CH$_3$–Br $\xrightarrow{\text{CH$_3$OH/Nu}}$ CH$_3$OH
Correct order of rate of this reaction for given nucleophile:
CH$_3$–Br $\xrightarrow{\text{CH$_3$OH/Nu}}$ CH$_3$OH
Correct order of rate of this reaction for given nucleophile:
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- F$^-$>PhO$^-$>C$_2$H$_5$O$^-$>I$^-$
- C$_2$H$_5$O$^-$>PhO$^-$>I$^-$>F$^-$
- I$^-$>C$_2$H$_5$O$^-$>PhO$^-$>F$^-$
- PhO$^-$>C$_2$H$_5$O$^-$>F$^-$>I$^-$
The Correct Option is C
Solution and Explanation
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).
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