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Answer the following.
Which will react faster by S_N2 mechanism with –OH, and why: CH₃Br or CH₃I?
Which will react faster by $S_N2$ mechanism with –OH, and why: $\text{CH}_3\text{Br}$ or $\text{CH}_3\text{I}$?
Answer
Answer
Official answer key$\mathrm{CH_3I}$ will react faster because $\mathrm{I^-}$ is a better leaving group. It has lower bond dissociation energy. Halides show following order of reactivity: R-F < R-Cl < R-Br < R-I