Which is the correct reactivity order for E_2 | vedclass

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(A) For an $E_2$ reaction to occur,the leaving group $(-Cl)$ and the $\beta$-hydrogen must be in an anti-periplanar conformation.In compound $(i)$,the

Vedclass · Accepted Answer

$i > ii > iii$

Vedclass · Answer

(A) For an $E_2$ reaction to occur,the leaving group $(-Cl)$ and the $\beta$-hydrogen must be in an anti-periplanar conformation.In compound $(i)$,the $\beta$-hydrogen is anti to the $-Cl$ group,allowing for easy elimination.In compound $(ii)$,the $\beta$-hydrogen is also anti to the $-Cl$ group,but the presence of a methyl group $(Me)$ creates more steric hindrance compared to $(i)$,making it slightly less reactive.In compound $(iii)$,the $\beta$-hydrogen is anti to the $-Cl$ group,but there are two methyl groups $(Me)$ on the adjacent carbons,leading to significant steric hindrance and making it the least reactive.Therefore,the correct order of reactivity is $(i) > (ii) > (iii)$.

Which is the correct reactivity order for $E_2$ reaction with alcoholic $KOH$?

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