Q.Match the hydrocarbons in Column I with the boiling points given in Column II.
Column I

The key idea is Boiling Point Trends, specifically how branching in alkane isomers affects their boiling points.

1. All three hydrocarbons (n-pentane, iso-pentane, neo-pentane) are isomers with the same molecular formula ($\text{C}_5\text{H}_{12}$). Their boiling points are primarily determined by the strength of London Dispersion Forces (LDFs).
2. LDFs increase with increasing surface area available for intermolecular contact.
3. Branching in an alkane reduces its surface area, leading to weaker LDFs and consequently a lower boiling point. …

The boiling point of a substance is the temperature at which its vapor pressure equals the surrounding atmospheric pressure. For a liquid to boil, enough energy must be supplied to overcome the intermolecular forces holding the molecules together in the liquid phase. The stronger these intermolecular forces, the more energy is required, and thus the higher the boiling point.

For nonpolar molecules like hydrocarbons (alkanes), the primary intermolecular forces are London Dispersion Forces (LDFs), also known as van der Waals forces. These forces arise from temporary, instantaneous dipoles created by the random movement of electrons.

The strength of LDFs depends mainly on two factors:

1. Number of electrons (molecular size/mass): Larger molecules with more electrons are more polarisable, leading to stronger LDFs. All three pentane isomers (n-pentane, iso-pentane, neo-pentane) have the same molecular formula ($\text{C}_5\text{H}_{12}$) and thus the same molecular mass and number of electrons. Therefore, this factor does not differentiate their boiling points.
2. Surface area for contact: Molecules with larger, more extended surface areas can have more points of contact with neighbouring molecules, leading to stronger LDFs. Conversely, molecules that are more compact or spherical have smaller surface areas for contact, resulting in weaker LDFs.

This second factor, the surface area for contact, is crucial for understanding the boiling point differences among isomers.

Here's how we apply this concept to the pentane isomers:

1. Identify the structures of the hydrocarbons:

   • (i) n-Pentane: This is a straight-chain alkane. Its structure is $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$. It has a relatively elongated, cylindrical shape.
   • (ii) iso-Pentane: This is 2-methylbutane. Its structure is $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3$. It has one methyl branch, making it slightly more compact than n-pentane.
   • (iii) neo-Pentane: This is 2,2-dimethylpropane. Its structure is $\text{C}(\text{CH}_3)_4$. It has two methyl branches at the same carbon, giving it a highly compact, nearly spherical shape.

2. Compare their surface areas for intermolecular contact:

   • n-Pentane has the largest surface area because its straight-chain structure allows for maximum contact between molecules.
   • iso-Pentane has a smaller surface area than n-pentane due to its single branch, which reduces the effective contact area.
   • neo-Pentane has the smallest surface area among the three because its highly branched, spherical shape minimises the points of contact between molecules.

3. Relate surface area to the strength of London Dispersion Forces:

   • Largest surface area (n-Pentane) $\implies$ Strongest LDFs. …