Reactions between certain haloalkanes (alkyl halides) and...

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.

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.

Key Concepts

SN1 Reaction Mechanism

The SN1 mechanism is a unimolecular nucleophilic substitution which involves a two-step process: first, the formation of a carbocation intermediate in the rate-determining step, and second, the attack of this carbocation by a nucleophile. The overall kinetics of the reaction reflect only the first step, which is why the experimental rate law depends solely on the concentration of the haloalkane.

Reaction Intermediates

Reaction intermediates are transient species that are formed in one step of the reaction mechanism and consumed in a subsequent step. They do not appear in the overall balanced equation or the final rate law because their concentrations do not build up significantly during the reaction process.

Rate-Determining Step

In a multi-step reaction mechanism, the overall rate law is governed by the slow, rate?determining step. This concept means that only the species involved in the slow step appear in the experimental rate law, which is why, even if water participates in subsequent fast steps, it does not appear in the rate expression if it is not involved in the slowest step.

Elementary Reaction Rate Laws

Elementary steps are reactions that occur in a single event with a known molecularity and have rate laws that directly reflect the number of molecules involved. For example, a unimolecular step has a rate law that depends on the concentration of a single reactant, while a bimolecular step involves the product of the concentrations of two reactants.

Transcript

Please give Ace some feedback