Question

Suppose a water solution was made up initially to be \(0.01 \mathrm{M}\) in methyl bromide and \(1.0 \mathrm{M}\) in sodium acetate at \(50^{\circ}\). In water, the \(\mathrm{S}_{\mathrm{N}} 2\) rate constant for reaction of hydroxide ion with methyl bromide at \(50^{\circ}\) is \(30 \times 10^{-4}\) liters \(/ \mathrm{mole} / \mathrm{sec}\), whereas that of acetate ion at \(50^{\circ}\) is \(1.0 \times 10^{-}\) liters/mole/sec. The ionization constant of acetic acid at \(50^{\circ}\) is \(1.8 \times 10^{-5}\). In the following, assume perfect solutions and neglect the rates of reaction of methyl bromide with water or acetic acid by themselves and any further reactions of methyl acetate: (a) Calculate the hydroxide-ion concentration in the initial solution. (b) Calculate the initial rates of formation of methyl acetate and methanol. (c) What kind of information would be needed to predict what products would be expected from a solution of methyl bromide and sodium hydroxide in methanol? Explain.

Short answer

(a) The hydroxide-ion concentration in the initial solution is found using the equilibrium constant for the conjugate base of acetic acid (\(K_b = 5.56 \times 10^{-10}\)), giving: \[5.56 \times 10^{-10} = [x][CH_3COOH]\] Upon solving, we obtain the hydroxide-ion concentration as \(x\). (b) The initial rates of formation of methyl acetate (\(v_1\)) and methanol (\(v_2\)) are given by: \[v_1 = (30 \times 10^{-4})(x)(0.01)\] \[v_2 = (1.0 \times 10^{-})(x)(0.01)\] (c) To predict the products of a reaction between methyl bromide and sodium hydroxide in methanol, we would need: rate constants of the reactions, solubility and ionization of the compounds in methanol, relative reactivity of intermediate products or competing reactions, and influence of methanol as a solvent on the kinetics and equilibria.

Step-by-step solution

(a) Calculate the hydroxide-ion concentration

The ionization constant of acetic acid is given as \(1.8 \times 10^{-5}\), this can be used to determine the hydroxide-ion concentration in the initial solution. We know that, \[K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}\] Since we have sodium acetate solution, which is the conjugate base of acetic acid, the reaction becomes: \(CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)\) Now, in this case, the equilibrium constant expression is: \[K_b = \frac{[OH^-][CH_3COOH]}{[CH_3COO^-]}\] and we use the relation \(K_w = K_a \times K_b\), where \(K_w\) is the ion-product constant for water (\(1 \times 10^{-14}\) at \(25^{\circ}\) C), to find the value of \(K_b\). \[K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\] Now, let \(x\) be the concentration of hydroxide-ion formed. In equilibrium, we have: \[5.56 \times 10^{-10} = \frac{[x][CH_3COOH]}{1.0 - x}\] Since x will be very small, \(1.0 - x \approx 1.0\). This simplifies the equation to: \[5.56 \times 10^{-10} = [x][CH_3COOH]\] Now we can solve for x, the hydroxide-ion concentration.

(b) Calculate the initial rates of formation of methyl acetate and methanol

Now that we have the hydroxide-ion concentration from part (a), we can calculate the initial rates of formation of methyl acetate and methanol. For methyl acetate formation: \([OH^-] = x\), rate constant \(k_1 = 30 \times 10^{-4} \frac{L}{mol \cdot s}\) Initial rate of methyl acetate formation, \(v_1 = k_1[OH^-][CH_3Br]\) \[v_1 = (30 \times 10^{-4})(x)(0.01)\] For methanol formation: \([OH^-] = x\), rate constant \(k_2 = 1.0 \times 10^{-} \frac{L}{mol \cdot s}\) Initial rate of methanol formation, \(v_2 = k_2[OH^-][CH_3Br]\) \[v_2 = (1.0 \times 10^{-})(x)(0.01)\]

(c) Information needed to predict products of methyl bromide and sodium hydroxide in methanol

To predict the products of a reaction between methyl bromide and sodium hydroxide in methanol, we would need the following information: 1. Rate constants of the reactions involved. 2. Solubility and ionization of the compounds in methanol as a solvent. 3. Relative reactivity of intermediate products or competing reactions in methanol to understand which pathways are favored. 4. Influence of methanol as a solvent on the kinetics and equilibria of the reactions. With this information, we could analyze the reaction pathways and determine the likely products of the reaction between methyl bromide and sodium hydroxide in methanol.

Key concepts

Hydroxide-Ion Concentration Calculation

Understanding the concentration of hydroxide ions in a solution is crucial when studying reactions in chemistry, especially in the context of SN2 reactions. The hydroxide-ion concentration informs us about the basicity of a solution which can greatly impact the course of a reaction.

To calculate the hydroxide-ion concentration, we often use equilibrium constants such as the acid dissociation constant (\( K_a \) ) and the ion-product constant for water (\( K_w \) ). Given that \( K_w \) at 50°C is not provided, we must assume it's similar to that at 25°C (\( 1 \times 10^{-14} \) ), while keeping in mind that it slightly increases with temperature. To find the hydroxide ion concentration in a solution with sodium acetate, we first derive the base dissociation constant (\( K_b \) ) using the provided \( K_a \) of acetic acid and the \( K_w \) value. This step lays down the fundamental concept of conjugate acid-base pairs where \( CH_3COOH \) and \( CH_3COO^- \) form such a pair.

Using the equilibrium expression for \( K_b \) and the assumption that \( [CH_3COO^-] \) remains largely unchanged (since the change is very small compared to the starting concentration), we simplify the equation to calculate 'x', which represents the hydroxide-ion concentration. This step is vital to students because it helps understand how a seemingly simple calculation requires careful consideration of equilibrium and approximations.

Initial Rate of Reaction

The initial rate of a chemical reaction is a pivotal concept in kinetics; it determines how fast reactants are converted into products at the beginning of the reaction. To calculate the initial rate, you'll need the rate constant and the initial concentrations of the reactants. In the case of SN2 reactions, this involves using the rate law, which for a bimolecular reaction such as SN2 is rate = k[Reactant 1][Reactant 2].

For our given scenario, we calculate the initial rate of formation of methyl acetate using the concentration of hydroxide-ion previously determined and the rate constant for hydroxide ion reaction with methyl bromide. Since we're examining an SN2 reaction, the rate is directly proportional to the concentrations of both the nucleophile and the electrophile.

Importantly, when teaching this concept, it is beneficial to note that the 'initial' rate is measured immediately after the reactants are mixed, before any significant change in concentrations – this helps avoid any complications arising from the changes in reactant concentrations over time.

Nucleophilic Substitution Mechanisms

Nucleophilic substitution mechanisms represent a fundamental principle in organic chemistry. Specifically, the SN2 mechanism involves a nucleophile attacking an electrophilic center, leading to a substitution reaction. The SN2 (substitution, nucleophilic, bimolecular) mechanism is characterized by a one-step process where the bond formation and bond breaking occur simultaneously.

The rate of an SN2 reaction is dependent on the concentration of both the nucleophile and the substrate, hence the term 'bimolecular'. This is clearly illustrated in the solved exercise above, where the rates of reaction are calculated based on the concentrations of the hydroxide ion -- acting as the nucleophile -- and methyl bromide -- the substrate.

Additionally, SN2 reactions are highly sensitive to steric effects; as such, they proceed faster with primary alkyl halides and are hindered by bulkier secondary and tertiary centers. When teaching these concepts, emphasizing visual representations of the transition state can also greatly enhance student comprehension.

Chemical Equilibrium Constants

Chemical equilibrium constants are central to understanding the balance between products and reactants in reversible reactions. They are a quantitative measure of the extent of a reaction at equilibrium. Equilibrium constants like \( K_a \) for acids and \( K_b \) for bases allow us to predict the position of equilibrium and the concentrations of substances in a chemical system when it is at rest.

The exercise demonstrates the use of these constants in determining hydroxide-ion concentration by relating \( K_a \) of acetic acid and the \( K_w \) of water to calculate \( K_b \) of the acetate ion. This concept underpins the predictive power of chemistry because knowing these constants enables chemists to formulate reactions with desired product concentrations.

Furthermore, it is important to explain to students that the magnitude of the equilibrium constant provides insight into the favorability of a reaction; a large \( K \) value indicates a reaction that heavily favors products at equilibrium, while a small \( K \) suggests a reaction that does not proceed far towards the products.