Question

Which of the following compounds will react faster with ${CH}_3{CH}_{2}Br$ in ethanol ?

${{CH}_{3}S}^{-}$ or ${{CH}_{3}O}^{-}$

• A. ${{CH}_{3}S}^{-}>{{CH}_{3}O}^{-}$
• B. ${{CH}_{3}S}^{-}<{{CH}_{3}O}^{-}$
• C. ${{CH}_{3}S}^{-}={{CH}_{3}O}^{-}$
• D. None of these

Answer 1

Answer: (A)

${{CH}_{3}S}^{-}>{{CH}_{3}O}^{-}$

Ethanol is a polar protic solvent that can participate in hydrogen bonding with a nucleophile, creating a “shell” of solvent molecules around it. In this type of solvent the ability of nucleophiles to participate in hydrogen bonding decreases as we go down the periodic table. Hence $O^{-}$ is the stronger hydrogen bond acceptor, and $S^{-}$ is not. This means that the $C{H}_3{S}^{-}$ has greater nucleophilicity and will react faster than $C{H}_3{O}^{-}$. Greater the nucleophilicity faster the reactivity towards $C{H}_{3}C{H}_{2}Br$ in ethanol solvent. Moreover, the given reaction follows a $S{N}_2$ reaction mechanism.
Hence, the correct option is A.