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What is the major product formed when N -allylbut-2-enamide is treated with 2 equivalents of HBr ?

Options

  1. ACH ₃- CH ₂- CH ( Br )- C (= O )- NH - CH ₂- CH ( Br )- CH ₃
  2. BCH ₃- CH ( Br )- CH ₂- C (= O )- NH - CH ₂- CH ( Br )- CH ₃
  3. CCH ₃- CH ( Br )- CH ₂- C (= O )- NH - CH ₂- CH ₂- CH ₂ Br
  4. DCH ₃- CH ₂- CH ( Br )- C (= O )- NH - CH ₂- CH ₂- CH ₂ Br

Correct answer

B. CH ₃- CH ( Br )- CH ₂- C (= O )- NH - CH ₂- CH ( Br )- CH ₃

Step-by-step solution

The reactant N -allylbut-2-enamide has the structure CH ₃- CH = CH - C (= O )- NH - CH ₂- CH = CH ₂ . It contains two different carbon-carbon double bonds: an , -unsaturated amide and an isolated terminal alkene. For the , -unsaturated system ( CH ₃- CH = CH - C (= O )- ), the electron-withdrawing nature of the carbonyl group destabilizes any positive charge at the -position. Therefore, protonation occurs at the -carbon to form a more stable carbocation at the -position (which is also stabilized by resonance in the

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