Question
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 1
From the balanced equation, we can see that one molecule of $n$-butyl bromide reacts with one molecule of sodium hydroxide to form one molecule of $n$-butanol and one molecule of sodium bromide. Show more…
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