Exams › JEE Advanced › Chemistry
Correct answer: 3
Mellitic acid (benzene-1,2,3,4,5,6-hexacarboxylic acid) on strong heating undergoes decarboxylation. Each pair of adjacent COOH groups loses CO2 and condenses to form a cyclic anhydride linkage (-CO-O-CO-). There are 3 such adjacent pairs in mellitic acid, so 3 moles of CO2 are evolved per mole of substrate, giving mellitic trianhydride. Note: the original ASCII structure in the source was garbled; this reconstruction is based on the standard JEE mellitic acid decarboxylation problem.