Question

Give the structures of the free-radical intermediates in the peroxide-initiated reaction of HBr with the following alkene. Include all lone-pair electrons and unpaired electrons. Hint: the radicals do not coexist in the same mechanistic step.

Answer 1

Concepts and reason

Hydrogen halide addition to the alkene leads to the alkyl halide product. This addition reaction follows the anti-Markovnikov’s addition pathway in the presence of peroxides, in the absence of peroxides leads to the Markovnikov’s addition product.

Fundamentals

Markovnikov’s addition: Markovnikov’s addition of hydrogen halide to alkene produces the most stable alkyl halide.

Markovnikov’s rule says that the negatively-charged addendum in hydrogen halide goes to the carbon position, which is connected to fewer hydrogen atoms than the other end of the alkene.

Anti-Markovnikov’s addition rule says that the negatively-charged addendum in hydrogen halide goes to the carbon position, which is connected to a higher number of hydrogen atoms than the other end of the alkene.

Given alkene structure is drawn below:

${\rm{HBr}}$ addition to this alkyne follows the radical pathway due to the presence of peroxide.

Bromine radical reacts with the given alkene.

The reaction between alkyl radical and hydrogen bromide is as follows:

The overall reaction is as follows:

Ans:

The radicals produced during the hydro halogenation reaction are shown below:

Answer 2

Hint: One organic radical and one inorganic radical. Br H2 Br:

Note the secondary organic radical and the inorganic radical that forms from the Br₂ that splits.