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NEETChemistryHaloalkanes and Haloarenes

Consider the following reaction sequence: 3-methylbutan-2-ol PBr ₃ A (major) [ ] alc. KOH B (major) Identify the IUPAC names of the major products A and B.

Options

  1. AA = 2-bromo-3-methylbutane, B = 3-methylbut-1-ene
  2. BA = 2-bromo-2-methylbutane, B = 2-methylbut-2-ene
  3. CA = 2-bromo-3-methylbutane, B = 2-methylbutan-2-ol
  4. DA = 2-bromo-3-methylbutane, B = 2-methylbut-2-ene

Correct answer

D. A = 2-bromo-3-methylbutane, B = 2-methylbut-2-ene

Step-by-step solution

In the first step, 3-methylbutan-2-ol reacts with PBr ₃ . This reagent converts alcohols to alkyl bromides via a nucleophilic substitution mechanism without carbocation rearrangement. The hydroxyl group is replaced by a bromine atom, yielding 2-bromo-3-methylbutane as product A. In the second step, 2-bromo-3-methylbutane is treated with alcoholic KOH and heated. Alcoholic KOH acts as a strong base and causes dehydrohalogenation ( -elimination). The -hydrogens are located on C1 (a primary carbon) and C3 (a tertiary

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