$C{H_2} = CH - C{H_3} + HBr \to C{H_3}CHBr - C{H_3}$
\(HBr \rightarrow H ^{+}+ Br ^{-}\)
Therefore, this reaction show electrophilic addition reaction.
\(CH _2= CH - CH _3+ HBr \longrightarrow CH _3-\stackrel{\oplus}{ C } H - CH _3 \stackrel{ Br ^{-}}{\longrightarrow} CH _3- CHBr - CH _3\)


