Answer the following questions for this \(S_{\mathrm{N}} 2\) reaction: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{NaOH} & \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+ \mathrm{NaBr} \end{aligned}$$ (a) What is the rate expression for the reaction? (b) Draw the reaction profile for the reaction. Label all parts. Assume that the products are lower in energy than the reactants. (c) What is the effect on the rate of the reaction of doubling the concentration of \(n\) -butyl bromide? (d) What is the effect on the rate of the reaction of halving the concentration of sodium hydroxide?

Step by step solution

01

Determine the rate expression

The rate expression for an \(S_{N}2\) reaction is typically written as the product of the concentration of the nucleophile and the concentration of the substrate (in this case, n-butyl bromide). So the rate expression is \(k[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}][\mathrm{NaOH}]\) where k is the rate constant.

02

Draw the reaction profile

The reaction profile for an \(S_N2\) reaction shows a single hump, representing the transition state. The reactants (\(n\)-butyl bromide and sodium hydroxide) are drawn on the left, higher in energy than the products (n-butyl alcohol and sodium bromide), which are drawn on the right. The difference in energy between the reactants and the products is the enthalpy change for the reaction. The peak of the hump is the transition state (highest energy point) and its difference from the reactants is the activation energy.

03

Effect of doubling the concentration of n-butyl bromide

In an \(S_N2\) reaction, doubling the concentration of the substrate (n-butyl bromide in this case) would double the reaction rate. This is because the rate depends on the concentration of the substrate, as shown in the rate expression derived in Step 1.

04

Effect of halving the concentration of sodium hydroxide

Similarly, if the concentration of the nucleophile (NaOH) is halved, the reaction rate would be halved as well. This is because the reaction rate directly depends on the concentration of the nucleophile.

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