See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The reaction of benzaldehyde (Ph-CHO) with benzene in the presence of H2SO4 proceeds via electrophilic aromatic substitution (Friedel-Crafts type / acid-catalyzed condensation). Step 1: Under acidic conditions (H2SO4), benzaldehyde is protonated at the carbonyl oxygen to generate a highly electrophilic oxocarbenium ion (PhCH=OH+), which acts as an electrophile. Step 2: This electrophile attacks one molecule of benzene in an electrophilic aromatic substitution to give Ph-CH(OH)-Ph (benzhydrol intermediate), or more directly, the intermediate carbocation Ph-CH(+)-Ph forms after protonation and loss of water. Step 3: Since 2 equivalents of benzene are used and the reaction is driven by H2SO4, the benzhydrol intermediate (Ph2CHOH) loses water under acidic conditions to form the carbocation Ph2CH+. Step 4: This carbocation (Ph2CH+) then undergoes a second electrophilic aromatic substitution with the second equivalent of benzene, yielding Ph2CH-Ph = Ph3CH (triphenylmethane). Step 5: The overall reaction is: Ph-CHO + 2 PhH → Ph3CH + H2O (under H2SO4 catalysis). Why other options fail: - Option (b) triphenylmethanol would require a different reaction pathway (e.g., PhLi + benzophenone) and would not be the product of PhCHO + 2 benzene under H2SO4. - Option (c) represents a diarylmethyl-substituted benzene, which is not consistent with the reagent stoichiometry and reaction mechanism. - Option (d) Ph2CH2 (diphenylmethane) would be the product if only 1 equivalent of benzene reacted fully, but with 2 equivalents and excess acid, the reaction goes further to give the trityl product. Therefore, the correct answer is A.