Animated Solution for Chemistry - Organic Chemistry: The structure of product $C$, formed by the following sequence of reactions is ${\text{CH}}_3\text{COOH}+{\text{SOCl}}_2⟶A\underset{{\text{AlCl}}_3}{\overset{\text{Benzene}}{\to }}B\underset{-\text{OH}}{\overset{\text{KCN}}{\to }}C$

Our goal is to identify the final product $C$. This sequence tests our knowledge of carboxylic acid derivatives, electrophilic aromatic substitution, and nucleophilic addition to carbonyls.

Formation of Acid Chloride The first step involves the reaction of acetic acid (${\text{CH}}_3\text{COOH}$) with thionyl chloride (${\text{SOCl}}_2$). This is a standard method for converting carboxylic acids into acid chlorides. The hydroxyl group ($-\text{OH}$) is replaced by a chlorine atom ($-\text{Cl}$), yielding acetyl chloride (${\text{CH}}_3\text{COCl}$), which is our intermediate $A$.

Thionyl chloride is often preferred for this transformation because the byproducts, sulfur dioxide and hydrogen chloride, are gases that easily escape the reaction mixture, driving the reaction forward according to Le Chatelier's principle.

Friedel-Crafts Acylation In the second step, acetyl chloride ($A$) reacts with benzene in the presence of anhydrous aluminum chloride (${\text{AlCl}}_3$). This is the classic Friedel-Crafts acylation. The Lewis acid ${\text{AlCl}}_3$ complexes with the chlorine atom of acetyl chloride, generating a highly reactive acylium ion (${\text{CH}}_3{\text{C}}^{+}=\text{O}$).

This electrophile attacks the electron-rich benzene ring, leading to the formation of a ketone. The product is acetophenone (${\text{C}}_6{\text{H}}_5{\text{COCH}}_3$), which is our intermediate $B$.

Cyanohydrin Formation The final step involves treating acetophenone ($B$) with potassium cyanide ($\text{KCN}$) in a slightly basic medium ($-\text{OH}$). The cyanide ion (${\text{CN}}^{-}$) acts as a strong nucleophile and attacks the electrophilic carbonyl carbon of acetophenone.

As the cyanide ion bonds to the carbon, the pi electrons of the carbon-oxygen double bond shift onto the oxygen atom, creating an alkoxide intermediate. This intermediate quickly picks up a proton from the solvent to form a hydroxyl group ($-\text{OH}$). The resulting product is a cyanohydrin, specifically 2-hydroxy-2-phenylpropanenitrile, which is our final product $C$.

Looking at the given options, the structure of 2-hydroxy-2-phenylpropanenitrile perfectly matches option (a).