JEE Main20266 April 2026Morning ShiftChemistryHydrocarbonsActual
Given below are two statements: Statement I: 3 -phenylpropene reacts with HBr and gives secondary alkyl bromide having a chiral carbon atom as the major product. Statement II: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction. In the light of the above statements, choose the correct answer from the options given below
Options
- ABoth Statement I and Statement II are true
- BBoth Statement I and Statement II are false
- CStatement I is true but Statement II is false
- DStatement I is false but Statement II is true
Correct answer
C. Statement I is true but Statement II is false
Step-by-step solution
Statement I: When 3 -phenylpropene reacts with HBr , protonation of the alkene occurs to form a secondary carbocation. C₆H₅-CH₂-CH=CH₂ + H^+ C₆H₅-CH₂-CH^+-CH₃ This secondary carbocation undergoes a 1,2 -hydride shift to form a more stable, resonance-stabilized benzylic carbocation. C₆H₅-CH₂-CH^+-CH₃ C₆H₅-CH^+-CH₂-CH₃ The bromide ion then attacks this benzylic carbocation to form the major product, 1 -bromo- 1 -phenylpropane. C₆H₅-CH^+-CH₂-CH₃ + Br^- C₆H₅-CH(Br)-CH₂-CH₃ The carbon atom bonded to the bromine is attac