Addition of HBr to propene takes place as per the Markownikoff’s rule via the formation of most stable carbocation as follows : Step 1 : Step 2 : Thus, it is an example of electrophilic addition reaction. But, when the same reaction is carried out in the presence of peroxide, which is a free-radical generator, the mechanism follows free radical mechanism where the intermediate is a carbon free radical. Step 1 : $\underset{\text{Benzoyl peroxide}}{\mathrm{C_6H_5}-\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\overset{\text{↶}}{\mathrm{O}}-\underset{\text{↺}}{\mathrm{O}}-\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\mathrm{C_6H_5}}$ → $2\mathrm{C_6H_5}-\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\underset{\text{..}}{\overset{\cdot}{\mathrm{O}}:}$ Step 2 : Generation of $\overset{\cdot}{\mathrm{Br}}$ $\overset{\cdot}{\mathrm{C_6H_5}}+\overset{\text{↶}}{\mathrm{H}}-\underset{\text{↺}}{\mathrm{Br}}$ → $\mathrm{C_6H_6}+\overset{\cdot}{\mathrm{Br}}$ Step 3 : $\underset{\text{more stable }}{\mathrm{CH_3}-\overset{\text{↶}}{\mathrm{CH}}\overset{\text{↷}}{=}\mathrm{CH_2}+\overset{\cdot}{\mathrm{Br}}}$ → $\mathrm{CH_3}-\overset{\cdot}{\mathrm{CH}}-\mathrm{CH_2}-\mathrm{Br}$ Here, the free radical is the 2° free radical and therefore it is the one which finally forms 1-bromopropane. Step 4 : $\mathrm{CH_3}-\overset{\cdot}{\mathrm{CH}}-\mathrm{CH_2Br}$ + $\overset{\text{↶}}{\mathrm{H}}-\overset{\text{↷}}{\mathrm{Br}}$ → $\mathrm{CH_3CH_2CH_2Br}+\overset{\cdot}{\mathrm{Br}}$ Step 5 : $\overset{\cdot}{\mathrm{Br}}+\overset{\cdot}{\mathrm{Br}}$ → $\mathrm{Br}_2$ This explains why different products are formed under different reaction conditions.