See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Heat of combustion depends on the energy content (enthalpy) of a compound. More strained molecules have higher potential energy and thus higher heats of combustion per mole. Additionally, for isomers or compounds with the same molecular formula, greater ring strain leads to higher heat of combustion. Branching in alkanes generally lowers the heat of combustion relative to straight-chain isomers. Step 1: Identify the compounds. - Compound (i): 1,1,2,2-tetramethylcyclopropane — a cyclopropane ring with four methyl groups (two on each of two adjacent ring carbons). Cyclopropane has significant angle strain (~115 kJ/mol total ring strain). The gem-dimethyl substitution adds additional torsional/steric strain. - Compound (ii): A bicyclic compound (two fused cyclopentane rings sharing a bond, i.e., bicyclo[2.2.1] type or a fused bicyclopentyl system) with methyl substituents. Bicyclic systems have combined ring strains from both rings plus the fusion strain. - Compound (iii): An open-chain alkane (no ring strain). The zigzag skeletal structure with terminal CH3 indicates a simple acyclic alkane. Step 2: Compare molecular formulas and strain. All three compounds appear to be isomers or have comparable carbon counts (C7 or C8 hydrocarbons). For isomers with the same molecular formula: - No ring strain (open chain) < monocyclic strain < bicyclic strain in terms of stability (higher strain = higher heat of combustion). - Compound (i) with cyclopropane ring has high angle strain (cyclopropane ring strain ~115 kJ/mol). - Compound (ii) with a bicyclic system (two five-membered rings) has strain from both rings and the fusion, but five-membered rings have lower individual strain than cyclopropane. However, bicyclic systems accumulate strain. - Compound (iii) is acyclic — lowest strain, lowest heat of combustion among isomers. Step 3: Rank the heats of combustion. The compound with the most strain (highest potential energy) releases the most heat upon combustion. - (i) cyclopropane-based: very high ring strain → highest heat of combustion - (ii) bicyclic cyclopentane-based: moderate ring strain → intermediate heat of combustion - (iii) open-chain alkane: no ring strain → lowest heat of combustion Thus the order is: (i) > (ii) > (iii) Step 4: Why other options fail. - (a) (iii) > (ii) > (i): Wrong — open chain would have lowest, not highest, heat of combustion among isomers. - (b) (ii) > (i) > (iii): Wrong — cyclopropane strain exceeds bicyclopentane strain per ring, making (i) higher than (ii). - (c) (iii) > (i) > (ii): Wrong — again places open chain highest, which is incorrect. Therefore, the correct answer is D.