Question

part 2 (1 point)
pentane and hexane
choose one or more:
a. london dispersion
b. dipole - dipole
c. hydrogen bonding
d. ion - induced dipole
part 3 (1 point)
pentanol and water
choose one or more:
a. london dispersion
b. dipole - dipole
c. hydrogen bonding
d. ion - induced dipole

Explanation:

• Part 2: Pentane (\(C_5H_{12}\)) and hexane (\(C_6H_{14}\)) are non - polar hydrocarbons. Non - polar molecules interact mainly through London dispersion forces. There are no permanent dipoles (so no dipole - dipole), no \(H\) attached to \(N\), \(O\) or \(F\) (so no hydrogen bonding), and no ions (so no ion - induced dipole).
• Part 3: Pentanol (\(C_5H_{11}OH\)) has an \( - OH\) group. Water (\(H_2O\)) has \( - OH\) groups. Both are polar molecules (so dipole - dipole interactions exist). Also, the \(H\) in \( - OH\) (in pentanol) can interact with the \(O\) in \(H_2O\) (and vice - versa) through hydrogen bonding. London dispersion forces are present in all molecules (even polar ones, as they have electrons that can disperse). There are no ions, so no ion - induced dipole.

Answer:

• Part 2: A. London dispersion
• Part 3: A. London dispersion, B. dipole - dipole, C. hydrogen bonding