The main product of the following reaction (1 | vedclass

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(A) The reaction is the electrophilic addition of $HBr$ to an alkene $(allylbenzene)$.According to Markovnikov's rule,the electrophile $(H^+)$ adds to

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$C_6H_5-CH_2-CH(Br)-CH_3$

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(A) The reaction is the electrophilic addition of $HBr$ to an alkene $(allylbenzene)$.According to Markovnikov's rule,the electrophile $(H^+)$ adds to the carbon atom of the double bond that has more hydrogen atoms,forming the more stable carbocation.In $C_6H_5-CH_2-CH=CH_2$,the carbocation formed after the addition of $H^+$ to the terminal $CH_2$ group is $C_6H_5-CH_2-CH^+-CH_3$.This is a secondary $(2^{\circ})$ carbocation,which is relatively stable.Then,the nucleophile $(Br^-)$ attacks this carbocation to form the product $C_6H_5-CH_2-CH(Br)-CH_3$ ($1$-phenyl$-2-$bromopropane).Therefore,the correct option is $A$.

The main product of the following reaction $(1 \ mol)$ will be:
$C_6H_5-CH_2-CH=CH_2 + HBr \rightarrow \text{Products}$

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The main product of the following reaction $(1 \ mol)$ will be:$C_6H_5-CH_2-CH=CH_2 + HBr \rightarrow \text{Products}$