Step 1: Understanding the Concept:
The boiling point of a molecular compound is primarily dictated by the strength of its intermolecular forces. Stronger forces, such as hydrogen bonding, require significantly more thermal energy to overcome during the transition from liquid to gas, resulting in a higher observed boiling point.
Step 2: Key Formula or Approach:
Approach: Identify the dominant types of intermolecular forces present in each compound. Specifically, look for the presence or absence of hydrogen bonding, a particularly strong dipole-dipole interaction that occurs only when a Hydrogen atom is directly covalently bonded to a highly electronegative atom (N, O, or F).
Step 3: Detailed Explanation:
Let's evaluate the intermolecular forces in each given compound:
(A) $(\text{C}_2\text{H}_5)_2\text{NH$ (Diethylamine):} This is classified as a secondary ($2^\circ$) amine. It contains one polar $\text{N}-\text{H}$ covalent bond. Therefore, discrete molecules of diethylamine can form intermolecular hydrogen bonds with each other, leading to a relatively high boiling point.
(B) $\text{C_2\text{H}_5\text{N}(\text{CH}_3)_2$ (Ethyldimethylamine):} This is classified as a tertiary ($3^\circ$) amine. The central nitrogen atom is bonded exclusively to three carbon groups and has zero $\text{N}-\text{H}$ bonds. Without a hydrogen atom directly bonded to an electronegative atom, it cannot act as a hydrogen bond donor to itself. Its molecules are held together only by much weaker dipole-dipole interactions and London dispersion forces, leading to a significantly lower boiling point compared to its isomers.
(C) $n-\text{C_4\text{H}_9\text{OH}$ (1-Butanol):} This is a primary alcohol. It contains a highly polar $\text{-OH}$ group, which engages in extensive and strong intermolecular hydrogen bonding. Alcohols generally exhibit higher boiling points than amines of comparable molar mass because oxygen is more electronegative than nitrogen, making the $\text{O}-\text{H}$ hydrogen bond stronger than the $\text{N}-\text{H}$ bond.
(D) $\text{C_2\text{H}_5\text{COOH}$ (Propanoic acid):} This is a carboxylic acid. It can form very strong hydrogen bonds, often pairing up to form stable, discrete dimers even in the liquid state. This extensive and robust hydrogen bonding gives carboxylic acids the highest boiling points among the common organic classes listed here.
Comparing all options, the tertiary amine stands out as it entirely lacks hydrogen bonding capability, giving it the weakest intermolecular forces and unequivocally the lowest boiling point.
Step 4: Final Answer:
The compound with the lowest boiling point is $\text{C}_2\text{H}_5\text{N}(\text{CH}_3)_2$.