Question

How can I illustrate antimarkovnikov's rule by the addition of hydrogen bromide to propene in the presence of benzoyl peroxide?

Answer 1

The reaction proceeds in three steps, mainly initiation, propagation and termination. A free radical is generated and is required for the reaction to continue.

Complete answer:
In order to answer the question, let us get to know about the chemical properties of alkenes. We have discussed the structure of alkenes based on hybridisation in unit-4 and are quite aware of the fact that the carbon atoms involved in $C=C$bond are $s{{p}^{2}}$ hybridised. Moreover, the double bond is made up of a stable sigma bond surrounded by the pi-electron cloud containing a pi bond. No doubt, double bond is stronger (bond dissociation energy 610 kJ per mole than single bond (bond dissociation energy = 347 kJ per mole) but at the same time it is more reactive since it has a weak pi-bond formed by lateral or sidewise overlap.
From a mechanism point of view, the addition in alkenes is generally electrophilic in nature which means that attacking reagent which carries the initial attack is an electrophile. The mechanism proceeds in two steps.
Now, let us come to the question. The reaction can be divided into three steps, known as initiation, propagation and termination.
Initiation: Here, the chain is initiated by the free radicals which is generated as the $O-O$ bond in the benzoyl peroxide gets cleaved. As a result, bromine radical is generated.
$PhCOO\bullet +H-Br\to PhCOOH+\bullet Br$
Propagation: The bromine radical gets added to the propene and forms the stable ${{2}^{0}}$radical, which in turn reacts with HBr to produce another bromine radical.
Termination: The radicals combine together and form the final molecule.

& Br\bullet +\bullet Br\to B{{r}_{2}} \\\ & C{{H}_{3}}CHC{{H}_{2}}-Br+\bullet Br\to C{{H}_{3}}CHBrC{{H}_{2}}-Br \\\ & 2C{{H}_{3}}CHC{{H}_{2}}Br\to BrC{{H}_{2}}CH(C{{H}_{3}})-CH(C{{H}_{3}})-C{{H}_{2}}-Br \\\ \end{aligned}$$ This is how antimarkovnikov's rule by the addition of hydrogen bromide to propene in the presence of benzoyl peroxide is carried out. **Note:** It is to be noted that the process stops in the above reaction, because no new free radicals are formed and free radicals are required for reaction to continue.