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

S H^{-} > O H^{-} > H_{2} O > A c O^{-} > P h O^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

S H^{-} > O H^{-} > P h O^{-} > A c O^{-} > H_{2} O

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

S H^{-} > P h O^{-} > O H^{-} > H_{2} O > A c O^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

O H^{-} > S H^{-} > P h O^{-} > A c O^{-} > H_{2} O

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $S H^{-} > O H^{-} > P h O^{-} > A c O^{-} > H_{2} O$
Nucleophilicity $α$ $n e g a t i v e c h a r g e i o n s$

$α$ $\frac{1}{d e l o c a l i s a t i o n}$

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

$S H^{-}, O H^{-}, O P h^{-}, O A c^{-} > H_{2} O$

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

$S H^{-} > O H^{-}, O P h^{-}, O A c^{-} > H_{2} O$

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

$S H^{-} > O H^{-} > O P h^{-} > O A c^{-} > H_{2} O$

Suggest Corrections