Animated Solution for Chemistry - Organic Chemistry: $[\text{P}]$ on treatment with ${\text{Br}}_2/{\text{FeBr}}_3$ in ${\text{CCl}}_4$ produced a single isomer ${\text{C}}_8{\text{H}}_7{\text{O}}_2\text{Br}$ while heating $[\text{P}]$ with sodalime gave toluene. The compound $[\text{P}]$ is

Welcome to a fantastic organic chemistry puzzle! We are presented with a mystery compound, denoted as $[\text{P}]$, which undergoes two very revealing reactions.

First, it reacts with bromine and iron(III) bromide (${\text{Br}}_2/{\text{FeBr}}_3$) to give exactly one single isomer with the formula ${\text{C}}_8{\text{H}}_7{\text{O}}_2\text{Br}$. Second, heating it with sodalime produces toluene. To solve this, we must decode these clues step by step.

The First Clue

Let's focus on the second reaction first. What exactly does sodalime do? Sodalime is a mixture of sodium hydroxide ($\text{NaOH}$) and calcium oxide ($\text{CaO}$), and it is famous for a process called decarboxylation.

It literally rips off a carboxyl group ($-\text{COOH}$) from an aromatic ring and replaces it with a hydrogen atom, releasing carbon dioxide in the process.

Now, look at the product of this decarboxylation. It's toluene (${\text{C}}_6{\text{H}}_5-{\text{CH}}_3$)! This is a massive clue. If removing a carboxyl group gives us toluene, then our original mystery compound $[\text{P}]$ must have been a benzene ring with both a methyl group and a carboxyl group attached to it. In other words, $[\text{P}]$ is a toluic acid.

So, we've narrowed it down to toluic acid. But wait, there are three different ways to arrange a methyl group and a carboxyl group on a benzene ring. They can be ortho, meta, or para to each other. Which one is our compound $[\text{P}]$? To find out, we need to look at the first reaction.

The first reaction is an electrophilic aromatic substitution using bromine. The problem explicitly states that this reaction produces a single isomer. This is the golden key! For a benzene ring with two substituents to yield only one mono-brominated product, the molecule must possess a high degree of symmetry. Let's test our three candidates.

Let's start with ortho-toluic acid. Look at the available positions on the ring. Because the methyl and carboxyl groups are right next to each other, the molecule is asymmetrical. Positions 3, 4, 5, and 6 are all chemically distinct environments. Brominating this would give us a messy mixture of several different isomers, not just one.

What about meta-toluic acid? It's a similar story. The positions 2, 4, 5, and 6 are all unique. There is no axis of symmetry that maps one available carbon onto another. So, just like the ortho isomer, brominating meta-toluic acid would result in multiple products. This can't be our compound $[\text{P}]$ either.

Now, let's examine para-toluic acid (4-methylbenzoic acid). Notice how the methyl and carboxyl groups are directly opposite each other. This creates a beautiful vertical axis of symmetry right down the middle of the molecule. This symmetry is exactly what we are looking for!

Because of this symmetry, the left side of the ring is a mirror image of the right side. The two positions next to the methyl group are chemically identical. The methyl group is an activating, ortho-directing group, so it will strongly direct the incoming bromine electrophile to these exact two equivalent positions.

Whether the bromine attaches to the left or the right of the methyl group, you get the exact same molecule: 3-bromo-4-methylbenzoic acid. A single, unique isomer!

This perfectly matches all the conditions given in our problem. Therefore, compound $[\text{P}]$ must be 4-methylbenzoic acid. Looking at our options, this corresponds to option (b). Always remember to look for symmetry when a reaction yields a single isomer!