Animated Solution for Chemistry - Organic Chemistry: In the following reaction sequence, P, Q, S and T are the major products.

Organic chemistry is often like solving a beautiful puzzle. In this problem, we are presented with two completely independent reaction sequences that eventually converge to form a final, complex molecule. Our goal is to trace these pathways, identify the major products $\mathbf{P}$, $\mathbf{Q}$, $\mathbf{S}$, and $\mathbf{T}$, and then evaluate the given statements.

Let's embark on this journey by breaking down the sequences step by step.

From Aniline to a Diazonium Salt

We begin the first sequence with aniline, a primary aromatic amine. The first reagent is acetic anhydride (${\text{Ac}}_2\text{O}$) in the presence of pyridine. Why do we need this step? The amino group (${\text{-NH}}_2$) is highly activating and strongly ortho/para directing. If we were to perform nitration directly on aniline, the highly oxidizing nature of nitric acid would lead to a messy mixture of oxidation products and poly-nitrated rings.

To tame this reactivity, we protect the amine by converting it into an amide, specifically acetanilide. The acetyl group withdraws some electron density from the nitrogen, moderating its activating power.

Next, we subject acetanilide to a nitrating mixture (Conc. ${\text{HNO}}_3$ and Conc. ${\text{H}}_2{\text{SO}}_4$). The bulky acetyl group provides significant steric hindrance at the ortho positions. Consequently, the incoming electrophile (the nitronium ion, ${\text{NO}}_2^{+}$) predominantly attacks the less hindered para position. This yields p-nitroacetanilide, which is our major product $\mathbf{P}$.

Now, we need to remove the protective group. Treating $\mathbf{P}$ with aqueous acid (${\text{H}}_3{\text{O}}^{+}$) hydrolyzes the amide back to the free amine, giving us p-nitroaniline.

Finally, we treat p-nitroaniline with sodium nitrite (${\text{NaNO}}_2$) and hydrochloric acid ($\text{HCl}$) at ice-cold temperatures ($273-278\text{K}$). This is the classic diazotization reaction. The primary amine is converted into a highly reactive diazonium salt. Thus, our product $\mathbf{Q}$ is p-nitrobenzenediazonium chloride.

From Cumene to Salicylic Acid

Let's shift our focus to the second sequence, starting with cumene (isopropylbenzene). The first step involves treating cumene with oxygen followed by acid hydrolysis. This is the renowned industrial method for synthesizing phenol. The reaction proceeds via a cumene hydroperoxide intermediate, which undergoes an elegant rearrangement to yield phenol and a valuable byproduct, acetone.

Next, the phenol is subjected to the Kolbe-Schmitt reaction. It is treated with sodium hydroxide to form the more reactive phenoxide ion, which then undergoes electrophilic aromatic substitution with carbon dioxide (${\text{CO}}_2$). Subsequent acidification yields ortho-hydroxybenzoic acid, universally known as salicylic acid. This is our product $\mathbf{S}$.

Diazo Coupling

The grand finale of our synthesis brings $\mathbf{Q}$ and $\mathbf{S}$ together in an alkaline medium (aqueous $\text{NaOH}$). The diazonium ion $\mathbf{Q}$ is a weak electrophile, and it seeks out an electron-rich aromatic ring. Salicylic acid $\mathbf{S}$ fits the bill perfectly, thanks to its strongly activating phenolic hydroxyl group.

The coupling reaction occurs at the para position relative to the hydroxyl group, as it is sterically more accessible than the ortho position (which is also partially blocked by the carboxyl group). The resulting molecule, $\mathbf{T}$, is an azo dye characterized by the $-\text{N}=\text{N}-$ linkage connecting the two aromatic rings.

With all our products identified, let's scrutinize the given options:

Option (A): It claims that treating $\mathbf{Q}$ with ethanol generates an aromatic aldehyde. Ethanol acts as a mild reducing agent towards diazonium salts. It reduces the diazonium group to a hydrogen atom, converting $\mathbf{Q}$ into nitrobenzene. In the process, ethanol is oxidized to acetaldehyde (${\text{CH}}_3\text{CHO}$), which is an aliphatic aldehyde, not aromatic. Thus, Option (A) is incorrect.

Option (B): It states that $\mathbf{S}$ gives a positive phthalein dye test. This test is characteristic of phenols. Since $\mathbf{S}$ (salicylic acid) possesses a free phenolic $-\text{OH}$ group, it will condense with phthalic anhydride in the presence of concentrated sulfuric acid to form a phthalein dye. Therefore, Option (B) is absolutely correct.

Option (C): It suggests that $\mathbf{P}$ is a dinitro compound. As we deduced earlier, $\mathbf{P}$ is p-nitroacetanilide, which contains only one nitro group on the benzene ring. It is a mononitro compound. Hence, Option (C) is false.

Option (D): It claims that $\mathbf{T}$ is a colored compound. Compound $\mathbf{T}$ is an azo compound. The extended conjugation of alternating single and double bonds across the two benzene rings and the azo linkage significantly lowers the energy gap for electronic transitions. This allows the molecule to absorb wavelengths in the visible spectrum, making it a brightly colored dye. Option (D) is correct.

In conclusion, the correct statements are (B) and (D).