Arrange the following in increasing order of their nucleophilicity in a polar aprotic solvent.
${{C H}_{3} {C H}_{2} O}^{-}, {{C H}_{3} {C H}_{2} S}^{-}, {C H}_{3} {C S}_{2}^{-}$

{C H}_{3} {C S}_{2}^{-} < {{C H}_{3} {C H}_{2} O}^{-} < {{C H}_{3} {C H}_{2} S}^{-}

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{{C H}_{3} {C H}_{2} O}^{-}

{C H}_{3} {C S}_{2}^{-} < {{C H}_{3} {C H}_{2} S}^{-}

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{{C H}_{3} {C H}_{2} S}^{-} < {C H}_{3} {C S}_{2}^{-} < {{C H}_{3} {C H}_{2} O}^{-}

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None of these

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Solution

The correct option is A ${C H}_{3} {C S}_{2}^{-} < {{C H}_{3} {C H}_{2} O}^{-} < {{C H}_{3} {C H}_{2} S}^{-}$
The increasing order of nuclophilicity in polar aprotic solvent is ${C H}_{3} {C S}_{2}^{-} < {{C H}_{3} {C H}_{2} O}^{-} < {{C H}_{3} {C H}_{2} S}^{-}$ .
${{C H}_{3} {C H}_{2} O}^{-}$ is less nucleophilic than ${{C H}_{3} {C H}_{2} S}^{-}$ as ${{C H}_{3} {C H}_{2} O}^{-}$ contains negative charge on electronegative $O$ atom whereas ${{C H}_{3} {C H}_{2} S}^{-}$ contains negative charge on electropositive $S$ atom.
${C H}_{3} {C S}_{2}^{-}$ is least nucleophilic due to resonance stablization.

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