Question

Arrange in the order of nucleophilicity:

${HO}^{-},{CH}_3{O}^{-},{CH}_3\equiv C^{-},{CH}_3{COO}^{-}$

• A. ${HO}^{-}>{CH}_3{O}^{-}>{CH}_3\equiv C^{-}>{CH}_3{COO}^{-}$
  
• B. ${CH}_3{COO}^{-}>{HO}^{-}>{CH}_3{O}^{-}>{CH}_3\equiv C^{-}$
• C. ${HO}^{-}<{CH}_3{O}^{-}<{CH}_3\equiv C^{-}<{CH}_3{COO}^{-}$
  
• D. ${CH}_3{COO}^{-}<{HO}^{-}<{CH}_3{O}^{-}<{CH}_3\equiv C^{-}$

Answer 1

The correct option is D ${CH}_3{COO}^{-}<{HO}^{-}<{CH}_3{O}^{-}<{CH}_3\equiv C^{-}$

In the polar aprotic solvent, the increasing order of nucleophilicity is ${CH}_3{COO}^{-}<{HO}^{-}<{CH}_3{O}^{-}<{CH}_3\equiv C^{-}$.

Hydroxide ion is stronger nucleophile than acetate ion as in acetate ion, the negative charge is involved in resonance with carboxylic group.

Methoxide ion is more nucleophilic than hydroxide ion due to +I effect of methyl group.

Acetylnide ion is more nucleophilic than methoxide ion as negative charge is present on less electronegative C atom.

Option D is correct.