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rank the following compounds in order of increasing strength of intermolecular forces.
ch₃ch₂oh ch₃ch₂ch₃ ch₃och₃
ch₃ch₂ch₃ < ch₃ch₂oh < ch₃och₃
ch₃och₃ < ch₃ch₂ ch₃ < ch₃ch₂oh
ch₃ch₂oh < ch₃och₃ < ch₃ch₂ch₃
ch₃ch₂ch₃ < ch₃och₃ < ch₃ch₂oh

Explanation:

Step1: Analyze intermolecular forces for each compound

• \(CH_3CH_2CH_3\) (propane): It is a non - polar hydrocarbon. The only intermolecular forces are London dispersion forces (the weakest among the three types of van der Waals forces).
• \(CH_3OCH_3\) (dimethyl ether): It is a polar molecule (due to the bent shape around the oxygen atom, which creates a dipole moment). So, it has dipole - dipole interactions in addition to London dispersion forces. Dipole - dipole forces are stronger than London dispersion forces.
• \(CH_3CH_2OH\) (ethanol): It is a polar molecule and can form hydrogen bonds (because of the \(-OH\) group). Hydrogen bonds are stronger than dipole - dipole interactions.

Step2: Compare the strength of intermolecular forces

Since London dispersion forces (\(CH_3CH_2CH_3\)) < dipole - dipole forces (\(CH_3OCH_3\)) < hydrogen bonds (\(CH_3CH_2OH\))

Answer:

\(CH_3CH_2CH_3