See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step-by-step analysis of each row: (a) Reactant is toluene (C6H5-CH3) and product is benzyl chloride (C6H5-CH2Cl). The CH3 group has been converted to CH2Cl, meaning one hydrogen on the methyl group has been replaced by chlorine. This is a free-radical (benzylic) halogenation, not electrophilic aromatic substitution. The reagent required is Cl2 under UV light (hv) or alternatively SO2Cl2 under hv. These conditions favor radical halogenation at the benzylic position rather than ring chlorination. Therefore A = Cl2/hv or SO2Cl2/hv. (b) Reactant is 4-nitroaniline. Treatment with NaNO2/dilute H2SO4 at 0-5 degrees C forms a diazonium salt. Subsequent heating (boiling) of the diazonium salt causes hydrolysis (Sandmeyer-type thermal decomposition with water), replacing the diazonium group (-N2+) with -OH. The NO2 group remains intact. The product is 4-nitrophenol: a benzene ring bearing -OH where -NH2 was and -NO2 at the para position. Therefore B = 4-nitrophenol (benzene ring with -OH and ortho/para -NO2 as drawn). (c) Reactant is toluene. Reagent is SO3/conc. H2SO4 (fuming sulfuric acid), which is the electrophilic aromatic sulfonation reagent. The electrophile SO3 attacks the ring. The CH3 group is an ortho/para director. The major product is predominantly the para isomer due to steric reasons: 4-methylbenzenesulfonic acid (toluene-4-sulfonic acid), a benzene ring with CH3 at top and SO3H at para (bottom). Therefore C = 4-methylbenzenesulfonic acid (benzene ring with -CH3 top and -SO3H bottom, para). (d) The products given are 3-methylphenol (m-cresol) and 4-methylphenol (p-cresol), produced by NaOH fusion at 330 degrees C followed by acid workup. This is the alkali fusion of an aryl sulfonate (or equivalently, this is consistent with the Dow process / fusion of a chlorotoluene). Since the products are a mixture of meta and para cresols, the reactant must be a chlorotoluene that undergoes nucleophilic aromatic substitution or fusion. Specifically, NaOH at 330 degrees C with a chlorotoluene (benzyl chloride type won't give phenols this way; it must be chlorobenzene derivative). The mixture of 3-methyl and 4-methylphenol suggests the reactant is either 3-chlorotoluene or 4-chlorotoluene undergoing rearrangement, but more precisely the reactant that gives both meta and para cresols upon NaOH fusion is 4-chlorotoluene (p-chlorotoluene): benzene ring with CH3 at top and Cl at bottom (para). NaOH fusion can give both isomers due to a benzyne intermediate mechanism. Therefore D = 4-chlorotoluene (benzene ring with -CH3 top and -Cl bottom, para). (e) Reactant is 1-chloro-2,4-dinitrobenzene. Treatment with aqueous NaOH at 60 degrees C (mild nucleophilic aromatic substitution) followed by acid workup: the two electron-withdrawing NO2 groups (ortho and para to Cl) strongly activate the ring toward nucleophilic aromatic substitution. OH- displaces Cl- to give 2,4-dinitrophenol: a benzene ring with -OH at position 1, -NO2 at position 2, and -NO2 at position 4. Therefore E = 2,4-dinitrophenol (benzene ring with -OH and two -NO2 substituents as drawn). Therefore, the correct answer is {"A": "Cl2/hv or SO2Cl2/hv", "B": "benzene ring with -OH and ortho -NO2 (as drawn, p.585)", "C": "benzene ring with -CH3 (top) and -SO3H (bottom, para) (as drawn, p.585)", "D": "benzene ring with -CH3 (top) and -Cl (bottom, para) (as drawn, p.585)", "E": "benzene ring with -OH and -NO2 substituents (as drawn, p.585)"}.