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

a. If the concentration of NaOH is doubled, how is the rate of the reaction affected? b. If the concentrations of both NaOH and 1-bromopropane are doubled, how is the rate of the reaction affected? c. If the volume of the solution is doubled, how is the rate of the reaction affected?

Step-by-step solution

Case (a)

If the concentration of NaOH is doubled, the nucleophile concentration increases, so there will be a higher chance of the nucleophile colliding with the substrate. Remember that the rate of an SN2 reaction depends on the concentration of both the nucleophile and the substrate. Thus, the rate of the reaction would double if the concentration of NaOH is doubled.

Case (b)

If the concentrations of both NaOH and 1-bromopropane are doubled, the chances of collisions between the nucleophile and the substrate will also increase. Since the rate depends on the concentrations of both reactants, it will be affected by the product of the change in their concentrations. In this case, their concentrations are both doubled, so the rate will increase by a factor of 2 * 2 = 4. Therefore, the rate of the reaction will be quadrupled.

Case (c)

If the volume of the solution is doubled, the concentrations of both reactants will be halved, since the amount of substance remains the same. This means that the probability of collisions between the reactants will decrease. The rate of the reaction depends on the concentrations of both reactants, so it will be affected by the product of the change in their concentrations. In this case, their concentrations are both halved, so the rate will decrease by a factor of 0.5 * 0.5 = 0.25. Therefore, the rate of the reaction will be one-fourth of its initial rate.