Statements on Hydrohalogenation and Preparation of Aryl Compounds

To determine the correctness of both statements, let us analyze them step-by-step:

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Evaluation of Statement I:

1. Structure of Reactant: 3-phenylpropene has the molecular formula $\text{C}_6\text{H}_5\text{-CH}_2\text{-CH=CH}_2$.

2. Electrophilic Addition Mechanism ($\text{HBr}$):

   • Step 1: Protonation The double bond undergoes electrophilic addition with $\text{H}^+$ to form a secondary carbocation: $\text{C}_6\text{H}_5\text{-CH}_2\text{-CH=CH}_2 + \text{H}^+ \longrightarrow \text{C}_6\text{H}_5\text{-CH}_2\text{-}\overset{+}{\text{C}}\text{H-CH}_3$

   • Step 2: Rearrangement via 1,2-Hydride Shift The secondary carbocation rearranges via a $1,2$-hydride shift to form a more stable benzylic carbocation, which is stabilized by resonance with the aromatic ring: $\text{C}_6\text{H}_5\text{-CH}_2\text{-}\overset{+}{\text{C}}\text{H-CH}_3 \xrightarrow{1,2\text{-H shift}} \text{C}_6\text{H}_5\text{-}\overset{+}{\text{C}}\text{H-CH}_2\text{-CH}_3$

   • Step 3: Nucleophilic Attack by $\text{Br}^-$ The bromide ion attacks the benzylic carbocation to form the major product: $\text{C}_6\text{H}_5\text{-}\overset{+}{\text{C}}\text{H-CH}_2\text{-CH}_3 + \text{Br}^- \longrightarrow \text{C}_6\text{H}_5\text{-CH(Br)-CH}_2\text{-CH}_3$

3. Product Characteristics:

   • The major product is 1-bromo-1-phenylpropane.
   • The carbon attached to the bromine atom is bonded to two other carbon atoms (the benzylic ring carbon and the ethyl group carbon), making it a secondary alkyl bromide.
   • The carbon atom ($\text{C}^*$) in $\text{C}_6\text{H}_5\text{-}\text{C}^*\text{H(Br)-CH}_2\text{CH}_3$ is bonded to four distinctly different groups: $-\text{H}$, $-\text{Br}$, $-\text{C}_6\text{H}_5$, and $-\text{CH}_2\text{CH}_3$. Hence, it is a chiral carbon atom.

Therefore, Statement I is true.

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Evaluation of Statement II:

1. Sandmeyer Reaction: Benzenediazonium chloride reacts with cuprous salts like $\text{Cu}_2\text{Cl}_2/\text{HCl}$, $\text{Cu}_2\text{Br}_2/\text{HBr}$, and $\text{CuCN}/\text{KCN}$ to give aryl chlorides ($\text{Ar-Cl}$), aryl bromides ($\text{Ar-Br}$), and aryl cyanides ($\text{Ar-CN}$), respectively.

2. Gattermann Reaction: In the Gattermann reaction, copper powder in the presence of the corresponding halogen acid ($\text{Cu}/\text{HCl}$ or $\text{Cu}/\text{HBr}$) is used to prepare aryl chlorides and aryl bromides: $\text{Ar-N}_2^+\text{X}^- \xrightarrow{\text{Cu/HCl}} \text{Ar-Cl} + \text{N}_2 + \text{CuX}$ However, aryl cyanides ($\text{Ar-CN}$) cannot be prepared by the Gattermann reaction.

Therefore, Statement II is false.

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Conclusion:

• Statement I is true
• Statement II is false

This corresponds to Option C.