JEE MainChemistryHydrocarbons
What is the major product formed when N -allylbut-2-enamide is treated with 2 equivalents of HBr ?
Options
- ACH ₃- CH ₂- CH ( Br )- C (= O )- NH - CH ₂- CH ( Br )- CH ₃
- BCH ₃- CH ( Br )- CH ₂- C (= O )- NH - CH ₂- CH ( Br )- CH ₃
- CCH ₃- CH ( Br )- CH ₂- C (= O )- NH - CH ₂- CH ₂- CH ₂ Br
- 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