What are the products of the following antimarkovnikov reaction: #CH_3-C(CH_3)=CH-CH_3 + HBr + H_2O_2# ?

Answer 1

Ellie Adams

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

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Answer 2

Easton Jones

The products of the given antimarkovnikov reaction are 2-bromo-2-methylpropane and water.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.