Arrange the leaving group ability of the foll | vedclass

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(B) The leaving group ability is inversely proportional to the basicity of the group. $A$ weaker base is a better leaving group. Among the given optio

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$II, I, IV, III$

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(B) The leaving group ability is inversely proportional to the basicity of the group. $A$ weaker base is a better leaving group. Among the given options,the sulfonate ions $(I)$ and $(II)$ are excellent leaving groups due to resonance stabilization of the negative charge across three oxygen atoms. Between $(I)$ and $(II)$,the methyl group in $(I)$ is electron-donating ($+I$ and $+H$ effect),which increases the electron density and basicity of the anion,making $(II)$ a better leaving group than $(I)$. $Br^-$ $(IV)$ is a better leaving group than $N_3^-$ $(III)$ because $HBr$ is a much stronger acid than $HN_3$. Thus,the decreasing order of leaving group ability is $II > I > IV > III$.

Arrange the leaving group ability of the following groups in decreasing order:
$(I)$ $CH_3-C_6H_4-SO_3^-$,$(II)$ $C_6H_5-SO_3^-$,$(III)$ $N_3^-$,$(IV)$ $Br^-$

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Arrange the leaving group ability of the following groups in decreasing order: $(I)$ $CH_3-C_6H_4-SO_3^-$,$(II)$ $C_6H_5-SO_3^-$,$(III)$ $N_3^-$,$(IV)$ $Br^-$