Correct order of reactivity of the given alco | vedclass

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(B) The reaction of alcohols with $(NaBr + H_2SO_4)$ proceeds via the formation of a carbocation intermediate ($S_N1$ mechanism). Greater the stabilit

Vedclass · Accepted Answer

$iv > ii > i > iii$

Vedclass · Answer

(B) The reaction of alcohols with $(NaBr + H_2SO_4)$ proceeds via the formation of a carbocation intermediate ($S_N1$ mechanism). Greater the stability of the carbocation formed,higher is the reactivity of the alcohol. The carbocation formed is a substituted benzyl carbocation $(Ar-CH_2^+)$. Electron-donating groups $(EDG)$ increase the stability of the carbocation,while electron-withdrawing groups $(EWG)$ decrease it. The substituents at the para-position are: $(iv)$ $-N(CH_3)_2$ (Strong $EDG$,$+M$ effect) $(ii)$ $-CH_3$ (Weak $EDG$,$+I$ and hyperconjugation) $(i)$ $-H$ (Reference) $(iii)$ $-NO_2$ (Strong $EWG$,$-M$ and $-I$ effect) Thus,the stability order of the carbocations is $(iv) > (ii) > (i) > (iii)$. Therefore,the reactivity order is $(iv) > (ii) > (i) > (iii)$.

Correct order of reactivity of the given alcohols towards the substitution with $(NaBr + H_2SO_4)$ is:

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