• # question_answer 30) Addition of $HBr$ to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Addition of $HBr$ to propene is an ionicelectrophilic addition reaction, in which the electrophile in the step 1 is$HBr$. Step 1.$HBr\to \underset{Attacking\,electrophile}{\mathop{{{H}^{+}}}}\,+_{\centerdot }^{\centerdot }B{{r}^{-}}$ Step 2. The proton $({{H}^{+}})$ attacks the $\pi$-bond to give a stable carbocation. Step 3. The nucleophile bromide ion attacks the more stable $2{}^\circ$carbocation to give 2-bromopropane(major product). $C{{H}_{3}}-\overset{+}{\mathop{CH}}\,-C{{H}_{3}}+_{\centerdot }^{\centerdot }B{{r}^{-}}\xrightarrow{Fast}C{{H}_{3}}-\underset{\underset{\underset{2-Bromopropane}{\mathop{\text{Br}}}\,}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{3}}$ In presence of benzoylperoxide, the reaction is freeradical electrophilic addition reaction, in which the electrophile here is $B{{r}^{\centerdot }}$ free radical which is obtained by the action of benzoylperoxide on$HBr$. Step 1. Peroxide undergoes fission to give free radicals (Initiation). ${{C}_{6}}{{H}_{5}}CO-O-O-CO{{C}_{6}}{{H}_{5}}--\xrightarrow[(Homolytic\,Fission)]{Heat}$ $\underset{\begin{smallmatrix} Benzoyl\,radical \\ (unstable) \end{smallmatrix}}{\mathop{2{{C}_{6}}{{H}_{5}}CO{{O}^{\centerdot }}}}\,\to \underset{Phenyl\,free\,readical}{\mathop{[2\overset{\centerdot }{\mathop{{{C}_{6}}{{H}_{5}}}}\,+2C{{O}_{2}}]}}\,$ Step 2.$~HBr$ combines with free radical to formbromine free radical. Step 3. Br attacks the double bond of alkene to form a more stable free radical (Propagation). Step 4. More stable free radical attacks the $HBr$(Termination). $\overset{\centerdot }{\mathop{Br}}\,+\overset{\centerdot }{\mathop{Br}}\,\to B{{r}_{2}}$