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Q7.4:
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

Solution :

1. Molecular Structure and Intermolecular Forces

To analyze the difference in boiling points between propan-1-ol ($CH_3CH_2CH_2OH$) and butane ($CH_3CH_2CH_2CH_3$), we must evaluate the dominant intermolecular forces in each molecular species.

Molecule Chemical Formula Molecular Mass Dominant Intermolecular Force
Propan-1-ol $C_3H_8O$ 60.1 g/mol Hydrogen Bonding & Dipole-Dipole
Butane $C_4H_{10}$ 58.1 g/mol London Dispersion Forces

2. Analysis of Intermolecular Interactions

A. Butane ($C_4H_{10}$): As a non-polar hydrocarbon, the only attractive forces present between butane molecules are weak London dispersion forces (induced dipole-induced dipole interactions). These forces are relatively weak and depend primarily on surface area and molecular mass [Per the theory of Van der Waals forces].

B. Propan-1-ol ($CH_3CH_2CH_2OH$): The presence of the hydroxyl ($-OH$) group imparts significant polarity to the molecule. The oxygen atom is highly electronegative ($3.44$ on the Pauling scale) compared to hydrogen ($2.20$), creating a permanent dipole. Furthermore, these dipoles facilitate the formation of intermolecular hydrogen bonds [Justified by the high electronegativity of Oxygen and the small atomic radius of Hydrogen].

CH₃-CH₂-CH₂ -OH CH₃-CH₂-CH₂ -OH H-bond

3. Energetic Requirements for Phase Transition

Boiling involves overcoming these intermolecular forces to transition from the liquid phase to the gas phase.

  • In butane, only minimal thermal energy is required to disrupt the weak London dispersion forces.
  • In propan-1-ol, a significantly larger amount of thermal energy is required to break the strong, directional hydrogen bonds alongside the dipole-dipole interactions.
[Per the Kinetic Molecular Theory, energy required for phase change is directly proportional to the strength of intermolecular forces].

4. Conclusion

Despite their comparable molar masses, the presence of hydrogen bonding in propan-1-ol creates a much stronger cohesive energy density within the liquid state compared to the weak van der Waals forces in butane. Consequently, propan-1-ol requires a higher temperature to achieve the vapor pressure necessary for boiling.

Final Solution: Propan-1-ol exhibits a higher boiling point than butane because it forms intermolecular hydrogen bonds, which are significantly stronger than the weak London dispersion forces present in butane, requiring more energy to disrupt.


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