Question

The correct order of boiling points of following molecules is:

• A. n-Hexane
• B. \(i>ii>iii\)
• C. \(iii>ii>i\)
• D. \(iii>i>ii\)

Hint:

For isomers, more branching leads to a lower boiling point due to a decrease in surface area and weaker London dispersion forces.

Answer 1

The Correct Option is A

Step 1: Understand the factors affecting boiling points of alkanes.
The boiling points of alkanes are primarily determined by the strength of their intermolecular van der Waals forces (specifically London dispersion forces). These forces depend on: % Option (A) Molecular weight (or molar mass): Larger molecules with more electrons have stronger London dispersion forces and thus higher boiling points. % Option (B) Surface area: For isomers with the same molecular weight, the molecule with a larger surface area will have stronger London dispersion forces and a higher boiling point because there is more area for intermolecular interactions. Branched alkanes have a more compact, spherical shape, leading to a smaller surface area compared to their straight-chain isomers.
Step 2: Analyze the given molecules. All three molecules have the same molecular formula, C$_6$H$_{14}$, so their molecular weights are the same. We need to consider their structures and surface areas. \begin{itemize} \item (i) n-Hexane: CH$_3$CH$_2$CH$_2$CH$_2$CH$_2$CH$_3$ - A straight-chain alkane with the largest surface area among the three isomers. \item (ii) 2-methylpentane: CH$_3$CH(CH$_3$)CH$_2$CH$_2$CH$_3$ - A branched alkane with one methyl group on the second carbon, resulting in a more compact shape and smaller surface area than n-hexane. \item (iii) 2,3-dimethylbutane: CH$_3$CH(CH$_3$)CH(CH$_3$)CH$_3$ - A more highly branched alkane with two methyl groups on adjacent carbons, resulting in the most compact shape and smallest surface area among the three isomers. \end{itemize}
Step 3: Determine the order of boiling points.
Since n-hexane has the largest surface area, it will have the strongest London dispersion forces and the highest boiling point. 2-methylpentane has a smaller surface area than n-hexane but a larger surface area than 2,3-dimethylbutane, so its boiling point will be intermediate. 2,3-dimethylbutane has the smallest surface area, leading to the weakest London dispersion forces and the lowest boiling point. Therefore, the correct order of boiling points is: n-hexane>2-methylpentane>2,3-dimethylbutane, which corresponds to i>ii>iii. Final Answer: \[ \boxed{\text{i>ii>iii}} \]