Question

Arrange the following in the order of leaving group ability.

${{NH}_2}^{-},{{NO}_2}^{-},{N{O}_3}^{-}$

• A. ${{NH}_2}^{-}<{{NO}_2}^{-}<{N{O}_3}^{-}$
• B. ${{NH}_2}^{-}>{{NO}_2}^{-}={N{O}_3}^{-}$
• C. ${{NH}_2}^{-}={{NO}_2}^{-}<{N{O}_3}^{-}$
• D. ${{NH}_2}^{-}>{{NO}_2}^{-}>{N{O}_3}^{-}$

Answer 1

The correct option is A ${{NH}_2}^{-}<{{NO}_2}^{-}<{N{O}_3}^{-}$

The increasing order of leaving group ability in a $S_{N}2$ reaction is ${{NH}_2}^{-}<{{NO}_2}^{-}<{N{O}_3}^{-}$.
Ammonia is the strongest base. Hence, amide ion is poorest leaving group.
Nitric acid is weakest base (in fact it is strong acid). Hence, nitrate ion is good leaving group.