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

Boiling points of organic compounds are primarily determined by the strength of their intermolecular forces. Let's rank the given compounds based on their molecular interactions:

1. n-butane (C): This is a non-polar alkane. It only exhibits weak Van der Waals (London dispersion) forces. Since these are the weakest intermolecular forces, n-butane has the lowest boiling point among the four.

2. Amines (B and D): Amines are polar and can form hydrogen bonds because they contain $N-H$ bonds. However, nitrogen is less electronegative than oxygen, so $N-H \dots N$ bonds are weaker than $O-H \dots O$ bonds.
Between n-butylamine (B, a primary amine) and diethylamine (D, a secondary amine), the primary amine generally has a higher boiling point. This is because primary amines have two hydrogen atoms bonded to nitrogen, allowing for a more extensive network of hydrogen bonding compared to secondary amines, which have only one. Thus, $(D)<(B)$.

3. n-butanol (A): This is an alcohol with an $O-H$ group. Oxygen is highly electronegative, making $O-H \dots O$ hydrogen bonds very strong. Alcohols therefore have much higher boiling points than amines or alkanes of similar molar mass. Thus, n-butanol has the highest boiling point.

Combining these observations, the increasing order of boiling points is:
n-butane (C)<diethylamine (D)<n-butylamine (B)<n-butanol (A).
Order: $(C)<(D)<(B)<(A)$.