Animated Solution for Chemistry - Organic Chemistry: Comprehension Passage

Welcome to a beautiful journey through organic synthesis! This problem is a masterpiece that tests your ability to string together multiple classic reactions.

Imagine you are a molecular sculptor, and your starting block of marble is compound $\mathbf{O}$, $o$-dipropylbenzene. Our goal is to chisel away at this molecule, step by step, until we reveal the hidden structures of compounds $\mathbf{R}$ and $\mathbf{T}$.

Don't let the long sequence intimidate you. We will break it down into logical, bite-sized pieces. Let's dive in!

Vigorous Oxidation

Our first tool is acidic potassium permanganate (${\text{KMnO}}_4/{\text{H}}^{+}$). This reagent is like a chemical chainsaw for alkylbenzenes.

Whenever you have an alkyl group attached to a benzene ring, ${\text{KMnO}}_4$ looks for one specific vulnerability: a benzylic hydrogen. If the carbon directly attached to the ring has at least one hydrogen, the entire alkyl chain is ruthlessly cleaved and oxidized.

In compound $\mathbf{O}$, we have two adjacent propyl groups, and both possess benzylic hydrogens. Therefore, the chainsaw cuts through both chains, leaving behind two carboxylic acid groups ($-\text{COOH}$).

This transforms our starting material into compound $\mathbf{P}$, which is benzene-1,2-dicarboxylic acid, universally known as phthalic acid.

Next, we treat phthalic acid with ammonia (${\text{NH}}_3$) and apply heat. Initially, this forms an acid-base salt, ammonium phthalate.

However, the heat acts as a dehydrating force, driving off water molecules. This dehydration converts the carboxylic acid groups into primary amide groups ($-{\text{CONH}}_2$).

The resulting molecule, compound $\mathbf{Q}$, is an open-chain diamide called phthalamide. Notice the 'a' in the name—it signifies the open, non-cyclic structure.

Now, the problem splits into two paths. Let's follow the first path to find compound $\mathbf{R}$.

Compound $\mathbf{Q}$ is treated with bromine (${\text{Br}}_2$) and sodium hydroxide ($\text{NaOH}$). This specific combination of reagents should immediately trigger a massive alarm in your brain!

It is the signature of the Hoffmann bromamide degradation. This reaction is famous for "stepping down" a carbon chain.

It surgically removes the carbonyl carbon of a primary amide, releasing it as a carbonate ion, and leaves behind a primary amine ($-{\text{NH}}_2$). Since compound $\mathbf{Q}$ has two amide groups, both undergo this degradation.

The result is compound $\mathbf{R}$, $o$-phenylenediamine. This perfectly matches option (A) for our first question!

Let's rewind to compound $\mathbf{Q}$ and follow the second path. The problem states that "strong heating" of $\mathbf{Q}$ produces compound $\mathbf{S}$.

While gentle heating formed the open-chain amide, strong heating provides the activation energy needed for the two adjacent amide groups to interact. They undergo a condensation reaction, eliminating a molecule of ammonia gas.

This forces the molecule to cyclize, forming a highly stable five-membered ring containing an imide group. This new cyclic compound $\mathbf{S}$ is called phthalimide. Notice the 'i'—it denotes the cyclic imide structure.

Compound $\mathbf{S}$ is the perfect starting material for the Gabriel phthalimide synthesis, a brilliant method for creating pure primary amines.

First, we treat phthalimide with potassium hydroxide ($\text{KOH}$). The strong base plucks the acidic proton off the nitrogen, creating a powerful, bulky nucleophile known as potassium phthalimide.

This nucleophile is then introduced to ethyl 2-bromopropanoate (${\text{CH}}_3\text{CH}(\text{Br})\text{COOEt}$). The nitrogen atom launches an ${\text{S}}_{\text{N}}2$ attack on the chiral carbon, kicking out the bromide leaving group.

This attaches the entire carbon chain to the nitrogen, forming an $N$-alkylated phthalimide intermediate. The bulky phthalimide group acts as a protective shield, preventing any unwanted over-alkylation.

The final step is to remove the protective shield via acidic hydrolysis (${\text{H}}_3{\text{O}}^{+}$). Boiling the intermediate in aqueous acid acts like a pair of chemical scissors.

It hydrolyzes the tough amide bonds of the phthalimide ring, releasing our original phthalic acid and freeing the nitrogen as a primary amine. But the acid doesn't stop there!

It also hydrolyzes the ester group ($-\text{COOEt}$) on the side chain, converting it into a carboxylic acid ($-\text{COOH}$).

Let's look at the final liberated molecule. It has a central carbon attached to an amino group ($-{\text{NH}}_2$), a carboxylic acid group ($-\text{COOH}$), and a methyl group ($-{\text{CH}}_3$).

This is the exact structure of the naturally occurring amino acid, Alanine! Therefore, compound $\mathbf{T}$ is Alanine, which corresponds to option (B) for our second question.

What a spectacular sequence! We started with a simple hydrocarbon derivative and elegantly sculpted it into a fundamental building block of life.