The correct order of increasing boiling points of the following compounds is:

Pentan-1-ol, n-Butane, Pentanal, Ethoxyethane

The correct option is: n-Butane, Ethoxyethane, Pentanal, Pentan-1-ol

To determine the correct order of increasing boiling points, we must examine the strength of the intermolecular forces (IMF) present in each compound. Stronger intermolecular forces require more thermal energy to break, resulting in a higher boiling point. Let us analyze each compound individually:

1. n-Butane (CH₃CH₂CH₂CH₃):
n-Butane is a non-polar hydrocarbon. The only intermolecular forces acting between its molecules are weak London dispersion forces. Due to these extremely weak forces, it has the lowest boiling point among the given compounds.

2. Ethoxyethane (CH₃CH₂OCH₂CH₃):
Ethoxyethane is an ether. Ethers are weakly polar due to the bent C-O-C bond geometry. The intermolecular forces present here are weak dipole-dipole interactions, which are slightly stronger than the dispersion forces in n-butane. Therefore, its boiling point is higher than that of n-butane.

3. Pentanal (CH₃CH₂CH₂CH₂CHO):
Pentanal is an aldehyde. It contains a highly polar carbonyl group (C=O). This polar group leads to strong dipole-dipole interactions between the molecules. These interactions are significantly stronger than the weak dipole-dipole forces in ethers, resulting in a higher boiling point than ethoxyethane.

4. Pentan-1-ol (CH₃CH₂CH₂CH₂CH₂OH):
Pentan-1-ol is an alcohol containing a polar -OH group. This allows the molecules to form strong intermolecular hydrogen bonds. Hydrogen bonding is the strongest type of intermolecular force among those present in these compounds, giving Pentan-1-ol the highest boiling point.

Combining these analyses, the order of increasing boiling points (from lowest to highest) is:
n-Butane < Ethoxyethane < Pentanal < Pentan-1-ol