\(3^{\circ}>2^{\circ}>1^{\circ}\)
The order of leaving group in \(E 2\) elimination reaction is shown below :
\(R - I > R - Br > R - Cl\)
Therefore, the correct order for reaction with alcoholic \(KOH\) is as follows :
\(a > c > b > d\)
${C{H_3} - C{H_2} - CH = C{H_2} + HBr\,\to \,CH _{3}- CH _{2}- CH _{2}- C^{+}H _{2}+ Br ^{-}} _{"A"}$
${C{H_3} - C{H_2} - CH = C{H_2} + HBr\, \to \,CH _{3}- CH _{2}- C^{+}H - CH _{3}+ Br ^{-}}_{"B"}$
${C_2}{H_5}OH + SOC{l_2}\xrightarrow{{{\text{Pyridine}}}}{C_2}{H_5}Cl + S{O_2} + HCl$