Question

The reaction of 1 -bromopropane and sodium hydroxide in ethanol occurs by an \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. What happens to the rate of this reaction under the following conditions? (a) The concentration of \(\mathrm{NaOH}\) is doubled. (b) The concentrations of both \(\mathrm{NaOH}\) and 1-bromopropane are doubled. (c) The volume of the solution in which the reaction is carried out is doubled.

Short answer

Answer: (a) Doubling the concentration of NaOH doubles the rate of the reaction. (b) Doubling the concentrations of both NaOH and 1-bromopropane quadruples the rate of the reaction. (c) Doubling the volume of the solution quartered the rate of the reaction.

Step-by-step solution

Recall the SN2 reaction mechanism

The SN2 reaction mechanism involves a nucleophile, in this case, the hydroxide ion (\(\mathrm{OH}^{-}\)), attacking the substrate with a leaving group, which is 1-bromopropane, leading to a one-step substitution. The main feature of the SN2 mechanism is that it follows a second-order kinetics, meaning the reaction rate depends on the concentration of both the nucleophile and the substrate.

Determine the rate equation for the reaction

The SN2 reaction follows second-order kinetics, so we can represent the rate of the reaction using the following rate equation: \(\text{Rate} = k[\mathrm{OH}^{-}][\text{1-bromopropane}]\) where \(k\) is the rate constant, and the concentrations of OH- and 1-bromopropane are denoted by [OH-] and [1-bromopropane], respectively.

Find the effect of doubling the concentration of NaOH (Scenario a)

If we double the concentration of \(\mathrm{NaOH}\), the concentration of \(\mathrm{OH}^{-}\) ions also doubles. To find the effect on the rate of the reaction, let's substitute the new concentration in the rate equation: New rate \(= k[2\cdot\mathrm{OH}^{-}][\text{1-bromopropane}] = 2k[\mathrm{OH}^{-}][\text{1-bromopropane}] = 2\cdot\text{Rate}\) Doubling the concentration of \(\mathrm{NaOH}\) doubles the rate of the reaction.

Find the effect of doubling the concentrations of NaOH and 1-bromopropane (Scenario b)

If we double the concentrations of both \(\mathrm{NaOH}\) and 1-bromopropane, we can again substitute the new concentrations in the rate equation: New rate \(= k[2\cdot\mathrm{OH}^{-}][2\cdot\text{1-bromopropane}] = 4k[\mathrm{OH}^{-}][\text{1-bromopropane}] = 4\cdot\text{Rate}\) Doubling the concentrations of both \(\mathrm{NaOH}\) and 1-bromopropane quadruples the rate of the reaction.

Find the effect of doubling the volume of the solution (Scenario c)

When we double the volume of the solution, the concentrations of both \(\mathrm{NaOH}\) and 1-bromopropane are halved. Examining the effect on the rate equation: New rate \(= k[\frac{1}{2}\cdot\mathrm{OH}^{-}][\frac{1}{2}\cdot\text{1-bromopropane}] = \frac{1}{4}k[\mathrm{OH}^{-}][\text{1-bromopropane}] = \frac{1}{4}\cdot\text{Rate}\) Doubling the volume of the solution quartered the rate of the reaction.