Question

CH$_3$–Br $\xrightarrow{\text{CH$_3$OH/Nu}}$ CH$_3$OH 
Correct order of rate of this reaction for given nucleophile: 
 

• A. F$^-$>PhO$^-$>C$_2$H$_5$O$^-$>I$^-$
• B. C$_2$H$_5$O$^-$>PhO$^-$>I$^-$>F$^-$
• C. I$^-$>C$_2$H$_5$O$^-$>PhO$^-$>F$^-$
• D. PhO$^-$>C$_2$H$_5$O$^-$>F$^-$>I$^-$

Hint:

The nucleophilicity of an ion is influenced by its size and electronegativity. Larger and less electronegative ions tend to be more nucleophilic.

Answer 1

The Correct Option is C

Step 1: Understanding nucleophilicity. 
Nucleophilicity refers to the ability of a nucleophile to donate electrons to form a new bond with an electrophile (in this case, CH$_3$–Br). The stronger the nucleophile, the faster the nucleophilic substitution reaction will occur. 

Step 2: Comparing the nucleophilicity of the given nucleophiles. 
- I$^-$ is the most nucleophilic due to its larger size and lower electronegativity, which makes it more willing to donate electrons. 
- C$_2$H$_5$O$^-$ (ethoxide) is a good nucleophile, but not as strong as I$^-$, because oxygen is more electronegative, making it less willing to donate electrons. 
- PhO$^-$ (phenoxide) is also a strong nucleophile, but the resonance stabilization of the phenoxide ion reduces its nucleophilicity compared to C$_2$H$_5$O$^-$ and I$^-$. 
- F$^-$ is the least nucleophilic because fluorine is highly electronegative, making it reluctant to donate electrons. 

Step 3: Conclusion. 
The correct order of nucleophilicity is I$^-$>C$_2$H$_5$O$^-$>PhO$^-$>F$^-$, which corresponds to option (3).