Product 'A' and 'B' are respectively: CH_3-C( | vedclass

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Question

(A) In the presence of peroxide,$HBr$ adds to alkenes via a free radical mechanism (Anti-Markovnikov addition). The $Br^\bullet$ radical adds to the c

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) In the presence of peroxide,$HBr$ adds to alkenes via a free radical mechanism (Anti-Markovnikov addition). The $Br^\bullet$ radical adds to the carbon atom to form the more stable radical. In this case,the benzylic radical is more stable than the tertiary radical,so $Br$ adds to the carbon with two methyl groups to form product '$A$': $CH_3-C(Br)(CH_3)-CH_2-C_6H_5$. In the absence of peroxide,$HBr$ adds via an electrophilic mechanism (Markovnikov addition). The $H^+$ adds to form the more stable carbocation. The benzylic carbocation is more stable than the tertiary carbocation,so $Br^-$ adds to the benzylic carbon to form product '$B$': $CH_3-CH(CH_3)-CH(Br)-C_6H_5$.

Product '$A$' and '$B$' are respectively:
$CH_3-C(CH_3)=CH-C_6H_5 \xrightarrow{HBr/Peroxide} A$
$CH_3-C(CH_3)=CH-C_6H_5 \xrightarrow{HBr} B$

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Product '$A$' and '$B$' are respectively: $CH_3-C(CH_3)=CH-C_6H_5 \xrightarrow{HBr/Peroxide} A$ $CH_3-C(CH_3)=CH-C_6H_5 \xrightarrow{HBr} B$