Ph-CH=C(CH_3)_2 xleftarrow{HBr} A and Ph-CH=C | vedclass

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(B) The reaction of $Ph-CH=C(CH_3)_2$ with $HBr$ follows Markovnikov's rule. The electrophilic $H^+$ adds to the carbon with more hydrogens,forming a

Vedclass · Accepted Answer

$A = Ph-CH(Br)-CH(CH_3)_2$ and $B = Ph-CH_2-C(Br)(CH_3)_2$

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(B) The reaction of $Ph-CH=C(CH_3)_2$ with $HBr$ follows Markovnikov's rule. The electrophilic $H^+$ adds to the carbon with more hydrogens,forming a stable benzylic carbocation,which then reacts with $Br^-$ to give $A = Ph-CH(Br)-CH(CH_3)_2$.In the presence of peroxide,the reaction follows the Anti-Markovnikov rule (Kharasch effect) via a free radical mechanism. The bromine radical adds to the terminal carbon to form a stable benzylic radical,which then abstracts a hydrogen atom to give $B = Ph-CH_2-C(Br)(CH_3)_2$.

$Ph-CH=C(CH_3)_2 \xleftarrow{HBr} A$ and $Ph-CH=C(CH_3)_2 \xrightarrow[\text{Peroxide}]{HBr} B$. Identify $A$ and $B$.

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