Increasing Order of Boiling Points for Organic Compounds

To determine the increasing order of boiling points for the given organic compounds, we evaluate the types and strengths of intermolecular forces present in each molecule:

1. $\text{n-C}_4\text{H}_{10}$ (C, n-butane): As a non-polar hydrocarbon, it experiences only weak London dispersion forces (van der Waals forces) between its molecules. Thus, it has the lowest boiling point among all given compounds ($\text{b.p.} \approx -0.5^\circ\text{C}$).

2. $\text{C}_2\text{H}_5\text{NHC}_2\text{H}_5$ (D, diethylamine): This is a secondary amine possessing a polar $\text{N-H}$ bond, allowing it to form hydrogen bonds. However, because it has only one hydrogen atom attached to the nitrogen atom and experiences steric hindrance from the two ethyl groups, the extent of intermolecular hydrogen bonding is weaker than in primary amines. Consequently, its boiling point is higher than that of the alkane, but lower than that of primary amines ($\text{b.p.} \approx 55^\circ\text{C}$).

3. $\text{n-C}_4\text{H}_9\text{NH}_2$ (B, 1-butanamine): This is a primary amine with two $\text{N-H}$ hydrogen bond donors per molecule, allowing for more extensive intermolecular hydrogen bonding compared to the secondary amine $\text{D}$. Therefore, its boiling point is higher than that of diethylamine ($\text{b.p.} \approx 77.8^\circ\text{C}$).

4. $\text{n-C}_4\text{H}_9\text{OH}$ (A, 1-butanol): Oxygen is more electronegative than nitrogen ($\text{EN of O} \approx 3.44$ vs $\text{EN of N} \approx 3.04$). The resulting $\text{O-H}$ bond is significantly more polar than the $\text{N-H}$ bond, leading to much stronger intermolecular hydrogen bonding in alcohols than in amines of comparable molecular mass. Thus, 1-butanol has the highest boiling point ($\text{b.p.} \approx 117.7^\circ\text{C}$).

Comparing the boiling points: $\text{n-C}_4\text{H}_{10} \text{ (C)} < \text{C}_2\text{H}_5\text{NHC}_2\text{H}_5 \text{ (D)} < \text{n-C}_4\text{H}_9\text{NH}_2 \text{ (B)} < \text{n-C}_4\text{H}_9\text{OH} \text{ (A)}$

Thus, the correct order is: $C < D < B < A$

Which corresponds to Option C.