The alkene shown reacts with $HBr$ either in the dark or with $ROOR$ . Select all correct statements.

AIn the dark the product is the tertiary benzylic bromide.
BWith $ROOR$ the product is the primary alkyl bromide formed by bromine radical adding to the $CH X_{2}$ carbon.
CWith $ROOR$ the intermediate radical is tertiary and benzylic.
DThe two conditions give the same constitutional isomer.

Answer: A, B, C

The alkene is alpha-methylstyrene. In the dark,

H X^{+}

adds to the

CH X_{2}

carbon to give the tertiary benzylic cation and then (2-bromopropan-2-yl)benzene. With peroxide,

Br

radical adds to the

CH X_{2}

carbon, giving the tertiary benzylic radical, and the product is (1-bromopropan-2-yl)benzene. Because the two products are constitutional isomers, statement D is false.

Practice Organic Reaction Mechanisms II: Electrophilic and Radical Reactions More on Markovnikov and anti-Markovnikov products of drawn alkenes

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The alkene shown reacts with HBr either in the dark or with ROOR. Select all correct statements.

Answer: A, B, C

More questions on Markovnikov and anti-Markovnikov products of drawn alkenes

The alkene shown reacts with $HBr$ either in the dark or with $ROOR$ . Select all correct statements.