Question

Identify the compound from following having lowest boiling point.

• A. $\mathrm{n-C_4H_9NH_2}$
• B. $\mathrm{C_2H_5CH(CH_3)_2}$
• C. $\mathrm{C_2H_5N(CH_3)_2}$
• D. $\mathrm{(C_2H_5)_2NH}$

Hint:

Alkanes are essentially non-polar hydrocarbons and can only form weak London dispersion forces. When compared against polar functional groups like amines or alcohols of similar molecular weight, alkanes will always have the lowest boiling point.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Four compounds with similar molecular masses—a primary amine, a secondary amine, a tertiary amine, and a branched alkane—are given. We must identify the one with the lowest boiling point.

Step 2: Key Formula or Approach:
Boiling point depends on intermolecular force strength. The ranking from strongest to weakest is: Hydrogen bonding (primary>secondary)>Dipole-dipole>London dispersion forces.

Step 3: Detailed Explanation:
Butylamine (A) has two N-H bonds, enabling strong H-bonding networks. Diethylamine (D) has one N-H bond, giving moderate H-bonding. Ethyldimethylamine (C) is a tertiary amine with no N-H bonds, so it cannot self-associate via H-bonds and relies only on dipole-dipole interactions. 2-Methylbutane (B) is a non-polar hydrocarbon depending entirely on weak London dispersion forces. Since dispersion forces are the feeblest among these, this alkane vaporizes most readily.

Step 4: Final Answer:
The lowest boiling point belongs to 2-methylbutane, option (B).