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(B) The reaction is the electrophilic addition of $HBr$ to $1-phenylprop-1-ene$ $(C_6H_5-CH=CH-CH_3)$.According to Markovnikov's rule,the proton $(H^+

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

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(B) The reaction is the electrophilic addition of $HBr$ to $1-phenylprop-1-ene$ $(C_6H_5-CH=CH-CH_3)$.According to Markovnikov's rule,the proton $(H^+)$ adds to the carbon atom with more hydrogen atoms,and the bromide ion $(Br^-)$ adds to the carbon atom with fewer hydrogen atoms.In this case,the carbocation formed at the benzylic position $(C_6H_5-CH^+-CH_2-CH_3)$ is highly stabilized by resonance with the phenyl ring.Therefore,the bromide ion attacks this stable benzylic carbocation to form $1-bromo-1-phenylpropane$ as the major product.The reaction is: $C_6H_5-CH=CH-CH_3 + HBr \rightarrow C_6H_5-CH(Br)-CH_2-CH_3$.

Identify the product of the following reaction.

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