Question

Arrange the following compounds according to increasing order of boiling points.
n-$C_4H_9OH$ (A), n-$C_4H_9NH_2$ (B), n-$C_4H_{10}$ (C) and $C_2H_5NHC_2H_5$ (D).

• A. (C)<(D)<(B)<(A)
• B. (C)<(B)<(D)<(A)
• C. (A)<(B)<(D)<(C)
• D. (D)<(C)<(B)<(A)

Hint:

Rank based on intermolecular forces: Alcohol (strongest H-bonds)>Primary Amine>Secondary Amine>Alkane (weakest forces).

Answer 1

The Correct Option is A

To determine the correct order of boiling points, we analyze the intermolecular forces present in each molecule:

1. Identify the types of forces:
- n-butane (C) is non-polar: Van der Waals forces only.
- n-butylamine (B) and diethylamine (D) are polar amines: Hydrogen bonding ($N-H \dots N$).
- n-butanol (A) is a polar alcohol: Strong hydrogen bonding ($O-H \dots O$).

2. Rank by force strength: Hydrogen bonds are much stronger than Van der Waals forces. Therefore, the alkane (C) will have the lowest boiling point. Comparing $O-H$ and $N-H$, Oxygen is more electronegative than Nitrogen, making the hydrogen bonds in alcohol (A) stronger than those in amines (B and D). Thus, (A) will have the highest boiling point.

3. Compare the amines (B and D): Both have similar molecular weights. Primary amines like (B) have two N-H bonds and less steric hindrance, allowing for a higher degree of association than secondary amines like (D). Consequently, n-butylamine (B) has a higher boiling point than diethylamine (D).

The final sequence is: (C)<(D)<(B)<(A).