Organic Pedagogical Electronic Network Developing Mathematical Relationships to Understand and Optimize Reaction Outcomes Elizabeth Bess & Matthew Sigman University of Utah

Identifying Quantitative Trends in Reaction Outcomes Wiki Pages: Other References: Hammett, L. P. Chem. Rev. 1935, 17, ; Taft, R. W., J. Am. Chem. Soc. 1952, 74, ; Wells, P. R., Chem. Rev. 1963, 63, ; Hansch, C.; Leo, A.; Taft, R. W., Chem. Rev. 1991, 91, Overview: Numerically describing the importance of reaction features on a reaction outcome is a tool for investigation of reaction mechanism and optimizing reaction systems. Classically, when parameters derived from relative-rate measurements (such as Hammett and Taft values) are correlated to the log(K), where K may be an equilibrium constant or measure of relative rates, this type of mechanistic study is a termed linear free-energy relationship (LFER) analysis. Early Examples Interpreting LFERs Hammett (σ) LFER σ R1R1 OMeHClCF 3 σ electron-richelectron-poor ionization reaction from which σ values were derived deriving σσ constants Taft LFER R2R2 HMeiPrtBu ESES hydrolysis reaction from which E S values were derived deriving E S E S constants smalllarge log(K) σ ρ = 1 0 < ρ < 1 ρ > 1 Template relationship describing Hammett LFER: log(K) = ρσ. ρ defines 1) sensitivity of evaluated reaction to Hammett- like changes and 2) the type of charge build-up in the transition state. ρ > 1ρ = 10 < ρ < 1 ρ < 0 ρ = 0 reaction’s sensitivity relative to benzoic acid ionization moreequallessN/A Not sensitive transition state charge build-up Neg. Pos.N/A

Applications of LFER in Asymmetric Catalysis Hammett LFER Analysis 2 Derivation of ∆∆G ‡ Equation: ∆G ‡ = -RTln(K eq ) ∆G ‡ S — ∆G ‡ R = -RT(lnK S – lnK R ) ∆∆G ‡ = -RTln(K S /K R ) = -2.3RTlog(K S /K R ) ∆∆G ‡ = -2.3RTlog(er) Theoretical Basis for Application: Curtin-Hammett Principle 1 References: 1. Curtin, D. Y., Rec. Chem. Prog. 1954, 15, ; Halpern, J. Science. 1982, 217, 401– Jacobsen, E. N.; Zhang, W.; Guler, M. L. J. Am. Chem. Soc. 1991, 113,

Applications of LFER in Asymmetric Catalysis References: 3. Knowles, R. R.; Jacobsen, E. N. Proc. Natl. Acad. Sci. USA 2010, 107, ; Knowles, R. R.; Lin, S.; Jacobsen, E. N. J. Am. Chem. Soc. 2010, 132, Harper, K. C.; Sigman, M. S. Science 2011, 333, ∆∆G ‡ = E S – 0.85S 2 – 3.79ES ES 2 2-Dimensional LFER 4 Quadropole Moments Predict Enatiomeric Ratio: Measuring Aromatic Systems‘ Electron Density 3

Free-Energy Relationship in C–H Functionalization σ +, a Measure of Resonance Stabilization, Parameterizes the Aryl Substrate 5 : References: 5. Bess, E. N.; DeLuca, R. J.; Tindall, D. J.; Oderinde, M. S.; Roizen, J. L.; Du Bois, J.; Sigman, M. S., J. Am. Chem. Soc. 2014, 136,

free-energy relationship: References: 5. Bess, E. N.; DeLuca, R. J.; Tindall, D. J.; Oderinde, M. S.; Roizen, J. L.; Du Bois, J.; Sigman, M. S., J. Am. Chem. Soc. 2014, 136, Infrared (IR) Molecular Vibrations Parameterize Sulfamate Ester Steric and Electronic Effects 5 : Free-Energy Relationship in C–H Functionalization

Problems 1. The transesterification of adenosines (1) and deoxyadenosines (2) is thought to proceed via differing mechanisms. Linear free-energy relationship studies have been performed to support these different reaction pathways by measuring the effect of benzoyl substituents on the two reaction types. Hammett plots for the two reactions resulted in ρ=3.19 and ρ=2.25. Based on your knowledge of LFER, determine which ρ corresponds with which reaction and justify your answer. (Tetrahedron Lett., 2007, 48, 2381–2384)

Problems 2. The below plot presents the results of the given asymmetric allylation that was carried out with varying substituents, G, on the chiral ligand. The alcohol product can be generated (from achiral starting materials) with a predominance of one stereoisomer. (J. Org. Chem., 2009, 74, 7633–7643) a) Based on your knowledge of LFER, interpret what the sensitivity constant of 1.51 indicates about the plotted asymmetric allylation. er is R/S. b) When the log (er) for the substituents CH(Pr) 2, CH( i Pr) 2, and CEt 3 are plotted for the same allylation reaction, a different sensitivity constant (0.72) results. What does the break in the trend line indicate about the reaction mechanism?

Solutions 1. In reaction 1, the oxyanion that removes the alcohol proton does so in close proximity to where the nucleophile attacks the electrophilic carbonyl, due to the conformation of the adenosine. The hydroxy and carbonyl groups are, by necessity of conformation, always relatively near each other, making reaction 1 more akin to an intramolecular reaction than an intermolecular reaction. In reaction 2, the base (DBU) removing the alcohol proton and the alcohol attacking the carbonyl are interactions that are not forced by any conformational constriants to occur near one another and have definite intermolecular character. Resultingly, the nucleophility of the alcohol is more enhanced in reaction 1 than in 2, and, therefore, a greater degree of bond formation between the nucleophile and the electrophile is seen in reaction 1. Greater bond formation results in a full negative charge on the oxygen of the tetrahedral center and makes the reaction sensitive to electron- withdrawing substituents on the phenyl ring to aid in the stabilization of this charge. An X substituent with greater eletron-withdrawing character would enhance the reaction rate, resulting in a ρ>0. In reaction 2, the termolecular nature of the transition state enhances the nucleophilicity of the alcohol to a lesser degree, resulting in less bond formation between the alcohol nucleophile and the carbonyl electrophile. Therefore, there is only a partial negative charge on the carbonyl oxygen, and the reaction would be less sensitive to the effects of electron-withdrawing substituents on the phenyl ring but would still result in ρ>0, due to the building of negative charge in the transition state. Because reaction 1 would be more sensitive to substituent effects due to the greater negative charge built in its transition state, it would have the higher sensitivity constant of Reaction 2 will have the lower sensitivity constant of 2.25.

Solutions 2. a) A sterically larger group in the G position of the ligand results in more R product and less S product. The larger the group, the larger the er. The allylation is sensitive to the size of the group in the G position of the ligand. b) In Hammett Plots, a break in the trend line can indicate a change in the mechanism. In the analogous Charton Plot, a break in the trend line indicates a similar change in the reaction. Because the Charton Plot analyzes the effect of sterics, when CH(Pr) 2, CH(iPr) 2, and CEt 3 groups are analyzed, the break in the trend line could indicate a different conformation of the ligand. The changed ligand conformation is less stereoselective. It has a lesser ability to differentiate between the pro-R and pro-S allyl faces, resulting in a lower er and a lower sensitivity constant. However, the sensitivity constant remains positive. This indicates that the R enantiomer is increasingly favored with larger substituents.