Question

Arrange them in increasing order of their leaving ability:

${CH}_3{CH}_2^{-},{CH}_3{CH}_2{O}^{-},{CH}_{3}CO{O}^{-}$

• A. ${CH}_3{CH}_2^{-}>{CH}_3{CH}_2{O}^{-}>{CH}_{3}CO{O}^{-}$.
• B. ${CH}_3{CH}_2{O}^{-}<{CH}_{3}CO{O}^{-}<{CH}_3{CH}_2^{-}$.
• C. ${CH}_3{CH}_2^{-}<{CH}_3{CH}_2{O}^{-}<{CH}_{3}CO{O}^{-}$.
• D. ${CH}_{3}CO{O}^{-}<{CH}_3{CH}_2^{-}<{CH}_3{CH}_2{O}^{-}<$.
  

Answer 1

The correct option is C ${CH}_3{CH}_2^{-}<{CH}_3{CH}_2{O}^{-}<{CH}_{3}CO{O}^{-}$.

The increasing order of the leaving group ability in a $S_{N}2$ reaction is ${CH}_3{CH}_2^{-}<{CH}_3{CH}_2{O}^{-}<{CH}_{3}CO{O}^{-}$.
In the acetate ion, the negative charge is stabilized by the resonance of the carboxylate group. Hence, it is good leaving group.
In the carbanion of ethane, there is no stabilization of negative charge. Hence, it is poorest leaving group.
In the ethoxide ion, the negative charge is on more electronegative oxygen atom. Hence, it is better leaving group than the carbanion of ethane.