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🧪 Chemistry  ·  Haloalkanes & Haloarenes  ·  NEET & JEE

The E2 elimination of (2R,3S)-2-bromo-3-phenylbutane gives predominantly:

Answer: cis (Z)-2-phenyl-2-butene due to anti periplanar arrangement requirement.

  • A trans (E)-2-phenyl-2-butene, formed from the corresponding diastereomeric bromide instead
  • B An equal mixture of cis and trans alkenes regardless of stereochemistry
  • C Only the terminal alkene 1-phenyl-2-butene via a different elimination pathway
  • D cis (Z)-2-phenyl-2-butene due to anti periplanar arrangement requirement

Correct answer: D. cis (Z)-2-phenyl-2-butene due to anti periplanar arrangement requirement

Explanation: E2 requires anti-periplanar arrangement of H and Br. In (2R,3S) configuration, the anti periplanar conformer has H and Br on adjacent carbons arranged so the resulting alkene has the phenyl and methyl on the same side: Z (cis) alkene is the major product.

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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