What are the products of the following antimarkovnikov reaction: #CH_3-C(CH_3)=CH-CH_3 + HBr + H_2O_2# ?
1 Answer
The product is 2-bromo-3-methylbutane, (CH₃)₂CH-CHBr-CH₃.
In a Markovnikov addition of HBr, the H adds to the alkene carbon that has more H atoms (C-3).
(CH₃)₂C=CH-CH₃ + HBr → (CH₃)₂CBr-CH₂-CH₃
In an anti-Markovnikov addition of HBr, the H adds to the alkene carbon that has fewer H atoms (C-2).
(CH₃)₂C=CH-CH₃ + HBr → (CH₃)₂CH-CHBr-CH₃
