Consider the following alcohols. The order of | vedclass

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(A) The reaction of alcohols with $HBr$ proceeds via the formation of a carbocation intermediate (usually $S_N1$ mechanism for benzylic alcohols).The

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$III > I > IV > II$

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(A) The reaction of alcohols with $HBr$ proceeds via the formation of a carbocation intermediate (usually $S_N1$ mechanism for benzylic alcohols).The rate of reaction is directly proportional to the stability of the carbocation formed in the rate-determining step.The carbocations formed are substituted benzyl carbocations: $X-C_6H_4-CH_2^+$.Electron-donating groups $(EDG)$ stabilize the carbocation,while electron-withdrawing groups $(EWG)$ destabilize it.- In $(III)$,the $-OCH_3$ group is a strong $EDG$ ($+M$ effect),making it the most reactive.- In $(I)$,there is no substituent.- In $(IV)$,the $-Br$ group is an $EWG$ ($-I$ effect),making it less reactive than $(I)$.- In $(II)$,the $-NO_2$ group is a very strong $EWG$ ($-M$ and $-I$ effects),making it the least reactive.Therefore,the order of decreasing reactivity is $III > I > IV > II$.

Consider the following alcohols. The order of decreasing reactivities of these alcohols towards substitution with $HBr$ is:
$(I)$ $C_6H_5CH_2OH$
$(II)$ $p-NO_2-C_6H_4CH_2OH$
$(III)$ $p-CH_3O-C_6H_4CH_2OH$
$(IV)$ $p-Br-C_6H_4CH_2OH$

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Consider the following alcohols. The order of decreasing reactivities of these alcohols towards substitution with $HBr$ is:$(I)$ $C_6H_5CH_2OH$$(II)$ $p-NO_2-C_6H_4CH_2OH$$(III)$ $p-CH_3O-C_6H_4CH_2OH$$(IV)$ $p-Br-C_6H_4CH_2OH$