Hammett Equation and pKa Calculation

The Hammett equation is a linear free-energy relationship that relates the reaction rate or equilibrium constant for a reaction involving a substituted benzene derivative to the nature and position of the substituent. For equilibrium reactions, such as the ionization of substituted benzoic acids, the equation is often expressed in terms of the equilibrium constant ($K$) or p$K$a. The basic form of the Hammett equation is: $$ \log \frac{K_X}{K_H} = \rho \sigma $$ where:

• $K_X$ is the equilibrium constant for the reaction of the substituted compound.
• $K_H$ is the equilibrium constant for the reaction of the unsubstituted compound (the parent compound).
• $\rho$ (rho) is the reaction constant, which depends on the specific reaction, conditions (like solvent and temperature), and reflects the susceptibility of the reaction to substituents.
• $\sigma$ (sigma) is the substituent constant, which depends only on the nature and position of the substituent and measures its electronic effect.

For the ionization of an acid, the equilibrium constant is the acid dissociation constant ($K_a$). Since $\text{p}K_a = -\log K_a$, we can rewrite the Hammett equation in terms of p$K$a: $$ \log K_X - \log K_H = \rho \sigma $$ $$ -\text{p}K_{a,X} - (-\text{p}K_{a,H}) = \rho \sigma $$ $$ \text{p}K_{a,H} - \text{p}K_{a,X} = \rho \sigma $$ Rearranging this to find the p$K$a of the substituted acid: $$ \text{p}K_{a,X} = \text{p}K_{a,H} - \rho \sigma $$

Calculating pKa of p-Anisic Acid

We are given the following information:

• Hammett substituent constant for the p-OMe group, $\sigma_{p-OMe} = -0.30$.
• p$K$a of benzoic acid (the parent compound), $\text{p}K_{a, \text{benzoic acid}} = 4.19$.

The reaction constant ($\rho$) for the ionization of benzoic acids in water at 25°C is approximately $+1.0$. We will use this standard value. Now we can substitute these values into the Hammett equation expressed in terms of p$K$a: $$ \text{p}K_{a, \text{p-anisic acid}} = \text{p}K_{a, \text{benzoic acid}} - \rho \sigma_{p-OMe} $$ $$ \text{p}K_{a, \text{p-anisic acid}} = 4.19 - (1.0) \times (-0.30) $$ $$ \text{p}K_{a, \text{p-anisic acid}} = 4.19 - (-0.30) $$ $$ \text{p}K_{a, \text{p-anisic acid}} = 4.19 + 0.30 $$ $$ \text{p}K_{a, \text{p-anisic acid}} = 4.49 $$ The calculated p$K$a of p-anisic acid is 4.49. A substituent with a negative $\sigma$ value, like the electron-donating OMe group, stabilizes the neutral acid more than the conjugate base (by destabilizing the negative charge on the carboxylate), thus making the acid weaker. A weaker acid has a higher p$K$a. Benzoic acid has a p$K$a of 4.19, and the p-OMe group makes it a weaker acid, resulting in a p$K$a higher than 4.19. Our calculated value of 4.49 is indeed higher than 4.19, which aligns with the electronic effect of the OMe group. Let's compare this result with the given options.

Option	p$K$a Value
1	4.79
2	3.89
3	3.59
4	4.49

The calculated value of 4.49 matches option 4.