MCQ
The following bimolecular elimination reaction $(E_2)$ is carried out with different  halogen leaving groups. The per cent yield of the two products ($2$ -hexene and $1$-hexene) for each leaving group is listed below.

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

Which of the following statement is (are) true concerning this series of $E_2$ reactions ?

  • A
    Based on the $pK_a$ 's of the conjugate acid, $I^-$ is the best leaving group and  $F^-$ is the poorest leaving group
  • B
    When $I^-, Br^-$ and $Cl^-$ are used as leaving groups, Zaitsev's rule is followed
  • C
    $F^-$ is the strongest base (and therefore the poorest leaving group) and the  transition state for reaction with fluoride as the leaving group has the least double  bond character
  • $a, b, c$ are true

Answer

Correct option: D.
$a, b, c$ are true
d
$(d)$ Informative question.

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