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?

Short answer

The rate expression for the reaction is \(k[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}][\mathrm{NaOH}]\). The reaction profile shows one transition peak, with reactants higher in energy than products. Doubling the concentration of n-butyl bromide doubles the reaction rate, while halving the concentration of sodium hydroxide halves the reaction rate.

Step-by-step solution

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.

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.

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.

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.

Key concepts

Reaction Profile

In an SN2 reaction, the reaction profile provides a visual representation of the energy changes that occur during the process. The profile generally features a single high peak, which symbolizes the transition state. As the reaction progresses,

• the reactants, represented here by n-butyl bromide and sodium hydroxide, start off higher in energy on the left.
• The transition state is at the peak of the curve, representing the highest energy point the system encounters. It's a temporary state during which the nucleophile attacks the substrate, replacing the leaving group.
• Finally, the products, n-butyl alcohol and sodium bromide, sit at a lower energy level on the right side, indicating they are more stable than the reactants.

The activation energy, seen as the energy difference between the reactants and the transition state's peak, must be overcome for the reaction to proceed. The energy difference between reactants and products is the enthalpy change.

Rate Expression

The rate expression for an SN2 reaction highlights the relationship between the reaction rate and the concentration of reactants. It's a crucial part of understanding the dynamics of these reactions. For the given reaction, the rate expression is written as \( k[ ext{n-butyl bromide}][ ext{NaOH}] \). This expression:

• reveals that the rate is directly proportional to the concentrations of both the nucleophile, here NaOH, and the substrate, n-butyl bromide.
• indicates that any change in the concentration of either reactant affects the overall rate of the reaction directly.

The constant \( k \) is the rate constant, a fixed value at a given temperature that reflects the reaction's inherent speed under those conditions.

Nucleophile

In SN2 reactions, the nucleophile plays a fundamental role in the mechanism. A nucleophile is an atom or molecule that donates an electron pair to form a new chemical bond. In our example:

• The nucleophile is sodium hydroxide (NaOH).
• It provides the hydroxide ion \( (OH^-) \) which attacks the carbon atom bonded to the bromine in n-butyl bromide, initiating the substitution process.
• This process occurs in a concerted manner, meaning the bond formation and bond breaking occur simultaneously.

The strength of the nucleophile can significantly influence the reaction rate, making it a pivotal feature in SN2 reactions.

Substrate

The substrate in an SN2 reaction is the molecule that undergoes substitution. It is pivotal to understand its nature and role. In this particular reaction:

• The substrate is n-butyl bromide \( ( ext{CH}_3 ext{CH}_2 ext{CH}_2 ext{CH}_2 ext{Br}) \).
• The carbon atom, bonded to the bromine leaving group, is the target for the nucleophile’s attack.
• The reaction occurs in a single kinetic step, meaning the substrate's structure affects how quickly the SN2 process unfolds.

The configuration of the substrate influences the outcome of the reaction significantly, impacting both the rate and the mechanism by which the nucleophile approaches it.