Animated Solution for Chemistry - Organic Chemistry: Identify correct A, B and C in the reaction sequence given below.

Imagine you are an architect, but instead of bricks and steel, you are building with atoms. Organic synthesis is exactly that—a strategic, step-by-step construction of complex molecules from simple starting materials.

In this problem, we are given a classic three-step reaction sequence starting from the simplest aromatic hydrocarbon: benzene. Our goal is to identify the intermediate products $A$ and $B$, and the final product $C$.

Let's break down this molecular assembly line!

The Nitration of Benzene

Our journey begins with benzene reacting with a mixture of concentrated nitric acid (${\text{HNO}}_3$) and concentrated sulfuric acid (${\text{H}}_2{\text{SO}}_4$) under heating ($\Delta$).

This specific combination of reagents is known as a nitrating mixture. Sulfuric acid, being the stronger acid, protonates the nitric acid. The protonated nitric acid then loses a water molecule to generate the highly reactive nitronium ion ($N{O}_2^{+}$).

This nitronium ion acts as a powerful electrophile. It attacks the electron-rich $\pi$-cloud of the benzene ring, undergoing an electrophilic aromatic substitution. A hydrogen atom is replaced by the nitro group, yielding nitrobenzene.

Thus, our product $A$ is nitrobenzene.

The Directing Power of the Nitro Group

Now we have nitrobenzene, and we subject it to chlorine gas (${\text{Cl}}_2$) in the presence of anhydrous aluminum chloride (${\text{AlCl}}_3$). This is a standard halogenation setup.

The anhydrous ${\text{AlCl}}_3$ acts as a Lewis acid catalyst, polarizing the $\text{Cl}-\text{Cl}$ bond to generate the chloronium ion electrophile ($C{l}^{+}$). But where will this chlorine attack the ring?

This is where the directing effect of the existing substituent comes into play. The nitro group ($-{\text{NO}}_2$) is strongly electron-withdrawing due to both its inductive ($-I$) and resonance ($-M$) effects. It pulls electron density away from the ring, particularly from the ortho and para positions.

Because the ortho and para positions are relatively electron-deficient, the incoming electrophile is forced to attack the meta position, which is comparatively less deactivated.

Therefore, the chlorine atom attaches at the meta position, giving us 1-chloro-3-nitrobenzene (or $m$-chloronitrobenzene).

This is our product $B$.

Selective Reduction

In the final step, we treat product $B$ with iron and hydrochloric acid ($\text{Fe/HCl}$).

This combination is a classic and highly specific reducing agent. It provides a source of electrons and protons that specifically target the nitro group ($-{\text{NO}}_2$), reducing it all the way down to an amino group ($-{\text{NH}}_2$).

Crucially, this mild reducing agent does not affect the carbon-chlorine bond on the aromatic ring. The halogen remains perfectly intact.

As a result, the meta-nitro group is converted to a meta-amino group, yielding 3-chloroaniline (or $m$-chloroaniline).

This is our final product $C$.

By carefully analyzing the reagents and the directing effects at each step, we have successfully mapped out the entire sequence: - $A=\text{Nitrobenzene}$ - $B=1\text{-chloro-}3\text{-nitrobenzene}$ - $C=3\text{-chloroaniline}$

Matching this with our given options, we can confidently conclude that option (a) is the correct answer. Always remember to evaluate the directing nature of the groups already present on the ring before predicting the next substitution!