Question

the university of manitoba, term test 1 chem 1110 a01, version a october 3, 2024 time: 70 minutes examiner: georg schreckenbach 1. predict which pure compound below has the highest melting point: a) (ch3)2chch3 b) ch3ch2ch2ch3 c) ch3ch2ch2oh d) ch3ch2ch2f e) ch3ch2ch2i 2. in which of the following pure substances will hydrogen bonding be an important inter force? 1) dichloromethane, ch2cl2, 2) ch3ch2oh, 3) methylamine, (ch3nh2), 4) trimethylamine n(ch3)3

Explanation:

Step1: Analyze intermolecular forces

Melting - point is related to intermolecular forces. The stronger the intermolecular forces, the higher the melting - point. The main types of intermolecular forces are London dispersion forces, dipole - dipole forces, and hydrogen bonding.

Step2: Identify intermolecular forces for each compound

• For a) \((CH_3)_2CHCH_3\) (2 - methylpropane) and b) \(CH_3CH_2CH_2CH_3\) (butane), they are non - polar hydrocarbons. They only have London dispersion forces.
• For c) \(CH_3CH_2CH_2OH\) (1 - propanol), it has a hydroxyl group (\(-OH\)). It can form hydrogen bonds in addition to London dispersion forces and dipole - dipole forces.
• For d) \(CH_3CH_2CH_2F\) (1 - fluoropropane), it has a polar \(C - F\) bond, so it has dipole - dipole forces and London dispersion forces.
• For e) \(CH_3CH_2CH_2I\) (1 - iodopropane), it has a polar \(C - I\) bond, so it has dipole - dipole forces and London dispersion forces.

Step3: Compare the strength of intermolecular forces

Hydrogen bonding is the strongest among these intermolecular forces. Since 1 - propanol (\(CH_3CH_2CH_2OH\)) can form hydrogen bonds, it has the strongest intermolecular forces among the given compounds.

Answer:

c) \(CH_3CH_2CH_2OH\)