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During the aqueous hydrolysis of 2‑bromo‑2‑methylpropane () with excess NaOH, which of the following rate laws correctly describes the reaction?

hard MCQ 2 marks Class 12 · Chemistry · Haloalkanes and Haloarenes
  1. A. Rate = k[ (CH₃)₃CBr ] (first order)
  2. B. Rate = k[NaOH] (first order in base)
  3. C. Rate = k[ (CH₃)₃CBr ][NaOH] (second order)
  4. D. Rate = k[ (CH₃)₃CBr ]² (second order in substrate)
Answer
A. Rate = k[ (CH₃)₃CBr ] (first order)

The correct answer is A. Rate = k[ (CH₃)₃CBr ] (first order) — The correct rate expression is , a first‑order law.

Step-by-step solution

  1. 2‑bromo‑2‑methylpropane is a tertiary alkyl halide; its hydrolysis follows an SN1 mechanism.
  2. In SN1, the rate‑determining step is the unimolecular dissociation of the C–Br bond to give a carbocation.
  3. Therefore the rate depends only on the concentration of the substrate, not on [OH⁻].
  4. The correct rate expression is , a first‑order law.
Common mistake: Writing a second‑order rate law assuming an SN2 mechanism

Concept tested: Rate law for SN1 hydrolysis of tertiary haloalkane

Published · How this solution was prepared and checked

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