The order of decreasing nucleophilicity is

H_{2} O > O H^{-} > C H_{3} C O O^{-} > C H_{3} O^{-}

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C H_{3} C O O^{-} > C H_{3} O^{-} > H O^{-} > H_{2} O

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H O^{-} > C H_{3} O^{-} > C H_{3} C O O^{-} > H_{2} O

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C H_{3} O^{-} > O H^{-} > C H_{3} C O O^{-} > H_{2} O

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Solution

The correct option is D $C H_{3} O^{-} > O H^{-} > C H_{3} C O O^{-} > H_{2} O$
1. Due to the presence of -ve charge $C H_{3} O^{-}$ , $O H^{-}, C H_{3} C O O^{-}$ , are more nucleophilic than $H_{2} O$ . Further due to + I effect of the $C H_{3}$ group, $C H_{3} O^{-}$ is more nucleophilic than $O H^{-}$ .

2. $C H_{3} C O O^{-}$ has the -ve charge on oxygen stabilized by resonance and therefore its nucloephilicity is less than those of $C H_{3} O^{-}$ and $O H^{-}$ .

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