In the above reaction,the maximum Saytzeff pr | vedclass

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(A) In an $E2$ elimination reaction,the nature of the leaving group $(X)$ significantly influences the product distribution between the Saytzeff produ

Vedclass · Accepted Answer

$X = -I$

Vedclass · Answer

(A) In an $E2$ elimination reaction,the nature of the leaving group $(X)$ significantly influences the product distribution between the Saytzeff product (more substituted alkene) and the Hoffmann product (less substituted alkene).When the leaving group is a good leaving group (like $I^-$,$Br^-$,or $Cl^-$),the transition state resembles the product alkene. The stability of the transition state is governed by the stability of the developing alkene. Since the Saytzeff product is more substituted and thus more stable,it becomes the major product.Among the halogens,the leaving group ability follows the order: $I^- > Br^- > Cl^- > F^-$.$I^-$ is the best leaving group among the options provided,leading to the most stable transition state for the formation of the more substituted alkene. Therefore,the maximum amount of Saytzeff product is obtained when $X = -I$.

In the above reaction,the maximum Saytzeff product will be obtained when

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