Question

Which of the following alkenes would produce a secondary alkyl halide upon reacting with HBr and H₂O₂?

a. This is an addition reaction that proceeds through radicals because of the peroxide.
b. Because the reaction is run with HBr and H₂O₂, the reaction is an Anti-Markovnikov addition. This means that the Br will add to the LEAST substituted side.
c. The products of each reactant are shown below. Note that this only occurs with HBr, not HCl nor HI.
d. This is because the propagation steps with HCl and HI are endothermic,
e. But the propagation steps with HBr are exothermic and favorable.

Answer 1

An alkene reacts with HBr and H₂O₂ via an Anti-Markovnikov addition to produce a secondary alkyl halide, with the bromine adding to the least substituted carbon.

Step-by-step explanation:

When an alkene reacts with HBr and H₂O₂, a secondary alkyl halide is produced through an Anti-Markovnikov addition reaction. This reaction proceeds via a free radical mechanism, initiated by the H₂O₂, which is a free radical initiator. During this reaction, the bromine (Br) adds to the least substituted side of the alkene because the propagation steps with HBr are exothermic and favorable, leading to the formation of a monosubstituted alkyl halide. It is important to note that this specific addition to an alkene to form an alkyl halide is unique to HBr and does not occur with HCl or HI due to their endothermic propagation steps.