Question

In the polar aprotic solvent, the increasing order of nucleophilicity is:

${CF}_3{{CH}_{2}O}^{-},{CH}_3{{CH}_{2}O}^{-},{CF}_3{{CH}_{2}S}^{-},{CH}_3{{CH}_{2}S}^{-}$

• A. ${CF}_3{{CH}_{2}O}^{-}={CH}_3{{CH}_{2}O}^{-}<{CF}_3{{CH}_{2}S}^{-}<{CH}_3{{CH}_{2}S}^{-}$
• B. ${CF}_3{{CH}_{2}O}^{-}<{CH}_3{{CH}_{2}O}^{-}<{CF}_3{{CH}_{2}S}^{-}<{CH}_3{{CH}_{2}S}^{-}$
• C. ${CF}_3{{CH}_{2}O}^{-}<{CH}_3{{CH}_{2}O}^{-}>{CF}_3{{CH}_{2}S}^{-}<{CH}_3{{CH}_{2}S}^{-}$
• D. ${CF}_3{{CH}_{2}O}^{-}>{CH}_3{{CH}_{2}O}^{-}>{CF}_3{{CH}_{2}S}^{-}>{CH}_3{{CH}_{2}S}^{-}$

Answer 1

Answer: (B)

${CF}_3{{CH}_{2}O}^{-}<{CH}_3{{CH}_{2}O}^{-}<{CF}_3{{CH}_{2}S}^{-}<{CH}_3{{CH}_{2}S}^{-}$

Oxygen is more electronegative than sulphur.
In the ethoxide ion, the negative charge is present on more electronegative O atom as compared to S atom in ethyl sulphide ion.

Hence, the negative charge can be less readily donated by ethoxide ion than by ethyl sulphide ion. Hence, ethyl sulphide ion is more nucleophilic than ethoxide ion.

Also when trifluoromethyl group is present, the nucleophilicity decreases due to the electron-withdrawing effect of trifluoromethyl group.

Hence, option $B$ is correct.