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(C) The elimination reaction (dehydration) of alcohols involves the formation of a carbocation or a transition state with a partial positive charge. T

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$II < III < I < IV$

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(C) The elimination reaction (dehydration) of alcohols involves the formation of a carbocation or a transition state with a partial positive charge. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ effect),which destabilizes the positive charge. The closer the $-NO_2$ group is to the reaction center (the carbon atoms involved in the double bond formation),the more it destabilizes the intermediate,thus decreasing the rate of reaction. The distance of the $-NO_2$ group from the $-OH$ group determines the extent of destabilization: $(II)$ has the $-NO_2$ group closest to the reaction site,followed by $(III)$,then $(I)$. In $(IV)$,there is no $-NO_2$ group,only the $+I$ effect of the alkyl chain,which stabilizes the carbocation,making it the most reactive. Therefore,the increasing order of reactivity is: $(II) < (III) < (I) < (IV)$.

Arrange the following alcohols in increasing order of reactivity towards elimination reaction:
$(I)$ $NO_2-CH_2-CH_2-CH_2-CH_2OH$
$(II)$ $CH_3-CH_2-CH(NO_2)-CH_2OH$
$(III)$ $CH_3-CH(NO_2)-CH_2-CH_2OH$
$(IV)$ $CH_3-CH_2-CH_2-CH_2OH$
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Arrange the following alcohols in increasing order of reactivity towards elimination reaction: $(I)$ $NO_2-CH_2-CH_2-CH_2-CH_2OH$ $(II)$ $CH_3-CH_2-CH(NO_2)-CH_2OH$ $(III)$ $CH_3-CH(NO_2)-CH_2-CH_2OH$ $(IV)$ $CH_3-CH_2-CH_2-CH_2OH$ Select the answer from the codes given below: