JEE Challenger
More from Hydrocarbons
ChemMedAdv 2014 • Paper 2

Boiling Point Comparison of Hexane Isomers Classed by Branching

Hexane isomers, based on their degree of branching, can be grouped into three distinct categories as shown below:

Select the correct trend regarding their boiling points.

Question Diagram 1

Options

A

I>II>III\text{I} > \text{II} > \text{III}

B

III>II>I\text{III} > \text{II} > \text{I}

Correct
C

II>III>I\text{II} > \text{III} > \text{I}

D

III>I>II\text{III} > \text{I} > \text{II}

Topics & Concepts

HydrocarbonsIsomerismAlkanes, Alkenes, Alkynes

Step-by-Step Solution

To determine the correct trend in boiling points for the given classes of hexane isomers, we analyze the effect of molecular structure on intermolecular forces:

  1. Intermolecular Forces and Surface Area: The boiling point of alkanes depends primarily on the strength of London dispersion (van der Waals) forces between molecules. The strength of these forces is proportional to the contact surface area of the molecule.

  2. Effect of Branching:

    • Straight-chain (unbranched) alkanes have an elongated shape, providing a large surface area for intermolecular contact, leading to stronger van der Waals forces and a higher boiling point.
    • Branching causes the molecule to become more compact and spherical. This reduces the surface area available for intermolecular interactions, weakening van der Waals forces and thereby lowering the boiling point.

  3. Classification of the Given Hexane Isomers:

    • Class III (Unbranched): Contains nn-hexane, which has zero branching. It possesses the maximum surface area, resulting in the strongest van der Waals forces and the highest boiling point.
    • Class II (Mono-branched): Contains 2-methylpentane and 3-methylpentane, which have one methyl branch. They have a smaller surface area than nn-hexane and thus an intermediate boiling point.
    • Class I (Di-branched): Contains 2,2-dimethylbutane and 2,3-dimethylbutane, which have two methyl branches. They are the most compact and spherical, having the least surface area and the lowest boiling point.

Therefore, the correct decreasing order of boiling points is: III>II>I\text{III} > \text{II} > \text{I}

Correct Option: B

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