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COMEDK · Chemistry · 25. Haloalkanes and Haloarenes

The reactions taking place with \(\mathrm{2- Phenyl-2-bromopropane}\) as the starting material is shown below. Identify [A] and [B] formed in the reaction.
\(
\mathrm{C}_6 \mathrm{H}_5-\mathrm{C}\left(\mathrm{CH}_3\right)_2^{-} \mathrm{Br} \xrightarrow[\Delta]{\text { KOH Alcoholic }}[\mathrm{A}]
\)
\(
[\mathrm{A}]+\mathrm{HBr} \xrightarrow{\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO})_2 \mathrm{O}\right.}[\mathrm{B}]\)

  1. A \([\mathrm{A}]=2\) - Phenylpropan-2-ol
    \([\mathrm{B}]=2\)-Phenyl-2-bromopropene
  2. B \([A]=4 \text {-Hydroxyphenyl-2-bromopropane } \quad[B]=4 \text {-Hydroxyphenylpropene }\)
  3. C \([\mathrm{A}]=2\)-Phenylpropene
    \([\mathrm{B}]=2\)-Phenyl-1- bromopropane
  4. D \([\mathrm{A}]=2\)-Bromopropene
    \([\mathrm{B}]=1\)-Bromopropane
Verified Solution

Answer & Solution

Correct Answer

(C) \([\mathrm{A}]=2\)-Phenylpropene
\([\mathrm{B}]=2\)-Phenyl-1- bromopropane

Step-by-step Solution

Detailed explanation

The starting material is 2-phenyl-2-bromopropane, which is a tertiary alkyl halide. Treatment with alcoholic KOH under heating conditions leads to a dehydrohalogenation reaction via the E2 mechanism. The elimination of HBr from 2-phenyl-2-bromopropane results in the formation of…
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