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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$(a)$ coordination number of $'N'$ is $3$ , and the structure is triagonal planar
$(b)$ formal charge on $N$ is $+1$
$(c)$ Average formal charge on $'O'$ is $-2/3$
$(d)$ Average bond order of $NO$ bond is $4/3$
$(e)$ All $NO$ bond lengths are identical
Correct code is