Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed contains

Imagine you are an architect, but instead of bricks and mortar, you are building with atoms and molecules. This problem is a classic example of a multi-step organic synthesis, where each reaction sets the stage for the next. Let's break down this chemical journey step-by-step.

The Directing Power of the Methyl Group

We start with toluene, a benzene ring adorned with a methyl group ($-C{H}_3$). This methyl group is not just a passive bystander; it is an electron-donating group. By pushing electron density into the aromatic ring via hyperconjugation and the inductive effect, it activates the ring towards electrophilic aromatic substitution.

Crucially, it directs incoming electrophiles to the ortho and para positions. When we treat toluene with a nitrating mixture (concentrated $HN{O}_3$ and concentrated $H_{2}S{O}_4$), the nitronium ion ($N{O}_2^{+}$) attacks these electron-rich spots.

As a result, we obtain a mixture of $o$-nitrotoluene and $p$-nitrotoluene.

Reduction to Amines

Now that we have our nitro compounds, we need to transform them. The problem states we use tin (Sn) and hydrochloric acid (HCl). This is a classic reducing environment specifically tailored to convert nitro groups ($-N{O}_2$) into primary amino groups ($-N{H}_2$).

This reduction step smoothly converts our $o$-nitrotoluene and $p$-nitrotoluene into $o$-toluidine and $p$-toluidine, respectively.

The Magic of Diazotisation

Primary aromatic amines are incredibly versatile because they can be converted into diazonium salts. By reacting our toluidine mixture with sodium nitrite ($NaN{O}_2$) and hydrochloric acid (HCl) at ice-cold temperatures ($0-5^{\circ }C$), we perform diazotisation.

The amino groups are transformed into the highly reactive diazonium group ($-N_2^{+}C{l}^{-}$). We now have a mixture of $o$-methylbenzenediazonium chloride and $p$-methylbenzenediazonium chloride.

The Sandmeyer Reaction

The final act of our synthesis involves heating these diazonium salts with cuprous bromide ($CuBr$). This is the famous Sandmeyer reaction. The diazonium group is an excellent leaving group because it departs as stable nitrogen gas ($N_2$).

The cuprous bromide facilitates the replacement of the diazonium group with a bromine atom. Consequently, our diazonium salts are converted into $o$-bromotoluene and $p$-bromotoluene.

Because the very first step (nitration) locked our substituents into the ortho and para positions relative to the methyl group, the final product is a mixture of $o$- and $p$-bromotoluenes. This perfectly aligns with option (a).