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(D) In a polar protic solvent,nucleophilicity is determined by the ability of the species to form hydrogen bonds with the solvent. Stronger hydrogen b

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$2 > 1 > 3$

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(D) In a polar protic solvent,nucleophilicity is determined by the ability of the species to form hydrogen bonds with the solvent. Stronger hydrogen bonding leads to greater solvation,which decreases the nucleophilicity of the species.$(1) \ CH_3CH_2CH_2O^{-}$ is a strong base and forms strong hydrogen bonds with the solvent.$(2) \ CH_3CH_2CH_2S^{-}$ is a larger,more polarizable ion with lower charge density,resulting in weaker hydrogen bonding and higher nucleophilicity in polar protic solvents.$(3) \ CH_3CH_2C(=O)O^{-}$ is a resonance-stabilized carboxylate ion,which makes it a weaker nucleophile compared to the alkoxide ion $(1)$.Thus,the order of decreasing nucleophilicity is $(2) > (1) > (3)$.

Rank the following species in order of decreasing nucleophilicity in a polar protic solvent (most $\to$ least nucleophilic):
$(1) \ CH_3CH_2CH_2O^{-}$,$(2) \ CH_3CH_2CH_2S^{-}$,$(3) \ CH_3CH_2C(=O)O^{-}$

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Rank the following species in order of decreasing nucleophilicity in a polar protic solvent (most $\to$ least nucleophilic):$(1) \ CH_3CH_2CH_2O^{-}$,$(2) \ CH_3CH_2CH_2S^{-}$,$(3) \ CH_3CH_2C(=O)O^{-}$