Chemistry 2000 Lecture 19: Organic acids Marc R. Roussel March 8, 2018 Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 1 / 22
Organic acids The acid dissociation constant, K a The strength of an acid A is measured by the acid dissociation constant, K a, the equilibrium constant for the dissociation of the acid into a proton and its conjugate base: A + + A It is sometimes useful to think of K a as the equilibrium constant for the reaction of an acid with water: Larger K a = stronger acid A + 2 O 3 O + + A Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 2 / 22
Organic acids pk a K a values range over many orders of magnitude, so they are not very convenient for some purposes (e.g. comparisons between acids). Define pk a = log 10 K a Smaller pk a = stronger acid Strong acids may have negative pk a values (dissociate completely in water). Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 3 / 22
Organic acids Organic acids Functional group pk a Stronger acid Carboxylic acids 3 5 Phenols 1 10 Water 14 (pk w ) Weaker acid Alcohols 15 18 Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 4 / 22
Organic acids Acidity of carboxylic acids O C O Carboxylic acids are among the most acidic organic compounds. Acid strength depends on the polarity of the bond to the dissociable hydrogen (in this case the O- bond), and the stability of the conjugate base. Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 5 / 22
Organic acids Acidity of carboxylic acids Polarity of the O- bond O C O The carbonyl group is electron withdrawing: Because of the large electronegativity of oxygen, electron density is pulled away from the carbon atom. This in turn pulls electron density away from the hydrogen atom, increasing the partial positive charge on this atom. Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 6 / 22
Organic acids Acidity of carboxylic acids Stability of carboxylate ions Carboxylate anions are resonance stabilized: O O C O The negative charge is therefore spread over a larger region of the molecule, which tends to stabilize the anion. ow would this manifest itself in an MO description? C O Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 7 / 22
Organic acids Substituents on the adjacent carbon to the carboxylic acid functional group can have a substantial effect on the pk a. Consider the following series: Molecule C 3 COO FC 2 COO F 2 CCOO F 3 CCOO pk a 4.67 2.66 1.24 0.23 The fluorine atom is very electronegative and pulls electron density to itself. This decreases the negative charge carried by the carboxylate group, which stabilizes the conjugate base. Base C 3 COO FC 2 COO F 2 CCOO F 3 CCOO Charge of O 0.77 0.754 0.739 0.711 This is called an inductive effect. Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 8 / 22
Organic acids Alcohols Unlike carboxylic acids, most alcohols are only about as acidic as water itself. Why? No electron-withdrawing group (in an ordinary alcohol) to increase the polarization of the O- bond No resonance stabilization of the negative charge of the conjugate base Inductive effects can increase the acidity of an alcohol. Example: Molecule C 3 C 2 O Cl 3 CC 2 O pk a 15.9 12.24 Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 9 / 22
Organic acids Phenols Phenols are much more acidic than ordinary alcohols. Phenol itself has a pk a of 9.95, vs 17 for cyclohexanol. Why? Resonance stabilization of the charge on the phenolate anion:.. :O: :O: :O: :O:...... Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 10 / 22
Organic acids Some phenols are even more acidic because the substituents can participate in charge delocalization. Example: p-nitrophenol has a pk a of 7.21... :O: :O: :O:.. + +.. N.... N.... O.. O:.. O.. O:.. O.... + N.. O:.. :O: :O:.... O.. + N.. O:.. + N.... :O.. O:.. Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 11 / 22
Acid dissociation equilibria Acid dissociation equilibria If we know the K a and concentration of an acid, we can calculate the p. Reminder: p = log 10 a + We usually don t need to take the autoionization of water into account unless the concentration of protons liberated from the acid is similar to the concentration of protons generated by autoionization. We often can treat the acid as if it s mostly undissociated. Note that we ll assume 25 C in all of the following calculations. Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 12 / 22
Acid dissociation equilibria Example: Calculating the p for a solution of a given concentration of acid Calculate the p of a a 0.32 M phenol solution. The pk a of phenol is 9.95. Answer: p = 5.22 Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 13 / 22
Acid dissociation equilibria Example: Calculate the p of a 4.2 10 5 M ethanoic (acetic) acid solution. The pk a of ethanoic acid is 4.76. Answer: p = 4.71 Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 14 / 22
Acid dissociation equilibria Balance of acid and conjugate base at given p Sometimes, we put an acid into a solution of fixed p (a buffer) and want to know how much is in the acid and how much in the conjugate base form. This is an easy problem because the p fixes a +, which immediately gives us the ratio of the conjugate base to the acid: K a a + = a A a A This can easily be converted to percentages of the two forms if we add the equation [A] + [A ] = 100% (with a slight abuse of notation). Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 15 / 22
Acid dissociation equilibria Example: ethanoic acid at p 4 Suppose that we want to calculate the proportions of ethanoic acid (K a = 1.74 10 5 ) and of the ethanoate ion (conjugate base) at p 4. K a a + = 1.74 10 5 10 4 = 0.174 = [C 3COO ] [C 3 COO] [C 3 COO ] = 0.174[C 3 COO] (1) and [C 3 COO] + [C 3 COO ] = 100% (2) Substituting equation (1) into (2), we get [C 3 COO] + 0.174[C 3 COO] = 1.174[C 3 COO] = 100% [C 3 COO] = 85% [C 3 COO ] = 15% Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 16 / 22
Acid dissociation equilibria Distribution curves If we repeat the above calculation at a number of different p values and plot the results, we obtain distribution curves for the acid and its conjugate base. Note: If p = pk a, we have a A a A = 10 pka 10 p = 1 In other words, 50% of the acid is undissociated, and 50% in the form of the conjugate base when p = pk a. Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 17 / 22
Acid dissociation equilibria Distribution curve of ethanoic acid 100 90 acid base 80 70 60 % 50 40 30 20 10 0 0 2 4 6 8 10 12 14 p Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 18 / 22
Acid dissociation equilibria Distribution curves for polyprotic acids The calculation is analogous for polyprotic acids except that there are two (or more) equilibria and three (or more) forms of the acid to consider. When the pk a s of a polyprotic acid differ by several units, the distribution curves look like a simple superposition of distribution curves for the monoprotic case. Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 19 / 22
Acid dissociation equilibria Example: Ethanedioic (oxalic) acid at p 3 We want to calculate the fraction (or percentage) of ethanedioic acid in each of the three possible ionization states (COO) 2, OOCCOO and (COO) 2 2 at p 3. The pk a s of the two protons are 1.27 and 4.27. Answer: [OOCCOO ] = 93%, [(COO) 2 2 ] = 5.0% and [(COO) 2 ] = 1.7% Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 20 / 22
Acid dissociation equilibria Distribution curves of ethanedioic acid 100 90 (COO) 2 OOCCOO - (COO) 2 2-80 70 60 % 50 40 30 20 10 0 0 2 4 6 8 10 12 14 p Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 21 / 22
Acid dissociation equilibria Take a log, have an equation named after you... We are now familiar with the equation K a = (a A )(a +) a A If we take the negative log of this equation, we get ( ) aa log 10 K a = log 10 a + log 10 ( ) aa pk a = p log 10 a A ( ) aa or p = pk a + log 10 a A a A This last equation is called the enderson-asselbalch equation. Marc R. Roussel Chemistry 2000 Lecture 19: Organic acids March 8, 2018 22 / 22