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Q7.22:
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

Solution :

1. Analysis of Molecular Structures

To understand the disparity in boiling points, we must compare the functional groups and intermolecular forces present in ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) and methoxymethane ($\text{CH}_3\text{OCH}_3$).

Property Ethanol ($\text{C}_2\text{H}_5\text{OH}$) Methoxymethane ($\text{CH}_3\text{OCH}_3$)
Functional Group Hydroxyl group ($-\text{OH}$) Ether linkage ($-\text{O}-$)
Molecular Weight $46.07 \text{ g/mol}$ $46.07 \text{ g/mol}$
Primary Intermolecular Force Hydrogen Bonding Dipole-Dipole Interactions

2. Theoretical Mechanism: Intermolecular Forces

Ethanol: The presence of the polar $-\text{OH}$ group allows for the formation of intermolecular hydrogen bonding. Because oxygen is highly electronegative, it develops a partial negative charge ($\delta^-$), and the hydrogen atom bonded to it develops a partial positive charge ($\delta^+$). This creates a strong electrostatic attraction between the hydrogen of one molecule and the oxygen of an adjacent molecule. [Per the theory of hydrogen bonding in protic solvents].

Methoxymethane: As an ether, methoxymethane lacks a hydrogen atom directly bonded to an electronegative atom. Consequently, it cannot engage in hydrogen bonding. Its intermolecular interactions are restricted to weak dipole-dipole interactions and London dispersion forces. [Per the study of intermolecular forces in ethers].

3. Visual Representation of Intermolecular Forces

Ethanol (H-Bonding) O H H-Bond Methoxymethane (Dipole-Dipole) CH3-O-CH3 CH3-O-CH3 Weak Dipole

4. Energy Requirement Analysis

Boiling is a phase transition that requires the input of energy to overcome the forces holding molecules together in the liquid phase. Since hydrogen bonds are significantly stronger than dipole-dipole interactions [per Coulomb's Law and molecular orbital interactions], a higher quantity of thermal energy is required to disrupt the hydrogen-bonded network of ethanol.

  • Ethanol ($\text{C}_2\text{H}_5\text{OH}$): Boiling Point $\approx 351 \text{ K}$
  • Methoxymethane ($\text{CH}_3\text{OCH}_3$): Boiling Point $\approx 248 \text{ K}$

Final Solution: The higher boiling point of ethanol compared to methoxymethane is due to the presence of strong intermolecular hydrogen bonding in ethanol, which is absent in methoxymethane. This necessitates significantly more thermal energy to break these bonds during the transition from liquid to gaseous phase.


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