Identify the correct reactivity order for E_2 | vedclass

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(B) The $E_2$ reaction involves the elimination of a proton and a leaving group to form a double bond. The rate of $E_2$ elimination is significantly

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$ii > i > iii$

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(B) The $E_2$ reaction involves the elimination of a proton and a leaving group to form a double bond. The rate of $E_2$ elimination is significantly enhanced if the resulting product is aromatic or anti-aromatic,as the transition state is stabilized by conjugation.$(i)$ Elimination in $8$-chlorocycloocta-$1,3,5$-triene leads to cyclooctatetraene,which is a stable $8\pi$ electron system (non-aromatic,but highly conjugated).(ii) Elimination in $3$-chlorocyclohexa-$1,4$-diene leads to benzene,which is highly aromatic and stable. This provides a very strong driving force for the reaction.(iii) Elimination in $3$-chlorocyclobutene leads to cyclobutadiene,which is anti-aromatic ($4\pi$ electrons) and highly unstable. This makes the reaction very slow.Therefore,the reactivity order is $(ii) > (i) > (iii)$.

Identify the correct reactivity order for $E_2$ reaction with alc. $KOH$ for the following compounds:
$(i)$ $8$-chlorocycloocta-$1,3,5$-triene
(ii) $3$-chlorocyclohexa-$1,4$-diene
(iii) $3$-chlorocyclobutene

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Identify the correct reactivity order for $E_2$ reaction with alc. $KOH$ for the following compounds:$(i)$ $8$-chlorocycloocta-$1,3,5$-triene(ii) $3$-chlorocyclohexa-$1,4$-diene(iii) $3$-chlorocyclobutene