Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

Phthalic Anhydride and Chlorobenzene

Welcome to the fascinating world of electrophilic aromatic substitution! In this problem, we are looking at a classic Friedel-Crafts acylation. Our reactants are phthalic anhydride and chlorobenzene, and the reaction is catalyzed by aluminum chloride ($AlC{l}_3$), followed by an aqueous workup ($H_{2}O$).

To predict the major product, we must understand how the electrophile is generated and how the aromatic ring directs the incoming attack.

Phthalic anhydride is a cyclic anhydride. On its own, it is not electrophilic enough to attack an aromatic ring. This is where our Lewis acid catalyst, $AlC{l}_3$, steps in.

The aluminum atom in $AlC{l}_3$ is electron-deficient. It happily accepts a lone pair of electrons from one of the oxygen atoms in the anhydride. This coordination weakens the adjacent carbon-oxygen bond, causing the ring to pop open. The result? A highly reactive, positively charged acylium ion ($E^{+}$) tethered to a carboxylate-aluminum complex.

Deactivating yet Ortho/Para Directing

Now, let's turn our attention to the substrate: chlorobenzene. Halogens are the famous anomalies of electrophilic aromatic substitution.

Because chlorine is highly electronegative, it pulls electron density away from the benzene ring through the inductive effect ($-I$). This makes the ring less reactive (deactivated) compared to plain benzene. However, chlorine also possesses lone pairs of electrons. Through resonance, it can donate these electrons back into the ring ($+R$ effect). This resonance donation specifically increases electron density at the ortho and para positions.

Therefore, chlorobenzene is a deactivating, yet ortho/para-directing group.

The Deciding Factor

Since the acylium ion can attack either the ortho or the para position, which one will be the major product?

This is where we must consider the physical size of our molecules. The acylium ion generated from phthalic anhydride is incredibly bulky. If it tries to attack the ortho position, it will crash into the large chlorine atom. This steric hindrance makes the ortho attack highly unfavorable.

Consequently, the bulky electrophile takes the path of least resistance and attacks the spacious para position.

After the acylium ion attacks the para position, the intermediate sigma complex loses a proton to restore aromaticity. Finally, the addition of water ($H_{2}O$) hydrolyzes the aluminum complex, freeing the carboxylic acid group.

The resulting major product is 2-(4-chlorobenzoyl)benzoic acid. Looking at our options, this perfectly matches option (d).