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(A) The reaction involves the nucleophilic substitution of the iodide group $(-I)$ in $2$-chlorobenzyl iodide with the methanethiolate ion $(CH_3S^-)$

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(A) The reaction involves the nucleophilic substitution of the iodide group $(-I)$ in $2$-chlorobenzyl iodide with the methanethiolate ion $(CH_3S^-)$,which is generated by the reaction of methanethiol $(CH_3SH)$ with sodium hydroxide $(NaOH)$.$CH_3SH + NaOH \rightarrow CH_3S^-Na^+ + H_2O$The $CH_3S^-$ ion is a strong nucleophile and will attack the benzylic carbon atom,which is attached to the iodine atom (a good leaving group),via an $S_N2$ mechanism.The aryl chloride bond is much less reactive towards nucleophilic substitution compared to the benzylic iodide bond. Therefore,the substitution occurs selectively at the benzylic position.The final product is $2$-chlorobenzyl methyl sulfide.

Similar Questions

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