Which of the following haloalkanes reacts with aqueous $KOH$ most easily? Explain giving reason.
$(i)$ $1$-Bromobutane
$(ii)$ $2$-Bromobutane
$(iii)$ $2$-Bromo-$2$-methylpropane
$(iv)$ $2$-Chlorobutane

(C) $(iii)$ $2$-Bromo-$2$-methylpropane reacts most easily with aqueous $KOH$.
Reason: The reaction follows the $S_N1$ mechanism. The rate-determining step involves the formation of a carbocation intermediate. Since $2$-Bromo-$2$-methylpropane is a $3^{\circ}$-alkyl halide,it forms a stable $3^{\circ}$-carbocation,which is more stable than the carbocations formed by $1^{\circ}$ or $2^{\circ}$ alkyl halides. Thus,it reacts most readily.