Given below are two statements:Statement I: 3 | vedclass

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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-$I$	Column-$II$
$A$. $S_N1$ reaction	$1$. vic-dibromides
$B$. Chemicals in fire extinguisher	$2$. gem-dihalides
$C$. Bromination of alkenes	$3$. Racemisation
$D$. Alkylidene halides	$4$. Saytzeff rule
$E$. Elimination of $HX$ from alkyl halide	$5$. Chlorobromocarbons

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.