In the above reaction,the maximum Saytzeff pr | vedclass

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(A) In an $E_2$ elimination reaction,the nature of the leaving group $X$ significantly influences the product distribution.For alkyl halides,the trans

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$-I$

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(A) In an $E_2$ elimination reaction,the nature of the leaving group $X$ significantly influences the product distribution.For alkyl halides,the transition state for the $E_2$ mechanism has significant carbanion character.Fluorine $(-F)$ is the most electronegative halogen,which makes the $\beta$-hydrogen more acidic and stabilizes the transition state leading to the less substituted alkene (Hoffmann product).Conversely,better leaving groups like $-I$,$-Br$,and $-Cl$ facilitate the formation of the more stable,more substituted alkene (Saytzeff product) because the transition state resembles the product more closely.Among the given options,$-F$ is the poorest leaving group,which promotes the formation of the Hoffmann product,while $-I$ is the best leaving group,which promotes the formation of the Saytzeff product.Therefore,the maximum Saytzeff product is obtained when $X = -I$.

In the above reaction,the maximum Saytzeff product will be obtained when $X$ is:

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