D. A. Evans Acid-Base Properties of Organic Molecules Chem 206 http://ww.courses.fasharvardedu/-chem206/ Articles on the Acidities of Organic Molecules Lowry Richardson: 3rd Edition, Chapter 3 Chemistry 206 Acids and Bases Advanced Organic Chemistry Equilibrium acidities in DMSO Solution", F. G Bordwell Acc.chem.Res.1988,21,456-463 Here is a web site containing Brodwell pKa data ecture Number 17 http://www.chem.wiscedu/areas/reich/pkatable/index.htm Acid-Base Properties of Organic Molecules ■ Problems of the Day Bronsted Acidity Concepts in the Activation of Organic Structures counterparts 2 and 4. (.Org. Chem. 1994, 59, 6A57 heir acyclic Explain why 1 and 3 are4 pKa units more acidic than a Medium Effects on Bronsted Acidity Substituent& Hybridization Effects on Bronsted Acidity Kinetic Thermodynamic Acidity of Ketones 3HNo 4 Me Kinetic Acidity. Carbon vs Oxygen Acids Tabulation of acid dissociation Constants in dmso Reading Assignment for this Lecture The thermodynamic acidities of phenol and nitromethane are both 10; however, using a common base, phenol is deprotonated 10 Carey& Sundberg: Part A; Chapter 7 times as fast. Rationalize Carbanions& Other Nucleophilic Carbon Species Equilibrium acidities in DMSO Solution", F.G. Bordwell pKa(H2O) Acc.Chem.Res.1988.21.456463 rel rate: 10*6 Matthew d. shair October 28. 2002 pKa(H2O) H3C
http://www.courses.fas.harvard.edu/~chem206/ http://www.chem.wisc.edu/areas/reich/pkatable/index.htm O H The thermodynamic acidities of phenol and nitromethane are both ~10; however, using a common base, phenol is deprotonated 10+6 times as fast. Rationalize H3C N O O H2C N O O O Base Base O O Me O O Et HN O O N Me O O Et H D. A. Evans Chem 206 Matthew D. Shair Monday, October 28, 2002 ■ Reading Assignment for this Lecture: Carey & Sundberg: Part A; Chapter 7 Carbanions & Other Nucleophilic Carbon Species Acid-Base Properties of Organic Molecules ■ Problems of the Day: Articles on the Acidities of Organic Molecules Chemistry 206 Advanced Organic Chemistry Lecture Number 17 Acid-Base Properties of Organic Molecules ■ Bronsted Acidity Concepts in the Activation of Organic Structures ■ Medium Effects on Bronsted Acidity ■ Substituent & Hybridization Effects on Bronsted Acidity ■ Kinetic & Thermodynamic Acidity of Ketones ■ Kinetic Acidity: Carbon vs. Oxygen Acids ■ Tabulation of Acid Dissociation Constants in DMSO "Equilibrium acidities in DMSO Solution", F. G. Bordwell. Acc. Chem. Res. 1988, 21, 456-463. Here is a web site containing Brodwell pKa data Explain why 1 and 3 are ~4 pKa units more acidic than their acyclic counterparts 2 and 4. (J. Org. Chem. 1994, 59, 6456) 1 2 3 4 Lowry & Richardson: 3rd Edition, Chapter 3 Acids and Bases "Equilibrium acidities in DMSO Solution", F. G. Bordwell. Acc. Chem. Res. 1988, 21, 456-463. rel rate: 1 rel rate: 10+6 pKa(H2O) ~10 pKa(H2O) ~10
D. A. Evans Acidity Concepts-1 Chem 206 Activation of Organic Molecules ■ Definition of ka Let H-x be any bronsted acid. In water ionization takes place ■ Base Activation H-X HOH H30+X base Nucleophile where HoH]=55.5 mol L H-baseo Ri where Keg: HH301X HX叶HOH Since [ HOH is, for all practical purposes, a constant value, the acid dissociation constant Ka is defined without regard to this entity. e. g pKa, describes quantitatively a molecule's propensity to act as an acid, ie to release a proton. H"+X- where H=H30 Medium effects Hence Structural effects (influence of substituents R1) From the above definitions, Ka is related to Keg by the relation Acid activation Ka(H-x)=55.5 Keq(H-x acid(protic or lewis acid R2 号 electrophile|■ Autoionizationofwater IOH+ HOH 5 H30+HO" X=e.gO, NR Eq C ■ The Aldol Example Since pKa is defined in the following equation base catalys Ka=-log1o[Kal The pKa of HOH iS 15.7 Keep in mind that the strongest base that can exist in water is HO Lets now calculate the acid dis constant for hydronium ion H3o H20 0 Ca 10*5 Activation SiMe obviously cid catalysis Ka[HOH]X Keg hence Ka=55.5 pKa=-log1o Ka=-1.7 The strongest acid that can exist in water is H3O
R1 R2 X C R1 R2 R3 H O R M R H R O O R SiMe3 R H R O M O R R H R O C R1 R2 R3 R1 R2 X acid R R O R OH R R O R O M R R O R O SiMe3 HOH H–X H3O + HOH H–X HOH [H3O + ] [X– ] [H–X] [HOH] H2O [H+ ] [X– ] H3O + H + H3O + X– HO– X– H3O + (B) H2O (C) (A) D. A. Evans Acidity Concepts-1 Chem 206 Activation of Organic Molecules base - H-base pKa , describes quantitatively a molecule's propensity to act as an acid, i.e. to release a proton. acid (protic or lewis acid) Nucleophile Electrophile X = e.g. O, NR ... - Medium effects - Structural effects (influence of substituents R1 ) ■ Base Activation ■ Acid Activation + ■ The Aldol Example + + base acid Let H–X be any Bronsted acid. In water ionization takes place: + + where Keq = where [HOH] = 55.5 mol L-1 Since [HOH] is, for all practical purposes, a constant value, the acid dissociation constant Ka is defined wilthout regard to this entity. e.g. + where H+ = H3O + Hence [H–X] Ka = From the above definitions, Ka is related to Keq by the relation: Ka (H–X) = 55.5 Keq(H–X) ■ Autoionization of water + + Keq = 3.3 X 10–18 From Eq C: Ka = 55.5 Keq = 55.5(3.3 X 10–18) Hence Ka = 1.8 X 10–16 Since pKa is defined in the following equation: pKa = – log10 [Ka] The pKa of HOH is + 15.7 Keep in mind that the strongest base that can exist in water is HO– . ■ Definition of Ka pKa = – log10 Ka = –1.7 Ka = [HOH] x Keq obviously: Keq = 1 + Lets now calculate the acid dissociation constant for hydronium ion. + Ka = 55.5 The strongest acid that can exist in water is H3O + . hence base catalysis acid catalysis Ca 10+6 Activation
D. A. Evans Acidity Trends for Carbonyl Related Compounds Chem 206 I The Gibbs Relationship a Medium Effects on the pKa of HOH AG°= RTIo K HOH pKa Medium 23RT=14 The gas phase ionization of HOH is HOH or△G°=-23RTg1K att= endothermic by 391 kcal/mol !! DMSO in kcal. mor1 △G298=-14l0g1 279(est) Vacuum Representative pKa Data △G298=14pk≈14pK Substrate HOH Hence, pKa is proportional to the free energy change 7. 7.7 0 290 13.7 C6H5OH 18.0 99 8.1 2. 8 kcal/mol Reaction coordinate O2N-CH3 17.2 7.2 I Medium Effects Consider the ionization process Ph-C-CH 246 7.6 H-A solvent= A:+ solvent(H) The change in pKa in going from water to DMSO is increasingly diminished as the conjugate base becomes resonance stabilized(Internal solvation) In the ionization of an acid in solution, the acid donates a proton to the medium. The more basic the medium, the larger the dissociation equilibrium. The ability of the medium to stabilize the conjugate base also plays an important role in the promotion of ionization. Let us consider two solvents, HOH and DMSO and the performance of ubstrate HOH these solvents in the ionization process. 160 2.1 The Protonated Solvent Conjug Base Stabiliz 0→H—A 164 DMSo HO-S+ No H-bonding Capacity 133 As shown ough HOH can stabilize anions via H-bonding, DMSO cannot NCCN 11.1 11.2 will show a greater propensity to dissociate in HOH. As illustrated below the acidity constants of water in HOH, DMSO and in a vacuum dramatically reflect this trend
31.2 14.7 29.0 18.0 17.2 24.6 17 10.0 9.9 15.3 7.0 15.7 18.1 16.0 13.3 8.9 16.4 13.3 11.1 11.2 H A H O + H H S + Me Me HO O H H A – HA A EtO OEt O O Me Me O O HOH C6H5OH NC CN HSH MeOH O2N–CH3 Ph C O CH3 DMSO DMSO HOH HOH HOH DMSO D. A. Evans Chem 206 D G° = - RT ln K or D G° = – 2.3 RT log10 K 2.3 RT = 1.4 at T = 298 K in kcal × mol-1 D G°298 = - 1.4 log10 Keq with pK = – log10 D G°298 = 1.4 pKeq » 1.4 pKa K Hence, pKa is proportional to the free energy change Keq pKeq D G° 1 10 100 0 - 1 - 2 0 - 1.4 - 2.8 kcal/mol Energy Reaction coordinate D G° ■ The Gibbs Relationship Consider the ionization process: + solvent + solvent(H+ A: ) – In the ionization of an acid in solution, the acid donates a proton to the medium. The more basic the medium, the larger the dissociation equilibrium. The ability of the medium to stabilize the conjugate base also plays an important role in the promotion of ionization. Let us consider two solvents, HOH and DMSO and the performance of these solvents in the ionization process. The Protonated Solvent Conjug. Base Stabiliz. Water DMSO No H-bonding Capacity As shown above, although HOH can stabilize anions via H-bonding, DMSO cannot. Hence, a given acid will show a greater propensity to dissociate in HOH. As illustrated below the acidity constants of water in HOH, DMSO and in a vacuum dramatically reflect this trend. ■ Medium Effects HOH pKa Medium 15.7 31 279 (est)** Vacuum ** The gas phase ionization of HOH is endothermic by 391 kcal/mol !!! Substrate D pKa 15.5 7.7 13.7 8.1 7.2 7.6 ■ Medium Effects on the pKa of HOH ■ Representative pKa Data Acidity Trends for Carbonyl & Related Compounds The change in pKa in going from water to DMSO is increasingly diminished as the conjugate base becomes resonance stabilized (Internal solvation!). Substrate D pKa 2.1 3.1 0 4.5
D. A. Evans Acidity Trends for Carbonyl Related Compounds Chem 206 Substituent Effects Electrons in 2S states see"a greater effective nuclear charge than electrons in 2P states Electronegativity e.g. Compare Carboxylic Acids vs. Ketones This becomes apparent when the radial probability functions for S and P-states are examined: The radial probability functions for the hydrogen atom s& p states are shown below 0- more stabile 100% 100% enolate because electronegative than C pKA=4.8 pKA≈19 (DMSO) O Orbital 2 S Orbital Hybridization -S-character of carbon hybridization 〓2 P Orbital sp-orbitals 25%s-character sp-orbitals 33%S-character sp-orbitals 50%S-character 3 S Orbital O> 3 POrbital C∧ Carbon acids S-states have greater radial penetration due to the nodal properties of the wave function. Electrons in s states see a higher nuclear charge The above observation correctly implies that the stability of nonbonding electron pairs is directly proportional to the of S-character in the doubly occupied orbital Hybridzation Carbanions (DCsP3 (DCsP2 DCsp Bond angle 180° 20° ≈120 Least stable Most stable pK(DMSO) 23 Carbenium ions Most stable <t Least stable The above trends indicate that the greater the of S-character at a give atom, the greater the electronegativity of that atom
sp3 -orbitals 25% s-character sp2 -orbitals 33% s-character sp-orbitals 50% s-character CSP3 CSP2 CSP CSP2 1 S Orbital 2 S Orbital 3 S Orbital 2 S Orbital 2 P Orbital 3 P Orbital CSP3 CSP R C O – CH2 H R R R R R H C O O – R H R R R C C O CH2–H H H H (DMSO) R H H C H H C O O H (DMSO) D. A. Evans Acidity Trends for Carbonyl & Related Compounds Chem 206 Substituent Effects Electronegativity e.g. Compare Carboxylic Acids vs. Ketones pKA = 4.8 pKA » 19 Carboxylate ion more stabile than enolate because O more electronegative than C Hybridization - S-character of carbon hybridization Remember: Hybridzation pKa (DMSO) Bond Angle sp sp2 » sp2 sp3 180° 120° 109° 23 32 » 39 50 » 120 Carbon Acids Carbenium ions Carbanions Most stable Least stable Least stable Most stable S-states have greater radial penetration due to the nodal properties of the wave function. Electrons in s states see a higher nuclear charge. The above observation correctly implies that the stability of nonbonding electron pairs is directly proportional to the % of S-character in the doubly occupied orbital. Electrons in 2S states "see" a greater effective nuclear charge than electrons in 2P states. Å Radial Probability 100 % This becomes apparent when the radial probability functions for S and P-states are examined: The radial probability functions for the hydrogen atom S & P states are shown below. 100 %Radial Probability Å The above trends indicate that the greater the % of S-character at a given atom, the greater the electronegativity of that atom
Evans Acidity Trends for Carbonyl Related Compounds Chem 206 Hybridization Vs Electronegativity Substituent Effects There is a linear relationship between %S character Pauling electronegativity I Alkyl Substituents on Localized Carbanions are Destabilizing: Steric hinderance of anion solvation JAcs1975,97,190) PhSOz- CH-H 29 31.1 383 a Heteroatom-Substituents:- 1st row elements of periodic table pKA (DMSO) 253035 Inductive stabilization versus %S-Character PhSO2-CH-OCH3 30.7 Lone Pair Repulsion l vS+M-Effect) There is a direct relationship between %S character H PhSo2-CH-OPh 27.9 PhSo2-CH-NMe3 19.4 Inductive Stabilization H Seomx a Heteroatom-Substituents:-2nd row elements of periodic table c6H(4 Strong carbanion stabilizing effect pKA ( DMSO) PhCC-H(29) PhSo2-CH-H PhSO2- CH-SO2Ph 12.2 PhSoz CH-SPh 20.5 PhSo2-CH-PPh2 20.5
2 2.5 3 3.5 4 4.5 5 Pauling Electronegativity 20 25 30 35 40 45 50 55 % S-Character CSP3 CSP2 CSP NSP3 NSP2 NSP 25 30 35 40 45 50 55 60 Pka of Carbon Acid 20 25 30 35 40 45 50 55 % S-Character CH4 (56) C6 H6 (44) PhCC-H (29) S S H H S S Me H PhSO2-CH-OCH3 H PhSO2 -CH–H H PhSO2 -CH–Me H PhSO2-CH-OPh H PhSO2-CH-NMe3 H PhSO2-CH-H H PhSO2-CH-SPh H PhSO2-CH-SO2Ph H PhSO2-CH-PPh2 H D. A. Evans Acidity Trends for Carbonyl & Related Compounds Chem 206 There is a linear relationship between %S character & Pauling electronegativity Hybridization vs Electronegativity There is a direct relationship between %S character & hydrocarbon acidity Substituent Effects ■ Alkyl Substituents on Localized Carbanions are Destabilizilng: Steric hinderance of anion solvation pKA (DMSO) 29 31 pKA (DMSO) 31.1 38.3 Compare: (JACS 1975, 97, 190) Inductive Stabilization versus Lone Pair Repulsion (-I vs +M -Effect) pKA (DMSO) 30.7 27.9 19.4 Inductive Stabilization ■ Heteroatom-Substituents: - 1st row elements of periodic table ■ Heteroatom-Substituents: - 2nd row elements of periodic table pKA (DMSO) 29 20.5 12.2 Strong carbanion stabilizing effect 20.5 pKA (DMSO)
D. A. Evans Acidity Trends for Carbonyl Related Compounds Chem 206 a Carbanion Stabilization by 2rd-Row Atoms: SR, SO2R, PR3 etc a Conjugative Stabilization of Conjugate Base 17.2 pKA(DMSO) S-CH H3C-S-CH 18.2 (DMSO) 26.5 H CH3 (JAcS1976,98,7498;:JAcs1977,99,5633:MAcs1978.100,200 C-C≡N C=C=N 31.5 The accepted explanation for carbanion stabilization in 3rd row elements is delocalization into vicinal antibonding orbitals For efficient conjugative stabilization, rehybridization of carbanion orbital from nsp3 to np is required for efficient overlap with low-lying t-orbital of stabilizing group. However, the cost of rehybridization must be considered sXo(empty I Stereoelectronic Requirement for Carbanion Overlap nolization of Carbonyl Stereoelectronic Requirements: The a-CH bond must be able to overlap with C-O This argument suggests a specific orientation requirement. This has been noted: Anti (or syn) periplanar orientation of Carbanion-orbital and orbital mandatory Rates for deprotonation with n-BuLi pKA(DMSO) ZIshe He Ha=8.6 JAcs1974,96,1811) 5.2 Ph3C-H He: Ha=30 Acs1978100,200) C-H
S He Ha H O O H He : Ha = 30 H Ph3C–H O O H S S Ha He Me Me H H Hb Hc R C O CH3 R O Hc Hb R O – Ha C S X C H H C N C H H C N P CH3 Ph Ph Ph C H H C O CH3 C H H C O CH3 S S H H H3C S CH3 O O S CH3 Me Me H3C S CH3 O C H H NO2 C H H N O O D. A. Evans Acidity Trends for Carbonyl & Related Compounds Chem 206 ■ Carbanion Stabilization by 2rd–Row Atoms: SR, SO2R, PR3 etc + 18.2 (DMSO) + 22.5 31 31 35 The accepted explanation for carbanion stabilization in 3rd row elements is delocalization into vicinal antibonding orbitals (JACS 1976, 98, 7498; JACS 1977, 99, 5633; JACS 1978, 100, 200). Cn S–Xs* E Cn (filled) S–Xs* (empty) This argument suggests a specific orientation requirement. This has been noted: Anti (or syn) periplanar orientation of Carbanion-orbital and s* orbital mandatory for efficient orbital overlap. He : Ha = 8.6 (JACS 1978, 100, 200) (JACS 1974, 96, 1811) pKA (DMSO) 17.2 26.5 31.5 For efficient conjugative stabilization, rehybridization of carbanion orbital from nsp3 to np is required for efficient overlap with low-lying p*-orbital of stabilizing group. However, the cost of rehybridization must be considered. ■ Conjugative Stabilization of Conjugate Base ■ Stereoelectronic Requirement for Carbanion Overlap: Enolization of Carbonyl Compounds pKA 5.2 C-H acidity not detectable pKA (DMSO) 31.5 47.7 Rates for deprotonation with n-BuLi Stereoelectronic Requirements: The a-C-H bond must be able to overlap with p* C–O – Ha + base p* C–O
D. A Evans Phenol Acidity: An Analysis of Resonance& Inductive Effects Chem 206 ■ Phenol Acidity a Is the benzene ring somehow special. i. e "larger resonance space Acetone enol acetone acetone enol Keg 10 a=109H2C his topic has a number of take-home lessons. Most importantly, isis a useful construct on which to discuss the role of FG's in influencing the acidity of this The surprising facts is that the acetone enol has nearby the same pKa as phenol How does one analyze the impact of structure on pKa of a weak Hence, the answer to the above question is no! acid ( pKa>0)? a How important are inductive effects in the stabilization of C6Hso ■ The Approach: vmm时x For equilibria such as that presented X-oH pKa(H2O) above, analyze the effect of stabilizing X一OH CH3-OH 155 more energetic constituent which △G° As the electronegativity of X increases this case is the conjugate base the acidity of X-OH increases CF3CH2-OH 12 Cl-OH 7.5 a Why is phenol so much more acidic than cyclohexanol? pKa of X-OH This argument suggests that the acidity of acetone enol is largely due to inductive stabilization not resonance H pKa(H2O)=10 D HOCI (7.5) H pKa(H2O)=17 acetone enol (10.9) Loudon(pg 730): The enhanced acidity of phenol is due largely to stabilization of its conjugate base by resonance phenol ( 10.0) 0→0 from previous discussion, A 298=-1.4 Log10 Keq =1.4 pKeq AG(stab)=1.4(Pkaphenol-pKa cyclohenanod=1. 4(-7)=9.8 kcal/mol a of X-oh
2 2.2 2.4 2.6 2.8 3 3.2 Electronegativity of X 6 8 10 12 14 16 pKa of X–OH HOCl (7.5) HOH (15.7) acetone enol (10.9) phenol (10.0) ■ Why is phenol so much more acidic than cyclohexanol? ■ The Approach: ■ Is the benzene ring somehow special. i.e "larger resonance space." ■ Acetone enol: How important are inductive effects in the stabilization of C6H5O – ■ ? Me H OH H2C O – 2C Me X O – X OH CH3 OH CF3CH2 OH Cl OH X OH Me Me O O FG O – O – FG O FG OH O – O – O OH FG OH OH O – ■ Phenol Acidity: D. A. Evans Phenol Acidity: An Analysis of Resonance & Inductive Effects Chem 206 + H+ This topic has a number of take-home lessons. Most importantly, is is a useful construct on which to discuss the role of FG's in influencing the acidity of this oxygen acid. ■ How does one analyze the impact of structure on pKa of a weak acid (pKa > 0) ? + solvent(H+ ) DG° Energy (1) DG° For equilibria such as that presented above, analyze the effect of stabilizing (or destabilizing) interactions on the more energetic constituent which in this case is the conjugate base. DG° + H+ pKa (H2O) = 10 + H pKa (H2O) = 17 + DG° Loudon (pg 730): "The enhanced acidity of phenol is due largely to stabilization of its conjugate base by resonance." – – – DG° (stab) = 1.4(Pkaphenol – pKacyclohenanol) = 1.4(-7) = 9.8 kcal/mol from previous discussion, D G˚298 = –1.4 Log10 Keq = 1.4 pKeq + H (1) + acetone acetone enol acetone enolate Keq = 10-8 pKa = 10.9 The surprising facts is that the acetone enol has nearly the same pKa as phenol. Hence, the answer to the above question is no! Consider the following general oxygen acid X–OH where X can only stabilize the conjugate base through induction: + H+ pKa(H2O) 15.5 12.4 7.5 As the electronegativity of X increases the acidity of X–OH increases. If you take the calculated electronegativity of an SP2 carbon (2.75) you can see that there is a linear correlation between the electronegativity of X and the pKa of X–OH. This argument suggests that the acidity of acetone enol is largely due to inductive stabilization, not resonance
D A. Evans Weak Acids: Impact of Structure on Acidity Chem 206 a The General Reaction: lonization of a weak acid (pKa s o) Case ll: Carboxylic Acids vs Ketones H solvent H-C- Ka=4.8 Ka~19 electronegative than C X= NH(amide) Case IV: Carboxy lic Acids, Esters, Amides Ketones X= CH2(Ketoneester) R=CR3 0 0 R=NR2 Eto-C a The Question: How does one analyze the impact of structure on pKa? CH2-H CH2-H CH2-H ■ The Approach pKa-26 Ka~30 Ka~34 pKa>34Me2N H3c but (0-?) Cl 0 Carboxylate ■ Resonance effect c-c stabilized by increased H electron-withdrawing CCl3 group he degree to which substituent X Ka=48 pKa=0.6 contributes"electron density into enolate represents a destabilizing interaction Trend: 0->Me 2N>OEt Case l: Carboxylic Acids: Inductive Effects Carbon Hybridization Resonance donation dominates inductive electron withdrawal as indicated by the data H-C-C-c→Hc〓c-c Carboxylate HC三C stabilized by increased Substituents on the a-carbon: Stabilization by either resonance induction or both is observed SP-hybridized carbon 0 pKa=4.9 Ph-C-CH2 CH3 Ph-C-CH2OCH3 Ph-C-CH2Ph Ph-C-CH2SPh pKa= 24.4 pKa= 22.9 a=177 Ka=17.1
■ The Approach: ■ Resonance Effect: The degree to which substituent X: "contributes" electron density into enolate represents a destabilizing interaction: X C O – CH2 Resonance donation dominates inductive electron withdrawal as indicated by the data. ■ – + ●● R C O – CH2 C O O – R C C O CH2–H H H H C H H H C OH O O – R X C CH2 O – X R C O CH2–H Ph C O CH2CH3 C CH2OCH3 O Ph Ph C O CH2Ph C CH2SPh O Ph R X O – O R XH C C O O – Cl Cl Cl HC C C O – O EtO C O CH2–H C CH2–H O Me C CH2–H O Me2N – O C O CH2–H Cl C Cl Cl C OH O C C O OH H C R X H O C C O OH H H H C C O OH H H C H H H Trend: O– > Me2N > OEt ●● ■ Inductive Effect: OEt > Me2N > H3C but (O–?) In this series of compounds, there are two variables to consider: The Analysis: pKa ~ 26 pKa ~ 30 pKa ~ 34 pKa > 34 < 40 Case IV: Carboxylic Acids, Esters, Amides & Ketones: pKa = 4.8 pKa ~ 19 Carboxylate ion more stabile than enolate because O more electronegative than C Case III: Carboxylic Acids vs Ketones: pKa = 4.9 pKa = 1.9 Case II: Carboxylic Acids: Inductive Effects & Carbon Hybridization Carboxylate ion stabilized by increased electron-withdrawing SP-hybridized carbon Carboxylate ion stabilized by increased electron-withdrawing CCl3 group. pKa = 4.8 pKa = 0.6 Case I: Carboxylic Acids: Inductive Effects ■ The Question: How does one analyze the impact of structure on pKa ? R = NR2 R = OR R = CR3 X = O (carboxylic acid) X = CH2 (Ketone/ester) X = NH (amide) + solvent(H+ ) DG° DG° Energy Variables: ■ The General Reaction: Ionization of a weak acid (pKa 0) + solvent(H+ + solvent ) D. A. Evans Weak Acids: Impact of Structure on Acidity Chem 206 Stabilization by either resonance, induction, or both is observed: Substituents on the a-carbon: pKa = 24.4 pKa = 22.9 pKa = 17.7 pKa = 17.1 For equilibria such as that presented above, analyze the effect of stabilizing (or destabilizing) interactions on the more energetic constituent which in this case is the conjugate base
D A. Evans Acidity of Carboxylic Acids, Esters, anf Lactones: Anomeric Effects Again? Chem 206 I Conformations: There are 2 planar conformations I Hyperconjugation: Let us now focus on the oxygen lone pair in the hybrid orbital lying in the sigma framework of the C=O plane ()Conformer () Conformer Conformer -o In the(z)conformation this lone pair is aligned to overlap Specific Case AG°=+2 kcal/mol withσ·C-0 Fomic Acid H The(E conformation of both acids and este stable by 2-3 kcal/mol. If (E)Conformer this equilibrium were governed only by ster one would predict that the (E conformation of formic acid would be m (H smaller than =O) Since this is not the case. there are electron which must also b :o In the(E)conformation this considered. These effects will be introduced shortly lone pair is aligned to overlap with C-R GC-R Rotational Barriers: There is hindered rotation about the =c-or bond These resonance structures sugges Since o*C-o is a better acceptor than GC-R hindered rotation about =c-or bond This is indeed observed (where R is a carbon substituent) it follows that the(z) conformation is stabilized by this interaction R Lone pair orientation Impact on pKa(DMSO Rotational barriers are 10 kcal/mol AG°~23 See Bordwell, J. Org. Chem. 1994, 59, 6456-6458 This is a measure of the strength of Meldrum's Acid the pi bond. 00 CH3CH2 Me a Lone Pair Conjugation: The oxygen lone pairs conjugate with the C=O pKa= 25.2 pKa= 15.9 pKa=7.3 - -o The filled oxygen p-orbital interacts with pi (and pi) A C=0 to form a 3-centered 4-electron bonding system Is this a dipole effect? see bordwell SP2 Hybridization a Oxygen Hybridization: Note that the alkyl oxygen is Sp2. Rehybridization pKa= 24.5 is driven by system to optimize pi-bonding Houk,JAcS1988,110,1870 supports the dipole argument E(re)=0 E(rel)=+3.8 kcal
O – O R' R + + + + + R O R' O O O H H R O R O O O R R O C O R R C R O R O R O R O R O O O Et CH3CH2 C O R O R O O O R HN O O N Me O O Et H O O O O Me Me O O O O Me Me O R Me O O Me O R O O R' R H O H O C O O R R R O R' O Lone pair orientation & Impact on pKa (DMSO) Since s* C–O is a better acceptor than s* C–R (where R is a carbon substituent) it follows that the (Z) conformation is stabilized by this interaction. (E) Conformer In the (E) conformation this lone pair is aligned to overlap with s* C–R. s* C–R s* C–O In the (Z) conformation this lone pair is aligned to overlap with s* C–O. (Z) Conformer ■ Hyperconjugation: Let us now focus on the oxygen lone pair in the hybrid orbital lying in the sigma framework of the C=O plane. ■ Oxygen Hybridization: Note that the alkyl oxygen is Sp2. Rehybridization is driven by system to optimize pi-bonding. The filled oxygen p-orbital interacts with pi (and pi*) C=O to form a 3-centered 4-electron bonding system. SP2 Hybridization ■ Lone Pair Conjugation: The oxygen lone pairs conjugate with the C=O. Rotational barriers are ~ 10 kcal/mol This is a measure of the strength of the pi bond. barrier ~ 10 kcal/mol DG° ~ 2-3 kcal/mol Energy These resonance structures suggest hindered rotation about =C–OR bond. This is indeed observed: ■ Rotational Barriers: There is hindered rotation about the =C–OR bond. The (E) conformation of both acids and esters is less stable by 2-3 kcal/mol. If this equilibrium were governed only by steric effects one would predict that the (E) conformation of formic acid would be more stable (H smaller than =O). Since this is not the case, there are electronic effects which must also be considered. These effects will be introduced shortly. DG° = +2 kcal/mol Specific Case: Formic Acid (Z) Conformer (E) Conformer ■ Conformations: There are 2 planar conformations. D. A. Evans Acidity of Carboxylic Acids, Esters, anf Lactones: Anomeric Effects Again? Chem 206 •• •• •• •• pKa ~ 30 pKa = 25.2 pKa = 24.5 pKa = 20.6 pKa = 15.9 pKa = 7.3 See Bordwell, J. Org. Chem. 1994, 59, 6456-6458 E(rel) = 0 E(rel) = +3.8 kcal Is this a dipole effect? See Bordwell Meldrum's Acid Houk, JACS 1988, 110, 1870 supports the dipole argument
D. A. Evans Kinetic Thermodynamic Acidity of Ketones Chem 206 a Kinetic Acidity: Rates of proton removal Consider enolization of the illustrated ketone under non-equilibrating conditions Kinetic Equilibrium Ratios of Enolates Resulting from Enolization with LDA Subsequent Equilibration HB ①Me KA K At Kinetic Ratios Ratios Kinetic Ratios Equilibrium Ratios LINR 87)(13) Equilibrium tic acidity refers to the rate of proton removal. e.g. k a vS kb. For example, (14 99)(1) ing the above energy diagram you would say that Ha has a lower kinetic acidity HB. As such, the structure of the base (hindered vs unhindered) employed Kinetic Ratios Equilibrium lays a role in determining the magnitude of k a and k b. For the case shown above, Ratios AG#A will increase more than a as the base becomes more hindered since the proton H a resides in a more sterically hindered environment. The example shown below shows the high level of selectivity which may be achieved with the sterically a Note that alkyl substitution stabilizes the enolate(Why??). This effect hindered base lithium diisopropylamide( LDA) shows up in the equilibrium ratios shown above hEnce, enolization under" kinetic control with LDA allows you to produce the less-substituted enolate while subsequent equilibration by simply heating the enolate mixture allows equilibration to the more substituted N一LH Kinetic Ratio 99:1 Equilibrium Ratio 10: 90
O O Me O O Me HB HB Me OLi LiNR2 O Me HB HA HB LiNR2 H H H Me O H H Me OLi OLi Me H H O H H H H O C3H7 CH3 O O C3H7 CH3 Ph CH3 O O CH3 Ph THF K A – K LiNR2 B – A – OLi Me HA HB B – N Me Me Me Me Li D. A. Evans Kinetic & Thermodynamic Acidity of Ketones Chem 206 ■ Kinetic Acidity: Rates of proton removal Consider enolization of the illustrated ketone under non-equilibrating conditions: kA kB Kinetic acidity refers to the rate of proton removal. e.g. k A vs k B . For example, in reading the above energy diagram you would say that HA has a lower kinetic acidity than H B . As such, the structure of the base (hindered vs unhindered) employed plays a role in determining the magnitude of k A and k B . For the case shown above, D G ‡ A will increase more than D G ‡ B as the base becomes more hindered since the proton H A resides in a more sterically hindered environment. The example shown below shows the high level of selectivity which may be achieved with the sterically hindered base lithium diisopropylamide (LDA). Reaction Coordinate Energy DG ‡ B DG ‡ A B ‡ A ‡ Kinetic Ratio 99 : 1 LDA Equilibrium Ratio 10 : 90 ■ Note that alkyl substitution stabilizes the enolate (Why??). This effect shows up in the equilibrium ratios shown above. Kinetic & Equilibrium Ratios of Enolates Resulting from Enolization with LDA & Subsequent Equilibration (99) (1) Kinetic Ratios Equilibrium Ratios (90) (10) (2) (98) Kinetic Ratios (34) (66) Equilibrium Ratios (13) (87) Kinetic Ratios (53) (47) Equilibrium Ratios (16) (84) Kinetic Ratios (87) (13) Equilibrium Ratios Equilibrium Ratios (99) (1) Kinetic Ratios (14) (86) ■ Hence, enolization under "kinetic control with LDA allows you to produce the less-substituted enolate while subsequent equilibration by simply heating the enolate mixture allows equilibration to the more substituted enolate. –78 °C