Lowest: Butane < 1-chloropropane < Propan-1-ol (Highest)

Explanation:

• Butane: Non-polar molecule. The only intermolecular forces present are induced dipole-dipole interactions (Van der Waals forces), which are weak and require little energy to overcome.
• 1-chloropropane: Polar molecule (due to the C-Cl bond). It has permanent dipole-dipole interactions in addition to Van der Waals forces, giving it a higher boiling point than butane.
• Propan-1-ol: Contains an -OH group allowing for Hydrogen Bonding between molecules. Hydrogen bonds are the strongest intermolecular force among the three, requiring the most energy to overcome.

A diagram should show a hydrogen bond (represented by a dashed line) forming between the lone pair on the oxygen of one molecule and the delta-positive hydrogen of the other. Either of the diagrams below is acceptable:

OR

Pentan-1-ol has a higher boiling point than ethanol.

Both molecules have an -OH group and can form hydrogen bonds. However, pentan-1-ol has a significantly longer carbon chain (more electrons). This results in stronger Van der Waals forces (induced dipole-dipole interactions) between the molecules, requiring more energy to separate them compared to the smaller ethanol molecules.

Propan-1-ol is a primary alcohol, whereas propan-2-ol is a secondary alcohol. While both have the same molecular formula and can form hydrogen bonds:

• The -OH group in propan-1-ol is at the end of the chain (more exposed), making it easier to form effective hydrogen bonds with neighboring molecules.
• The branched nature of propan-2-ol makes the molecule more spherical, reducing the surface area contact between molecules and slightly weakening the Van der Waals forces compared to the linear propan-1-ol.