Consider the nucleophilic attacks given below | vedclass

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(C) For $S_N2$ reactions,the rate depends on the strength of the nucleophile and the quality of the leaving group.$(A)$ In pair $A$,$I$ involves $Br^-

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$A-II; B-III; C-V; D-VIII$

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(C) For $S_N2$ reactions,the rate depends on the strength of the nucleophile and the quality of the leaving group.$(A)$ In pair $A$,$I$ involves $Br^-$ as a leaving group,while $II$ involves $CH_3COO^-$ as a leaving group. Since $Br^-$ is a better leaving group than $CH_3COO^-$,$I$ has a greater $S_N2$ rate.$(B)$ In pair $B$,$III$ uses $SH^-$ (a strong nucleophile) while $IV$ uses $CH_3SH$ (a neutral,weaker nucleophile). Thus,$III$ has a greater $S_N2$ rate.$(C)$ In pair $C$,both reactions occur in $DMSO$ (a polar aprotic solvent). In polar aprotic solvents,nucleophilicity increases as basicity increases. Since $Cl^-$ is more basic than $I^-$,$Cl^-$ is a better nucleophile in $DMSO$. Thus,$V$ has a greater $S_N2$ rate.$(D)$ In pair $D$,the reactions occur in methanol (a polar protic solvent). In polar protic solvents,nucleophilicity increases down the group due to decreased solvation. Thus,$I^-$ is a better nucleophile than $Cl^-$. Therefore,$VIII$ has a greater $S_N2$ rate.Conclusion: The correct choices are $I, III, V, VIII$,which corresponds to option $C$.

Consider the nucleophilic attacks given below. Select the reaction in each pair that shows the greater $S_N2$ reaction rate.

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