MCQ
Which of the following relation is correct for recativity$?\ \ce{CH_3COO^-}$ or $\ce{CH_CH_2​S^-}$ with $\ce{CH_3​I}$ in ethanol.
  • $ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~S}^{-}>\mathrm{CH}_3 \mathrm{COO}^{-} $
  • B
    $ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~S}^{-}<\mathrm{CH}_3 \mathrm{COO}^{-} $
  • C
    $ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~S}^{-}=\mathrm{CH}_3 \mathrm{COO}^{-} $
  • D
    None of these

Answer

Correct option: A.
$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~S}^{-}>\mathrm{CH}_3 \mathrm{COO}^{-} $
Higher electronegativity means lower nucleophilicity because the role of a nucleophile is to share $e^-.$ If the atom is more electronegative it is less willing to share its $e^-$ and wants to hold onto them.
In the polar protic solvent like ethanol $\ce{CH_3COO^-}$ participating in Hydrogen bonding, the nucleophile is considerably less reactive than $\ce{CH_3CH_2S^-}$ as it interacts less to hydrogen ion.
Thus the reaction will be faster for $\ce{CH_3CH_2S^-}$ than $\ce{CH_3COO^-}.$

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