Match Reaction Sequences with Resulting Phenolic Compounds

To determine the correct matching between the reaction sequences in List-I and the resulting phenolic compounds in List-II, we analyze each reaction sequence step-by-step:

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Reaction Sequence (P)

1. Step (i): Benzenesulfonic acid ($\text{C}_6\text{H}_5\text{SO}_3\text{H}$) undergoes alkali fusion with molten $\text{NaOH}$ followed by acidification with $\text{H}_3\text{O}^+$ to yield phenol ($\text{C}_6\text{H}_5\text{OH}$).
2. Step (ii): Nitration of phenol with $\text{Conc. HNO}_3$ results in electrophilic aromatic substitution at all activated ortho and para positions (2, 4, and 6) to yield 2,4,6-trinitrophenol (Picric acid).

$$\text{Ph-SO}_3\text{H} \xrightarrow{\text{molten NaOH, } \text{H}_3\text{O}^+} \text{Ph-OH} \xrightarrow{\text{Conc. HNO}_3} \text{2,4,6-trinitrophenol}$$

• Result: Matches with compound (3).

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Reaction Sequence (Q)

1. Step (i): Nitration of nitrobenzene ($\text{C}_6\text{H}_5\text{NO}_2$) using $\text{Conc. HNO}_3 / \text{Conc. H}_2\text{SO}_4$ directs the incoming nitro group to the meta-position, giving 1,3-dinitrobenzene.
2. Step (ii): Reduction with $\text{Sn/HCl}$ converts both nitro groups into amino groups, forming benzene-1,3-diamine ($m$-phenylenediamine).
3. Step (iii): Diazotization using $\text{NaNO}_2 / \text{HCl}$ at $0 - 5\ ^\circ\text{C}$ converts both amino groups into diazonium salts, yielding benzene-1,3-bis(diazonium) chloride.
4. Step (iv): Boiling with $\text{H}_2\text{O}$ hydrolyzes the diazonium groups to hydroxyl groups, yielding benzene-1,3-diol (Resorcinol).
5. Step (v): Nitration of resorcinol with $\text{Conc. HNO}_3 / \text{Conc. H}_2\text{SO}_4$ occurs at positions 2, 4, and 6, which are all strongly activated by both $-\text{OH}$ groups, yielding 2,4,6-trinitrobenzene-1,3-diol (Styphnic acid).

$$\text{Nitrobenzene} \xrightarrow{\text{Nitration}} \text{1,3-Dinitrobenzene} \xrightarrow{\text{Sn/HCl}} \text{Benzene-1,3-diamine} \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{Bis-diazonium salt} \xrightarrow{\text{H}_2\text{O}} \text{Resorcinol} \xrightarrow{\text{Nitration}} \text{2,4,6-Trinitrobenzene-1,3-diol}$$

• Result: Matches with compound (5).

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Reaction Sequence (R)

1. Step (i): Sulfonation of resorcinol (benzene-1,3-diol) with $\text{Conc. H}_2\text{SO}_4$ occurs preferentially at positions 4 and 6 due to steric hindrance at position 2 (between the two $-\text{OH}$ groups), yielding benzene-1,3-diol-4,6-disulfonic acid.
2. Step (ii): Nitration with $\text{Conc. HNO}_3$ introduces a nitro group at the remaining vacant, activated position 2, forming 2-nitrobenzene-1,3-diol-4,6-disulfonic acid.
3. Step (iii): Heating with acid ($\text{H}_3\text{O}^+, \Delta$) hydrolyzes and removes both sulfonic acid groups (desulfonation), selectively yielding 2-nitrobenzene-1,3-diol (2-nitroresorcinol).

$$\text{Resorcinol} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{Benzene-1,3-diol-4,6-disulfonic acid} \xrightarrow{\text{Conc. HNO}_3} \text{2-Nitrobenzene-1,3-diol-4,6-disulfonic acid} \xrightarrow{\text{H}_3\text{O}^+, \Delta} \text{2-Nitrobenzene-1,3-diol}$$

• Result: Matches with compound (4).

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Reaction Sequence (S)

1. Step (i): Oxidation of toluene ($\text{C}_6\text{H}_5\text{CH}_3$) with alkaline $\text{KMnO}_4, \Delta$ followed by acid workup ($\text{H}_3\text{O}^+$) yields benzoic acid.
2. Step (ii): Nitration of benzoic acid under vigorous conditions ($\text{Conc. HNO}_3 / \text{Conc. H}_2\text{SO}_4, \Delta$) introduces two nitro groups at both meta-positions, yielding 3,5-dinitrobenzoic acid.
3. Step (iii): Reaction with $\text{SOCl}_2$ followed by $\text{NH}_3$ converts the acid group to an amide, producing 3,5-dinitrobenzamide.
4. Step (iv): Hofmann bromamide degradation using $\text{Br}_2 / \text{NaOH}$ converts the amide group to a primary amine, yielding 3,5-dinitroaniline.
5. Step (v): Diazotization with $\text{NaNO}_2 / \text{HCl}$ at $0 - 5\ ^\circ\text{C}$ forms 3,5-dinitrobenzenediazonium chloride.
6. Step (vi): Hydrolysis by heating with $\text{H}_2\text{O}$ replaces the diazonium group with $-\text{OH}$, yielding 3,5-dinitrophenol.

$$\text{Toluene} \xrightarrow{\text{KMnO}_4/\text{KOH, } \text{H}_3\text{O}^+} \text{Benzoic acid} \xrightarrow{\text{Nitration, } \Delta} \text{3,5-Dinitrobenzoic acid} \xrightarrow{\text{SOCl}_2, \text{NH}_3} \text{3,5-Dinitrobenzamide} \xrightarrow{\text{Br}_2/\text{NaOH}} \text{3,5-Dinitroaniline} \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{Diazonium salt} \xrightarrow{\text{H}_2\text{O}} \text{3,5-Dinitrophenol}$$

• Result: Matches with compound (1).

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Conclusion:

• P $\rightarrow$ 3
• Q $\rightarrow$ 5
• R $\rightarrow$ 4
• S $\rightarrow$ 1

Thus, the correct option is C.