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(C) The leaving group ability is inversely proportional to the basicity of the group. $A$ weaker base is a better leaving group. The basicity is deter

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$I > II > IV > III$

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(C) The leaving group ability is inversely proportional to the basicity of the group. $A$ weaker base is a better leaving group. The basicity is determined by the strength of the conjugate acid. Conjugate acids are: $HBr$ $(pK_a \approx -9)$,$CH_3COOH$ $(pK_a \approx 4.76)$,$PhOH$ $(pK_a \approx 10)$,and $CH_3CH_2OH$ $(pK_a \approx 16)$. The order of acid strength is: $HBr > CH_3COOH > PhOH > CH_3CH_2OH$. Therefore,the order of basicity is: $Br^{-} Since the leaving group tendency is the reverse of basicity,the order is: $Br^{-} > CH_3COO^{-} > PhO^{-} > CH_3CH_2O^{-}$,which corresponds to $I > II > IV > III$.

Column-$I$	Column-$II$
$(i)$ $R-Cl + NaI \xrightarrow{\text{dry acetone}} R-I + NaCl$	$(a)$ Swarts reaction
$(ii)$ $CH_3-Br + AgF \xrightarrow{\Delta} CH_3-F + AgBr$	$(b)$ Wurtz reaction
$(iii)$ $R-X + Mg$ $\xrightarrow{\text{dry ether}} R-Mg-X$ $\xrightarrow{H_2O} RH + Mg(OH)X$	$(c)$ Finkelstein reaction
	$(d)$ Grignard reaction

Arrange the following in order of their leaving group tendency:
$(I) Br^{-}$,$(II) CH_3COO^{-}$,$(III) CH_3CH_2O^{-}$,$(IV) PhO^{-}$

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Arrange the following in order of their leaving group tendency: $(I) Br^{-}$,$(II) CH_3COO^{-}$,$(III) CH_3CH_2O^{-}$,$(IV) PhO^{-}$