Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: In the chemical reactions given above A and B respectively are

From Diazonium to Ethylbenzene

Organic chemistry is often like solving a puzzle. You are given a starting material and a final product, and you have to figure out the missing pieces. In this problem, we are presented with a two-step reaction sequence that transforms benzene diazonium chloride into ethylbenzene. Let's break down this elegant transformation step by step.

The Disappearing Act (Reduction of Diazonium)

Our journey begins with benzene diazonium chloride, a highly reactive intermediate. The first step involves reacting this compound with an unknown reagent 'A' and water to yield benzene as the major product. This is a classic reduction reaction.

The diazonium group ($-{\text{N}}_2^{+}{\text{Cl}}^{-}$) is an excellent leaving group because it can depart as stable nitrogen gas (${\text{N}}_2$). To replace this group with a hydrogen atom and form benzene, we need a mild reducing agent. The two most common reagents for this specific transformation are hypophosphorous acid (${\text{H}}_3{\text{PO}}_2$) and ethanol (${\text{CH}}_3{\text{CH}}_2\text{OH}$). Both of these can effectively donate a hydride equivalent to the benzene ring, reducing it while they themselves get oxidized. Therefore, reagent 'A' is likely ${\text{H}}_3{\text{PO}}_2$.

Molecular Lego (Friedel-Crafts Alkylation)

Now that we have our clean benzene ring, we move to the second step. Benzene reacts with an unknown reagent 'B' in the presence of anhydrous aluminum chloride (${\text{AlCl}}_3$) to form ethylbenzene.

This setup screams Friedel-Crafts alkylation! In this reaction, a Lewis acid catalyst like ${\text{AlCl}}_3$ is used to generate a strong electrophile from an alkyl halide. Since our final product has an ethyl group ($-{\text{CH}}_2{\text{CH}}_3$) attached to the benzene ring, the electrophile must be an ethyl carbocation (${\text{CH}}_3{\text{CH}}_2^{+}$). Consequently, reagent 'B' must be an ethyl halide, such as ethyl chloride (${\text{CH}}_3{\text{CH}}_2\text{Cl}$).

Putting the Pieces Together

By analyzing both steps, we have deduced that reagent 'A' is a reducing agent like hypophosphorous acid (${\text{H}}_3{\text{PO}}_2$), and reagent 'B' is an alkylating agent, specifically ethyl chloride (${\text{CH}}_3{\text{CH}}_2\text{Cl}$). Looking at our options, the pair that perfectly matches our deduction is ${\text{H}}_3{\text{PO}}_2$ and ${\text{CH}}_3{\text{CH}}_2\text{Cl}$. This problem beautifully illustrates how understanding the fundamental roles of different reagents allows us to navigate through multi-step organic syntheses with confidence.