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(A) The reaction with alcoholic $KOH$ is an $E2$ elimination reaction.The reactivity in $E2$ reactions depends on two main factors:$1$. The stability

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$a > c > b > d$

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(A) The reaction with alcoholic $KOH$ is an $E2$ elimination reaction.The reactivity in $E2$ reactions depends on two main factors:$1$. The stability of the alkene formed (more substituted alkenes are more stable).$2$. The nature of the leaving group $(I^- > Br^- > Cl^-)$.Analyzing the compounds:$(a)$ $2$-bromo-$2,3$-dimethylbutane is a $3^{\circ}$ alkyl bromide. It forms a highly substituted,stable alkene.$(c)$ $2$-bromobutane is a $2^{\circ}$ alkyl bromide.$(b)$ $2$-chlorobutane is a $2^{\circ}$ alkyl chloride. Since $Br^-$ is a better leaving group than $Cl^-$,$(c) > (b)$.$(d)$ $1$-bromopropane is a $1^{\circ}$ alkyl bromide,which is the least reactive towards $E2$ elimination.Thus,the correct order of reactivity is $(a) > (c) > (b) > (d)$.

Identify the correct order of reactivity for the reaction with alcoholic $KOH$ for the following compounds:
$(a)$ $2$-bromo-$2,3$-dimethylbutane
$(b)$ $2$-chlorobutane
$(c)$ $2$-bromobutane
$(d)$ $1$-bromopropane

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Identify the correct order of reactivity for the reaction with alcoholic $KOH$ for the following compounds:$(a)$ $2$-bromo-$2,3$-dimethylbutane$(b)$ $2$-chlorobutane$(c)$ $2$-bromobutane$(d)$ $1$-bromopropane