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(B) The reaction is the anti-Markovnikov addition of $HBr$ to an alkene in the presence of peroxide $(R_2O_2)$.The reactant is $3,4$-dimethylhex-$3$-e

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$4$

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(B) The reaction is the anti-Markovnikov addition of $HBr$ to an alkene in the presence of peroxide $(R_2O_2)$.The reactant is $3,4$-dimethylhex-$3$-ene.Upon addition of $HBr$ via a free radical mechanism,the bromine radical adds to one of the carbons of the double bond,and the hydrogen adds to the other.This creates two chiral centers in the product,$3$-bromo-$3,4$-dimethylhexane.Since there are two chiral centers,there are $2^n = 2^2 = 4$ possible stereoisomers (optical isomers) formed as products.These are the $(R,R)$,$(S,S)$,$(R,S)$,and $(S,R)$ configurations.

How many products will be formed in the reaction shown below?
$CH_3-CH_2-C(CH_3)=C(CH_3)-CH_2-CH_3 + HBr \xrightarrow{R_2O_2} \text{Products}$

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How many products will be formed in the reaction shown below?$CH_3-CH_2-C(CH_3)=C(CH_3)-CH_2-CH_3 + HBr \xrightarrow{R_2O_2} \text{Products}$