Arrange the following compounds in the increasing order of their boiling points: CH₃CH₂CH₂CHO, CH₃CH₂CH₂CH₂OH, H₅C₂-O-C₂H₅, CH₃CH₂CH₂CH₃
The increasing order of boiling points is: CH₃CH₂CH₂CH₃ < H₅C₂-O-C₂H₅ < CH₃CH₂CH₂CHO < CH₃CH₂CH₂CH₂OH.
The molecular masses of these compounds are in the range of 72 to 74. Butan-1-ol has the highest boiling point because its molecules are associated due to extensive intermolecular hydrogen bonding.
Butanal is more polar than Ethoxyethane; therefore, the intermolecular dipole-dipole attraction is stronger in the former. n-Pentane molecules have only weak van der Waals forces, resulting in the lowest boiling point.
Explanation
The textbook context explains that while the molecular masses are comparable, the boiling points differ due to the nature of intermolecular forces. Butan-1-ol exhibits strong hydrogen bonding, giving it the highest boiling point. Butanal has stronger dipole-dipole interactions compared to Ethoxyethane. The alkane (referred to as n-Pentane in the context explanation) relies on weak van der Waals forces, placing it at the lowest position in the order.
Solution Steps
Step 1: Compare the intermolecular forces of attraction for the given compounds.
Step 2: Identify that Butan-1-ol has the strongest forces (hydrogen bonding), followed by Butanal (dipole-dipole), Ethoxyethane (dipole-dipole, weaker), and n-Pentane (van der Waals forces).
Step 3: Arrange the compounds in increasing order of boiling points based on the strength of their intermolecular forces: n-Pentane < Ethoxyethane < Butanal < Butan-1-ol.