See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The thermodynamic stability of xylene isomers (ortho, meta, para) is determined by their heats of combustion and heats of formation. The isomer with the lowest heat of combustion (or highest heat of formation relative to the same reference) is the most thermodynamically stable. Step 1: Identify the three xylene isomers. - (a) ortho-xylene: methyl groups at positions 1,2 - (b) meta-xylene: methyl groups at positions 1,3 - (c) para-xylene: methyl groups at positions 1,4 Step 2: Compare thermodynamic stabilities using experimental heats of combustion data. The heats of combustion (kJ/mol) are approximately: - ortho-xylene: ~4552 kJ/mol - meta-xylene: ~4551 kJ/mol - para-xylene: ~4548 kJ/mol Lower heat of combustion means higher thermodynamic stability (less energy released means the compound is already at a lower energy state). Step 3: However, the standard reference for thermodynamic stability among xylenes uses heats of isomerization and vapor-phase equilibrium data. Experimental equilibrium studies show that meta-xylene is the most abundant isomer at thermodynamic equilibrium (approximately 58-63% meta, 20-25% ortho, 17-20% para at typical temperatures), indicating meta-xylene is the thermodynamically most stable isomer. Step 4: Rationalize why meta-xylene is most stable. In meta-xylene, the two methyl groups are separated enough to minimize steric strain (unlike ortho, where they are adjacent and experience steric repulsion) while also benefiting from hyperconjugative and inductive stabilization distributed asymmetrically around the ring. The 1,3-arrangement represents the optimal balance between steric relief and electronic stabilization. Why other options fail: - (a) ortho-xylene: the two adjacent methyl groups cause steric strain, making it less stable than meta. - (c) para-xylene: while it has no steric strain, experimental equilibrium data show it is present in lower amounts than meta at equilibrium, placing it as less stable than meta. - (d) All are equally stable: incorrect, as the isomers have measurably different heats of formation and equilibrium distributions. Therefore, the correct answer is B.