Question

Reactions between certain haloalkanes (alkyl halides) and water produce alcohols. Consider the overall reaction for \(t\) -butyl bromide ( 2 -bromo- 2 -methylpropane): \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}(a q)+\mathrm{H}^{+}(a q)+\mathrm{Br}^{-}(a q)\) The experimental rate law is rate \(=k\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\right] .\) The accepted mechanism for the reaction is \((1)\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}(a q) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(a q)+\mathrm{Br}^{-}(a q) \quad[\mathrm{slow}]\) (2) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}_{2}^{+}(a q) \quad[\) fast\(]\) (3) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}_{2}^{+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}(a q) \quad[\) fast\(]\) (a) Why doesn't \(\mathrm{H}_{2} \mathrm{O}\) appear in the rate law? (b) Write rate laws for the elementary steps. (c) What reaction intermediates appear in the mechanism? (d) Show that the mechanism is consistent with the experimental rate law.

Short answer

The rate-determining step involves only \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{CBr}\), explaining why \(\mathrm{H}_{2}\mathrm{O}\) doesn't appear in the rate law. Intermediates are \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}^{+}\) and \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{OH}_{2}^{+}\). The mechanism matches the experimental rate law.

Step-by-step solution

- Identify why \ \(\mathrm{H}_{2}\mathrm{O}\) doesn't appear in the rate law

The rate-determining step, or the slowest step in the reaction mechanism, dictates the overall rate law. Here, step 1 is the slow step: \[ \left(\mathrm{CH}_{3}\right)_{3}\mathrm{CBr}(a q) \longrightarrow \left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}^{+}(a q)+\mathrm{Br}^{-}(a q) \] Since \(\mathrm{H}_{2} \mathrm{O}\) is involved in a subsequent faster step, it doesn’t appear in the rate law. Thus, the rate law depends only on the concentration of \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{CBr}\).

- Write rate laws for the elementary steps

For step 1 (slow): \[ \text{rate = } k_1 \left[\left(\mathrm{CH}_{3}\right)_{3}\mathrm{CBr}\right] \] For step 2 (fast): \[ \text{rate = } k_2 \left[\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}^{+}\right] \left[\mathrm{H}_{2}\mathrm{O}\right] \] For step 3 (fast): \[ \text{rate = } k_3 \left[\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{OH}_{2}^{+} \right] \]

- Identify reaction intermediates

The intermediates in the mechanism are species that are formed in one step and consumed in a subsequent step. They do not appear in the overall reaction. Here, the intermediates are: 1. \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}^{+}\) (tert-butyl cation) 2. \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{OH}_{2}^{+}\) (protonated alcohol)

- Show that the mechanism is consistent with the experimental rate law

The experimentally determined rate law is: \[ \text{rate } = k\left[\left(\mathrm{CH}_{3}\right)_{3}\mathrm{CBr}\right] \] This matches the rate law of the slow step (step 1): \[ \text{rate } = k_1 \left[\left(\mathrm{CH}_{3}\right)_{3}\mathrm{CBr}\right] \] Since the overall rate law agrees with the rate law for the rate-determining step, the given reaction mechanism is consistent with the experimental rate law.

Key concepts

haloalkanes

Haloalkanes, also known as alkyl halides, are organic compounds containing halogen atoms (like bromine or chlorine) bonded to an alkyl group. In the case of the given reaction, \(\text{t-Butyl Bromide (} \left(\text{CH}_3\right)_3 \text{CBr)}\) is a haloalkane. These compounds are crucial in many chemical reactions because they possess a fairly reactive carbon-halogen bond, which is susceptible to breaking and forming new bonds. Breaking this bond is essential in SN1 reactions.
The carbon-bromine bond in \(\text{t-Butyl Bromide}\) is broken in the first step of the reaction, creating a carbocation and a bromide ion.

rate-determining step

In a multi-step chemical reaction, the rate-determining step is the slowest step. This step dictates the overall reaction rate and thus determines the rate law. For the reaction involving \(\text{t-Butyl Bromide}\), the rate-determining step is:

\[ \left(\text{CH}_3\right)_3\text{CBr} \rightarrow \left(\text{CH}_3\right)_3\text{C}^+ + \text{Br}^- \]

This step is slow because it involves the formation of a highly unstable intermediate—a carbocation \(\text{(}\text{CH}_3\right)_3\text{C}^+\) and a bromide ion. Since this is the slowest step, it essentially 'bottlenecks' the reaction, meaning that no matter how fast the subsequent steps are, the overall reaction can't proceed any faster than this rate-determining step allows.

• This explains why \(\text{H}_2\text{O}\) doesn't appear in the overall rate law, as it is involved in later, faster steps.

reaction intermediates

Reaction intermediates are species formed in one step of a reaction mechanism and consumed in subsequent steps without appearing in the overall balanced equation. In this reaction mechanism, the intermediates are:

\- \(\text{(}\text{CH}_3\right)_3\text{C}^+\) (tert-butyl cation): This forms in the first step when \((\text{CH}_3\right)_3text{CBr}\) dissociates.

- \(\text{(}\text{CH}_3\right)_3\text{C-OH}_2^+\) (protonated alcohol): This forms in the second step when the carbocation reacts with water. Both intermediates are transient and quickly transformed into products in the subsequent steps.

• Intermediates provide clues to the reaction pathway and help us understand how the overall products are formed from the reactants.

rate law

The rate law provides an equation that relates the reaction rate to the concentration of reactants. For the given reaction, the experimentally determined rate law is: \[ \text{Rate} = k\left[ \left( \text{CH}_3\right)_3\text{CBr} \right] \]

This means that the rate of the reaction depends solely on the concentration of \( \left( \text{CH}_3\right)_3\text{CBr}\), and not on \(\text{H}_2\text{O} \). This confirms that the rate-determining step is the dissociation of \((\text{CH}_3\right)_3\text{CBr} \rightarrow \left(\text{CH}_3\right)_3\text{C}^+ + \text{Br}^- \). Which is consistent with the mechanism suggested:

• The first step (slow) determines the rate law.
• The rate constant k incorporates all factors affecting the rate that aren't concentrations, like temperature and catalyst presence.