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:

For Statement I, 3-phenylpropene ((C_6H_5CH_2CH=CH_2)) undergoes electrophilic addition with (HBr) according to Markovnikov’s rule. Protonation of the double bond generates a resonance-stabilized benzylic carbocation, which is then attacked by (Br^-) to form 1-phenyl-1-bromopropane.

The product formed is

$$C_6H_5CH(Br)CH_2CH_3.$$

The carbon bearing the bromine atom is attached to four different groups ((Br,\ H,\ CH_2CH_3,\ C_6H_5)), making it a chiral carbon. Hence, Statement I is true.

For Statement II, the Sandmeyer reaction is used to prepare aryl chlorides and aryl cyanides from diazonium salts using cuprous salts. Although the Gattermann reaction can be employed for the preparation of aryl chlorides and aryl bromides, it is not the standard method for the preparation of aryl cyanides.

Therefore, Statement II is false.

Hence, Statement I is true, Statement II is false, and the correct answer is Option C.