See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 3-methyleneisobenzofuran-1(3H)-one (3-methylenephthalide), which has a benzene ring fused to a five-membered lactone ring bearing an exo-methylene (=CH2) at C3 and a carbonyl at C1. Step 2 - Reaction with C6H6/AlCl3 (Friedel-Crafts): The exo-methylene of the lactone undergoes Friedel-Crafts alkylation with benzene. The AlCl3 activates the electrophilic alkylation at the methylene position, introducing a phenyl group. This gives compound (A): 3-benzylidenephthalide or 3-benzyl-3H-isobenzofuranone derivative. More specifically, the methylene reacts with benzene via Friedel-Crafts to give 3-benzylphthalide or 3-benzylidenephthalide. Given the exo-methylene, electrophilic addition places Ph at the exo-carbon, giving 3-benzylideneisobenzofuranone (a benzylidene phthalide) as compound (A). Step 3 - Reaction (i) PCl5: PCl5 converts the lactone carbonyl to a gem-dichloro group (or acid chloride), converting C=O of the lactone to CCl2, giving a dichloride. This opens or modifies the ring: the lactone C-O bond and C=O are converted. PCl5 on a lactone typically gives an acid chloride with ring opening, producing an ortho-(chlorocarbonyl)benzyl chloride type intermediate, or it converts C=O to CCl2. With a cyclic lactone (phthalide), PCl5 gives the corresponding imido chloride or dichloro compound. Specifically, PCl5 converts the lactone to a gem-dichloro intermediate: the ring opens to give an ortho-substituted benzene with -CCl2- and -CHPh- (or =CHPh) groups as an open-chain dichloride. Step 4 - Reaction (ii) H2/Pd-BaSO4 (Rosenmund-type, partial hydrogenation): This is a controlled partial reduction. The vinylic/allylic C-Cl bonds can be selectively reduced to C-H under Rosenmund conditions (Pd-BaSO4 is a poisoned catalyst for selective reduction). The gem-dichloride (-CCl2-) is reduced to -CHCl- or further to -CH2-, but under Rosenmund conditions one Cl is removed. Alternatively, considering the structure: PCl5 converts C=O to C(Cl)= and H2/Pd-BaSO4 reduces one C-Cl. The net result of (i)+(ii) on the phthalide with benzylidene group gives compound (B): an ortho-disubstituted benzene bearing an aldehyde (-CHO) and a -CH=CHPh or similar group - actually, the combination converts the lactone to a dialdehyde. More precisely: PCl5 on lactone gives iminium chloride type, then H2/Pd-BaSO4 gives a partial reduction to aldehyde. The net result is phthalaldehyde derivative - specifically 2-(2-oxoethyl)benzaldehyde or ortho-phenylenedicarboxaldehyde bearing a Ph group. Compound (B) is likely 2-formyl-cinnamaldehyde derivative or an ortho-(PhCH=CH-)benzaldehyde. Step 5 - Reaction with NH2-NH2 (hydrazine): Hydrazine reacts with two carbonyl groups (dialdehydes) to form a cyclic dihydrazone or a phthalazine. With an ortho-dialdehyde bearing a Ph group, hydrazine would condense with both aldehyde groups to form a cyclic structure: phthalazine derivative. The two C=O groups react with the two NH2 groups of hydrazine (H2N-NH2) to form a five-membered ring containing -N=N- (or -CH=N-N=CH-) fused to benzene. This gives a phthalazine-type product with the Ph substituent on one carbon: specifically 1-phenylphthalazine or 1-phenyl-1,2-dihydrophthalazine. Step 6 - Identify compound (C): The product of hydrazine with an ortho-dialdehyde (one being Ph-CH=, the other being CHO on the ring) gives a cyclic hydrazone. The structure has the benzene ring with two adjacent positions bearing C(Ph)=N and CH=N connected through N-N (hydrazine), forming a five-membered diazacyclic ring fused to benzene. This matches option (b): ortho-disubstituted benzene where C(Ph)=N and CH=N are connected by N-N bond, with Ph on the C=N carbon. Why other options fail: - (a) has H instead of Ph on the C=N carbon - incorrect as Ph comes from Friedel-Crafts step - (c) has Ph and OH on the ring carbon - does not match the reaction mechanism - (d) is simply cinnamaldehyde/styrene derivative - does not account for the ring-forming reaction with hydrazine Therefore, the correct answer is B.