MCQ
Correct order for reaction with alcoholic $KOH$


- ✓$a > c > b > d $
- B$a > b > c > d$
- C$d > b > c > a$
- D$a > d > b > c$

$3^{\circ}>2^{\circ}>1^{\circ}$
The order of leaving group in $E 2$ elimination reaction is shown below :
$R - I > R - Br > R - Cl$
Therefore, the correct order for reaction with alcoholic $KOH$ is as follows :
$a > c > b > d$
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${{\text{[Fe(F)(Cl)(Br)(I)(N}}{{\text{H}}_3}{\text{)(}}{{\text{H}}_2}{\text{O)]}}^\Theta }$
