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
b. i.

Explanation

a. As discussed in Q. No. 18, greater the number 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 gt less branching gt more branching. So, the order is as follows:
Hexane gt Pentane gt 2-Methyl butane (isobutane) gt 2,2-Dimethly propane (neopentane). More is the branching, lesser is the heat of combustion, and greater is the stability of compound.

All the isomers to give the same number of moles of `CO_(2)` ans `H_(2)O`, so a direct comparison is not possible. Structure (iii) releases more heat on combustion than other, 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.) Higher the potential energy of a coompound, less stable or more recative 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 releases more energy. Since structure (i) is an eclipes structure, wheres (ii) is Gauche structure, hence, (i) is less stable than (ii).