Animated Solution for Chemistry - Organic Chemistry: Comprehension Passage

Don't Panic

When you first look at this problem, the starting material—1,3,5-tris(4-nitrophenyl)benzene—looks absolutely massive. It is easy to get intimidated by its size.

But here is the secret to organic chemistry: focus on the functional groups.

The central benzene ring is just a structural scaffold. It is highly sterically hindered and lacks strong activating groups, so it will just sit there as a spectator. All the exciting chemistry is going to happen at the three peripheral $-{\text{NO}}_2$ groups. Let's break it down step by step.

Reduction of Nitro Groups

Our first reagent is excess $\text{Sn/HCl}$. This is a classic, powerful reducing agent.

Its job is simple but crucial: it reduces nitro groups ($-{\text{NO}}_2$) to primary amine groups ($-{\text{NH}}_2$).

Because we have an excess of the reagent, all three nitro groups on the periphery of our giant molecule will be reduced simultaneously. We now have an intermediate with three amine groups.

The Magic of Diazotization

Next, we treat our amine intermediate with excess ${\text{NaNO}}_2/\text{HCl}$ at $0^{\circ }\text{C}$.

This is one of the most important reactions in organic chemistry: Diazotization.

The primary amine groups react with nitrous acid (generated in situ) to form diazonium salts. All three $-{\text{NH}}_2$ groups are converted into $-{\text{N}}_2^{+}{\text{Cl}}^{-}$ groups.

This magnificent new molecule is our Compound P: 1,3,5-tris(4-diazoniophenyl)benzene trichloride.

Hydrolysis to Phenol

Now, we take Compound P and treat it with excess water (${\text{H}}_2\text{O}$) at room temperature.

Diazonium salts are highly reactive. When warmed with water, they undergo hydrolysis, releasing nitrogen gas (${\text{N}}_2$) and leaving behind a hydroxyl group ($-\text{OH}$).

All three diazonium groups are replaced by $-\text{OH}$ groups. We have successfully synthesized a triphenol derivative.

This is our Compound Q: 1,3,5-tris(4-hydroxyphenyl)benzene.

Polybromination in Aqueous Medium

The problem states that Compound Q undergoes bromination in an aqueous medium to furnish Compound R.

Recall the behavior of phenol in bromine water. The $-\text{OH}$ group is a strongly activating, ortho/para-directing group. In an aqueous medium, phenol ionizes slightly to form the highly nucleophilic phenoxide ion, leading to polybromination at all available ortho and para positions.

In our molecule, the para position of each phenolic ring is already attached to the central benzene scaffold. It is blocked!

Therefore, the incoming bromine atoms have no choice but to attack the two ortho positions on each of the three phenolic rings.

Since there are two ortho positions per ring, and three rings in total, exactly six bromine atoms will be added to the molecule.

This gives us Compound R: 1,3,5-tris(3,5-dibromo-4-hydroxyphenyl)benzene.

Verifying with Molar Mass

The problem provides a brilliant built-in checkpoint: the molar mass difference between Q and R is $474\text{ g/mol}$. Let's verify our structure!

In going from Q to R, we replaced 6 hydrogen atoms with 6 bromine atoms.

The math matches perfectly! Our proposed structure for R is absolutely correct.

Counting Heteroatoms in R (Question 4)

Question 4 asks for the number of heteroatoms in one molecule of R.

A heteroatom is simply any atom that is not carbon or hydrogen.

Looking at Compound R, we have: - 3 Oxygen atoms (from the three $-\text{OH}$ groups) - 6 Bromine atoms (from the polybromination)

The answer to Question 4 is 9.

The Azo Coupling Reaction

Now, let's rewind and go back to Compound P (our diazonium salt). The problem states that P is treated with an excess of phenol under basic conditions to give Compound S.

What happens when a diazonium salt meets phenol in a basic medium? Azo Coupling!

In a basic medium, phenol is converted into the phenoxide ion, which is incredibly electron-rich. The diazonium group acts as an electrophile and attacks the para position of the phenoxide ring.

This reaction happens at all three diazonium sites, creating three beautiful azo linkages ($-\text{N}=\text{N}-$).

This massive, highly conjugated molecule is Compound S: 1,3,5-tris(4-(4-hydroxyphenylazo)phenyl)benzene.

Verifying S with Molar Mass

Let's use the second checkpoint. The molar mass difference between P and S is given as $172.5\text{ g/mol}$.

Let's calculate the total molar mass of both compounds.

Compound P has the formula ${\text{C}}_{24}{\text{H}}_{15}{\text{N}}_6{\text{Cl}}_3$.

Compound S has the formula ${\text{C}}_{42}{\text{H}}_{30}{\text{N}}_6{\text{O}}_3$.

Once again, the math is flawless. Our structure for S is confirmed!

Counting the Atoms (Question 5)

Question 5 asks for the total number of carbon atoms and heteroatoms in one molecule of S. Let's count them carefully.

Carbon Atoms: - Central scaffold ring: 6 carbons - 3 inner phenyl rings: $3\times 6=18$ carbons - 3 outer phenol rings: $3\times 6=18$ carbons - Total Carbon = $6+18+18=42$

Heteroatoms: - 6 Nitrogen atoms (from the three azo linkages) - 3 Oxygen atoms (from the three phenol groups) - Total Heteroatoms = $6+3=9$

The answer to Question 5 is 51. We have successfully conquered this giant molecule!