives information about how the potential energy of water molecule changes when the bond lengths are varied at a constant angle a A series of crosssectional plots at different values of a give information of how the potential energy depends on the angle a 7 The representation of the potential energy surface for the molecules with more than 3 atoms is much more complicated CHEM 6320 Fall 2005 Chemical Bonding and Structure A Chemical Bonding 1 Review of Quantum Mechanics a Any moving entity behave like a wave with an associated wavelength xl given by 1221 p mv b The wave treatment of the matter is necessary for light particles like electrons atoms andor molecules c The properties and behavior of a particle are specified by the wavefunction which is obtained by solving the Schrodinger equation ah ah ah 8nbw 7 7 2 2 3x 3y 62 h where M is the mass of particle E is the total energy and Vis the potential energy d Wave equation can be solved exactly for the H atom and a few other systems but approximations are needed for multielectron systems and especially for polyatomic molecules w mwzo e For polyatomic molecules BornOppenheimer approximation allows the separation of the total wavefunction l molecular rR into an electronic wavefunction we rR and a nuclear wavefunction zN R LIJmolecular r R Z We r RN f The electronic wavefunction is obtained by solving the Schrodinger equation at a fixed nuclear arrangement 3211 3211 3211 87r2m 7 7 7 2 3x 3y 62 h where m is the mass of electron E is the total electronic energy and Vis the potential energy E Vv0 87 2 m h E Vv0 Rewrite using Laplacian operator sz CHEM 6320 Fall 2005 2 Rearrange 87 Z V2V 91ny m Introduce Hamiltonian operator the sum of the kinetic and potential energies in operator form H 1 E 1 g The solutions are equations that represent set of orbitals or eigenfunctions of increasing energy or eigenvalues orbitals that have an increasing number of nodes h Eigenfunctions must be Orthogonal Normalized i An important property of the electronic wavefunction is that 112 x yz gives the 2 Valence Bond VB Theory 3 Consider wave equation for each various possible Lewis electronic configuration canonical forms Total electronic wave function is a linear combination of these each with weighting factor We 2 Gill1 0212 0313 b Example for H2 3 Hybrid Orbitals and Hybridization a Hybrid orbitals are formed by miXing atomic orbitals the wavefunction for the hybrid orbitals is obtained as a linear combination of s and 7 atomic wavefunctions Vhybrid orbital 02sV2s 02px szx czpy szy CZPZ szz The s character of the hybrid orbital is given by 0202 The p character of the hybrid orbital is given by csz gt2 czpy 2 czpz 2 b A hybrid orbital that is x S character and 100 x 19 character is said to be an hybrid orbital CHEM 6320 Fall 2005 c An spy hybrid orbital is S character and 19 character d Hybrid orbitals give directionality for bonding due to a more effective overlap SP 2 SP 3 SP 00 SP e A hybrid atomic orbital can be written as spz As Z increases f The angle 6 between any two hybrid orbitals sp l and spg g Rule Carbon and other elements uses more 19 character in hybrid orbitals bonding to more electronegative atoms and less 19 character in bonding to less electronegative atoms Hybridization Examples a Carbon in methane Electronic configuration of C atom One electron is easily promoted from 2s to 2 p promotion energy is small CHEM 6320 Fall 2005 The carbon valence becomes four Four Sp3 hybrid orbitals are formed orbitals that are one part S and three parts 19 25 S character and 75 19 character 1 3 4 12 4gtltZSZp ZSTp S3p The in equation above represents the S character of the hybrid orbital and is not the 023 coefficient The shape of Sp3 hybrid orbitals The orientation of Sp3 hybrid orbitals b Nitrogen in ammonia Electronic configuration of N The promotion energy of one electron from 2S2 to 3S that would increase the valence to five is much higher than the energy of the two additional bonds so the valence remains 3 Lewis structure Nitrogen can use in principle the three 19 orbitals for bonding to three H 1S orbitals but that would put the nonbonding electron pair in 2S orbital situation that is not favored because it wastes S orbital character on nonbonding Nitrogen can use in principle three sz orbitals for bonding to three H 1S orbitals but that would make 90 angles between the CHEM 6320 Fall 2005 nonbonding electron pair in 2 p orbital and the three Sp2 orbitals also not desired situation The preferred case is making Sp3 orbitals three used for bonding to three H 1s orbitals and one for the lone pair c Carbon in di uoromethane The angles are not exactly 1095 degrees The carbon uses hybrid orbitals with less 19 character than sp3 to bind to H and more 19 character to bind to F A consequence Experimental d Carbon in ethylene Assume it comes from ethane from which two H were removed sp3 hybridization for C This structure is possible buy not energetically favored because contains two bad 039overlaps In reality each C is Sp2 hybridized CHEM 6320 Fall 2005 This structure contains one 039 C C bond with strong or good overlap and one 7 C C bond with weak or bad overlap e Carbon in acetylene Each C is hybridized An Sp3 hybridization would produce in an ineffective overlap f Carbon in allene The molecule is not planar The central C hybridized while the other two are hybridized This is a case of a cumulative double bond g Carbon in cyclopropane If C is Sp3 hybridized there would be a large angle strain 1095 60 495 Assuming C uses pure 1 orbitals to bond with the other C the angle strain would be only 30 the C H bonds would involve Sp hybrids so the H C H angle would be 180 a H H Make a compromise and the result is carbon atoms being sp33963 hybridized to form C C bonds 106 interorbital angle CHEM 6320 Fall 2005 What kind of hybrid orbitals are used to form the C H bonds 5 Resonance a The actual molecule is a resonance hybrid weighed average of contributing resonance forms Contribution of individual resonance forms increases As number of covalent bond increases minimizing unpaired electrons With minimization of charge separation unlike charges With placement of or on electropositive or electronegative atoms respectively Note All resonance forms must have the same spin multiplicity Resonance energy is the extra stability of resonance hybrid as a result of resonance possibility difficult to determine experimentally Note that resonance involves only the distribution of electrons in a delocalized TE system There is no movement of atomic nuclei implying a fixed geometry and fixed hybridization if rehybridization is necessary to allow for resonance this must be done before considering resonance forms No resonance for methyl amine N is sp3 hybridized CH3 ix39fH2 CHEM 6320 Fall 2005 Acetamide O 0 CH3C NH2 lt gt CH3CNH2 8 Resonance 0 8 hybrid CH3C NH2 d Another way to think about resonance forms is to consider all possible ways of distributing the given number of TE electrons in the TI system Example butadiene N WCJDCJJJDC HgccgHgccHgggg Resonance hybrid If resonance energy turns out to be more than predicted on the basis of simple resonance the system is aromatic eg a 4n 2 system like benzene 0H0 If resonance energy turns out to be less than predicted on the basis of simple resonance the system is antiaromatic eg a 4n system like cyclobutadiene DH Other examples NOE No CHEM 6320 Fall 2005 Pyridine Pyrrole h Steric inhibition of resonance Example of uninhibited resonance 0 O OO N N lt gt Example of inhibited resonance Nitro group should be in the ring plane to be able to write the resonance structures Rule of siX The can be steric hindrance in between atoms connected through 4 other atoms for a total of 6 atoms Resonance is even more inhibited for CHEM 6320 Fall 2005 B Bond Energy Polarity and Polarizability 1 Bond Length It is mostly a re ection of hybridization not just the normal notion of single gt double gt triple S172 S172 S173 S173 S172 S172 S172 S172 CH2CH CH2 CH3 CH2CH CH CH2 134 150 154 134 146 134 Bond Energy a Homolytic bond energy BE re ects stability of two radicals produced by bond scission PhCH2 CH3 gt PhCH2 CH3 AH 70 kcalmol CH3CH2 CH3 gt CH3CH2 CH3 AH 85 kcalmol b Heats of formation and heats of hydrogenation can be used to estimate relative stability between related compounds through a thermodynamic cycle Bond Polarity a Bond polarity in A B is based on the relative electronegativities of the attached atomsgroups b Molecular dipole is the vector sum of all the bond dipoles and electron pair dipoles Example NH3 has dipole moment CCl4 has no dipole moment c C is slightly electronegative of H in a normal Sp3 bond d The electronegativity increases Sp3 lt Sp2 lt Sp explaining the acidity of terminal acetylenes 8 8 R CEC H R CEC H CHEM 6320 Fall 2005 4 Acid Dissociation 1 H 111 a CH3COOH CH3COO H Cl3CCOOH Cl3CCOO H CH3CH2COOH CH3CH2COO H II is more acidic than I on account of the fact that the inductive effect of the electronwithdrawing Cl selectively stabilizes the conjugate base One typically explains the lower solution acidity of 111 over I on the notion that the extra methyl group is inductively electron donating However in gas phase III is more acidic than 1 Actually the intrinsic inductive effect of alkyl substitution is either electron withdrawing or electrondonating depending on the electron demand of the reaction center Why is 111 less acidic that I in solution The reason is the steric inhibition of solvation H 0 HIquot 0 I CH34lt H 1 O o The same dichotomy between solution and gasphase acidities applies to the alcohol series MeOH EtOH iPrOH t BuOH Example I BuOH is less acid in liquid phase and more acid in gas phase than MeOH t BuO has a larger volume than MeO so a better redistribution of the charge a better stability in gas phase CHEM 6320 Fall 2005 C Molecular Symmetry 1 Symmetry Elements 3 Identity E b nFold axis of symmetry Cn The axis with the highest value of n is called principal axis c Plane of symmetry 039 0V the plane of symmetry is parallel to a unique axis or to a principal axis ah the plane of symmetry is perpendicular to a unique axis or to a principal axis 0d the plane of symmetry bisects the angle between C2 axes that are perpendicular to a principal axis ad is a special type of a UV plane d Center of symmetry 139 e nFold rotation re ection axis of symmetry improper rotation S 7 Symmetry Operations 3 E no change b Cquot Rotation about the axis by 36014 degrees 6 Re ection in a plane 5 Re ection through the center of symmetry Sn Rotation about the axis by 36014 degrees followed by re ection through a plane perpendicular to the axis Point Groups 3 A set of symmetry operations constitutes a point group b Each point group consists of a number of symmetry elements c A symmetry element may have more than one symmetry operation associated with it d The total number of symmetry operations which may be greater than the total number of symmetry elements is called the order of the point group CHEM 6320 e Examples Group C1 CsCh C1Sz C2 C2v C3v C4 V Fall 2005 Symmetry Elements Examples E E0h E C2 E C22jV E C330V E C4 20V 20d E3C2 0111 E3C23039z39 CHEM 6320 Fall 2005 e Examples continuation D311 E3C333C2 01130 S3 E C4 4C2 0h 20V 20d D4h S4z39 D EC63C20h3av30d 6h S ai Dzd E3S433C2320d Td E4C33C23S460d Coov E3 Cooaooav D0011 E CooaooC23 Sooaooav 0113i f Some helpful rules The CW group contains an nfold axis and n 0V mirror planes The CM group contains an nfold axis and a mirror plane perpendicular to the nfold axis The Dquot group contains an nfold axis and n 2fold axes perpendicular to 14fold axis The Dnh group contains an nfold axis and n 2fold axes perpendicular to 14fold axis plus a plane perpendicular to 14fold axis The Dnd group contains same symmetry elements as Dquot plus n dihedral mirror planes 14 CHEM 6320 Fall 2005 4 Summary of the Shapes Corresponding to Different Point Groups I 6 m aquot V 6 w Cone 0 o 15

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#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

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

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

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

#### STUDYSOUP REFUND POLICY

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

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

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

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