Physical Organic Chemistry (CHEM. 2035)  Correlation of structure with reactivity; linear free energy relationships  energetics, kinetics and methods.

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Presentation on theme: "Physical Organic Chemistry (CHEM. 2035)  Correlation of structure with reactivity; linear free energy relationships  energetics, kinetics and methods."— Presentation transcript:

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2 Physical Organic Chemistry (CHEM. 2035)  Correlation of structure with reactivity; linear free energy relationships  energetics, kinetics and methods of establishing reaction mechanisms  the chemistry of reactive intermediates: carbocations, carbanions, free radicals, carbenes and nitrenes  pericyclic reactions  applications of Frontier Orbital Theory in electrocyclic reactions, cycloaddition and sigmatropic rearrangements  Structure Elucidation of Organic Molecules using molecular spectroscopy. 7/25/20221

3 Atomic Orbitals : Quantum Mechanics view Erwin Schrödinger derived a complex mathematical formula to incorporate the wave and particle characteristics of electrons. Wave behavior is described with the wave function ψ. The probability of finding an electron in a certain area of space is proportional to ψ 2 and is called electron density. Unit one: Structure, Reactivity and Mechanism 7/25/20222

4 1.1 Atomic Orbitals All s orbitals are spherical in shape but differ in size: 1s < 2s < 3s 2s2s angular momentum quantum number ( l = 0) m l = 0; only 1 orientation possible principal quantum number (n = 2) 7/25/20223

5 The Schrödinger equation specifies possible energy states an electron can occupy in a hydrogen atom. The energy states and wave functions are characterized by a set of quantum numbers. 7/25/20224 Quantum numbers are required to describe the distribution of electron density in an atom. There are three quantum numbers necessary to describe an atomic orbital.  The principal quantum number (n) – designates size  The angular moment quantum number ( l ) – describes shape  The magnetic quantum number (m l ) – specifies orientation

6 The angular moment quantum number ( l ) describes the shape of the orbital. The values of l are integers that depend on the value of the principal quantum number The allowed values of l range from 0 to n – 1.  Example: If n = 2, l can be 0 or 1. A collection of orbitals with the same value of n and l is referred to as a subshell. l 0123 Orbital designationspdf 7/25/20225 The principal quantum number (n) designates the size of the orbital. Larger values of n correspond to larger orbitals. The allowed values of n are integral numbers: 1, 2, 3 and so forth. The value of n corresponds to the value of n in Bohr’s model of the hydrogen atom. A collection of orbitals with the same value of n is frequently called a shell.

7 The magnetic quantum number (m l ) describes the orientation of the orbital in space. The values of m l are integers that depend on the value of the angular moment quantum number: – l,…0,…+ l 7/25/20226

8 Quantum numbers designate shells, subshells, and orbitals. 7/25/20227

9 Worked Example 3.8 Strategy Recall that the possible values of m l depend on the value of l, not on the value of n. What are the possible values for the magnetic quantum number (m l ) when the principal quantum number (n) is 3 and the angular quantum number ( l ) is 1? Solution The possible values of m l are -1, 0, and +1. Setup The possible values of m l are – l,…0,…+ l. Think About It Consult Table 3.2 to make sure your answer is correct. Table 3.2 confirms that it is the value of l, not the value of n, that determines the possible values of m l. 7/25/20228

10 The electron spin quantum number (m s ) is used to specify an electron’s spin. There are two possible directions of spin. Allowed values of m s are +½ and −½. A beam of atoms is split by a magnetic field. Statistically, half of the electrons spin clockwise, the other half spin counterclockwise 7/25/20229

11 1.2 Hybridization &1.3 Bonding in Carbon compound (Single, Double, Triple bonds)  The atomic orbitals are converted into a new set of orbitals called Hybrid Orbitals.  Hybridization: process of combining two or more atomic orbitals to create new orbitals(hybrids) that explain the geometry of the compound  When determining the types of hybrid orbitals that an atom has, you must to first determine the number of groups around that atom much like VSEPR theory “sp 3 ” Hybridization: All 4 Region Species 7/25/202210

12 Ground-state electron configuration (lowest energy arrangement) of an atom lists orbitals occupied by its electrons. Rules: 1. Lowest-energy orbitals fill first: 1s  2s  2p  3s  3p  4s  3d (Aufbau (“build-up”) principle) 2. Electrons act as if they were spinning around an axis. Electron spin can have only two orientations, up  and down . Only two electrons can occupy an orbital, and they must be of opposite spin (Pauli exclusion principle) to have unique wave equations 3. If two or more empty orbitals of equal energy are available, electrons occupy each with spins parallel until all orbitals have one electron (Hund's rule). 7/25/202211

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14 Hybridization of C in CH 4 Atomic C : 1s 2 2s 2 2p 2 Valence e’s Hybrid sp 3 orbitals: 1 part s, 3 parts p 7/25/202213

15 FORMATION OF CH 4 : Each hydrogen atom, 1s 1, has one unshared electron in an s orbital. The half filled s orbital overlaps head-on with a half full hybrid sp 3 orbital of the carbon to form a sigma bond. 7/25/202214

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17 16 sp 2 Orbitals and the Structure of Ethylene sp 2 hybrid orbitals: 2s orbital combines with two 2p orbitals, giving 3 orbitals (spp = sp 2 ). This results in a double bond. sp 2 orbitals are in a plane with120° angles Remaining p orbital is perpendicular to the plane 7/25/2022

18 17 Bonds From sp 2 Hybrid Orbitals Two sp 2 -hybridized orbitals overlap to form a  bond p orbitals overlap side-to-side to formation a pi (  ) bond sp 2 –sp 2  bond and 2p–2p  bond result in sharing four electrons and formation of C-C double bond Electrons in the  bond are centered between nuclei Electrons in the  bond occupy regions are on either side of a line between nuclei 7/25/2022

19 18 Structure of Ethylene H atoms form  bonds with four sp 2 orbitals H–C–H and H–C–C bond angles of about 120° C–C double bond in ethylene shorter and stronger than single bond in ethane Ethylene C=C bond length 134 pm (C–C 154 pm) 7/25/2022

20 19 1.9 sp Orbitals and the Structure of Acetylene C-C a triple bond sharing six electrons Carbon 2s orbital hybridizes with a single p orbital giving two sp hybrids – two p orbitals remain unchanged sp orbitals are linear, 180° apart on x-axis Two p orbitals are perpendicular on the y-axis and the z- axis 7/25/2022

21 20 Orbitals of Acetylene Two sp hybrid orbitals from each C form sp–sp  bond p z orbitals from each C form a p z –p z  bond by sideways overlap and p y orbitals overlap similarly 7/25/2022

22 21 Bonding in Acetylene Sharing of six electrons forms C  C Two sp orbitals form  bonds with hydrogens 7/25/2022

23 1.4 The breaking and forming of bond HETEROLYTIC BOND CLEAVAGES 7/25/202222

24 HETEROLYTIC BOND FORMATION 7/25/202223

25 HOMOLYTIC BOND BREAKING Homolytic cleavage of a bond does not result in the formation of charge but does result in the formation of unpaired electron intermediates called free radicals. 7/25/202224

26 HOMOLYTIC BOND FORMING 7/25/202225

27 Electronegativity Values 1.5 Factors influencing electron availability 1.5.1 Inductive effects  The differences in electronegativity between bonded atoms leads to a polarisation of the covalent bond such the most electronegative atom has a partial negative charge and the other atom has a partial positive charge.  The polarisation is transmitted to adjacent bonds, but the effect drops rapidly.  Inductive effects are caused by differences in electronegativity between bonded atoms, which leads to a polarisation of the bond. 7/25/202226

28 - Bond Polarisation and Inductive Effects -I Inductive Effects +I Inductive Effects ++ ++ ++ -- -- -- ++ ++ ++ -- -- -- 7/25/202227

29 Inductive Effects are Short Range In Contrast to Resonance Effects ++ ++ ++ ++ -- The polarised C-Cl bond transmits further polarisation through the  -bond framework, But effect drops off quickly… This proton is acidic. See Elimination reactions and alkene formation. 7/25/202228

30 We have two kinds of substituent inductive effect: Summary of Common Subsituents according to their Inductive effects are: Inductive EWG Groups Inductive EDG Groups (+I Groups) Inductive EDG Groups (+I Groups ) 7/25/202229

31 Inductive Effects and Carbocation Stability Methyl Carbocation Primary Carbocation Secondary Carbocation Tertiary Carbocation LEAST STABLE MOST STABLE The methyl groups have +I inductive effects. Carbon atom is electron deficient (only has 6 electrons in its outer valence). Thus, extra electron density is ‘pushed’ onto the carbocation, which stabilises the carbocation. Important when considering substitution reactions in part 4 of this course 7/25/202230

32 1.5.2 Mesomeric/Resonance effects (i)Resonance is the process whereby (generally) p-electrons can be delocalised by exchanging double bonds and single bonds. (ii)Resonance can be used to delocalise both lone pairs of electrons and cationic charges which are adjacent to double bonds. (iii)Delocalisation of positive and negative charges lead to relatively stable cations and anions, respectively. Benzene Bond Lengths 7/25/202231

33 The 6  -electrons are able to flow (or resonate) continually around the  - molecular orbital formed from the six p atomic orbitals on each of the 6 carbon atoms on the ring structure. This is represented by the two resonance structures below (which are identical or degenerate). This flow of electrons leads to a very stable electronic structure, which accounts for benzenes low reactivity relative to alkenes. This stability is referred to as the Resonance Stablisation Energy. Canonical structurtes The  -electrons are referred to as being conjugated. 7/25/202232

34 Resonance Imparts Stability to Anionic Structures (and Cationic Structures) CH 2 Replaced by C=O Relatively difficult to form Relatively easy to form No adjacent double bond to the oxygen lone pair 7/25/202233

35 The Resonance Arrow and its Physical Meaning The resonance arrow is not an equilibrium arrow The resonance arrow shows only the distribution of electrons. Thus, for the two degenerate structures above, the implication is that there is an even distribution of the two electrons between the two oxygen atoms, at all times. Experimentally it is found that both C-O bonds are the same length and are intermediate in length between the C-O single and double bond, as are the C-C bonds in benzene. 7/25/202234

36 Some Important Aromatic Resonance Structures Nitro Group: An Electron Withdrawing Group Repels Attracts Canonical structurtes 7/25/202235

37 Methoxy Group: An Electron Donating Group Canonical structurtes …Note in a reaction mechanism we would not show the lone pairs on the carbons carrying the –ve charge… 7/25/202236

38 Attracts Repels These resonance structures allow us to rationalise (and predict) reactivity 7/25/202237

39 Question 1: Resonances On structure A draw on the curly-arrows that will lead to the bonding in the resonance structure B. Then place charges on structure B. 7/25/202238

40 Answer 1: Resonances On structure A draw on the curly-arrows that will lead to the bonding in the resonance structure B. Then place charges on structure B. EDG Methoxy EWG Nitro 7/25/202239

41 Question 2: Resonances Cyclopentadiene can be deprotonated to the anion A. Anion A, the cyclopentadienyl anion, has 4 degenerate resonance structures. Complete the arrow pushing in C and identify structures D and E. Cyclopentadiene 7/25/202240

42 Answer 2: Resonances Cyclopentadiene can be deprotonated to the anion A. Anion A, the cyclopentadienyl anion, has 4 degenerate resonance structures. Complete the arrow pushing in C and identify structures D and E. Cyclopentadiene The cyclopentadienyl anion is an aromatic species, and is isoelectronic to benzene (i.e. has 6 p-electrons in a continuos cyclic array). All C-C bonds are the same length. 7/25/202241

43 Question 3: Resonances Benzyl bromide undergoes C-Br bond cleavage to generate the benzyl cation A. Through resonance the positive charge can be delocalised through the ring. Identify resonance structures C and D and draw in the curly-arrows. Benzyl Bromide 7/25/202242

44 Answer 3: Resonances Benzyl bromide undergoes C-Br bond cleavage to generate the benzyl cation A. Through resonance the positive charge can be delocalised through the ring. Identify resonance structures C and D and draw in the curly-arrows. Benzyl Bromide Delocalisation of charges (positive or negative) into an aromatic ring results in a relatively stable ion. 7/25/202243

45 Question 4: Resonances The 4-methoxybenzylic bromide A undergoes C-Br bond cleavage much more easily to generate the benzyl cation B, than does the C-Br bond in benzyl bromide in question 3. Indentify B and then draw in the curly arrows that lead to the resonance structure C. Comment on why A undergoes C-Br cleavage more readily than the parent benzyl bromide from Q3. 7/25/202244

46 Answer 4: Resonances The benzylic bromide A undergoes C-Br bond cleavage very easily to generate the benzyl cation B. Indentify B and then draw in the curly arrows that lead to the resonance structure C. The reason that the C-Br bond is cleaved so readily, is due to the positive charge being able to delocalise through the aromatic ring and onto the oxygen atom. i.e. the charge is spread out over many atoms leading to a stable electronic structure. 7/25/202245

47 1.5.3 Steric effects or Steric Hindrance Steric hindrance occurs when the large size of groups within a molecule prevents chemical reactions that are observed in related molecules with smaller groups. Groups that block the path from the nucleophile to the electrophilic atom produce steric hindrance This results in a rate differences or no reaction at all methyl halide ethyl halide isopropyl halide t-butyl halide 7/25/202246

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50 In a solution phase reaction, the solvent plays a large role in how the reaction will occur Solvents are categorized as either protic or aprotic Protic Solvents Protic solvents has a H bonded to a N or O – It is a H bonder – Examples: H 2 O, CH 3 OH, NH 3, etc – Solvent is attracted to the Nucleophile and hinders its ability to attack the electrophile 1.5.4 Effects of the medium/solvent effect 7/25/202249

51 Aprotic Solvents Use an aprotic solvent Solvates cations Does not H bond with anions (nucleophile free) Partial + charge is on inside of molecule Negative charge on surface of molecule (solvates cation) Examples include: – DMSO (dimethyl sulfoxide) – DMF (dimethyl formamide) – Acetone (CH 3 COCH 3 ) DMSO DMF 7/25/202250

52 Nucleophiles In the organic solvent phase, INVERSE relationship between basicity and nucleophilicity with a protic solvent Question… 7/25/202251

53 Nucleophiles Solvents can solvate the nucleophile –Usually this is NOT good because the nucleophile is “tied up” in the solvent and LESS REACTIVE. Ion-dipole interactions 7/25/202252

54 Nucleophiles Solvents can solvate the nucleophile (Methanol is a polar protic solvent.) 7/25/202253

55 Role of the Solvent In an S N 1, a carbocation and halide ion are formed – Solvation provides the energy for X - being formed – In S N 1 the solvent “pulls apart” the alkyl halide – S N 1 cannot take place in a nonpolar solvent or in the gas phase – Increasing the polarity of the solvent will INCREASE the rate of Rx if none of the REACTANTS are charged. – If reactants are charged it will DECREASE the rate. 7/25/202254

56 So…. In an S N 1 reaction, the reactant is RX. The intermediate is charged and is STABILIZED by a POLAR solvent A POLAR solvent increases the rate of reaction for an S N 1 reaction. (However, this is true only if the reactant is uncharged.) 7/25/202255 Role of the Solvent Role of the Solvent In S N 2 In an S N 2 reaction, one of the reactants is the nucleophile (usually charged). The POLAR solvent will usually stabilize the nucleophile. A POLAR solvent decreases the rate of reaction for an S N 2 reaction. (However, this is true only if the nucleophile is charged.)

57 Polar Aprotic Solvents include: – DMFN,N-dimethylformamide – DMSOdimethylsulfoxide – HMPAhexamethylphosphoramide – THFTetrahydrofuran – And even… acetone Polar Aprotic Solvents  Polar Aprotic Solvents do not H bond solvate cations well do NOT solvate anions (nucleophiles) well good solvents for S N 2 reactions 7/25/202256

58 1.6 Correlation of structures with reactivity. 1.6.1 Electron demand 7/25/202257

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61 1.6.2 The Hammett equation  The Hammett equation in organic chemistry describes a linear free-energy relationship relating reaction rates and equilibrium constants for many reactions involving benzoic acid derivatives with meta- and para-substituents to each other with just two parameters: a substituent constant and a reaction constant.organic chemistryfree-energy relationshipreaction rates equilibrium constantsbenzoic acidmeta- and para-substituents  The basic idea is that for any two reactions with two aromatic reactants only differing in the type of substituent, the change in free energy of activation is proportional to the change in Gibbs free energy.free energy of activationGibbs free energy 7/25/202260

62 T he basic equation is:  relating the equilibrium constant, K, for a given equilibrium reaction with substituent Requilibrium constant  K 0 constant when R is a hydrogen atom to the substituent constant σ which depends only on the specific substituent Rσ  and the reaction constant ρ which depends only on the type of reaction but not on the substituent used.ρ  Key principle: All reactions that correlate to the Hammett equation will use the same set of substituent constants.  That is, a structural modification will produce a proportional change in reaction rate based on the " values. 7/25/202261

63  The equation also holds for reaction rates k of a series of reactions with substituted benzene derivatives :reaction rates  In this equation k 0 is the reference reaction rate of the unsubstituted reactant, and k that of a substituted reactant.  A plot of log(K/K 0 ) for a given equilibrium versus log(k/k 0 ) for a given reaction rate with many differently substituted reactants will give a straight line.  This equation was developed and published by Louis Plack Hammett in 1937as a follow-up to qualitative observations in a 1935 publication.equationLouis Plack Hammett 7/25/202262

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