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🧪 Chemistry  ·  Haloalkanes & Haloarenes  ·  NEET & JEE

Dehydrohalogenation of 2-bromo-2-methylbutane with alcoholic KOH gives, as the major product:

Answer: 2-methylbut-2-ene.

  • A 2-methylbut-2-ene
  • B 2-methylbut-1-ene
  • C 3-methylbut-1-ene
  • D 2-methylbutan-1-ol

Correct answer: A. 2-methylbut-2-ene

Explanation: By Saytzeff's rule the more substituted (more stable) alkene, 2-methylbut-2-ene, predominates.

SN2: one stepNu-CX-backside attackNuC+ X-Inversion of configuration(Walden inversion)SN1: two stepsCX-slow: leaving group exits+CcarbocationNu-Planar intermediate givesracemisation (both faces attacked)

SN2 proceeds in a single step with backside attack and inversion of configuration, while SN1 forms a planar carbocation intermediate first, leading to racemisation.

Concept context

Study carbon-halogen compounds: how they are made, how they react via SN1/SN2 and elimination, their stereochemistry, and their environmental impact.

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