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(A) Statement $I$: $3$-phenylpropene $(C_6H_5CH_2CH=CH_2)$ reacts with $HBr$ via electrophilic addition following Markovnikov's rule.The intermediate

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Both Statement $I$ and Statement $II$ are true

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(A) Statement $I$: $3$-phenylpropene $(C_6H_5CH_2CH=CH_2)$ reacts with $HBr$ via electrophilic addition following Markovnikov's rule.The intermediate carbocation formed at $C-2$ $(C_6H_5CH_2CH^+CH_3)$ is more stable due to the inductive effect and hyperconjugation compared to the primary carbocation at $C-1$.The major product is $2$-bromo-$3$-phenylpropane $(C_6H_5CH_2CH(Br)CH_3)$, which contains a chiral center at $C-2$.Thus, Statement $I$ is true.Statement $II$: The Sandmeyer reaction uses $Cu_2Cl_2/HCl$ or $Cu_2(CN)_2/KCN$ to convert diazonium salts to aryl chlorides or aryl cyanides.The Gattermann reaction is a modification of the Sandmeyer reaction that uses copper powder $(Cu)$ in the presence of the corresponding acid ($HCl$ or $HCN$).Both reactions are standard methods for the preparation of aryl halides and cyanides.Thus, Statement $II$ is true.

Column-$I$ (Reactants)	Column-$II$ (Type of Reaction)
$(A)$ $C_6H_5NH_2 \xrightarrow{NaNO_2, HX, 273-278 K} C_6H_5N_2^+X^-$	$(i)$ Wurtz-Fittig reaction
$(B)$ $C_6H_5N_2^+X^- \xrightarrow{Cu_2X_2} C_6H_5X$	$(ii)$ Diazotization
$(C)$ $C_6H_6 + 2Na + RX \xrightarrow{\text{dry ether}} C_6H_5R$	$(iii)$ Sandmeyer reaction
$(D)$ $2C_6H_6 + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5$	$(iv)$ Fittig reaction

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.

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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.