$(I)\,\,CH_2 =CH-CH_2Cl$ $(II)\,\,CH_2=CH-CH(CH_3) Cl$
$(III)\,\,CH_2 =C(CH_3)CH_2Cl$ $(IV)\,\, CH_3CH = CH-CH_2Cl$

${C{H_3} - C{H_2} - CH = C{H_2} + HBr\,\to \,CH _{3}- CH _{2}- CH _{2}- C^{+}H _{2}+ Br ^{-}} _{"A"}$
${C{H_3} - C{H_2} - CH = C{H_2} + HBr\, \to \,CH _{3}- CH _{2}- C^{+}H - CH _{3}+ Br ^{-}}_{"B"}$
