Which compound in each of the following pairs | vedclass

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(C) $(i)$ In $S_N2$ reactions,the rate depends on the leaving group ability. The order of leaving group ability is $I^{-} > Br^{-} > Cl^{-} > F^{-}$.

Vedclass · Accepted Answer

$(i)$ $CH_3-I$,$(ii)$ $CH_3-Cl$

Vedclass · Answer

(C) $(i)$ In $S_N2$ reactions,the rate depends on the leaving group ability. The order of leaving group ability is $I^{-} > Br^{-} > Cl^{-} > F^{-}$. Since $I^{-}$ is a better leaving group than $Br^{-}$,$CH_3-I$ reacts faster than $CH_3-Br$.$(ii)$ $S_N2$ reactions are highly sensitive to steric hindrance. The reactivity order for $S_N2$ is $	ext{methyl} > 	ext{primary} > 	ext{secondary} > 	ext{tertiary}$. $CH_3-Cl$ is a methyl halide,while $CH_3-C(CH_3)_2-Cl$ is a tertiary halide. Therefore,$CH_3-Cl$ reacts faster due to significantly less steric hindrance.

Which compound in each of the following pairs will react faster in $S_N2$ reaction with $OH^{-}$?
$(i)$ $CH_3-Br$ or $CH_3-I$
$(ii)$ $CH_3-C(CH_3)_2-Cl$ or $CH_3-Cl$

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Which compound in each of the following pairs will react faster in $S_N2$ reaction with $OH^{-}$?$(i)$ $CH_3-Br$ or $CH_3-I$$(ii)$ $CH_3-C(CH_3)_2-Cl$ or $CH_3-Cl$