Arrange the following compounds according to increasing order of boiling points. $\mathrm{n}-\mathrm{C}_4\mathrm{H}_9\mathrm{OH}(\mathrm{A})$, $\mathrm{n}-\mathrm{C}_4\mathrm{H}_9\mathrm{NH}_2(\mathrm{B})$, $\mathrm{n}-\mathrm{C}_4\mathrm{H}_{10}(\mathrm{C})$ and $\mathrm{C}_2\mathrm{H}_5\mathrm{NHC}_2\mathrm{H}_5(\mathrm{D})$.

(C) $\mathrm{C}<\mathrm{D}<\mathrm{B}<\mathrm{A}$

(A) $\mathrm{C}<\mathrm{B}<\mathrm{A}<\mathrm{D}$

(B) $\mathrm{D}<\mathrm{C}<\mathrm{B}<\mathrm{A}$

(D) $\mathrm{D}<\mathrm{B}<\mathrm{A}<\mathrm{C}$

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Chemical Bonding and Molecular Structure

Arrange the following compounds according to increasing order of boiling points.

n - C_{4} H_{9} OH (A)

n - C_{4} H_{9} NH_{2} (B)

n - C_{4} H_{10} (C)

and

C_{2} H_{5} NHC_{2} H_{5} (D)

Answer: (C)

Boiling point rises with the strength of intermolecular attraction.

n

-Butane has only weak van der Waals forces and boils lowest. Among the hydrogen-bonding molecules, the secondary amine

C_{2} H_{5} NH C_{2} H_{5}

has one

N - H

available, the primary amine

n - C_{4} H_{9} N H_{2}

has two

N - H

bonds and so hydrogen bonds more strongly, and

n - C_{4} H_{9} OH

forms the strongest hydrogen bonds because oxygen is more electronegative than nitrogen.

The increasing order of boiling point is

C < D < B < A

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option