The increasing order of the rate of reaction | vedclass

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Question

(A) The reaction with $Br_2 / AlCl_3$ is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of

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$iii < i < ii < iv$

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(A) The reaction with $Br_2 / AlCl_3$ is an electrophilic aromatic substitution reaction. The rate of this reaction depends on the electron density of the aromatic ring. Higher electron density increases the rate of reaction.$(i)$ The ring has a weak electron-withdrawing group attached.(ii) The ring has a strong electron-donating group $(-OH)$ and a weak electron-donating group.(iii) The ring has a strong electron-withdrawing carbonyl group $(C=O)$.(iv) The ring has a strong electron-donating group $(-OH)$ and an oxygen atom that provides additional electron density via resonance.Comparing the electronic effects:- (iii) has a strong electron-withdrawing group,making it the slowest.- $(i)$ has a weak electron-withdrawing effect.- (ii) has strong electron-donating groups.- (iv) has the strongest electron-donating effect due to the additional oxygen atom.Thus,the increasing order of the rate of reaction is $iii < i < ii < iv$.

The increasing order of the rate of reaction with $Br_2 / AlCl_3$ is:

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