Question

The decreasing order of relative nucleophilicity of the following nucleophiles in a protic solvent is
$S{H}^{-},Ac{O}^{-},Ph{O}^{-},O{H}^{-},H_{2}O$

• A. $S{H}^{-}>O{H}^{-}>H_{2}O>Ac{O}^{-}>Ph{O}^{-}$
• B. $S{H}^{-}>O{H}^{-}>Ph{O}^{-}>Ac{O}^{-}>H_{2}O$
• C. $S{H}^{-}>Ph{O}^{-}>O{H}^{-}>H_{2}O>Ac{O}^{-}$
• D. $O{H}^{-}>S{H}^{-}>Ph{O}^{-}>Ac{O}^{-}>H_{2}O$

Answer 1

The correct option is B $S{H}^{-}>O{H}^{-}>Ph{O}^{-}>Ac{O}^{-}>H_{2}O$

Nucleophilicity $\alpha$ $negativechargeions$

$\alpha$ $\frac{1}{delocalisation}$

Nucleophilicity of the negative charged ions are more than that of neutral compounds because they have more electron density therefore,

$S{H}^{-},O{H}^{-},OP{h}^{-},OA{c}^{-}>H_{2}O$

Nucleophilicity of the ions which have more size (along the group) will be more in case of protic solvents

$S{H}^{-}>O{H}^{-},OP{h}^{-},OA{c}^{-}>H_{2}O$

More the delocalisation less is the nucleophilicity because of the less availability of electron density. As in $OA{c}^{-}$ there is equivalent resonance hence it is weaker nucleophile than $OP{h}^{-}$.

$S{H}^{-}>O{H}^{-}>OP{h}^{-}>OA{c}^{-}>H_{2}O$