MCQ
The correct reactivity order for $E_2$ reaction with alcoholic $KOH$ will be


- ✓$i > ii > iii$
- B$ii > iii > i$
- C$i > iii > ii$
- D$iii > i > ii$

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Given : $N -N =159\,kJ\,mol^{-1};\,\,H-H = 436\,kJ\,mol^{-1}$
$N \equiv N = {\text{ }}941{\text{ }}kJ{\text{ }}mo{l^{ - 1}}{\text{ }};{\text{ }}N-H{\text{ }} = {\text{ }}398{\text{ }}kJ{\text{ }}mo{l^{ - 1}}$ .....$kJ\,mol^{-1}$

