Question

Decreasing order of relative nucleophilicity of the following nucleophiles in protic solvent is -
$\stackrel{\ominus }{S}H,AC\stackrel{\ominus }{O},Ph\stackrel{\ominus }{O},\stackrel{\ominus }{O}H,H_{2}O$

• A. $\stackrel{\ominus }{S}H>\stackrel{\ominus }{O}H>H_{2}O>AC\stackrel{\ominus }{O}>Ph\stackrel{\ominus }{O}$
• B. $\stackrel{\ominus }{S}H>\stackrel{\ominus }{O}H>Ph\stackrel{\ominus }{O}>AC\stackrel{\ominus }{O}>H_{2}O$
• C. $\stackrel{\ominus }{S}H>Ph\stackrel{\ominus }{O}>\stackrel{\ominus }{O}H>H_{2}O>AC\stackrel{\ominus }{O}$
• D. $\stackrel{\ominus }{O}H>\stackrel{\ominus }{S}H>Ph\stackrel{\ominus }{O}>AC\stackrel{\ominus }{O}>H_{2}O$

Answer 1

The correct option is B $\stackrel{\ominus }{S}H>\stackrel{\ominus }{O}H>Ph\stackrel{\ominus }{O}>AC\stackrel{\ominus }{O}>H_{2}O$

Neutral species have less nucleophilicity than negatively charged species. So, $H_{2}O$ has least nucleophilicity.
In these two compounds, negative charge is in decolalisation. So, less electron density will be available for bonding, hence nucleophilicity will be less.
In $Ph{O}^{⊝}$, negative charge will come on ${}^{\prime }{C}^{\prime }$ while in $AC{O}^{⊝},$ negative charge will come on ${}^{\prime }{O}^{\prime }$ which is more stable.

Nucleophilicity is directly proportional to size atom in a group.
So, $\stackrel{\ominus }{S}H>\stackrel{\ominus }{O}H$