In our discussion of acid-base equilibria, we saw how to calculate the equilibrium constant for any reaction involving two acids of known acid strength. By comparing differences in acidity within carefully selected groups of structurally related compounds, it is possible to gain insight into the relationship between molecular structure and chemical reactivity. In this topic we will make correlations between changes in pKa values and various atomic, molecular, and experimental variables:
We'll start by looking at HF and its relatives, H2O, NH3, and CH4.
Equation 1 describes the general reaction that we want to consider.
Exercise 1 The following table lists four different compounds that fit the formula HA along with some additional information. Complete the table. Enter an exponential value such as 10-6 as 10e-6.
Exercise 3 The ion -:A is most stable when A equals . Of the four compounds listed in the table, is the strongest acid. This is because ion is the most stable conjugate base.
Exercise 4 The ion -:A is least stable when A equals . Of the four compounds listed in the table, is the weakest acid. This is because ion is the least stable conjugate base.
In aqueous solution the relative acidities of the hydrogen halides is HI > HBr > HCl > HF.
Exercise 5 Complete the following table. Enter an exponential value such as 10-6 as 10e-6.
Exercise 6 In the hydrogen halides, there is adirect inverse correlation between the atomic radius of the halogen atom and the strength of the hydrogen-halogen bond.
Exercise 7 In the hydrogen halides, there is a direct inverse correlation between the bond strength and the pKa value.
Exercise 8 In the hydrogen halides, there is a direct inverse correlation between the H-A bond strength and the acidity of HA.
Figure 1 compares the dissociation reactions of methanol and phenol. As you can see, phenol is approximately 1,000,000 times more acidic than methanol. Clearly differences in bond strengths or electronegativity are not responsible for this difference in acidity since an O-H bond is broken in both cases and the negative charge resides on an oxygen atom in both conjugate bases.
In our discussion of resonance theory we saw that when an orbital containing a lone pair of electrons can overlap with a p orbital of an adjacent multiple bond as shown in Figure 2, the energy of the system decreases.
As Figure 3 indicates, the phenolate ion, C6H5O:-, meets this structural requirement.
Notice that Figure implies that the orbital overlap leads to changes in the hybridization of the oxygen atom. This change decreases the separation between the orbitals on the oxygen and the adjacent carbon and increases their overlap.
This orbital overlap leads to charge delocalization as indicated by the resonance structures shown in Figure 4.
As you can see, the O-H hydrogen in acetic acid is 11 orders of magnitude more acidic than that in methanol. Conversely, the methoxide ion, CH3O:-, is 11 orders of magnitude more basic than the acetate ion. Clearly the carbonyl group affects the acidity of the adjacent O-H group dramatically. We attribute this increase in acidity to the stabilization that the double bond affords the negative charge as it develops in the conjugate base of acetic acid. The interaction between a lone pair of electrons on the oxygen and the adjacent carbonyl group is shown in Figure 6.
Remember, the formalism shown in Figure 6 is our way of indicating that the negative charge in the acetate ion can experience the nuclear attraction of both oxygen atoms. This increase in Coulombic attraction reduces the potential energy of the acetate ion relative to that of the methoxide ion where the charge is localized on one oxygen atom.
Exercise 12 The pKa of ethane is approximately 50 while that of acetone is about 20. Write equations similar to Equation 1 for each of these compounds. Explain why acetone is 30 powers of 10 more acidic than ethane. Draw resonance structures and orbital diagrams to illustrate your answer.
Exercise 13 What is the equilibrium constant for the following reaction?