(A) i. Pentane ii. Hexane iii. 2–Methyl butane iv. 2, 2–Dimethyl propane

Question

Arrange the following compounds according to the decreasing order of heat of combustion.

(A) i. Pentane

ii. Hexane

iii. 2–Methyl butane

iv. 2, 2–Dimethyl propane

Answer 1

Greater the number of C atoms, more is the heat of combustion ; but when the number of C atoms is same, the order is straight chain > less branching > more branching:

Hexane > Pentane > 2-Methyl butane (isobutane) > 2, 2-Dimethyl propane (neopentane).

More is the branching, lesser is the heat of combustion, and greater is the stability of compound

All the isomers react to give the same number of moles of CO₂ and H₂ O, so a direct comparision is not possible. Structure

(iii) releases more heat on combustion than others, thus, it contains relatively more potential energy and must be thermodynamically less stable. (Higher the energy in a compound, less stable is that compound compared to the products and to its isomers.) or more reactive it is and, hence, more heat of combustion will be released .

In structures  (i) and (ii), (i) has higher potential energy than (ii) and, thus is less stable and release more energy. Since structure (i) is an eclipsed structure, whereas (ii) is Gauche structure, hence, (i) is less stable than (ii).