Arrange the following compounds in the order of their increasing boiling point: Pentan-1-ol, n-butane, pentanal, ethoxy ethane

n-butane <ethoxy ethane < pentanal < pentan-1-ol

The correct answer is option 1. n-butane <ethoxy ethane < pentanal < pentan-1-ol.

To explain the boiling points of the given compounds— n-butane, ethoxy ethane, pentanal, and pentan-1-ol—in detail, we need to understand how intermolecular forces and molecular size affect boiling points.

Key Factors Affecting Boiling Points

Intermolecular Forces: Stronger intermolecular forces lead to higher boiling points because more energy is required to separate the molecules.

van der Waals forces: Present in non-polar molecules, these forces are relatively weak.

Dipole-dipole interactions: Present in polar molecules, these are stronger than Van der Waals forces.

Hydrogen bonding: A strong type of dipole-dipole interaction, occurring when hydrogen is directly bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine. Hydrogen bonding leads to significantly higher boiling points.

Molecular Size and Surface Area: Larger molecules or those with greater surface area experience stronger Van der Waals forces due to increased contact between molecules, which increases boiling points.

Analyzing Each Compound:

n-Butane (\(C_4H_{10}\)):

A straight-chain alkane with no polar groups.

Intermolecular Forces: Only weak van der Waals forces are present because it's a non-polar molecule.

Boiling Point:Very low (approximately -0.5°C) because it has the weakest intermolecular forces among the four compounds. No significant dipole moment or hydrogen bonding is present to increase its boiling point.

Ethoxy ethane (Diethyl ether, \(C_2H_5OC_2H_5\)):

A polar molecule with an ether group \(-O-\) between two ethyl groups.

Intermolecular Forces: It has dipole-dipole interactions because of the polar C-O bond, but no hydrogen bonding occurs between molecules (as the oxygen is not bonded directly to a hydrogen atom). The dipole-dipole interactions are stronger than Van der Waals forces, resulting in a higher boiling point than non-polar compounds like n-butane.

Boiling Point: Approximately 34.6°C.

Pentanal (\(C_5H_{10}O\)):

A five-carbon chain aldehyde with a polar carbonyl group \((C=O)\).

Intermolecular Forces: Dipole-dipole interactions due to the polar carbonyl group. Although it cannot form hydrogen bonds (no -OH or -NH groups), the carbonyl group contributes to significant dipole-dipole interactions, which are stronger than those in ethoxy ethane.

Boiling Point: Higher than ethoxy ethane due to stronger dipole interactions but lower than alcohols that can hydrogen bond. The boiling point of pentanal is approximately 103°C.

Pentan-1-ol (\(C_5H_{11}OH\)):

A five-carbon alcohol with a hydroxyl group (-OH).

Intermolecular Forces:

Hydrogen bonding: The hydroxyl group (-OH) allows for strong hydrogen bonding between molecules, which greatly increases the boiling point. Hydrogen bonding is much stronger than dipole-dipole interactions or van der Waals forces, resulting in a significantly higher boiling point than the other compounds.

Boiling Point: Highest of all the given compounds, approximately 138°C.

Summary of Boiling Points (from lowest to highest):

n-Butane (Weak Van der Waals forces) → Lowest boiling point.

Ethoxy ethane (Dipole-dipole interactions) → Higher boiling point than n-butane, but no hydrogen bonding.

Pentanal (Dipole-dipole interactions with a carbonyl group) → Higher boiling point than ethoxy ethane due to stronger dipole interactions.

Pentan-1-ol (Hydrogen bonding) → Highest boiling point due to strong hydrogen bonding between hydroxyl groups.

Correct Order of Increasing Boiling Points: \(\text{n-butane} < \text{ethoxy ethane} < \text{pentanal} < \text{pentan-1-ol}\)