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

Solution :

1. Analysis of Molecular Structure and Intermolecular Forces

To understand the disparity in boiling points between propan-1-ol ($CH_3CH_2CH_2OH$) and butane ($CH_3CH_2CH_2CH_3$), we must analyze the nature of the intermolecular forces (IMFs) acting within the bulk liquid phase of each substance.

Property Propan-1-ol Butane
Molecular Formula $C_3H_8O$ $C_4H_{10}$
Molar Mass $60.1 \text{ g/mol}$ $58.1 \text{ g/mol}$
Primary IMF Hydrogen Bonding & Dipole-Dipole London Dispersion Forces (LDF)

2. Intermolecular Forces in Butane ($C_4H_{10}$)

Butane is a non-polar alkane. The only significant intermolecular force present between butane molecules is the London Dispersion Force (induced dipole-induced dipole interactions). These forces arise due to temporary fluctuations in electron distribution, creating instantaneous dipoles [Per the London dispersion theory of Van der Waals forces]. These forces are relatively weak and scale with molecular surface area and polarizability.

3. Intermolecular Forces in Propan-1-ol ($CH_3CH_2CH_2OH$)

Propan-1-ol contains a hydroxyl group ($-OH$). The oxygen atom is significantly more electronegative than the hydrogen atom ($\Delta EN \approx 1.24$), resulting in a permanent dipole moment. This facilitates intermolecular hydrogen bonding, where the partially positive hydrogen ($\delta^+$) of one molecule is strongly attracted to the lone pair of electrons on the oxygen ($\delta^-$) of a neighboring molecule [Per the Coulombic interaction model].

4. Visualization of Hydrogen Bonding

CH3-CH2-CH2-O H ... O-CH2-CH2-CH3 H-bond Fig 1: Schematic representation of intermolecular Hydrogen bonding in Propanol.

5. Energy Requirement for Phase Transition

The boiling point is the temperature at which the vapor pressure of the liquid equals the external atmospheric pressure. For a liquid to transition to a gas, the molecules must overcome the intermolecular forces holding them in the condensed phase.

  • Butane: Only requires energy to overcome weak London dispersion forces.
  • Propan-1-ol: Requires significant additional energy to break the strong, directional hydrogen bonds alongside standard dipole-dipole and dispersion forces.

6. Conclusion

Because the energy required to disrupt the hydrogen-bonded network of propan-1-ol is significantly greater than the energy required to disrupt the weak dispersion forces of butane, propan-1-ol exhibits a much higher boiling point ($370 \text{ K}$) compared to butane ($272 \text{ K}$).

Final Solution: Propanol has a higher boiling point than butane because it possesses the capacity for intermolecular hydrogen bonding, which is a significantly stronger force than the London dispersion forces present in the non-polar butane molecule.


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