Step-by-Step Solution:

1. Identify the Core Concept: The question asks to compare the boiling points of isomers of hexane. The boiling point of non-polar molecules like alkanes is determined by the strength of the intermolecular forces, which are primarily London dispersion forces (a type of van der Waals force).

2. Relate Intermolecular Forces to Molecular Structure: The strength of London dispersion forces depends on the surface area of the molecule. A larger surface area allows for more points of contact between adjacent molecules, leading to stronger temporary dipoles and thus stronger intermolecular attractions. Stronger intermolecular forces require more energy (higher temperature) to overcome, resulting in a higher boiling point.

3. Analyze the Isomer Classes: The hexane isomers are grouped into three classes based on their degree of branching:

   • Class I: This class represents the straight-chain isomer, n-hexane. It has no branching. Its elongated, linear shape provides the maximum possible surface area for intermolecular contact among all hexane isomers. CH₃-CH₂-CH₂-CH₂-CH₂-CH₃ (n-hexane)

   • Class II: This class includes isomers with a single branch, such as 2-methylpentane and 3-methylpentane. The presence of a branch makes the molecule more compact than the straight-chain isomer, reducing its overall surface area. CH₃-CH(CH₃)-CH₂-CH₂-CH₃ (2-methylpentane)

   • Class III: This class includes isomers with two branches, such as 2,2-dimethylbutane and 2,3-dimethylbutane. These molecules are even more compact and closer to a spherical shape than the single-branched isomers. This high degree of branching significantly reduces the surface area available for intermolecular contact. CH₃-C(CH₃)₂-CH₂-CH₃ (2,2-dimethylbutane)

4. Establish the Trend for Branching vs. Boiling Point:

   • As the number of branches in an alkane isomer increases, the molecule becomes more spherical and compact.
   • This decrease in linearity leads to a smaller surface area.
   • A smaller surface area results in weaker London dispersion forces between molecules.
   • Weaker intermolecular forces lead to a lower boiling point.

5. Order the Classes by Boiling Point:

   • Class I (n-hexane): Zero branches -> Largest surface area -> Strongest intermolecular forces -> Highest boiling point.
   • Class II (methylpentanes): One branch -> Intermediate surface area -> Intermediate intermolecular forces -> Intermediate boiling point.
   • Class III (dimethylbutanes): Two branches -> Smallest surface area -> Weakest intermolecular forces -> Lowest boiling point.

   Therefore, the correct order of boiling points is I > II > III.

6. Evaluate the Options:

   • A: I > II > III - This matches our derived order.
   • B: III > II > I - This is the reverse of the correct order.
   • C: II > III > I - This is incorrect.
   • D: III > I > III - This option is logically inconsistent.

Conclusion

The correct order of boiling points is Class I > Class II > Class III, which corresponds to option A.