Question

Which of the following is the correct order of the reactivity of given nucleophiles when treated with \( \text{CH}_3\text{Br} \) in methanol? \( \text{F}^- \), \( \text{I}^- \), \( \text{C}_2\text{H}_5\text{O}^- \), \( \text{C}_6\text{H}_5\text{O}^- \)

• A. \( \text{I}^->\text{C}_2\text{H}_5\text{O}^->\text{C}_6\text{H}_5\text{O}^->\text{F}^- \)
• B. \( \text{I}^->\text{F}^->\text{C}_2\text{H}_5\text{O}^->\text{C}_6\text{H}_5\text{O}^- \)
• C. \( \text{C}_2\text{H}_5\text{O}^->\text{F}^->\text{I}^->\text{C}_6\text{H}_5\text{O}^- \)
• D. \( \text{C}_6\text{H}_5\text{O}^->\text{I}^->\text{F}^->\text{C}_2\text{H}_5\text{O}^- \)

Hint:

The reactivity of nucleophiles is determined by their size and polarizability. Larger, more polarizable anions like \( \text{I}^- \) are more reactive than smaller anions like \( \text{F}^- \).

Answer 1

The Correct Option is A

The question tests the understanding of nucleophilic substitution reactions, specifically focusing on the reactivity of nucleophiles towards \( \text{CH}_3\text{Br} \) in methanol. This reaction involves a nucleophile attacking an electrophilic carbon, displacing the leaving group (bromide ion in this case).

To determine the reactivity order, we must consider the factors affecting nucleophilicity, which include:

1. Charge: Negatively charged nucleophiles are generally more reactive than their neutral counterparts.
2. Electronegativity: Less electronegative atoms tend to share electrons more readily, increasing nucleophilicity.
3. Solvent Effects: In polar protic solvents like methanol, larger anions are more nucleophilic because they are less solvated than smaller anions.

Given the nucleophiles in the question:

• \( \text{F}^- \): Small, highly electronegative, and strongly solvated in polar protic solvents like methanol.
• \( \text{I}^- \): Larger anion, less electronegative, and weakly solvated, thus more reactive.
• \( \text{C}_2\text{H}_5\text{O}^- \): Larger than \( \text{F}^- \), negatively charged, and less solvated due to its alkoxide nature.
• \( \text{C}_6\text{H}_5\text{O}^- \): Phenoxide ion, resonant stabilized and therefore less reactive than the alkoxide.

Comparing the reactivity in methanol (polar protic solvent), the order is determined by the interplay of these factors:

• \( \text{I}^- \) is the most reactive due to its size and less solvation, leading to higher nucleophilicity.
• \( \text{C}_2\text{H}_5\text{O}^- \) is next as alkoxides are strong nucleophiles and larger than fluoride.
• \( \text{C}_6\text{H}_5\text{O}^- \) comes next due to resonance stabilization reducing its reactivity compared to the ethoxide.
• \( \text{F}^- \) is the least reactive due to strong solvation and high electronegativity.

Thus, the correct reactivity order when reacting with \( \text{CH}_3\text{Br} \) in methanol is:

\( \text{I}^- > \text{C}_2\text{H}_5\text{O}^- > \text{C}_6\text{H}_5\text{O}^- > \text{F}^- \)

This corresponds to the answer option:

\( \text{I}^- > \text{C}_2\text{H}_5\text{O}^- > \text{C}_6\text{H}_5\text{O}^- > \text{F}^- \)