Question

Which of the following is the correct order of nucleophilic nature for the following reaction? \[ \text{CH}_3\text{Br} + \text{Nu}^- \longrightarrow \text{CH}_3\text{Nu} + \text{Br}^- \]

• A. \( \text{HO}^→ \text{PhO}^→ \text{CH}_3\text{COO}^→ \text{ClO}_4^- \)
• B. \( \text{PhO}^→ \text{HO}^→ \text{CH}_3\text{COO}^→ \text{ClO}_4^- \)
• C. \( \text{CH}_3\text{COO}^→ \text{HO}^→ \text{PhO}^→ \text{ClO}_4^- \)
• D. \( \text{HO}^→ \text{ClO}_4^→ \text{PhO}^→ \text{CH}_3\text{COO}^- \)

Hint:

In SN2 reactions, nucleophilicity decreases with increasing resonance stabilization. More localized negative charge means a stronger nucleophile.

Answer 1

The Correct Option is A

Concept: The given reaction is an SN2 reaction, as methyl bromide (\(\text{CH}_3\text{Br}\)) undergoes nucleophilic substitution via a single-step backside attack mechanism. For nucleophilicity in SN2 reactions:
Higher electron density increases nucleophilicity
Resonance stabilization decreases nucleophilicity
Weakly basic and highly stabilized anions are poor nucleophiles
Step 1: Analyze each nucleophile.
\(HO^-\): Small size, high charge density, and no resonance stabilization. Hence, it is a very strong nucleophile.
\(PhO^-\) (Phenoxide ion): The negative charge is delocalized over the aromatic ring through resonance, reducing its availability for nucleophilic attack. Thus, it is weaker than \(\text{HO}^-\).
\(CH_3COO^-\) (Acetate ion): The negative charge is delocalized over two oxygen atoms due to resonance, making it less nucleophilic than phenoxide.
\(ClO_4^-\) (Perchlorate ion): Highly resonance-stabilized with extensive charge delocalization. It is an extremely weak nucleophile.
Step 2: Compare nucleophilic strength based on availability of lone pair. \[ \text{HO}^→ \text{PhO}^→ \text{CH}_3\text{COO}^→ \text{ClO}_4^- \]
Step 3: Match the order with the given options. This order corresponds to Option (1).