Arrange the following compounds in the increasing order of their boiling points

1. \(CH_3CH_2CH_2COOH\)
2. \(CH_3CH_2CH_2CH_2OH\)
3. \(CH_3CH_2OCH_2CH_3\)
4. \(CH_3CH_2CH_2CH_3\)
5. \(CH_3CH_2CH_2CHO\)

Choose the correct answer from the options given below:

D < C < E < B < A

The correct answer is option  2 - D < C < E < B < A

Alkanes (D): n-Pentane consists of non-polar molecules held together only by very weak Van der Waals forces (specifically London dispersion forces). Because these attractions are easily broken with minimal thermal energy, alkanes have the lowest boiling points in this group.

Ethers (C): Ethoxyethane (diethyl ether) possesses a small dipole moment due to the C-O bond. While this results in slightly stronger dipole-dipole interactions compared to the purely non-polar alkanes, the boiling point remains relatively low because ethers cannot form hydrogen bonds with themselves.

Aldehydes (E): Butanal is a polar molecule with a significant dipole moment resulting from the carbonyl group ($C=O$). These dipole-dipole attractions are stronger than the weak interactions in ethers or alkanes, leading to a higher boiling point, though they still lack the strength of hydrogen bonding.

Alcohols (B): Butan-1-ol has a much higher boiling point because of the presence of the polar $-OH$ group, which allows molecules to form strong intermolecular hydrogen bonds. These bonds require significantly more energy to break compared to dipole-dipole or Van der Waals forces.

Carboxylic Acids (A): Butanoic acid has the highest boiling point of all. This is because carboxylic acid molecules can form two hydrogen bonds with each other, creating stable structures called dimers. Even in the vapor phase, these dimers often stay together, requiring the highest amount of energy to separate the molecules.

So,  increasing order of boiling points:

D (Alkane) < C (Ether) < E (Aldehyde) < B (Alcohol) < A (Carboxylic Acid)