Question

take a look at the two molecules below. what generally accounts for the difference in the boiling point of these two molecules? methanol (ch₃ - o - h) bp:64.7 °c molar mass: 32.04 g/mol ethane (c₂h₆) bp:-88 °c molar mass: 30.07 g/mol london dispersion forces dipole - dipole forces hydrogen bonding ionic bonding none of the above

Explanation:

• London dispersion forces: These are weak intermolecular forces present in all molecules. However, they are generally weaker than hydrogen bonding and do not account for the large difference in boiling points here. Methanol has a much higher boiling point than would be expected from London dispersion forces alone.
• Dipole - dipole forces: Methanol is a polar molecule (due to the \(O - H\) bond) and has dipole - dipole forces. Ethane is non - polar (symmetrical structure) and has no dipole - dipole forces. But hydrogen bonding in methanol is a stronger intermolecular force than dipole - dipole forces in polar molecules (in the absence of hydrogen bonding).
• Hydrogen bonding: Methanol (\(CH_3OH\)) has an \(O - H\) bond. Hydrogen bonding occurs when a hydrogen atom is bonded to a highly electronegative atom (\(N\), \(O\), or \(F\)). In methanol, the oxygen - hydrogen bond allows for hydrogen bonding between methanol molecules. Ethane (\(C_2H_6\)) has no such \(N - H\), \(O - H\), or \(F - H\) bonds and thus no hydrogen bonding. Hydrogen bonding is a relatively strong intermolecular force that requires more energy (higher temperature) to break, which is why methanol has a higher boiling point.
• Ionic bonding: Both methanol and ethane are covalent compounds. Methanol has covalent bonds within the molecule (\(C - H\), \(C - O\), \(O - H\)) and intermolecular forces (including hydrogen bonding) between molecules. Ethane has only covalent bonds (\(C - C\), \(C - H\)) and London dispersion forces (as it is non - polar). There is no ionic bonding in either compound.

Answer:

Hydrogen bonding