Write the mechanism involved in the reaction between tertiary butyl bromide and
aqueous KOH. Mention its order.
Answers
Explanation:
Mechanism for the formation of tert-Butyl Chloride ... chloride is treated with aqueous bicarbonate. ... Initially, the KOH removes the hydrogen bonded to the N
Answer:
The alkaline hydrolysis to tert-butyl bromide with aqueous alkali KOH is as follows.
(CH₃)₃C-Br+ NaOH(aq)⇒(CH₃)₃C-OH + NaBr.
tert-butyl bromide tert-butyl alcohol
The amount of alkali employed has no effect on the rate of this reaction, which is exclusively regulated by the concentration of tert-butyl bromide.
Rate ∝[(CH₃)₃C-Br]
Rate=K[(CH₃)C₃-Br]
This first order reaction because rate of hydrolysis of (CH₃)C₃-Br is independent of the concentration of alkali or OH⁻ ions. This can be explained by two steps mechanism. Each step is a simple reaction with a rate constant of its own.
Step 1. (CH₃)₃C-Br⇒(CH₃)C⁺+Br⁻
Rate of reaction=k₁[(CH₃)C-Br]
This type of mechanism is SN¹ mechanism.
Step 2. (CH₃)CH⁺+OH⁻⇒(CH₃)C-OH
Rate of reaction=k₂[(CH₃)₃C⁺]OH⁻
The second involves the attacks of OH⁻ ion. This is the fast step since it is a bond formation.
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