Question

Write an equation for the acid-base reaction between 2,4-pentanedione and sodium ethoxide and calculate its equilibrium constant, \(K_{\mathrm{eq}}\). The \(\mathrm{p} K_{\mathrm{a}}\) of 2,4 -pentanedione is 9 ; that of ethanol is \(15.9\).

Short answer

Answer: The equilibrium constant for the reaction between 2,4-pentanedione and sodium ethoxide is \(K_{\mathrm{eq}} =10^{-6.9}\).

Step-by-step solution

Identify the acidic and basic components

In the reaction between 2,4-pentanedione and sodium ethoxide, the acidic component is the 2,4-pentanedione as it loses a proton (H+), while the basic component is the ethoxide anion (CH\(_3\)CH\(_2\)O\(^-\)) that results from sodium ethoxide, which gains a proton (H+).

Write the reaction equation

The acid-base reaction will involve the transfer of a proton (H+) from the acidic component to the basic component. The equation for the reaction is: 2,4-Pentanedione + Ethoxide \( \rightarrow \) 2,4-Pentanedionate + Ethanol

Calculate the equilibrium constant

To calculate the equilibrium constant, we will use the given pKa values: \(\mathrm{p} K_{\mathrm{a}}\) of 2,4-pentanedione = 9 \(\mathrm{p} K_{\mathrm{a}}\) of ethanol = 15.9 First, we need to convert these to Ka values: \(K_{\mathrm{a1}} = 10^{-\mathrm{p} K_{\mathrm{a1}}} = 10^{-9}\) \(K_{\mathrm{a2}} = 10^{-\mathrm{p} K_{\mathrm{a2}}} = 10^{-15.9}\) Now, we can calculate the equilibrium constant by dividing the Ka of the conjugate acid (ethanol) by the Ka of the acid (2,4-pentanedione): \(K_{\mathrm{eq}} = \frac{K_{\mathrm{a2}}}{K_{\mathrm{a1}}} = \frac{10^{-15.9}}{10^{-9}}\) \(K_{\mathrm{eq}} = 10^{-15.9+9}\) \(K_{\mathrm{eq}} = 10^{-6.9}\) So, the equilibrium constant for the reaction between 2,4-pentanedione and sodium ethoxide is \(K_{\mathrm{eq}} =10^{-6.9}\).

Key concepts

2,4-pentanedione Acidity

Understanding the acidity of 2,4-pentanedione is crucial in the context of acid-base chemistry. Acidity refers to the capacity of a molecule to donate a proton to a base. In organic chemistry, the acidity of a compound is influenced by the stability of the resulting anion after a proton is donated. For 2,4-pentanedione, also known as acetylacetone, this stability comes from the resonance structures that spread the negative charge across the molecule after it loses a proton.

Specifically, the presence of electron-donating carbonyl groups adjacent to the acidic hydrogen enhances the molecule's acidity. This characteristic makes 2,4-pentanedione an ideal candidate to react with strong bases, such as sodium ethoxide, in acid-base reactions. Since it has a relatively low pKa of 9, it is considered to be a relatively strong acid within the field of organic compounds.

Sodium Ethoxide Reactivity

Sodium ethoxide, known chemically as sodium ethanolate, is the sodium salt of ethanol. It is highly reactive due to its strong basic nature and its ability to act as a nucleophile. In an acid-base reaction, the ethoxide ion (\( CH_3CH_2O^- \)) acts as the base, readily accepting protons from acidic compounds.

Its reactivity is also enhanced by the negative charge on the oxygen which makes it a potent nucleophile capable of attacking electrophilic centers in various organic molecules. Notably, in solution, sodium ethoxide is typically present as ethanol and sodium ions, but upon encountering a suitable acidic proton, the ethoxide ion swiftly reacts to form the corresponding alcohol and its conjugate base.

Equilibrium Constant Calculation

Calculating the equilibrium constant, denoted as \( K_{\mathrm{eq}} \), for an acid-base reaction involves understanding the concept of acid dissociation constants, \( K_a \). The equilibrium constant expresses the extent to which a reaction proceeds to form products under standard conditions.

Given the \( pK_a \) values of the acid and its conjugate acid after the reaction, we can calculate the \( K_a \) by taking the negative logarithm base ten of these values. The \( K_{\mathrm{eq}} \) is then determined by dividing the \( K_a \) of the conjugate base by the \( K_a \) of the acid. This calculation essentially quantifies the position of equilibrium between products and reactants, with a higher \( K_{\mathrm{eq}} \) indicating a reaction that favors the formation of products.

pKa Values in Organic Chemistry

The pKa value is a quantitative measure of the strength of an acid in solution. It is the logarithmic representation of the acid dissociation constant (Ka) and provides a comparative metric for the acidity of different substances. In organic chemistry, pKa values are indispensable for predicting the outcome of acid-base equilibria.

A lower pKa value means a stronger acid that is more willing to donate protons. The difference in pKa values between reacting substances can determine the direction of the reaction. For instance, when an acid with a lower pKa reacts with the corresponding base of an acid with a higher pKa (as in our original exercise with 2,4-pentanedione and sodium ethoxide), the equilibrium will usually favor the formation of the weaker acid and base.