Question

How can I explain the bromination of alkanes?

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, we need to learn about the chemical properties of alkanes. Alkanes have low chemical reactivity and are, therefore, known as paraffins. This is due to the presence of carbon carbon and carbon hydrogen bonds in their molecules which are difficult to break since these are sigma bonds in nature. These are, therefore, saturated in nature and are reluctant to take part in chemical reactions. Let us know about substitution reactions now:
Halogenation of alkanes: Halogenation-reactions of alkanes can be carried by treating with a suitable halogen in the presence of either ultra-violet light or by heating to 520-670 K. The reactions taking place in the presence of ultraviolet light are called photochemical reactions. The order of reactivity of different halogens in these reactions is :
${{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}$
Now, during the chlorination of alkanes, all the four hydrogen atoms present in the molecule get replaced by one by to form a mixture of different substituted products. The bromination of alkanes also occurs in a similar fashion, however, it is more selective in nature and is slower too. The reason being that the bromine atom has a lesser tendency to abstract a hydrogen atom, which is reflected by the higher bond energy of $H-Cl$ than $H-Br$.
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.