Multi Step Reaction Sequence Involving Aniline Derivatives and Phenols

To determine the correct statements regarding the products $\mathbf{P}$, $\mathbf{Q}$, $\mathbf{S}$, and $\mathbf{T}$, let us analyze the reaction steps sequentially:

1. Analysis of Reaction Sequence for $\mathbf{P}$ and $\mathbf{Q}$

• Step 1: Aniline ($\text{C}_6\text{H}_5\text{NH}_2$) reacts with acetic anhydride ($(\text{CH}_3\text{CO})_2\text{O}$) in pyridine to form acetanilide ($\text{C}_6\text{H}_5\text{NHCOCH}_3$) by protecting the $-\text{NH}_2$ group.

• Step 2: Nitration of acetanilide with concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ at $288\text{ K}$ yields $p$-nitroacetanilide as the major product $\mathbf{P}$, because the bulky $-\text{NHCOCH}_3$ group directs incoming electrophiles predominantly to the less sterically hindered para-position. $\mathbf{P} = p\text{-nitroacetanilide } (\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{NHCOCH}_3)$ Since product $\mathbf{P}$ contains only one nitro group, it is a mononitro compound. Thus, Option (C) is incorrect.

• Step 3: Hydrolysis ($\text{H}_3\text{O}^+$) of $\mathbf{P}$ removes the acetyl protecting group to give $p$-nitroaniline ($\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2$).

• Step 4: Diazotization of $p$-nitroaniline with $\text{NaNO}_2/\text{HCl}$ at $273-278\text{ K}$ converts the amino group into a diazonium salt: $\mathbf{Q} = p\text{-nitrobenzenediazonium chloride } (\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{N}_2^+\text{Cl}^-)$

• Reaction of $\mathbf{Q}$ with Ethanol: When $p$-nitrobenzenediazonium chloride ($\mathbf{Q}$) is treated with ethanol ($\text{CH}_3\text{CH}_2\text{OH}$), it undergoes reduction to form nitrobenzene, while ethanol is oxidized to acetaldehyde ($\text{CH}_3\text{CHO}$): $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{N}_2^+\text{Cl}^- + \text{CH}_3\text{CH}_2\text{OH} \longrightarrow \text{C}_6\text{H}_5\text{NO}_2 + \text{N}_2 + \text{HCl} + \text{CH}_3\text{CHO}$ Since acetaldehyde is an aliphatic aldehyde (not an aromatic aldehyde), Option (A) is incorrect.

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2. Analysis of Reaction Sequence for $\mathbf{S}$ and $\mathbf{T}$

• Step 1: Cumene ($\text{C}_6\text{H}_5\text{CH}(\text{CH}_3)_2$) undergoes aerobic oxidation followed by acidic workup ($\text{O}_2 / \text{H}_3\text{O}^+$) via the cumene hydroperoxide process to produce phenol ($\text{C}_6\text{H}_5\text{OH}$) and acetone.

• Step 2: Phenol reacts with $\text{NaOH}$ and $\text{CO}_2$ followed by acidification ($\text{H}_3\text{O}^+$) via the Kolbe-Schmitt reaction to form salicylic acid ($2$-hydroxybenzoic acid) as the major product $\mathbf{S}$: $\mathbf{S} = \text{Salicylic acid}$

• Phthalein Dye Test for $\mathbf{S}$: Salicylic acid ($\mathbf{S}$), being a phenolic compound containing an active aromatic ring with a free phenolic group, condenses with phthalic anhydride in the presence of concentrated $\text{H}_2\text{SO}_4$ to form a phthalein dye, giving a positive phthalein dye test. Thus, Option (B) is correct.

• Formation of Product $\mathbf{T}$: Electrophilic aromatic substitution (diazo coupling reaction) of $p$-nitrobenzenediazonium chloride ($\mathbf{Q}$) with salicylic acid ($\mathbf{S}$) in aqueous alkaline medium ($\text{NaOH}$) yields an azo dye $\mathbf{T}$: $\mathbf{S} + \mathbf{Q} \xrightarrow{\text{aqueous NaOH}} \mathbf{T} \text{ (an Azo Dye)}$ Azo compounds possess an extended conjugated $\pi$-electron system across the $-\text{N}=\text{N}-$ linkage, causing strong light absorption in the visible spectrum. Hence, $\mathbf{T}$ is an intensely coloured compound. Thus, Option (D) is correct.

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Conclusion

The correct statements are (B) and (D).