Animated Solution for Chemistry - Organic Chemistry: Which of these will produce the highest yield in Friedel-Crafts reaction?

The Friedel-Crafts reaction is a cornerstone of electrophilic aromatic substitution, allowing us to attach alkyl or acyl groups to a benzene ring. However, it comes with a strict requirement: a strong Lewis acid catalyst, most commonly anhydrous aluminum chloride (${\text{AlCl}}_3$). This catalyst is essential for generating the highly reactive electrophile. But what happens when the catalyst interacts with the substituents already present on the ring? This question explores that exact trap.

When we look at aniline and phenol, our first instinct is to declare them the winners. The $-{\text{NH}}_2$ and $-\text{OH}$ groups are powerful activating groups. Through their $+\text{M}$ (mesomeric) effect, they pump electron density into the benzene ring, making it highly attractive to electrophiles.

However, there is a massive catch. The nitrogen in aniline and the oxygen in phenol possess lone pairs of electrons. Aluminum chloride, being a strong Lewis acid, is electron-deficient and desperately seeks out these lone pairs. The moment ${\text{AlCl}}_3$ is introduced, it forms a strong coordinate bond (an adduct) with the $-{\text{NH}}_2$ and $-\text{OH}$ groups.

This adduct formation transforms the electron-donating $-{\text{NH}}_2$ into a positively charged $-{\text{NH}}_2^{+}{\text{AlCl}}_3^{-}$ group. Suddenly, a powerful activating group becomes a strongly deactivating group, pulling electron density away from the ring and completely halting the Friedel-Crafts reaction.

Next, we consider benzamide. The amide group ($-{\text{CONH}}_2$) is intrinsically a strongly deactivating group. The carbonyl oxygen pulls electron density away from the ring through a strong $-\text{M}$ effect. Because the ring is already severely electron-deficient, it is highly unreactive towards electrophilic substitution. Benzamide is out of the race from the very beginning.

Chlorobenzene

Finally, we arrive at chlorobenzene. Halogens are unique; they are deactivating because their electronegativity ($-\text{I}$ effect) dominates their ability to donate electrons through resonance ($+\text{M}$ effect). Therefore, chlorobenzene is less reactive than benzene itself.

So, why does it win? The key lies in its interaction with the catalyst. The lone pairs on the chlorine atom are highly delocalized and much less basic than those on nitrogen or oxygen. Consequently, chlorine does not form a strong adduct with ${\text{AlCl}}_3$.

While the ring is weakly deactivated, the catalyst remains free to generate the electrophile, and the reaction can proceed. Thus, among the given options, chlorobenzene is the only compound that will undergo the Friedel-Crafts reaction to a reasonable extent, yielding the highest relative amount of product.