The following bimolecular elimination reactio | vedclass

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(D) $1$. The leaving group ability is inversely proportional to the basicity of the leaving group,which is directly related to the $pK_a$ of its conju

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$a$,$b$,and $c$ are true.

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(D) $1$. The leaving group ability is inversely proportional to the basicity of the leaving group,which is directly related to the $pK_a$ of its conjugate acid. Since $HI$ has the lowest $pK_a$ $(-10)$,$I^{-}$ is the best leaving group,and $HF$ has the highest $pK_a$ $(3.2)$,$F^{-}$ is the poorest leaving group. Thus,statement $(a)$ is true.$2$. Zaitsev's rule states that the more substituted alkene is the major product. For $I^{-}$,$Br^{-}$,and $Cl^{-}$,the yield of $2$-hexene (more substituted) is greater than $1$-hexene (less substituted),so they follow Zaitsev's rule. Thus,statement $(b)$ is true.$3$. $F^{-}$ is the strongest base among the halides,making it the poorest leaving group. In $E_2$ reactions with poor leaving groups,the transition state is more 'E1cB-like',meaning the $C-H$ bond is broken more than the $C-X$ bond,resulting in less double bond character in the transition state. Thus,statement $(c)$ is true.$4$. Since $(a)$,$(b)$,and $(c)$ are all true,the correct option is $(d)$.

Leaving group	Conj. Acid $pK_a$	$\%$-yield of $2$-hexene	$\%$-yield of $1$-hexene
$X = I$	$-10$	$81\%$	$19\%$
$X = Br$	$-9$	$72\%$	$28\%$
$X = Cl$	$-7$	$67\%$	$33\%$
$X = F$	$3.2$	$30\%$	$70\%$

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