Consider the three aromatic molecules (P, Q a | vedclass

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Question

(A) The reaction is an electrophilic aromatic substitution (diazo coupling). The reactivity is determined by the electron density on the aromatic ring

Vedclass · Accepted Answer

$P > Q > R$

Vedclass · Answer

(A) The reaction is an electrophilic aromatic substitution (diazo coupling). The reactivity is determined by the electron density on the aromatic ring, which is enhanced by electron-donating groups like $-N(Me)_2$. Steric hindrance at the ortho position to the $-N(Me)_2$ group significantly reduces the reactivity by preventing the approach of the electrophile. Molecule $P$ has no ortho substituents, molecule $Q$ has one ortho methyl group, and molecule $R$ has two ortho methyl groups. Therefore, the steric hindrance increases in the order $P  Q > R$. Wait, let us re-evaluate: $P$ has no ortho substituents, $Q$ has one ortho methyl group, and $R$ has two ortho methyl groups. The reactivity order is $P > Q > R$.

Consider the three aromatic molecules ($P$, $Q$ and $R$) whose structures are given below. The correct order regarding the reactivity of these compounds with $Ph-N\equiv N^{(+)}Cl^{(-)}$ under optimum but slightly acidic medium is:

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