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(A) The reaction is the electrophilic addition of $HCl$ to $1$-phenylpropene $(C_6H_5-CH=CH-CH_3)$.According to Markovnikov's rule,the $H^+$ ion attac

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$C_6H_5CHClCH_2CH_3$

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(A) The reaction is the electrophilic addition of $HCl$ to $1$-phenylpropene $(C_6H_5-CH=CH-CH_3)$.According to Markovnikov's rule,the $H^+$ ion attaches to the carbon atom with more hydrogen atoms,and the $Cl^-$ ion attaches to the carbon atom with fewer hydrogen atoms.However,the stability of the intermediate carbocation is the deciding factor.The addition of $H^+$ to the $C_2$ position (the carbon adjacent to the phenyl ring) forms a resonance-stabilized benzylic carbocation: $C_6H_5-CH^+-CH_2-CH_3$.The $Cl^-$ ion then attacks this carbocation to form $1$-phenyl-$1$-chloropropane,which is $C_6H_5CHClCH_2CH_3$.

What is the product formed by the addition of $HCl$ to $1$-phenylpropene?

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Observe the following reactions:
$1$. $C_6H_5-CH=CH_2 + HBr \rightarrow X$
$2$. $C_6H_5-C(CH_3)=CH_2 + HBr \rightarrow Y$
$3$. $C_6H_5-CH=CH_2 + HBr \xrightarrow{(C_6H_5COO)_2} Z$
The correct order of reactivity of $X, Y, Z$ towards $S_N1$ reaction is:

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