anti-Markovnikov

Problem # 345

For the reaction below, draw the structures of the radical intermediate and the final product.

When HBr reacts with in an alkene in the presence of peroxides (which are free radical initiators), the reaction goes through a free radical mechanism and adds in an anti-Markovnikov manner. The product is an alkyl bromide.

The reaction goes anti-Markovnikov because the more substituted radical intermediate is only formed if the bromine adds to the less substituted carbon.

Problem Details

MendelSet practice problem # 345 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Reactions and Markovnikov's Rule

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): alkene addition, anti-Markovnikov

Problem # 344

For the reaction below, draw the structures of the borane intermediate and the final product.

Solution

This is a hydroboration reaction (also called hydroboration-oxidation in some textbooks). The product is an alcohol.

Boron bonds to the less substituted carbon, and is replaced with an oxygen after the borane intermediate is treated with peroxide, resulting in an anti-Markovnikov addition.

Note that one equivalent of BH₃ reacts with three equivalents of alkene; the borane intermediate is not BH₂(alkyl) but rather B(alkyl)₃.