Arrange the following compounds in decreasing order of their boiling points

A. Ethoxyethane

B. n-Butane

C. Pentanal

D. Pentan-1-ol

Choose the correct answer from the options given below:

D > C > A > B

The correct answer is option 2. D > C > A > B.

Boiling point is the temperature at which a liquid transitions into a gas. It's influenced by the strength of the intermolecular forces (forces between molecules) holding the liquid together. The stronger these forces, the more energy is required to overcome them and vaporize the liquid, resulting in a higher boiling point.

Let us analyze the intermolecular forces in each compound and arrange them in decreasing order of boiling point:

D. Pentan-1-ol \((CH_3(CH_2)_4OH)\):

Pentan-1-ol possesses a hydroxyl (OH) group. This allows for hydrogen bonding, a particularly strong intermolecular force where a hydrogen atom bonded to a highly electronegative atom (like oxygen in this case) is attracted to the lone pair electrons of another electronegative atom in a nearby molecule.

Hydrogen bonding creates a significant force of attraction between Pentan-1-ol molecules, requiring a high amount of energy to break them apart during boiling. This translates to the highest boiling point among the options.

C. Pentanal \((CH_3(CH_2)_3CHO)\):

Pentanal has a carbonyl group \((C=O)\) present. While the carbonyl group can participate in weak dipole-dipole interactions due to the difference in electronegativity between carbon and oxygen, its primary contribution isn't as strong as hydrogen bonding.

However, a hydrogen atom bonded to the oxygen in the carbonyl group can still form weaker hydrogen bonds with nearby molecules compared to van der Waals forces.

This combination of dipole-dipole interactions and weaker hydrogen bonding leads to a higher boiling point than compounds relying solely on van der Waals forces.

A. Ethoxyethane \((CH_3CH_2OCH_2CH_3)\):

Ethoxyethane (also called diethyl ether) contains two ether \((C-O-C)\) functional groups.

The oxygen atom creates a polar molecule with a partial negative charge, resulting in weak dipole-dipole interactions between molecules.

While these interactions are stronger than van der Waals forces, they are significantly weaker than hydrogen bonding.

B. n-Butane \((CH_3CH_2CH_2CH_3)\):

n-Butane is a simple hydrocarbon molecule with only carbon and hydrogen atoms.

These atoms have minimal electronegativity difference, resulting in weak van der Waals forces as the only intermolecular interaction.

Van der Waals forces are the weakest intermolecular force out of the ones mentioned here.

Order of Boiling Point:

Since the strength of intermolecular forces dictates the boiling point, here's the decreasing order:

Pentan-1-ol (D): Strongest hydrogen bonding

Pentanal (C): Weaker hydrogen bonding and dipole-dipole interactions

Ethoxyethane (A): Dipole-dipole interactions (weaker than hydrogen bonding)

n-Butane (B): Weakest van der Waals forces

Therefore, the correct answer is 2. D > C > A > B.