Q.Which of the following should be most volatile? (I) CH3CH2CH2NH2 (propan-1-amine); (II) (CH3)3N (trimethylamine); (III) CH3CH2–NH–CH3 (N-methylethanamine, a secondary amine); (IV) CH3CH2CH3 (propane).

Concept

Boiling point rises with stronger intermolecular attraction. Ranking of forces: hydrogen bonding (N–H···N) > dipole–dipole > London dispersion. Lower boiling point = higher volatility.

Comparing the four (similar molar masses)

• (I) propan-1-amine: two N–H bonds → extensive hydrogen bonding → highest boiling, least volatile.
• (III) N-methylethanamine: one N–H bond → some hydrogen bonding → fairly high boiling. …

Method: Predicting Volatility from Intermolecular Forces (Hydrogen Bonding vs. Dipole vs. Dispersion)

Core Concept

Volatility (a low boiling point) is highest when intermolecular attractive forces are weakest; for molecules of comparable size, the strength ranking is hydrogen bonding greater than dipole-dipole interactions greater than London dispersion forces alone.

Steps

1. For each candidate, count how many N-H (or O-H) bonds are present, since these enable hydrogen bonding between molecules.
2. A compound with two N-H bonds (a primary amine) hydrogen-bonds most extensively, giving the highest boiling point and least volatility.
3. A compound with one N-H bond (a secondary amine) still hydrogen-bonds, but less extensively, giving an intermediate boiling point.
4. A compound with no N-H bonds but a polar C-N linkage (a tertiary amine) cannot hydrogen-bond with itself, but is still held together by dipole-dipole forces, giving a lower boiling point than the primary/secondary amines.
5. A non-polar hydrocarbon with none of the above has only weak London dispersion forces, giving the lowest boiling point of the set and making it the most volatile.
6. Select the compound with the weakest total intermolecular attraction as the most volatile. …