Animated Solution for Chemistry - Organic Chemistry: Consider the following reactions : $\text{[P]}\stackrel{{\text{Br}}_2/{\text{H}}_2\text{O}}{\to }{\text{C}}_7{\text{H}}_6{\text{NBr}}_3$ $\text{[P]}\stackrel{\text{(i) }{\text{NaNO}}_2/\text{HCl},0-5^{\circ }\text{C}\text{ (ii) }\beta \text{-naphthol/NaOH}}{\to }\text{Coloured solid}$ The compound [P] is

Imagine you are a chemical detective presented with an unknown compound, [P]. This compound is quite expressive and reveals its identity through two very classic and distinct chemical reactions.

First, it reacts with bromine water to form a highly brominated product. Second, it undergoes diazotization followed by coupling with $\beta$-naphthol to form a brilliantly coloured solid. Let's break down these clues one by one to unmask the true identity of [P].

Let's start with the second reaction. Compound [P] is treated with sodium nitrite (${\text{NaNO}}_2$) and hydrochloric acid ($\text{HCl}$) at a chilling $0-5^{\circ }\text{C}$, followed by the addition of $\beta$-naphthol in a basic medium.

This sequence is the legendary Azo Dye Test. It is a hallmark reaction exclusively given by primary aromatic amines. When a primary aromatic amine reacts with nitrous acid at low temperatures, it forms a stable diazonium salt. This electrophilic diazonium ion then couples with an electron-rich aromatic ring like $\beta$-naphthol to form a brightly coloured azo dye.

Because [P] successfully forms this dye, we can definitively conclude that it must be a primary aromatic amine. This immediately rules out any secondary amines, such as N-methylaniline, which would instead form a yellow, oily N-nitroso compound.

Now, let's turn our attention to the first reaction. Compound [P] reacts with bromine water (${\text{Br}}_2/{\text{H}}_2\text{O}$) to yield a product with the molecular formula ${\text{C}}_7{\text{H}}_6{\text{NBr}}_3$.

Looking at our options, we see they are all toluidines (methylanilines) with the starting molecular formula ${\text{C}}_7{\text{H}}_9\text{N}$. Let's do some quick chemical math:

This mass difference clearly indicates that exactly three hydrogen atoms on the benzene ring have been replaced by three bromine atoms. We are dealing with a tribromination reaction!

The $-{\text{NH}}_2$ group is a powerhouse. It is a strongly activating group that directs incoming electrophiles strictly to the ortho and para positions via resonance.

For tribromination to occur, the incoming bromine atoms need space. Specifically, all three activated positions (both ortho positions and the single para position relative to the $-{\text{NH}}_2$ group) must be completely unsubstituted and free.

Let's evaluate the candidates: p-toluidine (4-methylaniline): The para position is blocked by a methyl group. It can only undergo dibromination at the two ortho positions. o-toluidine (2-methylaniline): One of the ortho positions is blocked by a methyl group. It can also only undergo dibromination. m-toluidine (3-methylaniline):* Here, the methyl group sits quietly at the meta position. Both ortho positions and the para position are perfectly free and waiting for electrophilic attack!

Therefore, the only compound that fits both criteria—being a primary aromatic amine and having the structural freedom to undergo tribromination—is 3-methylaniline. The mystery is solved!