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Image: http: //www. chimica. unipd. it/alessandro. bagno/pubblica/prs. html Solvation Effects on p. Ka Values Organic Pedagogical Electronic Network Jamie Allen Jacqueline Pasek-Allen Sarah Lefave
Solvation effects on charged molecules When substrates are put in solution, the solvent molecules can organize themselves around a charged species to stabilize it. Solvents can stabilize a charge most effectively when the charge on the substrate is easy to get to. That being said, sterically hindered ions are harder for solvents to stabilize then non-sterically hindered molecules. Conjugate base stabilization p. Ka = 16 R-AH-> R-A- + H+ neutral -> negative Stabilization of the conjugate base leads to a lower p. Ka. When addressing conjugate base anions, the stronger acids are seen in less sterically-hindered molecules. This is because more solvent molecules can surround the negative charge, thus higher conjugate base stabilization drive the reaction to the right p. Ka = 18 ↑ R group size = ↑ p. Ka Acid stabilization p. Ka =10. 62 p. Ka =9. 76 R-A+H-> R-A +H+ positive -> neutral Stabilization of the acid leads to a higher p. Ka. When addressing acid cations, the stronger acids are seen in more sterically hindered molecules. This is because less solvent molecules can surround the positive charge, thus lower acid stabilization drives the reaction to the left. ↑ R group size = ↓ p. Ka table: http: //www. 3 rd 1000. com/chem 301/p 00405. htm Other References: Anslyn, Eric V. , and Dennis A. Dougherty. Modern Physical Organic Chemistry. Sausalito, CA: U Science, 2006. Print.
Polarizability effects on p. Ka A molecule is stabilized when it can spread out it’s charge over the greatest amount of area. When predicting p. Kas based on polarizability effects, one measures the amount of charge that the anion/cation itself can spread out via its orbital size and substituents. Pure polarization effects are seen in the ideal gas phase where solvent (therefore, solvation effects) are absent. However, if polarization effects are strong enough, they can be seen in the presence of solvent as well. Conjugate base stabilization p. Ka = 3 R-AH-> R-A- + H+ neutral -> negative p. Ka = -8 p. Ka = -9 When comparing halogens, iodine has the lowest p. Ka because the conjugate base’s anion is able to spread out the most charge through iodine’s large size. p. Ka = -10 ↑ orbital size = ↓ p. Ka Acid stabilization p. Ka = 11 p. Ka = 9 R-A+H-> R-A +H+ positive -> neutral Triethylammonium is more stable than ammonium because the alkyl groups aid in spreading out the positive charge. Thus, triethylammonium has the highest p. Ka because the acid is stabilized. ↑ substituent size = ↑ p. Ka References: http: //employees. csbsju. edu/cschaller/Principles%20 Chem/acidity/acid%20 local. htm
The curious case of Ammonium ion derivatives p. Ka trend based on a mixture of polarizability and solvation effects Some substrates can exhibit a mixture of both polarization effects and solvation effects, thus creating a unique trend in acidity. An example is shown in the following p. Ka trends of ammonium derivatives compared to predicted trends. Predicted p. Ka Trends: Lowest p. Ka Polarizability Experimental p. Ka Trends: Highest p. Ka Polarizability predicts (CH 3)3 NH+ as the weakest acid because more alkyl groups spread out the positive charge better. Lowest p. Ka ACTUAL Highest p. Ka Actual p. Ka values indicate that both polarizability and solvation effects act in concert in this case, with (CH 3)3 NH+ in the middle of the p. Ka trend. Lowest p. Ka Solvation Highest p. Ka Solvation predicts (CH 3)3 NH+ as the strongest acid because it is the hardest for the solvent to get to and stabilize the positive charge. References: Anslyn, Eric V. , and Dennis A. Dougherty. Modern Physical Organic Chemistry. Sausalito, CA: U Science, 2006. Print.
Case Study 1: Experimental Solvation Effects on Relative Acidity of Alcohols The above figure visually shows the effects of adding a solvent-like particle to a mixture of ethanol and methanol. References: Mc. Iver, R. T. ; Scott, J. A. ; Riveros, J. M. Effect of Sovation on the intrinsic elative acidity of methanol and ethanol. J. Am. Chem. Soc. , 1973, 95 (8), pp 2706– 2708
Case Study 2: A modern approach to p. Ka estimates: Absolute p. Ka Determinations of Substituted Phenols Summary: As previously discussed, one simple method, such as solvation or polarization, rarely can accurately describe the p. Ka trends of acids. In the more modern times, a computational approach using continuum CPCM solvation calculations can been used. Specifically this method uses computations that combine both electronic and steric (polarization and solvation) effects, and chemists have used it to accurately predict the p. Kas in the solvent-phase of phenol derivative structures. Below are 2 trends that have been shown, along with their computationally predicted and experimental p. Ka values respectively. < < Predicted p. Kas: Experimental p. Kas: 7. 66 8. 56 7. 49 7. 57 9. 84 9. 38 9. 88 9. 97 10. 08 9. 88 9. 98 < < Predicted p. Kas: Experimental p. Kas: < < 10. 45 10. 21 10. 80 10. 30 Note: It still holds that this method is not accurate for p. Ka predictions of gas-phase phenol derivative structures as they deviate significantly from those in solvent phase. References: Liptak, M. D. ; Gross, K. C. ; Seybold, P. G. ; Feldgus, S. ; Shields, G. C. Absolute p. Ka Determinations for Substituted Phenols. J. Am. Chem. Soc. , 2002, 124 (22), pp 6421– 6427
Questions 1. Would the following molecule be more stable in the gas phase or in a polar solvent solution? a) Polar Solvent b) Gas-phase c) Same 5. If only solvent effects were at play in the following p. Ka trend, what order would the following phenols be in, in order of increasing p. Ka. 1 d)Neither 2. Is the following trend because of polarization effects, solvation effects, neither, or both? 7. 49 7. 57 a) 1, 2, 3, 4 3. What is the order of p. Ka lowest to highest due to only polarization effects? 1 2 3 4 2) 3) 4) a) 1, 2, 3, 4 b) 2, 1, 3, 4 c) 3, 1, 2, 4 d) 3, 4, 2, 1 4. How would you expect the K of the following reaction to change upon addition of C 6 H 5 O-⋯H⋯OCH 3 a) Increase + b) Decrease c) Same 2 < 3 < 4 9. 88 9. 97 7. 66 9. 98 10. 08 8. 56 b) 4, 3, 2, 1 c) 2, 1, 3, 4 d) 3, 2, 4, 1 Answers 1) b) In the gas phase; the many alkyl chains are able to spread a) Polarization b) Solvation c) Neither d) Both + < K >> 1 d) Cannot predict 5) out the positive charge. A polar solvent would be unfavorable because solvation effects would cause it to be unstable. c) Neither. This trend follows the solvation trend; however, the effect is mainly due to induction. d)the size and amount of halogen and R groups increases polarization and p. Ka of the molecule b) decrease – the original equation in the gas phase shows that p-ethylphenoxide is much more reactive than phenoxide, upon addition of a solvent-like molecule the K will decrease indicating that the relative acidity of the p-ethylphenoxide compared to phenoxide is decreasing. d) the expected trend with solvation effects will have the least sterically hindered alcohol with the lowest p. Ka and the most hindered alcohol with the highest p. Ka.
Contributed by: Jamie Allen, Jacqueline Pasek-Allen, Sarah Lefave (Undergraduates) University of Utah, 2016 This work is licensed under a Creative Commons Attribution. Share. Alike 4. 0 International License.
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