Rank the following anions in order of decreasing nucleophilicity in a polar protic solvent (most $\rightarrow$ least nucleophilic).
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In polar protic solvents (like water or alcohols):
- Down a group, nucleophilicity increases because larger atoms are less solvated ($\text{S}^- > \text{O}^-$).
- Among similar nucleophilic atoms, localized charges are much more nucleophilic than resonance-stabilized (delocalized) charges ($\text{R-O}^- > \text{R-COO}^-$).
Step 1: Understanding the Question:
We need to arrange three given anions in order of decreasing nucleophilicity in a polar protic solvent.
The anions are:
1. $\text{CH}_3\text{CH}_2\text{CH}_2-\text{O}^-$ (propoxide ion, an alkoxide)
2. $\text{CH}_3\text{CH}_2\text{CH}_2-\text{S}^-$ (propanethiolate ion, a thiolate)
3. $\text{CH}_3\text{CH}_2-\text{C}(=\text{O})-\text{O}^-$ (propanoate ion, a carboxylate)
Step 2: Detailed Explanation:
Nucleophilicity in polar protic solvents is influenced by several factors:
- Solvation effects: Polar protic solvents can form strong hydrogen bonds with small, highly electronegative anions.
Oxygen ($\text{O}^-$) is smaller and more electronegative than sulfur ($\text{S}^-$), making it heavily solvated (shielded by a cage of solvent molecules).
On the other hand, sulfur ($\text{S}^-$) is larger, less electronegative, and much less solvated because it forms very weak hydrogen bonds.
Thus, the less-solvated thiolate ion (2) is a much stronger nucleophile than the alkoxide ion (1) in polar protic solvents ($2 > 1$).
- Resonance and charge localization:
In the alkoxide ion (1), the negative charge is localized on a single oxygen atom, making it strongly basic and a strong nucleophile.
In the carboxylate ion (3), the negative charge on the oxygen is delocalized over two oxygen atoms via resonance, which makes it highly stable, weakly basic, and consequently a very poor nucleophile ($1 > 3$).
- Combining these factors gives the overall decreasing order of nucleophilicity in a polar protic solvent:
\[ 2 > 1 > 3 \]
