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(C) In reaction $(i)$,the addition of $HBr$ to styrene $(C_6H_5-CH=CH_2)$ follows Markovnikov's rule. The electrophile $H^+$ adds to the terminal carb

Vedclass · Accepted Answer

For $(i)$,$X = C_6H_5-CH(Br)-CH_3$; For (ii),$Y = C_6H_5-CH_2-CH_2-CH_2Br$

Vedclass · Answer

(C) In reaction $(i)$,the addition of $HBr$ to styrene $(C_6H_5-CH=CH_2)$ follows Markovnikov's rule. The electrophile $H^+$ adds to the terminal carbon to form the more stable benzylic carbocation $(C_6H_5-CH^+-CH_3)$,which then reacts with $Br^-$ to give $X = C_6H_5-CH(Br)-CH_3$.In reaction (ii),the addition of $HBr$ to allylbenzene $(C_6H_5-CH_2-CH=CH_2)$ in the presence of peroxide follows the anti-Markovnikov addition (Kharasch effect). The bromine radical adds to the terminal carbon to form the more stable radical,resulting in $Y = C_6H_5-CH_2-CH_2-CH_2Br$.Thus,the correct products are $X = C_6H_5-CH(Br)-CH_3$ and $Y = C_6H_5-CH_2-CH_2-CH_2Br$.

Identify the major products $X$ and $Y$ in the following reactions:
$(i)$ $C_6H_5-CH=CH_2 + HBr \rightarrow X$
(ii) $C_6H_5-CH_2-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} Y$

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Identify the major products $X$ and $Y$ in the following reactions:$(i)$ $C_6H_5-CH=CH_2 + HBr \rightarrow X$(ii) $C_6H_5-CH_2-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} Y$