Question

Addition of hydrogen bromide to propene yields $2 - bromopropane$, while in the presence of benzoyl peroxide, the same reaction yields $1 - bromopropane$. Explain and give mechanism.

Answer 1

Hint: We know that Markovnikov’s rule is employed to predict regioselectivity of electrophilic addition reactions of alkenes and alkynes. It states that, in hydrohalogenation of an unsymmetrical alkene, the atom within the hydrogen halide forms a bond with the doubly bonded atom within the alkene, bearing the greater number of hydrogen atoms.

Complete step by step answer: Let we see details about the formation of 1-bromopropane and 2-bromopropane.
Formation of 2-Bromopropane:
Addition to HBr to propene (unsymmetrical alkene) follows Markovnikov Rule according to which the negative part of the addition gets attached to that C atom which possesses a lesser number of hydrogen atoms. The reaction proceeds via an ionic mechanism and forms carbocation as intermediate. Secondary carbocation is more stable than primary carbocation.

We know that the secondary carbocations are more stable than primary carbocation. Hence, in the next step, bromine attacks the secondary carbocation to form $2 - bromopropane$as the major product.

Formation of 1-Bromopropane:
In the presence of benzoyl peroxide, addition to\[\;HBr\] to propene gives\[1{\text{ }} - {\text{ }}bromopropane\] via anti Markovnikov’s rule. This happens in the presence of peroxide and with \[\;HBr\] only. The reaction proceeds through a free radical mechanism.
Step-1: Formation of radical

Step-2: Formation of Br radical

Step-3: Formation of mixture of 1-Bromopropane and 2-Bromopropane

We know that,
Secondary free radicals are more stable than primary radicals and thus, $1 -bromopropane$ is obtained as the major product.

Note: We must remember that in the presence of peroxide,$Br$ free radical acts as an electrophile. Hence, two different products are obtained in addition to \[HBr\] to propene in the absence and presence of peroxide.