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
C. Statement I is true but Statement II is false
Statement I: When 3-phenylpropene reacts with HBr, protonation of the alkene occurs to form a secondary carbocation. C_6H_5-CH_2-CH=CH_2 + H^+ C_6H_5-CH_2-CH^+-CH_3 This secondary carbocation undergoes a 1,2-hydride shift to form a more stable, resonance-stabilized benzylic carbocation. C_6H_5-CH_2-CH^+-CH_3 C_6H_5-CH^+-CH_2-CH_3 The bromide ion then attacks this benzylic carbocation to form the major product, 1-bromo-1-phenylpropane. C_6H_5-CH^+-CH_2-CH_3 + Br^- C_6H_5-CH(Br)-CH_2-CH_3 The carbon atom bonded to the bromine is attached to four different groups: a phenyl group (-C_6H_5), an ethyl group (-CH_2CH_3), a hydrogen atom (-H), and a bromine atom (-Br). Thus, it is a chiral carbon. The product is a secondary alkyl bromide. Therefore, Statement I is true. Statement II: The Sandmeyer reaction involves treating a diazonium salt with Cu(I) salts (CuCl, CuBr, CuCN) to prepare aryl chlorides, aryl bromides, and aryl cyanides, respectively. The Gattermann reaction involves treating a diazonium salt with copper powder (Cu) and halogen acids (HCl, HBr) to prepare aryl chlorides and aryl bromides. Aryl cyanides are not prepared using the Gattermann reaction. Therefore, Statement II is false. Conclusion: Statement I is true but Statement II is false. Answer: Statement I is true but Statement II is false
Related: Chemistry — Hydrocarbons · All PYQ Banks