NEET UG 2004 — Chemistry Organic Chemistry

Reaction of $\mathrm{HBr}$HBr with propene in the presence of peroxide gives $n$n-propyl bromide. This addition reaction is an example of anti-Markownikoffs addition reaction.

Related Theory Step 1: Peroxide undergoes fission to give free radicals. $$\mathrm{R}-\mathrm{O}-\mathrm{O}-\mathrm{R} \longrightarrow 2 \mathrm{R}-\dot{\mathrm{O}}$$R−O−O−R⟶2R−O˙ Step 2: $\mathrm{HBr}$HBr combines with free radical to form bromine free radical. $$R-\dot{O}+\mathrm{HBr} \longrightarrow \mathrm{R}-\mathrm{OH}+\dot{\mathrm{B}}$$R−O˙+HBr⟶R−OH+B˙ Step 3: $\dot{B}$B˙ attacks the double bond of the alkene to form a more stable free radical.

Step 4: More stable free radical attacks the $\mathrm{HBr}$HBr. $$\begin{aligned} \mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{CHH}_2 \mathrm{Br}+\mathrm{HBr} \longrightarrow & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\dot{\mathrm{B}} \\ & \text { n-propylbromide } \end{aligned}$$CH3​C˙CHH2​Br+HBr⟶​CH3​CH2​CH2​Br+B˙ n-propylbromide ​ Step 5: $\dot{B} r+\dot{B} r \longrightarrow B r_2$B˙r+B˙r⟶Br2​