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

Based on the given information, the net ionic equation for the acid-base reaction between 2,4-pentanedione and sodium ethoxide is: \((CH_3)_2CHCOCH_2COCH_3 + CH_3CH_2O^{-} \rightarrow (CH_3)_2CHCOCH_2COCH_2CH_3 + OH^{-}\) The equilibrium constant, \(K_{eq}\), for this reaction is approximately \(10^{6.9}\).

Step-by-step solution

Write the equation for the acid-base reaction

Since we are given 2,4-pentanedione (\((CH_3)_2CHCOCH_2COCH_3\)) and sodium ethoxide (\((C_2H_5O)^{-}\)) for the reaction, we begin by writing their molecular structures: 2,4-Pentanedione: \(H_3C-CO-CH_2-CO-CH_3\) Sodium ethoxide: \(CH_3CH_2-O^{-}\) In the acid-base reaction, the sodium ethoxide, which is the base, will act as a nucleophile and attack the acidic hydrogen of the 2,4-pentanedione to form a new species and a sodium ion. The equation for the reaction is: \((CH_3)_2CHCOCH_2COCH_3 + CH_3CH_2O^{-}Na^{+} \rightarrow (CH_3)_2CHCOCH_2COCH_2CH_3 + Na^+ + OH^{-}\) Note that sodium ions are spectators in this reaction, so the net ionic equation is: \((CH_3)_2CHCOCH_2COCH_3 + CH_3CH_2O^{-} \rightarrow (CH_3)_2CHCOCH_2COCH_2CH_3 + OH^{-}\)

Calculate the equilibrium constant, \(K_{\mathrm{eq}}\)

Next, we will calculate the equilibrium constant using the relationship between the pKa values of the two species. The pKa of the acid is given as 9 and that of ethanol is given as 15.9. We will use the following formula: \(\Delta pK_{\mathrm{a}} = pK_{\mathrm{a}_{\mathrm{base}}} - pK_{\mathrm{a}_{\mathrm{acid}}} = \log K_{\mathrm{eq}}\) Substituting the given pKa values, we obtain: \(\Delta pK_{\mathrm{a}} = 15.9 - 9 = \log K_{\mathrm{eq}}\) Now, we can find \(K_{\mathrm{eq}}\) by taking the antilogarithm of the difference: \(K_{\mathrm{eq}} = 10^{\Delta pK_{\mathrm{a}}} = 10^{15.9 - 9} = 10^{6.9}\) Therefore, the equilibrium constant \(K_{\mathrm{eq}}\) for this acid-base reaction is approximately \(10^{6.9}\).

Key concepts

2,4-Pentanedione Reactivity

2,4-Pentanedione, also known as acetylacetone, is a beta-diketone with interesting reactivity due to the presence of two carbonyl (CO) groups separated by a methylene (CH2) spacer. This molecular structure allows the hydrogen atom positioned between the carbonyl groups to display acidic properties, making it susceptible to nucleophilic attack by bases.

In the presence of a strong base such as sodium ethoxide, it undergoes an acid-base reaction, resulting in deprotonation. The hydrogen is transferred to the base, producing an enolate anion of 2,4-pentanedione and an alcohol molecule, which in the case of sodium ethoxide, would be ethanol. It's key to understand this reactivity when delving into mechanisms of reactions involving 2,4-pentanedione, particularly in synthesizing heterocyclic compounds and in keto-enol tautomerism.

Sodium Ethoxide Reaction

Sodium ethoxide is the sodium salt of ethanol and acts as a strong base and nucleophile in reactions. It's commonly used to deprotonate weak acids like 2,4-pentanedione. In a typical reaction with a beta-diketone such as 2,4-pentanedione, sodium ethoxide abstracts the acidic alpha-hydrogen, generating an enolate ion alongside the corresponding sodium ion and hydroxide ion. The presence of solvents can influence the reaction's direction and outcome.

The reactiveness of sodium ethoxide makes it a staple in organic synthesis, often utilized for condensation reactions, such as Claisen condensation, and in the preparation of various carbon-carbon bonds. Understanding its function in reactions helps one process why certain conditions lead to specific products.

pKa and Equilibrium Relationship

The pKa of a substance is a quantitative measure of its acid strength: the lower the pKa value, the stronger the acid. In the context of an acid-base reaction, knowing the pKa values of the reactants can provide insights into the equilibrium position. The pKa difference (\f\(\fDelta\f pK_{\fmathrm{a}}\f\)) between an acid and a base can determine the equilibrium constant (\f\(K_{\fmathrm{eq}}\f\)) for the reaction.

\f\(\fDelta\f pK_{\fmathrm{a}}\f\) is calculated by subtracting the pKa of the acid from that of the base. A larger difference implies that the reaction will have a greater tendency to move forward, favoring the formation of products. In the case of the acid-base reaction between 2,4-pentanedione and sodium ethoxide, the large pKa difference reflects a strong driving force toward the products, yielding a large equilibrium constant.

Chemical Equilibrium Calculation

Calculating the chemical equilibrium constant (\f\(K_{\fmathrm{eq}}\f\)) is essential for understanding the extent to which a reaction proceeds. \f\(K_{\fmathrm{eq}}\f\) reflects the ratio of product concentrations to reactant concentrations at equilibrium, raised to the power of their stoichiometric coefficients. For the reaction between 2,4-pentanedione and sodium ethoxide, the \f\(K_{\fmathrm{eq}}\f\)is determined from the pKa values using the formula \f\(\fDelta\f pK_{\fmathrm{a}} = \flog\f K_{\fmathrm{eq}}\f\).

By rearranging this formula to solve for \f\(K_{\fmathrm{eq}}\f\) and taking the antilogarithm, as done in the step by step solution provided in the textbook, we obtain the value of the equilibrium constant. This calculated value tells us that at equilibrium, there is a significant favoring of product formation, indicating that this particular reaction is highly product-favored under standard conditions.