JEE Advanced levelSingle correct optionDifficulty 3 of 3o4161
1-Bromopropane is heated with alcoholic KOH to give Y. Y is treated with HBr (no peroxide) to give Z, and Z is then treated with NaNX3 in DMF. The final product is
A1-azidopropane
Bpropene
Cpropan-2-ol
D2-azidopropane
Answer: D
Alcoholic KOH gives Y = propene. HBr adds by Markovnikov's rule to give Z = 2-bromopropane. Azide is a good nucleophile and weak base in a polar aprotic solvent, so SN2 gives (CHX3)X2CH−NX3, 2-azidopropane.