Question

The correct order of nucleophilicity (nucleophilic reactivity) of the given species is :
A. $EtO^-$
B. $Me_3N$
C. $CN^-$
D. $OH^-$
E. $PhO^-$
Choose the correct answer from the options given below :

• A. C > D > E > A > B
• B. C > A > D > E > B
• C. B > E > D > A > C
• D. B > A > C > E > D

Hint:

A simple ranking strategy: Anions with less electronegative donor atoms (like $C^-$) $>$ Localized Anions with highly electronegative atoms ($O^-$) $>$ Resonance-stabilized Anions ($ArO^-$) $>$ Neutral molecules. Alkyl groups increase nucleophilicity slightly due to $+I$ effects.

Answer 1

The Correct Option is B

Step 1: Concept:
The problem requires ranking five different chemical species according to their nucleophilic strength (nucleophilicity), which is a measure of how quickly a species can attack an electrophilic carbon atom.

Step 2: Key Formula or Approach:
Nucleophilicity depends on several factors:
1. Charge: Anions are generally much better nucleophiles than neutral molecules.
2. Electronegativity: Less electronegative atoms are more willing to share their electrons, making them better nucleophiles (e.g., C > N > O > F).
3. Polarizability: Larger, more polarizable atoms are better nucleophiles.
4. Resonance Delocalization: If a lone pair or negative charge is delocalized via resonance, its availability to attack an electrophile drops significantly, lowering nucleophilicity.
5. Inductive Effects: Electron-donating groups (like alkyls) increase electron density and thus nucleophilicity.

Step 3: Step-by-step Explanation:

• C. $CN^-$ (Cyanide ion): This anion has the negative charge located primarily on the Carbon atom (though delocalized somewhat to Nitrogen). Carbon is less electronegative than Oxygen or Nitrogen, making its lone pair highly available and highly reactive. Cyanide is an exceptionally strong nucleophile. (Highest)

• Next, we compare the localized oxygen anions: $EtO^-$ (A) and $OH^-$ (D). The ethyl group in ethoxide ($EtO^-$) provides an electron-donating inductive effect ($+I$), pushing extra electron density onto the oxygen atom, making it a stronger nucleophile than the hydroxide ion. So, $EtO^- > OH^-$.

• E. $PhO^-$ (Phenoxide ion): The negative charge on the oxygen atom is highly delocalized into the aromatic benzene ring via resonance. Because the electron density is spread out and less available for attack, phenoxide is a significantly weaker nucleophile than localized alkoxides or hydroxide.

• B. $Me_3N$ (Trimethylamine): This is a neutral molecule. While nitrogen is less electronegative than oxygen, neutral amines are generally less nucleophilic than negatively charged oxygen anions (especially considering the steric hindrance of the three methyl groups in $S_N2$ reactions). Thus, it ranks at the bottom in this specific grouping. (Lowest)

• Combining these observations, the full sequence from most to least nucleophilic is: $CN^- > EtO^- > OH^- > PhO^- > Me_3N$.

• Translating to the letters provided: C > A > D > E > B.

Step 4: Final Answer:
The correctly derived sequence matches option (B).