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JEE Advanced levelSingle correct optionDifficulty 3 of 3o4205

Under identical conditions (/EtOH, same temperature) the E2 elimination of 2-phenylethyl bromide, , is much faster than that of 1-bromopropane, . The best explanation is:

  1. Athe phenyl group makes bromide a much better leaving group by an inductive effect
  2. Ba benzylic carbocation forms first, so the reaction becomes E1
  3. Cphenyl relieves steric strain by making the beta-hydrogens easier to remove
  4. Dthe beta-hydrogens are more acidic and the developing double bond conjugates with the ring, stabilising the transition state

Answer: D

Phenyl is electron-withdrawing and can delocalise negative charge, so the benzylic beta-hydrogens are more acidic; the incipient C=C is conjugated with the ring in the transition state. The carbon bearing Br is primary, so a carbocation does not form, and phenyl does not affect the leaving group directly.
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